Elementos de Maquinas SOLUCIONARIO - Bernard Hamrock - 1ed Solucionario

Prévia do material em texto

www.elsolucionario.net
http://elsolucionario.blogspot.com/feeds/posts/default
http://www.facebook.com/pages/EL-SOLUCIONARIO/185607668178277
http://twitter.com/#!/elsolucionario
http://www.elsolucionario.net/
Chapter 1 • Introduction
1.1 Design transport containers for milk in 1 gallon and 1 liter sizes.
Notes: There are a number of design considerations that can be included in this problem. Some
examples are considered here, but the student should not be restricted in the types of solutions
pursued. This type of problem is an excellent opportunity for brainstorming, where a group is
encouraged to provide all possible answers to the problem regardless of feasibility. For example,
the materials considered could include wood barrels, ceramic jugs or foil or plastic bags, as in
intraveneous fluid containers.
Solution:
The transport containers for milk should be easy to handle and store as well as being easy to pour
from without spilling. The material should be recyclable or be able to be burned leaving harmless
combustion products. The obvious material choices are paper or plastic materials, although glass
bottles and metal cans can be considered as possibilities. For a one liter container the standard
type Pure Pac or Tetra Pac Bric made of plastic carton is a good design.
A one gallon container has to be compact to be easily stored in a refrigerator. The width,
depth, and height should be similar in size, which would be about 156 mm x 156 mm x 156 mm.
That size is too large to grip with one hand. Therefore, a one gallon container is preferably made
of a plastic with an integral handle.
1.2 Design a simple coat hanger to be used by dry cleaners. It should be able to accommodate
both a winter coat and a pair of trousers. It should function well but also be as
inexpensive as possible.
Notes: See the notes to problem 1.
Solution:
If they are not hung simultaneously, a steel wire frame of triangular shape with a hook at
the top could be used. The horizontal portion could be covered with a high friction coating or tape
or glue to keep the trousers from sliding off.
For coats, a sturdier frame is needed, requiring either a larger gage of wire or pursuit of
another alternative such as wood or plastic hangers.
1.3 Journal bearings on train boxcars in the early 19th century used a “stink additive” in their
lubricant. If the bearings got too hot, they would attain a noticeable odor, and an oiler
would give the bearing a squirt of lubricant at the next train stop. What design philosophy
does this illustrate? Explain.
Solution:
The most obvious design philosophy is the Doctrine of Manifest Danger. This philosophy
suggests that systems be designed so that a failure can be detected before it occurs. In this case, a
dry bearing becomes hot, and the bad smell indicates a loss of lubricant before the bearing failure
is catastrophic.
It should be noted that other answers are possible. For example, the principal of uniform
safety suggests that the level of safety of all components be kept at the same level. If failure of
www.elsolucionario.net
one component, in this case a bearing, is imminent, then extra maintenance on this bearing is
justified by the principal of uniform safety.
1.4 A safety chain is used in automotive towing applications. Which design philosophy is
being incorporated into the towing system design?
Solution:
Since the safety chain is used in addition to the towbar, it is an example of redundancy,
and it is a passive system. Furthermore, it is an example of a fail-safe design, because if the
towbar fails, the towed vehicle will follow the car and will not swerve into other lanes and collide
with other vehicles.
1.5 A hand-held drilling machine has a bearing to take up radial and thrust load from the
drill. Depending on the number of hours the drill is expected to be used before it is
scrapped, different bearing arrangements will be chosen. A rubber bushing has a 50-hr
life. A small ball bearing has a 300-hr life. A two bearing combination of a ball bearing
and a cylindrical roller bearing has a 10,000 hr life. The cost ratios for the bearing
arrangements are 1:5:20. What is the optimum bearing type for a simple drill, a
semiprofessional drill, and a professional drill?
Notes: This problem is fairly subjective, but there should be a realization that there is an increase
in expected life for a professional drill versus a simple drill. In fact, for certain users, a 50 hour
life is probably sufficient.
Solution:
Notice that the cost per time is highest for the rubber bushing and lowest for the ball bearing
cylindrical bearing combination. However, the absolute cost is higher for the longer lasting
components, so the expected life must be assessed. Rough estimates based on expected lives are
as follows.
a) Simple drill - use rubber bushingbearings.
b) Semi-professional drill - either two ball bearings or a ball and cylindrical roller bearing
combination.
c) Professional drill - use a ball and cylindrical roller bearing combination.
1.6 Using the hand-held drill described in Problem 1.5, if the solution with the small ball
bearing was chosen for a semiprofessional drill, the bearing life could be estimated to be
300 hr until the first spall forms in a race. The time from first spall to when the whole
rolling-contact surface is covered with spalls is 200hr, and the time from then until a ball
cracks is 100 hr. What is the bearing life
a) If high precision is required?
b) If vibrations are irrelevent?
c) If an accident can happen when a ball breaks?
Notes: This problem is fairly subjective, but there should be a realization that there is a decrease
in expected life for the high precision drill, etc.
Solution:
a) For high precision, no spall should be allowed so the life is 300 hrs.
b) If vibrations are irrelevent, the drill can be used until a ball breaks, or 600hours.
www.elsolucionario.net
c) Depending on the severity of the accident, it is possible that an entirely different bearing
should be used. However, safety allowances of 300-500 hours may be reasonable.
1.7 The dimensions of skis used for downhill competition need to be determined. The
maximum force transmitted from one foot to the ski is 2500N, but the snow conditions
are not known in advance, so the bending moment acting on the skis is not known. The
weight of the skier and equipment is 100 kg. Estimate the safety factor needed.
Notes: There are two different approaches, that of section 1.5.1.1, and that of a worst case
scenario.
Solution:
Using the approach of section 1.5.1.1, the largest safety factor from Table 1.1 is nsx=3.95,
and the largest value of nsy from Table 1.2 is 1.6. Therefore, the largest safety factor, if there is
poor control of materials and loading, and if the consequences are sever, is given by Equation 1.2
as:
ns=nsxnsy=(3.95)(1.6)=6.32
Another approach is to perform a worst case scenario analysis. If the weight of skier and
equipment is 100 kg and the maximum load on the ski is 2500N, then the centrigugal acceleration
is 23 m/s2. If the speed of the skier is 35m/s (126km/hr), the radius from the ski track will be 53
m. If the speed is 20 m/s, the radius is 17.4m. This means that no overstressing should ever occur
due to bending of the ski, so the safety factor could be chosen just above one.
1.8 A crane has a loading hook that is hanging on a steel wire. The allowable normal tensile
stress in the wire gives an allowable force of 100,000N. Find the safety factor that should
be used.
a) If the wire material is not controlled, the load can cause impact, and fastening the hook
in the wire causes stress concentrations. (If the wire breaks, people can be seriously hurt
and expensive equipment can be destroyed.)
b) If the wire material is extremely well controlled, no impact loads are applied, and the
hook is fastened in the wire without stress concentrations. (If the wire breaks, no people
or expensive equipment can be damaged.)
Notes: Equation (1.2) is needed to solve this problem, withdata obtained from Tables 1.1 and 1.2.
Solution:
a). If the quality of materials is poor (A=p), the control over the load to the part is poor (B=p), and
the accuracy of the stress analysis is poor (C=p), then nsx=3.95 as given in Table 1.1. If the
economic impact and danger are both very serious (D=E=vs), then from Table 1.2, nsy=1.6.
Therefore, the required safety factor is obtained from Equation (1.2) as
ns=nsxnsy=(3.95)(1.6)=6.3
b) If the quality of materials, the control over the load to the part, and the accuracy of the stress
analysis are all very good (A=B=C=vg), then nsx=1.1 as given in Table 1.1. If the economic
impact and danger are both not serious (D=E=ns), then from Table 1.2, nsy=1.0. Therefore, the
required safety factor is obtained from Equation (1.2) as
ns=nsxnsy=(1.1)(1.0)=1.1
www.elsolucionario.net
1.9 Give three examples of fail-safe and fail-unsafe products.
Notes: This is an open-ended problem, and many different solutions are possible.
Solution:
Examples of fail safe products are: 1) electric motors, which fail to move and cause no hazards
when they fail. 2) Bowling pins, because of their distance from bowlers, do not present any
hazards when they shatter. 3) Furniture upholstery, which causes aesthetic objection long before
structural failure occurs. Examples of fail unsafe products include: 1) Bungee cords, as in bungee
cord jumping from bridges and the like; 2) parachutes, which cause a fatal injury in the vast
majority of failures, and 3) A fan on an exhaust hood over a fume producing tank, such as an
electroplating tank. Failure of the forced air circulation system exposes plant personnel to harmful
fumes.
1.10 An acid container will damage the environment and the people around if it leaks. The
cost of the container is proportional to the container wall thickness. The safety can be
increased either by making the container wall thicker or by mounting a reserve tray under
the container to collect the leaking acid. The reserve tray costs 10% of the thick-walled
container cost. Which is less costly, to increase the wall thickness or to mount a reserve
tray under the container?
Notes: This problem uses the approach described by Equation (1.2) and Tables 1.1 and 1.2.
Solution:
If the safety factor is unchanged, then for a thicker container wall, we see that with danger to
personnel rated very serious and economic impact very serious (D=E=vs), then nsy=1.6. If a
reserve tray is used, the required safety factor for the container (D=E=ns) is nsy=1.0. Therefore, if
A is the cost of the container with thicker walls, then the cost of the container with a reserve tray
is
A
1.6
+ 0.1( )A = 0.725A
Therefore, with the same safety factor, it is less costly to have the reserve tray.
1.11 The unit for dynamic viscosity in the SI system is newton-seconds per square meter, or
pascal-seconds (N-s/m
2
 = Pa-s). How can that unit be rewritten by using the basic
relationships described by Newton’s law for force and acceleration?
Notes: The only units needed are kg, m, and s.
Solution:
By definition, a Newton is a kilogram-meter per square second. Therefore, the unit for dynamic
viscosity can be written as:
Ns
m2
= kgm
s2
 
  
 
  
s
m2
= kg
sm
www.elsolucionario.net
1.12 The unit for dynamic viscosity in Problem 1.11 is newton-second per square meter
(Ns/m
2
) and the kinematic viscosity is defined as the viscosity divided by the fluid
density. Find at least one unit for kinematic viscosity.
Notes: The only units needed are m and s.
Solution:
The units for kinematic viscosity can be written as:
ηk =
η
ρ
=
Ns
m2
 
  
 
  
kg
m3
 
  
 
  
= Nsm
kg
=
kgm
s2
 
  
 
  
sm
kg
= m
2
s
1.13 A square surface has sides 1m long. The sides can be split into decimeters, centimeters,
or millimeters, where 1m=10dm, 1dm=10cm, and 1cm=10mm. How many millimeters,
centimeters, and decimeters equal 1m? Also, how many square millimeters, square
centimeters and square decimeters equal one square meter?
Notes: This is fairly straightforward; Table 1.3(b) is useful.
Solution:
From Table 1.3(b), one meter equals 10 dm, 100 cm and 1000 mm. Therefore, a square meter is:
1m
2
=(1000mm)
2
=1 million (mm)
2
.
1m
2
=(100cm)
2
=10,000 (cm)
2
1m
2
=(10dm)
2
=100(dm)
2
1.14 A volume is 1 tera (mm3) large. Calculate how long the sides of a cube must be to contain
that volume.
Notes: From Table 1.3 (b), tera is 1 x 10
12
.
Solution:
A tera (mm
3
) is 10
12
 (mm
3
). Therefore, this volume is
1012 mm3
1m
1000mm
 
  
 
  
3
= 1000m3
A cube with this volume has sides with length of 10m.
1.15 A ray of light travels at a speed of 300,000 km/s=3 x 108m/s. How fast will it travel in 1
ps, 1 ns, and 1µs?
Notes: An interesting assignment is to have students submit string of the required length for the
picosecond and nanosecond cases.
Solution:
The distance traveled in a picosecond is:
www.elsolucionario.net
x = v∆t = 3 ×108m / s( ) 10−12s( ) = 3 ×10−4m = 0.3mm
In one nanosecond, light travels:
x = v∆t = 3 ×108m / s( ) 10−9s( ) = 0.3 m = 300mm
In one microsecond, light travels:
x = v∆t = 3 ×108m / s( ) 10−6s( ) = 300m
1.16 Two smooth flat surfaces are separated by a 10-µm thick lubricant film. The viscosity of
the lubricant is 0.100 Pa-s. One surface has an area of 1 dm
2
 and slides over the plane
surface with a velocity of 1 km/hr. Determine the friction force due to shearing of the
lubricant film. Assume the friction force is the viscosity times the surface area times
velocity of the moving surfaces and divided by the lubricant film thickness.
Notes This problem is simple if consistent units are used.
Solution:
Rewriting terms into consistent units, 10µm=10
-5
m, 0.100Pas=0.100Ns/m
2
,
1dm
2
=(0.1m)
2
=0.01m
2
, and
1
km
hr
= 1000m
3600s
= 0.2778m / s
Therefore the friction force is given by:
F = η0Aub
h
=
0.1
Ns
m2
 
  
 
  
0.001m2( ) 0.2778m / s( )
10−5m
= 27.78N
1.17 A firefighter sprays water on a house. The nozzle diameter is small relative to the hose
diameter, so the force on the nozzle from the water is
F = v dma
dt
where v is the water velocity, and dma/dt is the water mass flow per unit time. Calculate
the force the firefighter needs to hold the nozzle if the water mass flow rate is 3 tons/hr
and the water velocity is 100km/hr.
Notes: A ton can be misperceived; in metric units, a ton is 1000kg, while in English units it is
2000 lbf. Since the units in this problem are metric, the metric interpretation is used. This
problem is not difficult as long as consistent units are used.
Solution:
The velocity can be written as:
100
km
hr
= 100,000m
3600s
= 27.78m / s
The mass flow rate can be rewritten as:
dma
dt
= 3tons
hr
= 3000kg
3600s
= 0.8333kg / s
Therefore, the force needed is
www.elsolucionario.net
F = v dma
dt
= 27.78m / s( ) 0.8333kg / s( ) = 23.15 N
1.18 The mass of a car is 1346 kg. The four passengers in the car weigh 65.55 kg, 75.23 kg,
88.66kg, and 91.32 kg. It is raining and the additional weight due to the water on the car
is 1.349 kg. Calculate the total weight of the car, including the weight of the passengers
and the weight of the water, using four significant figures.
Solution:
The weight is
Wtot=1346kg+65.55kg+75.23kg+88.66kg+91.32kg+1.349kg±0.5kg=1668.109kg±0.5kg
Therefore the total weight is 1668kg.
1.19 During an acceleration test of a car, the acceleration was measured to be 1.4363 m/s2.
Because slush and mud adhered to the bottom of the car, the weight was estimated to be
1400±100kg. Calculate the force driving the car and indicate the accuracy.
Solution:
Since the force is given by P=maa, the force is given by
P = maa = 1400kg ±100kg( ) 1.4363m / s2( ) = 2010.82 ±143.6 N = 2154.45N1867.19N
 
 
 
www.elsolucionario.net
Page 2-1
Chapter 2 • Load, Stress and Strain
2.1 The stepped shaft A-B-C shown in sketch a is loaded with the forces P1 and/or P2. Note
that P1 gives a tensile stress σ in B-C and σ/4 in A-B and that P2 gives a bendingstress σ
at B and 1.5σ at A. What is the critical section
a) If only P1 is applied?
b) If only P2 is applied?
c) If both P1 and P2 are applied?
Notes: This solution will ignore advanced concepts such as stress concentrations that would exist
at the fillet, since this material is covered later in the text. Similarly, the location in the cross
section where the maximum stress occurs could be identified with the information in Chapter 4,
but will be left for the student.
Solution:
If only P1 is applied, the critical section is the length B-C, since the tensile stress is highest in this
section.
If only P2 is applied, the largest stress occurs at A.
If both P1 and P2 are applied, the maximum stress in section A-B occurs at point A and is
σA=σ/4+1.5σ=1.75σ
The maximum stress in section B-C occurs at point B and is
σB=σ+σ=2σ
Therefore, the critical section is B-C, in particular point B.
2.2 The stepped shaft in Problem 2.1 (sketch a) has loads P1 and P2. Find the load
classification if P1’s variation is sinuous and P2 is the load from a weight
a) If only P1 is applied.
b) If only P2 is applied.
c) If both P1 and P2 are applied.
Notes: This solution uses the terminology on pages 30-32.
Solution:
If only P1 is applied, then the load is cyclic with respect to time (see page 30), and is a
normal load (see page 31).
If only P2 is applied, the load is sustained with respect to time, and is both a shear load
and a bending load. However, for long columns, bending is usually more important than shear, so
it is reasonable to classify P2 as a bending load.
If both P1 and P2 are applied, then the load is cyclic with respect to time (although the
mean stress is not zero), and the loading is combined (see page 32).
2.3 The sign convention for bending moments can be derived from the differential equation
for the elastic line:
www.elsolucionario.net
Page 2-2
EIy”=-M
Where
E=modulus of elasticity, Pa
I=area moment of inertia, m4
M=bending moment, N-m
y"=d2y/dx2=second derivative of y with respect to x, m-1
y=deflection, m
This gives a positive (tensile) stress in the fibers on the beam in the positive (y) direction.
Find how the slope of the beam changes with increasing x coordinate for positive and
negative bending.
Notes: The slope of the beam is the first derivative of the deflection, so one must integrate the
given equation to obtain the solution.
Solution:
The slope of the beam is given by θ=dy/dx. Integrating the given equation,
dy
dx
= θ = −
Mx
EI
+ C
Where C is a constant of integration, and reflects the effects of boundary conditions. If we ignore
C, we can see that if the bending moment is negative, the slope of the beam is positive for
positive x and negative for negative x. If the moment is positive, the slope is negative for positive
x and positive for negative x.
2.4 A bar hangs freely from a frictionless hinge. A horizontal force P is applied at the bottom
of the bar until it inclines 30° from the vertical direction. Calculate the horizontal and
vertical components of the force on the hinge if the acceleration due to gravity is g, the
bar has a constant cross section along its length, and the mass is ma.
Notes: This problem is straightforward once a free body diagram is drawn. Force and moment
equilibrium is then sufficient to solve the problem.
Solution:
The free body diagram of the bar is as shown to the right.
Since the bar cross section is constant, we can put the center of
gravity at the geometric center of the bar. The reactions at the
hinge have been drawn in red. Notice the absence of a moment
reaction, since the hinge was expressly described as frictionless
(this is a common assumption for hinges).
Most of the time, it saves time to perform moment
equilibrium first, before applying force equilibrium. This is true
here as well. Therefore, summing moments about the hinge point,
Equation (2.2) gives:
M = 0∑ = mag l2 sin30°
 
  
 
  
− Pl cos30°; mag =
2P
tan30°
Taking equilibrium in the x-direction, Equation (2.1) gives:
Px = 0 = −Rx + P; Rx = P∑ = gma2 t a n 3 0°
Similarly, in the y-direction,
www.elsolucionario.net
Page 2-3
Py = 0 = Ry − gma ; Ry = gma∑
2.5 Sketch b shows the forces acting on a rectangle. Is the rectangle in equilibrium?
Notes: To satisfy equilibrium, Equations (2.3) must be satisfied.
Solution:
Taking the sum of forces in the x-direction,
ΣFx=30N-20N-10N=0
Therefore, equilibrium in the x-direction is satisfied. Taking equilibrium in the y-direction gives
ΣFy=-18N+20N-20N+18N=0
Therefore, equilibrium in the y-direction is satisfied. Taking moments about point A,
ΣMA=(20N)(7cm)+(10N)(5cm)-18N(7cm)=64Ncm
Therefore, moment equilibrium is not satisfied, and the rectangle is not in equilibrium.
2.6 Sketch c shows the forces acting on a triangle. Is the triangle in equilibrium?
Notes: To satisfy equilibrium, Equations (2.3) must be satisfied.
Solution:
Taking the sum of forces in the x-direction,
ΣFx=65.37N-40N(cos10°)-30N(cos30°)=0.00
Therefore, equilibrium in the x-direction is satisfied. Taking equilibrium in the y-direction gives
ΣFy=39.76N-17.81N-40N(sin10°)-30N(sin30°)=0.00
Therefore, equilibrium in the y-direction is satisfied. Taking moments about point A,
ΣMA=(39.76N)(2.5cm)-40N(cos10°)(5cm)+30N(sin30°)(6.5cm)=0.00Ncm
Therefore, the triangle is in equilibrium.
www.elsolucionario.net
Page 2-4
2.7 Sketch d shows a cube with side lengths a and eight forces acting at the corners. Is the
cube in equilibrium?
Notes: For three dimensional equilibrium, Equations (2.1) and (2.2) must be satisfied.
Solution:
Taking the sum of forces in the x-direction,
ΣFx=P+P-P-P=0
Therefore, equilibrium in the x-direction is satisfied. Taking equilibrium in the y-direction gives
ΣFy=P+P-P-P=0
Therefore, equilibrium in the y-direction is satisfied. There are no forces in the z-direction, so
equilibrium is automatically satisfied. Taking moments about the z-axis,
ΣMoz=Pa+Pa-Pa-Pa=0
Therefore, equilibrium about the z-axis is satisfied. Taking moments about the x-axis,
ΣMoz=Pa-Pa=0
Therefore, equilibrium about the x-axis is satisfied. Taking moments about the y-axis,
ΣMoy=Pa-Pa=0
Therefore, the cube is in equilibrium.
2.8 A downhill skier stands on a slope with a 5° inclination. The coefficient of friction
between the skis and the snow is 0.10 when the ski is stationary and 0.07 when the ski
starts to slide. Is the skier in static equilibrium when he is stationary and when he slides
down the slope?
Notes: For equilibrium, the force Equations in (2.3) needs to be satisfied.
www.elsolucionario.net
Page 2-5
Solution:
A free-body diagram of the skier is shown to
the right. The skier’s weight is mag, and the
slope has a normal force as a reaction and a
friction force which develops tangent to the
slope. Summing forces in the direction normal
to the slope yields:
ΣF=N-magcos5°; N=magcos5°
Taking forces tangent to the slope,
ΣF=magsin5°-µN=magsin5°-µmagcos5°; µ=tan5°=0.08749
If the skier is stationary, the coefficient of friction is 0.10, and there is enough friction to keep the
skier stationary. If the skier is moving, then the coefficient of friction is 0.07, so the friction force
developed is not enough to maintain equilibrium. The skier will accelerate.
2.9 Given the components shown in sketches e and f, draw the free-body diagram of each
component and calculate the forces.
Notes: For equilibrium, the force Equations in (2.3) needs to be satisfied. Assume there is no
friction in sketch (e). In sketch f, moment equilibrium shows that the only possible force in each
member is an axial force.
Solution:
The free body diagram for sketch e is shown to the left. Flat surfaces give normal reaction forces
as shown. From Equation (2.3), and summing forces in the y-direction,
ΣFy=0=-W+N1cos30°+N2cos60°; N1=W/cos30°-N2(cos60°/cos30°)
Summing forces in the x-direction,
ΣFx=0=N1sin30°-N2sin60°=(W/cos30°-N2(cos60°/cos30°))sin30°-N2sin60°; N2=0.5W
Substituting into the x-direction equilibrium equation gives N1=0.866W.
The free bodydiagram for sketch f is shown to the right. For equilibrium in the x-direction,
ΣFx=0=P-P2sin30°; P2=2P=2.4kN
Taking equilibrium in the y-direciton.
ΣFy=-P1+P2cos30°=-P1+2Pcos30°; P1=1.732P=2.08kN
www.elsolucionario.net
Page 2-6
2.10 A 5-m-long bar is loaded as shown in sketch g. The bar cross section is constant along its
length. Draw the shear and moment diagrams and locate the critical section.
Notes: In such problems, one must first obtain the reactions before generating the shear and
moment diagrams. Also, it is useful to consider moment equilibrium first. In this problem, the
shear and moment diagrams are obtained through the method of sections.
Solution:
The reaction forces have been added to the figure in red. Summing moments about the origin
gives
ΣMo=0=10kN(2m)-Ay(3m)+3.2kN(5m);Ay=12kN
Summing forces in the y-direction yields
ΣFy=0=-Oy+10kN-Ay+3.2kN; Oy=10kN-12kN+3.2kN=1.2kN
The method of sections described on page 39 is used to obtain the shear and moment diagrams.
This proceeds as follows:
a) 0≤x<2m: ΣFy=0=-Oy+V; V=Oy=1.2kN
ΣMO=0=M-V(x); M=Vx=(1.2kN)x
Note: at x=2m, M=2.4kNm
b) 2m≤x<3m: ΣFy=0=-Oy+10kN+V;V=Oy-10kN=-8.8kN
ΣMO=0=M-V(x)-10kN(2m); M=20kNm-(8.8kN)x
c)3m≤x: ΣFy=0=-Oy+10kN-12kN+V;V=3.2kN
ΣMO=0=M-V(x)-10kN(2m)+12kN(3m);
M=-16kNm+3.2x
The shear and moment diagrams are plotted as follows:
From these shear and moment diagrams, it can be seen that the critical section is at point A.
2.11 A beam is supported by a wire in one end and by a pin in the other end. The beam is
loaded by a force parallel with the wire. What is the direction of the reaction force in the
pin?
Notes: This merely requires an understanding of the equations of equilibrium. One of the
fundamental notions is that “you can’t push on a rope”, so that the wire can only support an axial,
tensile load.
www.elsolucionario.net
Page 2-7
Solution:
Force equilibrium gives that the force in the pin has to be parallel with the wire since the wire can
only transmit an axial force. Therefore, the force in the pin has to be in the same direction as in
the wire.
2.12 Sketch h shows a simple bridge. The midpoint of the bridge is held up by a wire leading
to the top hinge of the two equal beams. When a truck runs over the bridge, the force in
the wire is measured to be 500,000 N. Determine the forces in the beams and the
horizontal force component in the middle of the bridge.
Notes: This problem can be easily solved by drawing a free body diagram of one of the beams
and then applying the equations of equilibrium.
Solution:
A free body diagram of the left beam is shown to the right. It is assumed that each beam takes
one-half the load, as shown. From moment equilibrium, about the pin, it can be seen that
ΣM=0=(250,000N)(lcos45°)-PH(lsin45°); PH=250,000N
Therefore, the load carried by the beam is
Pbeam = 250kN( )2 + 250kN( )2 = 353.6kN
Taking force equilibrium in the horizontal direction gives
ΣFh=0=Rx-PH; Rx=PH=250kN
This is the horizontal force component in the middle of the bridge.
2.13 Sketch i shows a 0.05-m-diameter steel shaft supported by self-aligning bearings (which
can provide radial but not bending loads on the shaft). A gear causes each force to be
applied as shown. All length dimensions are in meters. The shaft can be considered
weightless. Determine the forces at A and B and the maximum bending stress. Draw
shear and moment diagrams.
www.elsolucionario.net
Page 2-8
Notes: In such problems, one must first obtain the reactions before generating the shear and
moment diagrams. Also, it is useful to consider moment equilibrium first. In this problem, the
shear and moment diagrams are obtained through direct integration, as suggested by Equations
(2.4) and (2.5). The maximum stress is given by σ=Mc/I where I is the moment of inertia, given
in Table 4.1 on page 148.
Solution:
Taking moment equilibrium about point A yields:
ΣMA=0=(1kN)(0.2m)-By(0.5m)-(0.5kN)(0.64m); By=-0.24kN=-240N
Vertical force equilibrium gives Ay:
ΣFy=0=Ay-1kN+By+0.5kN; Ay=1kN-By-0.5kN=0.74kN=740N
The shear and moment diagrams are constructed directly from the applied load, using Equations
(2.4) and (2.5). Since the loads are simple, this approach allows simple integration. The diagrams
are as follows:
As can be seen, the maximum moment is at x=0.2 and has the value of
Mmax=(740N)(0.2m)=148Nm. This is the area under the shear force curve, indicated in red.
The moment of inertia of a round cross section is (see Table 4.1 on page 148, or the inside back
cover):
I = πd
4
64
= π 0.05m( )
4
64
= 3.068× 10−7 m4
The maximum bending stress in the shaft is:
σ = Mc
I
= 148Nm( ) 0.025m( )
3.068× 10−7 m4
= 12.06MPa
2.14 Sketch j shows a simply supported bar loaded with a force P at a position one-third of the
length from one of the support points. Determine the shear force and bending moment in
the bar. Also, draw the shear and moment diagrams.
Notes: In such problems, one must first obtain the reactions before generating the shear and
moment diagrams. Also, it is useful to consider moment equilibrium first. In this problem, the
shear and moment diagrams are obtained through direct integration, as suggested by Equations
(2.4) and (2.5).
www.elsolucionario.net
Page 2-9
Solution:
Moment equilibrium about point A yields:
ΣMA=0=P(l/3)-By(l); By=P/3
Vertical force equilibrium gives:
ΣFy=0=Ay-P+By; Ay=P-By=P-P/3=2P/3
The shear and moment diagrams are obtained through direct integration and are shown below.
2.15 Sketch k shows a simply supported bar loaded by two equally large forces P at a distance
l/4 from its ends. Determine the shear force and bending moment in the bar, and find the
critical section with the largest bending moment. Also, draw the shear and moment
diagrams.
Notes: In such problems, one must first obtain the reactions before generating the shear and
moment diagrams. Also, it is useful to consider moment equilibrium first. In this problem, the
shear and moment diagrams are obtained through direct integration, as suggested by Equations
(2.4) and (2.5).
Solution:
Moment equilibrium about point A yields:
ΣMA=0=P(l/4)+P(3l/4)-By(l); By=P
Vertical force equilibrium gives
ΣFy=0=Ay-P-P+By; Ay=P
The shear and moment diagrams are obtained through direct integration and are shown below.
The largest bending moment occurs between the two applied forces and has the magnitude of
M=Pl/4.
2.16 The bar shown in sketch l is loaded by the force P. Determine the shear force and
bending moment in the bar. Also, draw the shear and moment diagrams.
www.elsolucionario.net
Page 2-10
Notes: In such problems, one must first obtain the reactions before generating the shear and
moment diagrams. Also, it is useful to consider moment equilibrium first. In this problem, the
shear and moment diagrams are obtained through direct integration, as suggested by Equations
(2.4) and (2.5).
Solution:
Moment equilibrium about point A yields:
ΣMA=0=Pl+Byl; By=-P
Vertical force equilibrium gives:
ΣFy=0=-P+Ay+By; Ay=2P
The shear and moment diagrams are obtained through direct integration and are shown below.
2.17 Sketch m shows a simply supported bar with a constant load per unit length w0 imposed
over its entire length. Determine the shear force and bending moment as functions of x.
Draw a graph of these functions. Also, find the critical section with the largest bending
moment.
Notes: In such problems, one must first obtain the reactions before generating the shear and
moment diagrams. Also, it is useful to consider moment equilibrium first. In this problem, the
shear and moment diagrams are obtained through direct integration, as suggested by Equations
(2.4) and (2.5).
Solution:
For the purposes of equilibrium, the distributed load can be replaced by a point load of magnitude
w0l at the midpoint of the span. Therefore, moment equilibrium about point A yields:
MA = 0 = w0l
l
2
 
  
 
  − Byl∑ ; By =
w0 l
2
Force equilibrium in the vertical direction gives:
Fy = 0 = Ay − w0l∑ +By ; Ay =
w0 l
2
The shearand moment diagrams are obtained through direct integration and are shown below.
www.elsolucionario.net
Page 2-11
It can be seen that the maximum moment occurs at mid-span. The magnitude of this force is the
area under either triangle on the shear curve, as suggested by Equation (2.5). This gives
Mmax =
1
2
wol
2
 
  
 
  
l
2
 
  
 
  =
w0l
2
8
2.18 Sketch n shows a simply supported beam loaded with a ramp function over its entire
length, the largest value being 2P/l. Determine the shear force and the bending moment
and the critical section with the largest bending moment. Also, draw the shear and
moment diagrams.
Notes: One must first obtain the reactions before generating the shear and moment diagrams.
Also, it is useful to consider moment equilibrium first. In this problem, the shear and moment
diagrams are obtained through the method of sections.
Solution:
The loading is described by
q x( ) = − 2P
l
x
l
 
  
 
  = −
2Px
l2
In analyzing equilibrium, the loading is replaced by a force of magnitude P at a location of
x=2l/3. Therefore, moment equilibrium about point A gives:
MA = 0 = P
2 l
3
 
  
 
  − By l( ); By =
2P
3
∑
Vertical force equilibrium gives:
Fy = 0 = Ay − P + By ; Ay = P − By =
P
3
∑
Taking a section at a location x in the span,
Fy = 0 = Ay −
1
2
 
  
 
  
2Px
l2
 
  
 
  
x( )∑ −V ; V = P
3
− Px
2
l2
MA = 0 =
1
2
 
  
 
  
2Px
l2
 
  
 
  
x( ) 2
3
x
 
  
 
  ∑ + Vx − M;
 M = 2Px
3
3l2
+
P
3
− Px
2
l2
 
 
  
 
 
  x =
Px
3
− Px
3
3l2
The shear and moment diagrams are shown below:
www.elsolucionario.net
Page 2-12
 The maximum moment occurs when V=0, which occurs at:
0 = P
3
− Px
2
l2
; x = l
3
Substituting this into the moment equation gives
Mmax =
P
3
l
3
 
  
 
  
− P
3l2
l
3
 
  
 
  
3
= Pl
3 3
− Pl
9 3
= 2Pl
9 3
2.19 Draw the shear and moment diagrams, and determine the critical section for the loading
conditions shown in sketch o. Neglect the weight of the bar. Let w0=100kN/m, P=5kN,
M=2x10
4
N-m, and l=300mm.
Notes: One must first obtain the reactions before generating the shear and moment diagrams.
Also, it is useful to consider moment equilibrium first. In this problem, the shear and moment
diagrams are obtained through singularity functions. Table 2.2 on page 43 is useful for writing
the load distribution in terms of singularity functions.
Solution:
For equilibrium reasons, the distributed loads can be replaced by point loads as shown below:
Taking moment equilibrium about point A gives:
MA = 0 = 5kN( )
0.200m
3
 
  
 
  +10kN 0.15m( )+ 5kN 0.233 m( ) − By 0.2m( )+ 5kN 0.3m( ) + 20kNm∑
or By=22.6kN. Taking force equilibrium in the vertical direction gives:
Fy = 0 = Ay − 5kN −10kN − 5kN + By − 5kN ;Ay = −2.4kN∑
Therefore, the load distribution can be written in terms of singularity functions as (see Table 2.2
on page 43):
−q x( ) = 2.4kN x −1 − 1kN / m x 1 − x − 0.1 m 1[ ] + 22.6 kN x − 0.2 m −1
+1kN / m x − 0.2m 1 − x − 0.3 m 1[ ]− 5kN x −0.3m −1 + 20kNm x − 0.3m −2
Therefore, the shear distribution is given by Equation (2.4) as
V x( )= 2.4kN x 0 − 1kN / m
2
x 2 − x − 0.1m 2 − x − 0.2m 2 + x − 0.3m 2[ ]
+22.6kN x − 0.2m 0 − 5kN x − 0.3m 0 + 20kNm x − 0.3m −1
And the moment distribution is given by Equation (2.5) as
www.elsolucionario.net
Page 2-13
M x( ) = 2.4 kN x 1 − 1kN / m
6
x 3 − x − 0.1 m 3 − x − 0.2 m 3 + x − 0.3m 3[ ]
+22.6kN x − 0.2 m 1 − 5kN x − 0.3m 1 + 20kNm x − 0.3m 0
The shear and moment diagrams are sketched below:
2.20 Redo problem 2.19 while considering the weight of the bar per unit length to be
w0=5N/m.
Notes: This problem is useful after Problem 2.19 has been completed, since the statics then
suggests the answer - that the beam weight is negligible.
Solution:
Taking moment equilibrium about point A gives:
MA = 0 = 5kN( )
0.200m
3
 
  
 
  
− 5N / m( ) 0.3m( ) 0.15m( )∑
+10kN 0.15m( ) + 5kN 0.233m( ) − By 0.2m( ) + 5kN 0.3m( ) + 20kNm
or By=22.601kN. Taking force equilibrium in the vertical direction gives:
Fy = 0 = Ay − 5kN −10kN − 5kN + By − 5kN ;Ay = −2.4004kN∑
These reactions are so close to the reactions in Problem 2.19 that the difference in the shear and
moment diagrams is negligible. The derivation is identical to Problem 2.19, but the load function
has two terms associated with the weight.
2.21 Find the expressions for the shear force, bending
moment, and axial load P for the curved member shown
in sketch p. Neglect the weight of the bar. Let P=10N
and r=1m.
Notes: The reactions must be determined before the problem can
be solved. The method of sections can be used to obtain a
solution for this problem.
Solution:
Taking moments about the base of the curved member gives:
ΣM=0=M0-(Pcos30°)r-(Psin30°)r; M0=1.366Pr=13.7Nm
Taking vertical force equilibrium:
ΣFy=0=Ry-Psin30°; Ry=Psin30°=5N
Taking horizontal force equilibrium,
ΣFx=0=Rx+Pcos30°; Rx=-8.66N
Taking a section of the member at an angle θ gives the free body diagram shown below.
Equilibrium in the x-direction gives:
www.elsolucionario.net
Page 2-14
ΣFx=0=Rx+Vcosθ-Psinθ
Equilibrium in the y-direction gives:
ΣFy=0=Ry-Vsinθ-Pcosθ
These two equations yield
P=Rxsinθ+Rycosθ=(-8.66N)sinθ+(5N)cosθ
V=-Rxcosθ+Rysinθ=(8.66N)cosθ+(5N)sinθ
Moment equilibrium about the curved member end gives:
ΣM=0=M0+Rx(rsinθ)-Ry(r-rcosθ)+M; M=-M0-rRxsinθ+rRy(1-cosθ)
M=-13.7Nm-(1m)(-8.66N)sinθ+(1m)(5N)(1-cosθ)=-13.7Nm+8.66Nmsinθ+5Nm(1-cosθ)
2.22 Sketch q shows a sinusoidal distributed force applied to a beam. Determine the shear
force and bending moment for each section of the beam.
Notes: The reactions must be determined before the problem can be solved. This solution uses
singularity functions to obtain the shear and bending moment diagrams.
Solution:
The sinusoidal distribution is symmetric, so we know that the statically equivalent load can be
placed at the center of the beam, and has a magnitude of
P = w0 sin
πx
l
 
  
 
  dx0
l∫ = − lw0π cos
πx
l
 
  
 
  x=0
x= l
= 2 lw0
π
Taking moment equilibrium about point A,
MA = 0 = P
l
2
 
  
 
  − By l( ); By =
P
2
= lw0
π∑
Vertical force equilibrium gives
Fy = Ay − P + By ;Ay = P − By =
2lw0
π
−∑ lw0π =
lw0
π
The load function is given by
−q x( )= − lw0
π
x −1 + w0 sin
πx
l
 
  
 
  −
lw0
π
x − l −1
Equation (2.4) gives
V x( )= − q x( )dx∫ = − lw0π x
0 − lw0
π
cos
πx
l
 
  
 
  +
lw0
π
− lw0
π
x − l 0
Equation (2.5) gives
M x( ) = V x( )dx∫ = − lw0π x
1 − l
2 w0
π2
sin
πx
l
 
  
 
  +
lw0x
π
− lw0
π
x − l 1
2.23 Find the length c that gives the smallest maximum bending moment for the load
distribution shown in sketch r.
www.elsolucionario.net
Page 2-15
Notes: The reactions must be determined before the problem can be solved. This solution uses
method of sections to obtain the shear and bending moment diagrams. Note that large negative
bending stresses are still objectionable. For small values of c, note that the moment in the center
part will be large; for large values of c, the moment towards the edges will be larger. Therefore,
the solution requires that the absolute value of the largest negative moment equal the largest
positive moment.
Solution:
Because of symmetry, both R1 and R2 will have the value of w0l/4. Note that the shear and
bending moment diagrams will be symmetric. Therefore, using the method of sections between c
and l/2, and considering only moment equilibrium,
Mx = 0 =
w0x
2
2 l
x
3
 
  
 
  
−
w0l
4
x − c( ) + M; M = w0l
4
x − c( ) −∑ w0 x
3
6 l
At x=c, this gives a value of
Mx=c =
w0l
4
c − c( ) − w0c
3
6l
= −
w0c
3
6l
while at x=l/2,
Mx= l / 2 =
w0 l
4
l
2
− c
 
  
 
  
−
w0
l
2
 
  
 
  
3
6l
=
w0l
2
8
−
w0lc
4
−
w0l
2
48
=
5w0 l
2
48
−
w0lc
4
Using the requirement that Mx=l/2=-Mx=c gives:
w0c
3
6l
=
5w0 l
2
48
−
w0lc
4
; 
c3
6l
=
5l2
48
−
lc
4
This is solved numericallyas c=0.380l.
2.24 Draw the shear and moment diagrams for the load distribution shown in sketch s and the
load intensity of q(x)=<x>
n
 when n=2 and 3. Also, calculate the reaction forces.
Notes: The statics is difficult in this problem, since the load distribution requires one to obtain the
centroid through integration. Once the reactions are found, the shear and moment diagrams can be
found through the use of singularity functions.
www.elsolucionario.net
Page 2-16
Solution:
To obtain the forces, the load distribution is replaced by a concentrated force. The magnitude of
the force is given by:
P = kxn dx
0
1
∫ =
k
n + 1
xn+1
 
  
 
  0
l
=
kln+1
n + 1
The position of the concentrated force for this statically equivalent system is given by:
x =
xkxndx
0
l
∫
kx ndx
0
l
∫
=
kxn+1dx
0
l
∫
kxndx
0
l
∫
=
k
n + 2
ln+2
 
  
 
  
k
n + 1
ln+1
 
  
 
  
=
n + 1
n + 2
l
Moment equilibrium about point 1 yields R2 as follows:
M = 0∑ = R2 l − Px ; R2 = Px l =
1
l
kln+1
n +1
 
 
  
 
 
  
n + 1
n + 2
l
 
  
 
  
=
kln+1
n + 2
Taking vertical force equilibrium,
Fy = 0 = R1 + R2 − P; R1 = P − R2 =
kln+1
n +1
−
kln+1
n + 2
=
kln+1
n +1( ) n + 2( )∑
Therefore, the load distribution function for n=2 is given by
q x( ) = −R1 x −1 + k x 2 − k x − l 2 − 2kl x − l 1 − kl2 x − l 0 − R2 x − l −1
Using Equation (2.4), the shear in terms of singularity functions is
−V x( ) = −R1 x 0 +
k
3
x 3 − x − l 3( ) − kl x − 1 2 − kl2 x − l 1 − R2 x − l 0
Equation (2.5) gives the moment distribution in terms of singularity functions as
−M x( ) = −R1 x 1 +
k
12
x 4 − x − l 4( ) − kl3 x −1
3 −
kl 2
2
x − l 2 − R2 x − l
1
These are sketched below.
2.25 Draw the shear and moment diagrams and give the reaction forces for the load
distribution shown in sketch t.
www.elsolucionario.net
Page 2-17
Notes: One must first obtain the reactions before generating the shear and moment diagrams.
Also, it is useful to consider moment equilibrium first. In this problem, the shear and moment
diagrams are obtained through direct integration, as suggested by Equations (2.4) and (2.5).
Solution:
Due to symmetry, the reaction forces are equal, and have the value Ay=By=w0a. The shear and
moment diagrams are obtained through direct integration and are as follows:
2.26 The loading acting on a beam consists of a distributed load of 2500 N/m over the entire
beam, and three concentrated loads, each of 5000N, located at l/4, l/2, and 3l/4. The beam
length is 8m. Choose the type of beam support and the position. Also, draw the shear and
moment diagrams.
Notes: There is much freedom in this problem to choose the support. This solution uses a beam
which is simply supported at its ends, so that the problem is illustrated as follows.
Solution:
From symmetry, the reactions are simply calculated as Ay=By=17.5kN. Therefore, the shear and
moment diagrams, obtained through direct integration, are as follows:
2.27 The maximum load for a bridge crossing a river occurs when a fully loaded truck passes
over. The maximum load is 550,000N distributed over 22m (the farthest distance between
the wheels of a vehicle). The weight of the 100m long bridge is 210 tons. Choose the
position of the bridge pillars and the type of support.
www.elsolucionario.net
Page 2-18
Solution:
This is an open-ended problem. Since the number of pillars is not restricted, any number can be
chosen, but due to accuracy in the building process it should be assumed that each pillar carries
its own load independent of the other pillars. Thus the bridge works as if it is hinged at each
pillar. Choosing the distance between pillars to be 20 or 25m with the support type hinged at one
end of the bridge and simply supported, e.g., on rollers, on the other pillars to accommodate the
thermal expansion due to temperature variation.
2.28 The simply supported bar shown in sketch u has P1=5kN, P2=8kN, w0=4kN/m, and
l=12m. Use singularity functions to determine the shear force and bending moments as
functions of x. Also, draw your results.
Notes: One must first obtain the reactions before generating the shear and moment diagrams.
Also, it is useful to consider moment equilibrium first. Table 2.2 on page 43 helps in writing the
load in terms of singularity functions.
Solution:
Summing moments about A gives:
MA = 0 = P1
l
4
 
  
 
  
− w0l
2
 
  
 
  
l
2
 
  
 
  
+ P2
3l
4
 
  
 
  
− w0l
8
 
  
 
  
3 l
4
+ l
12
 
  
 
  
+ By l( )∑
By = −
5kN
4
+ 4kN / m( ) 12m( )
4
− 3 8kN( )
4
+ 5 4kN / m( ) 12m( )
48
= 9.75kN
Vertical force equilibrium gives
Fy∑ = 0 = Ay + P1 − w0 l2 + P2 −
w0 l
8
+ By = 0 ; Ay = 7.25kN
Using the information in Table 2.2 on page 43, the load using singularity functions is:
−q x( )= Ay x −1 + P1 x −
l
4
−1
+ P2 x −
3l
4
−1
+ By x − l
−1 − w0 x −
l
4
0
+ 4w0
l
x − 3l
4
1
− 4w0
l
x − l 1
Using Equation (2.4), the shear in terms of singularity functions is
V x( )= −q x( )dx∫
= Ay x
0 + P1 x −
l
4
0
+ P2 x −
3l
4
0
+ By x − l
0 − w0 x −
l
4
1
+ 2w0
l
x − 3l
4
2
− 2w0
l
x − l 2
Equation (2.5) gives the moment distribution in terms of singularity functions as
M x( ) = V x( )dx∫
= Ay x
1 + P1 x −
l
4
1
+ P2 x −
3 l
4
1
+ By x − l
1 − w0
2
x − l
4
2
+ 2w0
3l
x − 3l
4
3
− 2w0
3l
x − l 3
These are sketched below.
www.elsolucionario.net
Page 2-19
2.29 Use singularity functions for the force system shown in sketch v to determine the load
intensity, the shear force, and the bending moment. From a force analysis determine the
reaction forces R1 and R2. Also, draw the shear and moment diagrams.
Notes: One must first obtain the reactions before generating the shear and moment diagrams.
Also, it is useful to consider moment equilibrium first. Table 2.2 on page 43 helps in writing the
load in terms of singularity functions.
Solution:
Taking moment equilibrium about point O,
MO∑ = 0 = 40 lb( ) 4in( ) − 30lb( ) 8in( )− R2 14in( )+ 60lb( ) 18in( ); R2 = 71.43lb
Taking force equilibrium in the vertical direction,
ΣFy=0=R1-40lb+30lb+R2-60lb; R1=-1.43lb
Using the information in Table 2.2 on page 43, the load distribution can be written in terms of
singularity functions as:
−q x( )= −1.43lb x −1 − 40lb x − 4in −1 + 30lb x − 8in −1 + 71.43lb x −14in −1 − 60lb x −18in −1
Using Equation (2.4), the shear distribution in terms of singularity functions is
V x( )= −1.43lb x 0 − 40lb x − 4in 0 + 30lb x −8in 0 + 71.43 lb x −14in 0 − 60lb x −18in 0
Using Equation (2.5), the moment distribution in terms of singularity functions is:
M x( ) = −1.43lb x 1 − 40lb x − 4in 1 + 30lb x − 8in 1 + 71.43 lb x −14in 1 − 60lb x −18in 1
The shear and moment diagrams are sketched below.
www.elsolucionario.net
Page 2-20
2.30 Use singularity functions for the force system shown in sketch w to determine the load
intensity, the shear force, and the bending moment. Draw the shear and moment
diagrams. Also, from a force analysis, determine the reaction forces R1 and R2.
Notes: This problem is very similar to problem 12.29. One must first obtain the reactions before
generating the shear and moment diagrams. Also, it is useful to consider moment equilibrium
first. Table 2.2 on page 43 helps in writing the load in terms of singularity functions.
Solution:
Taking moment equilibrium about point O,
ΣMO=0=(35lb)(3in)+(40lb)(7in)+(60lb)(8in)-R2(18in); R2=48.06lb
Taking force equilibrium in the vertical direction,
ΣFy=0=R1-35lb-40lb-60lb+R2; R1=86.94lb
Using the information in Table 2.2 on page 43, the load distribution can be written in terms of
singularity functions as:
−q x( )= 86.94lb x −1 − 35lb x − 3in −1 − 40lb x − 7in −1 − 60lb x − 8in −1 + 48.06lb x −18in −1
Using Equation (2.4), the shear distribution in terms of singularity functions is
V x( )= 86.94lb x 0 − 35lb x − 3in 0 − 40 lb x − 7in 0 − 60lb x − 8in 0 + 48.06 lb x −18in 0
Using Equation (2.5), the moment distribution in terms of singularityfunctions is:
M x( ) = 86.94lb x 1 − 35lb x − 3in 1 − 40lb x − 7in 1 − 60lb x − 8in 1 + 48.06lb x −18in 1
The shear and moment diagrams are sketched below.
2.31 Draw a free-body diagram of the forces acting on the simply supported bar shown in
sketch x, with w0=6kN/m and l=12m. Use singularity functions to draw the shear force
and bending moment diagrams.
www.elsolucionario.net
Page 2-21
Notes: One must first obtain the reactions before generating the shear and moment diagrams. The
statics is greatly simplified by the symmetry of the problem. Table 2.2 on page 43 helps in
writing the load in terms of singularity functions.
Solution:
Because of symmetry,
Ay = By =
1
2
2w0 l
3
 
  
 
  =
w0l
3
= 6kN / m( ) 12m( )
3
= 24kN
Using the information in Table 2.2 on page 43, the load distribution can be written in terms of
singularity functions as:
−q x( )= 24kN x −1 − 3w0
l
x 1+ 3w0
l
x − l
3
1
+ 3w0
l
x − 2 l
3
1
+ 24kN x − l −1 − 3w0
l
x − l 1
Using Equation (2.4), the shear distribution in terms of singularity functions is
V x( )= 24kN x 0 − 3w0
2 l
x 2 + 3w0
2 l
x − l
3
2
+ 3w0
2 l
x − 2 l
3
2
+ 24kN x − l 0 − 3w0
2 l
x − l 2
Using Equation (2.5), the moment distribution in terms of singularity functions is:
M x( ) = 24kN x 1 − w0
2l
x 3 + w0
2l
x − l
3
3
+ w0
2l
x − 2 l
3
3
+ 24kN x − l 1 − w0
2l
x − l 3
= 24kN x 1 − 1kN
4m2
x 3 + 1kN
4m2
x − l
3
3
+ 1kN
4m2
x − 2 l
3
3
+ 24kN x − l 1− 1kN
4m2
x − l 3
The shear and moment diagrams are sketched below.
2.32 Sketch y shows a simply supported bar with w0=5kN/m and l=12m. Draw a free-body
diagram of the forces acting along the bar as well as the coordinates used. Use singularity
functions to determine the shear force and the bending moment.
www.elsolucionario.net
Page 2-22
Notes: One must first obtain the reactions before generating the shear and moment diagrams. The
statics is greatly simplified by the symmetry in the problem. Table 2.2 on page 43 helps in writing
the load in terms of singularity functions.
Solution:
Because of symmetry,
Ay = By =
1
2
w0l
2
 
  
 
  =
w0l
4
= 5kN / m( ) 12m( )
4
= 15kN
Using the information in Table 2.2 on page 43, the load distribution can be written in terms of
singularity functions as:
−q x( )= 15kN x −1 − 5 kN
m
x
0 + 5kN
6m2
x
1 − 5kN
3m2
x − 6m 1 + 5kN
6m2
x −12m 1 + 5kN / m x − 12 0 + 15 x −12 −1
Using Equation (2.4), the shear distribution in terms of singularity functions is
V x( )= 15kN x 0 − 5 kN
m
x 1 + 5kN
12m2
x 2 − 5kN
6m2
x − 6m 2 + 5kN
12m2
x −12m 2 + 5kN / m x −12 1 + 15 x −12 0
Using Equation (2.5), the moment distribution in terms of singularity functions is:
M x( ) = 15kN x 1 − 5kN
2m
x
2 + 5kN
36m2
x
3 − 5kN
18m2
x − 6m 3 + 5kN
36m2
x −12m 3 + 5kN
2m
x −12 2 +15 x −12 1
The shear and moment diagrams are sketched below.
2.33 Redo the case study 2.1 problem if the 400-N force is evenly distributed along the width
of the roller and a unit step is used to represent the loading. The width of the roller is 30 mm. Do
both (a) and (b) portions of the case study while considering a unit step representation.
Notes: The approach is very similar to that in the case study, except that the singularity functions
are slightly different to reflect periodic distributed loads instead of periodic point loads.
Solution:
The new load distribution function is
www.elsolucionario.net
Page 2-23
−q x( )= −800 N x −1 + 400N
0.03m
x − 0.060m 0 − 400 N
0.03m
x − 0.090m 0 + 400 N
0.03m
x − 0.13m 0 − 400N
0.03m
x − 0.16m 0
+ 400 N
0.03m
x − 0.20m 0 − 400 N
0.03m
x − 0.23m 0 + 400N
0.03m
x − 0.27m 0 − 400 N
0.03m
x − 0.30m 0 − 800 N x − 0.36m −1
Using Equation (2.4), the shear distribution in terms of singularity functions is
V x( )= −800N x 0 + x − 0.36m 0[ ]+ 400N0.03m x − 0.060m 1 − x − 0.090m 1 + x − 0.13m 1[
− x − 0.16m 1 + x − 0.20m 1 − x − 0.23m 1 + x − 0.27m 1 − x − 0.30m 1]
Using Equation (2.5), the moment distribution in terms of singularity functions is:
M x( ) = −800N x 1 + x − 0.36m 1[ ]+ 400N0.06m x − 0.060m 2 − x − 0.090m 2 + x − 0.13m 2[
− x − 0.16m 2 + x − 0.20m 2 − x − 0.23m 2 + x − 0.27m 2 − x − 0.30m 2]
The shear and moment diagrams are sketched below.
The maximum moment can be evaluated at x=0.18m as
M 0.18m( ) = −800N 0.18m 1[ ]+ 400 N0.06m 0.18m − 0.060m 2 − 0.18m − 0.090m 2[
+ 0.18m − 0.13m 2 − 0.18m − 0.16m 2] = −88Nm
The general expressions are similar to those in the case study, and are:
−q x( )= − nP
2
x −1 + x − 2a − n −1( )b −1[ ]+ Pc x − a − b i −1( )− c2
0
− x − a − b i−1( ) + c
2
0 
 
 
 
 
 
 
 i=1
n
∑
V x( )= − nP
2
x 0 + x − 2a − n −1( )b 0[ ]+ Pc x − a − b i −1( ) − c2
1
− x − a − b i−1( ) + c
2
1 
 
 
 
 
 
 
 i=1
n
∑
M x( ) = − nP
2
x 1 + x − 2a − n −1( )b 1[ ]+ P2c x − a − b i − 1( ) − c2
2
− x − a− b i−1( ) + c
2
2 
 
 
 
 
 
 
 i=1
n
∑
where n is the number of rollers, a is the distance from bearing to first roller, b is the center
distance between rollers and c is the roller width.
2.34 A simply supported beam has a parabolic load distribution beginning at x=0. Use
singularity functions to draw the shear and moment diagrams.
Notes: The maximum value is not specified, so it is given the value of qmax. Equilibrium is slightly
more complex in this case, since the geometry has no simple centroid expression.
www.elsolucionario.net
Page 2-24
Solution:
Moment equilibrium about point A gives
MA = 0 =∑ qmaxx
2
l2
x( )dx − P2l ;P2 =0
l∫ 1l
qmax x
2
l2
x( )dx = qmaxl
40
l∫
Force equlibrium allows determination of P1 as:
FA = 0 =∑ P1 + P2 − qmax x
2
l2
dx; P1 =
−qmaxl
4
+ qmax l
3
= qmax l
120
l∫
Therefore, the load function is
−q x( )= qmax
l
12
x −1 − x
2
l2
+ l
4
x − l −1
 
 
 
 
 
 
 
 
Using Equation (2.4), the shear distribution in terms of singularity functions is
V x( )= qmax
l
12
x 0 − x
3
3l2
+ l
4
x − l 0
 
 
 
 
 
 
 
 
Using Equation (2.5), the moment distribution in terms of singularity functions is:
M x( ) = qmax
l
12
x 1 − x
4
12 l2
+ l
4
x − l 1
 
 
 
 
 
 
 
 
The shear and moment diagrams are sketched below.
2.35 Draw a free-body diagram of the forces acting on the simply suported beam shown in
sketch x and used in Problem 2.31. The central section (l/3<x<2l/3) has the load acting
upward (in the negative y direction) instead of as shown. Use singularity functions to
determine the shear force and bending moments and draw the diagram.
Notes: This is straightforward if Problem 2.31 has already been completed. One must first obtain
the reactions before generating the shear and moment diagrams. The statics is greatly simplified
by the symmetry of the problem. Table 2.2 on page 43 helps in writing the load in terms of
singularity functions.
Solution:
www.elsolucionario.net
Page 2-25
One can see that the areas where the load acts downwards and upwards are equal, so there is no
net load applied. Because of symmetry, Ra=Rb=0. Using the information in Table 2.2 on page 43,
the load distribution can be written in terms of singularity functions as:
−q x( )= 3w0
l
x 1 − 3w0
l
x − l
3
1
− 2w0 x −
l
3
0
+ 2w0 x −
2 l
3
0
− 3w0
l
x − 2 l
3
1
+ 3w0
l
x − l 1
Using Equation (2.4), the shear distribution in terms of singularity functions is
V x( )= 3w0
2l
x 2 − x − l
3
2
− x − 2l
3
2
+ x − l 2
 
 
 
 
 
 
 
 
− 2w0 x −
l
3
1
− x − 2l
3
1 
 
 
 
 
 
 
 
Using Equation (2.5), the moment distribution in terms of singularity functions is:
M x( ) = w0
2l
x 3 − x − l
3
3
− x − 2l
3
3
+ x − l 3
 
 
 
 
 
 
 
 
− w0 x −
l
3
2
− x − 2l
3
2 
 
 
 
 
 
 
 
The shear and moment diagrams are sketched below.
2.36 An extra concentrated force with an intensity of 60kN is applied downward at the center
of the simply supported bar shown in sketch y. Draw a free-body diagram of the forces
acting on the bar. Assume l=12m and w0=5kN/m. Use singularity functions to determine
the shearforce and bending moments and draw the diagrams.
Notes: This problem is straightforward if Problem 2.32 has been completed. One must first obtain
the reactions before generating the shear and moment diagrams. The statics is greatly simplified
by the symmetry in the problem. Table 2.2 on page 43 helps in writing the load in terms of
singularity functions.
Solution:
Because of symmetry,
Ay = By =
1
2
w0l
2
 
  
 
  +
1
2
P = 45kN
www.elsolucionario.net
Page 2-26
Using the information in Table 2.2 on page 43, the load distribution can be written in terms of
singularity functions. Here we decide to ignore the terms acting at the right end of the beam, since
these do not affect the shear and moment diagrams.
−q x( )= 45kN x −1 − 5 kN
m
x 0 + 5kN
6m2
x 1 − 60kN x − 6m −1 − 5kN
3m2
x − 6m 1
Using Equation (2.4), the shear distribution in terms of singularity functions is
V x( )= 45kN x 0 − 5 kN
m
x 1 + 5kN
12m2
x 2 − 60kN x − 6m 0 − 5kN
6m2
x − 6m 2
Using Equation (2.5), the moment distribution in terms of singularity functions is:
M x( ) = 45kN x 1 − 5kN
2m
x 2 + 5kN
36m2
x 3 − 60kN x − 6m 1− 5kN
18m2
x − 6m 3
The shear and moment diagrams are sketched below.
2.37 Use singularity functions to determine the shear and bending moments for the loading
situation shown in sketch z.
Notes: One must first obtain the reactions before generating the shear and moment diagrams.
Table 2.2 on page 43 helps in writing the load in terms of singularity functions.
Solution:
Moment equilibrium about point A gives:
MA = 0 = 2Pl +
w0l
2
 
  
 
  
l
3
 
  
 
  + 2w0 l( ) 2 l( ) +∑
w0 l
2
 
  
 
  
11l
3
 
  
 
  + P 3l( ) − By 4l( ); By =
5P
4
+ 3w0l
2
Force equilibrium gives:
Fy = 0 = Ay + By − 3P − 3w0l ; Ay = 3P + 3w0 l∑ − By = 3w0 l2 +
7P
4
Therefore, the load distribution is given by
−q x( ) = −Ay x −1 + w0 x 0 −
w0
l
x 1 + 2 P x − l −1 + 3w0
l
x − l 1
− 4w0
l
x − 2 l 1 + P x − 3l −1 + 3w0
l
x − 3l 1
−q x( ) = 3w0l
2
+ 7P
4
 
  
 
  
x −1 + w0
l
l x 0 − x 1 + 3 x − l 1 − 4 x − 2 l 1 +3 x − 3l 1[ ]+ P 2 x − l −1 + x − 3l −1[ ]
www.elsolucionario.net
Page 2-27
Note that we have ignored the terms which become active at x=4l, since these do not affect the
diagram or calculations on the beam. The shear distribution is obtained from Equation (2.4) as:
V x( )= 3w0 l
2
+ 7P
4
 
  
 
  x
0 + w0
2l
2l x 1 − x 2 + 3 x − l 2 − 4 x − 2 l 2 + 3 x −3 l 2[ ]+ P 2 x − l 0 + x − 3l 0[ ]
Using Equation (2.5), the moment distribution in terms of singularity functions is:
M x( ) = 3w0l
2
+ 7P
4
 
  
 
  x
1 + w0
6l
3l x 2 − x 3 + 3 x − l 3 − 4 x − 2 l 3 + 3 x − 3l 3[ ]+ P 2 x − l 1 + x − 3l 1[ ]
2.38 Obtain the shear force and moment expressions using singularity functions for a pinned
beam at both ends with loading conditions described in sketch aa. Draw a free-body
diagram of the forces acting on the bar. Draw the shear force and moment along the
length of the bar and give tabular results. Assume w0=4kN/m, P=3kN, and l=12m.
Notes: One must first obtain the reactions before generating the shear and moment diagrams.
Table 2.2 on page 43 helps in writing the load in terms of singularity functions.
Solution:
Moment equilibrium about point A gives:
MA = 0 = P
l
3
− w0l
6
 
  
 
  
2l
9
 
  
 
  ∑ + Byl − P
2l
3
 
  
 
  +
w0l
6
 
  
 
  
7l
9
 
  
 
  ; By = −
5l
9
 
  
 
  
w0
6
 
  
 
  +
P
3
= −3.444kN
Force equilibrium gives:
Fy = 0∑ = Ay + By + P −P + w0 l6 −
w0l
6
; Ay = −By = 3.444kN
Therefore, the load distribution is given by
−q x( )= Ay x −1 −
3w0
l
x 1 − x − l
3
1
+ x − 2l
3
1 
 
 
 
 
 
 
 
+ w0 x −
l
3
0
+ x − 2l
3
0 
 
 
 
 
 
 
 
+ P x − l
3
−1
− x − 2l
3
−1 
 
 
 
 
 
 
 
Note that we have ignored the terms which become active at x=l, since these do not affect the
diagram or calculations on the beam. The shear distribution is obtained from Equation (2.4) as:
V x( )= Ay x 0 −
3w0
2l
x 2 − x − l
3
2
+ x − 2l
3
2 
 
 
 
 
 
 
 
+ w0 x −
l
3
1
+ x − 2l
3
1 
 
 
 
 
 
 
 
+ P x − l
3
0
− x − 2l
3
0 
 
 
 
 
 
 
 
Using Equation (2.5), the moment distribution in terms of singularity functions is:
M x( ) = Ay x 1 −
w0
2l
x 3 − x − l
3
3
+ x − 2l
3
3 
 
 
 
 
 
 
 
+ w0
2
x − l
3
2
+ x − 2l
3
2 
 
 
 
 
 
 
 
+ P x − l
3
1
− x − 2l
3
1 
 
 
 
 
 
 
 
The shear and moment diagrams are sketched below:
www.elsolucionario.net
Page 2-28
2.39 A steel bar is loaded by a tensile force P=25kN. The cross section of the bar is circular
with a radius of 7mm. What is the normal stress in the bar?
Notes: Equation (2.7) is needed to solve this problem.
Solution:
The cross sectional area of the bar is:
A = πr2 = π 0.007m( )2 = 1.539×10−4 m2
Equation (2.7) gives
σ = P
A
= 25kN
1.539 ×10−4 m2
= 162MPa
2.40 A stainless steel bar of square cross section has a tensile force of P=15kN acting on it.
Calculate how large the side l of the cross-sectional area must be to give a tensile stress in
the bar of 120MPa.
Notes: Equation (2.7) is needed to solve this problem.
Solution:
The cross sectional area of the bar is l2. Therefore, from Equation (2.7),
σ = P
A
; A = l2 = P
σ
; l = P
σ
= 15kN
120MPa
= 0.0112m = 11.2 mm
2.41 What is the maximum length lmax that a copper wire can have if its weight should not give
a higher stress than 75 MPa when it is hanging vertically? The density of copper is 8900
kg/m3, and the density of air is so small relative to that of copper that it may be neglected.
The acceleration of gravity is 9.81 m/s2.
Notes: Equation (2.7) is needed to solve this problem.
Solution:
If the wire has a cross sectional area of A, then its volume is V=Almax. The weight of this wire is
P=ρgAlmax
From Equation (2.7),
www.elsolucionario.net
Page 2-29
σ = P
A
= ρgAlmax
A
= ρglmax ; lmax =
σ
ρg
= 75MPa
8900kg / m3( ) 9.81m / s2( ) = 859m
2.42 A machine weighing 5 tons will be lifted by a steel rod with an ultimate tensile strength
of 860 MPa. A safety factor of 5 is to be used. Determine the diameter needed for the
steel rod.
Notes: Equations (1.1) and (2.7) are needed to solve this problem. Note that a ton is 1000 kg
when the problem is stated in metric units.
Solution:
The design stress is determined from Equation (1.1) as:
ns =
σall
σd
; σd =
σall
ns
= 860MPa
5
=172MPa
The applied load is P=(5000kg)(9.81m/s2)=49.05kN. Therefore, the required diameter is
calculated from Equation (2.7) as:
σ = P
A
= P
πd 2
4
 
 
  
 
 
  
; d = 4P
πσ
= 4 49.05kN( )
π 172MPa( ) = 0.0191m =19.1mm
2.43 The largest measured ocean depth (near the Philippines in the Pacific) is 11 km. What
ultimate strength is required for a steel wire to reach the bottom without being torn apart
by its own weight? The density of steel is 7860kg/m
3
 and that of water is 1000kg/m
3
.
Notes: The approach is very similar to problem 2.41, but now the buoyancy force is subtracted
from the load.
Solution:
If the wire has a cross sectional area of A, then its volume is V=Almax. The load on the top of the
wire is the weight of the wire minus the buoyancy force which tends to support the wire.
Therefore,
P=(ρsteel-ρwater)gAlmax
From Equation (2.7),
σ = P
A
=
ρsteel − ρwater( )gAlmax
A
= ρsteel − ρwater( )glmax
= 7860kg / m3 −1000kg / m3( ) 9.81m / s2( ) 11000m( ) = 740 MPa
Reviewing the properties of steel on the inside front cover, there is no steel that can reach to this
depth without failing at the surface.
2.44 In Problem 2.43 a steel wire is used to measure the ocean depth. Find a way to do the
measurements without risking breaking the wire due to tensile stresses being too high.
www.elsolucionario.net
Page 2-30
Notes: It is very helpful to complete Problem 2.43 before attempting this problem. This is an
open-ended problem, with many possible solutions. Students should beencouraged to develop
their own creative solutions to this problem.
Solution:
The obvious solution is to make the wire of a polymer or other material with exactly the same
density as water and with the same compressibility as water. Each part of the wire is then buoyant
and the tensile stress is zero along the whole wire. Another way is to use a steel wire but attach
floats at regular intervals, this producing buoyancy of the wire parts.
2.45 A string on a guitar is made of nylon and has a cross-sectional diameter of 0.6mm. It is
tightened to a force P=15N. What is the stress in the string?
Notes: Equation (2.7) is needed to solve this problem.
Solution:
The cross sectional area of the string is:
A = πd
2
4
= π 0.0006m( )
2
4
= 2.827×10−7 m2
Equation (2.7) gives
σ = P
A
= 15N
2.827 ×10−7m2
= 53.1MPa
2.46 Determine the normal and shear stresses in sections A and B of sketch bb. The cross-
sectional area of the rod is 0.025m2. Ignore bending and torsional effects.
Notes: To solve this problem, one must be capable of solving for the reactions at A, be familiar
with the method of sections, and be able to use Equation (2.7).
Solution:
From equilibrium, the reactions at the wall include the forces Rx=0, Ry=10kNcos30°=8.66kN, and
Rz=10kNsin10°=5kN. Taking a section at A, and applying Equation (2.7),
σ = −
Py
A
= − 8.66kN
0.025m2
= −346.4kPa
The maximum shear stress is discussed in Chapter 5, but we can estimate the average shear stress
as
τave =
Pz
A
= −5.0kN
0.025m2
= −200kPa
www.elsolucionario.net
Page 2-31
The negative signs are consistent with the sign convention shown on page 33. At location B, there
is no normal force, so σ=0. Since we are told to ignore torsional effects, we can estimate the
average shear stress as:
τave =
P
A
= 10kN
0.025m2
= 400kPa
2.47 Determine the normal and shear stresses due to axial and shear forces at sections A and B
in sketch cc. The cross sectional area of the rod is 0.025m2.
Notes: To solve this problem, one must be capable of solving for the reactions at A, be familiar
with the method of sections, and be able to use Equation (2.7). Also, use θ=30°. Finally, calculate
the average shear stress, not the maximum for a round cross section as is discussed in Chapter 4.
Solution:
Taking a section at AA, there is an axial force and a shear force. While one can work in the x and
y directions, the answer is more easily obtained by applying equilibrium to the coordinate system
which is defined by the section. The free body diagram of the forces (moments are ignored for
this problem) at section AA are:
Therefore, taking force equilibrium in the direction normal to the surface:
ΣF=0=Pn-Psinθ; Pn=Psinθ=(10kN)(sin30°)=5kN
Therefore the normal stress is:
σ =
Pn
A
=
5kN
0.025m2
= 200kPa
Taking force equilibrium in the direction of the surface yields:
ΣF=0=Ps-Pcosθ; Ps=Pcosθ=(10kN)(cos30°)=8.66kN
Therefore the average shear stress is
τ =
Ps
A
=
8.66kN
0.025m2
= 346.4kPa
www.elsolucionario.net
Page 2-32
At section BB, force equilibrium gives Pn=0 and Ps=10kN. Therefore the normal stress is σ=0
and the shear stress is
τ =
Ps
A
=
10kN
0.025 m2
= 400kPa
2.48 For the stress element given in Figure 2.13 determine how large the shear stress will
become if all three normal stresses are doubled.
Notes: This problem is deceptively simple.
Solution:
The normal stresses and shear stresses are independent of each other; if the normal stresses are
doubled, there is no effect on the shear stresses, so they remain the same.
2.49 The normal stresses in the x, y, and z directions are all equal to 100MPa, and the shear
stresses are all zero. Find the stresses acting on a plane having the normal in the direction of the
space diagonal
1
3
,
1
3
,
1
3
 
  
 
  .
Notes: This problem uses Equations (B.4) through (B.9).
Solution:
Refer to the directions in the new plane as x’, etc. These stresses are given by Equations (B.4)
through (B.9) in Appendix B. As long as the directions are orthogonal,
σ ′ x =
1
3
σx + σ y + σz( )+ 0 + 0 + 0 = 100MPa
σ ′ y =
1
3
σ x + σ y + σz( )+ 0 + 0 + 0 = 100MPa
σ ′ z =
1
3
σx + σ y + σz( )+ 0 + 0 + 0 = 100MPa
The shear stresses are:
τ ′ z , ′ x =
σ
3
cos ′ x ,z( )+ cos ′ x ,x( )+ cos ′ x , y( )[ ]+ 0 + 0 + 0
If x’ is perpendicular to y’ and z’, then this gives τz’,x’=0. Similarly, τx’,y’=τy’,z’=0.
2.50 If the stresses are such that σx=-σy and σz=τxy=τxz=τyz=0, find the shear stress acting on
the plane diagonal between the x and y axes and parallel to the z direction.
Notes: Equation (B.8) solves this problem.
Solution:
In this new orientation, note that
www.elsolucionario.net
Page 2-33
cos ′ x ,x( ) = 1
2
, c o s′ x ,y( ) = 1
2
, c o s ′ x ,z( ) = 0
cos ′ y , x( ) = − 1
2
, c o s ′ y ,y( ) = 1
2
, c o s ′ y ,z( ) = 0
cos ′ z ,x( ) = 0 , c o s ′ z , y( ) = 0, cos ′ z , z( ) =1
Using Equation (B.8) on page 906 yields the following, keeping only non-zero terms:
τ ′ x ′ y =
σx
2
− 1
2
 
  
 
  
+
σy
2
1
2
 
  
 
  
+ 0 + 0+ 0 + 0 = − σx
2
+
σ y
2
Since σx=-σy, one can write that τx’y’=-σx=σy.
2.51 A stressed element with shear stress τyx acting in the x-direction on the surface has the
normal in the y direction. Determine the shear stress acting in the y direction on the
surface having a normal in the x direction.
Notes: This merely requires an understanding of equilibrium.
Solution:
Refer to Figure 2.14 (b) to visualize the stresses. Applying moment equilibrium about the z-axes
yields:
M = 0 =∑ τ xydA1 dx2 2( )− τyxdA2
dy
2
2( )
Since dA1=dydz and dA2=dxdz,
τxydxdydz − τyxdxdydz = 0 ; τ xy = τ yx
2.52 Sketch dd shows a distributed load on a
semi-infinite plane. The stress in polar
coordinates based on the plane stress
assumption is
σr = −
2w0 cosθ
πr
σθ=τrθ=τθr=0
Determine the expressions σx, σy, and τxy
in terms of r and θ.
Notes: Since x,y and r,θ are orthogonal coordinates, Equations (2.12) and (2.14) can be used to
solve the problem.
Solution:
The derivation on pages 51 and 52 applies equally well if the desired stress in Equation (2.12) is
σx in terms of σr, σθ, and τrθ. Therefore,
σx = 2τrθ sinθ cosθ + σr cos2 θ +σθ sin 2θ = 0 −
2w0 cosθ
πr
 
  
 
  cos
2θ + 0 = − 2w0 cos
3θ
πr
For σy, use θ’=90°-θ and apply Equation (2.12):
www.elsolucionario.net
Page 2-34
σ y = 2τrθ sin ′ θ cos ′ θ + σ r cos
2 ′ θ + σθ sin
2 ′ θ 
= 2τrθ sinθ cosθ + σr sin
2 ′ θ +σ θ cos
2 ′ θ = 0 − 2w0 cosθ
πr
 
  
 
  
sin2 θ + 0 = −2w0 cosθ sin
2θ
πr
τxy is given by Equation (2.14) as
τxy = τrθ cos2θ −
σ r − σθ
2
s i n 2θ = −σr
sin2θ
2
= −σr sinθ cosθ
2.53 Sketch ee shows loading of an extremely thin and infinitely long plane. Determine the
angle θ needed so that the stress element will have no shear stress. For plate thickness tw,
modulus of elasticity E and Poisson’s ratio ν, find the reduction in thickness of the plate.
Notes: Equation (2.14) gives the required shear stress equation. The thickness reduction is given
by Hooke’s Law, the proper form of which is given in Table B.2 on page 920.
Solution:
The applied stress state is σx, σy, τxy=0. Equation (2.14) gives the shear stress at an orientation φ
as:
τφ = τ xy cos2φ −
σx −σ y
2
sin2 φ = 0 ; s i n 2φ = 0; φ = 0
From Table B.2 on page 920, Hooke’s Law for a biaxial stress state gives:
εz =
σz
E
− ν
E
σx +σ y( ) = − νE σx +σ y( )
From Equation (2.33),
εz = −
ν
E
σ x + σ y( ) = ∆ twtw ; ∆tw = −
νtw
E
σx + σ y( )
2.54 The stress state in a machine element is given by
S =
167 40 −30
40 25 12
−30 12 −25
 
 
 
 
 
 
 
 
 
 
where all the values are in megapascals. Determine the stress tensor if the coordinates are
rotated such that xà y=x’;yà-x=y’; and zàz=z’.
Notes: Equation (B.4) through (B.9) in Appendix B are needed for this problem.
Solution:
Note for this problem, cos(x’,x)=0, cos(x’,y)=1, cos(x’,z)=0, cos(y’,x)=-1, cos(y’,y)=0,
cos(y’,z)=0, cos(z’,x)=0, cos(z’,y)=0, cos(z’,z)=1. From Equation (B.5) on page 905,
www.elsolucionario.netPage 2-35
σx’=(σx)0+σy(1)+σz(0)+2τxy(0)(1)+2τyz(1)(0)+2τzx(0)(0)=σy
Therefore, σx’=25MPa. Similarly, σy’=σx=167MPa, σz’=σz=-25MPa, τx’y’=-τxy=-40MPa, τy’z’=-
τzx=30MPa, τz’x’=τyz=12MPa. Therefore, the stress tensor in the rotated coordinate system is:
S =
25 −40 12
−40 167 30
12 30 −25
 
 
 
 
 
 
 
 
 
 
2.55 A stress tensor is given by
S =
178 −83 0
−83 12 0
0 0 0
 
 
 
 
 
 
 
 
 
 
where all values are in megapascals. The stress state is applied to a machine element
made of AISI 1020 steel. Calculate the principal normal stresses and the principal shear
stresses. Alsom find the stress tensor if the coordinate system is rotated such that
xà z=x’;yày=y’; and zà-x=z’.
Notes: Equations (2.16) and (2.18) are used in this problem in addition to (B.4) through (B.9) as
in Problem 12.54.
Solution:
Note that τxz=τyz=σz=0, so this is a plane stress field. Therefore, Equation (2.16) gives
σ1,σ2 =
σx +σ y
2
± τxy2 +
σ x − σy( )2
4
=
178MPa +12MPa
2
± −83MPa( )2 + 178MPa −12MPa( )
2
4
or σ1=212.4MPa and σ2=-22.38MPa. From Equation (2.18), the principal shear stress is:
τ1 ,τ2 = ± τ xy2 +
σ x −σ y( )2
4
= ± −83MPa( )2 + 178MPa −12MPa( )
2
4
or τ1=117.4MPa and τ2=-117.4MPa. Note for this case cos(x’,x)=0, cos(x’,y)=0, cos(x’,z)=1,
cos(y’,x)=0, cos(y’,y)=1, cos(y’,z)=0, cos(z’,x)=-1, cos(z’,y)=0, cos(z’,z)=0. Therefore, from
Equations (B.4) to (B.9), σx’=σz=0, σy’=σy=12MPa, σz’=σz=178MPa, τx’y’=τyz=0, τy’z’=-
τxy=83MPa, τz’x’=-τzx=0. Therefore the stress tensor is
S =
0 0 0
0 12 83
0 83 178
 
 
 
 
 
 
 
 
 
 
2.56 A thin, square steel plate has normals to the sides in the x and y directions. A tensile
stress σ acts in the x direction, and a compressive stress -σ acts in the y direction.
Determine the normal and shear stresses on the diagonal of the square.
Notes: Equations (2.20) and (2.21) give the stresses at the desired angle.
Solution:
Note that there are no shear stresses on the sides of the square, so the applied normal stresses are
the principal stresses, and also this is a plane stress case. Therefore, σ1=σ and σ2=-σ. Also, the
www.elsolucionario.net
Page 2-36
diagonals of the square occur at angles φ=45° and 135° from the reference coordinate system.
From Equation (2.20),
σ45° =
σ1 +σ2
2
+ σ1 − σ2
2
cos2φ = σ −σ
2
+ σ − −σ( )
2
cos90° =0
σ135° =
σ1 + σ2
2
+ σ1 −σ2
2
cos2φ = σ − σ
2
+ σ − −σ( )
2
cos270 °= 0
From Equation (2.21),
τ45° = −
σ1 − σ 2
2
sin2φ = −
σ − −σ( )
2
sin90° = −σ
τ135° = −
σ1 − σ2
2
s i n 2φ = −
σ − −σ( )
2
s in270° =σ
2.57 A thin, rectangular brass plate has normals to the sides in the x and y directions. A tensile
stress σ acts on the four sides. Determine the principal normal and shear stresses.
Notes: The principal normal stresses are given by (2.16) and the principal shear stresses are given
by Equation (2.18).
Solution:
From Equation (2.16), the principal normal stresses are:
σ1,σ2 =
σx +σ y
2
± τxy2 +
σ x − σy( )2
4
= σ + σ
2
± 0 + 0 = σ
From Equation (2.18), the principal shear stresses are:
τ1 ,τ2 = ± τ xy2 +
σ x −σ y( )2
4
= 0
2.58 Given the thin, rectangular brass plate in Problem 2.57, but with the stress in y-direction
being σy=-σ instead of +σ, determine the principal normal and shear stresses and their
directions.
Notes: The solution approach is the same as Problem 2.57.
Solution:
From Equation (2.16), the principal normal stresses are:
σ1,σ2 =
σx +σ y
2
± τxy2 +
σ x − σy( )2
4
= σ − σ
2
± 0 + σ +σ
2
 
  
 
  
2
= ±σ
From Equation (2.18), the principal shear stresses are:
τ1 ,τ2 = ± τ xy2 +
σ x −σ y( )2
4
= ± 0+
σ − −σ( )( )2
4
= ±σ
2.59 For the following stress states, draw the appropriate Mohr’s circle, determine the
principal stresses and their directions, and show the stress elements:
www.elsolucionario.net
Page 2-37
a) σx=30, σy=-20, and τxy=10
b) σx=30, σy=-30, and τxy=10
c) σx=50, σy=-50, and τxy=0
d) σx=σy=-10
All stresses are in megapascals.
Notes: The Mohr’s Circle approach described on pages 55-56 is used to solve this problem.
Solution:
a) For this stress state, the center of the circle is placed at (5,0) per Equation (2.22). Using a point
on the circle of (σy,τxy)=(-20,10), the circle is drawn as follows:
The radius of the circle is given by Equation (2.23) as
r = 30 − −20( )
2
 
  
 
  
2
+ 10( )2 = 26.93
Therefore the principal stresses are 5±26.93, so σ1=31.93 and σ2=-21.93. The angle of the stress
element is given by Equation (2.15) as
t a n 2φσ =
2τ xy
σ x − σy
= 20
50
= 0.4; φσ = 10.9°
b) For this stress state, the center of the circle is placed at (0,0) per Equation (2.22). Using a point
on the circle of (σy,τxy)=(-30,10), the circle is drawn as follows:
The radius of the circle is given by Equation (2.23) as
r =
σx −σ y
2
 
 
 
 
 
 
2
+ τxy2 =
30 − −30( )
2
 
  
 
  
2
+ 10( )2 = 31.62
Therefore the principal stresses are 0±31.62, so σ1=31.62 and σ2=-31.62. The angle of the stress
element is given by Equation (2.15) as
www.elsolucionario.net
Page 2-38
t a n 2φσ =
2τ xy
σ x − σy
= 20
60
= 0.333; φσ = 9.22°
c) For this stress state, the center of the circle is placed at (0,0) per Equation (2.22). Using a point
on the circle of (σy,τxy)=(-50,0), the circle is drawn as follows:
The radius of the circle is given by Equation (2.23) as
r =
σx −σ y
2
 
 
 
 
 
 
2
+ τxy2 =
50 − −50( )
2
 
  
 
  
2
+ 0( )2 = 50
Therefore the principal stresses are 0±50, so σ1=50 and σ2=-50. The angle of the stress element is
given by Equation (2.15) as
t a n 2φσ =
2τ xy
σ x − σy
= 0
100
= 0; φσ = 0°
d) For this stress state, the center of the circle is placed at (-10,0) per Equation (2.22). Using a
point on the circle of (σy,τxy)=(-10,0), the circle is drawn as follows:
The radius of the circle is given by Equation (2.23) as
r =
σx −σ y
2
 
 
 
 
 
 
2
+ τxy2 =
−10 − −10( )
2
 
  
 
  
2
+ 0( )2 = 0
Mohr’s circle for this case is a point, the element is under hydrostatic pressure. The angle to the
principal stresses is undefined, since all orientations give the same stress state. The principal
stresses are -10±0, so σ1=-10 and σ2=-10.
2.60 Repeat Problem 2.59 for
a) σx=55, σy=-15, and τxy=40
b) σx=0, σy=30, and τxy=20
www.elsolucionario.net
Page 2-39
c) σx=-20, σy=40, and τxy=-40
d) σx=30, σy=0, and τxy=-20
All stresses are in megapascals.
Notes: The Mohr’s Circle approach described on pages 55-56 is used to solve this problem.
Solution:
a) For this stress state, the center of the circle is placed at (20,0) per Equation (2.22). Using a
point on the circle of (σy,τxy)=(-15,40), the circle is drawn as follows:
The radius of the circle is given by Equation (2.23) as
r =
σx −σ y
2
 
 
 
 
 
 
2
+ τxy2 =
55 − −15( )
2
 
  
 
  
2
+ 40( )2 = 53.15
Therefore the principal stresses are 20±53.15, so σ1=73.15 and σ2=-55.15. The angle of the stress
element is given by Equation (2.15) as
t a n 2φσ =
2τ xy
σ x − σy
= 80
70
= 1.1428; φσ = 24.4°
b) For this stress state, the center of the circle is placed at (15,0) per Equation (2.22). Using a
point on the circle of (σy,τxy)=(30,20), the circle is drawn as follows:
The radius of the circle is given by Equation (2.23) as
r =
σx −σ y
2
 
 
 
 
 
 
2
+ τxy2 =
0 − 30( )
2
 
  
 
  
2
+ 20( )2 = 25
Therefore the principal stresses are 15±25, so σ1=40 and σ2=-10. The angle of the stress element
is given by Equation (2.15) as
www.elsolucionario.net
Page 2-40
t a n 2φσ =
2τ xy
σ x − σy
=
40
−30
= −1.33333; φσ = −26.56°
c) For this stress state, the center of the circle is placed at (10,0) per Equation (2.22). Using a
point on the circle of (σy,τxy)=(40,-40), the circle is drawn as follows:
The radius of the circle is given by Equation (2.23) as
r =
σx −σ y
2
 
 
 
 
 
 
2
+ τxy2 =
−20 − 40
2
 
  
 
  
2
+ −40( )2 = 50
Therefore the principal stresses are 10±50, so σ1=60 and σ2=-40.The angle of the stress element
is given by Equation (2.15) as
t a n 2φσ =
2τ xy
σ x − σy
= −80
−60
= 1.3333; φσ = 26.56°
d) For this stress state, the center of the circle is placed at (15,0) per Equation (2.22). Using a
point on the circle of (σy,τxy)=(0,-20), the circle is drawn as follows:
The radius of the circle is given by Equation (2.23) as
r =
σx −σ y
2
 
 
 
 
 
 
2
+ τxy2 =
30 − 0
2
 
  
 
  
2
+ −20( )2 = 25
Therefore the principal stresses are 15±25, so σ1=40 and σ2=-10. The angle of the stress element
is given by Equation (2.15) as
t a n 2φσ =
2τ xy
σ x − σy
= −40
−30
= 1.3333; φσ = 26.56°
www.elsolucionario.net
Page 2-41
2.61 Given the state of stresses shown in the two parts of sketch ff determine the principal
stresses and their directions by using a Mohr’s circle and the stress equations. Show the
stress elements. All stresses in sketch ff are in megapascals.
Notes: Equations (2.13) and (2.14) are needed to solve this problem.
Solution:
a) Note that σ45°=28MPa and τ45°=-20MPa. From Equation (2.13),
σ45° = 28MPa =
σ x + σy
2
+
σx −σ y
2
cos2φ +τ xy sin2φ =
σ x +σ y
2
+ 30MPa;σx + σy = −4MPa
From Equation (2.14),
τ45° = −20MPa = τxy c o s 9 0° −
σ x −σ y
2
sin90° = −
σx − σy
2
; σx − σy = 40MPa
These two equations yield σx=18MPa and σy=-22MPa. The center of the Mohr’s circle is placed,
per Equation (2.22), at ((σx+σy)/2,0)=(-2MPa,0). Using a point on the circle of (-22MPa, 30MPa),
the circle is drawn as follows:
The radius of the circle is given by Equation (2.23) as:
r =
σx −σ y
2
 
 
 
 
 
 
2
+ τxy2 =
18MPa + 22MPa
2
 
  
 
  
2
+ 30MPa( )2 = 36.0MPa
Therefore the principal stresses are σ1=-2MPa+36MPa=34MPa and σ2=-2MPa-36MPa=-38MPa.
b) For this case, the known values are σφ=40MPa, σy=60MPa, τφ=20MPa and φ=-30°. Therefore,
Equations (2.13) and (2.14) give the following two simultaneous algebraic equations:
σ−30° = 40MPa =
σ x + σ y
2
+
σ x − σ y
2
cos2 φ + τxy s i n 2φ
=
σ x +60MPa
2
+
σx − 60MPa
2
cos −60°( ) + τxy sin −60°( ); 0.75σ x − 0.866τxy = 25MPa
and
τ−30° = 20MPa = τ xy c o s 2φ −
σ x − σy
2
sin2 φ = τxy cos −60°( ) −
σx − 60MPa
2
sin −60°( );
0.5τxy + 0.433σx = 46.0MPa
These equations are solved as σx=69.8MPa and τxy=31.57MPa. The center of the Mohr’s circle is
placed, per Equation (2.22), at ((σx+σy)/2,0)=(64.9MPa,0). Using a point on the circle of
(-60MPa, 31.57MPa), the circle is drawn as follows:
www.elsolucionario.net
Page 2-42
From the circle, the radius is given by:
r = 64.9MPa − 60MPa( )2 + 31.57MPa( )2 = 31.95MPa
Therefore the principal stresses are σ1=64.9MPa+31.95MPa=96.8MPa and σ2=64.9MPa-
31.95MPa=33MPa.
2.62 Given the normal and shear stresses σx=12 ksi, σy=6 ksi and τxy=-4ksi, draw the Mohr’s
circle diagram and the principal normal and shear stresses on the x-y axis. Determine the
triaxial stresses and give the corresponding Mohr’s circle diagram. Also, determine the
octahedral stresses.
Notes: Mohr’s circle is obtained in the same fashion as in Problems 2.59 and 2.60. The triaxial
stresss are ordered according to σ1≥σ2≥σ3, and it is known that this is a plane stress state so one of
these triaxial normal stresses is zero. The three dimensional Mohr’s circle diagram is then easily
drawn. The octahedral stresses are obtained from Equations (2.26) and (2.27).
Solution:
The center of the Mohr’s circle is placed, per Equation (2.22), at ((σx+σy)/2,0)=(9ksi,0). Using a
point on the circle of (σy,τxy)=(6ksi,-4ksi), the circle is drawn in the left sketch as follows:
The radius of the circle is calculated from the circle as:
r = 9ksi − 6ksi( )2 + 4ksi( )2 = 5ksi
Therefore the principal stresses are σ1=9ksi+5ksi=14ksi and σ2=9ksi-5ksi=4ksi. The third
principal stress is zero since this is a plane stress circumstance. The three dimensional Mohr’s
circle can be drawn using these stresses as shown in green. Note that the radii of the circles are
easily calculated to give the principal shear stresses (see Equation (2.25)) as τ1/2=5ksi, τ2/3=2ksi
and τ1/3=7ksi. Therefore, from Equation (2.26), the octahedral normal stress is
www.elsolucionario.net
Page 2-43
σoct =
σ 1 + σ 2 + σ3
3
=
14ksi + 4ksi + 0
3
= 6ksi
The octahedral shear stress is given by Equation (2.27) as
τoct =
2
3
τ1 / 22 + τ2 /32 + τ1 / 32[ ]1 / 2= 23 5ksi( )2 + 2ksi( )2 + 7ksi( )2 = 5.888ksi
2.63 Given the normal and shear stresses σx=16 ksi, σy=9 ksi and τxy=5ksi, draw the Mohr’s
circle diagram and the principal normal and shear stresses on the x-y axis. Determine the
triaxial stresses and give the corresponding Mohr’s circle diagram. Also, determine the
octahedral stresses.
Notes: Mohr’s circle is obtained in the same fashion as in Problems 2.59 and 2.60. The triaxial
stress are ordered according to σ1≥σ2≥σ3, and it is known that this is a plane stress state so one of
these triaxial normal stresses is zero. The three dimensional Mohr’s circle diagram is then easily
drawn. The octahedral stresses are obtained from Equations (2.26) and (2.27).
Solution:
The center of the Mohr’s circle is placed, per Equation (2.22), at ((σx+σy)/2,0)=(12.5ksi,0). Using
a point on the circle of (σy,τxy)=(9ksi,5ksi), the circle is drawn in the left sketch as follows:
The radius of the circle is calculated from the circle as:
r = 12.5ksi − 9ksi( )2 + 5ksi( )2 = 6.10ksi
Therefore the principal stresses are σ1=σc+r=12.5ksi+6.1ksi=18.6ksi and σ2=σc-r=12.5ksi-
6.1ksi=6.4ksi. The third principal stress is zero since this is a plane stress circumstance. The three
dimensional Mohr’s circle can be drawn using these stresses as shown in green. Note that the
radii of the circles are easily calculated to give the principal shear stresses (see Equation (2.25))
as τ1/2=6.1ksi, τ2/3=3.2ksi and τ1/3=9.3ksi. Therefore, from Equation (2.26), the octahedral normal
stress is
σoct =
σ 1 + σ 2 + σ3
3
=
18.6ksi + 6.4ksi + 0
3
= 8.33ksi
The octahedral shear stress is given by Equation (2.27) as
τoct =
2
3
τ1 / 22 + τ2 /32 + τ1 / 32[ ]1 / 2= 23 6.1ksi( )2 + 3.2ksi( )2 + 9.3ksi( )2 = 7.71ksi
2.64 Given the normal and shear stresses σx=10 ksi, σy=24 ksi and τxy=-6ksi, draw the Mohr’s
circle diagram and the principal normal and shear stresses on the x-y axis. Determine the
www.elsolucionario.net
Page 2-44
triaxial stresses and give the corresponding Mohr’s circle diagram. Also, determine the
octahedral stresses.
Notes: Mohr’s circle is obtained in the same fashion as in Problems 2.59 and 2.60. The triaxial
stresss are ordered according to σ1≥σ2≥σ3, and it is known that this is a plane stress state so one of
these triaxial normal stresses is zero. The three dimensional Mohr’s circle diagram is then easily
drawn. The octahedral stresses are obtained from Equations (2.26) and (2.27).
Solution:
The center of the Mohr’s circle is placed, per Equation (2.22), at ((σx+σy)/2,0)=(17ksi,0). Using a
point on the circle of (σy,τxy)=(24ksi,-6ksi), the circle is drawn in the left sketch as follows:
The radius of the circle is calculated from the circle as:
r = 24ksi− 17ksi( )2 + 6ksi( )2 = 9.22ksi
Therefore the principal stresses are σ1=σc+r=17ksi+9.22ksi=26.22ksi and σ2=σc-r=17ksi-
9.22ksi=7.78ksi. The third principal stress is zero since this is a plane stress circumstance. The
three dimensional Mohr’s circle can be drawn using these stresses as shown in green. Note that
the radii of the circles are easily calculated to give the principal shear stresses (see Equation
(2.25)) as τ1/2=9.22ksi, τ2/3=3.89ksi and τ1/3=13.11ksi. Therefore, from Equation (2.26), the
octahedral normal stress is
σoct =
σ 1 + σ 2 + σ3
3
=
26.22ksi + 7.78ksi + 0
3
= 11.3ksi
The octahedral shear stress is given by Equation (2.27) as
τoct =
2
3
τ1 / 22 + τ2 /32 + τ1 / 32[ ]1 / 2= 23 9.22ksi( )2 + 3.89ksi( )2 + 13.11ksi( )2 = 11.0ksi
2.65 In a three-dimensional stress field used to describe the stresses below the surface of a ball
in a ball bearing, the principal stresses are given as σ1=300MPa, σ2=200MPa and σ3=-
2800MPa. Calculate and drawthe Mohr’s circle diagram.
Notes: Since the principal stresses are given, the circle can be drawn directly.
Solution:
Note from Figure 2.19 that the principle stresses are the locations where the individual circles
intersect with the σ-axis. Therefore, the circle can be drawn as follows:
www.elsolucionario.net
Page 2-45
2.66 Given the normal and shear stresses σx=-10ksi, σy=15ksi, and τxy=5ksi, determine or draw
the following:
a) Two-dimensional Mohr’s circle diagram
b) Normal principal stresses on x-y axis
c) Shear principal stress on x-y axis
d) Three-dimensional Mohr’s circle diagram and corresponding principal normal
and shear stresses.
e) Octahedral normal and shear stresses.
Notes: Mohr’s circle is obtained in the same fashion as in Problems 2.59 and 2.60. The triaxial
stresss are ordered according to σ1≥σ2≥σ3, and it is known that this is a plane stress state so one of
these triaxial normal stresses is zero. The three dimensional Mohr’s circle diagram is then easily
drawn. The octahedral stresses are obtained from Equations (2.26) and (2.27).
Solution:
The center of the Mohr’s circle is placed, per Equation (2.22), at ((σx+σy)/2,0)=(2.5ksi,0). Using a
point on the circle of (σy,τxy)=(15ksi,5ksi), the circle is drawn in the left sketch as follows:
The radius of the circle is calculated from the circle as:
r = 15ksi − 2.5ksi( )2 + 5ksi( )2 = 13.5ksi
Therefore the principal stresses are σ1=σc+r=2.5ksi+13.5ksi=16ksi and σ3=σc-r=2.5ksi-13.5ksi=
-11ksi. The second principal stress is zero since this is a plane stress circumstance, and we’ve
numbered the principal stresses according to σ1≥σ2≥σ3. The three dimensional Mohr’s circle can
be drawn using these stresses as shown below:
www.elsolucionario.net
Page 2-46
Note that the radii of the circles are easily calculated to give the principal shear stresses (see
Equation (2.25)) as τ1/2=8ksi, τ2/3=5.5ksi and τ1/3=13.5ksi. Therefore, from Equation (2.26), the
octahedral normal stress is
σoct =
σ 1 + σ 2 + σ3
3
=
16ksi + 0 − 11ksi
3
= 1.67ksi
The octahedral shear stress is given by Equation (2.27) as
τoct =
2
3
τ1 / 22 + τ2 /32 + τ1 / 32[ ]1 / 2= 23 8ksi( )2 + 5.5ksi( )2 + 13.5ksi( )2 = 11.1ksi
2.67 A thin plate is loaded with stresses σx=σ, σy=-σ, σz=0 (perpendicular to the plate),
τxy=τyx=σ/2
1/2
 and τxz=τzx=τyz=τzy=0. Calculate the octahedral normal and shear stresses.
Notes: Equations (2.26) and (2.28) are evaluated to solve the problem.
Solution:
Equation (2.26) gives the octahedral normal stress as:
σoct =
σ x + σy + σz
3
=
σ − σ + 0
3
= 0
The octahedral shear stress is given by Equation (2.28) as:
τoct =
1
3
σx − σ y( )2 + σ y − σ z( )2 + σ z − σ x( )2 + 6 τ xy2 + τyz2 + τ xz2( )
= 1
3
σ + σ( )2 + −σ( )2 + −σ( )2 + 6 σ
2
2
+ 0 + 0
 
 
  
 
 
  =
1
3
9σ 2 = σ
2.68 The strain tensor in a machine element is
T =
0.0012 −0.0001 0.0007
−0.0001 0.0003 0.0002
0.0007 0.002 −0.0008
 
 
 
 
 
 
 
 
 
 
Find the strain in the x, y, and z directions, in the direction of space diagonal
1
3
,
1
3
,
1
3
 
  
 
  
, and in the direction εx, εy, εz.
Notes: This problem requires evaluation of direction cosines from Equations (B.4) through (B.9),
and evaluation of Equation (B.10) to get the desired strains.
www.elsolucionario.net
Page 2-47
Solution:
Note that the normal strains are εx=0.0012, εy=0.0003, and εz=-0.0008. From Equations (B.4) to
(B.9), cos(z’,x)=0.8147, cos(z’,y)=0.2037, and cos(z’,z)=-0.5431. Therefore, Equation (B.10)
gives εz’=0.0002245. The strains in direction 
1
3
,
1
3
,
1
3
 
  
 
  
 is given by
cos(z”,x)=cos(y”,x)=cos(x”,x)=0.577. Equation (B.10) then gives εz”=0.0005.
2.69 The strain tensor is
T =
0.0023 0.0006 0
0.0006 0.0005 0
0 0 0
 
 
 
 
 
 
 
 
 
 
Calculate the maximum shear strain and the principal strains.
Notes: This is a plane strain situation, so Equations (2.38) and (2.39) are needed to solve this
problem.
Solution:
Note that since εz=γzy=γzx=0, Equation (2.39) gives:
γ max = ±2
1
2
γ xy
 
  
 
  
2
+
εx − εy
2
 
 
  
 
 
  
2
= ±2 0.0006( )2 + 0.0023 − 0.0005
2
 
  
 
  
2
= ±0.00216
From Equation (2.38), the principal strains are:
ε1 ,ε2 =
εx + εy
2
±
1
2
γ xy
 
  
 
  
2
+
εx − εy
2
 
 
  
 
 
  
2
=
εx + εy
2
±
γ max
2
=
0.0023 + 0.0005
2
±
0.00216
2
or ε1=0.00248 and ε2=0.00032.
2.70 For the stresses in a stress element it is possible to find coordinate directions where the
shear stresses are zero and the normal stresses are parallel with the new coordinate
directions. Is it possible to determine the stress state in a machine element just by using
two strain gages?
Solution:
It is possible only if the element is in a state of plane stress. Since strain gages are placed on
surfaces, this is possible, unless the machine element is being compressed, such as if it was in a
pressure vessel.
www.elsolucionario.net
Page 3-1
Chapter 3 • Solid Materials
3.1 A 2-m long polycarbonate tensile rod has a cross-sectional diameter of 130 mm. It is used
to lift a tank weighing 65 tons from a 1.8 m deep ditch onto a road. The vertical motion
of the crane’s arc is limited to 4.2 m. Will it be possible to lift the tank onto the road?
Notes: There are two parts to this problem which must be considered: first, does the applied stress
exceed that allowed for polycarbonate, and second, can the tank clear the ditch after including the
deflection of the polycarbonate. Since metric units are used in this problem, a “ton” is interpreted
as 1000 kg.
Solution:
From Table A.4 for polycarbonate, Su=65.5 MPa, and the total percent elongation is 110%. The
load lifted is P=65(1000kg)(9.81m/s
2
)=637kN. The stress is given by Equation (2.7) as:
σ =
P
A
=
P
πd 2
4
 
 
  
 
 
  
=
637kN
π 0.13m( )2
4
 
 
  
 
 
  
= 48MPa
This stress is lower than the allowable stress for polycarbonate, so it is concluded that the tank
can be lifted.
3.2 Materials are normally classified according to their properties, processing routes, and
applications. Give examples of common metal alloys that do not show some of the
typical metal features in their applications.
Notes: There are many possible solutions to this problem, and students should be encouraged to
describe their own applications.
Solution:
In terms of properties, gray cast iron does not have any ductility due to the plate-formed graphite
distributed in the matrix. Thus it has much greater compressive strength than tensile strength and
is a good damping material, all of which are counter to usual metal trends. Often, powder metals
are sintered, using a processing route similar to ceramics. Mercury, being in molten form at room
temperature, is not used in typical metal applications. Sodium is used as a heat transfer medium in
some atomic power plants, where the metal is boiled at the hot end and condensed at the cool end.
3.3 Equation (B.56) gives the relationship between stresses and strains in isotropic materials.
For a polyurethane rubber the elastic modulus at 100% elongation is 7 MPa. When the
rubber is exposed to a hydrostatic pressure of 10 MPa, the volume shrinks 0.5%.
Calculate Poisson’s ratio for the rubber.
Notes: This problem is straightforward. Note that a hydrostatic pressure of p means that
σx=σy=σz=-p. The volume change can be derived as shown here, or else a hint can be given to the
student that the volume change is equal to:
∆V
V
= ε1 + ε2 + ε3
www.elsolucionario.net
Page 3-2
Solution:
Consider, for a box-shaped element of sides l1,l2, and l3, so the volume is l1l2l3. If under a stress
state the strains are ε1, ε2 and ε3, then the new lengths are l1’=(1+ε1)l1, l2’=(1+ε2)l2, and l3’=(1+ε3)l3.
The new volume is given by:
′ V = l1′ l2′l3′ = 1 + ε1( ) 1+ ε2( ) 1+ ε3( )l1l2l3
Therefore, the change in volume is given by:
∆V
V
= ′ V −V
V
=
1+ ε1( ) 1+ ε2( ) 1+ ε3( )l1l2l3 − l1l2l3
l1l2 l3
= 1+ ε1( ) 1+ ε2( ) 1+ ε3( )−1
If this is expanded, andif the strains are all small so that second order terms can be ignored, the
volume change is
∆V
V
= ε1 + ε2 + ε3
From Equations (B.56), if the stress state is σx=σy=σz=-10MPa and all other stresses are zero,
then
εx =
1
E
σx −ν σy +σz( )[ ]; εy = 1E σy − ν σz + σ z( )[ ]; εz =
1
E
σz −ν σx +σ y( )[ ]
Substituting into the expression for volume change,
∆V
V
= ε1 + ε2 + ε3 =
1
E
σx − ν σy + σz( )+ σ y − ν σ x + σ z( )+ σz −ν σx +σ y( )[ ] = 3 pE 1− 2ν( )
For p=10MPa, ∆V/V=0.5%, so
∆V
V
= 3p
E
1− 2ν( ); 0.005 = 3 1 0MPa( )
7MPa( ) 1− 2ν( ); ν = 0.4994
3.4 A fiber-reinforced plastic has fiber-matrix bond strength τf=10MPa and fiber ultimate
strength Su=1GPa. The fiber length is constant for all fibers at l=3mm. The fiber diameter
d=30µm. Find whether the fiber strength of the fiber-matrix bond will determine the
strength of the composite.
Notes: Equation (3.2) is needed to solve this problem.
Solution:
From Equation (3.2),
lcr =
Sud
2τ f
=
1GPa( ) 30µm( )
2 10MPa( ) = 0.0015m = 1.5mm
The actual fiber length is twice this critical length, so the fiber strength will determine the
strength of the composite.
3.5 Having the same material as in Problem 3.4, but with fiber length l=1mm, calculate if it is
possible to increase the fiber stress to Su=1GPa but making the fiber rectangular instead
of circular, maintaining the same cross sectional area of each fiber.
Notes: Equation (3.2) is needed to solve this problem.
Solution:
www.elsolucionario.net
Page 3-3
Equation (3.2) gives the critical length of a fiber, but this is derived from force equilibrium.
Equation (3.2) can be written as:
lcr =
Sud
2τ f
; 2τ f lcr = Sud ; πdτ f
lcr
2
 
  
 
  
= Su
πd2
4
; τ f πd( )
lcr
2
 
  
 
  
= SuA f
The first term in parentheses on the left side of the equation is the circumference; if the cross
section is square, then 2(wt+ht) should be substituted for πd, where wt is the width and ht is the
height of the cross section. Therefore, Equation (3.2) becomes:
τ f 2 wt + ht( )( ) lcr2
 
  
 
  
= Suw tht
Similarly, the cross-sectional area of the fiber is wtht, and since the cross sectional area is
constant, we know:
w tht =
π
4
30µm( )2 = 7.069 × 10−10m2
To answer the question, we use these two simultaneous equations, along with lcr=1mm=0.001m,
Su=1GPa, and τf=10MPa. This yields ht=12.06µm and wt=58.63µm or wt=12.06µm and ht=58.63
µm (orientation is arbitrary).
3.6 A copper bar is stressed to its ultimate strength Su=250MPa. The cross sectional area of
the bar before stressing is 100 mm
2
, and the area at the deformed cross section where the
bar starts to break at the ultimate strength is 60 mm
2
. How large a force is needed to reach
the ultimate strength?
Notes: To solve this problem, one must realize that the ultimate strength is based on maximum
load divided by initial area.
Solution:
The force needed to reach Su is given by
P=(Su)A=(250MPa)(100mm
2
)=25kN
3.7 AISI 440C stainless steel has ultimate strength Su=807MPa and fracture strength
Sfr=750MPa. At the ultimate strength the cross sectional area of a tension bar made of
AISI 440C is 80% of its undeformed value. At the fracture point the minimum cross-
sectional area has shrunk to 70%. Calculate the real stresses at the point of ultimate
strength and at fracture.
Notes: One needs to realize that Su and Sfr are defined for a tension test specimen from its initial
area.
Solution:
For an ultimate strength of 807MPa, the load that can be supported is (807MPa)(A). If the area is
0.8A, then the real stress at this point is given by Equation (2.7) as:
σ = P
A
= 807MPa( ) A
0.8A
=1009MPa
Similarly, at fracture, where the area is 0.70A, the real stress is:
σ = P
A
= 750MPa( )A
0.7A
= 1071MPa
www.elsolucionario.net
Page 3-4
3.8 According to sketch a a beam is supported at point A and at either B or C. At C the
silicon nitride tensile rod is lifting the beam end with force P=Stfr, where Ac is the cross
sectional area of the rod. Find the distance A-B such that the silicon nitride rod would be
crushed if it took up a compressive force at B instead of a tensile force at C. Note that
Scfr=15S
t
fr for silicon nitride. Also, find the reaction forces at A for the two load cases.
Notes: This is fairly straightforward. The reactions are found from equilibrium, and the reactions
yield the answer.
Solution:
Assume that the distance is such that there is a compressive load on rod B as shown. Since the
compressive fracture strength is fifteen times larger than the tensile fracture strength, we know
P’=15P. If the compressive rod exerts a force 15P and the tensile rod doesn’t exist, then moment
equilibrium about point A yields
MA = 0∑ = 15P ′ l − mag l − x( ); ′ l = mag l − x( )15P =
mag l − x( )
15 S ftAc( )
If there is a tensile rod at A and no compressive rod at B, and P=SftAc, then moment equilibrium
gives the load mag that causes failure as:
MA = 0∑ = P l( ) − mag l − x( ); mag = Pll − x
Vertical force equilibrium gives the reaction at A as:
Fy = 0∑ = Ay − mag + P; Ay = mag − P = Pll − x − P =
Px
l − x
=
Sft Acx
l − x
=
8P
7
However, if there is a compressive rod at B, then moment equilibrium yields:
MA = 0∑ = 15P ′ l ( ) − mag l − x( ); mag = 15P ′ l l − x
Vertical force equilibrium gives the reaction at A as:
Fy = 0∑ = Ay − mag − 15P; Ay = mag + 15P = 15P ′ l l − x + 15P =15S ft Ac
′ l + l − x
l − x
 
  
 
  
=
120P
7
The distance AB is l’=l/15.
3.9 Polymers have different properties depending on the relationship between the local
temperature and the polymer’s glass transition temperature Tg. The rubber in a bicycle
tire has Tg=-12°C. Could this rubber be used in tires for an Antarctic expedition at
temperatures down to -70°C?
Notes: This requires an understanding of the implications of glass transition temperature as
shown in Figure 3.10 on page 103. This problem should consider the rubber to be a thermoplastic.
Solution:
www.elsolucionario.net
Page 3-5
From Figure 3.10 on page 103, note that below the glass transition temperature, no permanent
deformation is allowed to the polymer. This should not be too great a concern, since a tire should
hopefully remain elastic. However, the elastic strains which can be withstood at low temperatures
are much lower than those near or above the glass transition temperature. Therefore, it is not
recommended that the tire be used below its glass transition temperature.
3.10 Given an aluminum bronze with 20wt% aluminum and 80wt% copper, find the density of
the aluminum bronze.
Notes: The rule of mixtures, described on Page 106, is needed to solve this problem.
Solution:
The density of aluminum is, from the inside front cover, 2710kg/m3. The density of copper is
8940kg/m3. The rule of mixtures (see page 106) suggests that the density can be calculated from a
linear interpolation between the densities of the alloy components. Therefore, the density of the
aluminum bronze is given by:
ρbronze = 0.2ρal + 0.8 ρcu = 0.2( ) 2710kg / m3( ) + 0.8( ) 8940kg / m3( ) = 7690kg / m3
3.11 The glass-fiber reinforced plastic in Example 3.8 (section 3.5.2) is used in an application
where the bending deformations, caused by the applied static load, will crack the plastic
by overstressing the fibers. Will a carbon-fiber reinforced plastic also crack if it has the
same elastic properties as the glass-reinforced plastic?
Notes: The following data is needed from Example 3.8: Ec=150GPa, Eg=70GPa, Sug=0.7GPa,
Suc=1GPa. Since the strains are constant, one merely needs to apply Hooke’s law to obtain the
stress in the carbon fibers.
Solution:
The fiber reinforced composite has the same elastic properties whether the reinforcement is glass
or carbon fiber (see Example 3.8 - volume percentages of reinforcement were calculated to ensure
this condition). Therefore, the total strain experienced by the composite is constant, as is the
strain seen by the fibers. Applying Hooke’s law,
σc
Ec
=
σg
Eg
;σc =
Ec
Eg
σg =
150GPa
70GPa
0.7GPa( )= 1.5GPa
Since this stress still exceeds the ultimate strength for the carbon fibers, they will crack in the
described loading.
3.12 In Problem 3.11, carbon fibers were used to reinforce a polymer matrix. The
concentration of fibers was decreased in Example 3.8 (section 3.5.2) to give the same
elastic properties for the carbon-fiber-reinforced polymer as for the glass-reinforced
polymer. If instead the fiber concentration were kept constant at 10% when the glass
fibers were changed to carbon fibers, how much smaller would the deformation be for the
same load, and would the fibers be overstressed or not? The material properties are the
same as in Example 3.8.
www.elsolucionario.net
Page 3-6
Notes: Assume that the loading is a given load, not deformation-based. To solve this problem,
one must determine the stiffness of the composite for the two cases, then express the strain ratio
in terms of the elastic modulus. Substituting for the material properties of the glass fibers at
fracture allows one to determine if the carbon fibers will survive. The following data is needed
from Example 3.8: Ec=150GPa, Eg=70GPa, Sug=0.7GPa, Suc=1GPa, Em=2GPa.
Solution:
For a fiber volume fraction of 10%, the stiffness of the carbon fiber reinforced polymer is given
by Equation (3.10) as:
Ecc=Emνm+Efνf=(2GPa)(0.9)+(150GPa)(0.1)=16.8GPa
The stiffness of the glass fiber reinforced polymer is similarly calulated as
Ecg=Emνm+Egνg=(2GPa)(0.9)+(70GPa)(0.1)=8.8GPa
Assuming a constant applied stress for the two cases, the fiber strain ratio is given by
δc
δg
=
Ecg
Ecc
=
8.8GPa
16.8 GPa
= 0.524
Therefore, the stress in the carbon fiber, assuming the loading is sufficient to just cause failure of
glass fibers, is:
σc = 0.524
Ec
Eg
Sug = 0.524
150GPa
70GPa
 
  
 
  
0.7GPa( ) = 0.786GPa
Since the strength of the carbon fibers is 1GPa, they will survive the loading.
3.13 A bent beam, shown in sketch b, is loaded with
P=128,000N. The beam has a square cross section a2. The
length of a side a=30mm. The length l1=50mm and
l2=100mm. The yield strength Sy=350MPa (medium-carbon
steel). Find whether the stresses in tension and shear are
below the allowable stresses. Neglect bending.
Notes: The stresses calculated are average stresses, and do not reflect
the distribution of shear stress in beams as covered in Chapter 4.
The allowable stresses given in Equations (3.13) and (3.14) are used
to solve this problem.
Solution:
The average shear stress over length l1 is given by
τave =
P
a2
=
128kN
0.03m( )2
= 142.2MPa
The allowable shear stress is given by Equation (3.14) as
τall=0.4Sy=0.4(350MPa)=140MPa
Therefore, the shear stress is larger than that allowed for the material.
The normal stress in length l2 is:
σ =
P
a2
= 142.2MPa
Since the lowest allowable stress is, from Equation (3.13), equal to 0.45Sy=157MPa, the normal
stresses are acceptable.
www.elsolucionario.net
Page 3-7
3.14 A tough material, such as soft stainless steel (AISI 316), has yield strength Sy=207MPa,
ultimate strength Su=552MPa, and 60% elongation. Find the ratio of the material
toughness to the resilience assuming that the stress-strain curve consists of two straight
lines according to sketch c.
Notes: The toughness is the ability to absorb energy up to fracture. Resilience is the amount of
energy available in elastic recovery after yield.
Solution:
Toughness is defined on page 113 as the ability to absorb energy up to fracture. Therefore, the
toughness is given by the area under the stress strain curve. For the material given, assuming the
curves are straight lines, this area is:
Toughness =
1
2
207MPa( ) 0.002( ) + 207MPa + 552MPa
2
 
  
 
  
0.60 − 0.002( ) = 220MPa = 220MNm / m3
The resilience is the energy available for elastic recovery. Based on yield strength, the resilience
is:
R =
1
2
207MPa( ) 0.002( ) = 0.207MNm / m3
The ratio of toughness to resilience is then 220/0.207=1062.
3.15 A steel cube has side length l=0.1m, modulus of elasticity E=206GPa, and Poisson’s ratio
ν=0.3. Find the compressive stresses needed on four of the cube faces to give the same
elongation perpendicular thereto as a stress σ in that perpendicular direction.
Notes: This problem uses Hooke’s law to obtain a solution.
Solution:
The elongation in the direction of stress is given by the strain and Hooke’s law as
δ = εσ l =
σ
E
l
The average transverse strain for a compressive stress σc is
εt = −ν
σc
E
The total transverse strain from the two perpendicular stresses is
εt ,total = 2ε t = −2ν
σc
E
The strain should be the same as if the stress σ was applied in the direction:
δ =
σ
E
l = −2ν
σc
E
l;σc = −
σ
2ν
= −
σ
2 0.3( )
= −1.67σ
www.elsolucionario.net
Page 3-8
3.16 For the stressed steel cube in Problem 3.15, calculate the volume ratio (vtension/vcompression)
when σ=500MPa.
Notes: Realize that the volume of the stressed cube is v=[l(1+ε1)][l(1+ε2)][l(1+ε3)]. This problem
uses the results of Problem 3.15.
Solution:
The volume of the cube in tension is given by
vt = l 1+ ε1( )l 1 + ε2( )l 1 + ε3( ) = l3 1 +
σ
E
 
  
 
  
1 − ν σ
E
 
  
 
  
2
= 0.1m( )3 1 + 500MPa
206GPa
 
  
 
  
1 − 0.3( ) 500MPa
206GPa
 
  
 
  
2
= 1.00097 ×10−3m3
The volume of the cube in compression is
vc = l 1 + ε1( )l 1 + ε2( ) l 1+ ε3( )= l3 1 − 2ν
σc
E
 
  
 
  
1 + σc
E
− ν σ c
E
 
  
 
  
2
= 0.1m( )3 1 − 2 0.3( ) −1.67( )500MPa
206GPa
 
  
 
  
1 + −1.67( ) 500 MPa( )
206GPa
− 0.3( ) 500MPa
206GPa
 
  
 
  
2
= 0.99675 ×10−3m3
Therefore, the ratio of the volumes is vt/vc=1.0097/0.99675=1.004235.
3.17 Hooke’s law describes the relationship between uniaxial stress and uniaxial strain. How
large is the ratio of deformation for a given load within the group of materials considered
in Chapter 3?
Notes: The materials in Table 3.2 should be considered. Therefore, as the stiff material, use
silicon carbide, and as the compliant material, use natural rubber.
Solution:
Reviewing Table 3.2 on page 108, we select silicon carbide as the stiffest material listed with
Esic=450GPa, and natural rubber is the least stiff with Er=0.004GPa. From Hooke’s law, the
deformation is inversely proportional to applied stress. Therefore,
δr
δsic
=
1/ Er( )
1/ Esic( )
=
Esic
Er
=
450GPa
0.004GPa
= 1.125 × 105
3.18 According to Archard’s wear equation, the wear depth is proportional to the sliding
distance and the contact pressure. How will the contact pressure be distributed radially
for a disk brake if the wear rate is the same for all radii?
Notes: Equation (3.26) is used to solve the problem.
Solution:
From Equation (3.26) the wear rate per unit width is
Wr
’=KAplA=KAplφr=constant
www.elsolucionario.net
Page 3-9
Therefore, solving for pl,
pl =
C
KAφr
=
C1
r
Therefore, the pressure is inversely proportional to the radius.
3.19 Given a brake pad for a disk brake on a car, and using Archard’s wear constant,
determine how the wear is distributed over the pad if the brake pressure is constant over
the pad.
Notes: Equation (3.26) is used to solve the problem.
Solution:
The wear rate per width is given by Equation (3.26) as:
Wr
’=KAplA=KAplφr
If p1 is constant, then this can be rewritted as Wr
’=C1r, or that the wear rate across the pad is
directly proportional to the radius.
www.elsolucionario.net
Page 4-1
Chapter 4 • Normal, Torsional, Bending, and
Transverse Shear Stresses and Strains
4.1 An area in the x-y coordinate system, as shown in sketch a, consists of a large circle
having radius r out of which are cut three smaller circles having radii r/3. Find the x and y
coordinates for the centroid. The radius r is 10 cm.
Notes: Equations (4.5) and (4.6) are used to solve the problem. It may be helpful to think of the
cutouts as “negative areas” in the application of Equations (4.5) and (4.6).
Solution:
Note that the area of the large circle is A=πr2=π(0.1m)2=3.14x10-2m2. The area of the smaller
circles is A=πr2=π(0.1m/3)2=3.49x10-3m2. The centroidlocations of the individual circles are
shown on the sketch. Therefore, the y-coordinate of the centroid is given by Equation (4.5) as:
y =
A1y 1 + A2 y 2 + ...
A1 + A2 + ...
=
3.14 × 10−2m2( ) 0.1m( ) − 3.49 ×10−3 m2( ) 0.1m + 0.1m + 0.167m( )
3.14 ×10−2 m2( ) − 3 3.49 ×10−3 m2( ) = 0.0889m
The x-coordinate is given by Equation (4.6) as:
x =
A1x 1 + A2x 2 + ...
A1 + A2 + ...
=
3.14 × 10−2m2( ) 0.1m( ) − 3.49 ×10−3 m2( ) 0.05m + 0.1m + 0.15m( )
3.14 ×10−2 m2( ) − 3 3.49 ×10−3 m2( ) = 0.10m
Therefore, the centroid is located at x=10cm and y=8.89cm.
4.2 The circular surface in sketch a has circular cutouts glued onto it diagonally below the
top cutout such that the centroids of the three cutouts are at x=r and y=r/3. Find the x and
y coordinates for the centroid (r=10cm).
Notes: The approach is similar to Problem 4.1. Equations (4.5) and (4.6) are used to solve the
problem. It may be helpful to think of the cutouts as “negative areas” in the application of
Equations (4.5) and (4.6).
www.elsolucionario.net
Page 4-2
Solution:
Note that the area of the large circle is A=πr2=π(0.1m)2=3.14x10-2m2. The area of the smaller
circles is A=πr2=π(0.1m/3)2=3.49x10-3m2. The centroid locations of the individual circles are
shown on the sketch. Therefore, the y-coordinate of the centroid is given by Equation (4.5) as:
y = A1y 1 + A2y 2 + ...
A1 + A2 + ...
=
3.14 × 10−2m2( ) 0.1m( ) − 3.49 × 10−3 m2( ) 0.1m + 0.1m + 0.167m( ) + 3 3.49 × 10−3m2( ) 0.0333m( )
3.14 ×10−2m2( ) − 3 3.49 ×10−3m2( ) + 3 3.49 ×10−3m2( )
or y=0.0704m=7.04cm. The x-coordinate is given by Equation (4.6) as:
x = A1x 1 + A2x 2 + ...
A1 + A2 + ...
=
3.14 × 10−2m2( ) 0.1m( ) − 3.49 × 10−3 m2( ) 0.5m + 0.1m + 0.15m( )+ 3 3.49 × 10−3 m2( ) 0.1m( )
3.14 ×10−2m2( ) − 3 3.49 ×10−3m2( ) + 3 3.49 ×10−3m2( )
or x=0.1m=10cm. Therefore, the centroid is located at x=10cm and y=8.89cm.
4.3 The rectangular area shown in sketch b is situated symmetrically in the x-y coordinate
system. The lengths of the rectangle’s sides are a in the x-direction and b in the y-
direction. Find the moments of inertia Ix and Iy and the polar moment of inertia J for the
rectangular surface.
Notes: The solution is obtained by evaluating Equations (4.7) and (4.9).
Solution:
The moments of inertia Ix and Iy are given by Equation (4.7):
Ix = y2dAA∫ = y
2ady−b /2
b / 2∫ = a3 y
3( )−b /2
b / 2
= ab
3
12
Iy = x2dAA∫ = x
2bdx−a /2
a / 2∫ = b3 x
3( )−a /2
a / 2
= ba
3
12
The polar moment of inertia, J, is obtained from Equation (4.9) as:
J = Ix + Iy =
ab3
12
+ ba
3
12
= ab
12
a2 + b2( )
4.4 Derive the area moment of inertia for the hollow circular area shown in Table 4.1.
www.elsolucionario.net
Page 4-3
Notes: The results from Example 4.2 on page 143 are used in this solution. An alternative
approach would be to evaluate the integrals in Equation (4.7).
Solution:
For the solid circle considered in Example 4.2 on page 143, the area moment of inertia was found
to be Ix=Iy=πr
4
/4. Therefore, the moments of inertia for the inner and outer circles are:
Ixi = Iyi =
πri4
4
; Ixo = Iyo =
πro4
4
Therefore, the moment of inertia for the hollow circular area is:
Ix = Iy =
πro4
4
− πri
4
4
4.5 Derive the area moment of inertia for the elliptical quadrant shown in Table 4.1.
Notes: This problem is solved by application of Equation (4.7) given the equation of an ellipse.
Solution:
Recall that the equation of an ellipse can be written as:
x2
a2
+ y
2
b2
= 1; x = a2 1 − y
2
b2
 
 
  
 
 
  ; y = b
2 1− x
2
a2
 
 
  
 
 
  
For the elliptical quadrant shown in Table 4.1, Equation (4.7) gives the following:
Ix = y2dAA∫ = y
2xdy0
b∫ = y2 a2 1 − y
2
b2
 
 
  
 
 
  dy0
b∫
This is solved with the help of an integral table as:
Ix =
a
b
b4
8
 
 
  
 
 
  
π
2
 
  
 
  
=
ab3π
16
Similarly, Iy is obtained from Equation (4.7) as:
Iy = x2dA
A
∫ = x2ydx0
b
∫ = x2 b2 1 −
x2
a2
 
 
  
 
 
  dx0
b
∫
which is solved as
Iy =
a3bπ
16
4.6 Derive the moment of inertia for the triangular section shown in Table 4.1.
Notes: This problem is solved by application of Equation (4.7).
Solution:
The sketch to the right helps to understand the definition of the geometry. The equation for x’ is
given by
′ x = b 1− y
h
 
  
 
  
and that for y’ is
www.elsolucionario.net
Page 4-4
′ y =
hx
a
, 0≤ x ≤ a
h
b− x
b − a
 
  
 
  
, a ≤ x ≤ b
 
 
  
 
 
 
Therefore, from Equation (4.7),
Ix = y2dAA∫ = y
2 ′ x dy0
h∫ = y2b 1− yh
 
  
 
  dy0
b∫ = b y
3
3
− y
4
4h
 
 
  
 
 
  
0
h
= bh
3
12
Iy = x
2dA
A
∫ = x2 ′ y dx0
a
∫ + x2 ′ y dxa
b
∫ = x2
hx
a
 
  
 
  
dx
0
a
∫ + x2
h b −x( )
b − a
 
  
 
  
dx
a
b
∫
= ha
3
4
+
h b4 − ba3( )
3 b − a( )
+
h a 4 − b4( )
4 b− a( )
=
hb a2 + ab + b2( )
12
4.7 Derive the area moment of inertia for a rectangular section with a cutout shown in sketch
c.
Notes: The moment of inertia should be calculated about the x- and y- axes, but an extension to
the problem could include calculating the moment of inertia about the new centroids. Note that
the circular cutout and the rectangular shape do not have the same centroid location in the y-
direction. Therefore, the Parallel-Axis Theorem given by Equations (4.11) and (4.12) is used to
solve this problem. Also, the problem is simplified by using data from Table 4.1.
Solution:
The area of the circle is πr2=π(0.025m)
2
=0.00196m
2
. Note that for the rectangular area, Table 4.1
gives
Ixr =
bh3
3
= 0.08m( ) 0.1m( )
3
3
= 2.667 ×10−5 m4
Iyr =
hb3
12
= 0.1m( ) 0.08m( )
3
12
= 4.267 ×10−6 m4
www.elsolucionario.net
Page 4-5
where the second subscript is intended to differentiate the rectangle from the circle. Also, it
should be noted that for the moments of inertia to be evaluated about the x-y axes, Ixr needs to be
taken for the base of the rectangle and Iyr needs to be taken from the centroid. For the circle,
Table 4.1 gives the moment of inertia about the circle’s centroid as:
Ix c = Iy c =
πr4
4
= π 0.025m( )
4
4
= 3.068×10−7 m4
Therefore, noting that the circle and the square have their centroids located at x=0, we can
immediately write
Iy=Iyr-Iyc=4.267x10
-6
m
4
-3.068x10
-7
m
4
=3.960x10
-6
m
4
However, the moment of inertia of the circle about a different axis than its centroid requires
application of the Parallel-Axis Theorem. Therefore, the moment of inertia of the circle about the
x-axis is given by Equation (4.11) as:
Ixc = Ix c + Ady2 = 3.068 ×10−7 + 0.00196m2( ) 0.06m( )2 = 7.375×10−6 m4
Therefore,
Ix=Ixr-Ixc=2.667x10
-5
m
4
-7.375x10
-6
m
4
=1.929x10
-5
m
4
.
4.8 Derive the area moment of inertia and the polar moment of inertia for the two
relationships shown in sketch d.
Notes: Equations (4.7) and (4.9) are needed to solve this problem.
Solution:
From Equation (4.7),
Ix = y2dA
A
∫ = y2dydxx2 /10
x
∫0
3
∫ =
1
3
x3 −
x2
10
 
 
  
 
 
  
3 
 
 
 
 
 
 
 
 
 
dx
0
3
∫ =
1
3
x4
4
−
x7
7000
 
 
 
 
 
 
 
 
0
3
= 6.646m4
Iy = x2dA
A
∫ = x2dydxx2 /10
x∫0
3∫ = x2 x − x
2
10
 
 
  
 
 
  
 
 
 
 
 
 
 
 
dx0
3∫ = x3 − x
4
10
 
 
 
 
 
 
 
 
dx0
3∫ = x
4
4
− x
5
50
 
 
 
 
 
 
 
 
0
3
=15.39m4
Therefore, from Equation (4.9),
J=Ix+Iy=6.646m
4
+15.39m
4
=22.04m
4
4.9 An elevator is hung by a steel rope. The rope has a cross sectional area of 200mm2 and a
modulus of elasticity of 70GPa. The upward acceleration when the elevator starts is 3
m/s
2
. The rope is 100m long and the elevator weighs 1000kg. Determine the stress in the
www.elsolucionario.net
Page 4-6
rope, the elongation of the rope due to the elevator’s weight, and the extra elongation due
to the acceleration.
Notes: The stress is obtained from Equation (4.22), and the elongation from (4.23) and (4.24).
Solution:
The stress due to the static weight alone is given by Equation (4.22)as:
σ = P
A
=
1000kg( ) 9.81m / s2( )
0.0002m2
= 49.05MPa
The stress due to the acceleration alone is given by the same equation as:
σ = P
A
=
1000kg( ) 3m / s2( )
0.0002m2
= 15MPa
Therefore the total stress is σtot=49.05MPa+15MPa=64.05MPa. The strain due to the elevator’s
weight is given by Equation (4.24), and the elongation by (4.23). Combining the equations gives:
ε = δ
l
= σ
E
; δ = σ l
E
= 49.05MPa( ) 100m( )
70GPa
= 0.0701m = 70.1mm
Similarly, the extra elongation due to the acceleration is:
ε = δ
l
= σ
E
; δ = σ l
E
= 15MPa( ) 100m( )
70GPa
= 0.0214m = 21.4mm
4.10 The 700-m-long cables in a suspension bridge are stressed to 200MPa tensile stress. The
total force in each cable is 10MN. Calculate the cross-sectional area, the total elongation
of each cable, and the spring rate when the modulus of elasticity is 70GPa.
Notes: The area is obtained from Equation (4.22). The elongation can be calculated as in Problem
4.9 or obtained directly from Equation (4.25). The spring rate is obtained from Equation (4.26).
Solution:
If σ=200MPa under a force of 10MN, then the area is obtained from Equation (4.22) as:
σ = P
A
; A = P
σ
= 10MN
200MPa
= 0.05m2
The elongation is obtained from Equation (4.25) as:
δ = Pl
AE
= 10MN( ) 700m( )
0.05m2( ) 70GPa( ) = 2m
The spring rate is obtained from Equation (4.26) as:
k = P
δ
= 10MN
2m
= 5MN / m
4.11 A 1.28-m long steel piston in a hydraulic cylinder exerts a force of 4000kgf. The piston is
made of AISI 1080 steel and has a diameter of 50mm. Calculate the stress in the piston,
the elongation, and the spring rate.
Notes: This problem is solved in the same manner as Problem 4.10. Some additional care must be
taken to ensure proper units, and the elastic modulus of AISI 1080 steel is obtained from the
www.elsolucionario.net
Page 4-7
inside front cover of the text. The stress is obtained from Equation (4.22). The elongation can be
calculated from Equation (4.25). The spring rate is obtained from Equation (4.26).
Solution:
From the inside front cover, the elastic modulus for AISI 1080 steel is 207GPa. The area of the
piston is A=πd
2
/4=π(0.05m)
2
/4=0.00196m
2
. The stress in the piston is obtained from Equation
(4.22) as:
σ = − P
A
= −
4000kg( ) 9.81m / s2( )
0.00196m2
= −20.0MPa
Note that the stress is negative because it is a compressive stress. The elongation is obtained from
Equation (4.25) as:
δ =
σl
E
=
−20MPa( ) 1.28m( )
207GPa
= −1.237 × 10−4 m = −0.124mm
The spring rate is obtained from Equation (4.26) as:
k = AE
l
=
0.00196m2( ) 207GPa( )
1.28m
= 317MN / m
4.12 A steel pillar supporting a highway bridge is 14m high and made of AISI 1040 steel
tubing having an outer diameter of 1.5m and a wall thickness of 30mm. The weight
carried by the pillar is 12MN. Calculate the deformation of the pillar, the spring rate, and
the stress in the pillar.
Notes: This problem uses a similar approach as Problem 4.11.
Solution:
The elastic modulus of AISI 1040 steel is obtained from the inside front cover as 207GPa. The
cross sectional area of the tube is:
A = π
4
1.5m( )2 − 1.5m − 0.06m( )2[ ] = 0.1385m2
The stress is then obtained from Equation (4.22), recalling that a compressive stress is negative:
σ = −
P
A
= −
12MN
0.1385m2
= −86.6MPa
The elongation is obtained from Equation (4.25) as:
δ = σl
E
= −86.6MPa( ) 14m( )
207GPa
= −0.00586m = −5.86mm
The spring rate is obtained from Equation (4.26) as:
k = AE
l
=
0.1385m2( ) 207GPa( )
14m
= 2.048GN / m
4.13 The foundation of a bronze statue is made of a 3m high conical tube of constant wall
thickness (8mm). The tube’s outer diameter is 200mm at the top (just under the statue)
and 400mm at the ground. The tube material is AISI 316 stainless steel. The statue
weighs 1630kg. Calculate the deformation of the tube, its spring rate, and the maximum
and minimum compressive stresses in the tube.
www.elsolucionario.net
Page 4-8
Notes: The first step is to calculate the area as a function of height of the tube, x. This then allows
calculation of strain as a function of x, which can be integrated to obtain the total deflection. The
stiffness is then the ratio of the load to the total deflection.
Solution:
From the inside front cover, the elastic modulus for AISI 316 stainless steel is 193 GPa. Let the
position, x, along the tube be measured from the top (just under the statue). The diameter as a
function of position is then
d = 0.2m + x 0.4m − 0.2m
3m
 
  
 
  = 0.2m 1 +
x
3m
 
  
 
  
The cross section area at a location x is:
A x( ) = πd
2
4
− π
4
d − 0.016m( )2 = π
4
0.032m( )d − 0.016m( )2[ ] = 0.0251m( ) d − 0.008m( )
Substituting for d,
A x( ) = 0.0251m( ) 0.2m( ) 1+ x
3m
 
  
 
  − 0.008m
 
  
 
  
= 0.00482m2 + 0.00167m( )x
The stress at any position x is given by Equation (4.22) as:
σ x( ) = − P
A x( ) = −
1630kg( ) 9.81m / s2( )
0.00482m2 + 0.00167m( )x
= − 9.575MN / m
2.88m + x
Note that the maximum stress occurs at x=0 and is -3.32MPa and the minimum stress occurs at
x=3m and is -1.63MPa. The strain as a function of position is then obtained from (4.24) as:
ε x( ) = σ x( )
E
= − 9.575MN / m
2.88m + x( ) 193GPa( ) = −
4.961× 10−5 m
2.88m + x
The deformation is then:
δ = ε x( )0
l∫ dx = − 4.961×10
−5 m
2.88m + x
dx0
l∫ = − 4.961× 10−5m( ) ln 5.882.88
 
  
 
  
 
  
 
  
= −35.3µm
The spring rate is calculated from Equation (4.26) as:
k = P
δ
=
−1630kg( ) 9.81m / s2( )
−35.3µm
= 453MN / m
4.14 Calculate the deflection at point A of the hanging cone shown in
sketch e. The cone has a density of 2000 kg/m
3
.
Notes: An expression must be obtained to find the volume as a function of
position. Then Equation (4.25) allows solution of the problem. Since E is
not given, the deflection can only be determined as a function of elastic
modulus.
Solution:
Note from Equation (4.25) that δ is given by:
www.elsolucionario.net
Page 4-9
δ =
Pxdx
AxE0
l
∫ =
ρvgdx
πrx
2 E0
l
∫ +
106dx
πrx
2E0
0.5
∫
But rx=0.05m at x=0, and rx=0.025m at x=1m. Therefore, rx can be solved as:
x
1
=
0.05 − rx
0.05 − 0.025
; rx = 0.05 −
x
40
where x is in meters. v can be expressed in terms of rx and x as:
v =
πrx
2 2 − x( )
3
−
π 0.025( )2
3
Substituting into the expression for δ,
δ =
2000
−π 0.025( )2
3
+ π 2 − x( )
3
0.05 − x
40
 
  
 
  
2 
 
 
 
 
 
 
 
9.807( )dx
π 0.05 − x
40
 
  
 
  
2
E
+ 10
6dx
π 0.05 − x
40
 
  
 
  
2
E
0
0.5
∫∫
= 20 ,540
E
+ 84,882,600
E
= 8.490 × 10
7
E
where δ is in meters for E in Pa.
4.15 An aluminum core having a diameter di of 30 mm is placed within a tubular steel shaft
having a diameter do of 50mm, see sketch f. A flange is welded to the end of the shaft,
and a pullet force of 100kN is applied. The shaft is 100mm long. Find the deflection at
the end of the shaft. Assume that the moduli of elasticity are 2x10
11
Pa for steel and
0.7x10
11
Pa for aluminum.
Notes: This is a statically indeterminate problem. To obtain the load in the steel and the
aluminum, two equations are needed. The first is force equilibrium, the second is that the
deflection of the steel and aluminum members is the same. Once the loads are obtained, the
stresses are calculated from Equation (4.22) and the deflections from (4.25) for either the
aluminum or steel member.
Solution:
Using an s subscript to refer to the steel tube and an al for the aluminum, the area of the
aluminum core is Aal=πd
2
/4 =7.069x10
-4
m
2
. The area of the steel tube is calculated as
As=0.001257m
2
. Note that the deflection of the tube and core must be the same, or, from Equation
(4.25),
δs =
Ps l
AsEs
= δal =
Pal l
AalEal
; Ps = Pal
AsEs
AalEal
= Pal
0.001257m2( ) 200GPa( )
0.0007069m2( ) 70GPa( ) = 5.08Pal
Also, from static equilibrium,
Pal+Ps=100kN;Pal+5.08Pal=100kN; Pal=16.44kN
Therefore, Ps=83.55kN. The deflection is obtained from Equation (4.25) for either aluminum or
steel, but is calculated using aluminum data here as:
www.elsolucionario.netPage 4-10
δs = δ al = δ =
Pal
AalEal
= 16.44kN
0.0007069m2( ) 70GPa( ) = 3.324 ×10
−4 m = 0.3324mm
The stress in the aluminum is obtained from Equation (4.22):
σal =
Pal
Aal
= 16.44kN
0.0007069m2
= 23.27MPa
The stress in the steel is similarly calculated as:
σs =
Ps
As
= 83.55kN
0.001257m2
= 66.47MPa
4.16 A bar of weight W is supported horizontally by three weightless rods as shown in sketch
g. Assume that the cross-sectional areas, the moduli of elasticity and the yield stresses are
the same for the three rods. What is the maximum weight that can be supported?
Notes: The load is determined as a function of yield strength Sy and rod area A. Just as in Problem
4.15, static equilibrium does not give sufficient information to solve the problem, so an additional
equation is obtained from deflection, or kinematic, constraints.
Solution:
Refer to the rods from left to right as bars 1, 2, and 3. Taking moment equlibrium about the point
where rod 2 attaches to the bar gives the requirement that P1=P3, or else the bar does not remain
horizontal. Also note that the deflection of bars 1, 2 and 3 must be equal; equating the deflection
of bars 1 and 2 and applying Equation (4.25) gives:
δ1 =
P1l
AE
= δ 2 =
P2 2l( )
AE
; P1 = 2P2
Clearly rods 1 and 3 are the more critical members, since rod 2 takes one-half their load. From
force equilibrium,
P1+P2+P3=W; P1+0.5P1+P1=W; P1=0.4W
Therefore, the stress in rods 1 and 3 is obtained from (4.22) as:
σ1 =
P1
A
= 0.4W
A
= Sy ;W = 2.5SyA
4.17 An electric motor transmits 100kW to a gearbox through a 50-mm diameter solid steel
shaft that rotates at 1000 rpm. Find the torque transmitted through the shaft and the
angular torsion of the 1-m-long shaft.
Notes: The torque is related to the power transmitted by Equation (4.42), while the angle of twist
is given by Equation (4.31). Some attention must be paid towards units.
Solution:
www.elsolucionario.net
Page 4-11
The angular velocity of the shaft is ω=1000rpm=104.7rad/s. Therefore, Equation (4.42) gives the
torque as:
T =
hp
ω
= 100kW
104.7rad / s
= 955Nm
From the inside front cover, the elastic modulus of steel is 207GPa, and Poisson’s ratio is 0.3.
Therefore, the shear modulus is, from Equation (B.53) on page 918:
G = E
2 1+ν( ) =
207GPa
2 1+ 0.3( ) = 79.6GPa
The polar moment of inertia for the round cross section is, from Table 4.1 on page 148,
J = πr
4
2
= π 0.025m( )
4
2
= 6.136×10−7 m4
The angle of twist is then obtained from Equation (4.31) as:
θ = Tl
JG
= 955Nm( ) 1m( )
6.136×10−7 m4( ) 79.6GPa( ) = 0.01955rad = 1.12°
4.18 The torque-transmitting shaft in Problem 4.17 is too heavy for the application, so it is
exchanged for a circular tube having a 50-mm outside diameter and a 40-mm inside
diameter. Find the angular torsion of the tube-formed shaft, which is 1m long, when
100kW is transferred at 1000 rpm. The shear modulus is 80,000N/mm
2
. Also, find the
maximum shear stress in the tube and the percentage of weight decrease from the solid
shaft.
Notes: The approach is similar to Problem 4.17, which is not needed to solve this problem. The
torque is related to the power transmitted by Equation (4.42), while the angle of twist is given by
Equation (4.31). The maximum shear stress is obtained from Equation (4.34). Some attention
must be paid towards units.
Solution:
The angular velocity of the shaft is ω=1000rpm=104.7rad/s. Therefore, Equation (4.42) gives the
torque as:
T =
hp
ω
= 100kW
104.7rad / s
= 955Nm
The polar moment of inertia for the tube is, from Table 4.1 on page 148,
J =
π ro
4 − ri
4[ ]
2
=
π 0.025m( )4 − 0.020 m( )4[ ]
2
= 3.623×10−7m4
The angle of twist is then obtained from Equation (4.31) as:
θ = Tl
JG
= 955Nm( ) 1m( )
3.623 ×10−7 m4( ) 80GPa( ) = 0.0329rad =1.89°
The weight ratio of the tube to the solid shaft is:
Wtube
Wsolid
= ρAt l
ρAsl
=
π ro
2 − ri
2( )
πro
2 =
0.025m( )2 − 0.020m( )2
0.025m( )2
= 0.36
Or in other words, there was a 64% weight savings.
www.elsolucionario.net
Page 4-12
4.19 A torque transmitting, hollow steel shaft with a circular cross section has an outer
diameter of 50mm and an inner diameter of 40mm. Find the maximum length possible
for the shaft if the torsion should be below 10° at a torque of 2000 Nm.
Notes: This is similar to problems 4.17 and 4.18. The length is obtained from Equation (4.31), but
the angle of twist must be expressed in radians.
Solution:
From the inside front cover, the elastic modulus of steel is E=207GPa, and Poisson’s ratio is
ν=0.3. Therefore, the shear modulus is, from Equation (B.53) on page 918:
G = E
2 1+ν( ) =
207GPa
2 1+ 0.3( ) = 79.6GPa
The polar moment of inertia for the round cross section is, from Table 4.1 on page 148,
J =
π ro
4 − ri
4[ ]
2
=
π 0.025m( )4 − 0.020 m( )4[ ]
2
= 3.623×10−7m4
The angle of twist is 10°=0.1745rad. Therefore, Equation (4.31) can be solved for the length as:
θ = Tl
JG
; l = θJG
T
=
0.1745rad( ) 3.623×10−7 m4( ) 79.6GPa( )
2000Nm
= 2.52m
4.20 The bronze statue described in Problem 4.13 is asymmetrical, so that when a gale force
wind blows against it, a twisting torque of 1000 Nm is applied to the tube. Calculate how
much the statue twists. The tube’s wall thickness, 6mm at the top and 12 mm at the
bottom, is assumed to be proportional to its diameter.
Notes: This problem requires the determination of the diameter as a function of tube height,
which then allows the determination of polar moment of inertia as a function of height. Equation
(4.31) gives the angle of twist.
Solution:
For the AISI 316 stainless steel, E=193GPa and ν=0.3 as found on the inside front cover.
Equation (B.53) on page 918 gives the shear modulus as:
G = E
2 1+ν( ) =
193GPa
2 1+ 0.3( ) = 74.23GPa
The outer diameter of the tube, in meters, is
d0 x( ) = 0.200 1 +
x
3
 
  
 
  
The inner diameter in meters is
di x( ) = 0.2− 0.012( ) 1+
x
3
 
  
 
  = 0.188 1+
x
3
 
  
 
  
The polar moment of inertia is, using the data from Table 4.1 on page 148,
J x( ) = π
32
do
4 − d i4( ) = π32 1+
x
3
 
  
 
  
4
0.24 − 0.1884( ) = 3.44 ×10−5( ) 1+ x3
 
  
 
  
4
Therefore, the angle of twist is
θ =
T
G
dx
J x( )0
3m
∫ =
1000Nm
74.23GPa
dx
3.44 × 10−5( ) 1+ x3
 
  
 
  
4 =0
3m
∫ 3.423× 10−4rad = 0.0196°
www.elsolucionario.net
Page 4-13
4.21 Determine the minimum diameter of a solid shaft used to transmit 500kW of power from
a 2000 rpm motor so that the shear stress does not exceed 50MPa.
Notes: The torque is related to the power transmitted by Equation (4.42). The maximum shear
stress is given by Equation (4.34). This should be expressed as a function of the diameter, set
equal to the allowable stress of 50MPa, and then solved for the diameter.
Solution:
The angular velocity of the shaft is ω=2000rpm=209rad/s. Therefore, the torque transmitted is
given by Equation (4.42) as:
T =
hp
ω
= 500kW
209rad / s
= 2390Nm
From Table 4.1, the polar moment of inertia for a round cross section is J=πd
4
/32. c from
Equation (4.34) is d/2. Therefore, the diameter is obtained form the maximum shear stress from
(4.34):
τmax =
cT
J
= d / 2( ) 2390Nm( )
πd4 / 3 2( ) =
38240Nm
πd3
= 50MPa; d3 = 38,240 Nm
π 50MPa( ) = 0.000243m
3
or d=0.0624m=62.4mm
4.22 A steel coupling is used to transmit a torque of 30,000Nm. The coupling is connected to
the shaft by a number of 5mm diameter bolts placed equidistant on a pitch circle of 0.3m
diameter. The inner diameter of the coupling is 0.1m. The allowable shear stress on the
bolts is 500MPa. Find the minimum number of bolts needed.
Notes: The bolts are assumed small enough and placed at a large enough diameter so that the
shear stress is assumed constant across a bolt section. One must merely apply torque equilibrium
to solve this problem.
Solution:
If the bolt diameter is 5mm, then the cross sectional area is A=πd2/4=1.963x10-5m2. If the
allowable shear stress on a bolt is 500MPa, then the force transmitted by each bolt is
F=τA=(500MPa)(1.963x10-5m2)=9817N.If there are n bolts on the 0.3m diameter, then the torque they can support is nF(0.15m). Equating
this to the applied torque yields:
T = nF 0.15m( ); 30 ,000Nm = n 9817N( ) 0.15m( ); n = 20.3
Therefore, 21 bolts are needed.
4.23 A shaft and a coupling are to transmit 50kW of power at an angular speed of 1000rpm.
The coupling is connected to the shaft by 10 bolts, 20mm in diameter, placed on a pitch
circle of 200mm. For an allowable stress on the bolts of 100MPa, are the bolts able to
transfer this power?
Notes: This problem is similar to Problem 4.22. The torque transmitted is calculated from
Equation (4.42). Torque equilibrium is then applied to solve the problem.
www.elsolucionario.net
Page 4-14
Solution:
The cross sectional area of each bolt is A=πd
2
/4=3.14x10
-4
m
2
. If the allowable shear stress on a
bolt is 100MPa, then the force transmitted by each bolt is
F=τA=(100MPa)(3.14x10-4m2)=31.4kN.
If there are 10 bolts on the 0.2m diameter, then the torque they can support is
10F(0.1m)=31.4kNm. The angular speed is 1000rpm=104.7rad/s. The applied torque is calculated
from Equation (4.42) as:
T =
hp
ω
= 50kW
104.7rad / s
= 477Nm
This is a much lower torque than the bolts can safely transfer.
4.24 A beam transmitting a bending moment M of 5000Nm has a square cross section 100 by
100mm. The weight is decreased by making either one or two axial circular holes along
the beam. Determine whether one or two holes gives the lowest weight for the beam at a
given bending stress. Neglect the transverse shear stress.
Notes: The ratio of moment of inertia of the section to the area can be compared for the two
alternatives; this is because the extreme fiber distance, the beam length and material are all
constant.
Solution:
The two alternatives are sketched above. The moments of inertia and areas for the two shapes are:
I1 =
a4
12
−
2π
4
r4 =
a4
12
−
π
4
a
4
 
  
 
  
4
2( ) = a
4
4
1
3
−
π
128
 
  
 
  
A1 = a2 − 2π
a
4
 
  
 
  
2
= a2 1 −
π
8
 
  
 
  
I2 =
a4
12
−
π
4
r 4 =
a4
12
−
π
4
a
2
 
  
 
  
4
=
a4
4
1
3
−
π
16
 
  
 
  
A2 = a2 − π
a
2
 
  
 
  
2
= a2 1−
π
4
 
  
 
  
Therefore, the following ratios can be written:
I1
A1
=
a4
4
1
3
− π
128
 
  
 
  
a2 1 − π
8
 
  
 
  
=
a2
1
3
− π
128
 
  
 
  
4 − π
2
 
  
 
  
www.elsolucionario.net
Page 4-15
I2
A2
=
a4
4
1
3
− π
16
 
  
 
  
a2 1 − π
4
 
  
 
  
=
a2
1
3
− π
16
 
  
 
  
4 − π
Therefore, comparing the two cases:
I1 / A1
I2 / A2
=
a2
1
3
− π
128
 
  
 
  
4 − π
2
 
  
 
  
 
 
 
 
 
  
 
 
 
 
 
  
4 − π
a2
1
3
− π
16
 
  
 
  
 
 
 
 
 
  
 
 
 
 
 
  
= 0.797
Therefore, the beam with two holes gives the lower weight at a given bending moment.
4.25 A straight beam is loaded at the ends by moments M: the area moment of inertia for the
beam is I=a
3
b/12. Find the bending stress distribution in the beam when M=1000Nm,
a=3cm, and b=6cm. Also, find the radius of curvature to which the beam is bent. The
beam’s modulus of elasticity is 2.05 x 10
11
 Pa.
Notes: Equation (4.49) gives the bending stress distribution; Equation (4.50) gives the radius of
curvature.
Solution:
From the given data, the moment of inertia is I=a
3
b/12=(0.03m)
3
(0.06m)/12=1.35x10
-7
m
4
.
Therefore, the bending stress distribution is given by Equation (4.49) as:
σ = My
I
= 1000Nm( )y
1.35 ×10−7 m3
= 7.41GN / m3( )y
The maximum bending stress occurs for y=a/2, or σmax=111MPa. The radius of curvature is
calculated from Equation (4.50) as:
1
r
= M
EI
; r = EI
M
=
205GPa( ) 1.35 ×10−7m4( )
1000Nm
= 28.1m
4.26 The beam in Problem 4.25 is bent in a perpendicular direction so that I=ab3/12. Find the
bending stress distribution and the radius of curvature to which the beam is bent.
Notes: This is solved in the same manner as Problem 4.25, except that the values of a and b have
changed.
Solution:
From the given data, the moment of inertia is I=ab
3
/12=(0.03m)(0.06m)
3
/12=5.4x10
-7
m
4
.
Therefore, the bending stress distribution is given by Equation (4.49) as:
σ = My
I
= 1000Nm( )y
5.4 ×10−7m3
= 1.85GN / m3( )y
The maximum bending stress occurs for y=b/2, or σmax=55.5MPa. The radius of curvature is
calculated from Equation (4.50) as:
1
r
= M
EI
; r = EI
M
=
205GPa( ) 5.4 ×10−7 m4( )
1000Nm
= 110.7m
www.elsolucionario.net
Page 4-16
4.27 A curved bar has a rectangular cross section with height h=r0-ri=50mm and width
b=100mm. Its inner radius is 100mm. Find the distance between the neutral axis and the
centroid.
Notes: This problem merely requires application of Equation (4.58).
Solution:
Note that ro=ri+h=150mm. The eccentricity is given by Equation (4.58) as:
e = ri + ro
2
− ro − ri
ln
ro
r1
 
 
  
 
 
  
= 100mm +150mm
2
− 150mm −100mm
ln
150
100
 
  
 
  
= 1.68mm
4.28 The curved bar is loaded with a bending moment of 3000 Nm. Find the stress at the
innermost and outermost radii.
Notes: These stresses are obtained from Equations (4.65) and (4.66). The results from Problem
4.27 are used to obtain this solution.
Solution:
As derived in Problem 4.27, e=1.68mm=0.00168m. With a height of 50mm and a width of
100mm, the cross sectional area is A=0.005m
2
. Therefore, the stress at the inner surface is, from
Equation (4.65),
σi = −
Mci
Aeri
= − 3000Nm( ) 0.025m − 0.00168m( )
0.005m2( ) 0.00168m( ) 0.1m( ) = −83.29MPa
The stress at the outer surface is, from Equation (4.66),
σo =
Mco
Aero
= 3000Nm( ) 0.025m + 0.00168m( )
0.05m2( ) 0.00168m( ) 0.15m( ) = 63.52MPa
4.29 Two beams with rectangular cross sections axb are placed on top of each other to form a
beam having height 2a and width b. Find the area moments of inertia Ix and Iy for the two
beams
a) When they are welded together along the length
b) When they are not welded together.
Notes: The implication is that if the beams are welded together they act as a single beam; if not
welded, they act as two individual beams.
Solution:
If the beams are welded together, it is as if the cross section is of one material, so from Table 4.1,
Ix =
bh3
12
= b 2a( )
3
12
= 2ba
3
3
Iy =
b3h
12
= b
3 2a( )
12
= b
3a
6
www.elsolucionario.net
Page 4-17
If they are not welded together, then the moments of inertia are:
Ix =
bh3
12
+ bh
3
12
= 2b a( )
3
12
= ba
3
6
Iy =
b3h
12
+ b
3h
12
= b
3 a( )
12
+ b
3 a( )
12
= b
3a
6
4.30 Two thin steel plates having width b/20 and height 5a are placed one at each side of the
two beams in Problem 4.29. See sketch h. Find the moments of inertia around the x and y
axes
a) When the plates are not welded together.
b) When they are welded together as shown in sketch h
c) When they are welded together to form a closed tube 5ax1.1b.
Notes: The approach is the same as in Problem 4.29.
Solution:
If the plates are not welded together, then the moments of inertia of the individual beams are
added together. Ix is calculated as
Ix = 2
1
12
 
  
 
  
b
20
 
  
 
  5a( )
3 + 2 ba
3
12
=1.21ba3
For Iy, note that even though the thin plates are not welded to the other plates, they still are
separated, so their moment of inertial about the y-axis needs to be evaluated using the Parallel-
Axis Theorem. Therefore,
Iy = 2
1
12
 
  
 
  ab
3 + 2 1
12
 
  
 
  5a( )
b
20
 
  
 
  
3
+ 2 5a( ) b
20
 
  
 
  
b
2
+ b
40
 
  
 
  
2
= 0.304ab3
If the plates are welded together, then the plates can be taken as one area. Ix is then calculated as:
Ix = 2
1
12
 
  
 
  
b
20
 
  
 
  5a( )
3 + 2 1
12
 
  
 
  b( ) 2a( )
3 = 2.375ba3
and by taking advantage of the geometry, Iy is
Iy =
1
12
5a( ) b + b
20
+ b
20
 
  
 
  
3
− 2 1
12
 
  
 
  1.5a( ) b( )
3 = 0.304ab3
Note that Iy has not changed appreciably; this is attributableto the very thin plates used at the
ends of the axb plates.
If they are welded to form a closed tube, then Ix is given by:
Ix =
1
12
1.1b( ) 5a( )3 − 1
12
b( ) 3a( )3 = 9.21ba3
and Iy is unchanged; it has the value 0.304ab
3
.
www.elsolucionario.net
Page 4-18
4.31 A cantilevered beam with a rectangular cross section is loaded by a force perpendicular to
the beam centerline at the free end. The cross section is 75 mm high and 25 mm wide.
The vertical load at the beam end is 20,000N. Calculate how long the beam should be to
give tensile and compressive stresses 10 times higher than the maximum shear stress.
Also, calculate these stresses.
Notes: Equation (4.48) gives the maximum bending stress, and Table 4.3 gives the maximum
shear stress.
Solution:
For the cross section defined, the moment of inertia is (0.025m)(0.075m)
3
/12=8.79x10
-7
m
4
. For a
cantilever with a load P at the free end, the maximum moment is M=Pl=(20,000N)l. The
maximum shear is V=P=20,000N. Therefore, the maximum shear stress is, from Table 4.3 on
page 170,
τmax =
3V
2A
= 3 20 ,000N( )
2 0.025m( ) 0.075m( ) =16MPa
Since the maximum normal stress is ten times the maximum shear stress, it is 160MPa. From
Equation (4.48),
σmax =
Mc
I
=
20,000 N( ) l( ) 0.075m
2
 
  
 
  
8.79 ×10−7m4
= 160MPa; l = 0.1875m = 187.5mm
4.32 A cantilevered beam with a circular-tube cross section has an outer diameter of 100mm
and a wall thickness of 10mm. The load perpendicular to the beam is 15,000N, and the
beam is 1.2m long from the point of force to the wall where the beam is fastened.
Calculate the maximum bending and shear stresses.
Notes: The approach is similar to Problem 4.31. Equation (4.48) gives the maximum bending
stress, and Table 4.3 gives the maximum shear stress.
Solution:
The maximum moment occurs at the wall and equals M=Pl=(15,000N)(1.2m)=18kNm. The
maximum shear force is V=P=15kN. The moment of inertia of the beam is (see Table 4.1 on page
148):
I = π
64
do
4 − di4( ) = π64 0.1m( )
4 − 0.08m( )4( ) = 2.898×10−6 m4
From Table 4.3 on page 170, the maximum shear stress is:
τmax =
2V
A
= 2 15kN( )π
4
0.1m( )2 − 0.08m( )2( )
= 10.61MPa
From Equation (4.48),
σ = Mc
I
= 18kNm( ) 0.05m( )
2.898×10−6 m4
= 310.5MPa
4.33 A fishing rod is made of glass-fiber-reinforced plastic in the form of a tube having an
outer diameter of 10 mm and a wall thickness of 1.5mm. The glass fibers are parallel to
www.elsolucionario.net
Page 4-19
the tube axis, so that the bending shear stress is carried by the plastic and the bending
stresses are carried by the fibers. The fishing rod is 2m long. Determine whether the rod
fails from tensile stresses in the fibers or from shear overstressing in the plastic. The
bending strength of the fiber-reinforced plastic is 800MPa, and its shear strength is
3.2MPa.
Notes: The bending normal stresses are calculated from Equation (4.48) and the maximum shear
stress is obtained from Table 4.3 on page 170. Comparing the critical load for the two failure
modes allows solution of the problem.
Solution:
The maximum moment occurs at the wall and equals M=Pl=(P)(2m). The maximum shear force
is V=P. The moment of inertia of the rod is (see Table 4.1 on page 148):
I = π
64
do
4 − di4( ) = π64 0.01m( )
4 − 0.007m( )4( ) = 3.73 ×10−10 m4
From Table 4.3 on page 170, the maximum shear stress is tmax=2V/A, so that the largest load
that can be applied at the end of the rod is:
τmax =
2P
A
= 3.2MPa; P =
3.2MPa( ) π
4
0.01m( )2 − 0.007m( )2( )   
 
  
2
= 64N
The maximum load that can be supported without causing bending failure is obtained from
Equation (4.48), recognizing the maximum allowable bending stress is 80MPa:
σmax =
Mc
I
= P 2m( ) 0.005m( )
3.73 ×10−10m4
= 800MPa ;P = 29.8N
Therefore, the rod fails first from overstressing the fibers.
4.34 Two aluminum beams, like the one shown in Fig. 4.6, are welded together to form a
closed cross section with dimensions 240 by 80mm. The weld was badly done, so that the
wall is only 2 mm thick instead of 8mm. The allowable shear stress in the weld is 50MPa,
whereas the maximum allowable bending stress is 150MPa. Find whether the 2-m-long
beam fails first at the outermost fibers or at the welds.
Notes: The moment of inertia and the first moment have to be calculated for the shape. The
maximum load is then calculated for the two failure modes as in Problem 4.33. The only
difference is that Equation (4.67) is needed to obtain the maximum shear stress.
Solution:
The moment of inertia for the cross section is:
I = b1h1
3
12
− b2h2
3
12
= 0.08m( ) 0.24m( )
3
12
− 0.064m( ) 0.224m( )
3
12
= 3.222 ×10−5 m4
For a cantilever, the maximum moment is M=Pl=(2m)P. Therefore, Equation (4.48) gives:
σmax =
Mc
I
= Plc
I
; P = σ maxI
lc
=
150MPa( ) 3.222× 10−5 m4( )
2m( ) 0.120m( ) = 20.14kN
The first moment of area is calculated as:
Q = 0.06m( ) 0.08m( ) 0.12m( )− 0.056m( ) 0.064 m( ) 0.112m( ) =1.746 ×10−4m3
Equation (4.67) then gives:
www.elsolucionario.net
Page 4-20
τmax =
VQ
Iwt
; V = τmaxIwt
Q
=
50MPa( ) 3.22 ×10−5 m4( ) 0.004m( )
1.746 ×10−4 m3
= 36.9kN
The weld fails first.
4.35 To decrease as much as possible the weight of beams subjected to bending, the center of
gravity of the cross section is placed as far away from the beam center of gravity as
possible. A beam supporting a floor has a cross section as shown in sketch i. The bending
moment acting on the beam is 3000Nm, and the shear force is 1000N. Calculate the
maximum bending stress and maximum shear stress.
Notes: The moment of inertia and the first moment have to be calculated for the shape. The
maximum load is then calculated for the two failure modes as in Problem 4.34.
Solution:
The moment of inertia for the cross section is:
I = b1h1
3
12
− b2h2
3
12
= 0.100m( ) 0.16m( )
3
12
− 2 0.048m( ) 0.14m( )
3
12
=1.218 ×10−5m4
Therfore, Equation (4.48) gives:
σmax =
Mc
I
= 3000Nm( ) 0.08m( )
1.218×10−5 m4
= 19.7MPa
The first moment of area is calculated as:
Q=(0.035m)(0.004m)(0.07m)+(0.01m)(0.1m)(0.075m)=8.48x10
-5
m
3
Equation (4.67) then gives:
τmax =
VQ
Iwt
=
1000N( ) 8.48 ×10−5m3( )
1.218×10−5 m( ) 0.004m( ) = 1.74MPa
www.elsolucionario.net
Page 5-1
Chapter 5
Deformation
5.1 A beam is loaded by a concentrated bending moment M at the free end. The size of the
moment equals the maximum moment, or Pl. Find the vertical and angular deformations
along the beam by using the equation of the elastic line, Eq. (5.3).
Notes: This problem requires integration of Equation (5.3) twice, and then application of
boundary conditions for a cantilever.
Solution:
Equation (5.3) gives
∂ 2y
∂x2
= − M
EI
= − Pl
EI
Integrating twice,
∂y
∂x
= − Pl
EI
x + C1
y = − Pl
2EI
x2 + C1x + C2
Referring to the sketch above, the boundary conditions are dy/dx=0 at x=0 and y=0 at x=0. The
first boundary condition gives C1=0. The second gives C2=0. Therefore, the angular deflection
and the vertical deflection along the beam are given by (see Equation (5.6)):
∂y
dx
= θ = − Pl
EI
x; y = − Pl
2EI
x2
The maximum vertical deflection occurs at x=l and is
ymax = −
Pl3
2EI
5.2 A simply supported beam of length l carries a force P. Find the ratio between the bending
stresses in the beam when P is concentrated in the middle of the beam and evenly
distributed along it. Use the moment-curvature relation given in Equation (5.3). Also,
calculate the ratio of deformations at the middle of the beam.
Notes: To determine the maximum stresses, the largest moment must be determined. The shear
and moment diagrams are found as in Chapter 2. Integration of Equation (5.3) then gives the
answer.
Solution:
The sketch and shear and moment diagrams for the two cases are shown below. In both cases, the
maximum stress and deflection are at the center.
www.elsolucionario.net
Page 5-2
I. Point Load Case
For the point load case, moment equation up to x=l/2 is M(x)=-Px/2. The maximum moment, at
x=l/2, is Mmax=-Pl/4. Therefore, from Equation (5.3), and integrating twice,d2y
dx 2
= − M
EI
= Px
2EI
; 
dy
dx
= Px
2
4EI
+ C1; y =
Px3
12EI
+ C1x + C2
The boundary conditions are that y=0 at x=0 and dy/dx=0 at x=l/2. The first boundary condition
gives C2=0. The second gives:
dy
dx
= Px
2
4EI
+ C1; 0=
Pl2
16EI
+ C1; C1 = −
Pl2
16EI
Therefore, the deflection is
y = Px
3
12EI
− Pl
2
16EI
x = Px
4EI
x2
3
− l
2
4
 
 
  
 
 
  
The maximum deflection occurs at x=l/2 and is:
ymax =
Pl
8EI
l2
12
− l
2
4
 
 
  
 
 
  = −
Pl3
48EI
II. Distributed load
For this case, the moment distribution is
M = P
2
x2
l
− x
 
 
  
 
 
  
The maximum moment, evaluated at x=l/2, is Mmax=-Pl/8. From Equation (5.3), and integrating
twice, gives
d2y
dx 2
= − M
EI
= P
2EI
x − x
2
l
 
 
  
 
 
  ; 
dy
dx
= P
2EI
x2
2
− x
3
3l
 
 
  
 
 
  + C3 ; y =
P
2 EI
x3
6
− x
4
12l
 
 
  
 
 
  + C3x + C4
The first boundary condition is that at x=0, y=0. Therefore, C4=0. The second boundary condition
is that dy/dx=0 at x=l/2, so that
dy
dx
= P
2 EI
x2
2
− x
3
3l
 
 
  
 
 
  + C3; 0 =
P
2EI
l2
8
− l
3
24l
 
 
  
 
 
  + C3; C3 = −
Pl2
24EI
Therefore the deflection, and the maximum value evaluated at x=l/2 are:
y =
P
2EI
x3
6
−
x4
12l
 
 
  
 
 
  −
Pl2
24EI
x =
P
12EI
x3 −
x4
2l
−
l2x
2
 
 
  
 
 
  ; ymax = yx=l /2 = −
5Pl3
384EI
www.elsolucionario.net
Page 5-3
III. Comparison
The ratio of the maximum stresses, from Equation (4.48) gives:
σconc
σ dist
=
Mconcc
I
 
  
 
  
Mdistc
I
 
  
 
  
= Mconc
Mdist
= −Pl / 4
−Pl / 8
= 2
The ratio of maximum deflections is:
yconc
ydist
=
− Pl
3
48EI
 
 
  
 
 
  
− 5Pl
3
384EI
 
 
  
 
 
  
= 384
240
= 1.6
5.3 A simply supported beam with length l is centrally loaded with a force P. How large a
moment needs to be applied at the ends of the beam
a) To maintain the slope angle of zero at the supports?
b) To maintain the midpoint of the beam without deflection when the load is applied?
Use the equation of the elastic line, Eq. (5.3).
Notes: The problem which must be solved is that of a simply supported beam with both a
concentrated load at the center and a moment applied at both ends. Once the elastic equation of
the beam is determined, the problem can be solved.
Solution:
The beam and moment diagrams are obtained as in Chapter 2. The moment distribution is given
by
M=M0-Px/2
Therefore, from Equation (5.3) and integrating twice,
d2y
dx 2
= − M x( )
EI
= Px
2EI
− M0
EI
; 
dy
dx
= Px
2
4EI
− M0x
EI
+ C1; y =
Px3
12EI
− M0x
2
2EI
+ C1x + C2
The boundary conditions are y=0 at x=0 and dy/dx=0 at x=l/2. The first boundary condition gives
C2=0. The second gives:
0 = Pl
2
16EI
− M0l
2EI
+ C1; C1 =
1
EI
M0l
2
− Pl
2
16
 
 
  
 
 
  
Therefore, the angular deflection and the vertical deflection are given by:
dy
dx
= Px
2
4EI
− M0x
EI
+ 1
EI
M0l
2
− Pl
2
16
 
 
  
 
 
  =
1
EI
P
x2
4
− l
2
16
 
 
  
 
 
  + M0
l
2
− x
 
  
 
  
 
 
 
 
 
 
 
 
y = Px
3
12EI
− M0x
2
2EI
+ 1
EI
M0l
2
− Pl
2
16
 
 
  
 
 
  x =
x
EI
P
x2
12
− l
2
16
 
 
  
 
 
  + M0
l
2
− x
2
 
  
 
  
 
 
 
 
 
 
 
 
To maintain a slope of zero at the supports, set dy/dx=0, x=0, and solve for M0 to give:
dy
dx
= 0 =
1
EI
P −
l2
16
 
 
  
 
 
  + M0
l
2
 
  
 
  
 
 
 
 
 
 
 
 ; M0 =
Pl
8
To have a deflection of zero at the midpoint, set y=0 at x=l/2 and solve for M0:
www.elsolucionario.net
Page 5-4
y = 0 = l
2EI
P
l2
48
− l
2
16
 
 
  
 
 
  + M0
l
2
− l
4
 
  
 
  
 
 
 
 
 
 
 
 
; M0 =
8Pl2
48l
= Pl
6
5.4 Find the relationship between P and w0 so that the slope of the deflected beam is zero at
the supports for the loading conditions shown in sketch a. Assume that E and A are
constant.
Notes: This solution uses singularity functions, although it could be solved through the other
methods in Chapter 2.
Solution:
The reactions are shown on the beam above in red. Due to symmetry, the reaction forces are
equal and are:
R = P
2
+ w0l
4
Using singularity functions, the load function is given by:
q x( ) = −w0 x 0 +
2w0
l
x 1 − 2w0
l
x − l
2
1
+ R x − l
2
−1
− P x − l −1
+R x − 3l
2
−1
− 2w0
l
x − 3l
2
1
+ 2w0
l
x − 2l 1 + w0 x − 2l
0
Therefore, From Equation (2.4),
−V x( ) = −w0 x 1+
w0
l
x 2 − w0
l
x − l
2
2
+ R x − l
2
0
− P x − l 0
+R x − 3l
2
0
− w0
l
x − 3l
2
2
+ w0
l
x − 2 l 2 + w0 x − 2l
1
And from Equation (2.5),
−M x( ) = − w0
2
x 2 + w0
3l
x 3 − w0
3l
x − l
2
3
+ R x − l
2
1
− P x − l 1
+R x − 3l
2
1
− w0
3l
x − 3l
2
3
+ w0
3l
x − 2l 3 + w0
2
x − 2 l 2
From Equation (5.3),
EI
d2y
dx2
= −M x( )
Continuing the integration,
EI
dy
dx
= − w0
6
x 3 + w0
12l
x 4 − w0
12 l
x − l
2
4
+ R
2
x − l
2
2
− P
2
x − l 2
+ R
2
x − 3l
2
2
− w0
12l
x − 3 l
2
4
+ w0
12l
x − 2l 4 + w0
6
x − 2l 3 + C1
www.elsolucionario.net
Page 5-5
EIy = − w0
24
x 4 + w0
60 l
x 5 − w0
60l
x − l
2
5
+ R
6
x − l
2
3
− P
6
x − l 3
+ R
6
x − 3 l
2
3
− w0
60l
x − 3l
2
5
+ w0
60l
x − 2 l 5 + w0
24
x − 2l 4 + C1x + C2
The boundary conditions are that y=0 at x=l/2 and y=0 at 3l/2. The first boundary condition gives:
EIy = −w0
24
l
2
 
  
 
  
4
+ w0
60l
l
2
 
  
 
  
5
+ C1
l
2
+ C2
The second boundary condition gives
EIy = −w0
24
3l
2
 
  
 
  
4
+ w0
60 l
3l
2
 
  
 
  
5
− w0
60l
l( )5 + R
6
l( )3 − P
6
l
2
 
  
 
  
3
+ C1
l
2
+ C2
Solving for C1 and C2 gives:
C1 =
55
960
w0l
3 − Pl
2
16
C2 = −
51w0l
4
1920
+ Pl
3
32
If the slope is zero at x=l/2,
EI
dy
dx
= 0 = − w0
6
l
2
 
  
 
  
3
+ w0
12l
l
2
 
  
 
  
4
+ C1 = −
w0
6
l
2
 
  
 
  
3
+ w0
12l
l
2
 
  
 
  
4
+ 55
960
w0l
3 − Pl
2
16
Or, simplifying the equation,
0 = 20w0 l
3
480
− Pl
2
16
; P = 2w0 l
3
5.5 Given a simply supported beam with two concentrated forces acting on it as shown in
sketch h, determine the expression for the elastic deformation of the beam for any x by
using singularity functions. Assume that E and I are constant. Also, determine the
location of maximum deflection and derive an expression for it.
Notes: This problem uses an approach similar to Problem 5.4.
Solution:
From symmetry, the reactions are equal and are equal to R1=R2=P. Therefore, the load function
in terms of singularity functions is:
q x( ) = P x −1 − P x − a −1 − P x − l + a −1 + P x − l −1 = P x −1 − x − a −1 − x − l + a −1 + x − l −1[ ]
From Equation (2.4),
−V x( ) = P x 0 − x − a 0 − x − l + a 0 + x − l 0[ ]
And from Equation (2.5),
www.elsolucionario.net
Page 5-6
−M x( ) = P x 1 − x − a 1 − x − l + a 1 + x − l 1[ ]
From Equation (5.3),
EI
d2y
dx2
= −M x( ) = P x 1 − x − a 1 − x − l + a 1 + x − l 1[ ]
Therefore, integrating twice,
EI
dy
dx
= P
2
x 2 − x − a 2 − x − l+ a 2 + x − l 2[ ]+ C1
EIy = P
6
x 3 − x − a 3 − x − l+ a 3 + x − l 3[ ]+ C1x + C2
The boundary conditions are y=0 at x=0 and y=0 at x=l. The first boundary condition gives C2=0.
The second gives:
EIy = 0 = P
6
l3 − l − a( )3 − a3[ ] + C1l; C1 = −P l − a( )a2
Therefore, the deflection is
EIy = P
6
x 3 − x − a 3 − x − l+ a 3 + x − l 3[ ]− P l − a( )a2 x
The maximum value is at x=l/2, and is:
EIymax =
P
6
l
2
 
  
 
  
3
− l
2
− a
 
  
 
  
3 
 
 
 
 
 
 
 
− P l − a( )al
4
= P
6
l
2
 
  
 
  
3
− l
2
− a
 
  
 
  
3
− 3al l − a( )
2
 
 
 
 
 
 
 
 
5.6 For the loading condition described in sketch c, obtain the internal shear force V(x) and
the internal moment M(x) by using singularity functions. Draw V(x), M(x), θ(x) and y(x)
as a function of x. Assume that w0=9kN/m and l=3m.
Notes: The statics of this problem are alittle more difficult than for Problem 5.4 and 5.5, but the
method of solution is similar. The reaction forces are shown in red in the sketch above.
Solution:
Summing moments about the left support,
M = 0 =∑ w0l
3
l
6
 
  
 
  +
1
2
w0l
3
 
  
 
  
4l
9
 
  
 
  − R2 l; R2 =
7w0 l
54
= 7 9kN / m( ) 3m( )
54
= 3.5kN
Taking vertical force equilibrium,
F = 0∑ = R1 + R2 − w0l3 −
w0l
6
; R1 =
w0l
2
− R2 =
9kN / m( ) 3m( )
2
− 3.5kN = 10kN
Therefore, the load distribution can be written in terms of singularity functions as:
www.elsolucionario.net
Page 5-7
q x( ) = R1 x −1 − w0 x 0 +
3w0
l
x − l
3
1
− 3w0
l
x − 2l
3
1
+ R2 x − l
−1
= R1 x
−1 − w0 x
0 + 3w0
l
x − l
3
1
− x − 2l
3
1 
 
 
 
 
 
 
 
+ R2 x − l
−1
From Equation (2.4),
−V x( ) = R1 x 0 − w0 x 1 +
3w0
2 l
x − l
3
2
− x − 2l
3
2 
 
 
 
 
 
 
 
+ R2 x − l
0
And from Equation (2.5),
−M x( ) = R1 x 1 −
w0
2
x 2 + 3w0
6l
x − l
3
3
− x − 2l
3
3 
 
 
 
 
 
 
 
+ R2 x − l
1
The shear and moment diagrams are sketched below:
From Equation (5.3),
EI
d2y
dx2
= −M x( ) = R1 x 1 −
w0
2
x 2 + 3w0
6l
x − l
3
3
− x − 2l
3
3 
 
 
 
 
 
 
 
+ R2 x − l
1
Therefore, integrating twice,
EI
dy
dx
= R1
2
x 2 − w0
6
x 3 + 3w0
24l
x − l
3
4
− x − 2l
3
4 
 
 
 
 
 
 
 
+ R2
2
x − l 2 + C1
EIy = R1
6
x 3 − w0
24
x 4 + 3w0
120 l
x − l
3
5
− x − 2l
3
5 
 
 
 
 
 
 
 
+ R2
6
x − l 3 + C1x + C2
Substituting for w0, R1, R2 and l,
EIy = 10kN
6
x 3 − 9kN / m
24
x 4 + 27kN / m
360m
x − l
3
5
− x − 2l
3
5 
 
 
 
 
 
 
 
+ 3.5kN
6
x − l 3 + C1x + C2
The boundary conditions are y=0 at x=0 and y=0 at x=l. The first boundary condition gives C2=0.
The second gives:
EIy = 0 = 10kN
6
3m( )3 − 9kN / m
24
3m( )4 + 27kN / m
360m
2m( )5 − 1m( )5[ ]+ C1 3m( );C1 = −5.65kNm2
Therefore, the angular deflection and vertical deflection relationships are:
EI
dy
dx
= EIθ = R1
2
x 2 − w0
6
x 3 + 3w0
24l
x − l
3
4
− x − 2l
3
4 
 
 
 
 
 
 
 
+ R2
2
x − l 2 − 5.65kNm2
EIy = 10kN
6
x 3 − 9kN / m
24
x 4 + 27kN / m
360m
x − l
3
5
− x − 2l
3
5 
 
 
 
 
 
 
 
+ 3.5kN
6
x − l 3 − 5.65kNm2x
These are sketched below.
www.elsolucionario.net
Page 5-8
5.7 A simply supported bar is shown in sketch d with w0=4kN/m and l=12m.
a) Draw the free body diagram of the bar.
b) Use singularity functions to determine shear force, bending moment, slope and
deflection.
Provide both tabular results and a diagram of the shear force, the bending moment, the
slope and the deflection.
Notes: This problem is very similar to Problem 5.6. The reaction forces are shown in red in the
sketch above.
Solution:
Summing moments about the left support,
M = 0 =∑ w0l
4
 
  
 
  
l
3
 
  
 
  +
w0l
8
 
  
 
  
7l
12
 
  
 
  − R2 l; R2 = w0l
1
12
+ 7
96
 
  
 
  = 4kN / m( ) 12m( )
5
32
 
  
 
  = 7.5kN
Taking vertical force equilibrium,
F = 0 =∑ − w0 l
4
 
  
 
  −
w0l
8
 
  
 
  + R1 + R2 ; R1 =
3w0 l
8
− R2 =
3 4kN / m( ) 12m( )
8
− 7.5kN = 10.5kN
Therefore, the load distribution can be written in terms of singularity functions as (note the
direction of positive y):
q x( ) = −R1 x −1 +
w0
l / 2( )
x 1 − w0
l / 2( )
x − l
2
1
− w0
l / 4( )
x − l
2
1
+ w0
l / 4( )
x − 3l
4
1
− R2 x − l
−1
= −R1 x
−1 + 2w0
l
x 1 − x − l
2
1
− 2 x − l
2
1
+ 2 x − 3l
4
1 
 
 
 
 
 
 
 
− R2 x − l
−1
From Equation (2.4),
−V x( ) = −R1 x 0 +
w0
l
x 2 − x − l
2
2
− 2 x − l
2
2
+ 2 x − 3l
4
2 
 
 
 
 
 
 
 
− R2 x − l
0
And from Equation (2.5),
−M x( ) = −R1 x 1 +
w0
3l
x 3 − x − l
2
3
− 2 x − l
2
3
+ 2 x − 3l
4
3 
 
 
 
 
 
 
 
− R2 x − l
1
The shear and moment diagrams are sketched below:
From Equation (5.3),
www.elsolucionario.net
Page 5-9
EI
d2y
dx2
= −M x( ) = −R1 x 1 +
w0
3l
x 3 − x − l
2
3
− 2 x − l
2
3
+ 2 x − 3l
4
3 
 
 
 
 
 
 
 
− R2 x − l
1
Therefore, integrating twice,
EI
dy
dx
= − R1
2
x 2 + w0
12 l
x 4 − x − l
2
4
− 2 x − l
2
4
+ 2 x − 3 l
4
4 
 
 
 
 
 
 
 
− R2
2
x − l 2 + C1
EIy = − R1
6
x 3 + w0
60l
x 5 − x − l
2
5
− 2 x − l
2
5
+ 2 x − 3 l
4
5 
 
 
 
 
 
 
 
− R2
6
x − l 3 + C1x + C2
Substituting for w0, R1, R2 and l,
EIy = −10.5kN
6
x 3 + 4kN / m
60 12m( ) x
5 − x − l
2
5
− 2 x − l
2
5
+ 2 x − 3l
4
5 
 
 
 
 
 
 
 
− 7.5kN
6
x − l 3 + C1x + C2
The boundary conditions are y=0 at x=0 and y=0 at x=l. The first boundary condition gives C2=0.
The second gives:
EIy = 0 = − 10.5kN
6
12m( )3 + 4kN / m
60 12m( ) 12m( )
5 − 12m
2
 
  
 
  
5
− 2 12m
2
 
  
 
  
5
+ 2 12m
4
 
  
 
  
5 
 
 
 
 
 
 
 
+ C1l; C1 =147.4kNm2
Therefore, the angular deflection and vertical deflection relationships are:
EI
dy
dx
= −5.25kN x 2 + 1kN / m
48m
x 4 − x − l
2
4
− 2 x − l
2
4
+ 2 x − 3 l
4
4 
 
 
 
 
 
 
 
− 3.75kN x − l 2 +147.4kNm2
EIy = −1.75kN x 3 + 1kN / m
1 5 1 2m( ) x
5 − x − l
2
5
− 2 x − l
2
5
+ 2 x − 3l
4
5 
 
 
 
 
 
 
 
−1.25kN x − l 3 +147.4kNm2
These are sketched below.
5.8 The simply supported beam in Problem 5.5 is altered so that instead of a concentrated
force P, a concentrated moment M is applied at the same location. The moments are
positive and act parallel with each other. Determine the deformation of the beam for any
position x along it by using singularity functions. Assume that E and I are constant. Also,
determine the location of the maximum deflection.
Notes: Finding the deformation equation uses the same approach as in Problems 5.4 through 5.7.
This problem illustrates that determination of the maximum deflection using singularity functions
is fairly difficult, unless it is numerically determined.
Solution:
Summing moments about the left support,
M∑ = 0 = M0 + M0 − R2 l; R2 = 2M0l
Taking vertical force equilibrium,
F∑ = 0 = R1 + R2; R1 = −R2 = −2M0l
www.elsolucionario.net
Page 5-10
Therefore, the load distribution can be written in terms of singularity functions as (note the
direction of positive y):
q x( ) = R1 x −1 + M0 x − a −2 + M0 x − l + a −2
Note that the terms at x=l have been ignored because they don’t affect the problem. From
Equation (2.4),
−V x( ) = R1 x 0 + M0 x − a −1 + M0 x − l + a −1
And from Equation (2.5),
−M x( ) = R1 x 1 + M0 x − a 0 + M0 x − l + a 0
From Equation (5.3),
EI
d2y
dx2
= −M x( ) = R1 x 1 + M0 x − a 0 + M0 x − l + a 0
Therefore, integrating twice,
EI
dy
dx
= R1
2
x 2 + M0 x − a
1 + M0 x − l + a
1 + C1
EIy = R1
6
x 3 + M0
2
x − a 2 + M0
2
x − l + a 2 + C1x + C2
The boundary conditions are that y=0 at x=0 and y=0 at x=l. The first boundary condition gives
C2=0. The second gives:
EIy = 0 = −2 M0
6l
l3 + M0
2
l − a( )2 + M0
2
a2 + C1l; C1 = M0 a−
a2
l
− l
6
 
 
  
 
 
  
Therefore, substituting for C1 and R1, the deformation of the beam is given by
EIy = −2M0
6 l
x 3 + M0
2
x − a 2 + M0
2
x − l + a 2 + M0 a −
a2
l
− l
6
 
 
  
 
 
  x
= M0 −
x 3
3 l
+ x − a
2
2
+ x − l + a
2
2
+ a − a
2
l
− l
6
 
 
  
 
 
  x
 
 
 
 
 
 
 
 
and dy/dx is
EI
dy
dx
= M0 −
x 2
l
+ x − a 1 + x − l + a 1 + a − a
2
l
− l
6
 
 
  
 
 
  
 
 
 
 
 
 
 
 
The maximum deflection occurs at dy/dx=0, but two intervals have to be examined. The first is
for x<a, so that the maximum deflection occurs at:
0 = − x
2
l
+ a− a
2
l
− l
6
 
 
  
 
 
  
 
 
 
 
 
 
 
 
; x = la − a2 − l
2
6
However, if a<x<l-a, setting dy/dx=0 yields:
EI
dy
dx
= 0 = M0 −
x2
l
+ x − a + a − a
2
l
− l
6
 
 
  
 
 
  
 
 
 
 
 
 
 
 
; x2 − xl + a2 + l
2
6
= 0
Using the quadratic formula and simplifying,
x =
l
2
1 −
1
3
−
2a
l
 
  
 
  
2 
 
 
  
 
 
 
  
The location of maximumdeflection is one of these values. One can substitute into the deflection
equation to see which is larger, but if a>l/4, the maximum deflection occurs at x<a, and it is
located at x>a for a<l/4,
www.elsolucionario.net
Page 5-11
5.9 The simply supported beam in Problem 5.8 has moments applied in opposite directions so
that the moment at x=a is M0 and at x=l-a is -M0. Find the elastic deformation of the beam
by using singularity functions. Also, determine the location and size of the maximum
deflection.
Notes: This is a much simpler problem than 5.8, although the approach is the same.
Solution:
Taking moment equilibrium about the left support,
ΣM=0=M0-M0-R2l; R2=0
Taking force equilibrium,
ΣF=0=R1+R2; R1=0
Therefore, the load intensity function is given by:
q x( ) = M0 x − a −2 − M0 x − l + a −2 = M0 x − a −2 − x − l + a −2[ ]
From Equation (2.4),
−V x( ) = M0 x − a −1 − x − l + a −1[ ]
And from Equation (2.5),
−M x( ) = M0 x − a 0 − x − l+ a 0[ ]
From Equation (5.3),
EI
d2y
dx2
= −M x( ) = M0 x − a 0 − x − l+ a 0[ ]
Therefore, integrating twice,
EI
dy
dx
= M0 x − a
1− x − l+ a 1[ ]+ C1
EIy = M0
2
x − a 2 − x − l+ a 2[ ]+ C1x + C2
The boundary conditions are y=0 at x=0 and y=0 at x=l. The first yields C2=0, the second yields
0 = M0
2
l − a( )2 − a2[ ]+ C1l; C1 = M0 2a − l( )2
Therefore, the deflection is
EIy = M0
2
x − a 2 − x − l+ a 2 + 2a − l( )x[ ]
The first derivative of this expression is:
EI
dy
dx
= M0 x − a
1 − x − l + a 1 + 2a − l( )
2
 
  
 
  
Setting this equal to zero yields, for x<a:
2a − l
2
= 0; a = l
2
Therefore, the deflection is a maximum only if the moments are both applied at the same location,
at the center of the beam. Otherwise, for a<x<l-a,
EI
dy
dx
= 0 = M0 x − a( )1 +
2a − l( )
2
 
  
 
  
; x − a + a − l
2
= 0; x = l
2
Therefore, the maximum deflection occurs at mid-span.
www.elsolucionario.net
Page 5-12
5.10 Given the loading condition shown in sketch e, find the deflection at the center and ends
of the beam. Assume that EI=750kNm
2
.
Notes: The statics are simplified by the symmetry of the problem. The approach is the same as in
Problems 5.4-5.9.
Solution:
The reactions have been added to the sketch in red. Due to the symmetry of the problem, R1=R2.
From vertical force equilibrium,
2R1-10kN-3kN=0; R1=R2=6.5kN
Therefore, the load distribution function is:
q x( ) = −5kN x −1[ ] + 6.5kN x − 2m −1 + x − 5m −1[ ] − 1kN / m x − 2m 0 − x − 5m 0[ ]
Note that the term associated with the point load at x=7m has been ignored since it does not affect
the problem. From Equation (2.4),
−V x( ) = −5kN x 0[ ] + 6.5kN x − 2m 0 + x − 5m 0[ ] −1kN / m x −2m 1 − x − 5m 1[ ]
From Equation (2.5),
−M x( ) = −5kN x 1[ ]+ 6.5kN x − 2m 1 + x − 5m 1[ ] − 1kN / m2 x − 2m 2 − x − 5m 2[ ]
Therefore, from Equation (5.3),
EI
d2y
dx2
= −M x( ) = −5kN x 1[ ]+ 6.5kN x − 2m 1 + x − 5m 1[ ] − 1kN / m2 x − 2m 2 − x − 5m 2[ ]
Integrating twice,
EI
dy
dx
= −
5kN
2
x 2[ ] + 6.5kN2 x − 2m 2 + x − 5m 2[ ]−
1kN / m
6
x − 2m 3 − x − 5m 3[ ]+ C1
EIy = −
5kN
6
x 3[ ]+ 6.5kN6 x − 2m 3 + x − 5m 3[ ] −
1kN / m
24
x − 2m 4 − x − 5m 4[ ]+ C1x + C2
The boundary conditions are dy/dx=0 at x=3.5m (from symmetry) and y=0 at x=2m. The first
boundary condition gives:
EI
dy
dx
= 0 = −
5kN
2
3.5m( )2[ ]+ 6.5kN2 1.5m( )2[ ] −
1kN / m
6
1.5m( )3[ ]+ C1; C1 = 23.875kNm2
The second boundary condition gives:
EIy = −
5kN
6
2m( )3[ ]+ 23.875kNm2( ) 2m( ) + C2; C2 = −41.08kNm3
The deflection at the ends of the beam can be obtained by evaluating y at x=0. All of the terms
except C2 vanish, so that
EIy = C2 ; y =
C2
EI
=
−41.08kNm3
750kNm2
= −0.0548m
At midspan, that is, at x=3.5m, the deflection is:
EIy = −
5kN
6
3.5m( )3[ ] + 6.5kN6 1.5m( )3[ ]−
1kN / m
24
1.5m( )4[ ]+ C1 3.5m( ) + C2 = 0.0136m
www.elsolucionario.net
Page 5-13
5.11 Given the loading condition shown in sketch f obtain an expression for the deflection at
any location on the beam. Assume that EI is a constant.
Notes: This is a statically indeterminate problem. The reactions are not known until a deflection
equation is obtained and statics is applied. The problem can be solved by using singularity
functions, and using the reactions as unknowns until the deflection equation is obtained.
Solution:
The reactions have been drawn on the sketch in red. Note that there are three forces and one
moment, so the problem is statically indeterminate. From statics, we obtain two equations, one for
force equilibrium, the other for moment equilibrium about point 1:
F∑ = 0 = −2w0a + R1 + R2 + kδ
M∑ = 0 = M1 − w0a2 + 2R2a + 4kδa − 103 w0a
2; M1 = −2R2a − 4kδa +
13
3
w0a
2
where δ is the deflection at the right side of the beam, or δ=y(x=4a). The load intensity function
can be written as:
q x( ) = −M1 x −2 + R1 x −1 − w0a x − a −1 + R2 x − 2a −1 −
w0
2a
x − 2a 1
Note that the terms which become active at x=4a have been ignored since they do not affect the
problem. Using Equation (2.4),
−V x( ) = −M1 x −1 + R1 x 0 − w0a x − a 0 + R2 x − 2a 0 −
w0
4a
x − 2a 2
From Equation (2.5),
−M x( ) = −M1 x 0 + R1 x 1 − w0a x − a 1 + R2 x − 2a 1 −
w0
12a
x − 2a 3
Therefore, from Equation (5.3),
EI
d2y
dx2
= −M x( ) = −M1 x 0 + R1 x 1 − w0a x − a 1 + R2 x − 2a 1 −
w0
12a
x − 2a 3
Integrating twice,
EI
dy
dx
= −M1 x
1 +
R1
2
x 2 −
w0a
2
x − a 2 +
R2
2
x − 2a 2 −
w0
48a
x − 2a 4 + C1
EIy = −
M1
2
x 2 +
R1
6
x 3 −
w0a
6
x − a 3 +
R2
6
x − 2a 3 −
w0
240a
x − 2a 5 + C1x + C2
We have three boundary conditions: 1) y=0 at x=0. 2) dy/dx=0 at x=0 3) y=0 at x=2a. The first
boundary condition yields C2=0. The second yields C1=0. The third yields
EIy = 0 = −
M1
2
2a( )2 + R1
6
2a( )3 − w0a
6
a3
This equation, along with the two equations from statics, allow solution for R1, R2 and M1 in terms
of w0, a and δ:
R1 = −
5
16
w0a +
3
2
kδ
R2 = −
5
2
kδ +
37
16
w0a
www.elsolucionario.net
Page 5-14
M1 = −
7
24
w0a
2 + kδa
δ is determined by evaluating y at x=4a:
EIy = EIδ = −
M1
2
4a( )2 + R1
6
4a( )3 − w0a
6
3a( )3 + R2
6
2a( )3 − w0
240a
2a( )5
This yields
δ =
2.55w0a
4
14
3
ka3 − EI
5.12 Given the loading condition and spring shown in sketch g determine the stiffness of the
spring so that the bending moment at point B is zero. Assume that EI is constant.
Notes: The reactions are determined through symmetry. The force needed to obtain a zero
moment can be obtained through statics, but to obtain the required stiffness of the spring, one
must calculate the deflection at mid-span. The approach is similar to previous problems, and uses
the method of singularity functions.
Solution:
From symmetry, the reactions are equal and are given by:
R1 = R2 = P −
1
2
kδ B
Therefore, the load distribution function is given by:
q x( ) = R1 x −1 − P x − l −1 + kδB x − 2l −1 − P x −3l −1 + R2 x − 4 l −1
From Equation (2.4),
−V x( ) = R1 x 0 − P x − l 0 + kδB x − 2l 0 − P x − 3l 0 + R2 x − 4l 0
From Equation (2.5),
−M x( ) = R1 x 1 − P x − l 1 + kδB x − 2l 1 − P x − 3l 1 + R2 x − 4 l 1
If M(x=2l)=0, then:
−M x = 2l( ) = R1 2l( )− Pl = 0; R1 =
P
2
= P − 1
2
kδ B; k =
P
δB
The moment expression becomes
−M x( ) = P
2
x 1 − P x − l 1 + P x − 2l 1 − P x − 3l 1+ P
2
x − 4 l 1
= P 1
2
x 1 − x − l 1 + x − 2l 1 − x − 3l 1 + 1
2
x − 4 l 1 
  
 
  
To obtain the stiffness, we need to determine δB. Therefore, from Equation (5.3),
EI
d2y
dx2
= −M x( ) = P 1
2
x 1 − x − l 1 + x − 2l 1 − x − 3l 1 + 1
2
x − 4 l 1
 
  
 
  
Integrating twice,
www.elsolucionario.net
Page 5-15
EI
dy
dx
= P
2
1
2
x 2 − x − l 2 + x − 2l 2 − x − 3l 2 + 1
2
x − 4l 2
 
  
 
  + C1
EIy = P
6
1
2
x 3 − x − l 3 + x − 2l 3 − x − 3l 3 + 1
2
x − 4 l 3
 
  
 
  + C1x + C2
The boundary conditions are that y=0 at x=0 and y=0 at x=4l. The first boundary condition gives
C2=0. The second boundary condition gives:
0 = P
6
1
2
4l( )3 − 3l( )3 + 2 l( )3 − l3   
 
  + C1 4l( ); C1 = −
Pl 2
2
Therefore, the deflection is:
EIy = P
6
1
2
x3 − x − l 3 + x − 2l 3 − x − 3l 3 + 1
2
x − 4 l 3 − 3l2x
 
  
 
  
At x=2l, the deflection is
EIy = EIδ B =
P
6
1
2
2l( )3 − l3 − 6l3   
 
  ; δB = −
Pl3
2EI
This deflection is downward, but the direction has no meaning for the spring; the stiffness is then
given by:
k = P
Pl 3
2EI
 
 
  
 
 
  
= 2EI
l3
5.13 When there is no load acting on the cantilevered beam shown in sketch h, the spring has
zero deflection. When there is a spring and a force of 20kN is applied at point C, a
deflection of 50mm occurs at the spring. When there is a spring, if a 50kN load is applied
at the location shown in sketch h, what will be the deflection of the bar? Assume that the
stiffness of the spring is 450kN/m.
Notes: The first load case allows calculation of EI through a similar approach as Problems 5.4
through 5.12. Once EI is known, the deflection of the bar as a function of position can be
determined similarly.
Solution:
Consider the first case, where a 20kN load is applied at the end and the beam deflection at the
spring is 0.050m. If the stiffness of the spring is 450kN/m, then the force applied at the spring is
(450kN/m)(0.050m)=22.5kN. Therefore, the load applied to the beam is as follows:
From vertical force equilibrium,
ΣF=0=R1+22.5kN-20kN; R1=-2.5kN
From Moment Equilibrium,
ΣM=0=-M1+(22.5kN)(2m)-(20kN)(4m); M1=-35kNm
Therefore, the load intensity function is given by:
q x( ) = −2.5kN x −1 − 35kNm x −2 + 22.5kN x − 2m −1 − 20kN x − 4m −1
www.elsolucionario.net
Page 5-16
From Equation (2.4)
−V x( ) = −2.5kN x 0 − 35kNm x −1 + 22.5kN x − 2m 0 − 20kN x − 4m 0
From Equation (2.5)
−M x( ) = −2.5kN x 1 − 35kNm x 0 + 22.5kN x − 2m 1 − 20kN x − 4m 1
Therefore, applying Equation (5.3),
EI
d2y
dx2
= −M x( ) = −2.5kN x 1− 35kNm x 0 + 22.5kN x − 2m 1 − 20kN x − 4m 1
Integrating twice,
EI
dy
dx
= −1.25kN x 2 − 35kNm x 1 +11.25kN x − 2m 2 −10kN x − 4m 2 + C1
EIy = −0.4167kN x 3 −17.5kNm x 2 +3.75kN x − 2m 3 − 3.333kN x − 4m 3 + C1x + C2
The boundary conditions are that y=0 at x=0 and dy/dx=0 at x=0. The first boundary condition
gives C2=0. The second gives C1=0. Therefore the deflection is
EIy = −0.4167kN x 3 −17.5kNm x 2 +3.75kN x − 2m 3 − 3.333kN x − 4m 3
It is known that at x=2m, y=-0.05m. Therefore, EI can be determined as:
EI −0.05m( ) = −0.4167kN 2m( )3 −17.5kNm 2m( )2 = −73.3333kNm3; EI = 1467kNm2
For the second part of the problem, the forces acting on the beam are as follows:
From Equilibrium,
ΣF=0=R1+kδB-50kN; R1=50kN-kδB
ΣM=0=-M1-kδB(2m)+50kN(3m); M1=150kNm-kδB(2m)
Therefore, the load intensity function is
q x( ) = R1 x −1 − M1 x −2 + kδB x − 2m −1 − 50kN x − 3m −1
From Equation (2.4)
−V x( ) = R1 x 0 − M1 x −1 + kδB x − 2m 0 − 50kN x − 3m 0
From Equation (2.5)
−M x( ) = R1 x 1 − M1 x 0 + kδB x − 2m 1 − 50kN x − 3m 1
Therefore, applying Equation (5.3),
EI
d2y
dx2
= −M x( ) = R1 x 1 − M1 x 0 + kδB x − 2m 1 − 50kN x − 3m 1
Integrating twice,
EI
dy
dx
= R1
2
x 2 − M1 x
1 + kδB
2
x − 2m 2 − 25kN x − 3m 2 + C1
EIy = R1
6
x 3 − M1
2
x 2 + kδB
6
x − 2m 3 − 8.333kN x − 3m 3 + C1x + C2
The boundary conditions are that y=0 at x=0 and dy/dx=0 at x=0. The first boundary condition
gives C2=0. The second gives C1=0. Therefore the deflection is
EIy = R1
6
x 3 − M1
2
x 2 + kδB
6
x − 2m 3 − 8.333kN x − 3m 3
At x=2m, the deflection is y=δB and is given by (after substituting for EI, R1 and M1):
1467kNm2( )δB = R16 2m( )
3 −
M1
2
2m( )2 = 50kN − (450kN / m)δB
6
8m3( ) − 2m2( ) 150kNm − 2kδBm( )
Solving for δB yields δB=-0.08749m. Therefore, M1=71.28kNm, and R1=10.63kN. Substituting in
the deflection equation, y is given by:
y = 1
1467kNm2
1.77kN x 3 − 35.63kNm x 2 + 6.565kN x − 2m 3 − 8.333kN x − 3m 3[ ]
www.elsolucionario.net
Page 5-17
5.14 Determine the deflection at point A and the maximum moment for the loading shown in
sketch i. Consider only the bending effect and assume that EI is constant.
Notes: The structure is two half rings hinged together; the shading can cause some students to
mistake it as a disk. This problem is solved by calculating the strain energy due to bending from
Equation (5.19), then applying Castigliano’s Theorem (Equation (5.30)).
Solution:
Due to symmetry, only one-half of the ring is considered. Consider the free body diagram of the
arbitrary section shown to the right above. From horizontal force equilibrium,
ΣF=0=R-P/2; R=P/2
From moment equilibrium about point O,
M∑ = 0 = M + P
2
rsinθ( ); M = − P
2
r sinθ( )
The maximum moment occurs for sinθ=1, which occurs at θ=π/2. The maximum moment is
M=Pr/2. From Equation (5.19), the strain energy due to bending is:
U = 2 M
2
2EI
rdθ0
π / 2∫ = 2 P
2r3 sin2θ
8EI
dθ = P
2r3
4EI
θ
2
− s i n 2θ
4
 
  
 
  0
π /2
= P
2r3π
16EI0
π /2
∫
Applying Castigliano’s Theorem in the form of Equation (5.30),
δH =
∂U
∂P
= ∂
∂P
P2r3π
16EI
 
 
  
 
 
  =
Pr3π
8EI
5.15 Determine the maximum deflection of the beam shown in sketch j.
Notes: The approach is very similar to the approach in Problems 5.5 through 5.14.
Solution:
The reactions have been added to the sketch in red. From vertical force equilibrium,
ΣF=0=R1-w0l; R1=w0l
From moment equilibrium,
ΣM=0=M1-w0l2/2; M1=w0l2/2
Therefore the load intensity function is:
www.elsolucionario.net
Page 5-18
q x( ) = M1 x −2 + R1 x −1 − w0 x 0 + w0 x − l 0
= w0 l
2
2
x −2 + w0 l x
−1 − w0 x
0 + w0 x − l
0
From Equation (2.4)
−V x( ) = w0l
2
2
x −1 + w0 l x
0 − w0 x
1 + w0 x − l
1
From Equation (2.5)
−M x( ) = w0 l
2
2
x 0 + w0 l x
1 − w0
2
x 2 + w0
2
x − l 2
Therefore, applying Equation (5.3),
EI
d2y
dx2
= −M x( ) = w0 l
2
2
x 0 + w0l x
1 − w0
2
x 2 + w0
2
x − l 2
Integrating twice,
EI
dy
dx
= w0l
2
2
x 1 + w0 l
2
x 2 − w0
6
x 3 + w0
6
x − l 3 + C1
EIy = w0l
2
4
x 2 + w0 l
6
x 3 − w0
24
x 4 + w0
24
x − l 4 + C1x + C2
The boundary conditions are that y=0 at x=0 and dy/dx=0 at x=0. The first boundary condition
gives C2=0. The second gives C1=0. Therefore the deflection is
EIy = w0l
2
4
x 2 + w0 l
6
x 3 − w0
24
x 4 + w0
24
x − l 4
The maximum deflection occurs at x=l+a, and is:
EIymax =
w0l
2
4
l + a( )2 + w0 l
6
l + a( )3 − w0
24
l + a( )4 + w0
24
a( )4
ymax =
w0l
2
24EI
9l2 + 20al +12a2[ ]
5.16 Determine the deflection at any point in the beam shown in sketch k. Use singularity
functions.
Notes: The approach is very similar to the approach in Problems 5.5 through 5.15. Note that this
problem is statically determinate.
Solution:
Summing moments about point B,
M∑ = 0 = kδ Al − Pl3 ; kδ A =
P
3
From vertical force equilibrium,
F∑ = 0 = kδA − P + RB ; RB = P− kδ A = P − P3 =
2P
3
Therefore, the load intensity function is:
www.elsolucionario.net
Page 5-19
q x( ) = P
3
x −1 − P x − 2l
3
−1
+ 2P
3
x − l −1 = P
3
x −1 −3 x − 2l
3
−1
+ 2 x − l −1
 
 
 
 
 
 
 
 
From Equation (2.4)
−V x( ) = P
3
x 0 − 3 x − 2 l
3
0
+ 2 x − l 0
 
 
 
 
 
 
 
 
From Equation (2.5)
−M x( ) = P
3
x 1− 3 x − 2l
3
1
+ 2 x − l 1
 
 
 
 
 
 
 
 
Therefore, applying Equation (5.3),
EI
d2y
dx2
= −M x( ) = P
3
x 1− 3 x − 2l
3
1
+ 2 x − l 1
 
 
 
 
 
 
 
 
Integrating twice,
EI
dy
dx
= P
6
x 2 − 3 x − 2l
3
2
+ 2 x − l 2
 
 
 
 
 
 
 
 
+ C1
EIy = P
18
x 3 −3 x − 2l
3
3
+ 2 x − l 3
 
 
 
 
 
 
 
 
+ C1x + C2
The boundary conditions are that y=-δA at x=0 and y=0 at x=l. The first boundary condition gives
C2
EI
= −δA = −
P
3k
The second boundary condition gives
0 =
P
18
l3 − 3
l
3
 
  
 
  
3 
 
 
 
 
 
 
 + C1l −
PEI
3k
; C1 =
PEI
3kl
−
4Pl2
81
Therefore, the deflection is:
EIy =
P
18
x 3 − 3 x −
2l
3
3
+ 2 x − l 3
 
 
 
 
 
 
 
 +
PEI
3kl
−
4Pl2
81
 
 
  
 
 
  x −
PEI
3k
5.17 Determine the deformation of a cantilevered beam with loading shown in sketch l as a
function of x. Also determine the maximum bending stress in the beam and the maximum
deflection. Assume E=207GPa, I=250cm
4
, P=1000N,w0=3000N/m, a=0.5m, b=0.15m,
c=0.45m. The distance from the neutral axis to the outermost fiber of the beam is 0.040m.
Notes: The method of superposition can be used for this problem, since the problem can be
broken down into two cases which appear in Table 5.1 on page 193. The maximum deflection
occurs at the free end of the cantilever, and the maximum moment occurs at the wall.
Solution:
The loading can be broken down into the first two cases in Table 5.1 are used, so the deflection of
the beam is the sum of the individual deflections, or:
www.elsolucionario.net
Page 5-20
y x( ) = − P
6EI
x − a + b( ) 3 − x 3 + 3x2 a + b( )[ ] + w0EI b + c( )x
3
6
−
b + c( )x2
2
a +
b + c
2
 
  
 
  
−
1
24
x − a 4
 
 
 
 
 
 
 
 
substituting for the variables,
y x( ) = − 3.221 ×10−4 m−2( ) x − 0.65m 3 − x3 + 3x2 0.65m( )[ ]
+ 5.797m−3( ) 0.1m( )x3 − 0.24m2( )x2 − 124 x − 0.5m
4 
  
 
  
The maximum deflection occurs at x=(a+b+c)=1.1m. Therefore,
ymax = − 3.221× 10−4m−2( ) 0.45m( )3 − 1.1m( )3 + 3 1 . 1m( )2 0.65m( )[ ]
+ 5.797m−3( ) 0.1m( ) 1.1m( )3 − 0.24m2( ) 1.1m( )2 − 124 0.6m( )
4 
  
 
  
= −1.304 × 10−3m = −1.304mm
The maximum bending moment occurs at the wall, and from statics
Mmax = P a + b( )+ b + c( )w0 a +
b + c
2
 
  
 
  
= 1000N( ) 0.5m + 0.15m( ) + 0.15m + 0.45m( ) 3000N / m( ) 0.5m + 0.15m + 0.45m
2
 
  
 
  
= 2090Nm
Therefore, from Equation (4.48),
σ =
Mc
I
=
2090Nm( ) 0.040m( )
250cm4
= 33.44MPa
5.18 Given the loading in sketch m let a=0.6m, b=0.7m, M=6500Nm and w0=20,000N/m. The
beam has a square cross section with sides of 75mm, and the beam material has a
modulus of elasticity of 207GPa. Determine the beam deformation by using the method
of superposition. Also, calculate the maximum bending stress and the maximum beam
deformation.
Notes: This problem is similar to Problem 5.17. The method of superposition can be used for this
problem, since the problem can be broken down into two cases which appear in Table 5.1 on page
193. The maximum deflection occurs at the free end of the cantilever, but the maximum moment
location is unknown; it is obtained by taking the derivative of the moment equation.
Solution:
Note that the moment of inertia for this cross section is 2.637x10
-6
m
4
. The problem can be broken
into two parts, characterized by the last two cases in Table 5.1 on page 193. Fortunately, the same
terminology is used for distances, so the expression for deflection can be written directly from the
principle of superposition:
www.elsolucionario.net
Page 5-21
y =
w0
EI
bx3
6
−
bx2
2
a +
b
2
 
  
 
  
−
1
24
x − a 4
 
 
 
 
 
 
 
 +
Mx2
2 EI
=
20kN / m( )
207GPa( ) 2.637 ×10−6m4( ) 0.1167m( )x
3 − 0.3325m2( )x2 − 124 x − 0.6m
4 
  
 
  
+
6500Nm
2 207GPa( ) 2.637 × 10−6 m4( ) x
2
= 0.004275m−2( )x3 − 0.00623m −1( )x2 − 0.001527m−3( ) x − 0.6m 4
The location of the maximum value is determined by setting the derivative equal to zero and
solving for x. We do this over two intervals because of the presence of a singularity function at
x=0.6m. Therefore, for x<0.6m,
dy
dx
= 3 0.004275m−2( )x2 − 2 0.00623m−1( )x = 0
This requires that x=0 or x=0.972m. At x=0, y=0 because it is a fixed support. x=0.972m is
outside the range considered for the form of dy/dx. Therefore, there is no local minimum over this
range, and the deflection at the edge of the interval is y=0. Over the range 0.6m<x<1.3m,
dy
dx
= 3 0.004275m−2( )x2 − 2 0.00623m−1( )x − 4 0.001527m−3( ) x − 0.6m( )3 = 0
This has roots of x=0 and x=1.002m. x=0 lies outside the bounds for which the equation for dy/dx
is valid. The deflection for x=1.002m is y=-0.001994m. At the end of the beam, y=-0.001503m.
Therefore, the maximum deflection is y=-0.001994m=-1.994mm.
The maximum moment location is unknown. From Equation (5.3),
M = −EI( )d
2y
dx 2
Therefore, Mmax will occur where the third derivative of y with respect to x equals zero. The first,
second and third derivatives of y(x) are:
dy
dx
= 0.01283m−2( )x2 − 0.01246m−1( )x − 0.006108m−3( ) x − 0.6m 3
d2y
dx 2
= 0.02566m−2( )x − 0.01246m−1( ) − 0.018324m−3( ) x − 0.6m 2
d3y
dx 3
= 0.02566m−2( ) − 0.03665m−3( ) x − 0.6m
Setting the last equation equal to zero yields x=1.3m. The moment is then a maximum either at
x=1.3 or at x=0, the other end of the beam. At x=1.3m, M=6500Nm. At x=0, M=6800Nm.
Therefore, Mmax=6800Nm. The maximum bending stress is, from Equation (4.48),
σ =
Mc
I
=
6800Nm( ) 0.075m
2
 
  
 
  
2.637 × 10−6 m4
= 96.71MPa
5.19 The cantilevered beam shown in sketch n has both a concentrated force and a moment
acting on it. Let a=1m, b=0.7m, P=8700N, and M=4000Nm. The beam cross section is
rectangular with a height of 80mm and a width of 35mm. Also, E=207GPa. Calculate the
beam deflection by using the method of superposition. Find how large M has to be to give
zero deflection at x=a.
www.elsolucionario.net
Page 5-22
Notes: The problem is broken into two parts, one where only a load is applied at x=a, and one
with a moment applied at the end of the beam. The deflection for the beam is then the sum of the
deflections from the two cases by the principle of superposition. Table 5.1 is used to obtain the
deflections for the two cases.
Solution:
The moment of inertia for the beam cross section is:
I =
bh3
12
=
0.035m( ) 0.080m( )3
12
= 1.493× 10−6m4
Referring to Table 5.1 on page 193, the problem can be separated into a point load case at x=a
(the first row in Table 5.1) and a moment applied at the free end (row 3 in Table 5.1). Therefore,
from the method of superposition and recognizing the moment is opposite that shown in Table
5.1, the deflection is:
y = − P
6 EI
x − a 3 − x 3 + 3x2a[ ]+ Mx
2
2EI
= −
8700N( )
6 207GPa( ) 1.493 × 10−6 m4( ) x −1m
3 − x 3 + 3x2 1m( )[ ]+ 4000Nm
2 207GPa( ) 1.493 ×10−6m4( ) x
2
= −0.00469m−2 x −1m 3 − x3 + 3m( )x2[ ] + 0.00647m−2( )x2
Note that if M is unknown, the deflection at x=a is given by:
y x = a( )= − P
6EI
−a3 + 3a3[ ] + Ma
2
2EI
; 
Ma2
2EI
=
2a3P
6EI
; M =
2
3
Pa =
2
3
8700N( ) 1m( ) = 5800Nm
5.20 A simply supported beam has loads as shown in sketch o. Calculate the beam
deformation by using the method of superposition. Also, calculate the maximum bending
stress and the maximum beam deformation and their locations. Assume that E=207GPa
and that the beam has a rectangular cross section with a height of 30mm and a width of
100mm. Also, P=1200N, w0=10,000N/m, a=0.2m, b=0.1m, c=0.4m, and d=0.2m.
Notes: This problem can be broken into the two cases given in Table 5.1(b). The approach is the
same as in Problem 5.19, but the relations for the deflection are slightly more complicated.
Solution:
The moment of inertia for the beam is
I =
bh3
12
=
0.100m( ) 0.030m( )3
12
= 2.25 × 10−7m4
So that EI=46.57kNm
2
. Note that this problem can be broken into the case of a point load and a
distributed load, and the two cases combined through the method of superposition. The two cases
are given in Table 5.1 (b) on page 193. Note that the variable labels in this problem statement do
not correlate exactly with the nomenclature in Table 5.1(b), so care must be taken to use the
proper expression. The deflection is:
www.elsolucionario.net
Page 5-23
y = P
6EI
b + c + d
l
x 3 − x − a 3 + 3a2x − 2alx − a
3x
l
 
 
 
 
 
 
 
 
+ w0c
24lEI
4 d + c
2
 
  
 
  
x 3 − l
c
x − a − b 4 − x − a − b − c 4[ ] + x c3 + 6cd2 + 4c2d + 4d 3 − 4 l2 d + c
2
 
  
 
  
 
  
 
  
 
 
  
 
 
  
Substituting the given values,
y = 1200N( )
6 46.57kNm2( )
0.7
0.9
x 3 − x − 0.2m 3 + 3 0.2m( )2x − 2 0.2m( ) 0.9m( )x − 0.2m( )
3 x
0.9m
 
 
 
 
 
 
 
 
+
10kN / m( ) 0.4m( )
24 0.9m( ) 46.57kNm2( ) 4 0.4m( ) x
3 −
0.9
0.4
x − 0.3m 4 − x − 0.7m 4[ ]+ x −0.976m 3[ ]   
 
 
 
= 0.00970m−2 x 3 − 0.004294m−2 x − 0.2m 3 − 0.004949x − 0.008946m−3 x − 0.3m 4 − x − 0.7m 4[ ]
To find the position of maximum deflection,set dy/dx=0 and solve for x:
dy
dx
= 0.0291m−2 x 2 − 0.01288m−2 x − 0.2m 2 − 0.004949 − 0.03578m−3 x − 0.3m 3 − x − 0.7m 3[ ]
For 0<x<0.2, this becomes
dy
dx
= 0 = 0.0291m−2 x( )2 − 0.004949; x = 0.41m
This is outside the interval considered, so there is no local minimum or maximum over 0<x<0.2.
Over 0.2<x<0.3,
dy
dx
= 0 = 0.0291m−2 x2 − 0.01288m−2 x − 0.2 m( )2 − 0.004949; x = 0.4429m
Again, there is no local minimum or maximum in the interval 0.2<x<0.3. For 0.3<x<0.7, dy/dx
becomes:
dy
dx
= 0.0291m−2x2 − 0.01288m−2 x − 0.2m( )2 − 0.004949 − 0.03578m−3 x − 0.3m( )3
This is solved as x=0.449m. Evaluating the deflection equation at x=0.449m,
ymax = 0.00970m
−2 0.449m( )3 − 0.004294m−2 0.249m( )3
−0.004949 0.449 m( ) − 0.008946m−3 0.149m( )4[ ]= −0.001415m = −1.415mm
The bending moment can be found by applying Equation (5.3), so that
−
M
EI
=
d2 y
dx2
= 0.0582m−2 x − 0.02576m−2 x − 0.2m − 0.10734m−3 x − 0.3m 2 − x − 0.7m 2[ ]
The maximum value is found by taking a derivative with respect to x, setting equal to zero, and
solving for x:
d
dx
M
EI
= 0.0582m−2 x 0 − 0.02576m−2 x − 0.2m 0 − 0.21468m−3 x − 0.3m 1 − x − 0.7m 1[ ]
or x=0.451m. The maximum moment is then:
− Mmax
EI
= 0.0582m−2 0.451m( ) − 0.02576m−2 0.251m( ) − 0.10734m−3 0.151m( )2[ ]
Mmax = EI −0.01733m−1( ) = 46.57kNm2( ) −0.01733m−1( ) = −807Nm
The maximum bending stress is, therefore,
σ =
Mc
I
=
807Nm( ) 0.015m( )
2.25 × 10−7m4
= 53.8MPa
www.elsolucionario.net
Page 5-24
5.21 The beam shown in sketch p is fixed at both ends and center loaded with a force of
2300N. The beam is 3.2m long and has a square tubular cross section with an outside
width of 130mm and a wall thickness of 10mm. The tube material is AISI 1080 high-
carbon steel. Calculate the deformation at any point along the beam by using the method
of superposition.
Notes: Symmetry should be used to transform the problem to a 1.6m long cantilever where the
slope is zero at the end of the cantilever. Then two cases can be taken from Table 5.1.
Solution:
The moment of inertia for the tube is:
I =
1
12
bo
4 − bi4( ) = 112 0.13m( )
4 − 0.11m( )4[ ] = 1.16 ×10−5m4
Applying symmetry, the beam can be taken as follows:
Note that one-half the load has been applied to the left half of the beam, and an unknown internal
moment exists at the mid-point of the beam, which becomes the end of the cantilever. Applying
Table 5.1,
y = −
P / 2( )
6 EI
x −
l
2
3
− x 3 +
3x2l
2
 
 
 
 
 
 
 
 −
M2 x
2
2EI
Note that M2 cannot be determined from statics. However, the end slope of the cantilever must be
zero to preserve symmetry. The slope of the cantilever is obtained by taking dy/dx:
dy
dx
= −
P( )
12EI
3 x −
l
2
2
− 3x2 + 3xl
 
 
 
 
 
 
 
 −
M2 x
EI
Since dy/dx=0 at x=l/2, we can solve for M2 as:
0 = −
P( )
12EI
−3
l
2
 
  
 
  
2
+ 3
l
2
 
  
 
  
l
 
 
 
 
 
 
 
 −
M2 l
2EI
; M2 = −
Pl
8
Therefore, substituting into the expression for y over 0<x<l/2:
y = −
P / 2( )
6 EI
−x3 +
3x2 l
2
 
 
 
 
 
 
 
 +
Plx2
16EI
= −
Px2 l
16EI
1−
4x
3 l
 
  
 
  
5.22 Bar A shown in sketch 1 is a 13-mm diameter aluminum bar; bar B is an 8 mm diameter
steel bar. The lower member is of uniform cross section and is assumed to be rigid. Find
the distance x if the lower member is to remain horizontal. Assume that the modulus of
elasticity for steel is three times that for aluminum.
www.elsolucionario.net
Page 5-25
Notes: The problem is statically indeterminate. There are three unknowns, namely the force in
members A and B and the distance x. The three equations which will be used to solve the problem
are force equilibrium, moment equilibrium and a deflection requirement - that for the beam to
remain horizontal, the deflection has to be the same at A and B.
Solution:
The reactions have been sketched in red. Note that the cross sectional area of bar A is
AA =
π
4
d2 = π
4
0.013m( )2 = 1.327 ×10−4m2
Similarly, the cross sectional area of bar B is 5.0265x10
-5
m
2
. Moment equilibrium about point A
yields
MA = 0∑ = 0.3m( )W − PBx; x = 0.3m( )WPB
Vertical force equilibrium yields:
F∑ = 0 = PA +PB − W ; PA = W − PB
From Equation (4.26), the deflections at A and B are, respectively:
δ A =
PA l
AA EA
; δB =
PBl
ABEB
For the bar to remain horizontal, δA=δB. Also, it is given that EB=3EA, and substituting for the
areas gives:
δ A = δB; 
PAl
AAEA
= PB l
ABEB
; 
PAl
1.327 ×10−4 m2( )EA =
PB l
5.0265×10−5 m2( ) 3EA( )
Solving for PA gives PA=0.8802PB. Substituting into the force equilibrium equation gives
PA = 0.8802PB = W − PB ; PB =
W
1.8802
Substituting into the moment equilibrium equation gives
x = 0.3m( )W
PB
= 0.3m( )W
W /1.8802( ) = 0.564m
5.23 An aluminum rod 3/4 in in diameter and 48 in long and a nickel steel rod 1/2 in in
diameter and 32 in long are spaced 60 in apart and fastened to a horizontal beam that
carries a 2000 lbf load, as shown in sketch r. The beam is to remain horizontal after load
is applied. Assume that the beam is weightless and absolutely rigid. Find the location x of
the load and determine the stresses in each rod.
www.elsolucionario.net
Page 5-26
Notes: The approach is the same as Problem 5.22. The problem is statically indeterminate. There
are three unknowns, namely the force in members A and B and the location of the applied force
W. The three equations which will be used to solve the problem are force equilibrium, moment
equilibrium and a deflection requirement - that for the beam to remain horizontal, the deflection
has to be the same at A and B.
Solution:
From the inside front cover, the stiffness of aluminum is EA=10Mpsi, and the stiffness of steel is
30Mpsi. The reactions have been sketched in red. Note that the cross sectional area of bar A is
AA =
π
4
d2 = π
4
0.75in( )2 = 0.4418in2
Similarly, the cross sectional area of bar B is 0.1963in
2
. Moment equilibrium about point A yields
MA = 0∑ = x 2000lb( ) − PB 60in( ); x = 60in( )PB2000lb
Vertical force equilibrium yields:
F∑ = 0 = PA +PB − 2000lb ; PA = 2000lb − PB
From Equation (4.26), the deflections at A and B are, respectively:
δ A =
PALA
AA EA
; δB =
PBLB
ABEB
For the bar to remain horizontal, δA=δB. Also, substituting for the lengths, areas, and elastic
moduli:
δ A = δB; 
PALA
AAE A
= PBLB
AB EB
; 
PA 48in( )
0.4418in2( ) 10Mpsi( ) =
PB 32 in( )
0.1963in2( ) 30Mpsi( )
Solving for PA gives PA=0.500PB. Substituting into the force equilibrium equation gives
PA = 2000lb − PB = 0.500PB ; PB =
2000lb
1.500
= 1333lb
Therefore, PA=0.516PB=680lb. Substituting into the moment equilibrium equation gives
x =
60in( ) 1333lb( )
2000lb
= 40.0 in
The stresses are obtained from Equation (4.22):
σA =
PA
AA
=
667lb
0.4418in2
=1510 psi
σB =
PB
AB
=
1333lb
0.1963in2
= 6741psi
www.elsolucionario.net
Page 5-27
5.24 Find the force on each of the vertical bars shown in sketch s. The weight is assumed to be
rigid and horizontal, implying that the three vertical bars are connected at the weight in a
straight line. Also, assume that the support at the top of the bars is rigid. The bar
materials and its circular cross-sectional area are given in the sketch.
Notes: This is a statically indeterminate problem. Fortunately, symmetry greatly simplifies the
statics. Just as in Problems 5.22 and 5.23, an equation based on deformations is used to obtain a
solution.
Solution:
From the inside front cover, Es=30Mpsi and Eb=16Mpsi. The internal forces in the bars are shown
in the sketch in red. Note that symmetry has been applied, so that the steel bars bear the same
load. Vertical force equilibrium on the weight gives:
ΣF=0=2Ps+Pb-5000lb; 2Ps+Pb=5000lb
From Equation (4.26), the deflections are:
δs =
Ps L
AsEs
; δb =
PbL
AbEb
Because of symmetry, the deflection of the steel and bronze must be equal. Also, substituting for
the lengths, areas, and elastic moduli:
δs = δ b; 
PsL
As Es
= Pb L
AbEb
; 
Ps 36in( )
0.2in2( ) 30Mpsi() =
Pb 36in( )
0.3in2( ) 16Mpsi( ) ; Ps = 1.25Pb
Substituting into the force equilibrium equation:
2 1.25Pb( ) + Pb = 5000lb; Pb =1430lb
Therefore, Ps=1.25Pb=1785lb.
5.25 Two solid spheres, one made of aluminum alloy 2014 and the other made of AISI 1040
medium-carbon steel, are lowered to the bottom of the sea at a depth of 8000 m. Both
spheres have a diameter of 0.3m. Calculate the elastic energy stored in the two spheres
when they are at the bottom of the sea if the density of water is 1000 kg/m
3
 and the
acceleration of gravity is 9.807m/s
2
. Also calculate how large the steel sphere has to be to
have the same elastic energy as the 0.3m aluminum sphere.
Notes: The hydrostatic stress state which is applied is σ1=σ2=σ3=-p. Equation (5.29) allows
calculation of the strain energy.
Solution:
The volume of the spheres is:
www.elsolucionario.net
Page 5-28
V = 4
3
πr3 = 4
3
π 0.13m( )3 = 0.0141m3
From the inside front cover, Eal=72GPa, νal=0.33, Es=207GPa, νs=0.3. Note that a pressure gives a
hydrostatic stress state of σ1=σ2=σ3=-p, where p is the pressure. p is calculated as
p = ρgh = 1000kg / m3( ) 9.807m / s2( ) 8000m( ) = 78.44MPa
From Equation (5.29), the strain energy is:
U = 1
2E
σ1
2 + σ2
2 +σ3
2( ) − νE σ1σ2 +σ2σ3 +σ 3σ1( )
 
  
 
  
dv
v∫
= 1
2E
3p2( ) − νE 3p2( )
 
  
 
  dvv∫ =
3Vp2
2E
1− 2ν( )
Substituting the elastic constants for aluminum,
Ual =
3Vp2
2E
1− 2ν( ) =
3 0 . 0 1 4 1m3( ) 78.44 MPa( )
2 72GPa( ) 1− 2 0.33( )( ) = 616.4Nm
and for the steel,
Us =
3Vp2
2 E
1− 2ν( ) =
3 0.0141m3( ) 78.44MPa( )
2 207GPa( ) 1− 2 0.3( )( ) = 252.2Nm
For the steel sphere to have the same energy as the aluminum sphere, that is, 616.4Nm, the
volume would have to be:
Us =
3Vp2
2 E
1− 2ν( ); V = 2Us E
3p2 1− 2ν( )
= 2 616.4 Nm( ) 207GPa( )
3 78.44MPa( )2 1− 2 0.3( )( )
= 0.0346m2
since V=4πr3/3, the radius would have to be 0.202m, so the diameter would need to be 0.404m.
5.26 Use Castigliano’s approach instead of singularity functions to solve Problem 5.5. Assume
that transverse shear is negligible.
Notes: To find the deflection at a location where no load is applied, apply a fictitious load Q at
that point. From statics, determine the moment in each section, then energy is obtained from
Equation (5.19). Equation (5.30) is then used to obtain the deflection.
Solution:
The reactions are shown in red, and a fictitious load Q is applied at the center of the beam where
the deflection is desired. From symmetry the reaction forces are equal. Therefore, from vertical
force equilibrium,
F∑ = 0 = 2R −2 P− Q; R1 = R2 = Q2 + P
Taking sections in the beam, and applying moment equilibrium, the bending moment in the beam
is found at the various sections:
MAB = P +
Q
2
 
  
 
  x
MBC = P +
Q
2
 
  
 
  x − P x − a( ) =
Qx
2
+ Pa
MCD = P +
Q
2
 
  
 
  x − P x − a( ) −Q x −
l
2
 
  
 
  = Q
l
2
− x
2
 
  
 
  + Pa
www.elsolucionario.net
Page 5-29
MDE = −
Qx
2
+ Pa + Ql
2
+ Q
2
+ P
 
  
 
  x − l + a( ) = Q
l
2
− x
2
 
  
 
  + P l − x( )
Note also from these expressions that
∂MAB
∂Q
= x
2
∂MBC
∂Q
= x
2
∂MCD
∂Q
= l
2
− x
2
 
  
 
  
∂MDE
∂Q
= l
2
− x
2
 
  
 
  
From Equation (5.19), the strain energy is:
U = MAB
2
2EI
dx0
a∫ +
MBC
2
2EI
dx
a
l /2∫ +
MCD
2
2EI
dx
l / 2
l−a∫ + MDE
2
2EI
dx
l−a
l∫
The deflection at C in the direction of Q is given by Equation (5.30) as:
δ = ∂U
∂Q
= MAB
EI
∂MAB
∂Q
dx0
a∫ +
MBC
EI
∂MBC
∂Q
dx
a
l /2∫ +
MCD
EI
∂MCD
∂Q
dx
l / 2
l−a∫ + MDEEI
∂MDE
∂Q
dx
l−a
l∫
Substituting for the moments and their derivatives,
EIδ = P + Q
2
 
  
 
  
x2
2
dx
0
a∫ +
Qx
2
+ Pa 
  
 
  
x
2
dx
a
l /2∫ + Q
l
2
− x
2
 
  
 
  
+ Pa
 
  
 
  
l
2
− x
2
 
  
 
  
dx
l / 2
l−a∫
+ Q l
2
− x
2
 
  
 
  
+ P l − x( )
 
  
 
  
l
2
− x
2
 
  
 
  
dx
l−a
l∫
Setting Q=0 yields
EIδ = P( ) x
2
2
dx0
a∫ + Pa( ) x2 dxa
l /2∫ + Pa( ) l2 −
x
2
 
  
 
  dxl / 2
l −a∫ + P l − x( )( ) l2 −
x
2
 
  
 
  dxl−a
l∫
This is evaluated as
δ = P
EI
a3
6
+ a
4
l2
4
− a2
 
 
  
 
 
  +
al
2
l
2
− a
 
  
 
  −
a
4
3l2
4
− 2la + a2
 
 
  
 
 
  +
1
6
l − l + a( )3
 
 
 
  
 
 
 
  
which can be reduced to
δ =
aP 3l2 − 4a2( )
24EI
5.27 Using Castigliano’s theorem, find the maximum deflection of the two-dimensional
cantilevered bar shown in sketch t. Neglect transverse shear.
Notes: The strain energy for bending is calculated with Equation (5.19). Since M=Px, then
Castigliano’s theorem (Equation (5.30)) is used to obtain the deflection at the end of the bar,
which is fortunately the location and in the direction of force application.
www.elsolucionario.net
Page 5-30
Solution:
From statics, the moment in the bar is M=Px. From Equation (5.19),
U = M
2
2EI
dx0
l∫ = P
2 x2
2EI
dx0
l∫
From Equation (5.30),
y = ∂U
∂P
= ∂
∂P
P2 x2
2 EI
dx0
l∫ = 2Px
2
2EI
dx0
l∫ = Px
2
EI
dx0
l∫
The integral has to be broken into two parts because of the change in cross section, so that:
y = Px
2
EI
dx0
l /2∫ + Px
2
2EI
dx
l / 2
l∫ = PEI
1
3
x3
 
  
 
  0
l /2
+ P
2EI
1
3
x3
 
  
 
  l / 2
l
= 3Pl
3
16EI
5.28 The right-angle-cantilevered bracket shown in sketch u is loaded with force P in the z-
direction. Derive an expression for the deflection of the free end in the z-direction by
using Castigliano’s theorem. Neglect transverse shear effects.
Notes: The total strain energy is calculated from Equations (5.19) and (5.23). Equation (5.30) is
then used to obtain the desired deflection equation.
Solution:
There are three sources of strain energy: bar ab is in bending, and bar bc is in bending and
torsion. For bar ab, the moment is Py, so equation (5.19) gives the strain energy as:
U1 =
M2
2EI
dy0
h∫ = P
2 y2
2 EI
dy0
h∫ = P
2
2EI
1
3
y3
 
  
 
  0
h
= P
2h3
6EI
For bar bc, the moment is M=Px (note the convenient definition of the coordinate system).
Therefore, Equation (5.19) gives:
U2 =
M2
2EI
dx0
l∫ = P
2 x2
2EI
dx0
l∫ = P
2
2EI
1
3
x3
 
  
 
  0
l
= P
2 l3
6EI
There is a constant torque of T=Ph on bar bc, so that Equation (5.23) gives:
U3 =
T2
2GJ
dx0
l∫ = P
2h2
2GJ
dx0
l∫ = P
2h2
2GJ
x( )0
l = P
2h2 l
2GJ
The total energy is then:
Utot = U1 + U2 +U3 =
P2h3
6EI
+ P
2 l3
6EI
+ P
2h2 l
2GJ
From Castigliano’s theorem (Equation (5.30)),
δ = ∂U
∂P
= ∂
∂P
P2h3
6EI
+ P
2 l3
6EI
+ P
2h2 l
2GJ
 
 
  
 
 
  =
Ph3
3EI
+ Pl
3
3EI
+ Ph
2l
GJ
= P
3EI
h3 + l3( ) + Ph
2l
GJ
www.elsolucionario.net
Page 5-31
5.29 A triangular cantilevered plate is shown in sketch v. Use Castigliano’s theorem to derive
an expression for the deflection at the free end, assuming that transverse shear is
neglected.
Notes: This problem is similar to Problems 5.26 through 5.28, except that the moment of inertia
of the cross section is a function of x. Once an expression is obtained for the moment of inertia,
Equation (5.19) gives the strain energy and application of Equation (5.30) (Castigliano’s
Theorem) gives the deflection equation.
Solution:
The width of the plate varies as a linear function of x. At x=0, w=0. At x=l, w=b. Therefore, the
width as a function of x is w=bx/l. The moment of inertia as a function of x is then:
I = wh
3
12
= 1
12
bx
l
 
  
 
  h
3 = bh
3x
12 l
The moment as a function of x is M=Px. Therefore, Equation (5.19) gives:
U = M
2
2EI
dx0
l∫ = Px( )
2
2E
bh3x
12l
 
 
  
 
 
  
dx0
l∫ = 6P
2lx
Ebh3
dx0
l∫ = 6P
2l
Ebh3
1
2
x2
 
  
 
  0
l
= 3P
2l3
Ebh3
Applying Equation (5.30),
δ = ∂U
∂P
= ∂
∂P
3P2l3
Ebh3
 
 
  
 
 
  =
6 Pl3
Ebh3
5.30 A right angle-cantilevered bracket with
concentrated load and torsional loading at the free
end is shown in sketch w. Using Castigliano’s
theorem, find the deflection at the free end in the z-
direction. Neglecttransverse shear effects.
Notes: This problem is similar to Problem 5.28, but
the loading is slightly more complicated. The total
strain energy is calculated from Equations (5.19)
and (5.23). Equation (5.30) is then used to obtain
the desired deflection equation.
www.elsolucionario.net
Page 5-32
Solution:
From statics, the loading in the vertical section is a torque of Tv=T and a bending moment Mv=Py.
For the horizontal section, the torque is Th=Pb, and the moment is Mh=Px+T. Using equations
(5.19) and (5.23), the total strain energy is:
U = Mv
2
2EI
dy
0
b∫ +
Tv
2
2GJ
dy
0
b∫ +
Mh
2
2EI
dx
0
a∫ +
Th
2
2GJ
dx
0
a∫
=
Py( )2
2EI
dy0
b∫ + T
2
2GJ
dy0
b∫ + Px + T( )
2
2EI
dx0
a∫ + Pb( )
2
2GJ
dx0
a∫
Applying Equation (5.30) (Castigliano’s theorem),
δ z =
∂U
∂P
= ∂
∂P
Py( )2
2EI
dy0
b
∫ +
T 2
2GJ
dy0
b
∫ +
Px + T( )2
2EI
dx0
a
∫ +
Pb( )2
2GJ
dx0
a
∫
 
 
  
 
 
  
= 2Py
2
2EI
dy0
b∫ + 2 Px + T( )x2 EI dx0
a∫ + 2Pb
2
2GJ
dx0
a∫ = PEI
1
3
y3
 
  
 
  0
b
+ 1
EI
Px3
3
+ Tx
2
2
 
 
  
 
 
  
0
a
+ Pb
2
GJ
x( )0
a
= Pb
3
3EI
+ 1
EI
Pa3
3
+ Ta
2
2
 
 
  
 
 
  +
Pb2a
2GJ
=
P b3 + a3( )
3EI
+ Ta
2
2EI
+ Pb
2a
GJ
5.31 A cantilevered I-beam has a concentrated load applied to the free end as shown in sketch
x. What upward force at point S is needed to reduce the deflection at S to zero? use
Castigliano’s theorem. Transverse shear can be neglected.
Notes: The bending moment must be determined through statics over the two intervals. The total
strain energy is then obtained through Equation (5.19) and Equation (5.30) is then used to obtain
the desired deflection equation.
Solution:
Taking a section at x<200mm, the moment in this section is M=Px. For 200mm<x<500mm, the
method of sections gives M=Px-Sy(x-0.2m). Therefore, the strain energy is given be Equation
(5.19) as:
U = M
2
2EI
dx
0
l∫ =
Px( )2
2EI0
0.2m∫ dx +
Px − Sy x − 0.2m( )( )2
2EI
dx
0.2m
0.5m∫
= Px( )
2
2EI0
0.2m∫ dx +
P − Sy( )x + Sy 0.2m( )( )2
2EI
dx
0.2m
0.5m∫
= P
2
2EI
1
3
x3
 
  
 
  0
0.2m
+
P− Sy( )2
2EI
1
3
x3
 
  
 
  0.2m
0.5m
+
P − Sy( )Sy 0.2m( )
EI
1
2
x2
 
  
 
  0.2m
0.5m
+
Sy 0.2m( )( )2
2EI
x( )0.2m
0.5m
= P
2 0.2m( )3
6EI
+
P− Sy( )2 0.5m( )3 − 0.2m( )3[ ]
6EI
+
P− Sy( )Sy 0.2m( ) 0.5m( )2 − 0.2m( )2[ ]
2EI
+
Sy 0.2m( )( )2 0.3m( )
2EI
www.elsolucionario.net
Page 5-33
Applying Equation (5.30), and recognizing the desired deflection is at location S,
δ = ∂U
∂Sy
= ∂
∂Sy
P2 0.2m( )3
6 EI
+
P − Sy( )2 0.5m( )3 − 0.2m( )3[ ]
6EI
+
P − Sy( )Sy 0.2m( ) 0.5m( )2 − 0.2m( )2[ ]
2 EI
+
Sy 0.2 m( )( )2 0.3m( )
2EI
 
 
 
 
 
 
 
 
 
 
= −
2 P − Sy( ) 0.5m( )3 − 0.2m( )3[ ]
6EI
+
P − 2Sy( ) 0.2m( ) 0.5m( )2 − 0.2m( )2[ ]
2EI
+
Sy 0.2m( )2 0.3m( )
EI
Setting δ=0, and substituting P=5kN,
δ = 0 = −
5kN − Sy( ) 0.117m3( )
3EI
+
5kN − 2Sy( ) 0.042m3( )
2 EI
+
Sy 0.012m
3
EI
; Sy = 10kN
5.32 Using Castigliano’s theorem calculate the horizontal and vertical deflections at point A
shown in sketch y. Assume that E and A are constant.
Notes: This problem is similar to the previous problems, but transverse shear is not negligible.
Therefore, the energy is calculated by combining axial, bending and transverse shear strain
energies. The equations are taken from Table 5.2 on page 200. Castigliano’s theorem (Equation
(5.30)) is used to obtain the deflection.
Solution:
The reactions have been drawn on the sketch in red. From statics, Rx=-Pcosθ=Px, Ry=Psinθ=Py,
and M0=-Prcosθ-Prsinθ=-r(Px+Py). If a section is taken at any angle φ, the shear and normal
forces can be determined as:
ΣFx=0=Rx+Vcosφ-Fsinφ
ΣFy=0=Ry-Vsinφ-Fcosφ
These equations are solved as
F=Rycosφ+Rxsinφ=Pycosφ-Pxsinφ
V=Rysinφ-Rxcosφ=Pysinφ+Pxcosφ
From Equation (2.5),
M = Vds∫ = Vrdφ∫ = Ry sinφ − Rx cosφ( )rdφ = r −Ry cosφ − Rx sinφ( ) + C∫
The boundary condition is that M=M0 at φ=0, so that
M0 = r − Ry c o s 0− Rx s i n 0( ) + C; C = M0 + rRy
Therefore, the moment is
M = r −Ry cosφ − Rx sinφ( ) + M0 + rRy = M0 + r 1− cosφ( )Ry − Rx sinφ[ ]
Referring to Table 5.2 on page 200, the total strain energy is:
U = M
2rdφ
2EI
+ F
2rdφ
2EA0
π /2∫ + 3V
2rdφ
5GA0
π /2∫0
π / 2∫ = M
2
2EI
+ F
2
2EA
+ 3V
2
5GA
 
 
  
 
 
  rdφ0
π / 2∫
Therefore, the horizontal deflection is obtained by applying Equation (5.30):
www.elsolucionario.net
Page 5-34
δx =
∂U
∂Px
= ∂
∂Px
M2
2EI
+ F
2
2EA
+ 3V
2
5GA
 
 
  
 
 
  rdφ0
π / 2
∫ =
M
EI
∂M
∂Px
+ F
EA
∂F
∂Px
+ 6V
5GA
∂V
∂Px
 
 
  
 
 
  rdφ0
π / 2
∫
= r
3
EI
0.5Py +0.356Px( ) + r20AGE E 12Py +18.85Px( ) − G 10Py − 15.71Px( )[ ]
The vertical deflection is similarly calculated from Equation (5.30):
δy =
∂U
∂Py
= ∂
∂Py
M2
2EI
+ F
2
2EA
+ 3V
2
5GA
 
 
  
 
 
  rdφ0
π / 2
∫ =
M
EI
∂M
∂Py
+ F
EA
∂F
∂Py
+ 6V
5GA
∂V
∂Py
 
 
  
 
 
  rdφ0
π / 2
∫
= r
3
4EI
πPy + 2Px( ) + r20AGE E 18.85Py +12Px( ) + G 15.71Py − 10Px( )[ ]
5.33 Calculate the deflection at the point of load application and in the load
direction for a load applied as shown in sketch z. Assume that E and I are
constant.
Notes: This problem is similar to Problem 5.32, in that the strain energy is fairly
complicated. Table 5.2 on page 200 gives the equations needed to calculate the
strain energy, and Castiglicano’s theorem (Equation (5.30)) gives the deflection.
Solution:
From statics, the horizontal portion has a bending moment Mh=Px and a transverse shear Vh=P.
the vertical portion has a bending moment Mv=Pl/4 and an axial force Pv=P. Therefore, using the
proper equations in Table 5.2 on page 200, the strain energy is:
U = Mh
2
2EI
dx
0
l / 4
∫ +
3Vh
2
5GA
dx + Pv
2
2EA
dx
0
l
∫ +0
l /4
∫
Mv
2
2 EI
dx
0
l
∫ =
P2x2
2EI
dx
0
l /4
∫ +
3P 2
5GA
dx + P
2
2EA
dx
0
l
∫ +0
l /4
∫
P2 l2
8EI
dx
0
l
∫
= P
2
2EI
1
3
x3
 
  
 
  0
l /4
+ 3P
2
5GA
x( )0
l /4 + P
2
2EA
x( )0
l + P
2 l2
8EI
x( )0
l = 13P
2 l3
384EI
+ 3P
2l
20GA
+ P
2 l
2EA
Therefore, from Equation (5.30),
δ = ∂U
∂P
= ∂
∂P
13P2 l3
384EI
+ 3P
2l
20GA
+ P
2 l
2EA
 
 
  
 
 
  =
13Pl3
192EI
+ 3Pl
10GA
+ Pl
EA
5.34 Using Castigliano’s theorem, determine the horizontal and vertical deflections at point A
of sketch aa. Assume that E and I are constant.
www.elsolucionario.net
Page 5-35
Notes: The sketch shows two links, not one structure. Once the loading on each link is determined
from statics, it can be shown that the problem reduces to that of Problem (5.32). This solution
uses the results from Problem (5.32) to obtain a solution.
Solution:
The reactions have been drawn on the sketch in red. Note that the force P acting at A has its line
of action through the left support. Therefore, taking moment equilibrium about point 1:
ΣM=0
From vertical force equilibrium,
ΣFy=0=R1y-Pcos45°; R1y=Pcos45°
From horizontal force equilibrium,
ΣFx=0=R1x-Psin45°; R1x=Psin45°
The right link has no load; obviously, there is no strain energy in the right link. Therefore, one
only needs to consider the left link. Note that this then reduces to Problem 5.32, where φ=-45°,
M0=0, and P=-P. Therefore, the deflections are, using the results from Equation (5.32),
δx =
r3
EI
0.5Py +0.356Px( ) + r20AGE E 12Py +18.85Px( ) − G 10Py −15.71Px( )[ ]
The vertical deflection is similarly calculated from Equation (5.30):
δy =
r3
4EI
πPy + 2Px( ) + r20AGE E 18.85Py +12Px( ) + G 15.71Py − 10Px( )[ ]
5.35 For the structure shown in sketch bb, find the force in each member and determine the
deflection at point A. Assume that E and A are the same in each member.
Notes: The forces in each member are obtained from statics, but in order to obtain the vertical
deflection at point A, include a load Q in the analysis as shown. The force in each member will be
obtained by substituting Q=0, but this allows application of Castigliano’s theorem to obtain the
vertical as well as horizontal deflection at A. Each membercontains only an axial load, so
Equation (5.18) gives the strain energy.
Solution:
I. Geometry Relations
The reactions have been added to the sketch in red. Note also the nomenclature for angles and
lengths which have been added to the sketch. From geometry, note that β+2θ=π/2. This can be
rewritten as
β+θ=π/2-θ, or cos(β+θ)=cos(π/2-θ)=sinθ
Also,
cosβ=cos(π/2-2θ)=sin(2θ)=2sinθcosθ
www.elsolucionario.net
Page 5-36
and from the law of sines,
a
sin θ
= b
sin
π
2
− β 
  
 
  
;b = 2acosθ
II. Determination of Reactions
The reactions are determined from statics. From moment equilibrium at point C,
M = 0∑ = 2Ba cosβ + P + Q( )acos β − Pbcos θ = 0
= B 2a sin2θ( )+ P +Q( )a sin2θ − P 2a cos2 θ( ) = 0; B = P2 cotθ −
P
2
− Q
2
From horizontal force equilibrium, Cx=-P. From vertical force equilibrium,
Fy = 0∑ = Cy − B − P − Q; Cy = B + P +Q = P2 cotθ −
P
2
− Q
2
+ P+ Q = P
2
cotθ + P
2
+ Q
2
III. Determination of Member Tensions
Taking a section at joint B gives the following free body diagram:
From horizontal force equilibrium,
Fx = 0 = FAB cos β +θ( ) − FDB cos β ;∑
FAB = FDB
cos β
cos β +θ( ) = FDB
2 s i nθ cos θ
sinθ
= 2FDB cosθ
From vertical force equilibrium,
Fy∑ = 0 = −B − FDB sinβ + FAB sin θ + β( )
= −B − FDB sinβ + 2 FDB cosθ sin θ + β( )
These yield
FDB =
PH
2
cot θ −
PV
2
−
Q
2
FAB=cosθ[(PHcotθ-PV)-Q]
Taking a section at Joint D gives:
From horizontal force equilibrium,
ΣFx=0=FCD-FDB; FCD=FDB
From vertical force equilibrium,
ΣFy=0=FCDsinβ+FDBsinβ-P-FAD
These yield
FCD = FDB =
PH
2
cotθ −
PV
2
−
Q
2
FAD = 2 FDB sin β − PV = cos2θ PH cotθ − PV − Q[ ] − PV
Taking a section at Joint A gives:
From horizontal force equilibrium,
FACsinθ+FABsinθ-PH=0
Therefore,
FAC =
PH
sinθ
− cosθ PH cotθ − PV − Q[ ]
The force in each member is obtained by substituting Q=0.
IV. Determination of Deflections
www.elsolucionario.net
Page 5-37
The total strain energy is obtained by taking the sum of the individual energies. Since each
member experiences only an axial load, Equation (5.17) is applied to each member. The total
strain energy is
U = FAB
2
2AE
dx0
b∫ +
FAC
2
2AE
dx0
b∫ + FAD
2
2AE
dx0
a∫ + FBD
2
2AE
dx0
a∫ +
FCD
2
2 AE
dx0
a∫
Applying Equation (5.30), the horizontal deflection is:
δx =
∂U
∂P
= FAB
AE
∂FAB
∂P
dx
0
b∫ +
FAC
AE
∂FAC
∂P
dx
0
b∫ + FAD
AE
∂FAD
∂P
dx
0
a∫ + FBD
AE
∂FBD
∂P
dx
0
a∫ +
FCD
AE
∂FCD
∂P
dx
0
a∫
Substituting the following:
∂FAB
∂PH
= cosθ cotθ
∂FAC
∂PH
=
1
sinθ
− cosθ cotθ =
1
sinθ
−
cos2 θ
sinθ
=
1− cos2θ
sinθ
= sinθ
∂FAD
∂PH
= cos2θ cotθ
∂FBD
∂PH
=
cotθ
2
∂FCD
∂PH
=
cotθ
2
δx is then given by
δx =
∂U
∂PH
= P
EA
b( )cos θ cotθ −1( ) cosθ cot θ + b sinθ( ) 1
sinθ
− cosθ cotθ − 1( ) 
  
 
  
 
 
 
+ac o s 2θ cot θ c o s 2θ cotθ − 1( ) −1( ) + 2a cotθ − 1
2
 
  
 
  
cot θ
2
 
  
 
  
 
 
 
or, simplifying,
δ x =
P
EA
b cotθ −1( )cos2θ cotθ +1 − cotθ − 1( ) cosθ sinθ( )[
+a cotθ − 1( )cos22θ cotθ − cos2θ cotθ + cotθ − 1( ) cotθ
2
 
  
 
  
 
 
 
The vertical deflection at point A is:
δy =
∂U
∂Q
= FAB
AE
∂FAB
∂Q
dx
0
b∫ +
FAC
AE
∂FAC
∂Q
dx
0
b∫ + FAD
AE
∂FAD
∂Q
dx
0
a∫ + FBD
AE
∂FBD
∂Q
dx
0
a∫ +
FCD
AE
∂FCD
∂Q
dx
0
a∫
Substituting:
∂FAB
∂PV
= −1
∂FAC
∂PV
= cosθ
∂FAD
∂PV
= − cos2θ
∂FBD
∂PV
= −
1
2
∂FCD
∂PV
= −
1
2
www.elsolucionario.net
Page 5-38
gives an expression for δy as:
δy =
∂U
∂PV
= bP
AE
cosθ cotθ −1( ) −1( ) + 1
sinθ
− cosθ cotθ − 1( )  
 
 
 
 
cosθ
 
 
 
 
 
 
+ aP
AE
cos2θ cotθ − 1{ } −1( ) 1− c o s 2θ( ) + cotθ −1
2
 
  
 
  
− 1
2
 
  
 
  
+ cotθ − 1
2
 
  
 
  
− 1
2
 
  
 
  
 
 
 
 
 
 
Simplifying,
δ y =
∂U
∂PV
=
bP
AE
1
sinθ
+ cot θ sinθ − cosθ( ) 1 − cosθ( ) 
  
 
  
+ aP
AE
1
2
1− cotθ( ) + cos2θ −2 + c o s 2θ + cot2θ 1− cos2θ( ){ }   
 
  
www.elsolucionario.net
Page 6-1
Chapter 6 • Failure Prediction for Static Loading
6.1 Given that the stress concentration factor is 3.12 for a machine element made of steel
with a modulus of elasticity of 207GPa, find the stress concentration factor for an
identical machine element made of aluminum instead of steel. The modulus of elasticity
for aluminum is 69GPa.
Solution:
Since the elastic stress concentration is entirely determined from the geometry of the machine
element, the stress concentration factor will remain the same. Therefore, Kc=3.12.
6.2 A flat part with constant thickness b is loaded in tension as shown in Fig. 6.3(a). The
height changes from 50 to 100mm with a radius r=10mm. Find how much higher a load
can be transmitted through the bar if the height increases from 50 to 70 mm and the
radius decreases from 10 to 3mm.
Notes: To answer this question, one must compare the stress concentration factors for the two
cases. The stress concentration factors are obtained from Figure 6.3(a) on page 233.
Solution:
Referring to the sketch in Figure 6.3(a), H=100mm, h=50mm and r=10mm for the first case.
Therefore, H/h=100mm/50mm=2 and r/h=10mm/50mm=0.2. From Figure 6.3(a), Kc for this case
is 1.8. For the second case, H/h=100/70=1.43 and r/h=3mm/70mm=0.043. Therefore, From
Figure 6.3(a), Kc is around 2.8.
The load that can be transmitted depends on the maximum stress. Therefore, for the first
case:
σmax = Kc1
P1
bh1
; P1 =
bh1σmax
Kc1
= b 50mm( )σ max
1.8
For the second case:
σmax = Kc 2
P2
bh2
; P2 =
bh2σmax
Kc2
=
b 70mm( )σ max
2.8
The ratio of the two forces is:
P1
P2
=
b 50mm( )σ max
1.8
 
  
 
  
b 70mm( )σ max
2.8
 
  
 
  
= 50mm( )2.8
70mm( )1.8 = 1.111
Therefore the load carrying capacity decreases even though the thickness of the bar increased
from 50 to 70 mm, mainly because the radius of the notch decreased.
6.3 A flat steel plate is axially loaded as shown in sketch a has two holes for electric cables.
The holes are situated beside each other and each has a diameter d. To make it possible to
draw more cables, the holes are placed with one hole having twice the diameter 2d, as
shown in sketch b. Assume that the ratio of diameter to width is d/b=0.2 for the two-hole
plate. Which plate will fail first?
www.elsolucionario.net
Page 6-2
Notes: The exact hole locations have not been specified, so some variation may occur with
assumed dimensions. Also, the two-hole case does not correspond to a particular chart in Figure
6.1 through 6.6. However, it is possible to obtain a reasonable solution from the existing data. For
a critical application, more advanced approaches, such as finite element analysis, would be
necessary. See also Problem 6.6.
Solution:
It will be assumed that the top and bottom halves of the plate are symmetric about the centerline,
and that the holes are placed in the center of each half of the plate. The locations where the largest
stress could occur are A and B in the sketch.
Considering the top half of the problem leads to:
Here the diameter to width ratio is 0.4, so Kc is
around 2.2 from Figure 6.2(a). Note that this is
true for point A but not point B in the figure
above.
Because of St. Venant’s Principal, we must be concerned about the stress concentrations
interacting between the two holes. For B, take a section through the hole diameters to yield:
Here H=b/2, h=b/2-d,
r/h=(d/2)/(b/2-d)=0.1b/(0.5b-0.2b)=0.33.
Therefore, Kc is just under 2.0 from Figure
6.4(a).
Therefore, the larger stress concentration is Kc=2.2 and point A is more important than point B.
For the single hole, Figure 6.2 (a) gives a stress concentration of Kc=2.2 (d/b=0.4). Therefore,
either design is expected to fail at the same stress.
6.4 A load-carrying beam is loaded with a bending moment M. The minor beam height is h
and the width in the perpendicular direction is b. This beam, which carries a balcony on
the wall of a house, is welded together with a beam in the house structure. That beam has
a major height H. Because of manufacturing problems the radius r of the connections
cannot be made largerthan rmax but can be made smaller, down to zero. Find how the
load-carrying beam should be placed on the house beam to give maximum strength.
Notes: Design recommendations arise from studying Figure 6.3 (b).
Solution:
From Figure 6.3 (b), the stick-out H/h should be small to give a low stress concentration factor.
Also, the radius to height ratio should be as large as possible. The optimum configuration is to
have the small beam flush with the top side of the large beam which gives zero stress
concentration.
www.elsolucionario.net
Page 6-3
6.5 A round bar has a fillet with r/d=0.15 and D/d=1.5. The bar transmits both bending
moment and torque. A new construction is considered to make the shaft stiffer and
stronger by making it equally thick on each side of the fillet or groove. Determine
whether that is a good idea.
Notes: Figures 6.5 and 6.6 are used to obtain the solution.
Solution:
For r/d=0.15 and D/d=1.5, the stress concentrations for bending is just over 1.5 for bending (from
Figure 6.5 (b)) and about 1.25 for torsion (from Figure 6.5(c)). If instead of a fillet the bar became
a groove, with the same root diameter, then the stress concentrations are obtained from Figure 6.6
(b) and (c). The new stress concentration factors are around 1.65 for bending and 1.325 for
torsion. Since the stress concentration factors are higher for the proposed redesign, it is not a good
idea.
6.6 A rectangular plate has a central hole. The width of the plate is b=50mm and the diameter
of the hole is d=10mm. The plate is axially loaded with a force P=1500N, and the minor
height of the plate is h=5mm. Is it possible to find the stress concentration factor for the
plate without using Figure 6.2?
Notes: Figure 6.2 can obviously be used to solve the problem, and serves as a useful comparison
for the alternate technique. The problem can be approximated by Figure 6.4 if the plate is
sectioned. This solution compares the results of both approaches.
Solution:
First of all, if Figure 6.2(a) is used with d/b=(10mm)/(50mm)=0.2, Kc=2.5. An alternative
approach is to cut the beam in half across the center of the hole, then reassembling the two
sections so the holes are on the exterior, thereby approximating the loading case in Figure 6.4(a).
Note that there are differences; there are not two matching fillets on opposite sides, and the
loading is complicated by the presence of a Poisson effect in the original geometry. However, the
two should be fairly close. For the new, sectioned geometry, r=d/2=5mm, H=b=50mm, h=(b-
d)=(50mm-10mm)=40mm. Therefore, H/h=(50mm)/(40mm)=1.25, and
r/h=(5mm)/(40mm)=0.125. From Figure 6.4(a), Kc is around 2.3. The error between the two cases
is:
%error =
2.5 − 2.3
2.5
×100% = 8%
6.7 A machine has three circular shafts, each with fillets giving stress concentrations. The
ratio of fillet radius to shaft diameter is 0.1 for all three shafts. One of the shafts transmits
a tensile force, one transmits a bending torque, and one transmits torsion. Because they
are stressed exactly to the stress limit (ns=1), a design change is proposed doubling the
notch radii to get a safety factor greater than 1. How large will the safety factors be for
the three shafts if the diameter ratio is 2 (D/d=2)?
Notes: Figure 6.5 is used to solve this problem.
www.elsolucionario.net
Page 6-4
Solution:
For the shaft under tension, Figure 6.5(a) is used to obtain the stress concentration
factors. For D/d=2 and r/d=0.1, Kc=2.0, and corresponds to the original case. If r is doubled, then
r/d=0.2 and Kc=1.65. Since the original design was fully stressed, the new safety factor is
ns =
2
1.65
= 1.21
For the shaft under bending, Figure 6.5(b) is used to obtain the stress concentration
factors. For D/d=2 and r/d=0.1, Kc=1.7, and corresponds to the original case. If r is doubled, then
r/d=0.2 and Kc=1.43. Since the original design was fully stressed, the new safety factor is
ns =
1.7
1.43
= 1.19
For the shaft under torsion, Figure 6.5(c) is used to obtain the stress concentration factors.
For D/d=2 and r/d=0.1, Kc=1.43, and corresponds to the original case. If r is doubled, then
r/d=0.2 and Kc=1.22. Since the original design was fully stressed, the new safety factor is
ns =
1.43
1.33
= 1.17
Therefore, the lowest safety factor is ns=1.17 and corresponds to the torsion-loaded shaft.
6.8 The shaft shown in sketch c is subjected to tensile, torsional, and bending loads.
Determine the principal stresses at the location of stress concentration.
Notes: This problem can be easily solved through the principal of superposition. The stress
concentration factors are obtained from Figures 6.5 (a) and (b).
Solution:
The rod will see normal stresses due to axial loads and bending, and a shear stress due to torsion.
Note that the shear stress due to shear is zero at the extreme fibers where the stresses are largest.
The critical location is at the bottom where the bending and axial stresses are both tensile. Assign
the x-axis to the rod axis. The normal stress is given by:
σx = Kc1
P
A
+ Kc2
Mc
I
= 2.0( ) 10kNπ
4
0.03m( )2
+ 1.75( ) 50kN( ) 0.130m( ) 0.015m( )π
64
0.030m( )4
= 4315MPa
where the stress concentration factors of 2.0 and 1.75 are obtained from Figure 6.5 (a) and (b).
The shear stress is
τ = Kc
Tc
J
= 1.45( ) 100Nm( ) 0.015m( )π
32
0.030m( )4
= 27.4MPa
Equation (2.16) gives the principal stresses.
σ1,σ2 =
σx + σ y
2
± τxy2 +
σx − σy
2
 
 
  
 
 
  
2
=
4315MPa
2
± 27.4MPa( )2 + 4315MPa
2
 
  
 
  
2
www.elsolucionario.net
Page 6-5
or σ1=4315MPa and σ2=-0.17MPa. Note that the shear stress is very small compared to the
normal stress; we could have taken σx as a principal direction.
6.9 A steel plate with dimensions shown in sketch d is subjected to a 150kN tensile force and
300Nm bending moment. The plate is made of AISI 1080 steel and is kept at 20°C. A
hole is to be punched in the center of the plate. What is the maximum diameter of the
hole for a safety factor of 1.5?
Notes: Equation (3.16) gives the allowable stress in bending. The normal stress is the sum of the
bending stress and the axial normal stress, and is equated to the allowable stress. This gives an
equation in terms of the hole diameter and the stress concentration factors in tension and bending
which can be solved iteratively.
Solution:
From the inside front cover, AISI steel has a yield strength of Sy=380MPa. Therefore, the
allowable stress is given by Equation (3.16) as:
σall=0.6Sy=0.6(380MPa)=228MPa
However, since the safety factor is 1.5, the allowable stress for this problem is 228MPa/1.5=
152MPa. The stress associated with axial tension is (see Figure 6.2(a)):
σa =
Kca P
b − d( )h
=
Kca 150kN( )
0.235m − d( ) 0.025m( )
=
6MN / m
0.235m − d
Kca
The stress associated with bending is (see Figure 6.2(b)):
σb =
6KcbM
b − d( )h2
=
6Kcb 300Nm( )
0.235m − d( ) 0.025m( )2
=
2.88MN / m
0.235m −d
Kcb
Therefore, the maximum stress is:
σmax = σa + σb =
6MN / m( )Kca + 2.88MN /m( )Kcb
0.235m − d
This should be equated to the maximum allowable stress, of σmax=152MPa. Note that Kca and Kcb
are functions of only d, since the other variables needed for their definition are fixed. Therefore,
this equation can be iteratively solved. Note that we can re-write the equation as:
6MN / m( )Kca + 2.88MN / m( )Kcb
0.235m − d
= 152MPa; d = 0.235m −
6MN / m( )Kca + 2.88MN / m( )Kcb
152MPa
Assume d=100mm, so that d/b=(100mm)/(235mm)=0.426, and d/h=100/25=4. From
Figure 6.2 (a), Kca=2.2 and From Figure 6.2 (b), Kcb=1.46. Therefore, this equation would predict
d = 0.235m −
6MN / m( ) 2.2( ) + 2.88MN /m( ) 1.46( )
152MPa
= 0.120m
Therefore, the initial value was too small. If we now use d=0.120m=120mm, then
d/h=120/25=4.8 and d/b=120/235=0.511. This gives Kca=2.16 and Kcb=1.4. Therefore,
d = 0.235m −
6MN / m( ) 2.16( ) + 2.88MN / m( ) 1.4( )
152MPa
= 0.123m
This is close to the assumed value, and closer agreement between assumed and calculated values
is difficult because of the resolution of Figures6.2(a) and 6.2(b).
www.elsolucionario.net
Page 6-6
6.10 A Plexiglass plate with dimensions 1m x 1m x 1cm is loaded by a nominal tensile stress
of 55MPa in one direction. The plate contains a small crack perpendicular to the load
direction. At this stress level a safety factor of 1.3 against crack propagation is obtained.
Find how much larger the crack can get before it grows catastrophically.
Notes: Equation (6.4) is used to solve this problem.
Solution:
For the first case, Equation (6.4) gives:
Kci
1.3
= Yσnom πa1 ; Kci = 1.3Yσnom πa1
For the second case, the safety factor would be unity so that:
Kci = Yσnom πa2
Substituting for Kci:
Yσnom πa2 = 1.3Yσnom πa1 ; a2 = 1.32a1 = 1.69a1
Therefore, the crack can be 69% larger before catastrophic failure occurs.
6.11 A pressure container is made of AISI 4340 steel. The wall thickness is such that the
tensile stress in the material is 1100MPa. The dimensionless geometry correction factor
Y=1 for the given geometry. Find how big the largest crack can be without failure if the
steel is tempered
a) at 260°C
b) at 425°C
Notes: The material properties as a function of temper temperature is obtained from Table 6.1.
Equation 6.4 is used to solve the problem.
Solution:
The nominal stress is given as σnom=1100MPa, and Y=1. The critical crack length is derived from
Eq. (6.4):
Kci = Yσnom πa;a =
1
π
Kci
Yσnom
 
 
  
 
 
  
2
From Table 6.1, at a temper of 260°C, the fracture toughness is 50MPam1/2. Therefore, the
critical crack length is 1.3 mm. At a temper of 425°C, The critical crack length is 4.0mm. (Note:
a is one-half the crack length.)
6.12 Two tensile test rods are made of AISI 4340 steel tempered at 260°C and aluminum alloy
2024-T351. The dimensionless geometry correction factor Y=1. Find how high a stress
each rod can sustain if there is a 2-mm crack half length in each of them.
Notes: The fracture toughness for these materials is obtained from Table 6.1 on page 232. The
nominal stress that can be sustained is then given by Equation (6.4).
Solution:
www.elsolucionario.net
Page 6-7
From Table 6.1 on page 232, Kci for AISI 4340 is 50.0MPam
1/2
. For Al 2024-T351, Kci is
36MPam
1/2
. Therefore, the stress in the steel is given by Equation (6.4) as:
Kci = Yσnom πa;σnom =
Kci
Y πa
=
50.0MPa
1( ) π 0.002( )
= 631MPa
For the Al 2024-T351,
Kci = Yσnom πa;σnom =
Kci
Y πa
=
36.0 MPa
1( ) π 0.002( )
= 454MPa
6.13 A plate made of titanium alloy Ti-6Al-4V has the dimensionless correction factor Y=1.
How large can the largest crack in the material be if it still should be possible to
plastically deform the plate in tension?
Notes: To plastically deform the plate, the nominal stress must exceed the yield strength of the
material. Therefore, Equation (6.4) solves the problem.
Solution:
From Table 6.1 on page 232, the fracture toughness for Ti-6Al-4V varies from 44-66 MPam
1/2
.
Also, the yield strength is 910MPa. To plastically deform the material, σnom>Sy, or, from Equation
(6.4),
Kci = Yσnom πa; σnom =
Kci
Y πa
≥ Sy
Solving for a,
Kci
Y πa
≥ Sy ; a =
1
π
Kci
YSy
 
 
 
 
 
 
 
 
2
=
1
π
Kci
1( ) 910 MPa( )
 
 
  
 
 
  
2
Since Kci has a value between 44-66MPam
1/2
, then a has a range of a≤0.744-1.67mm. Therefore,
if the largest crack is below 0.744mm in half-length (1.488mm in length), then the nominal stress
will be larger than the yield strength.
6.14 A plexiglass model of a gear has a 1-mm crack half-length formed in its fillet curve
(where the tensile stress is maximum). The model is loaded until the crack starts to
propagate. Y=1.5. How much higher a load can a gear made of AISI 4340 steel tempered
to 425°C carry with the same crack and the same geometry?
Notes: Equation (6.4) is used to solve the problem with data from Table 6.1. This solution makes
sure that the steel does not plastically deform before catastrophic crack propagation occurs.
Solution:
From Table 6.1 on page 232, the fracture toughnesses for plexiglass and steel are 1.0 and 87.4
MPam
1/2
, respectively. For a gear, the bending stress is directly proportional to the applied load,
so for a constant crack size a and correction factor Y the load possible is directly proportional to
Kci. Therefore, Kci for steel is 87.4 times larger than for plexiglass, so it may be possible to
support a load 87.4 times larger.
However, it is possible that the steel will plastically deform at a lower stress than that
needed to propogate the crack. With a=1mm=0.001m, Equation (6.4) predicts the nominal stress
level as:
www.elsolucionario.net
Page 6-8
Kci = Yσnom πa; σnom =
Kci
Y πa
=
87.4MPa m
1.5( ) π 0.001m( )
=1040MPa
Table 6.1 gives the yield strength for AISI 4340 tempered to 425°C as 1420MPa. Therefore, the
nominal stress level needed for crack propagation is not sufficient to cause plastic deformation,
and a load 87.4 times larger than for plexiglass can be carried.
6.15 A pressure container made of aluminum alloy 2024-T351 is manufactured for a safety
factor of 2 guarding against yielding. The material contains cracks through the wall
thickness with a crack half-length less than 3mm. Y=1. Find the safety factor when
considering crack propagation.
Notes: The safety factor guarding against crack propagation is obtained from the ratio of the
fracture toughness of the material to the stress intensity factor calculated by Equation (6.4).
Solution:
From Table 6.1 on page 232, Sy=325MPa and Kci=36MPa-m
1/2
. The safety factor guarding against
yielding is 2, therefore the nominal stress is one-half the yield strength, or σnom=162.5MPa. The
stress intensity factor is therefore calculated from Equation (6.4) as
Ki = Yσnom πa = 1( ) 162.5MPa( ) π 0.003m( ) =15.78MPa m
The safety factor against crack propagation is therefore
ns =
Kci
Ki
= 36MPa m
15.78MPa m
= 2.28
Since the safety factor for yielding is lower than the safety factor guarding against crack
propagation, the safety factor for the pressure vessel is still 2.
6.16 The clamping screws holding the top lid of a nuclear reactor are made of AISI 4340 steel
tempered at 260°C. They are stressed to a maximum level of 1250MPa during a
pressurization test before starting the reactor. Find the safety factor guarding against
yielding and the safety factor guarding against crack propagation if the initial cracks in
the material have Y=1 and a=1mm. Also, do the calculations for the same material but
tempered to 425°C.
Notes: This problem is similar to the previous problem. Equation (6.4) is used to solve this
problem.
Solution:
I. AISI 4340 Tempered at 260°C
From Table 6.1 on page 232, Sy=1640MPa and Kci=50MPa-m
1/2
. The safety factor against
yielding is therefore
ns =
Sy
σ
=
1640MPa
1250MPa
= 1.31
From Equation (6.4), the stress intensity factor is
Ki = Yσ nom πa = 1( ) 1250MPa( ) π 0.001m( ) = 70.06 MPa m
The safety factor guarding against crack propagation is therefore
www.elsolucionario.net
Page 6-9
ns =
Kci
Ki
= 50MPa m
70.06MPa m
= 0.714
Since the safety factor is less than 1, the bolts will fail.
II. AISI 4340 Tempered at 435°C
From Table 6.1 on page 232, Sy=1420MPa and Kci=87.4MPa-m
1/2
. Using the same equations, the
safety factor against yielding is ns=1.14, and the safety factor against crack propagation is
ns=1.25. Therefore, the bolts will not crack.
6.17 A glass tube used in a pressure vessel is made of aluminum oxide (sapphire) to make it
possible to apply 30MPa pressure and still have a safety factor of 2 guarding against
fracture. For a soda-lime glass of the same geometry only 7.5 MPa pressure can be
allowed if a safety factor of 2 is to be maintained. Find the size of the cracks the glass
tube can tolerate at 7.5MPa pressure and a safety factor of 2. Y=1 for both cases.
Notes: Material properties are obtained from Table A.3. As is shown in Chapter 9, the stress is
proportional to the pressure. Equation (6.4) is used to solve this problem.
Solution:
From Table A.3 on page 901, the fracture strength of soda-limeglass is 69MPa. The stresses in
the tube are directly proportional to the pressure, so the fracture strength of the aluminum oxide
tube is:
S fa
pa
=
Sfs
ps
;S fa =
S fs pa
ps
= 69MPa( ) 30MPa( )
7.5MPa
= 276MPa
Note that this is on the low end of the fracture strength values given in Table A.3 for aluminum
oxide (Al2O3).
For a safety factor of 2, the applied stress is 276/2=138MPa. From Table 6.1 on page
232, using a low value of fracture toughness for Al2O3, use Kcia=3.0MPa-m
1/2
. From Equation
(6.4),
Ki =
Kci
2
= Yσnom πa ; a =
1
π
Kci
2Yσnom
 
 
  
 
 
  
2
= 1
π
3.0MPa m
2 1( ) 138MPa( )
 
 
  
 
 
  
2
= 3.76 ×10−7 m = 37.6µm
For the soda lime glass, the lowest value of fracture toughness is Kci=0.7MPa-m
1/2
. The applied
stress is σnom=69MPa/2=34.5MPa. The stress intensity factor is
Ki =
Kci
2
= Yσnom πa ; a =
1
π
Kci
2Yσnom
 
 
  
 
 
  
2
= 1
π
0.7MPa m
2 1( ) 34.5MPa( )
 
 
  
 
 
  
2
= 3.28×10−7m = 32.8µm
Therefore, the largest crack in the aluminum oxide must be less than 2a or 75.2µm, while for
soda lime glass the largest crack must be smaller than 65.6µm.
6.18 A stress optic model used for demonstrating the stress concentrations at the ends of a
crack is made of polymethylmethacrylate. An artificially made crack 100mm long is
perpendicular to the loading direction. Y=1. Calculate the highest tensile stress that can be
applied to the model without propagating the crack.
Notes: Material properties are obtained from Table 6.1 and Table A-4. Equation (6.4) is used to
solve this problem.
www.elsolucionario.net
Page 6-10
Solution:
From Table 6.1 on page 232, the critical stress intensity factor for polymethylmethacrylate is
Kci=1.0MPa-m
1/2
. From Equation (6.4) the stress when the crack propagates catastrophically is:
Kci = Yσnom πa; σ nom =
Kci
Y πa
=
1.0MPa m
1( ) π 0.1m( )
= 1.78MPa
From Table A-4 on page 902, the ultimate strength is between 48 and 76 MPa. Therefore, the
crack will propagate at a stress far lower than the ultimate strength of the material.
6.19 A passengerless airplane requires wings that are lightweight and the prevention of cracks
more than 2mm long. The dimensionless geometry correction factor Y is usually 1.5 for a
safety factor of 2. What is the appropriate alloy for this application? If Y is increased to
4.5, what kind of alloy from Table 6.1 should be used?
Notes: Equation (6.4), combined with material properties from Table 6.1, allow solution of this
problem.
Solution:
To maintain a safety factor of 2, set σnom=Sy/2. Therefore, Equation (6.4) gives:
Kci = Yσnom πa = 1.5( )
Sy
2
π 0.002m( ) = 0.0594 m( )Sy
From Table 6.1 on page 232, the following data is obtained:
Material Kci/Sy (m
1/2
)
Aluminum alloy, 2020-T351
Aluminum alloy, 7075-T651
Alloy steel 4340, tempered at 260°C
Alloy steel 4340, tempered at 425°C
Titanium alloy Ti-6Al-4V
0.111
0.0574
0.0305
0.0615
0.0483
Therefore, either Aluminum alloy 2020-T351 or Alloy steel 4340 tempered at 425°C would work.
For weight savings, the aluminum alloy is probably the better choice.
If Y=4.5, Equation (6.4) gives:
Kci = Yσnom πa = 4.5( )
Sy
2
π 0.002m( ) = 0.178 m( )Sy
None of the materials in Table 6.1 would be acceptable for this case.
6.20 The anchoring of the cables carrying a suspension bridge are made of cylindrical AISI
1080 steel bars 210mm in diameter. The force transmitted from the cable to the steel bar
is 3.5 MN. Calculate the safety factor range guarding against yielding.
Notes: The material property is obtained from the inside front cover. Equation (3.13) gives a
range for allowable stresses in tension.
Solution:
From the inside front cover, the yield strength for AISI 1080 is Sy=380MPa. The stress in the steel
bar is
www.elsolucionario.net
Page 6-11
σ =
P
A
=
3.5MN
π
4
0.21m( )2
= 101MPa
Since the safety factor is ns=σall/σ, Equation (3.13) gives
0.45Sy ≤ σall ≤ 0.60Sy ; 
0.45 Sy
σ
≤ ns ≤
0.60Sy
σ
; 
0.45 380MPa( )
101MPa
≤ ns ≤
0.60 380MPa( )
101MPa
Therefore the safety factor is in the range of 1.69≤ns≤2.25
6.21 The arm of a crane has two steel plates connected with a rivet that transfers the force in
pure shear. The rivet is made of AISI 1040 steel and has a circular cross section with a
diameter of 25mm. The load on the rivet is 10kN. Calculate the safety factor.
Notes: Equation (3.14) gives the allowable stress in shear.
Solution:
The yield strength of AISI 1040 steel is obtained from the inside front cover as Sy=350MPa. From
Equation (3.14),
tall=0.4Sy=0.4(350MPa)=140MPa
The shear stress on the rivet is
τ =
P
A
=
10kN
π
4
0.025m( )2 
  
 
  
= 20.37MPa
Therefore, the safety factor is:
ns =
τall
τ
=
140MPa
20.37MPa
= 6.87
6.22 A telescope stands on four feet, each carrying a load of 200kN. The feet are made of AISI
1020 steel, and the floor is a thick plate of AISI 1080 steel on a concrete base. Calculate
how large the diameter of the feet needs to be if they are circular and flat. The safety
factor is 2.5.
Notes: This problem is straightforward. Equation (3.16) gives the allowable stress.
Solution:
From the inside front cover, the yield strength of AISI 1020 steel is Sy=295MPa, and for AISI
1080 the yield strength is 380MPa. Since the floor is stronger than the feet, the floor doesn’t have
to be analyzed - it will have a higher safety factor than the feet.
From Equation (3.16), the allowable stress is σall=0.9Sy=0.9(295MPa)=265.5MPa. Since
the safety factor is 2.5, the design stress is σd=σall/ns=(265.5MPa)/2.5=106.2MPa. Therefore, the
diameter of the feet is obtained from:
σ =
P
A
=
4P
πd2
;d =
4P
πσ
=
4 200kN( )
π 106.2MPa( )
= 0.049m = 49mm
www.elsolucionario.net
Page 6-12
6.23 A machine element is loaded so that the principal normal stresses at the critical location
for a biaxial stress state are σ1=20ksi and σ2=-15ksi. The material is ductile with a yield
strength of 60ksi. Find the safety factor according to
a) The maximum shear stress theory (MSST)
b) The distortion-energy theory (DET)
Notes: Equation (6.6) is used to obtain the safety factor for the MSST. Equation (6.11) gives the
safety factor for the DET after the von Mises stress is calculated from Equation (6.9). If the stress
is biaxial, then one principal stress is zero.
Solution:
First of all, since the stress state is biaxial, then one normal stress is zero. Therefore, the three
principle stresses are properly referred to as σ1=20ksi, σ2=0 and σ3=-15ksi, since σ1≥σ2≥σ3. For
the maximum shear stress theory, Equation (6.6) gives the safety factor as:
σ1 − σ3 =
Sy
ns
; ns =
Sy
σ1 − σ3
=
60ksi
20ksi +15ksi( )
= 1.714
The von Mises stress is obtained from Equation (6.9) as:
σe =
1
2
σ2 − σ1( )2 + σ3 − σ1( )2 + σ3 −σ2( )2[ ]1 / 2= 12 −20ksi( )2 + −35ksi( )2 + 15ksi( )2[ ]
1 / 2
or σe=30.4ksi. Therefore, the safety factor is, from Equation (6.11),
σe =
Sy
ns
; ns =
Sy
σe
=
60ksi
30.4ksi
= 1.97
6.24 A glass bottle is filled with a liquid having a compressibility similar to (~1% lower than)
the compressibility of the glass. To what depth within an ocean can the bottle be lowered
without cracking.
Notes: Equations (6.9) is used to solve the problem.
Solution.
If the pressure in the glass is hydrostatic, then σ1=σ2=σ3=-p, and Equation (6.10) gives
σe =
1
2
σ2 −σ1( )2 + σ3 −σ1( )2 + σ3 −σ2( )2[ ]1 / 2= 12 −p + p( )2 + − p + p( )2 + − p + p( )2[ ]
1 / 2
= 0
Therefore, there is no limit to the depth that the bottle can be lowered (and is also the reason
rocks are not pulverized at the bottom of the ocean or inside the earth’s crust).
6.25 A bolt is tightened, subjecting its shank to a tensile stress of 80ksi and a torsional shear
stress of 50ksi at a critical point. All the other stresses are zero. Find the safety factor tat
the critical point by the DET and the MSST. The material is high-carbon steel (AISI
1080). Will the bolt fail because of the static loading?
Notes: Equations (2.16), (6.6), (6.10), and (6.11) are used to solve this problem.
Solution:www.elsolucionario.net
Page 6-13
From the inside front cover, the yield stress for AISI 1080 steel is 55ksi. Directions are arbitrary;
let’s refer to the tensile stress as σx=80ksi and the shear stress as τxy=50ksi. Since all other
stresses are zero, Equation (2.16) gives the principal stresses as
σ1,σ2 =
σx + σ y
2
± τxy2 +
σx − σy
2
 
 
  
 
 
  
2
=
80ksi
2
± 50ksi( )2 + 80ksi
2
 
  
 
  
2
or σ1=104ksi, σ2=-24ksi. Note that the other stresses are zero, so the principal stress out of the
plane of the normal and shear stresses is zero. Putting the stresses in the proper order (σ1≥σ2≥σ3),
we assign them the values σ1=104ksi, σ2=0ksi, σ3=-24ksi. From Equation (6.6),
σ1 − σ3 =
Sy
ns
; ns =
Sy
σ1 − σ3
=
55ksi
104ksi − −24ksi( )
= 0.43
which is the safety factor for the maximum shear stress theory. Equation (6.9) gives
σe =
1
2
σ1 − σ2( )2 + σ1 − σ3( )2 + σ2 −σ 3( )2[ ]1 / 2
= 1
2
104ksi − 0( )2 + 104ksi + 24ksi( )2 + 0 + 24ksi( )2[ ]1 / 2= 118ksi
From Equation (6.11),
σe =
Sy
ns
; ns =
Sy
σe
=
55ksi
118ksi
= 0.47
Since the safety factor is less than one for both cases, both cases predict failure.
6.26 A torque is applied to a piece of chalk used in a classroom until the chalk cracks. Using
the maximum normal stress theory (MNST) and assuming the tensile strength of the
chalk to be small relative to its compressive strength, determine the angle of the cross
section at which the chalk cracks.
Notes: Given the loading condition, the angle of the largest tensile stress is obtained from
Equation (2.15). Based on the MNST, failure will occur at this angle.
Solution:
For pure torque, the stress state is τxy=τ and σx=σy=0. The angle of the largest tensile stress, φσ, is
given by Equation (2.15) as:
t a n 2φσ =
2τ xy
σ x − σy
=
2τ xy
0
= ∞
Therefore, φσ=45°. The chalk will crack along a 45° angle from its circumference.
6.27 A cantilevered bar 500mm long with square cross section has 25mm sides. Three
perpendicular forces are applied to its free end, a 100-N force is applied in the y and z-
directions, and a force of 1000N is applied in the x-direction (bar axis). Calculate the
equivalent stress at the clamped end of the bar by using the DET when the sides of the
square cross section are parallel with the y and z directions.
Notes: The stresses are largest at the corners, where the total stress is the sum of two bending
stresses and the axial stress. The effective stress is obtained from Equation (6.9).
Solution:
www.elsolucionario.net
Page 6-14
The moment of inertia for the cross section is:
I =
bh3
12
=
a4
12
=
0.025m( )4
12
= 3.255 × 10−8 m4
The cross sectional area is A=a
2
=(0.025m)
2
=6.25x10
-4
m
2
. Since the bar is cantilevered, the
loading is a combined situation of two bending moments and one axial load. The perpendicular
moments are M1=M2=Fl=(100N)(0.5m)=50Nm. The axial load is 1000N. Therefore, the
maximum stress occurs at a corner of the cross section, and is the sum of the stresses due to the
three loads. Therefore,
σx =
M1c
I
+
M2c
I
+
P
A
= 2
50Nm( ) 0.0125m( )
3.255 × 10−8m4
+
1000N
6.25 ×10−4 m2
= 40.00MPa
There is no stress in the y- or z-directions. Also, at the outside edge of the bar, the shear stress is
zero (see page 169). Therefore, σ1=40.00MPa, σ2=σ3=0. From Equation (6.9), the von Mises
stress is
σe =
1
2
σ1 − σ2( )2 + σ1 − σ3( )2 + σ2 −σ 3( )2[ ]1 / 2= 12 40MPa( )2 + 40MPa( )2 + 0( )2[ ]
1 / 2
This is evaluated as σe=40MPa.
6.28 A shaft transmitting torque from the gearbox to a rear axle of a truck is unbalanced, so
that a certrifugal load of 500N acts at the middle of the 3m-long shaft. The AISI 1040
tubular steel shaft has an outer diameter of 70mm and an inner diameter of 58mm.
Simultaneously, the shaft transmits a torque of 6000Nm. Use the DET to determine the
safety factor guarding against yielding.
Notes: The moment must be determined, which then allows for calculation of the bending stress.
The torque results in a shear stress; this combined stress state is then transformed to obtain the
principal stresses. Equations (6.9) and (6.11) are then used to solve the problem.
Solution:
From the inside front cover, the yield strength of AISI 1040 steel is Sy=350MPa. The moment of
inertia for the shaft is:
I =
π
64
do
4 − di4( ) = π64 0.070m( )
4 − 0.058m( )4( ) = 6.23 ×10−7 m4
Similarly, J=1.246x10
-6
m
4
. For a simply supported shaft, the maximum moment occurs at the
center of the shaft and has the value M=Pl/4=(500N)(3m)/4=375Nm. Therefore, the bending
stress is obtained from Equation (4.48) as:
σx =
Mc
I
=
375Nm( ) 0.035m( )
6.23 ×10−7m4
= 21.07MPa
The shear stress due to the torque is given by Equation (4.34) as:
τxy =
Tc
J
=
6000Nm( ) 0.035m( )
1.246 × 10−6m4
=168.5MPa
Also, σy=σz=τzx=τyz=0. Therefore, from Equation (2.16),
σ1,σ2 =
σx + σ y
2
± τxy2 +
σx − σy
2
 
 
  
 
 
  
2
=
21.07MPa
2
± 168.5MPa( )2 + 21.07MPa
2
 
  
 
  
2
Therefore, σ1=179.4MPa, σ2=0, and σ3=-158.3MPa. Note that the principal stresses have been
renumbered so that σ1≥σ2≥σ3. Equation (6.9) gives the effective stress as
www.elsolucionario.net
Page 6-15
σe =
1
2
σ1 − σ2( )2 + σ1 − σ3( )2 + σ2 −σ3( )2[ ]1 / 2
= 1
2
179.4MPa − 0( )2 + 179.4 MPa + 158.3MPa( )2 + 158.3MPa( )2[ ]1 / 2=292.6 MPa
From Equation (6.11),
σe =
Sy
ns
; ns =
Sy
σe
=
350MPa
292.6MPa
= 1.196
6.29 The right-angle-cantilevered bracket used in Problem 5.30, sketch w, has a concentrated
force of 1000N and a torque of 300Nm. Calculate the safety factor. Use the DET and
neglect transverse shear. Assume that the bracket is made of AISI 1040 steel and use the
following values: a=0.5m, b=0.3m, d=0.035m, E=205GPa, and ν=0.3.
Notes: The stresses must be determined using the approach described in Chapter 4. From the
stress state, the principal stresses are determined. Equation (6.6) gives the safety factor for the
Maximum Shear Stress Theory, and Equations (6.9) and (6.11) give the safety factor for the
Distortion-Energy Theory.
Solution:
From the inside front cover, Sy=350MPa for AISI 1040 steel. The moment of inertia of the
bracket cross section is:
I =
π
64
d4 =
π
64
0.035m( )4 = 7.366 × 10−8m4
Similarly, J=1.4732x10
-7
m
4
. The maximum stress for the bracket occurs at the wall (x=a). The
loading is a bending moment and a torque. The moment is due to the applied torque T and the
load P, and is
M=Pa+T=(1000N)(0.5m)+300Nm=800Nm
Therefore, the bending stress is obtained from Equation (4.48) as:
σx =
Mc
I
=
800Nm( ) 0.0175m( )
7.366 ×10−8m4
= 190MPa
At the wall, there is a torque of T=Pb=(1000N)(0.3m)=300Nm. The shear stress due to the torque
is given by Equation (4.34) as:
τxy =
Tc
J
=
300Nm( ) 0.0175m( )
1.4732 × 10−7 m4
= 35.6MPa
Also, σy=σz=τzx=τyz=0. Therefore, from Equation (2.16),
σ1,σ2 =
σx + σ y
2
± τxy2 +
σx − σy
2
 
 
  
 
 
  
2
=
190MPa
2
± 35.6MPa( )2 + 190 MPa
2
 
  
 
  
2
Therefore, σ1=196MPa, σ2=0, and σ3=-6.45MPa. Note that the principal stresses have been
renumbered so that σ1≥σ2≥σ3. From Equation (6.6), the safety factor for MSST is:
σ1 − σ3 =
Sy
ns
; ns =
Sy
σ1 − σ3
=
350MPa
196MPa + 6.45MPa
=1.73
Equation (6.9) gives the effective stress as
www.elsolucionario.net
Page 6-16
σe =
1
2
σ1 − σ2( )2 + σ1 − σ3( )2 + σ2 −σ 3( )2[ ]1 / 2
= 1
2
196MPa − 0( )2 + 196MPa + 6.45MPa( )2 + 6.45MPa( )2[ ]1/ 2 = 199MPa
From Equation (6.11), the safety factor for DET is:
σe =
Sy
ns
; ns =
Sy
σe
=
350MPa
199MPa
=1.76
6.30 A 10-cm diameter shaft is subjected to a 10kNm steady bending moment, an 8kN-m
steady torque, and a 150kN axial force. The yield strength of the shaft material is
600MPa. Use the MSST and the DET to determine the safety factors for the various types
of loading.
Notes: This is similar to problems 6.28 and 6.29, but now a stress due to the axial force must be
included. From the stress state, the principal stresses are determined. Equation (6.6) gives the
safety factor for the Maximum Shear Stress Theory,and Equations (6.9) and (6.11) give the
safety factor for the Distortion-Energy Theory.
Solution:
The moment of inertia of the shaft cross section is:
I =
π
64
d4 =
π
64
0.10m( )4 = 4.909 × 10−6 m4
Similarly, J=9.817x10
-6
m
4
. The area of the cross section is πd2/4=0.00785m2. The bending stress
is obtained from Equation (4.48) as:
σx =
Mc
I
=
10kNm( ) 0.05m( )
4.909 × 10−6 m4
=101.8MPa
The normal stress due to the axial load is
σx =
P
A
=
150kN
0.00785m2
= 19.10MPa
Therefore, the maximum normal stress is σx=101.8MPa+19.10MPa=120.9MPa. The shear stress
due to the torque is given by Equation (4.34) as:
τxy =
Tc
J
=
8kNm( ) 0.05m( )
9.817 ×10−6m4
= 40.7MPa
Also, σy=σz=τzx=τyz=0. Therefore, from Equation (2.16),
σ1,σ2 =
σx + σ y
2
± τxy2 +
σx − σy
2
 
 
  
 
 
  
2
=
120.9MPa
2
± 40.7MPa( )2 + 120.9 MPa
2
 
  
 
  
2
Therefore, σ1=133.3MPa, σ2=0, and σ3=-12.42MPa. Note that the principal stresses have been
renumbered so that σ1≥σ2≥σ3. From Equation (6.6), the safety factor for MSST is:
σ1 − σ3 =
Sy
ns
; ns =
Sy
σ1 − σ3
=
600MPa
133.3MPa + 12.42MPa
= 4.12
Equation (6.9) gives the effective stress as
σe =
1
2
σ1 − σ2( )2 + σ1 − σ3( )2 + σ2 −σ 3( )2[ ]1 / 2
= 1
2
133.3MPa − 0( )2 + 133.3MPa + 12.42MPa( )2 + 12.42 MPa( )2[ ]1 / 2=140MPa
www.elsolucionario.net
Page 6-17
From Equation (6.11), the safety factor for DET is:
σe =
Sy
ns
; ns =
Sy
σe
=
600MPa
140MPa
= 4.28
6.31 Use the MSST and the DET to determine the safety factor for 2024 aluminum alloys for
each of the following stress states:
a) σx=10MPa, σy=-60MPa
b) σx=σy=τxy=-30MPa
c) σx=-σy=20MPa and τxy=10MPa
d) σx=2σy=-70MPa, and τxy=40MPa
Notes: This problem does not require determination of the stresses as in Problems 6.28 through
6.30, but uses the same approach. From the stress state, the principal stresses are determined.
Equation (6.6) gives the safety factor for the Maximum Shear Stress Theory, and Equations (6.9)
and (6.11) give the safety factor for the Distortion-Energy Theory.
Solution:
From Table 6.1, the yield strength for 2024-T351 is Sy=325MPa.
For σx=10MPa, σy=-60MPa, note that there are no shear stresses. Therefore, we can
directly write the principal stresses as σ1=10MPa, σ2=0MPa and σ3=-60MPa. Note that the
principal stresses have been renumbered so that σ1≥σ2≥σ3. From Equation (6.6), the safety factor
for MSST is:
σ1 − σ3 =
Sy
ns
; ns =
Sy
σ1 − σ3
=
325MPa
10MPa + 60MPa
= 4.64
Equation (6.9) gives the effective stress as
σe =
1
2
σ1 − σ2( )2 + σ1 − σ3( )2 + σ2 −σ 3( )2[ ]1 / 2
= 1
2
10MPa − 0( )2 + 10MPa + 60MPa( )2 + 60MPa( )2[ ]1 / 2=65.57MPa
From Equation (6.11), the safety factor for DET is:
σe =
Sy
ns
; ns =
Sy
σe
=
325MPa
65.57MPa
= 4.96
For σx=σy=τxy=-30MPa, Equation (2.16) gives
σ1,σ2 =
σx + σ y
2
± τxy2 +
σx − σy
2
 
 
  
 
 
  
2
=
−30MPa − 30MPa
2
± 30MPa( )2 + 0( )2
Therefore, σ1=0MPa, σ2=0MPa and σ3=-60MPa. Note that the principal stresses have been
renumbered so that σ1≥σ2≥σ3. From Equation (6.6), the safety factor for MSST is:
σ1 − σ3 =
Sy
ns
; ns =
Sy
σ1 − σ3
=
325MPa
0MPa + 60MPa
= 5.42
Equation (6.9) gives the effective stress as
σe =
1
2
σ1 − σ2( )2 + σ1 − σ3( )2 + σ2 −σ 3( )2[ ]1 / 2
= 1
2
0 + 60MPa( )2 + 60MPa( )2[ ]1/ 2 = 60MPa
From Equation (6.11), the safety factor for DET is:
www.elsolucionario.net
Page 6-18
σe =
Sy
ns
; ns =
Sy
σe
=
325MPa
60MPa
= 5.42
For σx=-σy=20MPa and τxy=10MPa, Equation (2.16) gives
σ1,σ2 =
σx + σ y
2
± τxy2 +
σx − σy
2
 
 
  
 
 
  
2
=
20MPa − 20MPa
2
± 10MPa( )2 + 20MPa + 20MPa
2
 
  
 
  
2
Therefore, σ1=22.36MPa, σ2=0MPa and σ3=-22.36MPa. Note that the principal stresses have
been renumbered so that σ1≥σ2≥σ3. From Equation (6.6), the safety factor for MSST is:
σ1 − σ3 =
Sy
ns
; ns =
Sy
σ1 − σ3
=
325MPa
22.36MPa + 22.36MPa
= 7.27
Equation (6.9) gives the effective stress as
σe =
1
2
σ1 − σ2( )2 + σ1 − σ3( )2 + σ2 −σ3( )2[ ]1 / 2
= 1
2
22.36MPa − 0( )2 + 22.36 MPa + 22.36 MPa( )2 + 22.36MPa( )2[ ]1 / 2= 38.73MPa
From Equation (6.11), the safety factor for DET is:
σe =
Sy
ns
; ns =
Sy
σe
=
325MPa
38.73MPa
= 8.39
For σx=2σy=-70MPa, and τxy=40MPa, Equation (2.16) gives
σ1,σ2 =
σx + σ y
2
± τxy2 +
σx − σy
2
 
 
  
 
 
  
2
=
−70MPa − 35MPa
2
± 40MPa( )2 + −70MPa + 35MPa
2
 
  
 
  
2
Therefore, σ1=0MPa, σ2=-8.84MPa and σ3=-96.16MPa. Note that the principal stresses have
been renumbered so that σ1≥σ2≥σ3. From Equation (6.6), the safety factor for MSST is:
σ1 − σ3 =
Sy
ns
; ns =
Sy
σ1 − σ3
=
325MPa
0MPa + 96.16MPa
= 3.38
Equation (6.9) gives the effective stress as
σe =
1
2
σ1 − σ2( )2 + σ1 − σ3( )2 + σ2 −σ 3( )2[ ]1 / 2
= 1
2
8.84MPa( )2 + 96.16MPa( )2 + −8.84MPa + 96.16MPa( )2[ ]1 / 2= 92.06MPa
From Equation (6.11), the safety factor for DET is:
σe =
Sy
ns
; ns =
Sy
σe
=
325MPa
92.06MPa
= 3.53
6.32 Four different stress elements, each of the same material, are loaded as shown in sketches
e,f,g,h. Use the MSST and the DET to determine which element is the most critical.
www.elsolucionario.net
Page 6-19
Notes: Equations (2.16), (6.6), (6.10), and (6.11) are used to solve this problem.
Solution:
(e) σ1=21MPa, σ2=0, σ3=-21MPa. Therefore, from Eq. 6.6: σ1-σ3=42MPa. Also, from Eq. 6.9,
σe =
1
2
σ1 − σ2( )2 + σ2 − σ3( )2 + σ3 −σ1( )2[ ]1 / 2= 36MPa
(f) σ1=28.5 MPa, σ2=0, σ3=-7.5MPa. Thus, σ1-σ3=36MPa. Also,
σe =
1
2
σ1 − σ2( )2 + σ2 − σ3( )2 + σ3 −σ1( )2[ ]1 / 2= 33MPa
(g) σ1=30MPa, σ2=30MPa, σ3=0. Thus, σ1-σ3=30MPa. Also σe=30MPa.
(h) σx=30MPa, σy=0, τxy=-10MPa. Therefore, from Equation (2.16), σ1=33MPa, σ2=0, σ3=-
3.02MPa. Therefore, σ1-σ3=36.02MPa, σe=31.6 MPa.
This shows that the stress state in (e) is the largest.
6.33 The rod shown in sketch i is made of AISI 1040 steel
and has two 90° bends. Use the MSST and the DET to
determine the minimum rod diameter for a safety factor of 2 at
the most critical section.
Notes: Recognizing that the critical section is at the wall, the
component stresses can be expressed as functions of the rod
diameter. Applying MSST or DET gives an expression that can
be solved for d.
Solution:
The yield strength for AISI 1040 is obtained from the inside front cover as 350MPa. The critical
section is at the wall; the rod is slender so transverse shear effects will be ignored. The 8kN load
causes a torque equal to T1=8kN(0.75m)=6kNm, and bending moment Mx1=8kN(0.3m)=2.4kNm.
The 10kN load causes axial normal stress, a bending moment Mz1=10kN(0.75m)=7.5kNm and a
bending moment Mx2=10kN(0.05m)=500Nm, which is in the opposite direction as Mx1. The 75kN
www.elsolucionario.net
Page 6-20
load causes torque equal to T2=-75kN(0.05m)=3.75kNm (in the opposite direction as T1) and a
bending moment Mz2=75kN(0.3m)=22.5kNm, which is in the opposite direction as Mz1.
Therefore, the bar sees the following: Mx=Mx1-Mx2=1.9kNm, Mz=22.5kNm-7.5kNm=15kNm, and
T=T1-T2=6kNm-3.75kNm=2.25kNm. Therefore, the moment at the wall is
M = Mx2 + Mz2 = 1.9kNm( )2 + 15kNm( )2 = 15.12kNm
The normal stress is therefore
σ =
Mc
I
+
P
A
=
10kN
π
4
d2
+
15.12kNm( ) d / 2( )
π
64
d4
=
12.73 kN
d2
+
154kNm
d3
=
12.73kN( )d + 154kNm
d3
The shear stress is:
τ =
Tc
J
=
2.25kNm( ) d / 2( )
π
32
d4
=
11.46kNm
d 3
Equation (2.16) gives
σ1 ,σ2 =
σ x + σ y
2
± τ xy2 +
σ x − σy
2
 
 
  
 
 
  
2
=
1
d 3
6.365kN( )d + 77kNm[ ] ± 11.46kNm( )2 + 6.365kN( )d + 77kNm( )2   
 
 
 
It can be shown that unless d is in the millions of meters, that one of these stresses will be
positive, and the other negative. Therefore, for MSST, Equation (6.6) is:
σ1 − σ3 =
Sy
ns
=
350MPa
2
= 175MPa
Substituting for the stresses and solving yields d=0.0964m. Therefore, a 0.10m or 100mm
diameter cross section is a good design designation. For DET, Equations (6.9) and (6.11) give:
σe =
1
2
σ1 − σ2( )2 + σ1 − σ3( )2 + σ2 −σ3( )2[ ]1 / 2
= 1
2
σ1( )2 + σ1 − σ3( )2 + σ3( )2[ ]1 / 2= Syns = 175MPa
This is solved numericallyas d=0.0963m. Therefore, a 0.10m or 100mm diameter cross section is
still acceptable.
6.34 The shaft shown in shetch j is made of AISI 1020 steel. Determine the most critical
section by using the MSST and the DET. Dimensions of the various diameters shown in sketch j
are d=30mm, D=45mm, and d2=40mm.
Notes:
This problem requires the incorporation of stress concentration effects into the component
stresses before determining the principal stresses.
www.elsolucionario.net
Page 6-21
Solution:
First, considering the location of stress concentration 80 mm from the wall:
J =
π
32
d4 =
π
32
0.040m( )4 = 7.95 × 10−8 m4
I =
J
2
= 3.98 × 10−8m4 ,
A=πd2/4=7.07x10-4m2.
Also, from statics, V=10kN, M=0.4kNm, N=100kN, T=500Nm. The bottom location is critical,
since the bending and tensile stresses are additive at this location. Also, there is no shear stress
due to shear at the extreme location. The stress concentration due to bending is obtained from
Fig. 6.5(b) as 1.4, while for tension it is Kc=1.55 from Fig. 6.5(a). The stress concentration for
torsion is Kc=1.2 from Fig. 6.5(c). Therefore, σ1=458MPa and σ2=-28MPa.
For the location 40 mm from the wall, N=100kN, M=0.8kNm T=500Nm,
Kc(bending)=1.45, Kc(tension)=1.5, Kc(torsion)=1.2 (all from Fig. 6.6). Therefore, σ1=312MPa
and σ2=-7.6.
This means that the critical location is 80mm from the wall, as the stresses are higher.
www.elsolucionario.net
Page 7-1
Chapter 7
Failure Prediction for Cyclic and Impact Loading
7.1 A tuning fork is hit with a pencil and starts to vibrate with a frequency of 440Hz. The
maximum bending stress in the tuning fork is 4MPa at the end positions. Calculate the
mean stress, the range of stress, the stress amplitude, the stress ratio, and the amplitude
ratio. Also calculate how much stress the tuning fork can sustain without being plastically
deformed if it is made of AISI 1080 steel.
Notes: Equations (7.1) through (7.5) are used to obtain a solution.
Solution:
From Equation (7.1), the mean stress is:
σm =
σmax + σmin
2
= 4MPa − 4MPa
2
= 0
The stress range is given by Equation (7.2) as
σr = σmax −σmin = 4MPa − −4 MPa( ) = 8MPa
The stress amplitude is obtained from Equation (7.3) as:
σa =
σr
2
= 8MPa
2
= 4MPa
The stress ratio is obtained from Equation (7.4):
Rs =
σmin
σmax
= −4MPa
4MPa
= −1
The amplitude ratio is given by Equation (7.5):
Aa =
σa
σm
= 4MPa
0
= ∞
From the inside front cover, the yield strength of AISI 1080 steel is 380 MPa, and this is the
largest stress that the material can withstand without permanent deformation.
7.2 The jack for a Volvo consists of a mechanism in which the lift screw extends horizontally
through two corners of the mechanism while the other two corners apply a force between
the ground and the car to be lifted. The maximum compressive stress in the jack is 190
MPa when the Volvo is jacked up so high that both wheels on one side of the car are in
the air and the load on the jack is 8000N. How many times can the jack be used for a
small truck that weighs 6 tons and loads the jack to 17000N before the jack fails from
fatigue? The jack material is 1080 steel.
Notes: The first step is to determine which regime of fatigue is applicable by comparing the stress
to the fatigue strengths from Equations (7.7) and (7.8). The jack is in high-cycle, finite life, so
that Equation (7.15) is used to determine the number of cycles to failure.
Solution:
From the inside front cover for AISI 1080 steel, Su=615MPa and Sy=380MPa. Since the stress is
proportional to load, the stress encountered by the jack in raising the truck is:
www.elsolucionario.net
Page 7-2
σ = Ftσc
Fc
= 17000N( ) 190MPa( )
8000N
= 403.8MPa
From Equation (7.7), the stress endurance limit is:
Se
′ = 0.45Su = 0.45 615MPa( ) = 276.8MPa
While from Equation (7.8), the fatigue strength where high-cycle fatigue begins is:
Sl
’=0.75Su=0.75(615MPa)=461.3MPa
Since the applied stress is between these two levels, the jack encounters high-cycle, finite-life
fatigue. From Equations (7.12) and (7.13),
bs = −
1
3
log
Sl
′
Se
′
 
 
 
 
 
 
 
 = −
1
3
log
461.3MPa
276.8MPa
 
  
 
  = −0.07394
C = log
Sl
′( )2
Se
′
 
 
 
 
 
 
 
 
 
 
 
 
= log 461.3( )
2
276.8
 
 
 
 
 
 
 
 
= 2.8858
Therefore, the number of cycles until failure is, from Equation (7.15),
Nt
′ = S f ′ 10−C ( )1/ b s = 403.8( )10−2.8858( )1 /−0.07394( ) = 6052
This is the approximate number of cycles until failure.
7.3 The shaft shown in sketch a rotates at high speed while the imposed load remains static.
The shaft is machined from ground, high carbon steel (AISI 1080). If the loading is
sufficiently large to produce a fatigue failure after 1 million cycles, where would the
failure most likely occur? Show all necessary computations and reasoning.
Notes: The failure is expected to be due to bending, since the stresses are completely reversing.
Once must therefore investigate all likely locations to determine the maximum normal stress,
including the effects of stress concentrations.
Solution:
Note the locations a, b, c, and d in the figure, which are the probable failure locations,
either because of stress concentration (a, c, d) or because the loading is largest (b). Upon further
inspection, the difference between shoulder a and d is the radius. Since the radius is smaller for
the left shoulder than the right one, it will have a higher stress concentration factor, and the right
shoulder doesn’t have to be considered. In terms of the analysis to be followed, there are no
differences between the locations in the shaft in terms of manufacturing method, size effects,
temperature effects, reliability, etc, except that the fatigue stress concentration factor must be
calculated. Also, from the inside front cover, AISI 1080 steel has an ultimate strength of 89ksi.
www.elsolucionario.net
Page 7-3
I. Left Shoulder
Taking a section at point a, the moment is M=(P/2)(1in), with the diameter equal to 1 in. The
stress concentration is obtained from Figure 6.5 (b), using D/d=1.12in/1in=1.25 and
r/d=(1/16in)/1in=0.0625, so that Kc=1.75. From Figure 7.6 on page 272, qn is about 0.75.
Therefore, from Equation (7.19),
Kf=1+(Kc-1)qn=1+(1.75-1)(0.75)=1.56
The maximum stress at a is:
σ = K f
Mc
I
= 1.56( )
0.5in( )P( ) 0.5in( )
π
64
1in( )4
= 7.94in−2( )P
II. Midpoint of Shaft
At the midpoint, the bending moment is M=(P/2)(3in)=(1.5in)P. The diameter of the shaft is
1.25in at this location. There is no stress concentration here, so the maximum stress is given by
Equation (4.48) as:
σ =
Mc
I
=
1.5in( )P( ) 1.25in / 2( )
π
64
1.25in( )4
= 7.82in−2( )P
III. Groove
Taking a section at the groove, the moment is M=(P/2)(2in)=(1in)P. Although the diameter is not
specified at this location, it is reasonable to approximate it as 1in. Kc is obtained from Figure
6.6(b), using D/d=1.25in/1in=1.25 and r/d=(1/8in)/1in=0.125, so that Kc=1.7. From Figure 7.6, qn
is around 0.85. Therefore, from Equation (7.19),
Kf=1+(Kc-1)qn=1+(1.7-1)(0.85)=1.595
The maximum stress is
σ = K f
Mc
I
= 1.595( )
1in( )P( ) 0.5in( )
π
64
1in( )4
= 16.2in−2( )P
IV. Summary
Since the stress is largest at the groove, this is the most likely fatigue failure location.
7.4 For each of the bars shown in sketches b and c, which are made of AISI 1040 cold drawn
steel, determine
a) The static tensile load causing failure
b) The alternating (completely reversing) axial load ±P that would be just on the verge of
producing eventual fatigue failure.
Notes: The discussion on page 227 suggests that stress concentrations are relieved by plastic
deformation under static loads; certainly this mild steel will relieve the stresses so that they are
ignored. The stress concentration is not ignored for the fatigue failure. Also, a surface correction
factor kf is calculated from Equation (7.21).
Solution:
www.elsolucionario.net
Page 7-4
For AISI 1040 steel, the inside front cover gives Sut=75ksi=520MPa. Since the material is ductile,stress concentration are ignored (see discussion on page 227). Therefore, the critical load for
either bar is:
P=σA=(520MPa)(0.025m)2=325kN
For a fatigue loading, and the bar in sketch b (without a notch), we note from Equation
(7.7) for axial loading:
Se
’=0.45Su=0.45(520MPa)=234MPa
The bar may be cold drawn, but the notch must be machined; either case gives, from Table 7.3,
that e=4.51MPa, and f=-0.265. Therefore, from Equation (7.21),
k f = eSut
f = 4.51 520( )−0.265 = 0.86
The size factor is ks=1 for axial loading (see page 275). All other factors are ignored. Therefore,
from Equation (7.16),
Se=kfkskrktkmSe
’=(0.86)(1)(234MPa)=201MPa.
For the notched bar, from Figure 6.4(a) for H/h=30/25=1.2 and r/h=2.5/25=0.1, we get
Kc=2.35. From Figure 7.6 on page 272 we get for Sut=520MPa and r=2.5mm that qn=0.75
(approx). Therefore, Equation (7.19) gives
K f = 1 + Kc − 1( )qn = 1 + 2.35 −1( ) 0.75( ) = 2.01
For the unnotched bar, there is no stress concentration, so we set the applied stress equal
to the endurance limit:
sall=Pall/A=Se; Pall=SeA=(201MPa)(0.025m)2=126kN
For the notched bar, the endurance limit must be reduced by the factor Kf:
σall =
P
A
=
Se
K f
;P =
SeA
K f
=
201MPa( ) 0.025m( )2
2.01
= 62.5kN
Note that slightly different answers are acceptable; numbers read off charts can be slightly
different, and there can be small discrepancies between using Equation (7.21) versus Figure 7.7.
7.5 A stepped shaft, as shown in sketch d, has a major diameter of 2 in, a minor diameter of 1
in, and a notch radius of 0.1 in. It was machined from high-carbon steel (AISI 1080). The
loading is one of completely reversed torsion. During a typical 30s of operation under
overload conditions, the nominal stress in the 1-in diameter section was measured to be
as shown in sketch e. Estimate the life of the shaft when it is operating continually under
these conditions.
Notes: The modified endurance limit has to be calculated using a surface finish factor, fatigue
stress concentration factor, and size factor. The life at each stress is calculated from Equation
(7.15), and the total life is calculated from Miner’s Rule, Equation (7.24).
Solution:
www.elsolucionario.net
Page 7-5
From the inside front cover, Sy=55 ksi and Sut=89ksi for AISI 1080 steel. Kc is obtained from
Figure 6.5(c) using D/d=2in/1in=2 and r/d=0.1in/1in=0.1, so that Kc=1.45. From Figure 7.6 for
r=0.1 and Su=89ksi, qn is about 0.82. Therefore, Equation (7.19) gives
Kf=1+(Kc-1)qn=1+(1.45-1)(0.82)=1.37
From Equation 7.7, the stress endurance limit is Se
’=0.29Su=0.29(89ksi)=25.8ksi for torsion. To
calculate the modified endurance limit, we need to consider the surface finish effect and size
effect. Since the surface is machined, Table 7.3 gives e=2.70 for Sut in ksi and f=-0.265. Equation
(7.21) gives the surface finish factor as
k f = eSut
f = 2.70( ) 89( )−0.265 = 0.82
From Equation (7.22)
ks=0.869d
-0.112
=0.869(1in)
-0.112
=0.87
All other endurance limit modification factors will be ignored. Therefore the modified endurance
limit is, from Equation (7.16):
Se=kfksSe
’/Kf=(0.82)(0.87)(25.8ksi)/1.37=13.4ksi
This equation includes the fatigue stress concentration effects, although the stresses could have
been increased instead of incorporating the factor in the material strength. Since the stresses are
given in this problem, it is easier to incorporate Kf in the modified endurance limit.
Note that some of the applied stresses applied exceed this value, therefore, the life will be
finite. From Equation (7.8), Sl
’=0.72Su=0.72(89ksi)=64.1ksi. Therefore
Sl=kfksSl
’/Kf=33.4ksi
We conclude that the part will encounter high cycle, finite life fatigue. From Equations (7.12) and
(7.13),
bs = −
1
3
log
Sl
Se
 
 
  
 
 
  = −
1
3
log
33.4ksi
13.4ksi
 
  
 
  
= −0.1322
C = log
Sl( )2
Se
 
 
 
 
 
 
 
 = log
33400 psi( )2
13400psi
 
 
 
 
 
 
 
 = 4.920
Given the applied stress, Equation (7.15) gives the expected life. The applied stresses are 10ksi,
20ksi, 25ksi and 35 ksi. The lives at these stresses are:
For τ1=10ksi, N1=∞, because the stress is below the modified endurance limit
For τ2=20ksi,
N2 = τ210−C 
 
  
 
  
1/ bs
= 20000 psi( )10−4.92[ ]−1 / 0 . 1 3 2 2= 48,100 cycles
For τ3=25ksi,
N3 = τ310−C 
 
  
 
  
1/ bs
= 25000 psi( )10−4.92[ ]−1 / 0 . 1 3 2 2= 8890cycles
For τ4=35ksi,
N4 = τ410−C 
 
  
 
  
1/ bs
= 35000psi( )10−4.92[ ]−1/ 0.1322 = 698cycles
Note from the sketch of the loading that the number of stress cycles in a 30 second interval are
n1=14, n2=4, n3=2, n4=1. Therefore, from Equation (7.24), the life is predicted as:
L =
30s
n1
N1
+ n2
N2
+ n3
N3
+ n4
N4
 
 
 
 
 
 
 
 
=
30s
14
∞
+ 4
48,100
+ 2
8890
+ 1
698
 
 
 
 
 
 
= 17,200s = 4.79hr
www.elsolucionario.net
Page 7-6
7.6 A flood-protection dam gate is supposed to operate only once per week for 100 years, but
suddenly after 30 years of use it needs to be operated twice per day (each time the high
tide comes in). Find how much lower the bending stress must be from then on to still give
a total life of 100 years. The material being fatigued is medium-carbon steel (AISI 1040).
Notes: This problem is solved through application of Equation (7.14), requiring equations (7.12)
and (7.13).
Solution:
From the inside front cover, the ultimate strength of AISI 1040 steel is Su=520MPa. The original
intended life was (100years)(52weeks/year)(1cycle/week)=5200 cycles. The used life is
0.3(5200)=1560 cycles. If the gate will be operated twice a day for 70 years, the number of cycles
it must survive is (70years)(365.25days/year)(2cycles/day)=51100cycles. From Equation (7.8),
recognizing the loading is bending,
Sl
’=0.9Su=0.9(520MPa)=468MPa
Also, from Equation (7.7) for bending
Se
’=0.5Su=0.5(520MPa)=260MPa
From Equation (7.24), the required life is:
n1
N1
+
n2
N2
= 1=
1560
5200
+
51100
N2
; N2 = 73000cycles
From Equations (7.12) and (7.13),
bs = −
1
3
log
Sl
′
Se
′
 
 
 
 
 
 
 
 = −
1
3
log
468MPa
260 MPa
 
  
 
  
= −0.08509
C = log
′ S l( )2
′ S e
 
 
 
 
 
 
 
 = log
468MPa( )2
260 MPa
 
 
  
 
 
  = 8.9255
Therefore, the allowable stress level for N=73000 cycles is given by Equation (7.14) as:
S f
′ = 10C N( )bs = 108.9255 73000( )−0.08509 = 325MPa
Note that for the original design, the allowable stress level was:
S f
′ = 10C N( )bs = 108.9255 5200( )−0.08509 = 407MPa
so the new stress level is roughly 20% lower than the original case.
7.7 A hydraulic cylinder has a piston diameter D=100mm and a piston rod diameter
d=33mm. The hydraulic pressure alternately applied to each side of the piston is
p=300bars. The pressure induces in the rod a compressive force of πpD2/4 or a tensile
force of πp(D2-d2)/4, depending on which side of the piston the pressure is applied. The
piston rod material is martensitic stainless steel (AISI 410). Find the endurance life of the
piston rod.
Notes: One must calculate the stresses and compare them to the endurance strength of the
material to determine which regime of fatigue is appropriate. The life is obtained from Equation
(7.15).
Solution:
From the inside front cover, the ultimate strength of AISI 410 stainless steel is Su=483MPa and
the yield strength is Sy=275MPa. From equations (7.7) and (7.8) for axial loading,
www.elsolucionario.net
Page 7-7
Se
’=0.45Su=0.45(483MPa)=217MPa
Sl
’=0.75Su=0.75(483MPa)=362MPa
From the inside front cover, 1bar=10
5
 Pa. Therefore, the maximum stress exerted on the piston is:
σmax =
P
A
=
π
4
p D2 − d2( )
π
4
d2
=
p D2 − d2( )
d2
=
300 × 105 Pa( ) 0.1m( )2 − 0.033m( )2[ ]
0.033m( )2
= 245MPa
The minimum stress is (note that it is compressive):
σmin = −
P
A
= −
π
4
pD2
π
4
d2
= −
pD2
d2
= −
300 × 105 Pa( ) 0.1m( )2
0.033m( )2
= −275MPa
From Equation (7.1), the mean stress is:
σm =
σmax + σmin
2
=
245MPa + −275MPa( )
2
= −15MPa
From Equation (7.3), the stress amplitude isσa =
σmax − σmin
2
=
245MPa + 275MPa
2
= 260MPa
Note that σm is really very small, so it can be safely ignored. The stress amplitude is greater than
Se
’, but less than Sl
’, so the piston is subjected to high-cycle, finite life fatigue. From Equations
(7.12) and (7.13),
bs = −
1
3
log
Sl
′
Se
′
 
 
 
 
 
 
 
 = −
1
3
log
362MPa
217MPa
 
  
 
  
= −0.07408
C = log
Sl( )2
Se
 
 
 
 
 
 
 
 = log
362 MPa( )2
217MPa
 
 
  
 
 
  = 8.7810
From Equation (7.15) the endurance life of the piston rod is:
Nt
′ = S f ′10−C 
 
  
 
  
1/ bs
= 260MPa( ) 10−8.7810( )[ ]−1 / 0 . 0 7 4 0 8=87 ,290cycles
7.8 A 20mm diameter shaft transmits a variable torque of 500±400Nm. The frequency of the
torque variation is 0.1s
-1
. The shaft is made of high-carbon steel (AISI 1080). Find the
endurance life of the shaft.
Notes: Equation (7.15) gives the life.
Solution:
For AISI 1080, the inside front cover yields Sy=380 Mpa and Sut=615 Mpa. Then, from
Equations (7.7) and (7.8), one obtains:
Sl
′ = 0.72Su = 443MPa
Se
′ = 0.29Su =0.29 615MPa( ) =178MPa
The shear stress amplitude is
τa =
τmax − τmin
2
=
Mmax − Mmin( )r
2 J
=
900Nm −100 Nm( ) 0.01m( )
2
π
32
0.02m( )4
= 255MPa
www.elsolucionario.net
Page 7-8
Note that this value is indeed between Sl’ and Se’, so it will fail under a finite number of cycles.
The life is given by equation (7.15):
N = S f ′10−C( )1/ b
From Equations (7.12) and (7.13) as:
C = log
Sl ′( )2
Se
′
 
 
 
 
 
 
 
 
 
 
 
 
= log
443x106( )2
178x106
= 9.04
bs = −
1
3
log
Sl
′
Se
′
 
 
 
 
 
 
 
 = −0.132
So that the number of cycles for which 255MPa would cause failure, that is, for Sf’=255Mpa,
N = 255x106( ) 10−9.04( )( )−1 / 0 . 1 2 8 7= 62 ,900cycles
For a frequency of 0.1 cycles per second, this results in a life of 174 hrs.
7.9 Find for the shaft in Problem 7.8 how large the shaft diameter has to be to give infinite
life.
Notes: The stress amplitude must equal the fatigue endurance limit for this case. This solution
uses the results of Problem 7.8.
Solution:
From Problem 7.8, Se
’=178MPa. The shear stress amplitude is
τa =
τmax − τmin
2
=
Mmax − Mmin( )
d
2
J
=
900Nm − 100Nm( ) d
2
2
π
32
d4
=
2037Nm
d3
For infinite life, the shear stress amplitude must be less than or equal to Se
’. Therefore,
τa = 178MPa =
2037Nm
d 3
; d = 0.0225m = 22.5 mm
7.10 A notched bar has the diameter D=25mm and the notch radius r=0.5mm. It experiences
rotational bending fatigue. The bottom diameter of the notch is 24mm. Find which of the
steels AISI 1020, 1040, 1080, or 316 will give the highest allowable bending moment for
infinite life. Calculate that moment.
Notes: This problem requires calculation of endurance limit and fatigue stress concentration
factors for the metals under consideration.
Solution:
For the materials given, the following data can be obtained from the inside front cover:
Alloy Su (MPa) Sy (MPa) qn
1020
1040
1080
395
520
615
295
350
380
0.4
0.58
0.62
www.elsolucionario.net
Page 7-9
316 552 207 0.60
The notch sensitivity is read from Figure 7.6 on page 272 and placed in the last column above.
Also, from Figure 6.6 (b) on page 226, with D/d=25/24=1.04 and r/d=0.5/24=0.021, Kc=2.5.
The fatigue stress concentration factor for each alloy is then given by Eq. 7.19 and equals for
1020: 1.6; 1040: 1.87; 1080: 1.93; 316:1.9. To compare the alloys, one can reduce the
endurance limit by 1/Kf per the discussion on page 271. The allowable stresses after reducing the
endurance limit by the fatigue stress concentration factors are then, from (7.7): Se=0.5Su/Kf;
1020: 123MPa; 1040: 139 MPa; 1080: 159 MPa; 316: 145. Of these, the 1080 alloy gives the
highest allowable stress. The moment which is allowable in AISI 1080 is
Se =
Se
′
K f
= 159MPa =
Mc
I
= M
0.012m
π
64
0.024m( )4 
  
 
  
;M = 215Nm
7.11 A straight, circular rotating beam has a diameter of 30mm and a length of 1m. At the end
of the beam a stationary load of 500N is applied perpendicularly to the beam, giving a
bending deformation. Find which surface machining method can be used to give infinite
life to the rotating beam if it is made of:
a) AISI 1020 steel
b) AISI 1080 steel
Note that ks=kr=km=1
Notes: The bending stress is obtained from Equation (4.48). The endurance limit is obtained from
Equation (7.7), and the modified endurance limit from (7.16) with only the surface correction
factor (kf) incorporated.
Solution:
First of all, note that the bending stress is, from Equation (4.48):
σ =
Mc
I
=
M d / 2( )
π
64
d4
 
  
 
  
=
500N( ) 1m( ) 0.03m( ) /2
π
64
0.03m( )4 
  
 
  
=188.6MPa
For infinite life, the modified endurance limit must be greater than this stress.
I. AISI 1020 Steel
From the inside front cover, Sut=395MPa. Therefore, from Equation (7.7) for bending,
Se
′ = 0.5Su = 0.5 395MPa( ) = 197.5MPa
From Equation (7.16) and the requirement that Se≥σ,
Se = k f Se′ ≥ σ = 188.6MPa; k f ≥
188.6 MPa
197.5MPa
= 0.954
Reviewing Figure 7.7, the only acceptable surface finish is polished. Even a ground surface gives
kf of around 0.9, which would result in finite life.
II. AISI 1080 Steel.
From the inside front cover, Sut=615MPa. Therefore, from Equation (7.7) for bending,
Se
′ = 0.5Su = 0.5 615MPa( ) = 307.5MPa
From Equation (7.16) and the requirement that Se≥σ,
Se = k f Se′ ≥ σ = 188.6MPa; k f ≥
188.6 MPa
307.5MPa
= 0.61
Reviewing Figure 7.7, the acceptable finishes come from polishing, grinding, machining or cold
drawing. Hot rolling is marginal; it may be acceptable, but probably not.
www.elsolucionario.net
Page 7-10
7.12 The pedals on a bicycle are screwed into the crank with opposite threads, one left-hand
thread and one right-hand thread, to ensure that the pedals do not accidentally unscrew.
Just outside the thread is a 0.75-mm radius fillet connecting the 12.5mm-diameter
threaded part with an 11-mm diameter central shaft in the pedal. The shaft is made of
AISI 4340 alloy steel tempered at a low temperature to give an ultimate stress of 2GPa.
Find the fatigue stress concentration factor for the fillet, and calculate the maximum
allowable pedal force for infinite life if the force from the foot is applied 70mm from the
fillet.
Notes: The fatigue stress concentration factor must be calculated from Equation (7.19).
Insufficient information is given to obtain any endurance limit modification factors, so this
solution is based on maintaining a stress below Se
’.
Solution:
The stress concentration factor is obtained from Figure 6.5(b) using
r/d=0.75mm/11mm=0.068 and D/d=12.5mm/11mm=1.14, so that Kc=1.75. The notch sensitivity
is obtained from Figure 7.7 as qn=0.95. Therefore, Kf is obtained from Equation (7.19) as:
Kf=1+(Kc-1)qn=1+(1.75-1)(0.95)=1.71
Therefore, since the moment is P(0.07m), the bending stress is
σ = K f
Mc
I
= 1.71( ) P 0.07m( ) 0.011m / 2( )π
64
0.011m( )4
= 9.16 × 105m−2( )P
Note that the fatigue stress concentration factor has been used to increase the stress; it could have
been used to reduce the allowable strength to get the same result. For bending, the endurance
limit of the material is obtained from Equation (7.7) as Se’=0.5Su=0.5(2GPa)=1GPa. Therefore,
equating the stress to the strength:
σ = 9.16 × 105 m−2( )P = Se ′ = 1GPa; P =1090N
7.13 A saddle pole for a mountain bike has a fillet of 5-mm radius connecting the top part of
the pole (which has a diameter of 35mm) with the bottom part of the pole (which has a
diameter of 25 mm). The saddle pole is solid and made of 2024 aluminum. Find the
fatigue stress concentration factors for axial load and bending.
Notes: The theoretical stress concentration factor (Kc) can be obtained from the Figure 6.5 on
page 225. To get the fatigue stress concentration factors, one needs to use the notch sensitivity
from Figure 7.6, and Equation (7.19).
Solution:
For this pole, r/d=5/25=0.2 and D/d=35/25=1.4. Therefore, from Figure 6.5 on page 225,Kc for
axial loading is roughly 1.55 and for bending Kc is about 1.375. From Figure 7.6 with r=5 mm
and extrapolating off the chart, qn is approximately 0.8, so Kf is, from Equation (7.19), and axial
loading:
K f = 1 + Kc − 1( )qn = 1 + 1.55 −1( ) 0.8( ) = 1.44
For bending, Equation (7.19) becomes:
K f = 1 + Kc − 1( )qn = 1 + 1.375 − 1( ) 0.8( )= 1.3
www.elsolucionario.net
Page 7-11
7.14 During the development of a new submarine a 1-to-20 model was used. The model was
tested to find if there was any risk of fatigue failure in the full-scale submarine. To be on
the safe side, the stresses in the model were kept 15% higher than the stresses in the full-
size submarine. Is it possible to conclude that the full-size submarine will be safe if the
model was safe at a 15% higher stress level?
Notes: This involves consideration of the size factor, Equation (7.22).
Solution:
Some portions submarine may be effectively modeled with a scale model. However, consider a
bending or tension specimen, where all features are the same, but the size is smaller. From
Equation (7.22), the size factor for the scale model (m subscript) and the full-sized submarine (f
subscript) are:
ksm = 1.189dm−0.112
ksf = 1.189d f−0.112 =1.189 20dm( )−0.112 = 20( )−0.112ksm = 0.715ksm
Therefore, for the model has to be stressed at (1/0.715)=1.399 times the full size submarine stress
level. The model’s survival does not guarantee the success of the full-sized submarine.
7.15 During the development of a new car it was found that the power from the motor was too
high for the gearbox to handle. When maximum torque was transmitted through the
gearbox, 50% of the gearboxes failed within the required lifetime of 800 hours. How
much stronger should the gearbox be made to ensure that only one gearbox out of 1000
would fail within 800hr?
Notes: This problem requires application of the reliability factor.
Solution:
Referring to Table 7.4 on page 275, the reliability factor is kr=1.00 for 50% failures. If one out of
one thousand were to fail, that is, for a reliabilty of 99.9%, kr=0.75. Therefore the strength should
be increased by (1/0.75)=1.33. Therefore, the gearbox should be 33% stronger.
7.16 Truck gearboxes are dimensioned for infinite life at maximum torque regarding contact
stresses and bending stresses for all gear steps. Car gearboxes are dimensioned for finite
life according to typical running conditions. Maximum torque for the first gear can
typically be maintained only 3 to 6s for each acceleration before the maximum speed is
reached. If a driver accelerates at full power 20 times per day for 20 years on the first
gear, the required life is only 60 to 120hr. The normal load spectrum gives a life of
200,000 km for 99% of the gearboxes. A driver uses the car twice a year to move his 10-
ton boat 50km, the distance between his home and the harbor, during the 10-year life of
the car. Calculate how much of the 200,000km nominal life of the gearbox is consumed
by the boat moving if the life is inversely proportional to the load raised to the 3.5 power.
During the move the gearbox load is four times higher than normal.
www.elsolucionario.net
Page 7-12
Notes: This problem requires application of Miner’s Rule, Equation (7.24).
Solution:
Use an n subscript for the normal use condition and a b subscript for hauling the boat. The load
on the gear box is 4 times higher than normal during the 100km/year (50km to the harbor and
50km going back). At four times higher load, the life of the gearbox is
Nb =
200,000km
43.5
= 1563km
Therefore, from Equation (7.24)
nn
′
Nn
′ +
nb
′
Nb
′ = 1; nn
′ = Nn′ 1−
nb
′
Nb
′
 
 
 
 
 
 
 
 = 200,000( ) 1−
10yr( ) 100km / yr( )
1563km
 
 
 
 
 
 = 72,000km
Therefore, moving the boat consumes 64% of the gearbox life.
7.17 The hand brakes of a bicycle are often the same type for the front and rear wheels. The
high center of gravity (compared with the distance between the wheels) will increase the
contact force between the front wheel and the ground and will decrease the contact force
between the rear wheel and the ground when the brakes are applied. If equal force is
applied to each of the two brakes, the rear wheel will start sliding while the front wheel is
still rolling. The manufacturer of the brake wires did not know about this difference
between the front and rear wheels, so the wires were dimensioned as if the wheels had
equal contact force and thus needed equal force in the two brake wires. The wires were
originally dimensioned to withstand 20 years’ use with brake applications at a force
necessary to lock the two (equally loaded) wheels. How long will the life of the wires be
if the friction between the front wheels and the ground is just enough to lift the rear wheel
from the ground? Assume the endurance life is proportional to the stress raised to the -10
power.
Solution:
When both wheels brake the bike, the wire force on each brake is half of the force needed to
brake the bike with only one brake. Since the endurance life is proportional to the stress raised to
the -10, the life for the front brake is 2
-10
=9.766x10
-4
 times the calculated life, which was 20
years. The life is
L = 20years( ) 9.766 ×10−4( ) = 0.01953yr = 7.129days
Thus the life is only seven days instead of 20 years.
7.18 Ball bearings often run at varying load and speed. Life is inversely proportional to the
contact stress raised to the 31/3 power for a number of bearing types. For ball bearings
the contact stresses are approximately proportional to the third root of the load. A ball
bearing in a gearbox is dimensioned to have a life of 1752 million revolutions, half of
which are at half of the maximum motor torque, one-quarter at full motor torque, and
one-quarter at 75% of full motor torque. Calculate the bearing life if the motor torque is
kept constant at the maximum level.
Notes: This problem requires application of Miner’s Rule, Equation (7.24).
Solution:
www.elsolucionario.net
Page 7-13
Let the full motor torque give the life L1. 75% of the full motor load gives the life of
L75 =
L1
0.75( )3 1 / 9
= 2.694L1
and 50% of the full motor torque gives a life of
L50 =
L1
0.50( )31/9
= 10.89L1
For a total life of 1752 million revolutions, 876 million will be at maximum motor torque, and
438 million will take place at 75% and 50% torque. Therefore, Equation (7.24) gives:
n1
′
N1
′ +
n2
′
N2
′ +
n3
′
N3
′ = 1=
n100
′
N100
′ +
n50
′
N50
′ +
n75
′
N75
′ =
876×106
L1
+ 438×10
6
10.89 L1
+ 438× 10
6
2.694L1
Solving for L1 gives L1=681 million. Therefore, the life at maximum torque is 681 million
revolutions.
7.19 A circular cold-drawn steel bar with a solid cross section is subjected to a cyclic force
that ranges from a maximum of 80,000 lbf in tension to a minimum of 30,000 lbf in
compression. The ultimate strength is 158,000 psi and the yield strength is 133,000 psi.
The critical safety factor is 2. Determine the following:
a) The modified endurance limit
b) The cross-sectional area that will produce fatigue failure and the corresponding
diameter.
c) The region (a, b, c, or d) assuming modified Goodman criteria, and why.
Notes: A surface finish factor must be calculated, all other factors are proven or assumed to be
one. The region in the modified Goodman criteria follows from Table 7.6 on page 285.
Solution:
Since the bar is steel, the stress endurance limit can be obtained from Equation (7.7) as:
Se
’=0.45Su=0.45(158ksi)=71.1ksi
The modified endurance limit will be calculated from Equation (7.16), but correction factors need
to be calculated first. No information exists regarding stress concentrations, so we will assume
Kf=1. The surface finish factor is calculated from Equation (7.21), using e=2.70 and f=-0.265 as
obtained from Table 7.3 on page 274. Therefore, kf is:
k f = eSut
f = 2.70( ) 158( )−0.265 = 0.706
(Note that kf could also be read from Figure 7.7 as around 0.68, which is close to the value
calculated from Equation (7.21)).As discussed on page 275, ks=1 for axial loading. No
information is given to suggest reliability, temperatures or other effects are important, so we take
kr=kt=km=1. Therefore, Equation (7.16) gives:
Se=kfkskrktkmSe
’=(0.706)(71.1ksi)=50.20ksi
From Equation (7.1), the mean stress is:
σm =
σmax + σmin
2
= Pmax + Pmin
2A
=
80kip + −30kip( )
2A
= 25kip
A
We will use the modified Goodman relationship, as required in the problem statement. Note that
Se-Sy=50.2ksi-133ksi=--82.8ksi. Referring to Table 2.6 on page 285, region a therefore cannot be
the one corresponding to this stress state. Neither can region b, since σm>0. Therefore, either
region c or d is the proper one, and it will not be known which region is correct until after a
failure equation has been applied. Assume the stress state is described by region c. Therefore, the
failure equation from Table 7.6 gives:
www.elsolucionario.net
Page 7-14
σmax + σm
Se
K f Su
−1
 
 
  
 
 
  =
Se
nsK f
; 
80kip
A
+ 25kip
A
50.2ksi
158ksi
−1
 
  
 
  =
55ksi
2
Solving for A gives A=2.29in
2
, so d=1.71 in. The mean stress is then
σm =
25kip
A
= 25kip
2.29in2
= 10.9ksi
Referring to Table 7.6, and recognizing that Kf=1, Region c requires:
σm ≤
Sy − Se
1− Se
Su
= 133ksi − 50.2ksi
1− 50.2ksi
158ksi
= 121ksi
Since this is true, region c is the correct region of the modified Goodman relationship.
7.20 Both bars used in Problem 7.4 and shown in sketches b anc c are made from cold-drawn,
medium-carbon steel (AISI 1040). Determine the safety factor for each bar. The load
varies from 10 to 40 kN. Use the modified Goodman relationship.
Notes: This problem is similar to problem 7.19. The region in the modified Goodman criteria
follows from Table 7.6 on page 285.
Solution:
From the inside front cover, Su=520 MPa and Sy=350 MPa. From Equation (7.7), the endurance
limit is 0.45 Su=234 MPa. From Figure 7.7 (a) on page 273, the surface finish factor is kf=0.77.
For axial loading, the size factor is 1 (see text on page 275), and no further information is given.
Therefore the modified endurance limit is Se=(0.77)(234MPa)=180MPa.
I. Unnotched Bar:
Here the maximum stress is σmax=(40,000N)/(0.025m)2=64 MPa. The mean stress is
σm=(25,000N)/(0.025m)2=40MPa. Note that the mean stress is positive, so either region c or d
from Figure 7.10 on page 283 is active; we calculate
Sy −
Se
K f
1 −
Se
K f Su
=
350MPa −
180MPa
1
1 −
180MPa
(1)(520 MPa)
= 260MPa
Since the mean stress is less than this value, we conclude that region (c) is active (see Table 7.6,
page 285). Therefore, the failure equation is, from Table 7.6 on page 285,
σmax + σm
Se
K f Su
−1
 
 
 
 
 
 =
Se
nsK f
;64 MPa+ 40MPa
180MPa
(1)(520 MPa)
−1
 
 
  
 
=
180MPa
1(ns )
;ns = 4.75
II. Notched bar
Here there is a stress concentration factor of Kc=2.4 and a notch sensitivity of about 0.775.
Therefore, Kf=1+(2.4-1)(0.775)=2.085~2.1 The maximum and mean stresses are the same since
the root area is the same. Again, to determine if the active region is c or d we calculate
Sy −
Se
K f
1 −
Se
K f Su
=
350MPa − 180MPa
2.1
1 −
180MPa
(2.1)(520MPa)
= 316MPa
www.elsolucionario.net
Page 7-15
So again the active region is c, and the safety factor is calculated as before and equals 2.8.
7.21 The 7/16 in thick component in sketch f is designed with a fillet
and a hole. The load varies from 12000 to 2000 lbf. The
following strengths are given: Su=56ksi, Sy=41ksi. Using the
Goodman failure theory, determine the safety factor for the hole
as well as the fillet. At which location will failure occur first?
Will the component fail?
Notes: One must obtain the fatigue stress concentration factors from the
theoretical stress concentration factor, the notch sensitivity and
Equation (7.19). The endurance limit is obtained from Equation (7.7).
Insufficient information is given about the part to calculate any
correction factors for Equation (7.16) so Se is taken as Se
’. Equation
(7.28) gives the safety factor. Note that the errata suggests using a 3mm
radius; either approach gives full credit for the homework.
Solution:
Note that from Equation (7.7) for axial loading, Se
’=0.45Su=0.45(56ksi)=25.2ksi. Consider the
hole and the fillet separately.
I. Hole
Note that d/b=0.75/1.875=0.4, so from Figure 6.2(a) on page 222, Kc=2.22. From Figure 7.6,
qn=0.78. Therefore, from Equation (7.19),
K f = 1 + Kc − 1( )qn = 1 + 2.22 − 1( ) 0.78( ) =1.95
Note for the hole that the maximum and minimum nominal stresses are:
σmax =
Pmax
h w − d( )
=
12,000 lb
0.4375in( ) 1.875in − 0.75in( )
= 24.38ksi
σmin =
Pmin
h w − d( )
=
2,000 lb
0.4375 in( ) 1.875 in − 0.75in( )
= 4.063ksi
Therefore, the alternating and mean stresses are, from Equations (7.3) and (7.1), respectively:
σa =
σmax − σmin
2
=
24.38ksi − 4.063ksi
2
= 10.16ksi
σm =
σmax + σmin
2
=
24.38ksi + 4.063ksi
2
= 14.22ksi
Note that
Sy −
Se
K f
1 − Se
K fSu
=
41ksi − 25.2ksi
1.95
1− 25.2ksi
1.95 56ksi( )
= 36.5ksi
Since this is greater than σm, we conclude we are in region c of the modified Goodman diagram.
Therefore, the failure criterion is taken from Table 7.6 on page 285 as:
σmax + σm
Se
K f Su
− 1
 
 
 
 
 
 
 
 = 24.38ksi + 14.22ksi( )
25.2ksi
1.95 56ksi( )
− 1
 
 
  
 
 
  =
Se
nsK f
=
25.2
ns 1.95( )
or ns=0.96.
II. Fillet
www.elsolucionario.net
Page 7-16
For this geometry, r/h=0.25/1.375=0.18 and H/h=1.875/1.375=1.36. Therefore, from Figure
6.3(a) on page 223, Kc=1.6. From Figure 7.6 for r=0.25in and Su=56ksi, qn=0.78. Therefore, from
Equation (7.19),
K f = 1 + Kc − 1( )qn = 1 + 1.6 − 1( ) 0.78( ) = 1.468
The maximum and minimum stresses are:
σmax =
Pmax
hw
=
12,000 lb
1.375in( ) 0.4375( )
=19.95ksi
σmin =
Pmin
hw
=
2,000 lb
1.375in( ) 0.4375( )
= 3.325ksi
Therefore, the alternating and mean stresses are, from Equations (7.3) and (7.1), respectively:
σa =
σmax − σmin
2
=
19.95ksi − 3.325ksi
2
= 8.31ksi
σm =
σmax + σmin
2
=
19.95ksi + 3.325ksi
2
= 11.64ksi
Again, note that
Sy −
Se
K f
1 − Se
K fSu
=
41ksi − 25.2ksi
1.468
1− 25.2ksi
1.468 56ksi( )
= 34.37ksi
Since this is greater than σm, the failure equation is, from Table 7.6 on page 285,
σmax + σm
Se
K f Su
− 1
 
 
 
 
 
 
 
 = 19.95ksi + 11.64ksi( )
25.2ksi
1.468 56ksi( )
−1
 
 
  
 
 
  =
Se
nsK f
=
25.2
ns 1.468( )
Solving for ns gives ns=1.44. Therefore, the component will fail, and it will fail first at the hole.
7.22 The flat bar shown in sketch g is made of cold-drawn, high carbon steel (AISI 1080). The
cyclic, nonzero, mean axial load variation is from a minimum of 2kN to a maximum of
10kN. Using the modified Goodman failure theory, determine the safety factors for the
hole, the fillet and the groove. Also, indicate where the flat bar will first fail.
Notes: The hole, notch and fillet must each be analyzed separately. One must obtain the fatigue
stress concentration factors from the theoretical stress concentration factor, the notch sensitivity
and Equation (7.19). The endurance limit is obtained from Equation (7.7). Insufficient
information is given about the part to calculate any correction factors for Equation (7.16) so Se is
taken as Se
’. Equation (7.28) gives the safety factor.
Solution:
First of all, note that for AISI 1080 steel, Su=615MPa and Sy=380MPa.
Se
’=0.45Su=0.45(615MPa)=277MPa. The surface finish correction factor is, from Equation (7.21)
and data from Table 7.3 on page 274,
k f = eSut
f = 4.51( ) 615( )−0.265 = 0.82
www.elsolucionario.net
Page 7-17
Also, because the loading is axial, ks=1. Insufficient information is available to consider other
fatigue strength modification factors, so they will be ignored in this problem. Therefore, from
Equation (7.16),
Se=kfksSe
’=(0.82)(1)(277MPa)=227MPa
Consider the hole, groove and the fillet separately. Assume that the stress concentrations are far
enough apart that they can be taken independently.
I. Hole
For the hole, d/b=5mm/35mm=0.143,so from Figure 6.2(a) on page 222, Kc=2.6. From Figure
7.6 on page 272, qn=0.82. Therefore, from Equation (7.19),’
K f = 1 + Kc − 1( )qn = 1 + 2.6 − 1( ) 0.82( ) = 2.312
The maximum and minimum stresses are:
σmax =
Pmax
h w − d( )
=
10kN
0.005m( ) 0.035m − 0.005m( )
= 66.7MPa
σmin =
Pmin
h w − d( )
=
2kN
0.005m( ) 0.035m −0.005m( )
= 13.3MPa
Therefore, the alternating and mean stresses are, from Equations (7.3) and (7.1), respectively:
σa =
σmax − σmin
2
=
66.7MPa − 13.3MPa
2
= 26.67MPa
σm =
σmax + σmin
2
=
66.7MPa + 13.3MPa
2
= 40MPa
Now note the following:
Sy −
Se
K f
 
 
 
 
 
 
 
 
1 − Se
K fSu
 
 
 
 
 
 
 
 
=
380MPa − 227MPa
2.312
 
  
 
  
1 − 227MPa
2.312( ) 615MPa( )
 
 
  
 
 
  
= 335MPa
Since 0<σm<335, then the proper failure equation from Table 7.6 on page 285 is:
σmax + σm
Se
K f Su
− 1
 
 
 
 
 
 
 
 =
Se
nsK f
= 66.7MPa + 40MPa( ) 227MPa
2.312( ) 615MPa( )
−1
 
 
  
 
 
  =
227MPa
ns 2.312( )
or ns=2.97.
II. Fillet
For the fillet, r/h=0.160 and H/h=1.4, so from, so from Figure 6.3(a) on page 223, Kc=1.75. From
Figure 7.6 on page 272, qn=0.84. Therefore, from Equation (7.19),
Kf=1+(Kc-1)qn=1+(1.75-1)(0.84)=1.63
The maximum and minimum stresses are:
σmax =
Pmax
hw
=
10kN
0.005m( ) 0.025m( )
= 80MPa
σmin =
Pmin
hw
=
2kN
0.005m( ) 0.025m( )
= 16MPa
Therefore, the alternating and mean stresses are, from Equations (7.3) and (7.1), respectively:
σa =
σmax − σmin
2
=
80MPa −16MPa
2
= 32MPa
σm =
σmax + σmin
2
=
80MPa + 16MPa
2
= 48MPa
Now note the following:
www.elsolucionario.net
Page 7-18
Sy −
Se
K f
 
 
 
 
 
 
 
 
1 − Se
K fSu
 
 
 
 
 
 
 
 
=
380MPa − 227MPa
1.63
 
  
 
  
1 − 227MPa
1.63( ) 615MPa( )
 
 
  
 
 
  
= 311MPa
Since 0<σm<311, then the proper failure equation from Table 7.6 on page 285 is:
σmax + σm
Se
K f Su
− 1
 
 
 
 
 
 
 
 =
Se
nsK f
= 80MPa + 48MPa( ) 227MPa
1.63( ) 615MPa( )
− 1
 
 
  
 
 
  =
227MPa
ns 1.63( )
or ns=3.25.
III. Groove
For the groove, r/h=0.103 and H/h=1.2, so from, so from Figure 6.4(a) on page 224, Kc=2.3.
From Figure 7.6 on page 272, qn=0.83. Therefore, from Equation (7.19),
Kf=1+(Kc-1)qn=1+(2.3-1)(0.83)=2.08
Using the same approach as above the maximum and minimum stresses are σmax=68.97MPa,
σmin=13.79MPa, so that σa=27.59MPa and σm=41.38MPa. Now note the following:
Sy −
Se
K f
 
 
 
 
 
 
 
 
1 − Se
K fSu
 
 
 
 
 
 
 
 
=
380MPa − 227MPa
2.08
 
  
 
  
1 − 227MPa
2.08 615MPa( )
 
 
  
 
 
  
= 329MPa
Since 0<σm<329MPa, then the proper failure equation from Table 7.6 on page 285 is:
σmax + σm
Se
K f Su
− 1
 
 
 
 
 
 
 
 =
Se
nsK f
= 80MPa + 48MPa( ) 227MPa
2.08( ) 615MPa( )
− 1
 
 
  
 
 
  =
227MPa
ns 2.08( )
or ns=2.70.
7.23 A 10,000lbm elevator is supported by a stranded steel cable having a 2.5in2 cross section
and an effective modulus of elasticity of 12x10
6
 psi. As the elevator is descending at a
constant velocity of 400 ft/min, an accident causes the cable, 60 ft above the elevator, to
suddenly stop. Determine the static elongation of the cable, the impact factor, the
maximum elongation, and the maximum tensile stress developed in the cable.
Notes: The elongation is obtained from Equation (4.25). The impact factor is obtained from
Equation (7.46), which also gives the maximum deflection and force.
Solution:
Note that v=400ft/min=80in/s. The deflection is given by Equation (4.25) as:
δ = PL
AE
=
10,000 lb( ) 60 ft( ) 12in / ft( )
2.5 in2( ) 12Mpsi( ) = 0.24 in
The impact factor is obtained from Equation (7.44), but first it must be realized that Equation
(7.44) was derived using a dropped weight as the energy input, while in this case the input energy
is the kinetic energy associated with motion of the elevator car. Therefore, to apply Equation
(7.44), use
www.elsolucionario.net
Page 7-19
magh =
1
2
mv2 ; h = v
2
2g
= 80in / s( )
2
2 32.2 ft / s2( ) 12in /1 ft( ) = 8.28in
Therefore,
Im = 1+ 1 +
2h
δ
=1+ 1+ 2 8.28in( )
0.24in
= 9.37
The maximum deflection and load can also be obtained from Equation (7.46):
Im =
δmax
δ
; δmax = Imδ = 9.37( ) 0.24in( ) = 2.25in
Im =
Pmax
W
; Pmax = ImW = 9.37( ) 10,000 lb( ) = 93.7kip
Therefore, the maximum stress is:
σmax =
Pmax
A
= 93.7 kip
2.5in2
= 37.5ksi
7.24 A person is planning a bungee jump from a 40m high bridge. Under the bridge is a river
with crocodiles, so the person does not want to be submerged into the water. The rubber
rope fastened to the ankles has a spring constant of 3600N divided by the length of the
rubber rope. Thus one meter length of rope will have a spring constant of 3600 N/m. The
distance from the ankles to the top of the head is 175 cm, and the person weighs 80kg.
Calculate how long the rope should be.
Notes: Equation (7.42) is used to solve this problem. The maximum deflection is calculated so
that the person does not enter the water.
Solution:
The goal is to maximize the excitement without being eaten by crocodiles. Therefore, h=40m, and
so that the person doesn’t enter the water,
lr+dmax+lp=h
Where lr is the rope length, δmax is the maximum rope elongation, and lp is the person’s height.
Therefore,
40m=lr+δmax+1.75m; δmax=38.25-lr.
The weight of the person is 80kg(9.81m/s
2
)=784.6N. The stiffness of the rope is k=3600N/m/lr.
Substituting all of these variables into Equation (7.42) gives
δmax = 38.25m − lr =
W
k
+ W
k
 
  
 
  
2
+ 2 W
k
 
  
 
  h =
784.6N
3600N / lr( )
+ 784.6N
3600N / lr( )
 
 
  
 
 
  
2
+ 2 784.6N
3600N / lr( )
 
 
  
 
 
  40m( )
This is solved numerically as lr=16.96m or 60.07m. Since h is only 40 m, the only answer which
makes sense is lr=16.96m. Therefore, dmax=38.25-lr=21.29m.
7.25 A toy with a bouncing 50-mm diameter steel ball has a compression spring with a spring
constant of 100,000N/m. The ball falls from a 3m height down onto the spring, which can
be assumed to be weightless, and bounces away and lands in a hole. Calculate the
maximum force on the spring and the maximum deflection during the impact. The steel
ball density is 7840kg/m
3
.
www.elsolucionario.net
Page 7-20
Notes: The weight of the ball is calculated from its volume and density, and then Equation (7.42)
solves the problem.
Solution:
The weight of the ball is the product of its volume and density, or
W = 4
3
πr3
 
  
 
  ρg( ) =
4
3
π 0.025m( )3   
 
  7840kg / m
3( ) 9.81m / s2( ) = 5.03N
Therefore, Equation (7.42) gives:
δmax =
W
k
+ W
k
 
  
 
  
2
+ 2 W
k
 
  
 
  h =
5.03N
100,000 N / m
+ 5.03N
100,000 N / m
 
  
 
  
2
+ 2 5.03N
100,000 N / m
 
  
 
  
3m( )
or δmax=0.01743m=17.43mm. The maximum force is given by Equation (7.45) as:
Pmax=kδmax=(100,000N/m)(0.01743m)=1743N.
7.26 At a building site a 1-ton container
hangs from a crane in a wire and is
then placed on the floor so that the
wire becomes unloaded. The container
is pushed to the elevator shaft where it
is lowered as shown in sketch h. By
mistake there is a 1-m slack in the wire
from the crane when the container falls
into the elevator shaft. Calculate the
maximum force in the wire if it has a
cross-sectional steel surface of 500mm
2
and an effective modulus of elasticity
of 70GPa and is 25m long from the
crane to the container.
Notes: This problem is very similar to Problem 7.25. Equation (7.42) is used to calculate the
maximum deflection.
Solution:
The spring constant for the wire is given by:
k = AE
l
=
500mm2( ) 1m / 1 0 0 0mm( )2 70GPa( )
25m
= 1.4MN / m
The weight is W=(1000kg)(9.807m/s
2
)=9807N. Therefore, Equation (7.42) gives:
δmax =
W
k
+ W
k
 
  
 
  
2
+ 2 W
k
 
  
 
  h =
9807N
1.4MN / m
+ 9807N
1.4MN / m
 
  
 
  
2
+ 2 9807N
1.4MN / m
 
  
 
  
25m( )
or dmax=0.1256m=125.6mm. The maximum force is obtained fromEquation (7.45) as:
Pmax=kδmax=(1.4MN/m)(0.1256m)=175.8kN.
7.27 Modern kitchen drawers have small rubber springs mounted onto the sides of the inside
of the front plate to take up the force and stop the drawer when it is being closed. The
www.elsolucionario.net
Page 7-21
spring constant for each of the two rubber springs is 400kN/m. The drawer is full of
cutlery, which weighs 5 kg, and is closed with a speed of 0.5m/s, Calculate the maximum
force in each rubber spring if the drawer itself weighs 1 kg and
a) The cutlery is in a container that is fixed in the drawer so that it moves with the
drawer.
b) The cutlery is in a plastic container that can slide 8 cm with a coefficient of friction of
0.25 inside the drawer.
Notes: Part (a) can be solved by equating the kinetic energy to the stored energy in the spring, and
then applying Equation (7.45). In part (b), one must make certain the container is stopped by
friction before it hits the end of the drawer. The maximum force is then the friction force plus the
acceleration force for the empty drawer.
Solution:
I. Fixed Load
If the cutlery is fixed in the drawer, the mass which is to be stopped is 6kg. Equating the energy
stored in the spring to the kinetic energy, and recalling that there are two springs,
mau
2
2
= 2kδmax
2
2
; δ max =
mau
2
2k
The maximum force is then, from Equation (7.45),
P = kδmax = k
mau
2
2k
= mau
2k
2
=
6kg( ) 0.5m / s( )2 400kN / m( )
2
= 548N
II. Sliding Load
First it must be determined if the container is stopped by friction or if it hits the end of the drawer.
The work done against friction is magµl. Equating this to the kinetic energy gives the velocity
where the container strikes the back of the drawer as:
maumax
2
2
= ma gµl ;umax = 2gµl = 2 9.807m / s2( ) 0.25( ) 0.08m( ) = 0.626m / s
Therefore, since the drawer is closed at only 0.5m/s, the cutlery is stopped by friction. Therefore,
the maximum force in the rubber spring is the acceleration force for the empty drawer plus the
friction force from the cutlery. The acceleration force from the empty drawer is:
δa =
mau
2
2k
=
1kg( ) 0.5m / s( )2
2 400kN / m( ) = 5.59 ×10
−4m
The friction force is Fµ=µmg=(0.25)(5kg)(9.807m/s
2
)=12.26N. The deflection associated with
friction is then
δµ =
Fµ
2k
= 12.26N
2 400kNm( ) = 1.532 ×10
−5m
The total deflection is then δ=δa+δµ=5.59x10
-4
m+1.532x10
-5
m=5.743x10
-4
m. From Equation
(7.45),
Pmax=kδ=(400kN/m)(5.743x10
-4
m)=229.7N
7.28 Car doors are easy to slam shut but difficult to press shut by hand force. The door lock
has two latches, the first easily engaged, and the second requiring the rubber seal around
the door to be quite compressed before it can engage. The rubber seal gives a spring
constant of 50,000N/m at the lock position for the door, and the mass moment of inertia
for the door around its hinges is 2.5kg-m
2
. The distance between the lock and the hinges
www.elsolucionario.net
Page 7-22
is 1m. Calculate what force is needed to press the car door shut at the lock if a speed of
0.8m/s at the lock slams it shut.
Notes: This problem is similar to Problem 7.27. The deflection is obtained by equating kinetic
energy to the stored energy in the spring. Equation (7.45) gives the maximum force.
Solution:
The kinetic energy of the swinging door should overcome the elastic energy stored in the
compressed rubber seal. The kinetic energy of the rotating mass is Jω2/2 where J is the mass
moment of inertia and ω is the angular velocity. Equating this to the elastic energy in the rubber
seal:
Jω 2
2
≥ kδ max
2
2
; δmax =
Jω2
k
=
2.5kgm2( ) 0.8m / s1m
 
  
 
  
2
50,000N / m
= 0.00566m = 5.66mm
The force needed to overcome the spring for this deflection is obtained from Equation (7.45):
Pmax=kδmax=(50,000N/m)(0.00566m)=282.8N
www.elsolucionario.net
Page 8-1
Chapter 8
Lubrication, Friction and Wear
8.1 Describe the difference between conformal and nonconformal surfaces.
Solution:
Conformal surfaces fit snugly into each other, and they have very large contact areas.
Nonconformal surfaces have very small contact areas. At zero load, the contact area in a
conformal contact is still large, while in a non-conformal surface the contact at zero load is a
single point. The load carrying area in a conformal surface is relatively constant, but it increases
significantly with load for nonconformal surfaces.
8.2 Determine which of the following contact geometries is conformal and which is
nonconformal.
a) Meshing gear teeth
b) Ball and inner race of a ball bearing
c) Journal bearing
d) Railway wheel and rail contact
e) Car making contact with the road
f) egg and egg cup
g) Eye and socket
Notes: This problem requires an understanding of conformal and nonconformal surfaces
described in section 8.2 on p. 309.
Solution:
Conformal surfaces: journal bearing, egg and egg cup, eye and socket.
Nonconformal surfaces: meshing gear teeth, ball and inner race of a ball bearing, railway wheel
and rail contact, car making contact with the road.
8.3 Describe three applications for each of the four lubrication regimes: hydrodynamic
lubrication, elastohydrodynamic lubrication, boundary lubrication and partial lubrication.
Notes: The answers to this problem can be very different from those listed here. This assignment
is an excellent opportunity for a literature search.
Solution:
Examples of hydrodynamic lubrication include: journal bearing for a crankshaft in a car engine,
tilting pad bearings in hydroelectric power plants, magnetic head separated from a computer hard
disk, water skis, air hockey pucks, etc.
Examples of elastohydrodynamic lubrication include ball bearings, gear contacts, human
hip joints, bowling balls on lubricated lanes, etc.
Examples of boundary lubrication include rubbing bearings, door hinges, MEMS devices,
metal forming operations such as coining, etc.
Examples of partial lubrication include most metal forming operations, wet clutches
surfaces, artificial hips, lightly greased frying pan, etc.
www.elsolucionario.net
Page 8-2
8.4 A hydrodynamic journal bearing is loaded with a normal load W and rotating with a
surface velocity ub. According to Equation (8.1) find how much higher rotational speed
the bearing needs to maintain the same minimum film thickness if the load W is doubled.
Solution:
Equation (8.1) gives:
hmin( )HL ∝
ub
W
 
  
 
  
1 / 2
Therefore, if the load is doubled, the speed must be doubled to maintain the same film thickness.
8.5 A ball bearing lubricated with a mineral oil runs at 3000 rpm. The bearing is made of
AISI 52100 steel and is loaded to a medium-high level. Find how much thinner the oil
film will be if the load is increased 10 times. How much must the speed be increased to
compensate for the higher load while keeping the oil film thickness constant?
Notes: Equations (8.3) and (8.4) are used to solve this problem.
Solution:
If the load is increased tenfold, then W2=10W1. Therefore, from Equation (8.3)
hmin1
hm i n 2
= W1
W2
 
 
  
 
 
  
−0.073
= 1
10
 
  
 
  
−0.073
=1.183
or hmin1=1.183hmin2. If the speed is to be increased, and the new film thickness is the same as the
original case, then applying Equation (8.4),
hm i n 2
hmin1
= 1
1.183
= u2
u3
 
 
  
 
 
  
0.68
;
u2
u3
= 0.78
or the new speed has to be (1/0.78)=1.28
8.6 Two bearing types, one hydrodynamic journal bearing and one ball bearing, operate with
a film parameter of 8.3. Calculate how much lower the speed for each bearing has to be
to decrease the film parameter to 5.2.
Notes: The surface roughnesses do not change, so from Equation (8.8), the ratio of the film
parameters equals the ratios of film thicknesses. This then allows application of Equations (8.1)
and (8.4) to solve the problem.
Solution:
Note from Equation (8.8) that the film parameter for these cases is proportional to the film
thickness. Therefore, for either bearing,
Λ1
Λ2
= 8.3
5.2
= hmin1
hmin2
For the hydrodynamic journal bearing, Equation(8.1) is applicable. Therefore, hminαub
1/2
.
Substituting into the expression above yields
www.elsolucionario.net
Page 8-3
8.3
5.2
= hmin1
hmin2
= ub1
0.5
ub2
0.5 ; ub2 =
5.2
8.3
 
  
 
  
2
ub1 = 0.3925ub1
For a ball bearing, where elastohydrodynamic lubrication occurs, Equation (8.4) is applicable.
Therefore, hminαub
0.68
. Substituting,
8.3
5.2
= hmin1
hmin2
= ub1
0.68
ub2
0.68 ; ub2 =
5.2
8.3
 
  
 
  
1 / 0 . 6 8
ub1 = 0.5028ub1
8.7 A machined steel surface has roughness peaks and valleys. The roughness wavelength is
such that 400 peaks and 400 valleys are found during one roughness measurement. Find
how much the arithmetic average surface roughness Ra=0.1µm will change if one of the
surface roughness peaks and one of the valleys increases from 0.25 to 2.5µm.
Notes: Equation (8.5) is used to solve this problem. This solution is fairly simple, but a numerical
investigation with a prescribed surface function (sine wave, sawtooth, square wave, etc.) and
more measurement points on the interval could be performed.
Solution:
Equation (8.5) gives, for the original trace,
Ra1 =
1
N
z i
i=1
N
∑ = 1
800
zi
i=1
800
∑ = 0.1 µm
For the second case,
Ra2 =
1
N
zi
i=1
N
∑ = 1
800
zi
i=1
800
∑ = 1
800
z i
i=1
798
∑ + 2.5µm + 2.5µm
 
 
  
 
 
  =
1
800
800Ra1 + 5µm − 0.25µm − 0.25µm( )
Or Ra2=0.106µm.
8.8 A surface with a triangular sawtooth roughness pattern has a peak-to-valley height of
5µm. Find the Ra value and the Rq value.
Notes: This could be an interesting computer assignment, where the Ra and Rq values as a
function of the number of measurements could be compared to the exact values calculated here.
This solution uses Equations (8.5) and (8.6) in integral form.
Solution:
If there are an infinite number of samples, then Equation (8.5) can be written as an integral
instead of a summation. Therefore,
Ra =
1
N
zi
i=1
N
∑ = 1
l
z dx
0
l
∫
For a sawtooth shape, if the absolute value of the curve is to be considered, one can simplify the
integral by recognizing the integral can be broken into four equal parts. Therefore, we can write:
Ra =
1
l
z dx
0
l
∫ = 4l
4 2.5µm( )
l
xdx
 
 
 
 
 
 0
l /4∫ = 40µm
l2
1
2
x2
 
  
 
  0
l /4
= 20µm
l2
l2
16
 
 
  
 
 
  =1.25µm
Similarly, Equation (8.6) can be written as:
www.elsolucionario.net
Page 8-4
Rq =
1
N
zi
2
i=1
N
∑
 
 
  
 
 
  
1 / 2
= 1
l
z2dx
0
l
∫
 
 
 
 
 
 
 
 
1 / 2
For a triangular shape, this becomes
Rq =
1
l
z2dx
0
l
∫
 
 
  
 
 
  
1 / 2
= 4
l
10µm
l
x
 
  
 
  
2
dx
0
l / 4
∫
 
 
  
 
 
  
1 / 2
= 10µm
l
 
  
 
  
4
l
1
3
x3
 
  
 
  0
l /4 
 
  
 
 
  
1 / 2
= 10µm
l
 
  
 
  
4
3l
l3
64
 
 
  
 
 
  
 
 
  
 
 
  
1 / 2
= 10µm
l
 
  
 
  
1
3
l2
16
 
 
  
 
 
  
 
 
  
 
 
  
1 / 2
= 10µm
l
 
  
 
  
l
4
1
3
 
  
 
  
1 / 2
= 1.44 µm
8.9 A precision ball bearing has a race root-mean-square surface roughness of 0.07µm and a
ball root-mean-square roughness of 0.02µm. Changing the roughness of the components
may give a higher Λ value. Determine which components to smooth if it costs equally as
much to halve the roughness of the race as it costs to halve the roughness of the balls.
Note that hmin=0.2µm.
Notes: This problem is solved through evaluation of the film parameter Λ given by Equation
(8.8).
Solution:
For the initial case, taking the ball as surface a and the race as surface b, the film parameter is
given by Equation (8.8) as:
Λ =
hmin
Rqa
2 + Rqb2
=
0.2µm
0.02µm( )2 + 0.07µm( )2
= 2.75
If the ball is smoothed to a roughness of Rqa=0.01µm with Rqb still 0.07µm, then Λ becomes
Λ =
hmin
Rqa
2 + Rqb2
=
0.2µm
0.01µm( )2 + 0.07µm( )2
= 2.83
While if the race is smoothened to Rqb=0.035µm with Rqa=0.02µm,
Λ =
hmin
Rqa
2 + Rqb2
=
0.2µm
0.02µm( )2 + 0.035µm( )2
= 4.96
Therefore, the largest film parameter occurs if the race is polished, or if the rougher surface is
polished.
8.10 The minimum film thickness in a particular application is 10µm. Assume that the surface
roughnesses of the two surfaces being lubricated are identical. What lubrication regime
would you expect the application to be operating in? Also, what is the maximum surface
roughness allowed?
Notes: An understanding of the typical surface roughness for metal parts is necessary to answer
this problem, such as the summary in Table 8.1 on page 317. The lubrication regimes are defined
on page 318. Equation (8.8) is needed to solve this problem.
Solution:
www.elsolucionario.net
Page 8-5
Referring to Table 8.1 on page 317, typical surface roughnesses for machine components
are usually less than 2µm, except for some extremely large gears. Further, based on
manufacturing processes, a surface roughness around 1 µm is plausible. Therefore, the film
parameter, given by Equation (8.8), will be larger than 5, and therefore is the hydrodynamic
lubrication regime. (Note: it is possible, but very unlikely, that the partial lubrication regime may
be the proper one.)
From the definition on page 318, the film parameter must be greater than 5 for
hydrodynamic lubrication. Therefore, from Equation (8.8), recognizing the two surfaces have the
same roughness,
Λ = hmin
Rqa
2 + Rqb2( )1 / 2
= hmin
2Rq
2( )1 / 2
; Rq =
hmin
Λ 2
= 10µm
5 2
= 1.414µm
8.11 Two equally rough surfaces with rms surface roughness Rq are lubricated and loaded
together. The oil film thickness is such that Λ=2. Find the Λ value if one of the surfaces is
polished so that Rq→0 (i.e., the surface becomes absolutely smooth).
Notes: Equation (8.8) is needed to solve this problem.
Solution:
For the first case, Equation (8.8) gives:
Λ1 =
hmin
Rqa
2 + Rqb2( )1 / 2
= hmin
Rq
2 + Rq2( )1 / 2
= hmin
2 Rq
2( )1 / 2
= 2; hmin
Rq
= 2 2
For the second case,
Λ2 =
hmin
Rqa
2 + Rqb2( )1 / 2
= hmin
Rq
2 + 0( )1 / 2
= hmin
Rq
2( )1 / 2
= hmin
Rq
= 2 2
Or Λ2=2.83. Therefore, the film parameter increases from 2 to 2.83.
8.12 Two lubricated surfaces have rms surface roughnesses Rq of 0.23 and 0.04 µm,
respectively. By using a new honing machinery either surface can be made twice as
smooth. For good lubrication, which surface roughness is it more important to decrease?
Notes: See problem 8.9.
Solution:
The film thickness is determined by geometry and operating conditions, but not roughness in the
hydrodynamic regime. Therefore, from Equation (8.8) for the original surfaces,
Λ = hmin
Rqa
2 + Rqb2
= hmin
0.23µm( )2 + 0.04µm( )2
= 4.28 µm( )−1[ ]hmin
If the 0.23µm surface was smoothened to 0.115µm,
Λ = hmin
Rqa
2 + Rqb2
= hmin
0.115µm( )2 + 0.04µm( )2
= 8.21 µm( )−1[ ]hmin
while if the 0.04µm surface was smoothened to 0.02µm,
www.elsolucionario.net
Page 8-6
Λ = hmin
Rqa
2 + Rqb2
= hmin
0.23µm( )2 + 0.02µm( )2
= 4.33 µm( )−1[ ]hmin
Therefore, it is better to smoothen the rougher surface.
8.13 The absolute viscosity of a fluid at atmospheric conditions is 6 x 10-3kgf-s/m2. Give this
absolute viscosity in
a) Reyns
b) Poise
c) Pound force-seconds per square inch
d) Newton-seconds per square meter
Notes: Table 8.2 is useful for solving this problem.
Solution:
As discussed on page 320, a reyn is a pound-force second per square inch. Using the
conversions from Table 8.2 on page 320, the viscosity in lbf-s/in
2
 is
η = 6 ×10−3kgf − s / m2( ) 1.422 ×10−3 reyn
kgf − s / m2
 
 
  
 
 
  = 8.532µreyn
In poise, the conversion is as follows:
η = 6 ×10−3kgf − s / m2( ) 9.807×103 cP
kgf − s / m2
 
 
  
 
 
  = 58.84cP = 0.5884P
A pound force-second per square inch is the same as a reyn, so the viscosity is
η=8.532lbf-s/in2.
In Ns/m
2
, the conversion is as follows:
η = 6 ×10−3kgf − s / m2( ) 9.807 cP
kgf − s / m2
 
 
  
 
 
  = 0.05884Ns / m
2
8.14 Given a fluid with an absolute viscosity η between two 1m2 surfaces spaced 1 mm apart,find how fast the surfaces will move relative to each other if a 10-N force is applied in the
direction of the surfaces when η is
a) 0.001Ns/m
2
 (water)
b) 0.100 Ns/m
2
 (a thin oil at room temperature)
c) 10.0Ns/m
2
 (syrup or cold oil)
d) 108Ns/m
2
 (asphalt)
Notes: Equation (8.11) is needed to solve this problem.
Solution:
From Equation (8.11), the velocity can be calculated as:
η = F / A
ub / h
; ub =
Fh
ηA
= 10N( ) 0.001m( )
η 1m2( ) =
0.01N / m
η
Therefore, the following can be calculated:
η (Ns/m2) ub (m/s)
0.001 10
www.elsolucionario.net
Page 8-7
0.1
10
10
8
0.1
0.001
10
-10
8.15 For a single ball rolling in a groove determine the following:
a) Contact zone dimensions
b) Maximum elastic deformations
c) Maximum pressure
The steel ball has a 10mm diameter and a 4N normal force acting on it. The ball and the
groove have a 200GPa modulus of elasticity and a Poisson’s ratio of 0.3. Assume a
6.08mm radius groove in a semi-infinite steel block.
Notes: To solve this problem, calculate the following in order: 1) the radii from Equations (8.15)
through (8.18); 2) αr, ke, F and E (Equations (8.19), (8.24), and Table 8.6); 3) E’ (Equation
(8.28); 4) Dy and Dx (Equations (8.25) and (8.26)); 5) δmax (Equation (8.27)); 6) pmax (Equation
(8.23)).
Solution:
A sketch of the ball in the groove is shown below.
Note that the direction the ball rolls is the x-direction, so we can write from the given data that
rax=ray=5mm=0.005m; rbx=∞, rby=-6.08mm=-0.00608m. Therefore, from Equation (8.17),
1
Rx
= 1
rax
+ 1
rbx
= 1
0.005m
+ 1
∞
; Rx = 0.005 m
And from Equation (8.18),
1
Ry
= 1
ray
+ 1
rby
= 1
0.005m
− 1
0.00608m
; Ry = 0.02815m
Therefore, from Equation (8.15),
1
R
= 1
Rx
+ 1
Ry
= 1
0.005m
+ 1
0.02815m
; R = 0.00425m
From Equation (8.19), the radius ratio is
αr =
Ry
Rx
= 0.02815m
0.005m
= 5.630
Therefore, from Equation (8.24),
ke = αr2/ π = 5.630( )2/ π = 3.00
Note that αr≥1, so F and E are calculated as:
 
F = π
2
+ π
2
−1
 
  
 
  lnαr =
π
2
+ π
2
−1
 
  
 
  ln 5.630( ) = 2.557
www.elsolucionario.net
Page 8-8
 
E = 1+
π
2
−1
 
  
 
  
αr
=1+
π
2
−1
 
  
 
  
5.630
= 1.101
Note from Equation (8.28) that E’ is:
′ E = 2
1−νa
2
Ea
+ 1−ν b
2
Eb
= 2
1 −ν2
E
+ 1−ν
2
E
= E
1 −ν2
= 200GPa
1 − 0.32
= 220GPa
Therefore, the contact dimensions are obtained from Equations (8.25) and (8.26) as:
 
Dy = 2
6ke
2EWR
π ′ E 
 
 
  
 
 
  
1 / 3
= 2 6 3.00( )
2 1.101( ) 4N( ) 0.00425m( )
π 220GPa( )
 
 
  
 
 
  
1 / 3
= 2.27× 10−4 m = 0.227mm
 
Dx = 2
6EWR
πke ′ E 
 
 
  
 
 
  
1 / 3
= 2 6 1.101( ) 4N( ) 0.00425m( )
π 3.00( ) 220GPa( )
 
 
  
 
 
  
1 / 3
= 7.57× 10−5 m = 0.0757mm
The maximum deformation at the contact center is given by Equation (8.27) as:
 
δmax = F
9
2ER
W
πke ′ E 
 
 
  
 
 
  
2 
 
 
 
 
 
 
 
 
 
1 / 3
= 2.557( ) 9
2 1.101( ) 0.00425m( )
4N
π 3.00( ) 220GPa( )
 
 
  
 
 
  
2 
 
 
 
 
 
 
 
 
 
1 / 3
= 3.91 ×10−7 m
or δmax=0.391µm. The maximum pressure is given by Equation (8.23) as:
pmax =
6W
πDxDy
= 6 4 N( )
π 7.57 ×10−5 m( ) 2.27 ×10−4 m( ) = 0.445GPa
8.16 A solid cylinder rolls with a load against the inside of an outer cylinder race. The solid
cylinder radius is 20mm and the race internal radius is 150mm. The race and the roller
have the same axial length. What is the radius of a geometrically equivalent cylinder near
a plane? The cylinder is made of silicon nitride (E=314GPa, ν=0.26) and the race is made
of stainless steel (E=193 GPa, ν=0.30). If the normal applied load is per unit width is
12000N/m, determine
a) Contact semiwidth
b) Maximum surface stress
c) Maximum elastic deformation
Also, indicate what these values would be if the silicon nitride cylinder were replaced
with a stainless steel cylinder.
Notes: The solution is similar to Problem 8.15, but this is a rectangular contact. The radius is
obtained from Equation (8.15) after obtaining Rx and Ry from Equations (8.17) and (8.18). The
contact semiwidth is obtained from Equation (8.29), the surface stress from (8.32), and the
maximum deformation from Equation (8.31).
Solution:
I. Geometry
For the given geometry, rax=0.020m, rbx=-0.150m, ray=rby=∞. Therefore, from Equations (8.17)
and (8.18),
1
Rx
= 1
rax
+ 1
rbx
= 1
0.020 m
− 1
0.15m
; Rx = 0.02308m
www.elsolucionario.net
Page 8-9
1
Ry
= 1
ray
+ 1
rby
= 1
∞
− 1
∞
; Ry = ∞
Therefore, from Equation (8.15),
1
R
= 1
Rx
+ 1
Ry
= 1
0.02308m
+ 1
∞
; R = 0.02308m
II. Silicon Nitride on Steel
From Equation (8.28),
′ E = 2
1− νa
2( )
Ea
+
1−ν b
2( )
Eb
= 2
1− 0.262
314GPa
+ 1− 0.3
2
207GPa
= 272GPa
The dimensionless load is given by Equation (8.30) as:
′ W = ′ w 
′ E Rx
= 12,000 N / m
272GPa( ) 0.02308m( ) = 1.912 ×10
−6
so that the semi-contact width can be calculated from (8.29) as
b* = Rx
8 ′ W 
π
 
  
 
  
1 / 2
= 0.02308 m( )
81 .912×10−6( )
π
 
 
 
  
 
 
 
  
1 / 2
= 5.093 ×10−5 m = 50.93µm
The maximum deformation is, from Equation (8.31),
δmax =
2 ′ W Rx
π
ln
2π
′ W 
 
  
 
  − 1
 
  
 
  
=
2 1 . 9 1 2×10−6( ) 0.02308 m( )
π
ln
2π
1.912 ×10−6
 
  
 
  
−1
 
 
 
 
 
 = 0.393µm
The maximum pressure is given by Equation (8.32) as:
pmax = ′ E 
′ W 
2π
 
  
 
  
1 / 2
= 272GPa( ) 1.912 ×10
−6
2π
 
 
  
 
 
  
1 / 2
= 150MPa
III. Steel on Steel
If a steel cylinder is used instead of silicon nitride, the following equations are changed:
′ E = 2
1− νa
2( )
Ea
+
1−ν b
2( )
Eb
= 2
1 − 0.32
207GPa
+ 1− 0.3
2
207GPa
= 227GPa
′ W = ′ w 
′ E Rx
= 12,000 N / m
227GPa( ) 0.02308m( ) = 2.290 ×10
−6
b* = Rx
8 ′ W 
π
 
  
 
  
1 / 2
= 0.02308 m( )
8 2.290× 10−6( )
π
 
 
 
  
 
 
 
  
1 / 2
= 55.73µm
δmax =
2 ′ W Rx
π
ln
2π
′ W 
 
  
 
  − 1
 
  
 
  
=
2 2.290 ×10−6( ) 0.02308 m( )
π
ln
2π
2.290× 10−6
 
  
 
  
−1
 
 
 
 
 
 = 0.464µm
pmax = ′ E 
′ W 
2π
 
  
 
  
1 / 2
= 227GPa( ) 2.290 ×10
−6
2π
 
 
  
 
 
  
1 / 2
= 137MPa
8.17 A 10cm diameter shaft has a 2 cm diameter ball rolling around the outside. Find the
curvature sum, the maximum contact stress, the maximum deflection, and the contact
dimensions if the ball load is 1000N. The ball is made of silicon nitride (E=314GPa,
www.elsolucionario.net
Page 8-10
ν=0.26), and the shaft is made of steel (E=206GPa, ν=0.3). Also determine the values if
both ball and shaft are made of steel.
Notes: This is similar to Problems 8.15 and 8.16.
Solution:
I. Geometry
For the given geometry, rax=ray=0.010m, rbx=0.05m, rby=∞. Therefore, from Equations (8.17) and
(8.18),
1
Rx
= 1
rax
+ 1
rbx
= 1
0.010 m
+ 1
0.05m
; Rx = 0.00833m
1
Ry
= 1
ray
+ 1
rby
= 1
0.010m
− 1
∞
; Ry = 0.010m
Therefore, from Equation (8.15),
1
R
= 1
Rx
+ 1
Ry
= 1
0.00833m
+ 1
0.010
; R = 0.00454m
From Equation (8.19),
αr =
Ry
Rx
= 0.010m
0.00833m
=1.2
and from Equation (8.24),
ke=αr
2/π
=(1.2)
2/π
=1.123
Since 1<αr<100, Table 8.6 gives
 
F = π
2
+ π
2
−1
 
  
 
  lnαr =
π
2
+ π
2
−1
 
  
 
  l n 1 . 2( ) = 1.675
 
E = 1+
π
2
−1
αr
= 1+
π
2
−1
1.2
= 1.476
II. Silicon Nitride on Steel
From Equation (8.28),
′ E = 2
1− νa
2( )
Ea
+
1−ν b
2( )
Eb
= 2
1− 0.262
314GPa
+ 1− 0.3
2
207GPa
= 272GPa
Therefore, the dimensions of the contact are obtained from Equations (8.25) and (8.26):
 
Dy = 2
6ke
2EWR
π ′ E 
 
 
  
 
 
  
1 / 3
= 2 6 1.123( )
2 1.476( ) 1000N( ) 0.00454 m( )
π 272GPa( )
 
 
  
 
 
  
1 / 3
= 7.80 ×10−4 m
 
Dx = 2
6EWR
πke ′ E 
 
 
  
 
 
  
1 / 3
= 2 6 1.476( ) 1000N( ) 0.00454( )
π 1.123( ) 272GPa( )
 
 
  
 
 
  
1 / 3
= 6.95×10−4 m
The maximum contact pressure can be calculated from Equation (8.23) as:
pmax =
6W
πDxDy
= 6 1000N( )
π 7.80 ×10−4m( ) 6.95 ×10−4m() = 3.52GPa
The maximum deflection is obtained from Equation (8.27) as:
www.elsolucionario.net
Page 8-11
 
δmax = F
9
2ER
W
πke ′ E 
 
 
  
 
 
  
2 
 
 
 
 
 
 
 
 
 
1 / 3
= 1.675( ) 9
2 1.476( ) 0.00454m( )
1000N
π 1.123( ) 272GPa( )
 
 
  
 
 
  
2 
 
 
 
 
 
 
 
 
 
1 / 3
=15.1µm
II. Steel on Steel
If the ball is changed to steel, the following equations are changed:
′ E = 2
1− νa
2( )
Ea
+
1−ν b
2( )
Eb
= 2
1 − 0.32
207GPa
+ 1− 0.3
2
207GPa
= 227GPa
 
Dy = 2
6ke
2EWR
π ′ E 
 
 
  
 
 
  
1 / 3
= 2 6 1.123( )
2 1.476( ) 1000N( ) 0.00454 m( )
π 227GPa( )
 
 
  
 
 
  
1 / 3
= 8.28×10−4 m
 
Dx = 2
6EWR
πke ′ E 
 
 
  
 
 
  
1 / 3
= 2 6 1.476( ) 1000N( ) 0.00454( )
π 1.123( ) 227GPa( )
 
 
  
 
 
  
1 / 3
= 7.38×10−4 m
pmax =
6W
πDxDy
= 6 1 0 0 0N( )
π 8.28 ×10−4m( ) 7.38×10−4m( ) = 3.13GPa
 
δmax = F
9
2ER
W
πke ′ E 
 
 
  
 
 
  
2 
 
 
 
 
 
 
 
 
 
1 / 3
= 1.675( ) 9
2 1.476( ) 0.00454m( )
1000N
π 1.123( ) 227GPa( )
 
 
  
 
 
  
2 
 
 
 
 
 
 
 
 
 
1 / 3
=17.0µm
8.18 The ball-outer-race contact of a ball bearing has a 17mm ball diameter, an 8.84-mm outer
race groove radius, and a 44.52-mm radius from the bearing axis to the bottom of the
groove. The load on the ball is 25,000N. Calculate the dimensions of the contact ellipse
and the maximum deformation at the center of contact. The race and the ball are made of
steel.
Notes: The problem statement means that rax=8.5mm, rbx=-44.52mm, ray=8.5mm, rby=-8.84mm.
Then using Equations (8.15) through (8.21) and *8.25) through (8.27) gives the solution.
Solution:
Note that rax=8.5mm, rbx=-44.52mm, ray=8.5mm, rby=-8.84mm. Also, for steel it is known from
the inside front cover that E=207GPa and ν=0.30. Therefore, from Equation (8.28),
′ E =
2
1− νa
2
Ea
+ 1− ν b
2
Eb
=
2
2
1− ν 2
E
 
 
  
 
 
  
=
E
1 − ν2
=
207GPa
1 − 0.302
= 227.5GPa
From Equations (8.17) and (8.18):
1
Rx
=
1
rax
+
1
rbx
=
1
0.0085m
−
1
0.04452m
; Rx = 0.0105m
1
Ry
=
1
ray
+
1
rby
=
1
0.0085m
−
1
0.00884m
; Ry = 0.221m
Therefore, from Equation (8.15),
1
R
=
1
Rx
+
1
Ry
=
1
0.0105m
+
1
0.221m
; R = 0.0100m
www.elsolucionario.net
Page 8-12
Note that from Equation (8.19) that αr=Ry/Rx=0.221/0.0105=21.035. From Equation (8.24),
ke = αr2/ π = 21.035( )2 /π = 6.954
From Table 8.6 on page 331, and noting that 1<αr<100,
 
F =
π
2
+
π
2
−1
 
  
 
  
lnαr =
π
2
+
π
2
−1
 
  
 
  
ln21.035 = 3.310
 
E = 1+
π
2
−1
 
  
 
  
αr
=1 +
π
2
−1
 
  
 
  
21.035
= 1.0271
From Equation (8.25),
 
Dy = 2
6ke
2EWR
π ′ E 
 
 
  
 
 
  
1 / 3
= 2
6 6.954( )2 1.0271( ) 25000N( ) 0.01m( )
π 227.5GPa( )
 
 
  
 
 
  
1 / 3
= 0.00942m = 9.42mm
From Equation (8.20), (Note: Equation (8.26) gives the same answer)
ke =
Dy
Dx
; Dx =
Dy
ke
=
9.42mm
6.954
= 1.35mm
Finally, from Equation (8.27), the maximum deflection is
 
δmax = F
9
2ER
W
πke ′ E 
 
 
  
 
 
  
2 
 
 
 
 
 
 
 
 
 
1 / 3
= 3.310( ) 9
21.0271( ) 0.01m( )
25000N
π 6.954( ) 227.5GPa( )
 
 
  
 
 
  
2 
 
 
 
 
 
 
 
 
 
1 / 3
or δmax=73.7µm.
8.19 A rectangular box stands on an inclined plane with the slope angle α. The center of
gravity is in the geometric center of the box, which has a height h and length l. Find the α
at which the box starts to slide down.
Notes: This problem required only force equilibrium to be solved. Although this can be done in
the vertical and horizontal directions, it is more easily derived by considering force equilibrium
perpendicular and tangent to the surface.
Solution:
The free body diagram of the box on the plane is sketched below.
Consider force equilibrium perpendicular to the surface:
ΣFy’=0=-magcosα+N=0; N=magcosα
Force equilibrium along the angled surface gives:
Fα = 0 = −magsinα + µN∑ = −mag sinα + µ magcosα( ); µ = magsinαmag cosα
= tan α
Therefore, the block starts to slide as soon as α>tan-1µ.
www.elsolucionario.net
Page 8-13
8.20 Find the ratio h/l when the box in Problem 8.19 falls over rather than slides down, and
determine how high the coefficient of friction then has to be.
Notes: This solution uses the results from Problem 8.19.
Solution:
For the box to fall over, the vertical gravitational force mag must
be directed outside the bottom surface of the box. Therefore, the
box falls over if (see sketch):
tan α ≥ l /2
ht / 2
= l
ht
From Problem 8.19, the box will slide if µ<tanα. Therefore, for
the box to tip,
µ > tanα = l
ht
The box falls over if the ratio ht/l is greater than cotα, but slides (doesn’t tip) if the coefficient of
friction is less than tanα.
8.21 A polymer box stands on a slope. The box weighs 10kg and the slope angle is 25°. To
pull the box down the slope, a force F=5N is needed. The friction force is proportional to
the load raised to the 0.75. Find the extra weight needed in the box to make it slide down
the slope.
Notes: The non-linear proportionality between friction force and normal load is typical for soft
polymers. This solution needs only equilibrium. Consideration of the unloaded (10kg) box allows
determination of the proportionality constant between friction and force to the 0.75 power. Using
a load in the box then allows determination of the critical angle.
Solution:
Note that the friction relationship is F=CN
0.75
, where C is a
constant. A free body diagram of the box is shown.
Taking force equilibrium perpendicular to the surface gives:
ΣFp=0=N-magcos25°; N=magcos25°
Taking force equilibrium tangent to the surface gives:
F∑ t = 0 = −mags i n 2 5° −F + CN0.75 ; C =
mags i n 2 5° +F
N 0.75
Substituting for N and solving,
C =
mags i n 2 5°+ F
mag cos25°( )0.75
=
10kg( ) 9.807m / s2( )sin 25°+ 5N
10kg( ) 9.807m / s2( ) cos25°[ ]0.75
=1.60 N / N 0.75
If the weight in the box is unknown, so the weight of the box plus additional material is denoted
as W. For the box to slide down the plane, F=0, so that the force equilibrium equations give:
ΣFp=0=N-Wcos25°; N=Wcos25°
F∑ t = 0 = −W s i n 2 5° + 1.6N / N0.75( )N0.75 ; W =
1.6N / N0.75( )N 0.75
sin25°
www.elsolucionario.net
Page 8-14
Substituting for N gives an equation for W:
W =
1.6N / N0.75( ) W cos25°( )0.75
s i n 2 5°
; W 0.25 =
1.6 N0.25( ) c o s 2 5°( )0.75
sin25°
= 3.52N0.25
Therefore W=153N. Since the box itself weighs (10kg)(9.81m/s
2
)=98.1N, the extra weight of
material needed is 153N-98.1N=54.9N, or 5.60kg.
8.22 A cross-country skier who waxes the central part of the skis (under the foot) with a high-
friction wax can run up a 5.7° slope without slipping backward. The front and the back
parts of the skis are waxed with a low-friction wax having a 0.01 coefficient of friction.
Determine the downhill slope needed to continue sliding at constant speed.
a) If the skier is so light that only the low-friction parts of the skis contact the snow.
b) If the skier is so heavy that half the weight is carried by the central part of the skis.
Notes: It was shown in Problem 8.19 that sliding is imminent when µ=tanα, where α is the angle
of inclination. This result is used along with force equilibrium to solve this problem.
Solution:
For the high friction wax, we can use the results of Problem 8.19 to get the coefficient of friction
as:
µ = tanα = t a n 5 . 7° = 0.100
For the light skier, where only the low friction parts of the skis contact the snow, the same
equation predicts the slope where sliding is imminent as:
µ = tanα ; α = tan−1 µ( ) = tan−1 0.01( ) = 0.57°
For the heavy skier, the friction force is
F = 0.1( )W
2
+ 0.01( )W
2
= 0.055W
where W is the weight of the skier. Therefore, the coefficient of friction is µ=F/W=0.055, so the
critical slope is:
µ = tanα ; α = tan−1 µ( ) = tan−1 0.055( ) = 3.15°
8.23 A moose suddenly jumps out onto a dry asphaltroad 80m in front of a car running at 108
km/hr. The maximum coefficient of friction between the rubber tires and the road is 1.0
when they just start to slide and 0.8 when the locked wheels slide along the road. It takes
1s for the driver to apply the brakes after the moose jumps onto the road. Find whether
the car can stop before it hits the moose
a) With locked wheels sliding along the road
b) With an automatic braking system (ABS) that keeps the sliding speed so low that
maximum friction is maintained.
Notes: There are a number of ways of solving this problem. This solution considers the distance
needed to dissipate the initial kinetic energy through friction.
Solution
The velocity of the car is v=108km/hr=30m/s. Since it takes one second before the brakes are
applied, the car travels 30m before it begins to slow down. Therefore, the car must be able to stop
in 80m-30m=50m in order to avoid hitting the moose. The initial kinetic energy in the car is
www.elsolucionario.net
Page 8-15
KE =
1
2
mav
2 =
1
2
ma 30m / s( )2 = 450m2 / s2( )ma
The energy dissipated in friction is the product of the frictional force and the distance. For a
distance of 50m, the energy dissipated for locked wheels is
E=Fd=µmagd=(0.8)ma(9.807m/s
2
)(50m)=392m
2
/s
2
ma
This energy is less than the initial kinetic energy in the car. Therefore, the moose is hit. If an ABS
system is present so that µ=1,
E=Fd=µmagd=(0.8)ma(9.807m/s
2
)(50m)=490m
2
/s
2
ma
Since this is greater than the initial kinetic energy in the car, we conclude that the car stops short
of the moose.
8.24 Given a block on an incline as shown in sketch a, find the force required
a) To prevent motion downward
b) To cause motion upward
The coefficient of friction is 0.30. Draw a free body diagram for both situations showing
the forces involved.
Notes: Because of the direction of the applied force P, it is convenient in this case to perform
force equilibrium in the vertical and horizontal directions.
Solution:
I. Prevention of Downward Motion
The free body diagram for this case is as shown. The figure
already has applied the rule that F=µN for the friction force on
the surface. Note that the friction force opposes relative motion.
Applying vertical force equilibrium,
Fy = 0 = −600lb + Nc o s 3 5°+ µNsin 35°; N =
600 lb
cos35° + 0.3( )sin35°
= 605lb∑
Applying horizontal force equilibrium and solving for P gives:
Fx = 0 = P − N s i n 3 5°+ µN cos35°∑
P = N sin35° − µc o s 3 5°( ) = 605 lb( ) s i n 3 5°− 0.3cos35°( ) = 198lb
II. Initiation of Upward Motion
The free body diagram is as shown. Note that the friction force,
which opposes relative motion, is applied in the opposite
direction as in part I. Applying vertical force equilibrium,
Fy = 0 = −600lb + Nc o s 3 5°− µNsin 35°; N =
600 lb
cos35° − 0.3( )sin35°
= 927lb∑
www.elsolucionario.net
Page 8-16
Applying horizontal force equilibrium and solving for P gives:
Fx = 0 = P − N sin35° −µN cos35°∑
P = N sin35° + µc o s 3 5°( ) = 927lb( ) sin35° +0 . 3 c o s 3 5°( ) = 760lb
8.25 Three equal blocks of plastic are carrying the weight of a steel slider. At first each block
carries one-third of the weight, but the central block is made of polytetrafluoroethylene
(PTFE) and the outer two blocks are made of polyethylene, so they have different
adhesive wear constants (PTFE, k1=2x10
-5
; polyethylene, k1=2x10
-8
). Determine how the
load redistributes between the blocks if the hardnesses of the plastics are assumed to be
the same.
Notes: The load redistributes until the wear rate is the same for the two materials. The key
equation is (8.44).
Solution:
The load will redistribute until the wear rate is constant. If the wear volume is given by Equation
(8.44) as
v = k1
WL
3H
then the wear rate is
∂v
∂t
=
∂
∂t
k1
WL
3H
 
  
 
  
=
k1W
3H
∂L
∂t
=
k1Wu
3H
where u is the sliding velocity. For the wear rate of polyethylene (PE) to equal the wear rate of
polytetrafluoroethylene, (PTFE), and noting HPE=HPTFE=H and u=constant,
k1,PEWPEu
3H
=
k1,PTFEWPTFEu
3H
;WPE = WPTFE
k1,PTFE
k1,PE
= WPTFE
2 × 10−5
2 × 10−8
= 1000WPTFE
or that the polyethylene carries 1000 times the load of the PTFE.
8.26 Given the plastic blocks in Problem 8.25 and sharing the load so that the wear is equal on
each block, find the coefficient of friction.
Notes: The coefficients of friction for the two materials are obtained from Table 8.8 on page 344.
This solution uses the results from Problem 8.25.
Solution:
From Table 8.8 on page 344, the coefficient of friction of polyethylene on steel is 0.5, while the
coefficient of friction for polytetrafluoroethylene on steel is 0.2. Recall from Problem 8.25 that
WPE=1000WPTFE. Therefore, the total normal force, which equals the total weight, is
N=2WPE+WPTFE=2001WPTFE
The friction force developed is:
F=2(0.5)WPE+(0.2)WPTFE=2(0.5)(1000WPTFE)+0.2WPTFE=1000.2WPTFE
The coefficient of friction is then
µ =
F
N
=
1000.2WPTFE
2001WPTFE
= 0.4999
Note that the polytetrafluoroethylene has a negligible effect on friction because it carries such a
small portion of the load.
www.elsolucionario.net
Page 8-17
8.27 A 50kg copper piece is placed on a flat copper surface as shown in sketch b. Assume that
copper has a hardness of 275MPa. Calculate the shear stress, the abrasive wear volume,
and the adhesive wear volume.
Notes: Using a coefficient of friction from Table 8.8, the shear stress can be calculated from
Equation (8.39). The adhesive wear volume is calculated from Equation (8.44) while the abrasive
wear volume is obtained from Equations (8.45) and (8.46).
Solution:
Note from Table 8.8 on page 344 that for copper sliding against copper, µ=1.2 and k1=0.01-0.1.
The average value of the adhesive wear constant is 0.055, which will be used in this solution.
Note that the problem illustration matches Figure 8.19, with θ=6°. Therefore, Equation (8.39)
gives:
µ =
τ
H
+
2
π
tanθ ; τ = H µ −
2
π
tanθ
 
  
 
  
= 275MPa( ) 1.2 − 2
π
tan6°
 
  
 
  
= 312MPa
Using the value of adhesive wear constant k1=0.055, and using L=7.5m as shown, the adhesive
wear volume is obtained from Equation (8.44) as:
v =
k1WL
3H
=
0.055( ) 50kg( ) 9.807m / s2( ) 7.5m( )
3 275MPa( )
= 2.45 ×10−7m3
The abrasive wear constant (to be used in Equation (8.45)) is obtained from Equation (8.46) as:
k2 =
2tan θ
π
=
2 t a n 6°
π
= 0.0669
Therefore, the abrasive wear volume is (from Equation (8.45)),
vabr =
k1k2WL
H
=
0.055( ) 0.0669( ) 50kg( ) 9.807m / s2( ) 7.5m( )
275MPa
= 4.92 ×10−8m3
www.elsolucionario.net
Page 9-1
Chapter 9
Columns
9.1 A yo-yo begins to move at the origin of the x-y axis (at point A) and is at its farthest point
from the origin at point B, as shown in sketch a. The string is 0.75m long, and its weight
is small relative to the weight of the yo-yo. Find the equilibrium position for the yo-yo.
Notes: The weight of the yo-yo causes a tension in the string so that the loaded string consists of
two lines. This is mainly an advanced geometry problem.
Solution:
The sketch below shows the yo-yo at the equilibrium position. Note that the angle α has been
indicated, and it is the same angle AOC and BOC (this can be proven by taking horizontal force
equilibrium at point O). Point B’ is defined by extending the line AO up to the x-coordinate of
point B. Note that OB=OB’, and from similar triangles, the angle AB’D=α.
The length of the string is 0.75m, so that AO+OB=0.75, and since OB=OB’, AB’=0.75m.
Therefore,
sinα =
AD
A ′ B 
=
0.6m
0.75m
= 0.8; α = 53.13°
The distance BB’ can be calculated from the Pythagorean theorem for triangle ADB’:
A ′ B ( )2 = 0.75m( )2 = AD( )2 + D ′ B ( )2 = 0.6m( )2 + 0.2m + B ′ B ( )2 ; B ′ B = 0.25m
Therefore, the y coordinate of point O is
y = 0.2m +
0.25m
2
= 0.325m
Therefore, from triangle AOC, and recognizing that the x-coordinate of point O is the distance
AC,
x = OC tanα = y tanα = 0.325m( ) tan53.13° =0.433m
www.elsolucionario.net
Page 9-2
9.2 For the yo-yo in Problem 9.1, thecentral pin has a diameter of 5mm. The outside
diameter of the yo-yo is 50mm, and the diameter of the string is 1mm, which is the same
as the width of the slot of the yo-yo. Will the yo-yo roll from point A to point B if
tightened slowly?
Notes: The solution assumes that the yo-yo weight is small enough (or the string is strong
enough) so that the string does not break under the tension. This solution uses the solution of
Problem 9.1.
Solution:
As the string is shortened, the equilibrium position shifts. The main question is whether or not the
yo-yo rolls to A or jumps (i.e., whether or not the equilibrium is stable or unstable). If the string is
taught between A and B, its length is lAB=(0.2
2
+0.6
2
)=0.632m. Therefore, the length varies from
0.75m to 0.632m. Note that the angle α as a function of string length is:
sinα =
AD
l
=
0.6m
l
; α =sin −1
0.6m
l
 
  
 
  
The length BB’ as a function of string length l is given by:
A ′ B ( )2 = l2 = AD( )2 + D ′ B ( )2 = 0.6m( )2 + 0.2m + B ′ B ( )2 ; B ′ B = l2 − 0.6m( )2 − 0.2m
The y-coordinate as a function of string length is:
y = DB +
B ′ B 
2
= 0.2m +
l2 − 0.6m( )2 − 0.2m
2
So that the x-coordinate of the yo-yo (point O) as a function of string length is:
x = y tan α = 0.2m +
l2 − 0.6m( )2 − 0.2m
2
 
 
 
  
 
 
 
  
tan sin−1
0.6m
l
 
  
 
  
 
 
 
 
 
 
This is plotted below. Note that the x-coordinate of the yo-yo follows a smooth transition, without
bifurcations, to point B.
www.elsolucionario.net
Page 9-3
Alternate Solution:
Refer to the sketch given in Problem 9.1. For stable equilibrium,
tan β >
2.5 + 0.5
25
= 0.12; β > 6.84°
The angle γ is given by
tan γ =
0.2
0.6
; γ =18 .43°
Since β<γ. we conclude that the yo-yo will roll to point B.
9.3 A person rides a bike on a horizontally flat road. Is this a neutral, stable, or unstable
equilibrium position?
Notes: See the text on page 360 for the definitions of neutral, stable and unstable equilibrium.
Also, Figure 9.1 depicts these three states.
Solution:
This can be answered in two ways, depending on the variable of interest. In terms of position, a
flat plane where gravity is the only force field is neutrally stable, as shown in Figure 9.1.
In terms of the angle between the bike and the ground it is unstable equilibrium.
9.4 A golf ball is placed
a) On top of a small hill
b) On a horizontal flat plane
c) In a shallow groove
Find the type of equilibrium for each condition.
Notes: Refer to Figure 9.1 on page 360.
Solution:
a) On top of a small hill is unstable equilibrium, because any infinitesmally small perturbation
causes the ball to roll down the hill.
b) On a flat plane is neutral equilibrium, since a small perturbation places the ball in a new
equilibrium position.
c) In a shallow groove is a stable equilibrium position, since a small perturbation results in the
ball moving back to the lowest point in the groove.
9.5 A column has pinned ends and is axially loaded in compression. The length is 6m and the
weight is 200kg, but the form of the cross section can be changed. The column is made of
steel (AISI 1040). Find the Euler buckling load for
a) A solid circular section
b) A solid square section
c) A circular tube with outer diameter of 100mm.
d) A square tube with outer dimension of 100mm.
www.elsolucionario.net
Page 9-4
Notes: The cross sectional area is determined from the weight of the column and the known
density of steel. Moments of inertia are found on the inside back cover for the different cross
sections. The Euler load is obtained from Equation (9.7).
Solution:
From the inside front cover, note that for AISI 1040 steel that E=207GPa and ρ=7850kg/m3.
From the weight of the column, the area of the cross section can be determined as follows
W = ρV = ρAL; A =
W
ρL
=
200kg
7850kg / m3( ) 6m( ) = 0.00425m
2
For both ends pinned, Table 9.1 gives le=l.
I. Solid circular section.
For the solid circular section, the radius is found from the area as:
A = πr2 ; r =
A
π
=
0.00425m2
π
= 0.0367m
The moment of inertia for a solid circle is (see inside back cover) I=πr4/4=1.44x10-6m4.
Therefore, Equation (9.7) gives
Pcr =
π 2EI
l2
=
π2 207GPa( ) 1.44 ×10−6 m4( )
6m( )2
= 81.7kN
II. Solid square section
For the solid square section, the side dimension is calculated from the area as x
2
=A; x=0.0652m.
Therefore the moment of inertia is I=x
4
/12=1.506x10
-6
m
4
. The Euler load is obtained from
Equation (9.7) as:
Pcr =
π 2EI
l2
=
π2 207GPa( ) 1.506 × 10−6m4( )
6m( )2
= 85.5kN
III. Circular Tube
If the outer diameter is 100mm=0.1m, then the inner diameter is calculated from the area as:
A =
π
4
Do
2 − Di2( ) ; Di = Do2 − 4Aπ = 0.1m( )
2 −
4 0.00425m2( )
π
= 0.0677m
The moment of inertia is then (see inside back cover),
I =
π
4
ro
4 − ri4( ) = π4
Do
2
 
  
 
  
4
−
Di
2
 
  
 
  
4 
 
  
 
 
  =
π
4
0.05m( )4 − 0.0339m( )4( ) = 3.875 × 10−6m4
So that the Euler load is
Pcr =
π 2EI
l2
=
π2 207GPa( ) 3.875 ×10−6m4( )
6m( )2
= 220kN
IV. Square Tube
If the outer dimension is 100mm=0.1m, then the inner dimension is obtained from the area as:
A = xo2 − xi2; x i = xo2 − A = 0.1m( )2 − 0.00425m2 = 0.0758m
The moment of inertia is then
I =
xo
4 − xi
4
12
=
0.1m( )4 − 0.0758m( )4
12
= 5.58 ×10−6m4
The Euler load is then
www.elsolucionario.net
Page 9-5
Pcr =
π 2EI
l2
=
π2 207GPa( ) 5.58 × 10−6 m4( )
6m( )2
= 317kN
9.6 A column with both ends pinned is 3m long and has a section with an outer diameter of
30mm and wall thickness of 5mm. Find which material given in Tables A.1 through A.4
gives the highest buckling load. Also, give the buckling loads for AISI 1090 steel,
aluminum alloy 2014, and molybdenum.
Notes: The column will be considered to be buckling elastically, since plastic buckling is covered
in later sections. Therefore, Equation (9.7) is used to solve this problem.
Solution:
The moment of inertia for the cross section is:
I =
π
64
do
4 − di4( ) = π64 0.030m( )
4 − 0.020m( )4( ) = 3.19× 10−8 m4
Since we will used Equation (9.7) to obtain the elastic buckling load, the theoretical effective
column length for two pinned ends is, from Table 9.1 on page 366, le=l=3m. Therefore, from
Equation (9.7),
Pcr =
π 2EI
l2
=
π2 3.19 × 10−8m4( )
3m( )2
E = 3.50 × 10−8 m2( )E
Therefore, the material in Tables A.1 through A.4 with the highest elastic modulus will give the
largest buckling load. Reviewing these tables, the material with the highest elastic modulus is
titanium carbide.
For AISI 1090 steel, the elastic modulus is E=207GPa. Therefore, the buckling load is:
Pcr = 3.50 ×10−8m2( )E = 3.50 × 10−8m2( ) 207GPa( ) = 7245N
For aluminum alloy 2014, E=72GPa, so that the buckling load is:
Pcr = 3.50 ×10−8m2( )E = 3.50 × 10−8m2( ) 72GPa( ) = 2520N
For molybdenum, the elastic modulus is E=324GPa, so that
Pcr = 3.50 ×10−8m2( )E = 3.50 × 10−8m2( ) 324GPa( ) =11,340 N
9.7 Determine the critical stresses for the four columns considered in Problem 9.5. Let the
column be 8m long and the cross-sectional area be 50cm
2
. The column material is AISI
1080 steel.
Notes: The critical stress is obtained from Equation (9.10). Otherwise the solution follows the
form of Problem 9.5.
Solution:
From the inside front cover, note that for AISI 1040 steel that E=207GPa. The area of the cross
section is A=50cm
2
=0.005m
2
. For both ends pinned, Table 9.1 gives le=l=8m.
I. Solid circular section.
For the solid circular section, the radius is found from the area as:
www.elsolucionario.net
Page 9-6
A = πr2 ; r =
A
π
=
0.005m2
π
= 0.0399m
The moment of inertia for a solid circle is (see inside back cover) I=πr4/4=1.99x10-6m4.
Therefore, Equation (9.7) gives
Pcr =
π 2EI
l2
=
π2 207GPa( ) 1.99 ×10−6 m4( )
8m( )2
= 63.5kN
The critical stress is given by Equation (9.10) as:
σcr =
Pcr
A
=
63.5kN
0.005m2
= 12.7MPa
II. Solid square section
For the solid square section, the side dimension is calculated from the area as x
2
=A=0.005m2
;
x=0.0707m. Therefore the moment of inertia is I=x
4
/12=2.083x10
-6
m
4
. The Euler load is obtained
from Equation (9.7) as:
Pcr =
π 2EI
l2
=
π2 207GPa( ) 2.083 ×10−6m4( )
8m( )2
= 66.5 kN
The critical stress is given by Equation (9.10) as:
σcr =
Pcr
A
=
66.5kN
0.005m2
= 13.3MPa
III. Circular Tube
If the outer diameter is 100mm=0.1m, then the inner diameter is calculated from the area as:
A =
π
4
Do
2 − Di2( ) ; Di = Do2 − 4Aπ = 0.1m( )
2 −
4 0.005m2( )
π
= 0.0603m
The moment of inertia is then (see inside back cover),
I =
π
4
ro
4 − ri4( ) = π4
Do
2
 
  
 
  
4
−
Di
2
 
  
 
  
4 
 
  
 
 
  =
π
4
0.05m( )4 − 0.0301m( )4( ) = 4.261× 10−6m4
So that the Euler load is
Pcr =
π 2EI
l2
=
π2 207GPa( ) 4.261 ×10−6m4( )
8m( )2
= 136kN
The critical stress is given by Equation (9.10) as:
σcr =
Pcr
A
=
136kN
0.005m2
= 27.2 MPa
IV. Square Tube
If the outer dimension is 100mm=0.1m, then the inner dimension is obtained from the area as:
A = xo2 − xi2; x i = xo2 − A = 0.1m( )2 − 0.005m2 = 0.0707m
The moment of inertia is then
I =
xo
4 − xi
4
12
=
0.1m( )4 − 0.0707m( )4
12
= 6.25 ×10−6m4
The Euler load is then
Pcr =
π 2EI
l2
=
π2 207GPa( ) 6.25 × 10−6 m4( )
8m( )2
= 200kN
The critical stress is given by Equation (9.10) as:
www.elsolucionario.net
Page 9-7
σcr =
Pcr
A
=
200kN
0.005m2
= 40.0MPa
9.8 A column is axially loaded in compression. The ends were specified to be fixed, but
because of a manufacturing mistake they had
a) One end fixed and the other pinned
b) Both ends pinned
Find how much the critical elastic buckling load is decreased because of the mistake.
Also, calculate the buckling loads for the 4m long column made of AISI 1040 steel
having a solid, square cross section with 30-mm sides for the three end conditions.
Assume the theoretical effective column length.
Notes: This problem is solved with Equation (9.12), with effective column lengths obtained from
Table 9.1 on page 366.
Solution:
I. Effects of Boundary Conditions on Buckling Loads
For the column as-designed, with both ends fixed, the effective column length is le=0.5l, as
obtained from Table 9.1 on page 366. Therefore, Equation (9.12) gives the buckling load as:
Pcr =
π 2EI
le
2 = π
2EI
1
0.5 l( )2
 
 
  
 
 
  = 4
π 2EI
l2
If one end is fixed and the other pinned, Table 9.1 gives le=0.7l. Therefore, Equation (9.12) gives
Pcr =
π 2EI
le
2 = π
2EI
1
0.7l( )2
 
 
  
 
 
  = 2.04
π 2EI
l2
which is about 51% of the original column. If both ends are pinned, Table 9.1 gives le=l, so that
Equation (9.12) gives:
Pcr =
π 2EI
le
2 = π
2EI
1
l( )2
 
 
  
 
 
  =
π2 EI
l2
which is only 25% of the intended buckling load.
II. Calculation of Buckling Loads
From the inside front cover, the elastic modulus for AISI 1040 steel is 207GPa. The moment of
inertia for the cross section is:
I =
x4
12
=
0.03m( )4
12
= 6.75 ×10−8m4
Therefore, the buckling load for the fixed-fixed column is, using the results obtained above:
Pcr = 4
π 2EI
l2
= 4π2
207GPa( ) 6.75 × 10−8m4( )
4m( )2
= 34.5kN
Therefore, for the fixed-pinned case, the load is 51% of this load, or Pcr=17.6kN. For the pinned-
pinned column, the load is 25% of the fixed-fixed load, or 8.625kN.
9.9 An elastic column has one end pinned and the other end fixed in an elastic bushing so
that the values given in Table 9.1 do not apply. Instead, the effective length is le=0.836.
The cross section of the column is a circular tube with an outer diameter of 73mm and a
www.elsolucionario.net
Page 9-8
3.2 mm wall thickness. The column is 11.2 m long. Calculate the elastic buckling load if
the column is made of
a) AISI 1080 steel
b) Polycarbonate
Notes: The buckling load is calculated from Equation Equation (9.12).
Solution:
The moment of inertia for the cross section is:
I =
π
4
ro
4 − ri4( ) = π4 0.0365m( )
4 − 0.0333m( )4( ) = 4.28× 10−7 m4
From the inside front cover, the elastic modulus for AISI 1080 steel is E=207 GPa, so the elastic
buckling load is calculated from Equation (9.12) as:
Pcr =
π 2EI
L2
=
π2 207GPa( ) 4.28 ×10−7 m4( )
0.83( ) 11.2m( )[ ]2
= 10.1kN
For polycarbonate, E=2.38 GPa (from Table A.4 on page 902) so
Pcr =
π 2EI
L2
=
π2 2.38GPa( ) 4.28× 10−7 m4( )
0.83( ) 11.2m( )[ ]2
=116N
9.10 An elastic AISI 1020 steel column has both ends pinned. It is 12.5m long and has a
square tubular cross section with an outside dimension of 160mm and 4-mm wall
thickness. Its compressive axial load is 130kN.
a) Determine the safety factor guarding against buckling and yielding.
b) If the ends are changed to fixed and the material is changed to aluminum alloy 2014,
calculate the safety factors guarding against buckling and yielding.
Notes: Equation (9.12) is used to obtain the buckling load. The safety factor against buckling is
the ratio of the allowable stress to the applied stress, but since the area is constant, this is also the
ratio of allowable load to applied load. The safety factor against yielding is the ratio of allowable
stress to the applied stress.
Solution:
For this cross section the area and moment of inertia are:
A = x2 − x − 8mm( )2 = 0.160 m( )2 − 0.152m( )2 = 0.002496 m2
I =
1
12
x4 − x − 8mm( )4( ) = 112 0.160m( )
4 − 0.152 m( )4( ) = 1.013 × 10−5m4
I. AISI 1020 Steel Column
For AISI 1020 steel, the inside front cover gives E=207GPa and Sy=295MPa. With both ends
pinned, Table 9.1 gives le=l=12.5m. Therefore, Equation (9.12) gives the allowable elastic
buckling load as:
Pcr =
π 2EI
L2
=
π2 207GPa( ) 1.013 ×10−5m4( )
12.5m( )2
= 132.5kN
The safety factor against elastic buckling is (see Equation (1.1)):
ns =
σall
σd
=
Pcr / A( )
P / A( )
=
Pcr
P
=
132.5kN
130kN
= 1.02
www.elsolucionario.net
Page 9-9
Against yielding the safety factor is
ns =
σall
σd
=
Sy
P / A( )
=
ASy
P
=
0.002496m2( ) 295MPa( )
130kN
= 5.66
II. Aluminum 2014 Column
For aluminum 2014, the elastic modulus and yield strength are obtained from the inside front
cover as E=72GPa and Sy=97MPa. If both ends are fixed, then from Table 9.1, le=0.5l=6.25m.
Therefore, Equation (9.12) gives the allowable elastic buckling load as:
Pcr =
π 2EI
L2
=
π2 72GPa( ) 1.013 ×10−5 m4( )
6.25m( )2
= 184.3kN
The safety factor against elastic buckling is (see Equation (1.1)):
ns =
σall
σd
=
Pcr / A( )
P / A( )
=
Pcr
P
=
184.3kN
130kN
=1.42
Against yielding the safety factor is
ns =
σall
σd
=
Sy
P / A( )
=
ASy
P
=
0.002496m2( ) 97MPa( )
130kN
=1.86
9.11 A beam has both ends mounted in stiff rubber bushings giving bending moments in the
beam ends proportional to the angular displacement of the beam ends. Calculate the
effective beam length if the angular spring constant at the ends is ∂M/∂θ=105Nm/radian
in all directions. The beam is 3m long and has a solid circular cross section of AISI 1080
steel with a 24mm diameter.
Notes: This problem is solved by using Equation (9.2) and resolving for the critical load using the
same approach as on page 363, but with different boundary conditions. The effective length is
obtained from Equation (9.12) once the critical load is obtained.
Solution:
From Equation (9.2) and differentiating,
dy
dx
= θ = C1
P
EI
cos x
P
EI
 
 
  
 
 
  − C2
P
EI
sin x
P
EI
 
 
  
 
 
  
The boundary conditions are that y=0 at x=0 and θ=0 at x=l/2-x0. The first boundary condition
gives C2=0. The second gives:
0 = C1
P
EI
cos
l
2
− x0
 
  
 
  
P
EI
 
 
  
 
 
  ; 
l
2
− x0
 
  
 
  
P
EI
= ±
π
2
; x0 =
l
2
−
π
2
EI
P
From Equation (5.3), the bending moments at the loading points are:
M = − 105 Nm / rad( ) dydx = −EI
d2y
dx2
where d
2
y/dx
2
 is
d2y
dx 2
=
d
dx
dy
dx
 
  
 
  
=
d
dx
C1
P
EI
cos x
P
EI
 
 
  
 
 
  
 
 
  
 
 
  = −C1
P
EI
sin x
P
EI
 
 
  
 
 
  
Therefore, the moment equation becomes
www.elsolucionario.net
Page 9-10
105 Nm / rad( )C1 PEI cos x
P
EI
 
 
  
 
  = −C1
P
EI
sin −x0( )
P
EI
 
 
  
 
 
  ; t a nx0
P
EI
 
 
  
 
 
  =
105 Nm / rad
P / EI
Solving this expression numerically gives Pcr=14,145N.
The effective length is obtained from Equation (9.12) as:
Pcr =
π 2EI
le
2 ; le = π
EI
Pcr
= π
207GPa( )π 0.024m( )4 /64
14,145N
= 1.534m
9.12 Two circular columns are made of different materials, steel and aluminum. The cross-
sectional areas are the same and the moduli of elasticity are 207GPa for the steel and
72GPa for the aluminum. Find the ratio of the critical buckling lengths for the steel and
aluminum columns while assuming that the same buckling load is applied to both
columns.
Notes: Equation (9.10) is used to solve this problem.
Solution:
For the same buckling load, Equation (9.10) gives:
Pcr =
π 2EI
le
2 ; 
π 2EaI
lea
2 =
π 2EsI
les
2 ; 
les
lea
=
Es
Ea
=
207GPa
72GPa
=1.696
9.13 A television mast shown in sketch b consists of a circular tube with an outer diameter do
and a wall thickness th. Calculate how long the distance l between the anchoring points
for the cables can be if the television mast should deform plastically rather than buckle.
Notes: The purpose of the anchoring points is to force a higher buckling mode. This is a common
design approach for lattice structures such as booms and scaffolding. This problem is solved by
preserving the n from Equation (9.6) in the derivation of Equation (9.10).
Solution:
For this cross section, the area and moment of inertia are:
A =
π
4
do
2 − di2( )
www.elsolucionario.net
Page 9-11
I =
π
64
do
4 − di4( ) = π64 do2 − di2( ) do2 + di2( )
The anchoring points with the cables as shown force the column into a higher buckling mode.
The given geometry uses n=3. Therefore, from Equation (9.6),
P =
n2π 2EI
l2
If the mast deforms plastically, then
P
A
= Sy ; P = ASy
Substituting and solving for l gives:
ASy =
n2π 2EI
3l( )2
; l =
1
3
n2π2EI
ASy
=
πn
3
E
π
64
do
2 − di
2( ) do2 + di2( )   
 
  
π
4
do
2 − di
2( )Sy
=
π
4
E do
2 + di
2( )
Sy
or, substituting for di,
l =
π
4
E do
2 + do − 2th( )2( )
Sy
=
π
4
2E do
2 − 2do th + 2 th
2( )
Sy
9.14 A 3m long column of square, solid cross section with both ends fixed must sustain a
critical load of 3x10
5
N. The material is steel with a modulus of elasticity of 207GPA and
a yield stress of 700MPa. Determine the following:
a) What minimum dimensions of a square, solid cross section are permitted without
having failure? Should the Euler or Johnson equation be used?
b) If the critical load is increased by two orders of magnitude (Pcr=3 x 10
7
N), what
minimum dimensions of a square, solid cross section are permitted without having
failure? Also, indicate whether the Euler or Johnson equation should be used.
Notes: Equation (9.18) gives the critical column length. If le/rg from Euler analysis (Equation
(9.15) is greater than this value, the column is elastic. Otherwise, the column buckles plastically
and the Johnson equation (9.16) should be used. This problem specifies use of either the Euler or
Johnson equation, so the AISC recommended lengths from Table 9.1 are not used.
Solution:
(a) The critical column length for this material is given by Equation (9.18) as:
le
rg
 
 
 
 
 
 
 
 
T
=
2π2E
Sy
=
2π2 207GPa( )
700MPa( )
= 76.40
From Equation (4.13), for a square cross section, the radius of gyration is
rg =
I
A
=
w4
12w2
=
w
2 3
For both ends fixed, Table 9.1 gives le=0.5l=1.5m. Assume the buckling is elastic, so that the
Euler equation should be used. Then from Eq. (9.15),
le
rg
 
 
 
 
 
 
 
 
E
=
π 2EA
Pcr
; 
1.5m( )2 3
w
=
π2 207GPa( )w2
3× 105 N
; w = 0.045m
www.elsolucionario.net
Page 9-12
Therefore, rg=0.013m, and le/rg is 115. Since this is greater than the critical column length,
buckling is elastic and the minimum dimension is 45 mm.
(b) If the buckling load is 3 x 10
7
 N, again assuming the Euler equation holds,
le
rg
 
 
 
 
 
 
 
 
E
=
π 2EA
Pcr
; 
1.5m( )2 3
w
=
π2 207GPa( )w2
3× 107 N
; w = 0.140m
Then rg=.040m, le/rg=38. Since this is less than the critical aspect ratio, the column is plastic and
the Johnson equation should be used. Therefore, from Eq. (9.16):
σcr =
Pcr
w2
= Sy −
Sy
2
4π 2E
le
rg
 
 
 
 
 
 
 
 
2
; w = 0.220m
9.15 A solid, round column with a length of 2m and a diameter of 50mm is fixed at one end
and free at the other end. The material’s yield strength is 300MPa and its modulus of
elasticity is 207GPa. Assume concentric loading of the column. Determine the following:
a) The critical load on this column.
b) The critical load on the column if the free end is also fixed (i.e. both ends are fixed).
Notes: This problem asks for determination of a column’s critical load, so either the Euler or
Johnson equations are used for solution of this problem. The AISC requirements would be used
for a case where the design load for a column has been stated and the goal is to design a column
that will support this load without buckling.
Solution:
For this solid round column, the cross sectional area is A=πr2=π(0.05m/2)2=0.00196m2. The
moment of inertia is (see inside back cover) I=πr4/4=π(0.05m/2)4/4=0.3068x10-6m4. The radius of
gyration is calculated from Equation (4.14) as
rg =
I
A
=
0.3068 × 10−6m4
0.00196m2
= 0.0125m
I. Free-Fixed Column
Table (9.1) gives the theoretical effective column length as le=2l=2(2m)=4m. Therefore,
the slenderness ratio for the column is C=le/rg=4m/0.0125m=320. The critical slenderness ratio is
obtained from Equation (9.15) as
Cc =
2Eπ2
Sy
=
2 207GPa( )π 2
300MPa
= 116.7
Since the slenderness ratio of the column exceeds the critical slenderness ratio, the column
buckles elastically, and the Euler equation (9.7) can be used:
Pcr =
π 2EI
l2
=
π2 207GPa( ) 0.3068 × 10−6 m4( )
4m( )2
= 39.2kN
II. Fixed-Fixed Column
For this case, Table (9.1) gives the effective column length as le=0.5l=0.5(2m)=1m.
Therefore, the slenderness ratio for this columns is C=le/rg=1m/0.0277m=36.1. Since this is less
than the critical slenderness ratio, the column buckles plastically and the Johnson equation must
be used (Equation 9.16). The buckling force is then:
www.elsolucionario.net
Page 9-13
Pcr = σcrJ A = Sy −
Sy
2
4π2 E
le
rg
 
 
 
 
 
 
 
 
2 
 
 
 
 
 
 
 
 
 
A[ ] = 300MPa − 300MPa( )
2
4π 2 207GPa( )
80( )2
 
 
 
 
 
 
 
 0.00196m
2( ) = 450kN
9.16 A circular cross section bar with a diameter of 2 in and a length of 40 in is axially loaded.
The bar is made of medium carbon steel. Both ends are pinned. Determine
a) Whether the Johnson or Euler formula should be used
b) The critical load
Notes: The approach is similar to Problem 9.15. This problem asks for determination of a
column’s critical load, so either the Euler or Johnson equations are used for solution of this
problem. The AISC requirements would be used for a case where the design load for a column
has been stated and the goal is to design a column that will support this load without buckling.
Solution:
For medium carbon steel, the inside front cover gives for AISI 1040 steel, E=30Mpsi and
Sy=51ksi. Therefore, from Equation (9.18),
le
rg
 
 
 
 
 
 
 
 
T
=
2π2E
Sy
=
2π2 30Mpsi( )
51ksi
=108
The area and moment of inertia are:
A =
π
4
d2 =
π
4
2in( )2 = 3.14in2
I =
π
64
d4 =
π
64
2in( )4 = 0.7854in4
Therefore, the radius of gyration is obtained from Equation (4.13),
rg =
I
A
=
0.7854in4
3.14in2
= 0.500in
From Table 9.1, for pinned ends, le=l. Therefore, the slenderness ratio for this column is
le
rg
=
40 in
0.5 in
= 80
Since this is less than the critical slenderness ratio of 108, the column will buckle plastically, and
the Johnson equation (Equation 9.16) must be used. Therefore,
Pcr
A
= Sy −
Sy
2
4π 2E
le
rg
 
 
 
 
 
 
 
 
2
= 51ksi −
51ksi( )2
4π 2 30Mpsi( )
40in
0.5 in
 
  
 
  
2
= 36.9ksi
Since the area was calculatedas 3.14in
2
, the critical load is Pcr=(3.14in
2
)(36.9ksi)=116kip.
www.elsolucionario.net
Page 9-14
9.17 A low-carbon steel pipe, as shown in sketch c, has an outer
diameter of 2in and a thickness of 0.5in. If it is held in place by a
guy wire, determine the largest horizontal force P that can be applied
without causing the pipe to buckle. Assume that the ends of the pipe
are pin connected
Notes: This problem is similar to the previous problems, but the
axial compressive load must be determined from statics.
Solution:
I. Column Analysis
From the inside front cover for low carbon steel (AISI 1020),
Sy=43ksi and E=30Mpsi. Therefore, Equation (9.18) gives
le
rg
 
 
  
 
 
  
T
= 2π
2E
Sy
=
2π2 30Mpsi( )
43ksi
=117
For this cross section, the area and moment of inertia are:
A = π
4
do
2 − di2( ) = π4 2in( )
2 − 1in( )2( ) = 2.356in2
I = π
64
do
4 − di4( ) = π64 2in( )
4 − 1in( )4( ) = 0.7363in4
Therefore, the radius of gyration is obtained from Equation (4.13),
rg =
I
A
= 0.7363in
4
2.356in2
= 0.5590 in
From Table 9.1, le=l=144in. Therefore,
le
rg
= 144in
0.5590 in
= 257.6
Since this is larger than the critical slenderness ratio, the buckling is elastic and the Euler formula
(Equation (9.13)) will be used.
II. Statics
The sketch shows a section at the top of the column. The load F is the
compressive buckling load, the force T is the tension in the guy wire, and the
force P is the applied load. Taking horizontal force equilibrium gives
Fx = 0 =∑ 513 T − P; T =
13
5
P
From vertical force equilibrium,
Fy = 0 =∑ F − 1213 T = F −
12
13
13
5
P
 
  
 
  ; F =
12
5
P
III. Determination of Maximum Load
From Equation (9.13), the stress at buckling is:
σcr =
π 2E
le / rg( )2
=
π2 30Mpsi( )
257.6( )2
= 4460 psi
Therefore, the maximum value of F is:
F = σ crA = 4460 psi( ) 2.356in2( ) =10.5kip
www.elsolucionario.net
Page 9-15
Therefore, the largest value of P is:
F = 12
5
P; P = 5
12
F = 5
12
10.5kip( ) = 4.38kip
9.18 A 20-mm diameter, medium carbon steel (AISI 1040) rod is loaded as a column with
rounded ends. The critical load is 110kN. How long can the rod be and still carry the
following percentages of the critical load: a) 90%, b) 50%, and c) 2%? Also, indicate
whether the Johnson or Euler equation should be used.
Notes: The approach is similar to problems (9.14) to (9.17). For elastic buckling, the Euler
equation (9.13) is used. For plastic buckling, the Johnson equation (9.16) is used. Selecting which
equation is used depends on the slenderness ratio of the column compared to the critical
slenderness ratio (Equation (9.18)).
Solution:
I. Column Analysis
From the inside front cover for AISI 1040 steel, E=207GPa and Sy=350MPa. Therefore, from
Equation (9.18),
le
rg
 
 
  
 
 
  
T
= 2π
2E
Sy
= 2π
2 207GPa( )
350MPa
=108
The area and moment of inertia are:
A = π
4
d2 = π
4
0.02m( )2 = 3.14 ×10−4m2
I = π
64
d4 = π
64
0.02m( )4 = 7.854× 10−9 m4
Therefore, the radius of gyration is obtained from Equation (4.13),
rg =
I
A
= 7.854× 10
−9 m4
3.14 ×10−4 m2
= 0.00500m
From Table 9.1, for rounded or pinned ends, le=l.
II. 90% of Critical Load
For this case, the load that can be carried is (0.90)(110kN)=99kN. Assume that the buckling is
elastic. From Equation (9.13),
σcr =
Pcr
A
= π
2E
le / rg( )2
; 
le
rg
 
 
  
 
 
  =
π 2EA
Pcr
=
π 2 207GPa( ) 3.14 ×10−4 m2( )
99kN
= 80.5
Since this is less than the critical slenderness ratio, the column does not buckle elastically, and the
Johnson equation should be used. From Equation (9.16),
Pcr
A
= Sy −
Sy
2
4π 2E
le
rg
 
 
  
 
 
  
2
;
le
rg
= 4π
2 E
Sy
2 Sy −
Pcr
A
 
  
 
  
= 4π
2 207GPa( )
350MPa( )2
350MPa − 99kN
3.142 ×10−4 m2
 
  
 
  
= 48.2
Therefore, the maximum column length is
le
rg
= 48.2; le = 48.2rg = 48.2 0.005m( ) = 0.241m
www.elsolucionario.net
Page 9-16
III. 50% of Critical Load
For this case, the load that can be carried is (0.50)(110kN)=55kN. Assume that the buckling is
elastic. From Equation (9.13),
σcr =
Pcr
A
= π
2E
le / rg( )2
; 
le
rg
 
 
  
 
 
  =
π 2EA
Pcr
=
π 2 207GPa( ) 3.14 ×10−4 m2( )
55kN
=108
This is very close to the critical slenderness ratio, and the buckling mode can be taken as either
elastic or plastic to get the same answer. Since the slenderness ratio of the column is 108 to
achieve 55kN, the column length is
le
rg
=108;le = 108 rg( ) = 108 0.00500m( ) = 0.540m
IV. 2% of Critical Load
For this case, the load that can be carried is (0.02)(110kN)=2.2kN. Assume that the buckling is
elastic. From Equation (9.13),
σcr =
Pcr
A
= π
2E
le / rg( )2
; 
le
rg
 
 
  
 
 
  =
π 2EA
Pcr
=
π 2 207GPa( ) 3.14 ×10−4 m2( )
2.2kN
= 540
This is larger than the critical slenderness ratio, so the column buckles elastically. Since the
slenderness ratio of the column is 540 to achieve 2.2kN, the column length is
le
rg
=108;le = 540 rg( ) = 540 0.00500m( ) = 2.7m
9.19 Most types of steel have similar moduli of elasticity but can have widely different yield
strength properties depending on the alloying and the heat treatment. For a column
having a solid circular cross section with a diameter of 20mm, the yield strengths for
three steels are 300, 600, and 1000MPa. Find for each steel the critical column length
where buckling would not be a problem.
Notes: The critical slenderness ratio is given in Equation (9.15).
Solution:
The area and moment of inertia are:
A = π
4
d2 = π
4
0.02m( )2 = 3.14 ×10−4m2
I = π
64
d4 = π
64
0.02m( )4 = 7.854× 10−9 m4
Therefore, the radius of gyration is obtained from Equation (4.13),
rg =
I
A
= 7.854× 10
−9 m4
3.14 ×10−4 m2
= 0.00500m
The critical length can be obtained from Equation (9.18):
le = rg
2π 2E
Sy
= 0.5 ×10−2 m( ) 2π
2 207GPa( )
Sy
If Sy=300MPa, this becomes:
le = 0.5 ×10−2 m( ) 2π
2 207GPa( )
300MPa
= 0.5835m
For Sy=600MPa,
www.elsolucionario.net
Page 9-17
le = 0.5 ×10−2 m( ) 2π
2 207GPa( )
600MPa
= 0.4126m
For Sy=1000MPa,
le = 0.5 ×10−2 m( ) 2π
2 207GPa( )
1000MPa
= 0.3196m
9.20 For a fixed cross sectional area of a steel column, find the geometry that would give the
largest buckling load. The minimum column wall thickness is 5mm
Solution:
From Equation (9.12), the buckling load is:
Pcr =
π 2EI
le
2
The largest critical load is one which maximizes the moment of inertia. Example 9.3 showed that
a hollow circular area gives the largest moment of inertia and therefore the largest buckling loads.
Note, however, that boundary conditions can affect the optimum cross section. For
example, if a column is pinned in one direction but fixed in the perpendicular direction, buckling
is more likely in the pinned direction. A circle would no longer be the best geometry. The
extreme case is when the boundary conditions preclude buckling in one direction, in which case,
the optimum geometry is an ideal I-beam, with all of the material in the flanges and zero web.
9.21 A rectangular cross section tube is axially loaded in compression. The tube outside
dimensions are 70 and 90mm. The tube is 2.5m long, and both ends are pinned. The wall
thickness is 9mm. Find which of the steels in Appendix A.1 should be used so that plastic
buckling is not a problem.
Notes: Equation (9.18) is used to obtain a material property requirement in order to avoid plastic
buckling. The materials in Appendix A.1 are then evaluated with respect to this property.
Solution:
The cross section will buckle across the axis that minimizes the buckling load, so it will naturally
buckle along the axis that minimizes the moment of inertia. For the rectangular shape, this occurs
when the smaller dimension is the height of the cross section. Therefore, use ht=0.070m and
b=0.090m. The area and moment of inertia for the cross section are:
A = bht − b − 2t( ) h − 2t( ) = 0.07m( ) 0.09m( ) − 0.052m( ) 0.072m( ) = 0.002556m2
I =
1
12
bht
3( ) − b − 2 t( ) ht − 2t( )3[ ] = 112 0.090m( ) 0.07m( )3 − 0.072m( ) 0.052m()3[ ] =1.729 ×10−6 m4
Therefore the radius of gyration is obtained from Equation (4.13) as:
rg =
I
A
=
1.729 ×10−6 m4
0.002556m2
= 0.0260m
From Equation (9.18),
le
rg
=
2π 2E
Sy
; 
E
Sy
=
1
2π2
le
rg
 
 
 
 
 
 
 
 
2
=
1
2π 2
2.5m
0.0260m
 
  
 
  
2
= 468
www.elsolucionario.net
Page 9-18
Note that this is a maximum value for E/Sy; larger values would indicate the column is shorter
than the critical length needed to avoid plastic buckling. Therefore, a material with a value of
E/Sy>468 would buckle plastically. Using the material properties in Table A.1, the following table
can be constructed:
Material E (GPa) Sy (MPa) E/Sy
Iron
Nodular cast iron
Malleable cast iron
AISI 1020
AISI 1040
AISI 1080
Ferritic SS
Austenitic SS
Martensitic SS
207
165
172
207
207
207
200
193
200
130
275
220
295
350
380
345
207
275
1592
600
782
702
591
548
578
932
727
Therefore, none of the materials in Appendix A.1 will buckle elastically for the given column.
9.22 The column considered in Problem 9.21 is made of AISI 1020 steel. Find the critical
buckling load.
Notes: This solution relies upon the solution for Problem 9.21. Problem 9.21 showed that this
material buckles plastically. Therefore, the Johnson equation will be used without the usual
calculations to determine if the buckling mode is elastic or plastic.
Solution:
From the solution to Problem 9.21, A=0.002556m
2
, I=1.729x10
-6
m
4
, rg=0.0260m, and le=2.5m.
From the inside front cover for AISI 1020 steel, E=207GPa and Sy=295MPa. It was shown in
Problem 9.21 that the column buckles plastically. Therefore, from Equation (9.16),
Pcr
A
= Sy −
Sy
2
4π 2E
le
rg
 
 
 
 
 
 
 
 
2
= 295MPa( ) − 295MPa( )
2
4π 2 207GPa( )
2.5m
0.026m
 
  
 
  
2
= 196.5MPa
Since the area is 0.002556m
2
, the critical load is Pcr=(0.002556m
2
)(196.5MPa)=502kN.
9.23 A 2.5m long column with one end fixed and one end free is made of aluminum alloy
2014 and has a solid, round cross section. Determine the diameter of the column for the
following loads:
a) P=500kN
b) P=800kN
Notes: This problem uses a similar approach to the previous problems. For elastic buckling, the
Euler equation (9.13) is used. For plastic buckling, the Johnson equation (9.16) is used. Selecting
which equation is used depends on the slenderness ratio of the column compared to the critical
slenderness ratio (Equation (9.18)).
Solution:
From the inside front cover for aluminum alloy 2014, E=72GPa and Sy=97MPa. Note from Table
9.1 that for one end fixed and the other free, le=2l=2(2.5m)=5m. The critical slenderness ratio is
obtained from Equation (9.15) as
www.elsolucionario.net
Page 9-19
Cc =
2Eπ2
Sy
=
2 72GPa( )π 2
97MPa
=121
I. 500kN Load
Assume the buckling is elastic (this will be verified after a cross section has been determined).
Equation (9.12) then gives:
Pcr =
π 2EI
le
2 ; I =
Pcr le
2
π 2E
=
500kN( ) 5m( )2
π2 72GPa( )
= 1.759 × 10−5 m4
If the cross section is solid and round, the diameter must be:
I =
πd4
64
=1.759 × 10−5m4; d = 0.1376m
Therefore the area of the cross section is
A =
π
4
d2 =
π
4
0.1376m( )2 = 0.01487m2
The radius of gyration is obtained from Equation (4.13) as:
rg =
I
A
=
1.759 ×10−5 m4
0.01487m2
= 0.03440m
The slenderness ratio for this column is le/rg=(5m)/(0.03440m)=145. Since this is greater than the
critical slenderness ratio, the column buckles elastically.
II. 800kN Load
Assume the buckling is elastic (this will be verified after a cross section has been determined).
Equation (9.12) then gives:
Pcr =
π 2EI
le
2 ; I =
Pcr le
2
π 2E
=
800kN( ) 5m( )2
π2 72GPa( )
= 2.814 × 10−5 m4
If the cross section is solid and round, the diameter must be:
I =
πd4
64
= 2.814 × 10−5m4; d = 0.1547m
Therefore the area of the cross section is
A =
π
4
d2 =
π
4
0.1547m( )2 = 0.01881m2
The radius of gyration is obtained from Equation (4.13) as:
rg =
I
A
=
2.814 × 10−5 m4
0.01881m2
= 0.03868m
The slenderness ratio for this column is le/rg=(5m)/(0.03868m)=129. Since this is greater than the
critical slenderness ratio, the column buckles elastically.
9.24 A 1-m long column with a 60- by 100-mm rectangular cross section and made of
aluminum alloy 2014 is subjected to a compressive axial load. Determine the critical
buckling load
a) If both ends are fixed.
b) If one end is fixed and the other is free
c) If the load is applied eccentrically at a distance of 10mm for case b.
Notes: This problem asks for determination of a column’s critical load, so either the Euler or
Johnson equations are used for solution of this problem. The AISC requirements would be used
www.elsolucionario.net
Page 9-20
for a case where the design load for a column has been stated and the goal is to design a column
that will support this load without buckling. For this homework only, either approach is
acceptable. For the eccentric loading, the critical column load is determined by numerically
solving Equation (9.33)
Solution:
Note that for Al-2014, the inside front cover gives the values E=72GPa and Sy=97MPa.
Therefore, the critical slenderness ratio is, from Equation (9.15),
Cc =
2Eπ2
Sy
=
2 72GPa( )π 2
97MPa
=121
The cross sectional area is A=(0.06m)(0.10m)=0.006m
2
. The moment of inertia is (see inside back
cover) I=bh
3
/12=(0.1m)(0.06m)
3
/12=1.80x10
-6
m
4
. (Note that we use h=0.06 so that the moment of
inertia is minimized; the column will buckle on this axis before it buckles on the axis where the
height is 0.1m. This can be easily demonstrated with a yard or meter long ruler.) The radius of
gyration is then obtained from Equation (4.14) as
rg =
I
A
=
1.80 ×10−6m4
0.006m2
= 0.0173m
I. Fixed-Fixed Column
For this case, Table 9.1 gives le=0.5l=0.5(1m)=0.5m. Therefore, the slenderness ratio for this
column is C=le/rg=0.5m/0.0173m=28.9. Since this is below the critical slenderness ratio, the
column buckles plastically and the Johnson equation (Equation 9.16) is used. The buckling force
is then:
Pcr = σcrJA = Sy −
Sy
2
4π 2E
le
rg
 
 
 
 
 
 
 
 
2 
 
 
 
 
 
 
 
 
 
A[ ]= 97MPa − 97MPa( )
2
4π 2 72GPa( )
28.9( )2
 
 
 
 
 
 
 
 0.006m
2( ) = 565kN
II. Fixed-Free Column
For this case, Table 9.1 gives le=2l=2(1m)=2m. Therefore, the slenderness ratio for this column is
C=l/rg=2m/0.0173m=115.6. Since this is below the critical slenderness ratio (although not by
much), the column buckles plastically and the Johnson equation (Equation 9.16) is used. The
buckling force is then:
Pcr = σcrJA = Sy −
Sy
2
4π 2E
le
rg
 
 
 
 
 
 
 
 
2 
 
 
 
 
 
 
 
 
 
A[ ]= 97MPa − 97MPa( )
2
4π 2 72GPa( )
115.6( )2
 
 
 
 
 
 
 
 0.006m
2( ) = 316kN
III. Eccentric Column
For the eccentric column, the critical load is obtained from Equation (9.33) using e=0.01m and
c=h/2=0.03m:
P =
SyA / 2( )
1 + ec
rg
2 sec
le
2rg
P
EA
 
 
 
 
 
 
 
 
=
97MPa( ) 0.006m2( ) / 2
1 + 0.01m( ) 0.03m( )
0.0173m( )2
sec
115.6
2
P
72GPa( ) 0.006m2( )
 
 
 
 
 
 
 
 
 
 
This is solved numerically as P=109.6kN.
9.25 A column with pinned ends and solid rectangular cross section with dimensions of
35x60mm is made of AISI 1040 steel. The 4m long column is loaded with an eccentricity
www.elsolucionario.net
Page 9-21
of 15mm. Find the elastic deflection if a 100N compressive load is applied. Also,
calculate how large a load can be applied without permanent deformation occurring.
Notes: The maximum deflection is given by Equation (9.28). The largest load that can be applied
without plastic deformation is obtained by substituting the yield strength for σmax in Equation
(9.30).
Solution:
The largest deflection occurs when the 35 mm dimension is the height of the cross section. For
this case, I=bh
3
/12=(0.060m)(0.035m)
3
/12=2.14x10
-7
m
4
. The area of the cross section is
A=(0.035)(0.060)=0.0021m
2
. The radius of gyration is then r
2
=I/A=1.02x10
-4
m
2
. rg is then 0.0101
m. Ifthe load is 100 N, the deflection is given by Eq. (9.28) as:
ymax = e sec
l
2
P
EI
 
 
  
 
 
  − 1
 
 
 
 
 
 
 
 = 0.015m( ) sec
4m
2
100N
207GPa( ) 2.14 ×10−7m4( )
 
 
 
 
 
 
 
 
 
 
− 1
 
 
 
 
 
 
 
 
 
 
= 68µm
For 1040 steel, the yield stress is 350 MPa. Then from Equation. (9.30),
Sy =
P
A
1 + ec
rg
2 sec
l
2rg
P
EA
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
= P
0.0021m2
1+ 0.015m( ) 0.0175m( )
0.010m( )2
sec
4m
2 0.010m( )
P
207GPa( ) 0.0021m2( )
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
This is solved numerically to yield P=21.94 kN.
Alternate Solution:
As shown above, I=2.1438 x 10-7m4, A=0.021m2 and r=0.01010m. From Eq. (9.28),
ymax = 0.015 sec
4
2
100
207 ×109( ) 2.144 × 10−7( )
 
 
 
 
 
 
 
 
 
 
− 1
 
 
 
 
 
 
 
 
 
 
= 67.86 × 10−6 m = 67.86µm
Note that
ec
rg
=
0.015( ) 0.0175( )
1.021 ×10−4
= 2.571
From Equation (9.30),
350 × 106
2
 
 
  
 
 
  =
P
2.1 ×10−3
1= 2.571sec
4
2 0.0101( )
 
 
  
 
 
  
P
207 × 109( ) 2.1 ×10−3( )
 
 
 
 
 
 
 
 
 
 
This reduces to:
P =
367,500
1 + 2.571sec 9.498× 10−3( ) P[ ]
Successive iteration gives P=21,938N.
9.26 The column considered in Problem 9.25 will have a varying load in the range 0 to
24000N. Calculate and plot the deformation as a function of the axial load.
www.elsolucionario.net
Page 9-22
Notes: The deflection as a function of load is given by Equation (9.28).
Solution:
The largest deflection occurs when the 35 mm dimension is the height of the cross section. For
this case, I=bh
3
/12=(0.060m)(0.035m)
3
/12=2.14x10
-7
m
4
. The area of the cross section is
A=(0.035)(0.060)=0.0021m
2
. The radius of gyration is then r
2
=I/A=1.02x10
-4
m
2
. rg is then 0.010
m. The deflection is given by Eq. (9.28) as:
ymax = e sec
l
2
P
EI
 
 
  
 
 
  − 1
 
 
 
 
 
 
 
 = 0.015m( ) sec
4m
2
P
207GPa( ) 2.14 ×10−7 m4( )
 
 
 
 
 
 
 
 
 
 
− 1
 
 
 
 
 
 
 
 
 
 
Some typical calculations and a plot are as follows:
P (N) y (m)
0
2000
4000
6000
8000
10,000
12,000
14,000
16,000
18,000
20,000
22,000
24,000
0
.00146
.00319
.00524
.00773
.01080
.01468
.01974
.02661
.03644
.05167
.07837
.13726
www.elsolucionario.net
Page 10-1
Chapter 10 • Cylinders
10.1 A journal bearing is to be manufactured with optimum geometry for minimum power loss
for a given load and speed. The relative clearance c/r=0.001. The journal diameter is
100mm. Find the accuracy to which the bearing parts have to be manufactured so that
there is not more than ±10% error in the relative clearance.
Notes: This problem is straightforward, merely requiring an understanding of the terms
introduced in section 10.2.
Solution:
Since the radius is 50mm, and c/r=0.001, so that c=0.001(50mm)=0.050mm=50µm. The
allowable error is ±10% of 50 µm or ±5µm. The radial error, if shared equally between the
journal and the bearing, is one-half this value (±2.5µm), or the diametral error for each is ±5µm.
10.2 A press fit between a solid steel shaft and a steel housing is dimensioned to be of class 7.
By mistake the shaft is ground at 22°C higher temperature than originally anticipated, so
that when the shaft cools, the diameter is slightly too small. The shaft material is AISI
1040 steel. What will be the class of fit between the shaft and the housing because of this
mistake? Also, if the grinding temperature were 50°C higher rather than 22° higher, what
would the class of fit be?
Notes: The class of fit is obtained from the definitions in Table 10.3. The change in diameter is
obtained from thermal stresses, and can be calculated from Equation (10.56).
Solution:
The coefficient of thermal expansion for AISI 1040 steel is obtained from the inside front cover
as α=11.3x10-6 °C-1. Because of this unexpected temperature, the diametral change is calculated
from Equation (10.56) as
∆d = αd∆ t = 11.3 ×10−6°C−1( ) 22°C( )d = 2.486 ×10−4d
From Table 10.3 on page 388, the interference for a Class 7 fit is δ=0.0005d. For this shaft, upon
cooling, the interference decreases an additional amount due to the heating, so that
δ=0.0005d-2.486x10-4d=0.0002514d
From Table 10.3, it can be seen that the class of fit has changed to Class 6.
If the temperature change is 50° higher, then the diametral change is
∆d = αd∆ t = 11.3 ×10−6°C−1( ) 50°C( )d = 5.65×10−4 d
Therefore, the interference is now
δ=0.0005d-5.65x10-4d=-0.000065d
This means that the shaft is loose. The class of fit becomes meaningless for this case, but it could
reasonably be labeled as Class 4.
10.3 A cylinder with a 0.3-m inner diameter and a 0.40m outer diameter is internally
pressurized to 140 MPa. Determine the maximum shear stress at the outer surface of the
cylinder.
www.elsolucionario.net
Page 10-2
Notes: Equation (10.2) suggests that thick walled cylinder theory holds. Therefore, Equations
(10.23) and (10.24) give σr and σθ, respectively, while Equation (10.9) gives σz. The maximum
shear stress is, as can be seen from a three-dimensional Mohr’s circle as in Figure 2.19 on page
59, τmax=σ1-σ3.
Solution:
Note that the wall thickness for this case is 0.05m, and di/th=0.3m/0.05m=6. Therefore, from
Equation (10.2), we must use thick walled cylinder theory. Therefore, using Equation (10.23) and
r=ro,
σr =
piri
2 1− ro
2
r2
 
 
  
 
 
  
ro
2 − ri
2 =
piri
2 1− ro
2
ro
2
 
 
  
 
 
  
ro
2 − ri
2 = 0
From Equation (10.24),
σθ =
piri
2 1 + ro
2
r2
 
 
  
 
 
  
ro
2 − ri
2 =
piri
2 1+ ro
2
ro
2
 
 
  
 
 
  
ro
2 − ri
2 =
2piri
2
ro
2 − ri
2 =
2 140MPa( ) 0.3m / 2( )2
0.4m / 2( )2 − 0.3m / 2( )2
= 360MPa
From equilibrium, we can equate the force from the internal pressure acting on the area πri
2
 to the
stress acting on area π(ro
2
-ri
2
). Therefore,
piπri2 = σ zπ ro2 − ri2( );σ z = piπri
2
π ro
2 − ri
2( ) =
piri
2
ro
2 − ri
2( ) =
140MPa( ) 0.15m( )2
0.2m( )2 − 0.15m( )2
= 180MPa
Since σ1≥σ2≥σ3, the principal stresses are σ1=360MPa, σ2=180MPa and σ3=0. Therefore, from
the three dimensional Mohr’s circle of Figure 2.19, the maximum shear stress is
τmax =
σ1 −σ3
2
= 360MPa − 0
2
= 180MPa
10.4 A rubber balloon has the shape of a cylinder with spherical ends. At low pressure the
cylindrical part is 25cm long with a diameter of 2 cm. The rubber material has constant
thickness and is linearly elastic. How long will the cylindrical part of the balloon be when
it is inflated to a diameter of 10cm? Assume that the rubber’s modulus of elasticity is
constant and that the Poisson’s ratio is 0.5.
Notes: It will be assumed that the thickness, though not specified, is small enough to allow the
approximation of a thin-walled pressure vessel. Equations (10.10) and (10.11) give the stresses in
the hoop and axial directions. The stress is related to the strain by Equation (B.54) on page 918
with the radial stress equal to zero, or else the biaxial stress equations given in Table B.2 on page
920 can be used directly.
Solution:
From Equations (10.10) and (10.11):
σ1 =
pir
th
σ2 =
pir
2th
=
1
2
σ1
We know that for a thin-walled pressure vessel, the radial stress is negligible, so this is a biaxial
stress state. The strain ε1 occurs in the θ-direction and is (see Equation 4.23):
www.elsolucionario.net
Page 10-3
ε1 =
δ
l
=
π d f − di( )
πdi
=
10cm − 2cm
2cm
= 4
Hooke’s Law for this biaxial stress case is given in Table B.2 on page 920. In the first principal
direction, this yields:
ε1 =
σ1
E
−
νσ2
E
=
σ1
E
−
0.5( ) 1
2
σ1
 
  
 
  
E
=
3σ1
4E
= 4; σ1 =
16E
3
In the other principal direction, Table B.2 on page 920 gives:
ε2 =
σ2
E
−
νσ1
E
=
σ1
E
1
2
−
1
2
 
  
 
  
= 0
Therefore, the balloon does not stretch. Note, however, that any other value of Poisson’s ratio
than 0.5 would lead to balloon elongation.
10.5 A thin-walled