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MACHINE DESIGN - An Integrated Approach, 4th Ed. 1-1-1
PROBLEM 1-1
Statement: It is often said, "Build a better mousetrap and the world will beat a path to your door." Consider
this problem and write a goal statement and a set of at least 12 task specifications that you would
apply to its solution. Then suggest 3 possible concepts to achieve the goal. Make annotated,
freehand sketches of the concepts.
Solution:
Goal Statement: Create a mouse-free environment.
Task Specifications:
1. Cost less than $1.00 per use or application.
2. Allow disposal without human contact with mouse.
3. Be safe for other animals such as house pets.
4. Provide no threat to children or adults in normal use.
5. Be a humane method for the mouse.
6. Be environmentally friendly.
7. Have a shelf-life of at least 3 months.
8. Leave no residue.
9. Create minimum audible noise in use.
10. Create no detectable odors within 1 day of use.
11. Be biodegradable.
12. Be simple to use with minimal written instructions necessary.
Concepts and sketches are left to the student. There are an infinity of possibilities.
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 1-2-1
PROBLEM 1-2
Statement: A bowling machine is desired to allow quadriplegic youths, who can only move a joystick, to
engage in the sport of bowling at a conventional bowling alley. Consider the factors involved,
write a goal statement, and develop a set of at least 12 task specifications that constrain this
problem. Then suggest 3 possible concepts to achieve the goal. Make annotated, freehand
sketches of the concepts.
Solution:
Goal Statement: Create a means to allow a quadriplegic to bowl.
Task Specifications:
1. Cost no more than $2 000.
2. Portable by no more than two able-bodied adults.
3. Fit through a standard doorway.
4. Provide no threat of injury to user in normal use.
5. Operate from a 110 V, 60 Hz, 20 amp circuit.
6. Be visually unthreatening.
7. Be easily positioned at bowling alley.
8. Have ball-aiming ability, controllable by user.
9. Automatically reload returned balls.
10. Require no more than 1 able-bodied adult for assistance in use.
11. Ball release requires no more than a mouth stick-switch closure.
12. Be simple to use with minimal written instructions necessary.
Concepts and sketches are left to the student. There are an infinity of possibilities.
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 1-3-1
PROBLEM 1-3
Statement: A quadriplegic needs an automated page turner to allow her to read books without assistance.
Consider the factors involved, write a goal statement, and develop a set of at least 12 task
specifications that constrain this problem. Then suggest 3 possible concepts to achieve the goal.
Make annotated, freehand sketches of the concepts.
Solution:
Goal Statement: Create a means to allow a quadriplegic to read standard books with minimum assistance.
Task Specifications:
1. Cost no more than $1 000.
2. Useable in bed or from a seated position
3. Accept standard books from 8.5 x 11 in to 4 x 6 in in planform and up to 1.5 in thick.
4. Book may be placed, and device set up, by able-bodied person.
5. Operate from a 110 V, 60 Hz, 15 amp circuit or by battery power.
6. Be visually unthreatening and safe to use.
7. Require no more than 1 able-bodied adult for assistance in use.
8. Useable in absence of assistant once set up.
9. Not damage books.
10. Timing controlled by user.
11. Page turning requires no more than a mouth stick-switch closure.
12. Be simple to use with minimal written instructions necessary.
Concepts and sketches are left to the student. There are an infinity of possibilities.
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 1-4-1
PROBLEM 1-4
Statement: Convert a mass of 1 000 lbm to (a) lbf, (b) slugs, (c) blobs, (d) kg.
Units: blob
lbf sec
2
in
:=
Given: Mass M 1000 lb:=
Solution: See Mathcad file P0104.
1. To determine the weight of the given mass, multiply the mass value by the acceleration due to gravity, g.
W M g:= W 1000 lbf=
2. Convert mass units by assigning different units to the units place-holder when displaying the mass value.
Slugs M 31.081 slug=
Blobs M 2.59 blob=
Kilograms M 453.592 kg=
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 1-5-1
PROBLEM 1-5
Statement: A 250-lbm mass is accelerated at 40 in/sec2. Find the force in lb needed for this acceleration.
Given: Mass M 250 lb:= Acceleration a 40
in
sec
2
:=
Solution: See Mathcad file P0105.
1. To determine the force required, multiply the mass value, in slugs, by the acceleration in feet per second squared.
Convert mass to slugs: M 7.770 slug=
Convert acceleration to feet per second squared: a 3.333s
2-
ft=
F M a:= F 25.9 lbf=
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 1-6-1
PROBLEM 1-6
Statement: Express a 100-kg mass in units of slugs, blobs, and lbm. How much does this mass weigh?
Units: blob
lbf sec
2
in
Given: M 100 kg
Assumptions: The mass is at sea-level and the gravitational acceleration is
g 32.174
ft
sec
2
or g 386.089
in
sec
2
or g 9.807
m
sec
2
Solution: See Mathcad file P0106.
1. Convert mass units by assigning different units to the units place-holder when displaying the mass value.
The mass, in slugs, is M 6.85 slug
The mass, in blobs, is M 0.571 blob
The mass, in lbm, is M 220.5 lb
Note: Mathcad uses lbf for pound-force, and lb for pound-mass.
2. To determine the weight of the given mass, multiply the mass value by the acceleration due to gravity, g.
The weight, in lbf, is W M g W 220.5 lbf
The weight, in N, is W M g W 980.7 N
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 1-7-1
PROBLEM 1-7
Statement: Prepare an interactive computer program (using, for example, Excell, Mathcad, or TKSolver) from
which the cross-sectional properties for the shapes shown in the inside front cover can be
calculated. Arrange the program to deal with both ips and SI unit systems and convert the results
between those systems.
Solution: See the inside front cover and Mathcad file P0107.
1. Rectangle, let:
b 3 in h 4 in
Area A b h A 12.000 in
2
A 7742 mm
2
Moment about x-axis Ix
b h
3
12
Ix 16.000 in
4
Ix 6.660 10
6
mm
4
Moment about y-axis Iy
h b
3
12
Iy 9.000 in
4
Iy 3.746 10
6
mm
4
Radius of gyration about x-axis kx
Ix
A
kx 1.155 in
kx 29.329 mm
Radius of gyration about y-axis ky
Iy
A
ky 0.866 in
ky 21.997 mm
Polar moment of inertia Jz Ix Iy Jz 25.000 in
4
Jz 1.041 10
7
mm
4
2. Solid circle, let:
D 3 in
Area A
π D
2
4
A 7.069 in
2
A 4560 mm
2
Moment about x-axis Ix
π D
4
64
Ix 3.976 in
4
Ix 1.655 10
6
mm
4
Moment about y-axis Iy
π D
4
64
Iy 3.976 in
4
Iy 1.655 10
6
mm
4
Radius ofgyration about x-axis kx
Ix
A
kx 0.750 in
kx 19.05 mm
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 1-7-2
Radius of gyration about y-axis ky
Iy
A
ky 0.750 in
ky 19.05 mm
Polar moment of inertia Jz
π D
4
32
Jz 7.952 in
4
Jz 3.310 10
6
mm
4
3. Hollow circle, let:
D 3 in d 1 in
Area A
π
4
D
2
d
2
A 6.283 in2
A 4054 mm
2
Moment about x-axis Ix
π
64
D
4
d
4
Ix 3.927 in4
Ix 1.635 10
6
mm
4
Moment about y-axis Iy
π
64
D
4
d
4
Iy 3.927 in4
Iy 1.635 10
6
mm
4
Radius of gyration about x-axis kx
Ix
A
kx 0.791 in
kx 20.08 mm
Radius of gyration about y-axis ky
Iy
A
ky 0.791 in
ky 20.08 mm
Polar moment of inertia Jz
π
32
D
4
d
4
Jz 7.854 in4
Jz 3.269 10
6
mm
4
4. Solid semicircle, let:
D 3 in R 0.5 D R 1.5 in
Area A
π D
2
8
A 3.534 in
2
A 2280 mm
2
Moment about x-axis Ix 0.1098 R
4
Ix 0.556 in
4
Ix 2.314 10
5
mm
4
Moment about y-axis Iy
π R
4
8
Iy 1.988 in
4
Iy 8.275 10
5
mm
4
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 1-7-3
Radius of gyration about x-axis kx
Ix
A
kx 0.397 in
kx 10.073 mm
Radius of gyration about y-axis ky
Iy
A
ky 0.750 in
ky 19.05 mm
Polar moment of inertia Jz Ix Iy Jz 2.544 in
4
Jz 1.059 10
6
mm
4
Distances to centroid a 0.4244 R a 0.637 in
a 16.17 mm
b 0.5756 R b 0.863 in
b 21.93 mm
5. Right triangle, let:
b 2 in h 1 in
Area A
b h
2
A 1.000 in
2
A 645 mm
2
Moment about x-axis Ix
b h
3
36
Ix 0.056 in
4
Ix 2.312 10
4
mm
4
Moment about y-axis Iy
h b
3
36
Iy 0.222 in
4
Iy 9.250 10
4
mm
4
Radius of gyration about x-axis kx
Ix
A
kx 0.236 in
kx 5.987 mm
Radius of gyration about y-axis ky
Iy
A
ky 0.471 in
ky 11.974 mm
Polar moment of inertia Jz Ix Iy Jz 0.278 in
4
Jz 1.156 10
5
mm
4
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 1-8-1
PROBLEM 1-8
Statement: Prepare an interactive computer program (using, for example, Excell, Mathcad, or TKSolver) from
which the mass properties for the solids shown in the page opposite the inside front cover can be
calculated. Arrange the program to deal with both ips and SI unit systems and convert the results
between those systems.
Units: blob
lbf sec
2
in
Solution: See the page opposite the inside front cover and Mathcad file P0108.
1. Rectangular prism, let: a 2 in b 3 in c 4 in γ 0.28 lbf in
3
Volume V a b c V 24.000 in
3
V 393290 mm
3
Mass M
V γ
g
M 0.017 blob
M 3.048 kg
Moment about x-axis Ix
M a
2
b
2
12
Ix 0.019 blob in
2
Ix 2130.4 kg mm
2
Moment about y-axis Iy
M a
2
c
2
12
Iy 0.029 blob in
2
Iy 3277.6 kg mm
2
Moment about z-axis Iz
M b
2
c
2
12
Iz 0.036 blob in
2
Iz 4097.0 kg mm
2
Radius of gyration about x-axis kx
Ix
M
kx 1.041 in
kx 26.437 mm
Radius of gyration about y-axis ky
Iy
M
ky 1.291 in
ky 32.791 mm
Radius of gyration about z-axis kz
Iz
M
kz 1.443 in
kz 36.662 mm
2.Cylinder, let: r 2 in L 3 in γ 0.30 lbf in
3
Volume V π r
2
L V 37.699 in
3
V 617778 mm
3
Mass M
V γ
g
M 0.029 blob
M 5.13 kg
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 1-8-2
Moment about x-axis Ix
M r
2
2
Ix 0.059 blob in
2
Ix 6619.4 kg mm
2
Moment about y-axis Iy
M 3 r
2
L
2
12
Iy 0.051 blob in
2
Iy 5791.9 kg mm
2
Moment about z-axis Iz
M 3 r
2
L
2
12
Iz 0.051 blob in
2
Iz 5791.9 kg mm
2
Radius of gyration about x-axis kx
Ix
M
kx 1.414 in
kx 35.921 mm
Radius of gyration about y-axis ky
Iy
M
ky 1.323 in
ky 33.601 mm
Radius of gyration about z-axis kz
Iz
M
kz 1.323 in
kz 33.601 mm
3. Hollow cylinder, let:
a 2 in b 3 in L 4 in γ 0.28 lbf in
3
Volume V π b
2
a
2
L V 62.832 in3
V 1029630 mm
3
Mass M
V γ
g
M 0.046 blob
M 7.98 kg
Moment about x-axis Ix
M
2
a
2
b
2
Ix 0.296 blob in2
Ix 3.3 10
4
kg mm
2
Moment about y-axis Iy
M
12
3 a
2
3 b
2
L
2
Iy 0.209 blob in2
Iy 2.4 10
4
kg mm
2
Moment about z-axis Iz
M
12
3 a
2
3 b
2
L
2
Iz 0.209 blob in2
Iz 2.4 10
4
kg mm
2
Radius of gyration about x-axis kx
Ix
M
kx 2.550 in
kx 64.758 mm
Radius of gyration about y-axis ky
Iy
M
ky 2.141 in
ky 54.378 mm
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 1-8-3
Radius of gyration about z-axis kz
Iz
M
kz 2.141 in
kz 54.378 mm
4. Right circular cone, let:
r 2 in h 5 in γ 0.28 lbf in
3
Volume V
π r
2
h
3
V 20.944 in
3
V 343210 mm
3
Mass M
V γ
g
M 0.015 blob
M 2.66 kg
Moment about x-axis Ix
3
10
M r
2
Ix 0.018 blob in
2
Ix 2059.4 kg mm
2
Moment about y-axis Iy M
12 r
2
3 h
2
80
Iy 0.023 blob in
2
Iy 2638.5 kg mm
2
Moment about z-axis Iz M
12 r
2
3 h
2
80
Iz 0.023 blob in
2
Iz 2638.5 kg mm
2
Radius of gyration about x-axis kx
Ix
M
kx 1.095 in
kx 27.824 mm
Radius of gyration about y-axis ky
Iy
M
ky 1.240 in
ky 31.495 mm
Radius of gyration about z-axis kz
Iz
M
kz 1.240 in
kz 31.495 mm
5. Sphere, let:
r 3 in
Volume V
4
3
π r
3
V 113.097 in
3
V 1853333 mm
3
Mass M
V γ
g
M 0.082 blob
M 14.364 kg
Moment about x-axis Ix
2
5
M r
2
Ix 0.295 blob in
2
Ix 33362 kg mm
2
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 1-8-4
Moment about y-axis Iy
2
5
M r
2
Iy 0.295 blob in
2
Iy 33362 kg mm
2
Moment about z-axis Iz
2
5
M r
2
Iz 0.295 blob in
2
Iz 33362 kg mm
2
Radius of gyration about x-axis kx
Ix
M
kx 1.897 in
kx 48.193 mm
Radius of gyration about y-axis ky
Iy
M
ky 1.897 in
ky 48.193 mm
Radius of gyration about z-axis kz
Iz
M
kz 1.897 in
kz 48.193 mm
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 1-9-1
PROBLEM 1-9
Statement: Convert the template in Problem 1-7 to have and use a set of functions or subroutines that can be
called from within any program in that language to solve for the cross-sectional properties of the
shapes shown on the inside front cover.
Solution: See inside front cover and Mathcad file P0109.
1. Rectangle: Area A b h( ) b h
Moment about x-axis Ix b h( )
b h
3
12
Moment about y-axis Iy b h( )
h b
3
12
2. Solidcircle: Area A D( )
π D
2
4
Moment about x-axis Ix D( )
π D
4
64
Moment about y-axis Iy D( )
π D
4
64
3. Hollow circle: Area A D d( )
π
4
D
2
d
2
Moment about x-axis Ix D d( )
π
64
D
4
d
4
Moment about y-axis Iy D d( )
π
64
D
4
d
4
4. Solid semicircle:
Area A D( )
π D
2
8
Moment about x-axis Ix R( ) 0.1098 R
4
Moment about y-axis Iy R( )
π R
4
8
5. Right triangle:
Area A b h( )
b h
2
Moment about x-axis Ix b h( )
b h
3
36
Moment about y-axis Iy b h( )
h b
3
36
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 1-10-1
PROBLEM 1-10
Statement: Convert the template in Problem 1-8 to have and use a set of functions or subroutines that can be
called from within any program in that language to solve for the cross-sectional properties of the
shapes shown on the page opposite the inside front cover.
Solution: See the page opposite the inside front cover and Mathcad file P0110.
1 Rectangular prism:
Volume V a b c( ) a b c
Mass M a b c γ( )
V a b c( ) γ
g
Moment about x-axis Ix a b c γ( )
M a b c γ( ) a
2
b
2
12
Moment about y-axis Iy a b c γ( )
M a b c γ( ) a
2
c
2
12
Moment about z-axis Iz a b c γ( )
M a b c γ( ) b
2
c
2
12
2. Cylinder:
Volume V r L( ) π r
2
L
Mass M r L γ( )
V r L( ) γ
g
Moment about x-axis Ix r L γ( )
M r L γ( ) r
2
2
Moment about y-axis Iy r L γ( )
M r L γ( ) 3 r
2
L
2
12
Moment about z-axis Iz r L γ( )
M r L γ( ) 3 r
2
L
2
12
3. Hollow cylinder:
Volume V a b L( ) π b
2
a
2
L
Mass M a b L γ( )
V a b L( ) γ
g
Moment about x-axis Ix a b L γ( )
M a b L γ( )
2
a
2
b
2
Moment about y-axis Iy a b L γ( )
M a b L γ( )
12
3 a
2
3 b
2
L
2
Moment about z-axis Iz a b L γ( )
M a b L γ( )
12
3 a
2
3 b
2
L
2
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 1-10-2
4. Right circular cone:
Volume V r h( )
π r
2
h
3
Mass M r h γ( )
V r h( ) γ
g
Moment about x-axis Ix r h γ( )
3
10
M r h γ( ) r
2
Moment about y-axis Iy r h γ( ) M r h γ( )
12 r
2
3 h
2
80
Moment about z-axis Iz r h γ( ) M r h γ( )
12 r
2
3 h
2
80
5. Sphere:
Volume V r( )
4
3
π r
3
Mass M r γ( )
V r( ) γ
g
Moment about x-axis Ix r γ( )
2
5
M r γ( ) r
2
Moment about y-axis Iy r γ( )
2
5
M r γ( ) r
2
Moment about z-axis Iz r γ( )
2
5
M r γ( ) r
2
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-1-1
PROBLEM 2-1
Statement: Figure P2-1 shows stress-strain curves for three failed tensile-test specimens. All are plotted on
the same scale.
(a) Characterize each material as brittle or ductile.
(b) Which is the stiffest?
(c) Which has the highest ultimate strength?
(d) Which has the largest modulus of resilience?
(e) Which has the largest modulus of toughness?
Solution: See Figure P2-1 and Mathcad file P0201.
1. The material in Figure P2-1(a) has a moderate amount of strain beyond the yield point, P2-1(b) has very little, and
P2-1(c) has considerably more than either of the other two. Based on this observation, the material in Figure
P2-1(a) is mildly ductile, that in P2-1(b)is brittle, and that in P2-1(c) is ductile.
2. The stiffest material is the one with the grearesr slope in the elastic range. Determine this by dividing the rise by
the run of the straight-line portion of each curve. The material in Figure P2-1(c) has a slope of 5 stress units per
strain unit, which is the greatest of the three. Therefore, P2-1(c) is the stiffest.
3. Ultimate strength corresponds to the highest stress that is achieved by a material under test. The material in
Figure P2-1(b) has a maximum stress of 10 units, which is considerably more than either of the other two.
Therefore, P2-1(b) has the highest ultimate strength.
4. The modulus of resilience is the area under the elastic portion of the stress-starin curve. From observation of the
three graphs, the stress and strain values at the yield points are:
P2-1(a) σya 5:= εya 5:=
P2-1(b) σyb 9:= εyb 2:=
P2-1(c) σyc 5:= εyc 1:=
Using equation (2.7), the modulus of resiliency for each material is, approximately,
P21a
1
2
σya⋅ εya⋅:= P21a 12.5=
P21b
1
2
σyb⋅ εyb⋅:= P21b 9=
P21c
1
2
σyc⋅ εyc⋅:= P21c 2.5=
P2-1 (a) has the largest modulus of resilience
5. The modulus of toughness is the area under the stress-starin curve up to the point of fracture. By inspection,
P2-1 (c) has the largest area under the stress-strain curve therefore, it has the largest modulus of toughness.
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-2-1
PROBLEM 2-2
Statement: Determine an approximate ratio between the yield strength and ultimate strength for each material
shown in Figure P2-1.
Solution: See Figure P2-1 and Mathcad file P0202.
1. The yield strength is the value of stress at which the stress-strain curve begins to be nonlinear. The ultimate
strength is the maximum value of stress attained during the test. From the figure, we have the following values
of yield strength and tensile strength:
Figure P2-1(a) Sya 5:= Sua 6:=
Figure P2-1(b) Syb 9:= Sub 10:=
Figure P2-1(c) Syc 5:= Suc 8:=
2. The ratio of yield strength to ultimate strength for each material is:
Figure P2-1(a) ratioa
Sya
Sua
:= ratioa 0.83=
Figure P2-1(b) ratiob
Syb
Sub
:= ratiob 0.90=
Figure P2-1(c) ratioc
Syc
Suc
:= ratioc 0.63=
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-3-1
PROBLEM 2-3
Statement: Which of the steel alloys shown in Figure 2-19 would you choose to obtain
(a) Maximum strength
(b) Maximum modulus of resilience
(c) Maximum modulus of toughness
(d) Maximum stiffness
Given: Young's modulus for steel E 207 GPa
Solution: See Figure 2-19 and Mathcad file P0203.
1. Determine from the graph: values for yield strength, ultimate strength and strain at fracture for each material.
Steel Yield Strength Ultimate Strength Fracture Strain
AISI 1020: Sy1020 300 MPa Sut1020 400 MPa εf1020 0.365
AISI 1095: Sy1095 550 MPa Sut1095 1050 MPa εf10950.11
AISI 4142: Sy4142 1600 MPa Sut4142 2430 MPa εf4142 0.06
Note: The 0.2% offset method was used to define a yield strength for the AISI 1095 and the 4142 steels.
2. From the values of Sut above it is clear that the AISI 4142 has maximum strength.
3. Using equation (2-7) and the data above, determine the modulus of resilience.
UR1020
1
2
Sy1020
2
E
UR1020 0.22
MN m
m
3
UR1095
1
2
Sy1095
2
E
UR1095 0.73
MN m
m
3
UR4142
1
2
Sy4142
2
E
UR4142 6.18
MN m
m
3
Even though the data is approximate, the AISI 4142 clearly has the largest modulus of resilience.
4. Using equation (2-8) and the data above, determine the modulus of toughness.
UT1020
1
2
Sy1020 Sut1020 εf1020 UT1020 128 MN m
m
3
UT1095
1
2
Sy1095 Sut1095 εf1095 UT1095 88 MN m
m
3
UT4142
1
2
Sy4142 Sut4142 εf4142 UT4142 121 MN m
m
3
Since the data is approximate, there is no significant difference between the 1020 and 4142 steels. Because of
the wide difference in shape and character of the curves, one should also determine the area under the
curves by graphical means. When this is done, the area under the curve is about 62 square units for 1020
and 66 for 4142. Thus, they seem to have about equal toughness, which is about 50% greater than that for
the 1095 steel.
5. All three materials are steel therefore, the stiffnesses are the same.
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-4-1
PROBLEM 2-4
Statement: Which of the aluminum alloys shown in Figure 2-21 would you choose to obtain
(a) Maximum strength
(b) Maximum modulus of resilience
(c) Maximum modulus of toughness
(d) Maximum stiffness
Given: Young's modulus for aluminum E 71.7 GPa
Solution: See Figure 2-21 and Mathcad file P0204.
1. Determine, from the graph, values for yield strength, ultimate strength and strain at fracture for each material.
Alum Yield Strength Ultimate Strength Fracture Strain
1100: Sy1100 120 MPa Sut1100 130 MPa εf1100 0.170
2024-T351: Sy2024 330 MPa Sut2024 480 MPa εf2024 0.195
7075-T6: Sy7075 510 MPa Sut7075 560 MPa εf7075 0.165
Note: The 0.2% offset method was used to define a yield strength for all of the aluminums.
2. From the values of Sut above it is clear that the 7075-T6 has maximum strength.
3. Using equation (2-7) and the data above, determine the modulus of resilience.
UR1100
1
2
Sy1100
2
E
UR1100 0.10
MN m
m
3
UR2024
1
2
Sy2024
2
E
UR2024 0.76
MN m
m
3
UR7075
1
2
Sy7075
2
E
UR7075 1.81
MN m
m
3
Even though the data is approximate, the 7075-T6 clearly has the largest modulus of resilience.
4. Using equation (2-8) and the data above, determine the modulus of toughness.
UT1100
1
2
Sy1100 Sut1100 εf1100 UT1100 21 MN m
m
3
UT2024
1
2
Sy2024 Sut2024 εf2024 UT2024 79 MN m
m
3
UT7075
1
2
Sy7075 Sut7075 εf7075 UT7075 88 MN m
m
3
Even though the data is approximate, the 7075-T6 has the largest modulus of toughness.
5. All three materials are aluminum therefore, the stiffnesses are the same.
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-5-1
PROBLEM 2-5
Statement: Which of the thermoplastic polymers shown in Figure 2-22 would you choose to obtain
(a) Maximum strength
(b) Maximum modulus of resilience
(c) Maximum modulus of toughness
(d) Maximum stiffness
Solution: See Figure 2-22 and Mathcad file P0205.
1. Determine, from the graph, values for yield strength, ultimate strength, strain at fracture, and modulus of
elasticity for each material.
Plastic Yield Strength Ultimate Strength Fracture Strain Mod of Elasticity
Nylon 101: SyNylon 63 MPa SutNylon 80 MPa εfNylon 0.52 ENylon 1.1 GPa
HDPE: SyHDPE 15 MPa SutHDPE 23 MPa εfHDPE 3.0 EHDPE 0.7 GPa
PTFE: SyPTFE 8.3 MPa SutPTFE 13 MPa εfPTFE 0.51 EPTFE 0.8 GPa
2. From the values of Sut above it is clear that the Nylon 101 has maximum strength.
3. Using equation (2-7) and the data above, determine the modulus of resilience.
URNylon
1
2
SyNylon
2
ENylon
URNylon 1.8
MN m
m
3
URHDPE
1
2
SyHDPE
2
EHDPE
URHDPE 0.16
MN m
m
3
URPTFE
1
2
SyPTFE
2
EPTFE
URPTFE 0.04
MN m
m
3
Even though the data is approximate, the Nylon 101 clearly has the largest modulus of resilience.
4. Using equation (2-8) and the data above, determine the modulus of toughness.
UTNylon
1
2
SyNylon SutNylon εfNylon UTNylon 37 MN m
m
3
UTHDPE
1
2
SyHDPE SutHDPE εfHDPE UTHDPE 57 MN m
m
3
UTPTFE
1
2
SyPTFE SutPTFE εfPTFE UTPTFE 5 MN m
m
3
Even though the data is approximate, the HDPE has the largest modulus of toughness.
5. The Nylon 101 has the steepest slope in the (approximately) elastic range and is, therefore, the stiffest of the
three materials..
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-6-1
PROBLEM 2-6
Statement: A metal has a strength of 414 MPa at its elastic limit and the strain at that point is 0.002. What is
the modulus of elasticity? What is the strain energy at the elastic limit? Assume that the test
speimen is 12.8-mm dia and has a 50-mm gage length. Can you define the type of metal based on
the given data?
Given: Elastic limit: Strength Sel 414 MPa Strain εel 0.002
Test specimen: Diameter do 12.8 mm Length Lo 50 mm
Solution: See Mathcad file P0206.
1. The modulus of elasticity is the slope of the stress-strain curve, which is a straight line, in the elastic region.
Since one end of this line is at the origin, the slope (modulus of elasticity) is
E
Sel
εel
E 207 GPa
2. The strain energy per unit volume at the elastic limit is the area under the stress-strain curve up to the elastic
limit. Since the curve is a straight line up to this limit, the area is one-half the base times the height, or
U'el
1
2
Sel εel U'el 414
kN m
m
3
The total strain energy in the specimen is the strain energy per unit volume times the volume,
Uel U'el
π do
2
4
Lo Uel 2.7 N m
3. Based on the modulus of elasticity and using Table C-1, the material is steel.
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-7-1
PROBLEM 2-7
Statement: A metal has a strength of 41.2 kpsi (284 MPa) at its elastic limit and the strain atthat point is
0.004. What is the modulus of elasticity? What is the strain energy at the elastic limit?
Assume that the test speimen is 0.505-in dia and has a 2-in gage length. Can you define the
type of metal based on the given data?
Given: Elastic limit: Strength Sel 41.2 ksi Strain εel 0.004 Sel 284 MPa
Test specimen: Diameter do 0.505 in Length Lo 2.00 in
Solution: See Mathcad file P0207.
1. The modulus of elasticity is the slope of the stress-strain curve, which is a straight line, in the elastic region.
Since one end of this line is at the origin, the slope (modulus of elasticity) is
E
Sel
εel
E 10.3 10
6
psi E 71 GPa
2. The strain energy per unit volume at the elastic limit is the area under the stress-strain curve up to the elastic
limit. Since the curve is a straight line up to this limit, the area is one-half the base times the height, or
U'el
1
2
Sel εel U'el 82.4
lbf in
in
3
U'el 568
kN m
m
3
The total strain energy in the specimen is the strain energy per unit volume times the volume,
Uel U'el
π do
2
4
Lo Uel 33.0 in lbf
3. Based on the modulus of elasticity and using Table C-1, the material is aluminum.
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-8-1
PROBLEM 2-8
Statement: A metal has a strength of 134 MPa at its elastic limit and the strain at that point is 0.006. What is
the modulus of elasticity? What is the strain energy at the elastic limit? Assume that the test
speimen is 12.8-mm dia and has a 50-mm gage length. Can you define the type of metal based on
the given data?
Given: Elastic limit: Strength Sel 134 MPa Strain εel 0.003
Test specimen: Diameter do 12.8 mm Length Lo 50 mm
Solution: See Mathcad file P0208.
1. The modulus of elasticity is the slope of the stress-strain curve, which is a straight line, in the elastic region.
Since one end of this line is at the origin, the slope (modulus of elasticity) is
E
Sel
εel
E 45 GPa
2. The strain energy per unit volume at the elastic limit is the area under the stress-strain curve up to the elastic
limit. Since the curve is a straight line up to this limit, the area is one-half the base times the height, or
U'el
1
2
Sel εel U'el 201
kN m
m
3
The total strain energy in the specimen is the strain energy per unit volume times the volume,
Uel U'el
π do
2
4
Lo Uel 1.3 N m
3. Based on the modulus of elasticity and using Table C-1, the material is magnesium.
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-9-1
PROBLEM 2-9
Statement: A metal has a strength of 100 kpsi (689 MPa) at its elastic limit and the strain at that point is
0.006. What is the modulus of elasticity? What is the strain energy at the elastic limit?
Assume that the test speimen is 0.505-in dia and has a 2-in gage length. Can you define the
type of metal based on the given data?
Given: Elastic limit: Strength Sel 100 ksi Strain εel 0.006
Sel 689 MPa
Test specimen: Diameter do 0.505 in Length Lo 2.00 in
Solution: See Mathcad file P0209.
1. The modulus of elasticity is the slope of the stress-strain curve, which is a straight line, in the elastic region.
Since one end of this line is at the origin, the slope (modulus of elasticity) is
E
Sel
εel
E 16.7 10
6
psi E 115 GPa
2. The strain energy per unit volume at the elastic limit is the area under the stress-strain curve up to the elastic
limit. Since the curve is a straight line up to this limit, the area is one-half the base times the height, or
U'el
1
2
Sel εel U'el 300
lbf in
in
3
U'el 2 10
3
kN m
m
3
The total strain energy in the specimen is the strain energy per unit volume times the volume,
Uel U'el
π do
2
4
Lo Uel 120.18 in lbf
3. Based on the modulus of elasticity and using Table C-1, the material is titanium.
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-10-1
PROBLEM 2-10
Statement: A material has a yield strength of 689 MPa at an offset of 0.6% strain. What is its modulus of
resilience?
Units: MJ 10
6
joule
Given: Yield strength Sy 689 MPa
Yield strain εy 0.006
Solution: See Mathcad file P0210.
1. The modulus of resilience (strain energy per unit volume) is given by Equation (2.7) and is approximately
UR
1
2
Sy εy UR 2.067
MJ
m
3
UR 2.1 MPa
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-11-1
PROBLEM 2-11
Statement: A material has a yield strength of 60 ksi (414 MPa) at an offset of 0.2% strain. What is its
modulus of resilience?
Units: MJ 10
6
joule
Given: Yield strength Sy 60 ksi Sy 414 MPa
Yield strain εy 0.002
Solution: See Mathcad file P0211.
1. The modulus of resilience (strain energy per unit volume) is given by Equation (2.7) and is approximately
UR
1
2
Sy εy UR 60
in lbf
in
3
UR 0.414
MJ
m
3
UR 0.414 MPa
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-12-1
PROBLEM 2-12
Statement: A steel has a yield strength of 414 MPa, an ultimate tensile strength of 689 MPa, and an
elongation at fracture of 15%. What is its approximate modulus of toughness? What is the
approximate modulus of resilience?
Given: Sy 414 MPa Sut 689 MPa εf 0.15
Solution: See Mathcad file P0212.
1. Determine the modulus of toughness using Equation (2.8).
UT
Sy Sut
2
εf UT 82.7
MN m
m
3
UT 82.7 MPa
2. Determine the modulus of resilience using Equation (2.7) and Young's modulus for steel: E 207 GPa
UR
1
2
Sy
2
E
UR 414
kN m
m
3
UR 0.41 MPa
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-13-1
PROBLEM 2-13
Statement: The Brinell hardness of a steel specimen was measured to be 250 HB. What is the material's
approximate tensile strength? What is the hardness on the Vickers scale? The Rockwell scale?
Given: Brinell hardness of specimen HB 250
Solution: See Mathcad file P0213.
1. Determine the approximate tensile strength of the material from equations (2.10), not Table 2-3.
Sut 0.5 HB ksi Sut 125 ksi Sut 862 MPa
2. From Table 2-3 (using linear interpolation) the hardness on the Vickers scale is
HV
HB 241
277 241
292 253( ) 253 HV 263
3. From Table 2-3 (using linear interpolation) the hardness on the Rockwell C scale is
HRC
HB 241
277 241
28.8 22.8( ) 22.8 HRC 24.3
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-14-1
PROBLEM 2-14
Statement: The Brinell hardness of a steel specimen was measured to be 340 HB. What is the material's
approximate tensile strength? What is the hardness on the Vickers scale? The Rockwell scale?
Given: Brinell hardness of specimen HB 340
Solution: See Mathcad file P0214.
1. Determine the approximate tensile strength of the material from equations (2.10), not Table 2-3.
Sut 0.5 HB ksi Sut 170 ksi Sut 1172 MPa
2. From Table 2-3 (using linear interpolation) the hardness on the Vickers scale is
HV
HB 311
341 311
360 328( ) 328 HV 359
3. From Table 2-3 (using linear interpolation) the hardness on the Rockwell C scale is
HRC
HB 311
341 311
36.6 33.1( ) 33.1 HRC 36.5
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-15-1
PROBLEM 2-15
Statement: What are the principal alloy elements of an AISI 4340 steel? How much carbon does it have? Is it
hardenable? By what techniques?
Solution: See Mathcad file P0215.
1. Determine the principal alloying elements from Table 2-5 for 43xx steel..
1.82% Nickel
0.50 or 0.80% Chromium
0.25% Molybdenum
2. From "Steel Numbering Systems" in Section 2.6, the carbon content is
From the last two digits, the carbon content is 0.40%.
3. Is it hardenable? Yes, all of the alloying elements increase the hardenability. By what techniques? It can be
through hardened by heating, quenching and tempering; and it can also be case hardened (See Section 2.4).
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-16-1
PROBLEM 2-16
Statement: What are the principal alloy elements of an AISI 1095 steel? How much carbon does it have? Is it
hardenable? By what techniques?
Solution: See Mathcad file P0216.
1. Determine the principal alloying elements from Table 2-5 for 10xx steel.
Carbon only, no alloying elements
2. From "Steel Numbering Systems" in Section 2.6, the carbon content is
From the last two digits, the carbon content is 0.95%.
3. Is it hardenable? Yes, as a high-carbon steel, it has sufficient carbon content for hardening. By what
techniques? It can be through hardened by heating, quenching and tempering; and it can also be case
hardened (See Section 2.4).
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-17-1
PROBLEM 2-17
Statement: What are the principal alloy elements of an AISI 6180 steel? How much carbon does it have? Is it
hardenable? By what techniques?
Solution: See Mathcad file P0217.
1. Determine the principal alloying elements from Table 2-5 for 61xx steel..
0.15% Vanadium
0.60 to 0.95% Chromium
2. From "Steel Numbering Systems" in Section 2.6, the carbon content is
From the last two digits, the carbon content is 0.80%.
3. Is it hardenable? Yes, all of the alloying elements increase the hardenability. By what techniques? It can be
through hardened by heating, quenching and tempering; and it can also be case hardened (See Section 2.4).
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-18-1
PROBLEM 2-18
Statement: Which of the steels in Problems 2-15, 2-16, and 2-17 is the stiffest?
Solution: See Mathcad file P0218.
1. None. All steel alloys have the same Young's modulus, which determines stiffness.
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-19-1
PROBLEM 2-19
Statement: Calculate the specific strength and specific stiffness of the following materials and pick one for
use in an aircraft wing spar.
Given: Material Code Ultimate Strength Young's Modulus Weight Density
Steel st 0 Sut
st
80 ksi E
st
30 10
6
psi γ
st
0.28
lbf
in
3
Aluminum al 1 Sut
al
60 ksi E
al
10.4 10
6
psi γ
al
0.10
lbf
in
3
Titanium ti 2 Sut
ti
90 ksi E
ti
16.5 10
6
psi γ
ti
0.16
lbf
in
3
Index i 0 1 2
Solution: See Mathcad file P0219.
1. Specific strength is the ultimate tensile strength divided by the weight density and specific stiffness is the
modulus of elasticity divided by the weight density. The text does not give a symbol to these quantities.
Specific strength
Sut
i
γ
i
1
in
3286·10
3600·10
3563·10
Specific stiffness
E
i
γ
i
1
in
6107·10
6104·10
6103·10
Steel
Aluminum
Titanium
2. Based on the results above, all three materials have the same specific stiffness but the aluminum has the largest
specific strength. Aluminum for the aircraft wing spar is recommended.
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-20-1
PROBLEM 2-20
Statement: If maximum impact resistance were desired in a part, which material properties would you look for?
Solution: See Mathcad file P0220.
1. Ductility and a large modulus of toughness (see "Impact Resistance" in Section 2.1).
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-21-1
PROBLEM 2-21 _____
Statement: Refer to the tables of material data in Appendix A and determine the strength-to-weight ratios of
the following material alloys based on their tensile yield strengths: heat-treated 2024 aluminum,
SAE 1040 cold-rolled steel, Ti-75A titanium, type 302 cold-rolled stainless steel.
Given: Material Yield Strength Specific Weight
Mat
1
"2024 Aluminum, HT" Sy
1
290 MPa γ
1
0.10 lbf in
3
γ
1
27.14
kN
m
3
Mat
2
"1040 CR Steel" Sy
2
490 MPa γ
2
0.28 lbf in
3
γ
2
76.01
kN
m
3
Mat
3
"Ti-75A Titanium" Sy
3
517 MPa γ
3
0.16 lbf in
3
γ
3
43.43
kN
m
3
Mat
4
"Type 302 CR SS" Sy
4
1138 MPa γ
4
0.28 lbf in
3
γ
4
76.01
kN
m
3
i 1 2 4
Solution: See Mathcad file P0221.
1. Calculate the strength-to-weight ratio for each material as described in Section 2.1.
SWR
i
Sy
i
γ
i
SWR
i
10
4
m1.068
0.645
1.190
1.497
Mat
i
"2024 Aluminum, HT"
"1040 CR Steel"
"Ti-75A Titanium"
"Type 302 CR SS"
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-22-1
PROBLEM 2-22 _____
Statement: Refer to the tables of material data in Appendix A and determine the strength-to-weight ratios of
the following material alloys based on their ultimate tensile strengths: heat-treated 2024 aluminum
SAE 1040 cold-rolled steel, unfilled acetal plastic, Ti-75A titanium, type 302 cold-rolled stainless
steel.
Given: Material Tensile Strength Specific Weight
Mat
1
"2024 Aluminum, HT" Sut
1
441 MPa γ
1
0.10 lbf in
3
γ
1
27.14 kN m
3
Mat
2
"1040 CR Steel" Sut
2
586 MPa γ
2
0.28 lbf in
3
γ
2
76.01 kN m
3
Mat
3
"Acetal, unfilled" Sut
3
60.7 MPa γ
3
0.051 lbf in
3
γ
3
13.84 kN m
3
Mat
4
"Ti-75A Titanium" Sut
4
586 MPa γ
4
0.16 lbf in
3
γ
4
43.43 kN m
3
Mat
5
"Type 302 CR SS" Sut
5
1310 MPa γ
5
0.28 lbf in
3
γ
5
76.01 kN m
3
i 1 2 5
Solution: See Mathcad file P0222.
1. Calculate the strength-to-weight ratio for each material as described in Section 2.1.
SWR
i
Sut
i
γ
i
SWR
i
10
4
m
1.625
0.771
0.438
1.349
1.724
Mat
i
"2024 Aluminum, HT"
"1040 CR Steel"
"Acetal, unfilled"
"Ti-75A Titanium"
"Type 302 CR SS"
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-23-1
PROBLEM 2-23 _____
Statement: Refer to the tables of material data in Appendix A and calculate the specific stiffness of
aluminum, titanium, gray cast iron, ductile iron, bronze, carbon steel, and stainless steel.
Rank them in increasing order of this property and discuss the engineering significance of
these data.
Units: Mg 10
3
kg
Given: Material Modulus of Elasticity Density
Mat
1
"Aluminum" E
1
71.7 GPa ρ
1
2.8 Mg m
3
Mat
2
"Titanium" E
2
113.8 GPa ρ
2
4.4 Mg m
3
Mat
3
"Gray cast iron" E
3
103.4 GPa ρ
3
7.2 Mg m
3
Mat
4
"Ductile iron" E
4
168.9 GPa ρ
4
6.9 Mg m
3
Mat
5
"Bronze" E
5
110.3 GPa ρ
5
8.6 Mg m
3
Mat
6
"Carbon steel" E
6
206.8 GPa ρ
6
7.8 Mg m
3
Mat
7
"Stainless steel" E
7
189.6 GPa ρ
7
7.8 Mg m
3
i 1 2 7
Solution: See Mathcad file P0223.
1. Calculate the specific stiffness for each material as described in Section 2.1.
E'
i
E
i
ρ
i
E'
i
10
6
s
2
m
2
25.6
25.9
14.4
24.5
12.8
26.5
24.3
Mat
i
"Aluminum"
"Titanium"
"Gray cast iron"
"Ductile iron"
"Bronze"
"Carbon steel"
"Stainless steel"
2. Rank them in increasing order of specific stiffness.
Mat
5
"Bronze"
E'
5
10
6
s
2
m
2
12.8
Mat
3
"Gray cast iron"
E'
3
10
6
s
2
m
2
14.4
Mat
7
"Stainless steel"
E'
7
10
6
s
2
m
2
24.3
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-23-2
Mat
4
"Ductile iron"
E'
4
10
6
s
2
m
2
24.5
Mat
1
"Aluminum"
E'
1
10
6
s
2
m
2
25.6
Mat
2
"Titanium"
E'
2
10
6
s
2
m
2
25.9
Mat
6
"Carbon steel"
E'
6
10
6
s
2
m
2
26.5
3. Bending and axial deflection are inversely proportional to the modulus of elasticity. For the same shape and
dimensions, the material with the highest specific stiffness will give the smallest deflection. Or, put another
way, for a given deflection, using the material with the highest specific stiffness will result in the least
weight.
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-24-1
PROBLEM 2-24
Statement: Call your local steel and aluminum distributors (consult the Yellow Pages) and obtain current
costs per pound for round stock of consistent size in low-carbon (SAE 1020) steel, SAE 4340
steel, 2024-T4 aluminum, and 6061-T6 aluminum. Calculate a strength/dollar ratio and a
stiffness/dollar ratio for each alloy. Which would be your first choice on a cost-efficiency
basis for an axial-tension-loaded round rod
(a) If maximum strength were needed?
(b) If maximum stiffness were needed?
Solution: Left to the student as data will vary with time and location.
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-25-1
PROBLEM 2-25
Statement: Call your local plastic stock-shapes distributors (consult the Yellow Pages) and obtain current
costs per pound for round rod or tubing of consistent size in plexiglass, acetal, nylon 6/6, and
PVC. Calculate a strength/dollar ratio and a stiffness/dollar ratio for each alloy. Which would
be your first choice on a cost-efficiency basis for an axial-tension-loaded round rod or tube of
particular diameters.
(a) If maximum strength were needed?
(b) If maximum stiffness were needed?
Solution: Left to the student as data will vary with time and location.
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-26-1
PROBLEM 2-26
Statement: A part has been designed and its dimensions cannot be changed. To minimize its deflections
under the same loading in all directions irrespective of stress levels, which material woulod you
choose among the following: aluminum, titanium, steel, or stainless steel?
Solution: See Mathcad file P0226.
1. Choose the material with the highest modulus of elasticity because deflection is inversely proportional to
modulus of elasticity. Thus, choose steel unless there is a corrosive atmosphere, in which case, choose
stainless steel.
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-27-1
PROBLEM 2-27
Statement: Assuming that the mechanical properties data given in Appendix Table A-9 for some carbon
steels represents mean values, what is the value of the tensile yield strength for 1050 steel
quenched and tempered at 400F if a reliability of 99.9% is required?
Given: Mean yield strength Sy 117 ksi Sy 807 MPa
Solution: See Mathcad file P0227.
1. From Table 2-2 the reliability factor for 99.9% is Re 0.753 . Applying this to the mean tensile strength gives
Sy99.9 Sy Re Sy99.9 88.1 ksi Sy99.9 607 MPa
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-28-1
PROBLEM 2-28
Statement: Assuming that the mechanical properties data given in Appendix Table A-9 for some carbon
steels represents mean values, what is the value of the ultimate tensile strength for 4340 steel
quenched and tempered at 800F if a reliability of 99.99% is required?
Given: Mean ultimate tensile strength Sut 213 ksi Sut 1469 MPa
Solution: See Mathcad file P0228.
1. From Table 2-2 the reliability factor for 99.99% is Re 0.702 . Applying this to the mean ultimate tensile
strength gives
Sut99.99 Sut Re Sut99.99 150 ksi Sut99.99 1031 MPa
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-29-1
PROBLEM 2-29
Statement: Assuming that the mechanical properties data given in Appendix Table A-9 for some carbon
steels represents mean values, what is the value of the ultimate tensile strength for 4130 steel
quenched and tempered at 400F if a reliability of 90% is required?
Given: Mean ultimate tensile strength Sut 236 ksi Sut 1627 MPa
Solution: See Mathcad file P0229.
1. From Table 2-2 the reliability factor for 90% is Re 0.897 . Applying this to the mean ultimate
tensile strength gives
Sut99.99 Sut Re Sut99.99 212 ksi Sut99.99 1460 MPa
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-30-1
PROBLEM 2-30
Statement: Assuming that the mechanical properties data given in Appendix Table A-9 for some carbon
steels represents mean values, what is the value of the tensile yield strength for 4140 steel
quenched and tempered at 800F if a reliability of 99.999% is required?
Given: Mean yield strength Sy 165 ksi Sy 1138 MPa
Solution: See Mathcad file P0230.
1. From Table 2-2 the reliability factor for 99.999% is Re 0.659 . Applying this to the mean
tensile
strength gives
Sy99.9 Sy Re Sy99.9 109 ksi Sy99.9 750 MPa
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-31-1
PROBLEM 2-31
Statement: A steel part is to be plated to give it better corrosion resistance. Two materials are being
considered: cadmium and nickel. Considering only the problem of galvanic action, which would
you chose? Why?
Solution: See Mathcad file P0231.
1. From Table 2-4 we see that cadmium is closer to steel than nickel. Therefore, from the standpoint of reduced
galvanic action, cadmium is the better choice. Also, since cadmium is less noble than steel it will be the material
that is consumed by the galvanic action. If nickel were used the steel would be consumed by galvanic action.
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-32-1
PROBLEM 2-32
Statement: A steel part with many holes and sharp corners is to be plated with nickel. Two processes are
being considered: electroplating and electroless plating. Which process would you chose? Why?
Solution: See Mathcad file P0232.
1. Electroless plating is the better choice since it will give a uniform coating thickness in the sharp corners and in
the holes. It also provides a relatively hard surface of about 43 HRC.
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-33-1
PROBLEM 2-33
Statement: What is the common treatment used on aluminum to prevent oxidation? What other metals can
also be treated with this method? What options are available with this method?
Solution: See Mathcad file P0233.
1. Aluminum is commonly treated by anodizing, which creates a thin layer of aluminum oxide on the surface.
Titanium, magnesium, and zinc can also be anodized. Common options include tinting to give various colors to
the surface and the use of "hard anodizing" to create a thicker, harder surface.
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-34-1
PROBLEM 2-34
Statement: Steel is often plated with a less nobel metal that acts as a sacrificial anode that will corrode instead
of the steel. What metal is commonly used for this purpose (when the finished product will not be
exposed to saltwater), what is the coating process called, and what are the common processes used
to obtain the finished product?
Solution: See Mathcad file P0234.
1. The most commonly used metal is zinc. The process is called "galvanizing" and it is accomplished by
electroplating or hot dipping.
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-35-1
PROBLEM 2-35
Statement: A low-carbon steel part is to be heat-treated to increase its strength. If an ultimate tensile
strength of approximately 550 MPa is required, what mean Brinell hardness should the part
have after treatment? What is the equivalent hardness on the Rockwell scale?
Given: Approximate tensile strength Sut 550 MPa
Solution: See Mathcad file P0235.
1. Use equation (2.10), solving for the Brinell hardness, HB.
Sut 3.45 HB= HB
Sut
3.45 MPa
HB 159
2. From Table 2-3, the equivalent hardness on the Rocwell scale is 83.9HRB.
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-36-1
PROBLEM 2-36
Statement: A low-carbon steel part has been tested for hardness using the Brinell method and is found to
have a hardness of 220 HB. What are the approximate lower and upper limits of the ultimate
tensile strength of this part in MPa?
Given: Hardness HB 220
Solution: See Mathcad file P0236.
1. Use equation (2.10), solving for ultimate tensile strength.
Minimum: Sutmin 3.45 HB 0.2 HB( ) MPa Sutmin 715 MPa
Maximum: Sutmax 3.45 HB 0.2 HB( ) MPa Sutmax 803 MPa
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-37-1
PROBLEM 2-37Statement: Figure 2-24 shows "guide lines" for minimum weight design when failure is the criterion. The
guide line, or index, for minimizing the weight of a beam in bending is f
2/3/, where f is the
yield strength of a material and is its mass density. For a given cross-section shape the
weight of a beam with given loading will be minimized when this index is maximized. The
following materials are being considered for a beam application: 5052 aluminum, cold rolled;
CA-170 beryllium copper, hard plus aged; and 4130 steel, Q & T @ 1200F. The use of which of
these three materials will result in the least-weight beam?
Units: Mg kg
3
Given: 5052 Aluminum Sya 255 MPa ρa 2.8 Mg m
3
CA-170 beryllium copper Syb 1172 MPa ρb 8.3 Mg m
3
4130 steel Sys 703 MPa ρs 7.8 Mg m
3
Solution: See Mathcad file P0237.
1. The values for the mass density are taken from Appendix Table A-1 and the values of yield strength come
from from Tables A-2, A-4, and A-9 for aluminum, beryllium copper, and steel, respectively.
2. Calculate the index value for each material.
Index Sy ρ
Sy
0.667
ρ
Mg m
3
MPa
0.667
Aluminum Ia Index Sya ρa Ia 14.4
Beryllium copper Ib Index Syb ρb Ib 13.4
Steel Is Index Sys ρs Is 10.2
The 5052 aluminum has the highest value of the index and would be the best choice to minimize weight.
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-38-1
PROBLEM 2-38
Statement: Figure 2-24 shows "guide lines" for minimum weight design when failure is the criterion. The
guide line, or index, for minimizing the weight of a member in tension is f/, where f is the
yield strength of a material and is its mass density. The weight of a member with given
loading will be minimized when this index is maximized. For the three materials given in Problem
2-37, which will result in the least weight tension member?
Units: Mg kg
3
Given: 5052 Aluminum Sya 255 MPa ρa 2.8 Mg m
3
CA-170 beryllium copper Syb 1172 MPa ρb 8.3 Mg m
3
4130 steel Sys 703 MPa ρs 7.8 Mg m
3
Solution: See Mathcad file P0238.
1. The values for the mass density are taken from Appendix Table A-1 and the values of yield strength come
from from Tables A-2, A-4, and A-9 for aluminum, beryllium copper, and steel, respectively.
2. Calculate the index value for each material.
Index Sy ρ
Sy
ρ
Mg m
3
MPa
Aluminum Ia Index Sya ρa Ia 91.1
Beryllium copper Ib Index Syb ρb Ib 141.2
Steel Is Index Sys ρs Is 90.1
The beryllium copper has the highest value of the index and would be the best choice to minimize weight.
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-39-1
PROBLEM 2-39
Statement: Figure 2-23 shows "guide lines" for minimum weight design when stiffness is the criterion. The
guide line, or index, for minimizing the weight of a beam in bending is E1/2/, where E is the
modulus of elasticity of a material and is its mass density. For a given cross-section shape the
weight of a beam with given stiffness will be minimized when this index is maximized. The
following materials are being considered for a beam application: 5052 aluminum, cold rolled;
CA-170 beryllium copper, hard plus aged; and 4130 steel, Q & T @ 1200F. The use of which of
these three materials will result in the least-weight beam?
Units: Mg kg
3
Given: 5052 Aluminum Ea 71.7 GPa ρa 2.8 Mg m
3
CA-170 beryllium copper Eb 127.6 GPa ρb 8.3 Mg m
3
4130 steel Es 206.8 GPa ρs 7.8 Mg m
3
Solution: See Mathcad file P0239.
1. The values for the mass density and modulus are taken from Appendix Table A-1.
2. Calculate the index value for each material.
Index E ρ( )
E
0.5
ρ
Mg m
3
GPa
0.5
Aluminum Ia Index Ea ρa Ia 3.0
Beryllium copper Ib Index Eb ρb Ib 1.4
Steel Is Index Es ρs Is 1.8
The 5052 aluminum has the highest value of the index and would be the best choice to minimize weight.
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-40-1
PROBLEM 2-40
Statement: Figure 2-24 shows "guide lines" for minimum weight design when stiffness is the criterion. The
guide line, or index, for minimizing the weight of a member in tension is E/, where E is the
modulus of elasticity of a material and is its mass density. The weight of a member with given
stiffness will be minimized when this index is maximized. For the three materials given in Problem
2-39, which will result in the least weight tension member?
Units: Mg kg
3
Given: 5052 Aluminum Ea 71.7 GPa ρa 2.8 Mg m
3
CA-170 beryllium copper Eb 127.6 GPa ρb 8.3 Mg m
3
4130 steel Es 206.8 GPa ρs 7.8 Mg m
3
Solution: See Mathcad file P0240.
1. The values for the mass density and modulus are taken from Appendix Table A-1.
2. Calculate the index value for each material.
Index E ρ( )
E
ρ
Mg m
3
GPa
Aluminum Ia Index Ea ρa Ia 25.6
Beryllium copper Ib Index Eb ρb Ib 15.4
Steel Is Index Es ρs Is 26.5
The steel has the highest value of the index and would be the best choice to minimize weight.
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-1-1
PROBLEM 3-1
Statement: Which load class from Table 3-1 best suits these systems?
(a) Bicycle frame
(b) Flag pole
(c) Boat oar
(d) Diving board
(e) Pipe wrench
(f) Golf club.
Solution: See Mathcad file P0301.
1. Determine whether the system has stationary or moving elements, and whether the there are constant or
time-varying loads.
(a) Bicycle frame Class 4 (Moving element, time-varying loads)
(b) Flag pole Class 2 (Stationary element, time-varying loads)
(c) Boat oar Class 2 (Low acceleration element, time-varying loads)
(d) Diving board Class 2 (Stationary element, time-varying loads)
(e) Pipe wrench Class 2 (Low acceleration elements, time-varying loads)
(f) Golf club Class 4 (Moving element, time-varying loads)
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-2a-1
PROBLEM 3-2a
Statement: Draw free-body diagrams for the system of Problem 3-1a (bicycle frame).
Assumptions: 1. A two-dimensional model is adequate.
2. The lower front-fork bearing at C takes all of the thrust load from the front forks.
3. There are no significant forces on the handle bars.
Solution: See Mathcadfile P0302a.
1. A typical bicycle frame is shown in Figure 3-2a. There are five points on the frame where external forces and
moments are present. The rider's seat is mounted through a tube at A. This is a rigid connection, capable of
transmitting two force components and a moment. The handle bars and front-wheel forks are supported by the
frame through two bearings, located at B and C. These bearings are capable of transmitting radial and axial
loads. The pedal-arm assembly is supported by bearings at D. These bearings are capable of transmitting radial
loads. The rear wheel-sprocket assembly is supported by bearings mounted on an axle fixed to the frame at E.
dyF
ex
E
F
eyF
dxF D
eR
axF
ayF
a
aM
A
R
dR
bR
Rc
C
ct
br
B
F F
α
crF
FIGURE 3-2a
Free Body Diagram for Problem 3-2a
2. The loads at B and C can be determined by analyzing a FBD of the front wheel-front forks assembly. The loads
at D can be determined by analyzing a FBD of the pedal-arm and front sprocket (see Problem 3-3), and the loads at E
can be determined by analyzing a FBD of the rear wheel-sprocket assembly.
3. With the loads at B, C, D, and E known, we can apply equations 3.3b to the FBD of the frame and solve for Fax,
Fay, and Ma.
ΣΣΣΣ F
x
: Fax− Fbr cos α( )⋅− Fcr cos α( )⋅+ Fct sin α( )⋅− Fdx− Fex+ 0= (1)
ΣΣΣΣ F
y
: Fay− Fbr sin α( )⋅− Fcr sin α( )⋅+ Fct cos α( )⋅+ Fdy− Fey+ 0= (2)
ΣΣΣΣ M
z
: Ma Rbx Fby⋅ Rby Fbx⋅−( )+ Rcx Fcy⋅ Rcy Fcx⋅−( )+
Rdx Fdy⋅ Rdy Fdx⋅−( ) Rex Fey⋅ Rey Fex⋅−( )++
...�
�
�
�
�
�
0= (3)
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-2e-1
PROBLEM 3-2e
Statement: Draw free-body diagrams for the system of Problem 3-1e (pipe wrench).
Assumptions: A two-dimensional model is adequate.
Solution: See Mathcad file P0302e.
1. A typical pipe wrench with a pipe clamped in its jaw is shown in Figure 3-2e(a). When a force Fhand is applied on
the wrench, the piping system provides an equal and opposite force and a resisting torque, Tpipe.
T
F
pipe
hand
hand
a
F
(a) FBD of pipe wrench and pipe
bd
Fay
F
ax A
Fbt
Fbn
α
(b) FBD of pipe wrench only
FIGURE 3-2e
Free Body Diagrams for Problem 3-2e
2. The pipe reacts with the wrench at the points of contact A and B. The forces here will be directed along the
common normals and tangents. The jaws are slightly tapered and, as a result, the action of Fhand tends to drive the
wrench further into the taper, increasing the normal forces. This, in turn, allows for increasing tangential forces. It is
the tangential forces that produce the turning torque.
3. Applying equations 3.3b to the FBD of the pipe wrench,
ΣΣΣΣ F
x
: Fax− Fbn cos α( )⋅+ Fbt sin α( )⋅− 0= (1)
ΣΣΣΣ F
y
: Fay− Fbn sin α( )⋅+ Fbt cos α( )⋅+ Fhand− 0= (2)
ΣΣΣΣ M
A
: d Fbt cos α( )⋅ Fbn sin α( )⋅+( )⋅ d a+( ) Fhand⋅− 0= (3)
4. These equations can be solved for the vertical forces if we assume α is small so that sin(α) = 0 and cos (α) = 1.
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-3-1
PROBLEM 3-3
Statement: Draw a free-body diagram of the pedal-arm assembly from a bicycle with the pedal arms in the
horizontal position and dimensions as shown in Figure P3-1. (Consider the two arms, pedals and
pivot as one piece). Assuming a rider-applied force of 1500 N at the pedal, determine the torque
applied to the chain sprocket and the maximum bending moment and torque in the pedal arm.
Given: a 170 mm b 60 mm Frider 1.5 kN
Assumptions: The pedal-arm assembly is supprted by bearings at A and at B.
Solution: See Figure 3-3 and Mathcad file P0303.
1. The free-body diagram (FBD) of the pedal-arm assembly (including the sprocket) is shown in Figure 3-3a. The
rider-applied force is Frider and the force applied by the chain (not shown) is Fchain. The radial bearing
reactions are Fax, Faz, Fbx, and Fbz. Thus, there are five unknowns Fchain, Fax, Faz, Fbx, and Fbz. In general, we
can write six equilibrium equations for a three-dimensional force system, but in this system there are no forces
in the y-direction so five equations are available to solve for the unknowns.
F
rider
Pedal
x
b Arm
a
F
ax
Fbx
A
B Arm (sectioned)
y
Sprocket
Faz
Fbz
Fchain
z
(a) FBD of complete pedal-arm assembly
x
Pedal
b Arm
F
rider
a
c
F
Mc
y
z
Tc
(b) FBD of pedal and arm with section through the origin
FIGURE 3-3
Dimensions and Free Body Diagram of the pedal-arm assembly for Problem 3-3
2. The torque available to turn the sprocket is found by summing moments about the sprocket axis. From Figure
3-3a, it is
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-3-2
T
y-axis
: a Frider r Fchain a Frider Tsprocket= 0=
Tsprocket a Frider Tsprocket 255 N m
where r is the sprocket pitch radius.
3. In order to determine the bending moment and twisting torque in the pedal arm, we will cut the arm with a
section plane that goes through the origin and is parallel to the y-z plane, removing everything beyond that
plane and replacing it with the internal forces and moments in the pedal arm at the section. The resulting
FBD is shown in Figure 3-3b. The internal force at section C is Fc the internal bending moment is Mc, and
the internal twisting moment (torque) is Tc. We can write three equilibrium equations to solve for these
three unknowns:
Shear force in pedal arm at section C
F
z
: Fc Frider 0= Fc Frider Fc 1.5 kN
Bending moment in pedal arm at section C
M
y-axis
: a Frider Mc 0= Mc a Frider Mc 255 N m
Twisting moment in pedal arm at section C
M
x-axis
: b Frider Tc 0= Tc b Frider Tc 90 N m
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MACHINE DESIGN - An Integrated Approach,4th Ed. 3-4-1
PROBLEM 3-4
Statement: The trailer hitch from Figure 1-1 has loads applied as shown in Figure P3-2. The tongue weight
of 100 kg acts downward and the pull force of 4905 N acts horizontally. Using the dimensions
of the ball bracket in Figure 1-5 (p. 15), draw a free-body diagram of the ball bracket and find the
tensile and shear loads applied to the two bolts that attach the bracket to the channel in Figure
1-1.
Given: a 40 mm b 31 mm c 70 mm d 20 mm
Mtongue 100 kg Fpull 4.905 kN t 19 mm
Assumptions: 1. The nuts are just snug-tight (no pre-load), which is the worst case.
2. All reactions will be concentrated loads rather than distributed loads or pressures.
Solution: See Figure 3-4 and Mathcad file P0304.
1. The weight on the tongue is
Wtongue Mtongue g Wtongue 0.981 kN
2. The FBD of the hitch and bracket assembly is shown in Figure 3-4. The known external forces that act on the
ball are Fpull and Wtongue . The reactions onthe bracket are at points C and D. The bolts at C provide tensile
(Fc2x) and shear (Fc2y) forces, and the bracket resists rotation about point D where the reaction force Fd2 is
applied by the channel to which the bracket is bolted.
3. Solving for the reactions by summing the horizontal and vertical forces and the moments about D:
F
x
: Fpull Fc2x Fd2 0= (1)
F
y
: Fc2y Wtongue 0= (2)
M
D
: Fc2x d Fpull a t b d( ) Wtongue c 0= (3)
31 = b
20 = d
D Fd2
A
1
70 = c
C
2
B 19 = t
40 = a
Fc2x
C
F c2y
D
pull
2
B
F 1
tongueW
FIGURE 3-4
Dimensions and Free Body Diagram for Problem 3-4
4. Solving equation (3) for Fc2x
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MACHINE DESIGN - An Integrated Approach,4th Ed. 3-4-2
Fc2x
Fpull a t b d( ) Wtongue c
d
Fc2x 30.41 kN (4)
5. Substituting into (1) and solving for Fd2
Fd2 Fc2x Fpull Fd2 25.505 kN (5)
6. Solving (2) for Fc2y
Fc2y Wtongue Fc2y 0.981 kN (6)
7. The loads applied to the two bolts that attach the bracket to the channel are:
Axial force on two bolts Fc2x 30.4 kN
Shear force on two bolts Fc2y 0.98 kN
We assume that each bolt would carry one half of these loads.
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-5-1
PROBLEM 3-5
Statement: For the trailer hitch of Problem 3-4, determine the horizontal force that will result on the ball from
accelerating a 2000-kg trailer to 60 m/sec in 20 sec.
Given: Mass of trailer Mtrailer 2000 kg
Final velocity vf 60
m
sec
Time to reach velocity τ 20 sec
Assumptions: 1. Acceleration is constant.
2. The rolling resistance of the tires and the wheel bearings is negligible.
Solution: See Mathcad file P0305.
1. From elementary kinematics, the acceleration required is
a
vf
τ
a 3.00
m
sec
2
(1)
2. Using Newton's second law to find the force required to accelerate the trailer,
Fhitch Mtrailer a Fhitch 6.00 kN (2)
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-6-1
PROBLEM 3-6
Statement: For the trailer hitch of Problem 3-4, determine the horizontal force that will result on the ball from
an impact between the ball and the tongue of the 2000-kg trailer if the hitch deflects 2.8 mm
dynamically on impact. The tractor weighs 1000 kg. The velocity at impact is 0.3 m/sec.
Given: Mass of trailer Mtrailer 2000 kg
Dynamic deflection δi 2.8 mm
Mass of tractor Mtractor 1000 kg
Impact velocity vi 0.3
m
sec
Assumptions: 1. The tractor is the "struck member" because the hitch is on the tractor and it is the hitch that
deflects.
2. Equations (3.9) and (3.10) can be used to model the impact.
Solution: See Mathcad file P0306.
1. The weight of the trailer (the "striking member") is
Wtrailer Mtrailer g Wtrailer 19.613 kN
2. The correction factor, from equation (3-15), is
η
1
1
Mtractor
3 Mtrailer
η 0.857
3. Eliminating E from equations (3.9a) and (3.10) and solving for the horizontal force on the ball Fi yields
1
2
Fi δi η
1
2
Mtrailer vi
2
=
Fi
η Mtrailer vi
2
δi
Fi 55.1 kN
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-7-1
PROBLEM 3-7
Statement: The piston of an internal-combustion engine is connected to its connecting rod with a "wrist
pin." Find the force on the wrist pin if the 0.5-kg piston has an acceleration of 2 500 g.
Given: Mass of piston Mpiston 0.5 kg
Acceleration of piston apiston 2500 g
Assumptions: The force on the wrist pin due to the weight of the piston is very small compared with the
acceleration force.
Solution: See Mathcad file P0307.
1. The acceleration in m/s is apiston 2.452 10
4
m
sec
2
2. Using Newton's Second Law expressed in equation (3.1a), the force on the wrist pin is
Fwristpin Mpiston apiston Fwristpin 12.258 kN
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-8-1
PROBLEM 3-8 _____
Statement: A cam-follower system similar to that shown in Figure 3-15 has a mass m = 1 kg, a spring
constant k = 1000 N/m, and a damping coefficient d = 19.4 N-s/m. Find the undamped and
damped natural frequencies of this system.
Units: cps 2 π⋅ rad⋅ sec
1−
⋅:=
Given: Mass M 1 kg⋅:= , Spring constant k 1000 N⋅ m
1−
⋅:=
Damping coefficient d 19.4 N⋅ s⋅ m
1−
⋅:=
Solution: See Figure 3-15 and Mathcad file P0308.
1. Calculate the undamped natural frequency using equation 3.4.
ωn
k
M
:= ωn 31.6
rad
sec
= ωn 5.03 cps=
2. Calculate the undamped natural frequency using equation 3.7.
ωd
k
M
d
2 M⋅
�
�
�
�
�
�
2
−:= ωd 30.1
rad
sec
= ωd 4.79 cps=
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-9-1
PROBLEM 3-9
Statement: A ViseGrip plier-wrench is drawn to scale in Figure P3-3. Scale the drawing for dimensions. Find
the forces acting on each pin and member of the assembly for an assumed clamping force of P =
4000 N in the position shown. What force F is required to keep it in the clamped position shown?
Given: Clamping force P 4.00 kN
Dimensions a 50.0 mm e 28.0 mm α 21.0 deg
b 55.0 mm f 26.9 mm β 129.2 deg
c 39.5 mm g 2.8 mm
d 22.0 mm h 21.2 mm
Assumptions: Links 3 and 4 are in a toggle position, i.e., the pin that joins links 3 and 4 is in line with the pins
that join 1 with 4 and 2 with 3.
Solution: See Figure 3-9 and Mathcad file P0309.
1. The FBDs of the assembly and each individual link are shown in Figure 3-9. The dimensions, as scaled from
Figure P3-3 in the text, are given above and are shown on the link FBDs.
50.0 = a
22.0 = d
21.2 = h
F
26.9 = f
32
2.8 = g
28.0 = e
2
F12
P
14F
4
34F
F 4
2
1
P
P
F
55.0 = b
F
39.5 = c
3
F41
43F
23
F
129.2°
F21
1
F
3
P
FIGURE 3-9
Free Body Diagrams for Problem 3-9
2. Looking first at Part 3, we see that it is a three-force body. Therefore, the lines of action of the three forces
must intersect at a point. But, since Parts 3 and 4 are in a toggle position, F43 and F23 are colinear, which
means that their x- and y-components must be equal and opposite, leading to the conclussion that F = 0.
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MACHINE DESIGN - An Integrated Approach,4th Ed. 3-9-2
3. Now, looking at Part 1, we see that (for F = 0) it is also a three-force body, as is Part 2. In fact, the forces on
Part 1 are identical to those on Part 2. Solving for the unknown reactions on Parts 1 and 2,
F
x
: F41 cos 180 deg α( ) F21 cos β 180 deg( ) 0= (a)
F
y
: F41 sin 180 deg α( ) F21 sin β 180 deg( ) P 0= (b)
Solving equation (a) for F21
F21
F41 cos 180 deg α( )
cos β 180 deg( )
= (c)
Substituting equation (c) into (b)
F41 sin 180 deg α( )
F41 cos 180 deg α( )
cos β 180 deg( )
sin β 180 deg( ) P 0= (d)
Solving equation (d) for F41
F41
P
sin 180 deg α( )
cos 180 deg α( )
cos β 180 deg( )
sin β 180 deg( )
F41 5.1 kNF21
F41 cos 180 deg α( )
cos β 180 deg( )
F21 7.5 kN
Checking moment balance on Part 1,
F41 sin α( ) c F21 sin β 90 deg( ) d P a 0 kN m
The result is, within the accuracy of the scaled dimensions, zero as it must be.
4. The x and y components of the pin forces on Part 1 are
F41x F41 cos 180 deg α( ) F41x 4.749 kN
F41y F41 sin 180 deg α( ) F41y 1.823 kN
F21x F21 cos β 180 deg( ) F21x 4.749 kN
F21y F21 sin β 180 deg( ) F21y 5.823 kN
5. The forces on the pins at the ends of Part 4 are
F14 F41 F14 5.1 kN
F34 F14 F34 5.1 kN
6. The forces on the pins at the ends of Part 3 are
F43 F34 F43 5.1 kN
F23 F43 F23 5.1 kN
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-9-3
7. The forces on the pins at the ends of Part 2 are
F12 F21 F12 7.5 kN
F32 F23 F32 5.1 kN
Checking moment equilibrium on Part 2,
F12 e cos β 90 deg( ) g sin β 90 deg( )( )
F32 h cos α( ) f sin α( )( )
0 kN m
which is zero, as it must be.
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-10-1
PROBLEM 3-10
Statement: An overhung diving board is shown in Figure P3-4a. Find the reaction forces and construct the
shear and moment diagrams for this board with a 100 kg person standing at the free end.
Determine the maximum shear force, maximum moment and their locations.
700 = a
2000 = L
R
R
1
2
P
Given: Beam length L 2000 mm
Distance to support a 700 mm
Mass at free end M 100 kg
Assumptions: The weight of the beam is negligible
compared to the applied load and so can
be ignored.
Solution: See Figure 3-10 and Mathcad file P0310.
FIGURE 3-10A
Free Body Diagram for Problem 3-10
1. From inspection of Figure 3-10, write the load function equation
q(x) = -R
1
<x - 0>-1 + R
2
<x - a>-1 - P<x - L >-1
2. Integrate this equation from - to x to obtain shear, V(x)
V(x) = -R
1
<x - 0>0 + R
2
<x - a>0 - P<x - L >0
3. Integrate this equation from - to x to obtain moment, M(x)
M(x) = -R
1
<x - 0>1 + R
2
<x - a>1 - P<x - L >1
4. Determine the magnitude of the force, P P M g P 980.7 N
5. Solve for the reactions by evaluating the shear and moment equations at a point just to the right of x = L, where
both are zero.
At x = L+, V = M = 0 V R1 R2 P= 0= M R1 L R2 L a( )= 0=
R1 P
L a
a
R1 1821 N
R2 P R1 R2 2802 N
6. Define the range for x x 0 in 0.005 L L
7. For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than z,
and a value of one when it is greater than or equal to z.
S x z( ) if x z 1 0( )
8. Write the shear and moment equations in Mathcad form, using the function S as a multiplying factor to get the
effect of the singularity functions.
V x( ) R1 S x 0 in( ) R2 S x a( ) P S x L( )
M x( ) R1 S x 0 in( ) x R2 S x a( ) x a( ) P S x L( ) x L( )
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-10-2
9. Plot the shear and moment diagrams.
0 500 1000 1500 2000
2000
1000
0
1000
V x( )
N
x
mm
Shear
Diagram
0 500 1000 1500 2000
1500
1125
750
375
0
M x( )
N m
x
mm
Moment
Diagram
FIGURE 3-10B
Shear and Moment Diagrams for Problem 3-10
10. The maximum value of the shear force ocuurs throughout the distance from x = 0 to x = a and is
R1 1821 N
11. Find the maximum value of the bending moment by determining the value of x where the shear is zero.
Inspection of the shear diagram shows that this occurs at x = a.
Mmax M a( ) Mmax 1275 N m
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-11-1
PROBLEM 3-11
Statement: Determine the impact force and dynamic deflection that will result when the 100-kg person in
Problem 3-10 jumps up 250 mm and lands back on the board. Assume that the board weighs
29 kg and deflects 131 mm statically when the person stands on it. Find the reaction forces
and construct the shear and moment diagrams for this dynamic loading. Determine the
maximum shear force, maximum moment and their locations along the length of the board.
700 = a
2000 = L
R
R
1
2
F i
Given: Beam length L 2000 mm
Distance to support a 700 mm
Mass of person mpers 100 kg
Mass of board mboard 29 kg
Static deflection δst 131 mm
Height of jump h 250 mm
Assumptions: Equation (3.15) can be used to approximate a
mass correction factor.
FIGURE 3-11A
Free Body Diagram for Problem 3-11
Solution: See Figure 3-11 and Mathcad file P0311.
1. The person is the moving object and the board is the struck object. The mass ratio to be used in equation
(3.15) for the correction factor is
massratio
mpers
mboard
massratio 3.448
2. From equation (3.15), the correction factor is
η
1
1
1
3 massratio
η 0.912
3. The weight of the moving mass is Wpers mpers g Wpers 0.981 kN
4. The dynamic force is found by solving equation (3.14) for Fi.
Fi Wpers 1 1
2 η h
δst
Fi 3.056 kN
From this we see that the dynamic force ratio is
Fi
Wpers
3.12
5. From inspection of Figure 3-11, write the load function equation
q(x) = -R
1
<x - 0>-1 + R
2
<x - a>-1 - Fi<x - L >
-1
6. Integrate this equation from - to x to obtain shear, V(x)
V(x) = -R
1
<x - 0>0 + R
2
<x - a>0 - Fi<x - L >
0
7. Integrate this equation from - to x to obtain moment, M(x)
M(x) = -R
1
<x - 0>1 + R
2
<x - a>1 - Fi<x - L >
1
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-11-2
8. Solve for the reactions by evaluating the shear and moment equations at a point just to the right of x = L,
where both are zero.
At x = L+, V = M = 0 V R1 R2 Fi= 0= M R1 L R2 L a( )= 0=
R1 Fi
L a
a
R1 5676 N
R2 Fi R1 R2 8733 N
9. Define the range for x x 0 in 0.005 L L
10. For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than
z, and a value of one when it is greater than or equal to z.
S x z( ) if x z 1 0( )
11. Write the shear and moment equations in Mathcad form, using the functionS as a multiplying factor to get
the effect of the singularity functions.
V x( ) R1 S x 0 in( ) R2 S x a( ) Fi S x L( )
M x( ) R1 S x 0 in( ) x R2 S x a( ) x a( ) Fi S x L( ) x L( )
12. Plot the shear and moment diagrams.
Shear Diagram Moment Diagram
0 0.5 1 1.5 2
6
4
2
0
2
4
V x( )
kN
x
m
0 0.5 1 1.5 2
4
3
2
1
0
M x( )
kN m
x
m
FIGURE 3-11B
Shear and Moment Diagrams for Problem 3-11
13. The maximum value of the shear force ocuurs throughout the distance from x = 0 to x = a and is
R1 5676 N
14. Find the maximum value of the bending moment by determining the value of x where the shear is zero.
Inspection of the shear diagram shows that this occurs at x = a.
Mmax M a( ) Mmax 3973 N m
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-12-1
PROBLEM 3-12
Statement: An overhung diving board is shown in Figure P3-4b. Find the reaction forces and construct the
shear and moment diagrams for this board with a 100 kg person standing at the free end.
Determine the maximum shear force, maximum moment and their locations.
P
R 1
700
2000
1300 = L
M1
Given: Beam length L 1300 mm
Mass at free end M 100 kg
Assumptions: 1. The weight of the beam is negligible
compared to the applied load and so can
be ignored.
Solution: See Figure 3-12 and Mathcad file P0312.
1. From inspection of Figure 3-12, write the load function
equation FIGURE 3-12A
Free Body Diagram for Problem 3-12
q(x) = -M
1
<x - 0>-2 + R
1
<x - a>-1 - P<x - L >-1
2. Integrate this equation from - to x to obtain shear, V(x)
V(x) = -M
1
<x - 0>-1 + R
1
<x - a>0 - P<x - L >0
3. Integrate this equation from - to x to obtain moment, M(x)
M(x) = -M
1
<x - 0>0 + R
1
<x - a>1 - P<x - L >1
4. Determine the magnitude of the force, P P M g P 980.7 N
5. Solve for the reactions by evaluating the shear and moment equations at a point just to the right of x = L,
where both are zero.
At x = L+, V = M = 0 V R1 P= 0= M M1 R1 L= 0=
R1 P R1 981 N
M1 R1 L M1 1275 m N
6. Define the range for x x 0 in 0.005 L L
7. For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than z,
and a value of one when it is greater than or equal to z.
S x z( ) if x z 1 0( )
8. Write the shear and moment equations in Mathcad form, using the function S as a multiplying factor to get the
effect of the singularity functions.
V x( ) R1 S x 0 mm( ) P S x L( )
M x( ) M1 S x 0 mm( ) R1 S x 0 mm( ) x 0 mm( ) P S x L( ) x L( )
9. Plot the shear and moment diagrams.
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-12-2
0 0.5 1 1.5 2
0
200
400
600
800
1000
V x( )
N
x
m
Shear
Diagram
0 0.5 1 1.5 2
1500
1200
900
600
300
0
M x( )
N m
x
m
Moment
Diagram
FIGURE 3-12B
Shear and Moment Diagrams for Problem 3-12
10. The maximum value of the shear force ocuurs throughout the distance from x = 0 to x = L and is
R1 981 N
11. Find the maximum value of the bending moment by determining the value of x where the shear is zero.
Inspection of the shear diagram shows that this occurs at x = 0.
Mmax M 0 mm( ) Mmax 1275 N m
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-13-1
PROBLEM 3-13
Statement: Determine the impact force and dynamic deflection that will result when a 100-kg person jumps
up 25 cm and lands back on the board. Assume the board weighs 19 kg and deflects 8.5 cm
statically when the person stands on it. Find the reaction forces and construct the shear and
moment diagrams for this dynamic loading. Determine the maximum shear force, maximum
moment, and their locations along the length of the board.
Given: Total board length b 2000 mm
2000
1300 = L
R 1
700
M1
F i
Supported length a 700 mm
Mass of board mboard 19 kg
Static board deflection δstat 85 mm
Mass of person mperson 100 kg
Height of jump h 250 mm
Assumptions: 1. The board can be modelled as a cantilever
beam with maximum shear and moment at the
edge of the support.
FIGURE 3-13A
Solution: See Figure 3-13 and Mathcad file P0313.
Free Body Diagram for Problem 3-13
1. The person impacts the board upon landing. Thus, the board is the struck object and the person is the
striking object. To determine the force exerted by the person we will first need to know the impact correction
factor from equation (3.15).
η
1
1
mboard
3 mperson
η 0.94 (1)
2. We can now use equation (3.14) to determine the impact force, Fi,
Fi mperson g 1 1
2 η h
δstat
Fi 3.487 kN (2)
3. Write an equation for the load function in terms of equations 3.17 and integrate the resulting function twice
using equations 3.18 to obtain the shear and moment functions. Note use of the unit doublet function to
represent the moment at the wall. For the beam in Figure 3-13,
q(x) = -M
1
<x - 0>-2 + R
1
<x - 0>-1 - Fi<x - l>
-1 (3)
V(x) = -M
1
<x - 0>-1 + R
1
<x - 0>0 - Fi<x - l>
0 + C1 (4)
M(x) = -M
1
<x - 0>0 + R
1
<x - 0>1 - Fi<x - l>
1 + C1x+ C2 (5)
The reaction moment M1 at the wall is in the z direction and the forces R1 and Fi are in the y direction in
equation (4). All moments in equation (5) are in the z direction.
4. Because the reactions have been included in the loading function, the shear and moment diagrams both
close to zero at each end of the beam, making C1 = C2 = 0.
5. The reaction force R1 and the reaction moment M1 can be calculated from equations (4) and (5) respectively
by substituting the boundary conditions x = l+, V = 0, M = 0. Note that we can substitute l for l+ since their
difference is vanishingly small.
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-13-2
Unsupported beam length l b a l 1300 mm
V(l) = -M
1
<l - 0>-1 + R
1
<l - 0>0 - Fi<l - l>
0 = 0
V R1 Fi= 0= (6)
R1 Fi R1 3.487 kN
M(l) = -M
1
<l - 0>0 + R
1
<l - 0>1 - Fi<l - l>
1
= 0
M M1 R1 l Fi l l( )= 0= (7)
M1 R1 l M1 4533 N m
6. To generate the shear and moment functions over the length of the beam, equations (4) and (5) must be
evaluated for a range of values of x from 0 to l, after substituting the above values of C1, C2, R1, and M1 in them.
For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than the
dummy variable z, and a value of one when it is greater than or equal to z. It will have the same effect as the
singularity function.
Range of x x 0 in 0.005 l l
Unit step function S x z( ) if x z 1 0( )
Write the shear and moment equations in Mathcad form, using the function S as a multiplying factor to get
the effect of the singularity functions.
V x( ) R1 S x 0 in( ) x 0( )
0
Fi S xl( ) x l( )
0
(8)
M x( ) M1 S x 0 in( ) x 0( )
0
R1 S x 0 in( ) x 0( )
1
Fi S x l( ) x l( )
1
Plot the shear and moment diagrams (see below).
Shear Diagram Moment Diagram
0 0.5 1
0
1
2
3
V x( )
kN
x
m
0 0.5 1 1.5
5
4
3
2
1
0
M x( )
kN m
x
m
7. The graphs show that the shear force and the moment are both largest at x = 0. The function values of
these points can be calculated from equations (4) and (5) respectively by substituting x = 0 and evaluating the
singularity functions:
Vmax = V(0) = R1<0 - 0>
0 - Fi<0 - l>
0 = R1 (9)
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-13-3
Vmax R1 Vmax 3.49 kN
M.max = M(0) = -M1<0 - 0>
0 + R
1
<0 - 0>1 - Fi<0 - l>
1 = -M
1
(10)
Mmax M1 Mmax 4533 N m
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-14-1
PROBLEM 3-14
Statement: Figure P3-5 shows a child's toy called a pogo stick. The child stands on the pads, applying half
her weight on each side. She jumps off the ground, holding the pads up against her feet, and
bounces along with the spring cushioning the impact and storing energy to help each rebound.
Find the natural frequency of the system, the static deflection of the spring with the child
standing still, and the dynamic force and deflection when the child lands after jumping 2 in off
the ground.
P
F /2i F /2i
Units: blob
lbf sec
2
⋅
in
:=
Given: Child's weight Wc 60 lbf⋅:=
Spring constant k 100 lbf⋅ in
1−
⋅:=
Pogo stick weight Wp 5 lbf⋅:=
Height of drop h 2 in⋅:=
Assumptions: 1. An approximate energy method will be acceptable.
2. The correction factor for energy dissipation will be applied.
Solution: See Figure 3-14 and Mathcad file P0314.
1. Find the natural frequency of the (child/spring) system.
Mass of child (striker) m
Wc
g
:= m 0.155 blob⋅=
Mass of stick (struck) mb
Wp
g
:= mb 0.013 blob⋅=
Natural frequency ω
k
m
:= ω 25.367
rad
sec
⋅=
f
ω
2 π⋅
:= f 4.037 Hz⋅=
FIGURE 3-14
Free Body Diagram for Problem 3-14
2. The static deflection of the spring with the child standing still is
Static deflection of spring δst
Wc
k
:= δst 0.6 in⋅=
3. Determine the mass ratio correction factor from equation (3.15):
Correction factor η
1
1
mb
3 m⋅
+
:= η 0.973=
4. Using equation (3.14), determine the dynamic force.
Fi Wc 1 1
2 η⋅ h⋅
δst
++
�
�
�
�
�
�
⋅:= Fi 224 lbf⋅=
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-15-1
PROBLEM 3-15
Statement: A pen plotter imparts a constant acceleration of 2.5 m/sec2 to the pen assembly, which travels in
a straight line across the paper. The moving pen assembly weighs 0.5 kg. The plotter weighs 5
kg. What coefficient of friction is needed between the plotter feet and the table top on which it
sits to prevent the plotter from moving when the pen accelerates?
Given: Acceleration of pen ass'y a 2.5 m sec
2
Mass of pen ass'y mpen 0.5 kg
Mass of plotter mplot 5 kg
Solution: See Mathcad file P0315.
1. The force imparted to the pen assembly by the internal drive mechanism must be reacted at the table top by
the plotter feet. The horizontal force at the feet will be equal to the force on the pen assembly and must be
less than or equal to the maximum friction force, which is the product of the coefficient of friction and the
normal force, which is the weight of the plotter.
Horizontal driving
force on pen ass'y Fpen mpen a Fpen 1.25 N
Weight of plotter Wplot mplot g Wplot 49.033 N
Minimum coefficient
of friction μ
Fpen
Wplot
μ 0.025
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-16-1
PROBLEM 3-16
Statement: A track to guide bowling balls is designed with two round rods as shown in Figure P3-6. The
rods are not parallel to one another but have a small angle between them. The balls roll on the
rods until they fall between them and drop onto another track. The angle between the rods is
varied to cause the ball to drop at different locations. Each rod's unsupported length is 30 in
and the angle between them is 3.2 deg. The balls are 4.5 in dia and weigh 2.5 lb. The center
distance between the 1-in-dia rods is 4.2 in at the narrow end. Find the distance from the
narrow end at which the ball drops through and determine the worst-case shear and moment
maximum for the rods as the ball rolls a distance from the narrow end that is 98% of the distance
to drop. Assume that the rods are simply supported at each end and have zero deflection under
the applied loading. (Note that assuming zero deflection is unrealistic. This assumption will be
relaxed in the next chapter after deflection has been discussed.)
Given: Unsupported rod length L 30 in Bowling ball weight W 2.5 lbf
Half-angle between rods α 1.6 deg Rod diameter d 1.0 in
Bowling ball diameter D 4.5 in Half width of rod gap c 2.1 in
TOP VIEW
SECTION A-A
F
F
W/2
A A
Solution: See Figure 3-16 and Mathcad file P0316.
1. Calculate the distance between the ball and rod
centers.
Distance between centers h
D d
2
h 2.75 in
x
width(x)
c
u
(a) Distance between the roll axis and
the rod axis.
(b) Partial FBD of the bowling ball.
FIGURE 3-16
Dimensions and Free Body Diagrams for Problem 3-16
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-16-2
2. Let x be the distance along the roll axis, and u be the corresponding distance to the point of contact between
the ball and rods, measured along the rods. Then the distance from the center plane of the ball to the center
of a rod as shown in Figure 3-16(a) is,
width x( ) c cos α( ) x sin α( ) (1)
And the distance from the narrow end to the point at which the ball drops (assuming rigid rods) is
xdrop
h c cos α( )
sin α( )
xdrop 23.31 in
The distance along the rod corresponding to xdrop is
udrop
xdrop h sin α( )
cos α( )
udrop 23.24 in
3. The angle made by a line through the ball-rod centers and the horizontal plane (see Figure 3-16b) is
θ x( ) acos
width x( )
h
When x = 0, this is θ0 θ 0 in( ) θ0 40.241 deg
When x = 0.98xdrop, this is θ98% θ 0.98 xdrop θ98% 5.577 deg
4. The loading on the ball is symmetric about its center plane along the x-axis. Figure 3-16(b) shows a FBD of
one half of the ball with the internal forces along the plane of symmetry due to the reaction at the other rod
omitted. With these forces omitted we may only sum forces in the vertical direction.
F
y
: F sin θ( ) μ F cos θ( )
W
2
0= (2)
F
W
2 sin θ( ) μ cos θ( )( )
= (3)
5. The ball will drop through the rods when is zero. If there were no friction force present( = 0) then F would
become very large as approached zero. The presence of the friction term in the denominator of equation (3)
limits F to finite values. However, with the assumption that the rods are rigid, there is no way for the rods to
provide a normal force when reaches zero. Thus, we will need to limit the range of for this analysis.
Let μ 0 and θmin θ98%
Then xmax
h cos θmin c cos α( )
sin α( )
xmax 22.84 in
umax
xmax h sin α( )
cos α( )
umax 22.77 in
Fmax
W
2 sin θmin
Fmax 12.86 lbf (4)
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-16-3
6. Determine the worst-case shear and moment maximum for the rods as the ball rolls along their length from
Figure B-2(a) in Appendix B where a in the figure is umax. Then,
Mmax Fmax umax 1
umax
L
Mmax 70.6 in lbf (5)
For the shear, we must find the reactions, which are
R1 Fmax 1
umax
L
R1 3.10 lbf
R2 Fmax R1 R2 9.76 lbf
The maximum absolute value of shear is the larger of these two. Thus
Vmax R2 Vmax 9.8 lbf (6)
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-17-1
PROBLEM 3-17
Statement: A pair of ice tongs is shown in Figure P3-7. The ice weighs 50 lb and is 10 in wide across the
tongs. The distance between the handles is 4 in, and the mean radius r of a tong is 6 in. Draw
free-body diagrams of the two tongs and find all forces acting on them. Determine the bending
moment at point A.
Given: Weight of ice W 50 lbf
Distances ax 11.0 in ay 6.0 in
bx 5.0 in by 12.0 in
cx 2.0 in cy 3.5 in
Assumptions: Assume that the horizontal force at C (the handle) is zero, thus Fc 0 lbf (1)
Solution: See Figure 3-17 and Mathcad file P0317.
F F
W
F
W/2
B
FB
A
O
C
F
F
C
O
3.5 = cy
12.0 = by
2.0 = cx
5.0 = bx
11.0 = ax
FIGURE 3-17A
Free Body Diagrams for Problem 3-17
1. Summing forces and moments on a single tong (see FBD above right).
F
x
FO FB FC 0= (2)
F
y
W
2
F 0= (3)
M
C
FO cy FB by cy W
2
bx cx 0= (4)
2. From equations (1) and (2), FO FB= (5)
3. Eliminating FO from equations (4) and (5) and solving for FB
FB
W bx cx
2 by
FB 14.58 lbf
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-17-2
F
A
O
C
F
F
C
O
3.5 = cy
2.0 = cx11.0 = ax
D
FMF DnDDs
4. From equation (3), the vertical force
on one handle is
F
W
2
F 25 lbf
5. From Figure 3-17B we see that, at any
section that we might take through the
tong, there will be an internal moment,
shear force, and axial force present. The
bending moment will be a maximum at
point A because it is the fartherest point
from the centroid of the system.
Summing forces and moments:
F
x
-F
Ds
cos + F
Dn
sin
+ FO = 0 (6)
FIGURE 3-17B
Free Body Diagram with section at D for Problem 3-17
F
y
-F
Ds
sin - F
Dn
cos
+ F = 0 (7)
M
O
F c
x
- M
D
- (F
Ds
cos + F
Dn
sin )(a
y
+ r
c
sin )
+ (F
Ds
sin + F
Dn
cos )[a
x
- r
c
(1 - cos)] = 0
(8)
6. Solving equations (6) and (7) for FDs and FDn
FDn F cos θ( ) FO sin θ( )= FDs
FDn sin θ( ) FO
cos θ( )
=
7. The maximum value of MD will occur at = 0 deg. At = 0 deg,
FO FB
FDn F FDn 25 lbf
FDs FO FDs 14.58 lbf
MD F cx FDs ay FDn ax MD 237.5 lbf in
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-18-1
PROBLEM 3-18
Statement: A tractor-trailer tipped over while negotiating an on-ramp to the New York Thruway. The road
has a 50-ft radius at that point and tilts 3 deg toward the outside of the curve. The 45-ft-long by
8-ft-wide by 8.5-ft-high trailer box (13 ft from ground to top) was loaded 44 415 lb of paper rolls in
two rows by two high as shown in Figure P3-8. The rolls are 40-in-dia by 38-in-long and weigh
about 900 lb each. They are wedged against rolling backward but not against sliding sidewards.
The empty trailer weighed 14 000 lb. The driver claims that he was traveling at less than 15 mph
and that the load of paper shifted inside the trailer, struck the trailer sidewall, and tipped the
truck. The paper company that loaded the truck claims the load was properly stowed and would
not shift at that speed. Independent test of the coefficient of friction between similar paper rolls
and a similar trailer floor give a value of 0.43 +/- 0.08. The composite center of gravity of the
loaded trailer is estimated to be 7.5 ft above the road. Determine the truck speed that would
cause the truck to just begin to tip and the speed at which the rolls will just begin to slide
sidways. What do you think caused the accident?
Given: Weight of paper Wp 44415 lbf⋅:=
Weight of trailer Wt 14000 lbf⋅:=
Radius of curve r 50 ft⋅:=
Nominal coefficient of friction μnom 0.43:=
Coefficient of friction uncertainty uμ 0.08:=
Trailer width w 8 ft⋅:=
Height of CG from pavement h 7.5 ft⋅:=
Assumptions: 1. The paper rolls act as a monolith since they are tightly strapped together with steel bands.
2. The tractor has about 15 deg of potential roll freedom versus the trailer due to their relative angle
in plan during the turn combined with the substantial pitch freedom in the fifth wheel. So the trailer
can tip independently of the tractor.
3. The outside track width of the trailer tires is equal to the width of the trailer.
Solution: See Figure 3-18 and Mathcad file P0318.
7.500'
4.000'
ba
xbar
ybar
3°
1. First, calculate the location of the
trailer's CG with respect to the outside
wheel when it is on the reverse-banked
curve. From Figure 3-18A,
Tilt angle θ 3 deg⋅:=
a h tan θ( )⋅:= a 0.393 ft⋅=
b
w
2
a−:= b 3.607 ft⋅=
xbar b cos θ( )⋅:= xbar 3.602 ft⋅=
ybar b sin θ( )⋅
h
cos θ( )
+:=
ybar 7.699 ft⋅=
The coordinates of the CG of the
loaded trailer with respect to the lower
outside corner of the tires are:
FIGURE 3-18A
xbar 3.602 ft⋅= ybar 7.699 ft⋅= Location of CG for Problem 3-18
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-18-2
2. The trailer is on the verge of tipping over when the copule due to centrifugal force is equal to the couple formed
by the weight of the loaded trailer acting through its CG and the vertical reaction at the outside edge of the tires.
At this instant, it is assumed that the entire weight of the trailer is reacted at the outside tires with the inside tires
carrying none of the weight. Any increase in tangential velocity of the tractor and trailer will result in tipping.
Summing moments about the tire edge (see Figure 3-18B),
R y
xbar
ybar
R x
Fw
Fc
ΣΣΣΣ M Fw xbar⋅ Fc ybar⋅−0= (1)
where Fc is the centrifugal force due
to the normal acceleration as the tractor
and trailer go through the curve. The
normal acceleration is
atip
vtip
2
r
= (2)
and the force necessary to keep the
tractor trailer following a circular path
is
Fc mtot atip⋅= (3)
where mtot is the total mass of the
trailer and its payload. Combining
equations (2) and (3) and solving for
vtip, we have
vtip
Fc r⋅
mtot
= (4)
FIGURE 3-18B
FBD of Trailer on the Verge of Tipping
or,
vtip
Fc r⋅ g⋅
Fw
= (5)
3. Calculate the minimum tipping velocity of the tractor/trailer. From equations (1) and (5),
Total weight Fw Wt Wp+:= Fw 58415 lbf⋅=
Centrifugal force required
to tip the trailer
Fc
xbar
ybar
Fw⋅:= Fc 27329 lbf⋅=
Minimum tipping speed vtip
Fc r⋅ g⋅
Fw
:= vtip 18.7 mph⋅=
Thus, with the assumptions that we have made, the trailer would not begin to tip over until it reached a speed of
vtip 18.7 mph⋅=
4. The load will slip when the friction force between the paper rolls and the trailer floor is no longer sufficient to
react the centrifugal force on the paper rolls. Looking at the FBD of the paper rolls in Figure 3-18C, we see that
Normal force between
paper and floor Fn Wp cos θ( )⋅ Fcp sin θ( )⋅−=
(6)
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-18-3
Tangential force tending to slide the paper
F
F n
Wp
t
Fcp
Ft Wp sin θ( )⋅ Fcp cos θ( )⋅+= (7)
Centrifugal force on the paper
Fcp
Wp
g
as⋅=
Wp
g
vs
2
r
⋅= (8)
But, the maximum friction force is
FIGURE 3-18C
Ff μ Fn⋅= Ft= (9) FBD of Paper on the Verge of Sliding
Substituting equation (9) into (7), then combining (6) and (7) to eliminate Fn, and solving for Fcp yields
Fcp
Wp μ cos θ( )⋅ sin θ( )−( )⋅
μ sin θ( )⋅ cos θ( )+
= (10)
Substituting equation (10) into (8), to eliminate Fcp, and solving for vs yields
vs
μ cos θ( )⋅ sin θ( )−( )
μ sin θ( )⋅ cos θ( )+
�
�
�
�
�
�
r⋅ g⋅= (11)
5. Use the upper and lower limit on the coefficient of friction to determine an upper and lower limit on the speed
necessary to cause sliding.
Maximium coefficient μmax μnom uμ+:= μmax 0.51=
Minimium coefficient μmin μnom uμ−:= μmin 0.35=
Maximum velocity to cause sliding
vsmax
μmax cos θ( )⋅ sin θ( )−( )
μmax sin θ( )⋅ cos θ( )+
�
�
�
�
�
�
r⋅ g⋅:= vsmax 18.3 mph⋅=
Minimum velocity to cause sliding
vsmin
μmin cos θ( )⋅ sin θ( )−( )
μmin sin θ( )⋅ cos θ( )+
�
�
�
�
�
�
r⋅ g⋅:= vsmin 14.8 mph⋅=
6. This very rough analysis shows that , if the coefficient of friction was at or near the low end of its measured
value, the paper load could slide at a tractor/trailer speed of 15 mph, which would lead to the trailer tipping over.
In any case, it appears that the paper load would slide before the truck would tip with the load in place.
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-19-1
PROBLEM 3-19
Statement: Assume that the CG of the paper rolls in Problem 3-18 is 2.5 ft above the floor of the trailer. At what
speed on the same curve will the pile of rolls tip over (not slide) with respect to the trailer?
Given: Weight of paper Wp 44415 lbf⋅:=
Radius of curve r 50 ft⋅:=
Paper roll length L 38 in⋅:= L 3.167 ft⋅=
Height of CG from floor h 2.5 ft⋅:=
Assumptions: The paper rolls act as a single, lumped mass and tip about one corner where they are braced
against sliding. The brace provides no moment support.
Solution: See Figure 3-19 and Mathcad file P0319.
1. First, calculate the location of the paper's CG with
respect to the outside corner when it is on the
reverse-banked curve. From Figure 3-19,
2.500'
R x
R y
a
ybar
b
3.167'
xbar
Wp
FcpTilt angle θ 3 deg⋅:=
a h tan θ( )⋅:= a 0.131 ft⋅=
b L a−:= b 3.036 ft⋅=
xbar b cos θ( )⋅:= xbar 3.031 ft⋅=
ybar b sin θ( )⋅
h
cos θ( )
+:=
FIGURE 3-19
FBD of Paper on the Verge of Tipping
ybar 2.662 ft⋅=
The coordinates of the CG of the paper with respect to the lower outside corner are:
xbar 3.031 ft⋅= ybar 2.662 ft⋅=
2. The paper is on the verge of tipping over when the couple due to centrifugal force is equal to the couple formed
by the weight of the paper acting through its CG and the vertical reaction at the outside edge of the rolls. At this
instant, it is assumed that the entire weight of the paper is reacted at the outside corner. Any increase in
tangential velocity of the tractor and trailer will result in tipping. Summing moments about the outside corner
nearest the floor (see Figure 3-19),
ΣΣΣΣ M Wp xbar⋅ Fcp ybar⋅− 0= (1)
where Fcp is the centrifugal force due to the normal acceleration as the tractor and trailer go through the curve.
The normal acceleration is
atip
vtip
2
r
= (2)
and the force necessary to keep the tractor trailer following a circular path is
Fcp mp atip⋅= (3)
where mp is the mass of the paper. Combining equations (2) and (3) and solving for vtip, we have
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-19-2
vtip
Fcp r⋅
mp
= (4)
or,
vtip
Fcp r⋅ g⋅
Wp
= (5)
3. Calculate the minimum paper tipping velocity of the tractor/trailer. From equations (1) and (5),
Centrifugal force required
to tip the paper
Fcp
xbar
ybar
Wp⋅:= Fcp 50574 lbf⋅=
Minimum tipping speed vtip
Fcp r⋅ g⋅
Wp
:= vtip 29.2 mph⋅=
Thus, with the assumptions that we have made, the paper would not begin to tip over until the tractor/trailor
reached a speed of
vtip 29.2 mph⋅=
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-20-1
PROBLEM 3-20
Statement: Assume that the load of paper rolls in Problem 3-18 will slide sideways at a truck speed of 20 mph
on the curve in question. Estimate the impact force of the cargo against the trailer wall. The
force-deflection characteristic of the trailer wall has been measured as approximately 400 lb/in.
Given: Weight of paper Wp 44415 lbf⋅:=
Weight of trailer Wt 14000 lbf⋅:=
Speed of tractor/trailer vt 20 mph⋅:=
Radius of curve r 50 ft⋅:=
Trailer width w 8 ft⋅:=
Paper roll length L 38 in⋅:= L 3.167 ft⋅=
Trailer wall stiffness k 400
lbf
in
:=
Assumptions: 1. The paper rolls act as a monolith since they are tightly strapped together with steel bands.
2. The worst case will result if friction between the floor and the paper is neglected.
Solution: See Figure P3-8 and Mathcad file P0320.
1. Calculate the distance that the rolls will slide before impacting the wall.
s
1
2
w 2 L⋅−( )⋅:= s 10 in⋅=
2. Determine the centripetal acceleration at 20 mph.
ap
vt
2
r
:= ap 206.507
in
sec
2
⋅=
3. From elementary particle dynamics, estimate the velocity at impact due to the centripetal acceleration
vi 2 ap⋅ s⋅:= vi 64.266
in
sec
⋅=
4. With the paper as the moving mass and the trailer as the stationary or struck mass, calculate the correction factor
using equation (3.15)
η
1
1
Wt
3 Wp⋅
+
:= η 0.905=
5. Calculate the static deflection caused by the paper against the trailer wall.
δst
Wp
k
:= δst 111.037 in⋅=
6. Using equation (3.12), estimate the dynamic force of the paper rolls impacting the trailer wall.
Fi Wp vi⋅
η
g δst⋅
⋅:= Fi 13114 lbf⋅=
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-21-1
PROBLEM 3-21
Statement: Figure P3-9 shows an automobile wheel with two common styles of lug wrench being used to
tighten the wheel nuts, a single-ended wrench in (a), and a double-ended wrench in (b). In each
case two hands are required to provide forces respectively at A and B as shown. The distance
between points A and B is 1 ft in both cases. The wheel nuts require a torque of 70 ft-lb. Draw
free body diagrams for both wrenches and determine the magnitudes of all forces and moments
on each wrench. Is there any difference between the way these two wrenches perform their
assigned task? Is one design better than the other? If so, why? Explain.
Given: Distance between A and B dAB 1 ft⋅:=
Tightening torque T 70 ft⋅ lbf⋅:=
Assumptions: 1. The forces exerted by the user's hands lie in a plane through the wrench that is also parallel to
the plane of the wheel.
2. The applied torque is perpendicular to the plane of the forces.
3. By virtue of 1 and 2 above, this is a planar problem that can be described in a 2D FBD.
Solution: See Figure 3-21 and Mathcad file P0321.
F
F
T
12" = dAB
(a) Single-ended Wrench
F
F
T
12" = dAB
6"
(b) Double-ended Wrench
1. Summing moments about the left end of the
wrench (for either case)
T F dAB⋅− 0=
2. Solving for F
F
T
dAB
:= F 70 lbf⋅=
3. This result is the same for both wrenches.
Is there any difference between the way these
two wrenches perform their assigned task?
No, they both require the same
two-handed exertion of 70 lb from each
hand.
Is one design better than the other? If so,
why? Explain.
FIGURE 3-21
Design (b) has advantages over (a)
because it is balanced about the wheel
nut. This allows the user to spin the
wrench once the nut is loosened. It is
also slightly easier to apply the upward
and downward forces (F) in a plane with
design (b).
Free Body Diagrams for Problem 3-21
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-22-1
PROBLEM 3-22
Statement: A roller-blade skate is shown in Figure P3-10. The polyurethane wheels are 72 mm dia. The
skate-boot-foot combination weighs 2 kg. The effective "spring rate" of the person-skate
subsystem is 6000 N/m. Find the forces on the wheels' axles for a 100-kg person landing a 0.5-m
jump on one foot. (a) Assume all 4 wheels land simultaneously. (b) Assume that one wheel
absorbs all the landing force.
Given: Mass of struck member Msys 2 kg
Stiffness of struck member k 6000
N
m
Mass of striking member Mperson 100 kg
Height of drop h 0.5 m
Assumptions: Equation (3.14) applies in this case.
Solution: See Figure P3-10 and Mathcad file P0322.
1. The weight of the striking mass is
Wperson Mperson g Wperson 980.7 N
2. The static deflection of the subsystem is
δst
Wperson
k
δst 163.444 mm
3. The correction factor is
η
1
1
Msys
3 Mperson
η 0.993
4. From equation (3.14), the force of impact is
Fi 1 1
2 η h
δst
Wperson Fi 3.59 kN
(a) If this will be absorbed by 4 wheel axles, the force per axle is
Fa
Fi
4
Fa 897 N
(b) If one wheel absorbs all force Fb Fi Fb 3.59 kN
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-23a-1
PROBLEM 3-23a
Statement: A beam is supported and loaded as shown in Figure P3-11a. Find the reactions, maximum shear,
and maximum moment for the data given in row a from Table P3-1.
R 2
F
R 1
a
b
L
w
Given: Beam length L 1 m
Distance to distributed load a 0.4 m
Distance to concentrated load b 0.6 m
Distributed load magnitude w 200 N m
1
Concentrated load F 500 N
Solution: See Figures 3-23 and Mathcad file P0323a.
FIGURE 3-23A
Free Body Diagram for Problem 3-23
1. From inspection of Figure P3-11a, write the load function equation
q(x) = R
1
<x - 0>-1 - w<x - 0>0 + w<x - a>0 - F<x - b>-1 + R2<x - L>
-1
2. Integrate this equation from - to x to obtain shear, V(x)
V(x) = R
1
<x - 0>0 - w<x - 0>1 + w<x - a>1 - F<x - b>0 + R2<x - L>
0
3. Integrate this equation from - to x to obtain moment, M(x)
M(x) = R
1
<x - 0>1 - w<x - 0>2/2 + w<x - a>2/2 - F<x - b>1 + R2<x - L>
1
4. Solve for the reactions by evaluating the shear and moment equations at a point just to the right of x = L,
where both are zero.
At x = L+, V = M = 0
V R1 w L( ) w L a( ) F R2= 0=
M R1 L
w
2
L
2
w
2
L a( )
2
F L b( )= 0=
R1
w
2
L
F
L
L b( )
w
2 L
L a( )
2
R1 264 N
R2 w a F R1 R2 316 N
5. Define the range for x x 0 m 0.005 L L
6. For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than
z, and a value of one when it is greater than or equal to z.
S x z( ) if x z 1 0( )
7. Write the shear and moment equations in Mathcad form, using the function S as a multiplying factor to get
the effect of the singularity functions.
V x( ) R1 S x 0 m( ) w S x 0 m( ) x( ) w S x a( ) x a( ) F S x b( ) R2 S x L( )
M x( ) R1 S x 0 m( ) x
w
2
S x 0 m( ) x
2
w
2
S x a( ) x a( )
2
F S x b( ) x b( )
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-23a-2
8. Plot the shear and moment diagrams.
0 0.2 0.4 0.6 0.8
400
200
0
200
400
V x( )
N
x
m
Shear
Diagram
0 0.2 0.4 0.6 0.8
0
50
100
150
M x( )
N m
x
m
Moment
Diagram
FIGURE 3-23aB
Shear and Moment Diagrams for Problem 3-23a
9. Determine the maximum shear and maximum moment from inspection of the diagrams.
Maximum shear: Vmax V b( ) Vmax 316 N
Maximum moment occurs where V is zero, which is x = b:
Mmax M b( ) Mmax 126.4 N m
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-24a-1
PROBLEM 3-24a
Statement: A beam is supported and loaded as shown in Figure P3-11b. Find the reactions, maximum shear,
and maximum moment for the data given in row a from Table P3-1.
F
w
R 1
M1
L
a
Given: Beam length L 1 m
Distance to distributed load a 0.4 m
Distributed load magnitude w 200 N m
1
Concentrated load F 500 N
Solution: See Figures 3-24 and Mathcad file P0324a.
1. From inspection of Figure P3-11b, write the load function
equation
FIGURE 3-24A
Free Body Diagram for Problem 3-24
q(x) = -M1<x - 0>
-2 + R
1
<x - 0>-1 - w<x - a>0 - F<x - L>-1
2. Integrate this equation from - to x to obtain shear, V(x)
V(x) = -M1<x - 0>
-1 + R
1
<x - 0>0 - w<x - a>1 - F<x - L>0
3. Integrate this equation from - to x to obtain moment, M(x)
M(x) = -M1<x- 0>
0 + R
1
<x - 0>1 - w<x - a>2/2 - F<x - L>1
4. Solve for the reactions by evaluating the shear and moment equations at a point just to the right of x = L,
where both are zero.
At x = L+, V = M = 0
V R1 w L a( ) F[ ]= 0=
M M1 R1 L
w
2
L a( )
2
= 0=
R1 w L a( ) F R1 620 N
M1
w
2
L a( )
2
R1 L M1 584 N m
5. Define the range for x x 0 m 0.005 L L
6. For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than
z, and a value of one when it is greater than or equal to z.
S x z( ) if x z 1 0( )
7. Write the shear and moment equations in Mathcad form, using the function S as a multiplying factor to get
the effect of the singularity functions.
V x( ) R1 S x 0 m( ) w S x a( ) x a( ) F S x L( )
M x( ) M1 R1 S x 0 m( ) x
w
2
S x a( ) x a( )
2
F S x L( ) x L( )
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-24a-2
8. Plot the shear and moment diagrams.
0 0.2 0.4 0.6 0.8
0
200
400
600
V x( )
N
x
m
Shear
Diagram
0 0.2 0.4 0.6 0.8
600
450
300
150
0
M x( )
N m
x
m
Moment
Diagram
FIGURE 3-24aB
Shear and Moment Diagrams for Problem 3-24a
9. Determine the maximum shear and maximum moment from inspection of the diagrams.
Maximum shear: Vmax V 0 m( ) Vmax 620 N
Maximum moment occurs where V is zero, which is x = 0:
Mmax M 0 m( ) Mmax 584 N m
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-25a-1
PROBLEM 3-25a
Statement: A beam is supported and loaded as shown in Figure P3-11d. Find the reactions, maximum shear,
and maximum moment for the data given in row a from Table P3-1.
b
L
a
1R 2R
F
w
Given: Beam length L 1 m
Distance to distributed load a 0.4 m
Distance to concentrated load b 0.6 m
Distributed load magnitude w 200 N m
1
Concentrated load F 500 N
Solution: See Figures 3-25 and Mathcad file P0325a.
FIGURE 3-25A
Free Body Diagram for Problem 3-25
1. From inspection of Figure P3-11c, write the load function equation
q(x) = R
1
<x - 0>-1 - w<x - a>0 + R2<x - b>
-1 - F<x - L>-1
2. Integrate this equation from - to x to obtain shear, V(x)
V(x) = R
1
<x - 0>0 - w<x - a>1 + R2<x - b>
0 - F<x - L>0
3. Integrate this equation from - to x to obtain moment, M(x)
M(x) = R
1
<x - 0>1 - w<x - a>2/2 + R2<x - b>
1 - F<x - L>1
4. Solve for the reactions by evaluating the shear and moment equations at a point just to the right of x = L,
where both are zero.
At x = L+, V = M = 0
V R1 w L a( ) R2 F= 0=
M R1 L
w
2
L a( )
2
R2 L b( )= 0=
R1
1
b
w
2
L a( )
2
F L b( ) w L a( ) L b( )
R1 353 N
R2 w L a( ) F R1 R2 973 N
5. Define the range for x x 0 m 0.005 L L
6. For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than
z, and a value of one when it is greater than or equal to z.
S x z( ) if x z 1 0( )
7. Write the shear and moment equations in Mathcad form, using the function S as a multiplying factor to get
the effect of the singularity functions.
V x( ) R1 S x 0 m( ) w S x a( ) x a( ) R2 S x b( ) F S x L( )
M x( ) R1 S x 0 m( ) x
w
2
S x a( ) x a( )
2
R2 S x b( ) x b( )
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-25a-2
8. Plot the shear and moment diagrams.
0 0.2 0.4 0.6 0.8
1
0.5
0
0.5
1
V x( )
kN
x
m
Shear
Diagram
0 0.2 0.4 0.6 0.8
300
225
150
75
0
M x( )
N m
x
m
Moment
Diagram
FIGURE 3-25aB
Shear and Moment Diagrams for Problem 3-25a
9. Determine the maximum shear and maximum moment from inspection of the diagrams.
Maximum shear: Vmax V b( ) Vmax 580.0 N
Maximum moment occurs where V is zero, which is x = a:
Mmax M b( ) Mmax 216 N m
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-26a-1
PROBLEM 3-26a
Statement: A beam is supported and loaded as shown in Figure P3-11d. Find the reactions, maximum shear,
and maximum moment for the data given in row a from Table P3-1.
Given: Beam length L 1 m
b
L
a
1R
F
R 2
w
Distance to distributed load a 0.4 m
Distance to reaction load b 0.6 m
Distributed load magnitude w 200 N m
1
Concentrated load F 500 N
Solution: See Figures 3-26 and Mathcad file P0326a.
FIGURE 3-26A
Free Body Diagram for Problem 3-26
1. From inspection of Figure 3-26aA, write the load function equation
q(x) = R
1
<x - 0>-1 - w<x - a>0 + R2<x - b>
-1 - F<x - a>-1
2. Integrate this equation from - to x to obtain shear, V(x)
V(x) = R
1
<x - 0>0 - w<x - a>1 + R2<x - b>
0 - F<x - a>0
3. Integrate this equation from - to x to obtain moment, M(x)
M(x) = R
1
<x - 0>1 - w<x - a>2/2 + R2<x - b>
1 - F<x - a>1
4. Solve for the reactions by evaluating the shear and moment equations at a point just to the right of x = L,
where both are zero.
At x = L+, V = M = 0
V R1 w L a( ) R2 F= 0=
M R1 L
w
2
L a( )
2
R2 L b( ) F L a( )= 0=
R1
1
b
w
2
L a( )
2
F b a( ) w L a( ) L b( )
R1 147 N
R2 w L a( ) F R1 R2 473 N
5. Define the range for x x 0 m 0.005 L L
6. For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than
z, and a value of one when it is greater than or equal to z.
S x z( ) if x z 1 0( )
7. Write the shear and moment equations in Mathcad form, using the function S as a multiplying factor to get
the effect of the singularity functions.
V x( ) R1 S x 0 m( ) w S x a( ) x a( ) R2 S x b( ) F S x a( )
M x( ) R1 S x 0 m( ) x
w
2
S x a( ) x a( )
2
R2 S x b( ) x b( ) F S x a( ) x a( )
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-26a-2
8. Plot the shear and moment diagrams.
0 0.2 0.4 0.6 0.8
500
250
0
250
500
V x( )
N
x
m
Shear
Diagram
0 0.2 0.4 0.6 0.8
20
0
20
40
60
M x( )
N m
x
m
Moment
Diagram
FIGURE 3-26aB
Shear and Moment Diagrams for Problem 3-26a
9. Determine the maximum shear and maximum moment from inspection of the diagrams.
Maximum shear: Vmax V b 0.001 mm( ) Vmax 393 N
Maximum moment occurs where V is zero, which is x = a:
Mmax M a( ) Mmax 58.7 N m
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-27-1
PROBLEM 3-27Statement: A storage rack is to be designed to hold the paper roll of Problem 3-8 as shown in Figure
P3-12. Determine the reactions and draw the shear and moment diagrams for the mandrel that
extends 50% into the roll.
Given: Paper roll dimensions OD 1.50 m
ID 0.22 m
Lroll 3.23 m
Roll density ρ 984 kg m
3
Assumptions: 1. The paper roll's weight creates a concentrated load acting at the tip of the mandrel.
2. The mandrel's root in the stanchion experiences a distributed load over its length of engagemen
Solution: See Figure 3-27 and Mathcad file P0327.
M
W
L
R
m
1
1
1. Determine the weight of the roll
and the length of the mandrel.
W
π
4
OD
2
ID
2
Lroll ρ g
W 53.9 kN
Lm 0.5 Lroll
FIGURE 3-27
Lm 1.615 m Free Body Diagram for Problem 3-27
2. From inspection of Figure 3-27, write the load function equation
q(x) = -M1<x - 0>
-2 + R
1
<x - 0>-1 - W<x - L>-1
3. Integrate this equation from - to x to obtain shear, V(x)
V(x) = -M1<x - 0>
-1 + R
1
<x - 0>0 - W<x - L>0
4. Integrate this equation from - to x to obtain moment, M(x)
M(x) = -M1<x - 0>
0 + R
1
<x - 0>1 - W<x - L>1
5. Solve for the reactions by evaluating the shear and moment equations at a point just to the right of x = L,
where both are zero.
At x = L+, V = M = 0
V R1 W= 0= M M1 R1 L= 0=
R1 W R1 53.895 kN
M1 R1 Lm M1 87.040 kN m
6. Define the range for x x 0 m 0.005 Lm Lm
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-27-2
7. For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than
z, and a value of one when it is greater than or equal to z.
S x z( ) if x z 1 0( )
8. Write the shear and moment equations in Mathcad form, using the function S as a multiplying factor to get
the effect of the singularity functions.
V x( ) R1 S x 0 m( ) W S x Lm
M x( ) M1 R1 S x 0 m( ) x W S x Lm x Lm
9. Plot the shear and moment diagrams.
0 0.5 1 1.5 2
0
20
40
V x( )
kN
x
m
Shear
Diagram
0 0.5 1 1.5 2
100
70
40
10
20
M x( )
kN m
1.615
x
m
Moment
Diagram
FIGURE 3-27B
Shear and Moment Diagrams for Problem 3-27
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-28-1
PROBLEM 3-28
Statement: Figure P3-13 shows a forklift truck negotiating a 15 deg ramp to to drive onto a 4-ft-high loading
platform. The truck weighs 5 000 lb and has a 42-in wheelbase. Determine the reactions and draw
the shear and moment diagrams for the worst case of loading as the truck travels up the ramp.
Given: Ramp angle θ 15 deg
Platform height h 4 ft h 48 in
Truck weight W 5000 lbf
Truck wheelbase Lt 42 in
Assumptions: 1. The worst case is when the truck CG is located at the center of the beam's span.
2. Use a coordinate frame that has the x-axis along the long axis of the beam.
3. Ignore traction forces and the weight components along the x-axis of the beam.
4. There are two ramps, one for each side of the forklift.
5. The location of the CG in Figure P3-13 is 32 in from the front wheel and 10 in from the rear
wheel.
CGa 32 in CGb 10 in
Solution: See Figure 3-28 and Mathcad file P0328.
a
b
L
W
R1
Fa
y
xF
R 2
b
a
Wb
CGa
CGb
FIGURE 3-28A
Dimensions and Free Body Diagram for Problem 3-28
1. Determine the length of the beam between supports and the distances a and b.
Length of beam L
h
sin θ( )
L 15.455 ft
With the CG at midspan, we have a CGa
L
2
= a
L
2
CGa a 5.061 ft
and b
L
2
CGb b 8.561 ft
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-28-2
2. The weight distribution on the wheels is determined from the distance from the front wheel to the CG. Each
wheel weight is divided by 2 to get the weight on a single ramp.
Weight on front wheel Wa
CGb
Lt
W
2
Wa 595 lbf
Weight on rear wheel Wb
W
2
Wa Wb 1905 lbf
3. The normal force on the ramp at each wheel is adjusted for the ramp angle.
Load at front wheel Fa Wa cos θ( ) Fa 575 lbf
Load at rear wheel Fb Wb cos θ( ) Fb 1840 lbf
4. From inspection of Figure 3-28A, write the load function equation
q(x) = R
1
<x - 0>-1 - Fa<x - a>
-1 - Fb<x - b>
-1 + R2<x - L>
-1
5. Integrate this equation from - to x to obtain shear, V(x)
V(x) = R
1
<x - 0>0 - Fa<x - a>
0 - Fb<x - b>
0 + R2<x - L>
0
6. Integrate this equation from - to x to obtain moment, M(x)
M(x) = R
1
<x - 0>1 - Fa<x - a>
1 - Fb<x - b>
1 + R2<x - L>
1
7. Solve for the reactions by evaluating the shear and moment equations at a point just to the right of x = L,
where both are zero.
At x = L+, V = M = 0
V R1 Fa Fb R2= 0=
M R1 L Fa L a( ) Fb L b( )= 0=
R1
1
L
Fa L a( ) Fb L b( ) R1 1207 lbf
R2 Fa Fb R1 R2 1207 lbf
8. Define the range for x x 0 m 0.005 L L
9. For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than
z, and a value of one when it is greater than or equal to z.
S x z( ) if x z 1 0( )
10. Write the shear and moment equations in Mathcad form, using the function S as a multiplying factor to get the
effect of the singularity functions.
V x( ) R1 S x 0 m( ) Fa S x a( ) Fb S x b( ) R2 S x L( )
M x( ) R1 S x 0 m( ) x Fa S x a( ) x a( ) Fb S x b( ) x b( ) R2 S x L( ) x L( )
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-28-3
11. Plot the shear and moment diagrams.
0 2 4 6 8 10 12 14 16
2000
1000
0
1000
2000
V x( )
lbf
x
ft
Shear
Diagram
0 2 4 6 8 10 12 14 16
0
2000
4000
6000
8000
10000
M x( )
ft lbf
15.455
x
ft
Moment
Diagram
FIGURE 3-28B
Shear and Moment Diagrams for Problem 3-28
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-29-1
PROBLEM 3-29 _____
Statement: Run the TKSolver or Mathcad model for Case Study 1A and move the point of application of the
hand force along the lever by changing the values of Rb2, recalculate and observe the changes to
the forces and moments.
Problem: Determine the forces on the elements of the bicycle brake lever assembly shown in Figure 3-1
during braking.
Given: The geometry of each element is known. The average human's hand can develop a grip force of
about 267 N (60 lb) in the lever position shown.
Magnitude of handle force Fb2 Fb2 267 N⋅:=
Direction of handle force Fb2 θb2 270 deg⋅:=
Direction of cable force Fc2 θc2 184 deg⋅:=
Direction of cable force Fcable θcable 180 deg⋅:=
Position vectorcomponents (Change the value of Rb2x and note the results)
Rb2x 19 mm⋅:= Rc2x 25− mm⋅:= R12x 12− mm⋅:=
Rb2y 4− mm⋅:= Rc2y 0 mm⋅:= R12y 7− mm⋅:=
R21x 7 mm⋅:= Rb1x 47.5 mm⋅:= R31x 27− mm⋅:=
R21y 19 mm⋅:= Rb1y 14− mm⋅:= R31y 30 mm⋅:=
Assumptions: The accelerations are negligible. All forces are coplanar and two-dimensional. A class 1 load
model is appropriate and a static analysis is acceptable. The higher applied load will be used as a
worst case, assuming that it can be reached before bottoming the tip of the handle on the handgrip.
If that occurs, it will change the beam's boundary conditions and the analysis.
Solution: See Figures 3-1, 3-2, and Mathcad file P0329.
1. Figure 3-1 shows the hand brake lever assembly, which consists of three subassemblies: the handlebar (1), the
lever (2), and the cable (3). The lever is pivoted to the handlebar and the cable is connected to the lever. The
cable runs within a plastic-lined sheath (for low friction) down to the brake caliper assembly at the bicycle's
wheel rim. The user's hand applies equal and opposite forces at some point on the lever and handgrip. These
forces are transformed to a larger force in the cable by reason of the lever ratio of part 2.
Figure 3-1 is a free-body diagram of the entire assembly since it shows all the forces and moments acting on it
except for its weight, which is small compared to the applied forces and is thus neglected for this analysis. The
"broken away" portion of the handlebar provides internal x and y force components and a moment. These are
arbitrarily shown as positive in sign. Their actual signs will "come out in the wash" in the calculations. The
known applied forces are shown in their actual directions and senses.
2. Figure 3-2 shows the three subassembly elements separated and drawn as free-body diagrams with all relevant
forces and moments applied to each element, again neglecting the weights of the parts. The lever (part 2) has
three forces on it, Fb2, Fc2, and F12. The two-character subscript notation used here should be read as, force of
element 1 on 2 (F12) or force at B on 2 (Fb2), etc. This defines the source of the forces (first subscript) and the
element on which it acts (second subscript).
This notation will be used consistently throughout this text for both forces and position vectors such as Rb2,
Rc2, and R12 in Figure 3-2, which serve to locate the above three forces in a local, non rotating coordinate system
whose origin is at the center of gravity (CG) of the element or subassembly being analyzed. (See foot note on
page 83 of the text).
On this brake lever, Fb2 is an applied force whose magnitude and direction are known. Fc2 is the force in the
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-29-2
cable. Its direction is known but not its magnitude. Force F12 is provided by part 1 on part 2 at the pivot pin.
Its magnitude and direction are both unknown. We can write equations 3.3b for this element to sum forces
in the x and y directions and sum moments about the CG. Note that all unknown reactive forces and moments are
initially assumed positive in the equations. Their true signs will come out in the calculation. (See foot note on
page 84 of the text).
ΣFx Fb2x Fc2x+ F12x+= 0=
ΣFy Fb2y Fc2y+ F12y+= 0= (a)
ΣMz R12 F12×( ) Rb2 Fb2×( )+ Rc2 Fc2×( )+= 0=
The cross products in the moment equation represent the "turning forces" or moments created by the
application of these forces at points remote from the CG of the element. Recall that these cross products can be
expanded to
ΣMz R12x F12y⋅ R12y F12x⋅−( )
Rb2x Fb2y⋅ Rb2y Fb2x⋅−( )+
...
Rc2x Fc2y⋅ Rc2y Fc2x⋅−( )+
...
�
�
�
�
�
�
�
�
= 0=
(b)
We have three equations and four unknowns (F12x, F12y, Fc2x, Fc2y) at this point, so we need another equation. It
is available from the fact that the direction of Fc2 is known. (The cable can pull only along its axis). We can
express one component of the cable force Fc2 in terms of its other component and the known angle θc2 of the
cable.
Fc2y Fc2x tan θc2( )⋅= (c)
We will now use a Mathcad solve block to solve equations a through c.
Calculate components of Fb2
Fb2x Fb2 cos θb2( )⋅:= Fb2x 0− N⋅=
Fb2y Fb2 sin θb2( )⋅:= Fb2y 267− N⋅=
Guess F12x 1000 N⋅:= Fc2x 1000− N⋅:= F12y 1000 N⋅:= Fc2y 1000− N⋅:=
Given Fb2x Fc2x+ F12x+ 0=
Fb2y Fc2y+ F12y+ 0=
R12x F12y⋅ R12y F12x⋅−( )
Rb2x Fb2y⋅ Rb2y Fb2x⋅−( )+
...
Rc2x Fc2y⋅ Rc2y Fc2x⋅−( )+
...
�
�
�
�
�
�
�
�
0=
Fc2y Fc2x tan θc2( )⋅=
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-29-3
F12x
F12y
Fc2x
Fc2y
�
�
�
�
�
�
�
�
�
�
Find F12x F12y, Fc2x, Fc2y, ( ):=
Components of the unknown forces F12, and Fc2
F12x 1047 N⋅= Fc2x 1047− N⋅= F12y 340 N⋅= Fc2y 73.2− N⋅=
3. Part 3 in Figure 3-2 is the cable that passes through a hole in part 1. This hole is lined with a low friction material,
which allows us to assume no friction at the joint between parts 1 and 3. We will further assume that the three
forces F13, Fc3, and Fcable form a concurrent system of forces acting through the CG and thus create no moment.
With this assumption, only a summation of forces is necessary for this element.
ΣFx Fcablex F13x+ Fc3x+= 0=
(d)
ΣFy Fcabley F13y+ Fc3y+= 0=
Using Newton's third law, we have Fc3x Fc2x−:= and Fc3y Fc2y−:= .
We also assume that the cable entering from the left is horizontal and that the reaction F13 is vertical, thus
Fcabley 0 N⋅:= and F13x 0 N⋅:= (e)
We can now solve for the forces on part 3 directly,
Fcablex F13x− Fc3x−:= Fcablex 1047− N⋅=
F13y Fcabley− Fc3y−:= F13y 73.2− N⋅=
The assembly of elements labeled part 1 in Figure 3-2 has both force and moments on it (i.e., it is not a concurrent
system), so the three equations 3.3b are needed.
ΣFx F21x Fb1x+ F31x+ Px+ Fsheathx+= 0=
ΣFy F21y Fb1y+ F31y+ Py+= 0= (f)
ΣMz Mh R21 F21×( )+ Rb1 Fb1×( )+ R31 F31×( )+
Rp Fp×( ) Rd Fsheath×( )++
...= 0=
Expanding cross products in the moment equation gives the moment magnitude as
ΣMz Mh R21x F21y⋅ R21y F21x⋅−( )+
Rb1x Fb1y⋅ Rb1y Fb1x⋅−( )+
...
R31x F31y⋅ R31y F31x⋅−( )+
...
RPx FPy⋅ RPy FPx⋅−( )+
...
0 Rdy Fsheathx⋅−( )+
...
�
�
�
�
�
�
�
�
�
�
�
�
= 0=
(g)
Using Newton's third law, we have
F31x F13x−:= F21x F12x−:= Fb1x Fb2x−:=
(h)
F31y F13y−:= F21y F12y−:= Fb1y Fb2y−:=
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-29-4
Fsheathx Fcablex−:=
Given RPx 27− mm⋅:= RPy 0 mm⋅:=
Rdx 41− mm⋅:= Rdy 27 mm⋅:=
We will now use a Mathcad solve block to solve equations (f) through (h).
Guess Px 1000 N⋅:= Mh 100− N⋅ m⋅:= Py 0 N⋅:=
Given F21x Fb1x+ F31x+ Px+ Fsheathx+ 0=
F21y Fb1y+ F31y+ Py+ 0=
Mh R21x F21y⋅ R21y F21x⋅−( )+
Rb1x Fb1y⋅ Rb1y Fb1x⋅−( )+
...
R31x F31y⋅ R31y F31x⋅−( )+
...
RPx Py⋅ RPy Px⋅−( )+
...
0 N⋅ m⋅ Rdy Fsheathx⋅−( )+
...
�
�
�
�
�
�
�
�
�
�
�
�
0=
Px
Py
Mh
�
�
�
�
�
�
�
�
Find Px Py, Mh, ( ):=
Summarizing, the results obtained for a grip force Fb2 267 N⋅= are:
Handlebar (1) Fb1x 0 N⋅= Fb1y 267 N⋅=
F21x 1047− N⋅= F21y 340− N⋅=
F31x 0 N⋅= F31y 73.2 N⋅=
Px 1 10
6−
× N⋅= Py 0 N⋅=
Mh 0.0 N m⋅⋅=
Lever (2) Fc2x 1047− N⋅= Fc2y 73.2− N⋅=
F12x 1047 N⋅= F12y 340 N⋅=
Cable (3) Fc3x 1047 N⋅= Fc3y 73.2 N⋅=
F13x 0 N⋅= F13y 73.2− N⋅=
Fcablex 1047− N⋅= Fcabley 0 N⋅=
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-30-1
PROBLEM 3-30_____
Statement: Run the TKSolver or Mathcad model for Case Study 2A and move the point of application of the
crimp force along the jaw by changing the values of Rhand, recalculate and observe the changes to
the forces and moments.
Problem: Determine the forces on the elements of the crimping tool shown in Figure 3-3 during a crimp
operation.
Given: The geometry is known and the tool develops a crimp force of 2000 lb (8896 N) at closure in the
position shown.
Applied crimp force Fc4x 1956.30− lbf⋅:= Fc4y 415.82 lbf⋅:=
Position vector components (Change the value of Rhand and note the results)
Rc4x 0.454 in⋅:= R12x 1.399 in⋅:= R32x 2.199 in⋅:=
Rc4y 0.337 in⋅:= R12y 0.049 in⋅:= R32y 0.077 in⋅:=
R23x 0.602− in⋅:= R43x 0.602 in⋅:= R14x 0.161− in⋅:=
R23y 0.127 in⋅:= R43y 0.127− in⋅:= R14y 0.758− in⋅:=
R34x 0.161 in⋅:= R34y 0.758 in⋅:= Rhand 4.40− in⋅:=
Assumptions: The accelerations are negligible. All forces are coplanar and two-dimensional. A class 1 load
model is appropriate and a static analysis is acceptable.
Solution: See Figures 3-3, 3-4, and Mathcad file P0330.
1. Figure 3-3 shows the tool in the closed position, in the process of crimping a metal connector onto a wire. The
user's hand provides the input forces between links 1 and 2, shown as the reaction pair Fhand. The user can grip
the handle anywhere along its length but we are assuming a nominal moment arm of Rhand for the application of
the resultant of the user's grip force (see Figure 3-4). The high mechanical advantage of the tool transforms the
grip force to a large force at the crimp.
Figure 3-3 is a free-body diagram of the entire assembly, neglecting the weight of the tool, which is small
compared to the crimp force. There are four elements, or links, in the assembly, all pinned together. Link 1 can
be considered to be the "ground" link, with the other links moving with respect to it as the jaw is closed. The
desired magnitude of the crimp force Fc is defined and its direction will be normal to the surfaces at the crimp.
2. Figure 3-4 shows the elements of the crimping tool assembly separated and drawn as free-body diagrams with all
forces applied to each element, again neglecting their weights as being insignificant compared to the applied
forces. The centers of gravity of the respective elements are used as the origins of the local, non rotating
coordinate systems in which the points of application of all forces on the element are located. (See footnote on
page 116 of the text).
3. We will consider link 1 to be the ground plane and analyze the remaining moving links. Note that all unknown
forces and moments are initially assumed positive. Link 4 has three forces acting on it: Fc4 is the known (desired)
force at the crimp, and F14 and F34 are the reaction forces from links 1 and 3, respectively. The magnitudes of
these two forces are unknown as is the direction of F14. The direction of F34 will be the same as link 3, since it is
a two-force member. Writing equations 3.3b for this element:
ΣFx F14x F34x+ Fc4x+= 0=
ΣFy F14y F34y+ Fc4y+= 0= (a)
ΣMz R14x F14y⋅ R14y F14x⋅−( )
R34x F34y⋅ R34y F34x⋅−( )+
...
Rc4x Fc4y⋅ Rc4y Fc4x⋅−( )+
...
�
�
�
�
�
�
�
�
= 0=
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-30-2
We have three equations and four unknowns (F14x, F14y, F34x, F34y) at this point, so we need another equation.
It is available from the fact that the direction of F34 is known. We can express one component of the force F34 in
terms of its other component and the known angle θ3 of link 3.
(b)
F34y F34x tan θ3( )⋅=
(c)
where θ3 168 deg⋅:=
Guess F14x 500 lbf⋅:= F34x 1000 lbf⋅:= F14y 100− lbf⋅:= F34y 100− lbf⋅:=
Given F14x F34x+ Fc4x+ 0= F14y F34y+ Fc4y+ 0=
Rc4x Fc4y⋅ Rc4y Fc4x⋅−( )
R14x F14y⋅ R14y F14x⋅−( )+
...
R34x F34y⋅ R34y F34x⋅−( )+
...
�
�
�
�
�
�
�
�
0=
F34y F34x tan θ3( )⋅=
F14x
F14y
F34x
F34y
�
�
�
�
�
�
�
�
�
�
Find F14x F14y, F34x, F34y, ( ):=
Components of the unknown forces F14, and F34
F14x 442.9 lbf⋅= F14y 94.1− lbf⋅= F34x 1513.4 lbf⋅= F34y 321.7− lbf⋅=
4. Link 3 has two forces on it, F23 and F43. Because this is a two-force link, these two forces are equal in magnitude
and opposite in direction. Also, from Newton's third law, F43 = - F34. Thus,
F43x F34x−:= F43y F34y−:= F23x F43x−:= F23y F43y−:= (d)
F43x 1513.4− lbf⋅= F43y 321.7 lbf⋅= F23x 1513.4 lbf⋅= F23y 321.7− lbf⋅=
5. Link 2 has three forces acting on it: Fhand is the unknown force from the hand, and F12 and F32 are the reaction
forces from links 1 and 3, respectively. Force F12 is provided by part 1 on part 2 at the pivot pin and force F32 is
provided by part 3 acting on part 2 at their pivot pin. The magnitude and direction of F32 is known and the
direction of Fhand is known. Using equations 3.3b, we can solve for the magnitude of Fhand and the two
components of F12. From the third law,
F32x F23x−:= F32y F23y−:= F32x 1513.4− lbf⋅= F32y 321.7 lbf⋅=
ΣFx F12x F32x+= 0=
ΣFy Fhand F12y+ F32y+= 0= (e)
ΣMz R12 F12×( ) R32 F32×( )+
Rhand Fhand×( )+
...= 0=
Guess F12x 1500 lbf⋅:= F12y 100− lbf⋅:= Fhand 100 lbf⋅:=
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-30-3
Given F12x F32x+ 0=
F12y F32y+ Fhand+ 0=
R12x F12y⋅ R12y F12x⋅−( )
R32x F32y⋅ R32y F32x⋅−( )+
...
Rhand Fhand⋅+
...
�
�
�
�
�
�
�
�
0=
F12x
F12y
Fhand
�
�
�
�
�
�
�
�
Find F12x F12y, Fhand, ( ):=
F12x 1513.4 lbf⋅= F12y 373.4− lbf⋅= Fhand 51.7 lbf⋅=
6. The four forces on link 1 can now be determined using the third law.
F21x F12x−:= F21y F12y−:= F41x F14x−:= F41y F14y−:=
F21x 1513.4− lbf⋅= F21y 373.4 lbf⋅= F41x 442.9− lbf⋅= F41y 94.1 lbf⋅=
Fc1x Fc4x−:= Fc1y Fc4y−:= Fc1x 1956.3 lbf⋅= Fc1y 415.8− lbf⋅=
7. The solution to this problem for the scaled dimensions in Figure 3-3 assuming a 2000-lb (8896-N) force applied at
the crimp, normal to the crimp surface, is given above. The total forces at the pivot points are:
Pivot A F12 F12x
2
F12y
2
+�
�
0.5
:= F12 1559 lbf⋅=
Pivot B F32 F32x
2
F32y
2
+�
�
0.5
:= F32 1547 lbf⋅=
Pivot C F43 F43x
2
F43y
2
+�
�
0.5
:= F43 1547 lbf⋅=
Pivot D F14 F14x
2
F14y
2
+�
�
0.5
:= F14 453 lbf⋅=
The moment that must be applied to the handles to generate the crimp force of
Crimp force Fc4 Fc4x
2
Fc4y
2
+�
�
0.5
:= Fc4 2000 lbf⋅=
Moment Mh Rhand Fhand⋅:= Mh 227 lbf in⋅⋅=
This moment can be obtained with a force of Fhand 52 lbf⋅= applied at mid-handle. This force is within the
physiological grip-force capacity of the average human.
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-31-1
PROBLEM 3-31 _____
Statement: Run the TKSolver or Mathcad model for Case Study 2A and move the point of application of P
along the x direction by changing the values of Rpx, recalculate and observe the changes to the
forces and moments. What happens when the vertical force P is centered on link 3? Also, change
the angle of the applied force P to create an x component and observe the effects on the forces and
moments on the elements.
Problem: Determine the forces on the elements of the scissors-jack in the position shown in Figure 3-5.
Given: The geometry is known and the jack supports a force of 1000 lb (4448 N) in the position shown.
Support force Px 0.0 lbf⋅:= Py 1000− lbf⋅:=
Position vector components (Change the value of Rpx and note the results)
Rpx 0.50− in⋅:= R12x 3.12− in⋅:= R32x 2.08 in⋅:=
Rpy 0.87 in⋅:= R12y1.80− in⋅:= R32y 1.20 in⋅:=
R42x 2.71 in⋅:= R23x 0.78− in⋅:= R43x 0.78 in⋅:=
R42y 1.00 in⋅:= R23y 0.78− in⋅:= R43y 0.78− in⋅:=
R14x 3.12 in⋅:= R24x 2.58− in⋅:= R34x 2.08− in⋅:=
R14y 1.80− in⋅:= R24y 1.04 in⋅:= R34y 1.20 in⋅:=
Angle of gear teeth common normal θ 45.0− deg⋅:=
Assumptions: The accelerations are negligible. The jack is on level ground. The angle of the elevated car chassis
does not impart an overturning moment to the jack. All forces are coplanar and two-dimensional.
A class 1 load model is appropriate and a static analysis is acceptable.
Solution: See Figures 3-5 through 3-8, and Mathcad file P0331.
1. Figure 3-5 shows a schematic of a simple scissors jack used to raise a car. It consists of six links that are
pivoted and/or geared together and a seventh link in the form of a lead screw that is turned to raise the jack.
While this is clearly a three-dimensional device, it can be analyzed as a two-dimensional one if we assume that
the applied load (from the car) and the jack are exactly vertical (in the z direction). If so, all forces will be in the xy
plane. This assumption is valid if the car is jacked from a level surface. If not, then there will be some forces in
the yz and xz planes as well. The jack designer needs to consider the more general case, but for our simple
example we will initially assume two-dimensional loading. For the overall assembly as shown in Figure 3-5, we
can solve for the reaction force Fg, given force P, by summing forces: Fg = -P.
2. Figure 3-6 shows a set of free-body diagrams for the entire jack. Each element or subassembly of interest has
been separated from the others and the forces and moments shown acting on it (except for its weight, which is
small compared to the applied forces and is thus neglected for this analysis). The forces and moments can be
either internal reactions at interconnections with other elements or external loads from the "outside world." The
centers of gravity of the respective elements are used as the origins of the local, non rotating coordinate systems
in which the points of application of all forces on the element are located. In this design, stability is achieved by
the mating of two pairs of crude (non involute) gear segments acting between links 2 and 4 and between links 5
and 7. These interactions are modeled as forces acting along a common normal shared by the two teeth. This
common normal is perpendicular to the common tangent at the contact point.
There are 3 second-law equations available for each of the seven elements allowing 21 unknowns. An additional
10 third-law equations will be needed for a total of 31. This is a cumbersome system to solve for such a simple
device, but we can use its symmetry to advantage in order to simplify the problem.
3. Figure 3-7 shows the upper half of the jack assembly. Because of the mirror symmetry between the upper
and lower portions, the lower half can be removed to simplify the analysis. The forces calculated for this half will
be duplicated in the other. If we wished, we could solve for the reaction forces at A and B using equations 3.3b
from this free-body diagram of the half-jack assembly.
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-31-2
4. Figure 3-8a shows the free-body diagrams for the upper half of the jack assembly, which are essentially the same
as those of Figure 3-6. We now have four elements but can consider the subassembly labeled 1 to be the
"ground," leaving three elements on which to apply equations 3.3. Note that all forces and moments are initially
assumed positive in the equations.
5. Link 2 has three forces acting on it: F42 is the unknown force at the gear tooth contact with link 4; F12 and F32 are
the unknown reaction forces from links 1 and 3, respectively. Force F12 is provided by part 1 on part 2 at the
pivot pin and force F32 is provided by part 3 acting on part 2 at their pivot pin. The magnitudes and the
directions of these pin forces and the magnitude of F42 are unknown. The direction of F42 is along the common
normal shown in Figure 3-8b. Write equations 3.3b for this element to sum the forces in the x and y directions
and sum moments about the CG (with the cross products expanded).
ΣFx F12x F32x+ F42x+= 0=
ΣFy F12y F32y+ F42y+= 0= (a)
ΣMz R12x F12y⋅ R12y F12x⋅−( )
R32x F32y⋅ R32y F32x⋅−( )+
...
R42x F42y⋅ R42y F42x⋅−( )+
...
�
�
�
�
�
�
�
�
= 0=
6. Link 3 has three forces acting on it: P, F23 and F43. Only P is known. Writing equations 3.3b for this element
gives
ΣFx F23x F43x+ Px+= 0=
ΣFy F23y F43y+ Py+= 0= (b)
ΣMz R23x F23y⋅ R23y F23x⋅−( )
R43x F43y⋅ R43y F43x⋅−( )+
...
Rpx Py⋅ Rpy Px⋅−( )+
...
�
�
�
�
�
�
�
�
= 0=
7. Link 4 has three forces acting on it: F24 is the unknown force from link 2; F14 and F34 are the unknown reaction
forces from links 1 and 3, respectively.
ΣFx F14x F24x+ F34x+= 0=
ΣFy F14y F24y+ F34y+= 0= (c)
ΣMz R14x F14y⋅ R14y F14x⋅−( )
R24x F24y⋅ R24y F24x⋅−( )+
...
R34x F34y⋅ R34y F34x⋅−( )+
...
�
�
�
�
�
�
�
�
= 0=
8. The nine equations in sets a through c have 16 unknowns in them, F12x, F12y, F32x, F32y, F23x, F23y, F43x, F43y,
F14x, F14y, F34x, F34y, F24x, F24y, F42x, F42y. We can write the third-law relationships between action-reaction pairs
at each of the joints to obtain six of the seven additional equations needed:
F32x F23x−= F32y F23y−=
F34x F43x−= F34y F43y−= (d)
F42x F24x−= F42y F24y−=
9. The last equation needed comes from the relationship between the x and y components of the force F24 (or
F42) at the tooth/tooth contact point. Such a contact (or half) joint can transmit force (excepting friction force)
only along the common normal , which is perpendicular to the joint's common tangent as shown in Figure 3-8b.
The common normal is also called the axis of transmission.
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-31-3
The tangent of the angle of this common normal relates the two components of the force at the joint:
F24y F24x tan θ( )⋅= (e)
10. Equations (d) and (e) will be substituted into equations (a) through (c) to create a set of nine simultaneous
equations for solution.
Guess F12x 500 lbf⋅:= F12y 500 lbf⋅:= F14x 500− lbf⋅:= F14y 500 lbf⋅:=
F23x 500 lbf⋅:= F23y 500 lbf⋅:= F24x 500 lbf⋅:=
F43x 500− lbf⋅:= F43y 500 lbf⋅:=
Given R12x F12y⋅ R12y F12x⋅−( )
R32x− F23y⋅ R32y F23x⋅+( )+
...
R42x− F24x⋅ tan θ( )⋅ R42y F24x⋅+( )+
...
�
�
�
�
�
�
�
�
0= F12x F23x− F24x− 0=
F12y F23y− F24x tan θ( )⋅− 0=
R23x F23y⋅ R23y F23x⋅−( )
R43x F43y⋅ R43y F43x⋅−( )+
...
Rpx Py⋅ Rpy Px⋅−( )+
...
�
�
�
�
�
�
�
�
0= F23x F43x+ Px+ 0=
F23y F43y+ Py+ 0=
R14x F14y⋅ R14y F14x⋅−( )
R24x F24x⋅ tan θ( )⋅ R24y F24x⋅−( )+
...
R34x− F43y⋅ R34y F43x⋅+( )+
...
�
�
�
�
�
�
�
�
0= F14x F24x+ F43x− 0=
F14y F24x tan θ( )⋅+ F43y− 0=
F12x
F12y
F14x
F14y
F23x
F23y
F24x
F43x
F43y
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
Find F12x F12y, F14x, F14y, F23x, F23y, F24x, F43x, F43y, ( ):=
Results: F14x 877.8− lbf⋅= F14y 469.6 lbf⋅=
F24x 290.1 lbf⋅= F24y F24x tan θ( )⋅:= F24y 290.1− lbf⋅=
F34x F43x−:= F34y F43y−:=
F23x 587.7 lbf⋅= F23y 820.5 lbf⋅= F43x 587.7− lbf⋅= F43y 179.5 lbf⋅=
F12x 877.8 lbf⋅= F12y 530.4 lbf⋅= F32x F23x−:= F32y F23y−:=
F42x F24x−:= F42y F24y−:=
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-32-1
PROBLEM 3-32 _____
Statement: Figure P3-14 shows a cam-follower arm. If the load P = 200 lb,what spring force is needed at the
right end to maintain a minimum load between cam and follower of 25 lb? Find the maximum shear
force and bending moment in the follower arm. Plot the shear and moment diagrams.
Given: Load at left end of beam P 200 lbf⋅:=
Load at cam follower Pcam 25 lbf⋅:=
Distance from left end to: Pivot point a 10 in⋅:=
Cam follower b 22 in⋅:=
Spring c 29 in⋅:=
Solution: See Figure P3-14 and Mathcad file P0332.
1. Draw a FBD of the cam-follower arm (beam).
FPR
b
c
cam spring
C
P
2. From inspection of the FBD, write the load function equation
q(x) = -P<x - 0>-1 + R<x - a>-1 + P
cam
<x - b >-1 - F
spring
<x - 0>-1
3. Integrate this equation from -∞ to x to obtain shear, V(x)
V(x) = -P<x - 0>0 + R<x - a>0 + P
cam
<x - b >0 - F
spring
<x - 0>0
4. Integrate this equation from -∞ to x to obtain moment, M(x)
M(x) = -P<x - 0>1 + R<x - a>1 + P
cam
<x - b >1 - F
spring
<x - 0>1
5. Solve for the reactions by evaluating the shear and moment equations at a point just to the right of x = c, where
both are zero.
At x = c+, V = M = 0 V P− R+ Pcam+ Fspring−= 0=
M P− c⋅ R c a−( )⋅+ Pcam c b−( )⋅+= 0=
Fspring
P a⋅ Pcam b a−( )⋅+
c a−
:= Fspring 121.05 lbf⋅=
R Fspring P+ Pcam−:= R 296.05 lbf⋅=
6. Define the range for x x 0 in⋅ 0.002 c⋅, c..:=
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-32-2
7. For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than z,
and a value of one when it is greater than or equal to z.
S x z, ( ) if x z≥ 1, 0, ( ):=
8. Write the shear and moment equations in Mathcad form, using the function S as a multiplying factor to get the
effect of the singularity functions.
V x( ) P− S x 0 in⋅, ( )⋅ R S x a, ( )⋅+ Pcam S x b, ( )⋅+ Fspring S x c, ( )⋅−:=
M x( ) P− S x 0 in⋅, ( )⋅ x⋅ R S x a, ( )⋅ x a−( )⋅+ Pcam S x b, ( )⋅ x b−( )⋅+ Fspring S x c, ( )⋅ x c−( )⋅−:=
9. Plot the shear and moment diagrams and find the maximum shear force and bending moment.
0 10 20 30
300−
200−
100−
0
100
200
SHEAR DIAGRAM
V x( )
lbf
x
in
Vmax V 0 in⋅( ):= Vmax 200 lbf⋅=
0 10 20 30
2000−
1500−
1000−
500−
0
MOMENT DIAGRAM
M x( )
in lbf⋅
x
in
Mmax M a( ):= Mmax 2000 in lbf⋅⋅=
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-33-1
PROBLEM 3-33 _____
Statement: Write a computer program or equation solver model to calculate all the singularity functions listed
in equations 3.17. Set them up as functions that can be called from within any other program or
model.
Solution: See Mathcad file P0333.
1. No solution is provided for this programming problem.
© 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be
MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-34a-1
PROBLEM 3-34a
Statement: A beam is supported and loaded as shown in Figure P3-15. Find the reactions, maximum shear, and
maximum moment for the data given in row a from Table P3-2.
a
b
R 1 2R
P
Given: Beam length L 20 in⋅:=
Distance to RH bearing a 16 in⋅:=
Distance to concentrated load b 18 in⋅:=
Concentrated load P 1000 lbf⋅:=
FIGURE 3-34aA
Solution: See Figure 3-34 and Mathcad file P0334a. Free Body Diagram for Problem 3-34
1. From inspection of Figure 3-34, write the load function equation
q(x) = R
1
<x - 0>-1 + R2<x - b>
-1 - P<x - L>-1
2. Integrate this equation from -∞ to x to obtain shear, V(x)
V(x) = R
1
<x - 0>0 + R2<x - b>
0 - P<x - L>0
3. Integrate this equation from -∞ to x to obtain moment, M(x)
M(x) = R
1
<x - 0>1 + R2<x - b>
1 - P<x - L>1
4. Solve for the reactions by evaluating the shear and moment equations at a point just to the right of x = b, where
both are zero.
At x = b+, V = M = 0
V R1 R2+ P−= 0=
M R1 b⋅ R2 b a−( )⋅+= 0=
R1
P
a
a b−( )⋅:= R1 125− lbf⋅=
R2 P R1−:= R2 1125 lbf⋅=
5. Define the range for x x 0 m⋅ 0.002 L⋅, L..:=
6. For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than z,
and a value of one when it is greater than or equal to z.
S x z, ( ) if x z≥ 1, 0, ( ):=
7. Write the shear and moment equations in Mathcad form, using the function S as a multiplying factor to get the
effect of the singularity functions.
V x( ) R1 S x 0 in⋅, ( )⋅ R2 S x a, ( )⋅+ P S x b, ( )⋅−:=
M x( ) R1 S x 0 in⋅, ( )⋅ x⋅ R2 S x a, ( )⋅ x a−( )⋅+ P S x b, ( )⋅ x b−( )⋅−:=
8. Plot the shear and moment diagrams.
© 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be
MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-34a-2
0 5 10 15 20
500−
0
500
1000
SHEAR DIAGRAM
V x( )
lbf
x
in
0 5 10 15 20
3000−
2000−
1000−
0
1000
MOMENT DIAGRAM
M x( )
in lbf⋅
x
in
FIGURE 3-34aB
Shear and Moment Diagrams for Problem 3-34a
9. Determine the maximum shear and maximum moment from inspection of the diagrams.
Maximum shear: Vmax V a( ):= Vmax 1000 lbf⋅=
Maximum moment occurs where V is zero, which is x = a:
Mmax M a( ):= Mmax 2000 in lbf⋅⋅=
© 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be
MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-35a-1
PROBLEM 3-35a
Statement: A beam is supported and loaded as shown in Figure P3-15. Write a computer program or equation
solver model to find the reactions and calculate and plot the loading, shear, and moment functions.
Test the program with the data given in row a from Table P3-2.
a
b
1R 2R
F
Input data: Enter data in highlighted areas
Beam length L 20 in⋅:=
Distance to RH bearing a 16 in⋅:=
Distance to concentrated load b 18 in⋅:=
Concentrated load F 1000 lbf⋅:=
FIGURE 3-34aA
Solution: See Figures 3-35 and Mathcad file P0335a. Free Body Diagram for Problem 3-34
1. From inspection of Figure 3-35, write the load function equation
q(x) = R
1
<x - 0>-1 + R2<x - b>
-1 - F<x - L>-1
2. Integrate this equation from -∞ to x to obtain shear, V(x)
V(x) = R
1
<x - 0>0 + R2<x - b>
0 - F<x - L>0
3. Integrate this equation from -∞ to x to obtain moment, M(x)
M(x) = R
1
<x - 0>1 + R2<x - b>
1 - F<x - L>1
4. Solve for the reactions by evaluating the shear and moment equations at a point just to the right of x = b, where
both are zero.
At x = b+, V = M = 0
V R1 R2+ F−= 0=
M R1 b⋅ R2 b a−( )⋅+= 0=
R1
F
a
a b−( )⋅:= R1 125− lbf⋅=
R2 F R1−:= R2 1125 lbf⋅=
5. Define the range for x x 0 m⋅ 0.002 L⋅, L..:=
6. For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than z,
and a value of one when it is greater than or equal to z.
S x z, ( ) if x z≥ 1, 0, ( ):=
7. Write the shear and moment equations in Mathcad form, using the function S as a multiplying factor to get the
effect of the singularity functions.
V x( ) R1 S x 0 in⋅, ( )⋅ R2 S x a, ( )⋅+ F S x b, ( )⋅−:=
M x( ) R1 S x 0 in⋅, ( )⋅ x⋅ R2 S x a, ( )⋅ x a−( )⋅+ F S x b, ( )⋅ x b−( )⋅−:=
8. Plot the shear and moment diagrams.
© 2011 Pearson Education, Inc., Upper Saddle River,NJ. All rights reserved. This publication is protected by Copyright and written permission should be
MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-35a-2
0 5 10 15 20
500−
0
500
1000
SHEAR DIAGRAM
V x( )
lbf
x
in
0 5 10 15 20
3000−
2000−
1000−
0
1000
MOMENT DIAGRAM
M x( )
in lbf⋅
x
in
FIGURE 3-34aB
Shear and Moment Diagrams for Problem 3-35a
9. Determine the maximum shear and maximum moment from inspection of the diagrams.
Maximum shear: Vmax V a( ):= Vmax 1000 lbf⋅=
Maximum moment occurs where V is zero, which is x = a:
Mmax M a( ):= Mmax 2000 in lbf⋅⋅=
© 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be
MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-36a-1
PROBLEM 3-36a
Statement: A beam is supported and loaded as shown in Figure P3-16. Find the reactions, maximum shear, and
maximum moment for the data given in row a from Table P3-2.
Given: Beam length L 20 in⋅:=
b
L
p
R 1
a
2R
Distance to RH bearing L 20 in⋅:=
Distance to start of load a 16 in⋅:=
Distance to end of load b 18 in⋅:=
Distributed load p 1000
lbf
in
⋅:=
FIGURE 3-36aA
Solution: See Figures 3-36 and Mathcad file P0336a. Free Body Diagram for Problem 3-36
1. From inspection of Figure 3-36, write the load function equation
q(x) = R
1
<x - 0>-1 - p<x - a>0 + p<x - b>0 + R2<x - L>
-1
2. Integrate this equation from -∞ to x to obtain shear, V(x)
V(x) = R
1
<x - 0>0 - p<x - a>1 + p<x - b>1 + R2<x - L>
0
3. Integrate this equation from -∞ to x to obtain moment, M(x)
M(x) = R
1
<x - 0>1 - p<x - a>2/2 + p<x - b>2/2 + R2<x - L>
1
4. Solve for the reactions by evaluating the shear and moment equations at a point just to the right of x = L, where
both are zero.
At x = L+, V = M = 0
V R1 p L a−( )⋅− p L b−( )⋅+ R2+= 0=
M R1 L⋅
p
2
L a−( )
2
⋅−
p
2
L b−( )
2
⋅+ R2 L b−( )⋅+= 0=
R1
p
2 L⋅
2 b a−( )⋅ L⋅ a
2
+ b
2
−�� ��⋅:= R1 300 lbf⋅=
R2 p b a−( )⋅ R1−:= R2 1700 lbf⋅=
5. Define the range for x x 0 in⋅ 0.002 L⋅, L..:=
6. For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than z,
and a value of one when it is greater than or equal to z.
S x z, ( ) if x z≥ 1, 0, ( ):=
7. Write the shear and moment equations in Mathcad form, using the function S as a multiplying factor to get the
effect of the singularity functions.
V x( ) R1 S x 0 in⋅, ( )⋅ p S x a, ( )⋅ x a−( )⋅− p S x b, ( )⋅ x b−( )⋅+ R2 S x L, ( )⋅+:=
M x( ) R1 S x 0 in⋅, ( )⋅ x⋅
p
2
S x a, ( )⋅ x a−( )
2
⋅−
p
2
S x b, ( )⋅ x b−( )
2
⋅+ R2 S x L, ( )⋅ x L−( )⋅+:=
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-36a-2
8. Plot the shear and moment diagrams.
0 5 10 15 20
2000−
1000−
0
1000
SHEAR DIAGRAM
V x( )
lbf
x
in
0 5 10 15 20
0
1000
2000
3000
4000
5000
MOMENT DIAGRAM
M x( )
in lbf⋅
x
in
FIGURE 3-36aB
Shear and Moment Diagrams for Problem 3-36a
9. Determine the maximum shear and maximum moment from inspection of the diagrams.
Maximum shear: Vmax V b( ):= Vmax 1700 lbf⋅=
Maximum moment occurs where V is zero, which is x = c, where:
c
R1 b⋅ R2 a⋅+
R1 R2+
:= c 16.3 in⋅=
Mmax M c( ):= Mmax 4845 in lbf⋅⋅=
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-37a-1
PROBLEM 3-37a
Statement: A beam is supported and loaded as shown in Figure P3-16. Write a computer program or equation
solver model to find the reactions and calculate and plot the loading, shear, and moment functions.
Test the program with the data given in row a from Table P3-2.
Input data: Enter data in highlighted areas
b
L
p
R 1
a
2R
Beam length L 20 in⋅:=
Distance to RH bearing L 20 in⋅:=
Distance to start of load a 16 in⋅:=
Distance to end of load b 18 in⋅:=
Distributed load p 1000
lbf
in
⋅:= FIGURE 3-37aA
Free Body Diagram for Problem 3-37
Solution: See Figures 3-37 and Mathcad file P0337a.
1. From inspection of Figure 3-37, write the load function equation
q(x) = R
1
<x - 0>-1 - p<x - a>0 + p<x - b>0 + R2<x - L>
-1
2. Integrate this equation from -∞ to x to obtain shear, V(x)
V(x) = R
1
<x - 0>0 - p<x - a>1 + p<x - b>1 + R2<x - L>
0
3. Integrate this equation from -∞ to x to obtain moment, M(x)
M(x) = R
1
<x - 0>1 - p<x - a>2/2 + p<x - b>2/2 + R2<x - L>
1
4. Solve for the reactions by evaluating the shear and moment equations at a point just to the right of x = L, where
both are zero.
At x = L+, V = M = 0
V R1 p L a−( )⋅− p L b−( )⋅+ R2+= 0=
M R1 L⋅
p
2
L a−( )
2
⋅−
p
2
L b−( )
2
⋅+ R2 L b−( )⋅+= 0=
R1
p
2 L⋅
2 b a−( )⋅ L⋅ a
2
+ b
2
−�� ��⋅:= R1 300 lbf⋅=
R2 p b a−( )⋅ R1−:= R2 1700 lbf⋅=
5. Define the range for x x 0 in⋅ 0.002 L⋅, L..:=
6. For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than z,
and a value of one when it is greater than or equal to z.
S x z, ( ) if x z≥ 1, 0, ( ):=
7. Write the shear and moment equations in Mathcad form, using the function S as a multiplying factor to get the
effect of the singularity functions.
V x( ) R1 S x 0 in⋅, ( )⋅ p S x a, ( )⋅ x a−( )⋅− p S x b, ( )⋅ x b−( )⋅+ R2 S x L, ( )⋅+:=
M x( ) R1 S x 0 in⋅, ( )⋅ x⋅
p
2
S x a, ( )⋅ x a−( )
2
⋅−
p
2
S x b, ( )⋅ x b−( )
2
⋅+ R2 S x L, ( )⋅ x L−( )⋅+:=
© 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be
MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-37a-2
8. Plot the shear and moment diagrams.
0 5 10 15 20
2000−
1000−
0
1000
SHEAR DIAGRAM
V x( )
lbf
x
in
0 5 10 15 20
0
1000
2000
3000
4000
5000
MOMENT DIAGRAM
M x( )
in lbf⋅
x
in
FIGURE 3-37aB
Shear and Moment Diagrams for Problem 3-37a
9. Determine the maximum shear and maximum moment from inspection of the diagrams.
Maximum shear: Vmax V b( ):= Vmax 1700 lbf⋅=
Maximum moment occurs where V is zero, which is x = c, where:
c
R1 b⋅ R2 a⋅+
R1 R2+
:= c 16.3 in⋅=
Mmax M c( ):= Mmax 4845 in lbf⋅⋅=
© 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be
MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-38a-1
PROBLEM 3-38a
Statement: A beam is supported and loaded as shown in Figure P3-17. Find the reactions, maximum shear, and
maximum moment for the data given in row a from Table P3-2.
a
b
R 1
p
2R
P
Given: Beam length L 20 in⋅:=
Distance to RH bearing a 16 in⋅:=
Distance to concentrated load b 18 in⋅:=
Concentrated load P 1000 lbf⋅:=
Distributed load p 1000 lbf⋅ in
1−
⋅:=
FIGURE 3-38aA
Solution: See Figure 3-38 and Mathcad file P0338a. Free Body Diagram for Problem 3-38
1. Determine the distance from the origin to the left and right ends of the roller.
Distance to left end e 0.1 a⋅:= e 1.600in⋅=
Distance to right end f 0.9 a⋅:= f 14.400 in⋅=
2. From inspection of Figure 3-38, write the load function equation
q(x) = R
1
<x - 0>-1 - p<x - e>0 + p<x - f>0 + R2<x - a>
-1 - P<x - b>-1
3. Integrate this equation from -∞ to x to obtain shear, V(x)
V(x) = R
1
<x - 0>0 - p<x - e>1 + p<x - f>1 + R2<x - a>
0 - P<x - b>0
4. Integrate this equation from -∞ to x to obtain moment, M(x)
M(x) = R
1
<x - 0>1 - p<x - e>2/2 + p<x - f>2/2 + R2<x - a>
1 - P<x - b>1
5. Solve for the reactions by evaluating the shear and moment equations at a point just to the right of x = b, where
both are zero.
At x = b+, V = M = 0
V R1 p b e−( )⋅− p b f−( )⋅+ R2+ P−= 0=
M R1 b⋅
p
2
b e−( )
2
⋅−
p
2
b f−( )
2
⋅+ R2 b a−( )⋅+= 0=
R1
e
2
f
2
−
2 a⋅
f+ e−
�
�
�
�
�
�
p⋅
b a−
a
�
�
�
�
�
�
P⋅−:= R1 6275 lbf⋅=
R2 p f e−( )⋅ R1− P+:= R2 7525 lbf⋅=
6. Define the range for x x 0 m⋅ 0.002 L⋅, L..:=
7. For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than z,
and a value of one when it is greater than or equal to z.
S x z, ( ) if x z≥ 1, 0, ( ):=
8. Write the shear and moment equations in Mathcad form, using the function S as a multiplying factor to get the
effect of the singularity functions.
V x( ) R1 S x 0 m⋅, ( )⋅ p S x e, ( )⋅ x e−( )⋅− p S x f, ( )⋅ x f−( )⋅+ R2 S x a, ( )⋅+ P S x b, ( )⋅−:=
M x( ) R1 S x 0 m⋅, ( )⋅ x⋅
p
2
S x e, ( )⋅ x e−( )
2
⋅−
p
2
S x f, ( )⋅ x f−( )
2
⋅+
R2 S x a, ( )⋅ x a−( )⋅ P S x b, ( )⋅ x b−( )⋅−+
...:=
© 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be
MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-38a-2
9. Plot the shear and moment diagrams.
0 5 10 15 20
10000−
5000−
0
5000
10000
SHEAR DIAGRAM
V x( )
lbf
x
in
0 5 10 15 20
10000−
0
10000
20000
30000
MOMENT DIAGRAM
M x( )
in lbf⋅
x
in
FIGURE 3-38aB
Shear and Moment Diagrams for Problem 3-38a
9. Determine the maximum shear and maximum moment from inspection of the diagrams.
Maximum shear: Vmax V f( ):= Vmax 6525 lbf⋅=
Maximum moment occurs where V is zero, which is x = c:
c e−
R1
f c−
R2 P−
= c
f R1⋅ e R2⋅+ e P⋅−
R1 R2+ P−
:= c 7.875 in⋅=
Mmax M c( ):= Mmax 29728 in lbf⋅⋅=
© 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be
MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-39a-1
PROBLEM 3-39a
Statement: A beam is supported and loaded as shown in Figure P3-17. Write a computer program or equation
solver model to find the reactions and calculate and plot the loading, shear, and moment functions.
Test the program with the data given in row a from Table P3-2.
a
b
R 1
p
2R
P
Input data: Enter data in highlighted areas
Beam length L 20 in⋅:=
Distance to RH bearing a 16 in⋅:=
Distance to concentrated load b 18 in⋅:=
Concentrated load P 1000 lbf⋅:=
FIGURE 3-39aA
Distributed load p 1000 lbf⋅ in
1−
⋅:= Free Body Diagram for Problem 3-39
Solution: See Figure 3-39 and Mathcad file P0339a.
1. Determine the distance from the origin to the left and right ends of the roller.
Distance to left end e 0.1 a⋅:= e 40.64 mm⋅=
Distance to right end f 0.9 a⋅:= f 365.76mm⋅=
2. From inspection of Figure 3-39, write the load function equation
q(x) = R
1
<x - 0>-1 - p<x - e>0 + p<x - f>0 + R2<x - a>
-1 - P<x - b>-1
3. Integrate this equation from -∞ to x to obtain shear, V(x)
V(x) = R
1
<x - 0>0 - p<x - e>1 + p<x - f>1 + R2<x - a>
0 - P<x - b>0
4. Integrate this equation from -∞ to x to obtain moment, M(x)
M(x) = R
1
<x - 0>1 - p<x - e>2/2 + p<x - f>2/2 + R2<x - a>
1 - P<x - b>1
5. Solve for the reactions by evaluating the shear and moment equations at a point just to the right of x = b, where
both are zero.
At x = b+, V = M = 0
V R1 p b e−( )⋅− p b f−( )⋅+ R2+ P−= 0=
M R1 b⋅
p
2
b e−( )
2
⋅−
p
2
b f−( )
2
⋅+ R2 b a−( )⋅+= 0=
R1
e
2
f
2
−
2 a⋅
f+ e−
�
�
�
�
�
�
p⋅
b a−
a
�
�
�
�
�
�
P⋅−:= R1 6275 lbf⋅=
R2 p f e−( )⋅ R1− P+:= R2 7525 lbf⋅=
6. Define the range for x x 0 m⋅ 0.002 L⋅, L..:=
7. For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than z,
and a value of one when it is greater than or equal to z.
S x z, ( ) if x z≥ 1, 0, ( ):=
8. Write the shear and moment equations in Mathcad form, using the function S as a multiplying factor to get the
effect of the singularity functions.
V x( ) R1 S x 0 m⋅, ( )⋅ p S x e, ( )⋅ x e−( )⋅− p S x f, ( )⋅ x f−( )⋅+ R2 S x a, ( )⋅+ P S x b, ( )⋅−:=
© 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be
MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-39a-2
M x( ) R1 S x 0 m⋅, ( )⋅ x⋅
p
2
S x e, ( )⋅ x e−( )
2
⋅−
p
2
S x f, ( )⋅ x f−( )
2
⋅+
R2 S x a, ( )⋅ x a−( )⋅ P S x b, ( )⋅ x b−( )⋅−+
...:=
9. Plot the shear and moment diagrams.
0 5 10 15 20
10000−
5000−
0
5000
10000
SHEAR DIAGRAM
V x( )
lbf
x
in
0 5 10 15 20
10000−
0
10000
20000
30000
MOMENT DIAGRAM
M x( )
in lbf⋅
x
in
FIGURE 3-39aB
Shear and Moment Diagrams for Problem 3-39a
9. Determine the maximum shear and maximum moment from inspection of the diagrams.
Maximum shear: Vmax V f( ):= Vmax 6525 lbf⋅=
Maximum moment occurs where V is zero, which is x = c:
c e−
R1
f c−
R2 P−
= c
f R1⋅ e R2⋅+ e P⋅−
R1 R2+ P−
:= c 7.875 in⋅=
Mmax M c( ):= Mmax 29728 in lbf⋅⋅=
© 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be
MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-40a-1
PROBLEM 3-40a
Statement: A beam is supported and loaded as shown in Figure P3-18. Find the reactions, maximum shear, and
maximum moment for the data given in row a from Table P3-2.
b
L
a
R 1
P
1
2R
P
2Given: Distance to gear 2 L 20 in⋅:=
Distance to gear 1 a 16 in⋅:=
Distance to RH bearing b 18 in⋅:=
Concentrated load at gear 2 P2 1000 lbf⋅:=
Concentrated load at gear 1 P1 0.4 P2⋅:=
FIGURE 3-40a
Solution: See Figure 3-40 and Mathcad file P0340a. Free Body Diagram for Problem 3-40
1. From inspection of Figure 3-40, write the load function equation
q(x) = R
1
<x - 0>-1 - P1<x - a>
-1 + R2<x - b>
-1 - P2<x - L>
-1
2. Integrate this equation from -∞ to x to obtain shear, V(x)
V(x) = R
1
<x - 0>0 - P1<x - a>
0 + R2<x - b>
0 - P2<x - L>
0
3. Integrate this equation from -∞ to x to obtain moment, M(x)
M(x) = R
1
<x - 0>1 - P1<x - a>
1 + R2<x - b>
1 - P<x - L>1
4. Solve for the reactions by evaluating the shear and moment equations at a point just to the right of x = L, where
both are zero.
At x = L+, V = M = 0
V R1 P1− R2+ P2−= 0=
M R1 L⋅ P1 L a−( )⋅− R2 b a−( )⋅+= 0=
R1 P1 1
a
b
−
�
�
�
�
�
�
⋅ P2 1
L
b
−
�
�
�
�
�
�
⋅+:= R1 67− lbf⋅=
R2 P1 P2+ R1−:= R2 1467 lbf⋅=
5. Define the range for x x 0 m⋅ 0.002 L⋅, L..:=
6. For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than z,
and a value of one when it is greater than or equal to z.
S x z, ( ) if x z≥ 1, 0, ( ):=
7. Write the shear and moment equations in Mathcad form, using the function S as a multiplying factor to get the
effect of the singularity functions.
V x( ) R1 S x 0 in⋅,( )⋅ P1 S x a, ( )⋅− R2 S x b, ( )⋅+ P2 S x L, ( )⋅−:=
M x( ) R1 S x 0 mm⋅, ( )⋅ x 0 mm⋅−( )⋅ P1 S x a, ( )⋅ x a−( )⋅−
R2 S x b, ( )⋅ x b−( )⋅ P2 S x L, ( )⋅ x L−( )⋅−+
...:=
8. Plot the shear and moment diagrams.
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-40a-2
0 5 10 15 20
500−
0
500
1000
SHEAR DIAGRAM
V x( )
lbf
x
in
0 5 10 15 20
3000−
2000−
1000−
0
MOMENT DIAGRAM
M x( )
in lbf⋅
x
in
FIGURE 3-40aB
Shear and Moment Diagrams for Problem 3-40a
9. Determine the maximum shear and maximum moment from inspection of the diagrams.
Maximum shear: Vmax V b( ):= Vmax 1000 lbf⋅=
Maximum moment occurs where V is zero, which is x = b:
Mmax M b( ):= Mmax 2000 in lbf⋅⋅=
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-41a-1
PROBLEM 3-41a
Statement: A beam is supported and loaded as shown in Figure P3-18. Write a computer program or equation
solver model to find the reactions and calculate and plot the loading, shear, and moment functions.
Test the program with the data given in row a from Table P3-2.
b
L
a
R 1
P
1
2R
P
2
Input data: Enter data in highlighted areas
Distance to gear 2 L 20 in⋅:=
Distance to gear 1 a 16 in⋅:=
Distance to RH bearing b 18 in⋅:=
Concentrated load at gear 2 P2 1000 lbf⋅:=
Concentrated load at gear 1 P1 0.4 P2⋅:=
FIGURE 3-41aA
Solution: See Figure 3-41 and Mathcad file P0341a. Free Body Diagram for Problem 3-41
1. From inspection of Figure 3-41, write the load function equation
q(x) = R
1
<x - 0>-1 - P1<x - a>
-1 + R2<x - b>
-1 - P2<x - L>
-1
2. Integrate this equation from -∞ to x to obtain shear, V(x)
V(x) = R
1
<x - 0>0 - P1<x - a>
0 + R2<x - b>
0 - P2<x - L>
0
3. Integrate this equation from -∞ to x to obtain moment, M(x)
M(x) = R
1
<x - 0>1 - P1<x - a>
1 + R2<x - b>
1 - P<x - L>1
4. Solve for the reactions by evaluating the shear and moment equations at a point just to the right of x = L, where
both are zero.
At x = L+, V = M = 0
V R1 P1− R2+ P2−= 0=
M R1 L⋅ P1 L a−( )⋅− R2 b a−( )⋅+= 0=
R1 P1 1
a
b
−
�
�
�
�
�
�
⋅ P2 1
L
b
−
�
�
�
�
�
�
⋅+:= R1 67− lbf⋅=
R2 P1 P2+ R1−:= R2 1467 lbf⋅=
5. Define the range for x x 0 m⋅ 0.002 L⋅, L..:=
6. For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than z,
and a value of one when it is greater than or equal to z.
S x z, ( ) if x z≥ 1, 0, ( ):=
7. Write the shear and moment equations in Mathcad form, using the function S as a multiplying factor to get the
effect of the singularity functions.
V z( ) R1 S z 0 in⋅, ( )⋅ P1 S z a, ( )⋅− R2 S z b, ( )⋅+ P2 S z L, ( )⋅−:=
M z( ) R1 S z 0 mm⋅, ( )⋅ z 0 mm⋅−( )⋅ P1 S z a, ( )⋅ z a−( )⋅−
R2 S z b, ( )⋅ z b−( )⋅ P2 S z L, ( )⋅ z L−( )⋅−+
...:=
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-41a-2
8. Plot the shear and moment diagrams.
0 5 10 15 20
500−
0
500
1000
SHEAR DIAGRAM
V x( )
lbf
x
in
0 5 10 15 20
3000−
2000−
1000−
0
MOMENT DIAGRAM
M x( )
in lbf⋅
x
in
FIGURE 3-41aB
Shear and Moment Diagrams for Problem 3-41a
9. Determine the maximum shear and maximum moment from inspection of the diagrams.
Maximum shear: Vmax V b( ):= Vmax 1000 lbf⋅=
Maximum moment occurs where V is zero, which is x = b:
Mmax M b( ):= Mmax 2000 in lbf⋅⋅=
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-42-1
PROBLEM 3-42 _____
Statement: A 1000 kg speedboat reaches a speed of 16 kph at the instant it takes up the slack in a 100 m-long
tow rope attached to a surfboard carrying a 100 kg passenger. If the rope has k = 5 N/m, what is
the dynamic force exerted on the surfboard?
Given: Mass of speedboat ms 1000 kg⋅:=
Speed of boat vi 16 kph⋅:=
Mass of passenger mp 100 kg⋅:=
Rope stiffness k 5 N⋅ m
1−
⋅:=
Assumptions: 1. The water does not influence the dynamic force.
2. An impact model can be used to estimate the dynamic force.
Solution: See Mathcad file P0342.
1. For the impact model, the passenger is the "struck" mass and the speedboat is the "striking mass". Thus, from
equation 3.15, the energy correction factor is:
η
1
1
mp
3 ms⋅
+
:= η 0.97=
2. Use equation 3.11 to estimate the dynamic force on the surfboard/passenger.
Fi vi η ms⋅ k⋅⋅:= Fi 309 N⋅=
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-43-1
PROBLEM 3-43
Statement: Figure P3-19 shows an oil-field pump jack. For the position shown, draw free-body diagrams of the
crank (2), connecting rod (3) and walking beam (4) using variable names similar to those used in
Case Studies 1A and 2A. Assume that the crank turns slowly enough that accelerations can be
ignored. Include the weight acting at the CG of the walking beam and the crank but not the
connecting rod.
Assumptions: 1. A two-dimensional model is adequate.
2. Inertia forces may be ignored.
Solution: See Mathcad file P0343.
1. Isolate each of the elements to be analyzed, starting with the walking beam, since the external forces on it are
known. Place the known force, Fcable, at the point P and the known weight at the CG. Assume the forces at the
interfaces O4 and B to be positive. The position vectors R14, R34, and Rp will be known as will the angle, θ3,that
the connecting rod makes with the horizontal axis.
θ
F cable
P 4
head end
O4
F 14x
B
F 14y
R P
14R
F
34
W4
R
counterweigh
34
x
y
3
B
R
F
43
3
F
23
A
R
23
x
θ3
y
43
2. The connecting rod is a two-force member with
the forces acting at the interfaces A and B along
the line joining points A and B. The assumption
made in step 1 is that these are compressive
forces on link 3.
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-43-2
3. The crank is acted on by forces at A and O2, its weight at its CG, and a torque which we will assume to be
positive (CCW). As in step 1, assume that the unknown reaction force at O2 is positive.
F
12y
T
2
F
O2 F 12x
W2
A
2
x
θ3
y
32
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-44-1
PROBLEM 3-44
Statement: For the pump jack of Problem 3-43 and the data of Table P3-3, determine the pin forces on the
walking beam, connecting rod, and crank and the reaction torque on the crank.
Given: R12 13.2 in⋅:= θ12135 deg⋅:= R14 79.22 in⋅:= θ14 196 deg⋅:=
R32 0.80 in⋅:= θ32 45 deg⋅:= R34 32.00 in⋅:= θ34 169 deg⋅:=
Fcable 2970 lbf⋅:= W2 598 lbf⋅:= W4 2706 lbf⋅:=
θ3 98.5 deg⋅:= RP 124.44 in⋅:= θP 185 deg⋅:=
Solution: See Mathcad files P0343 and P0344.
1. Draw free-body diagrams of each element (see Problem 3-43).
θ
F cable
P 4
head end
O4
F 14x
B
F 14y
R P
14R
F
34
W4
R
counterweigh
34
x
y
3
R
F
43
3
B
F
23
A
R
23
x
θ3
y
43
F
12y
T
2
F
O2 F 12x
W2
A
2
x
θ3
y
32
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-44-2
2. Calculate the x- and y-components of the position vectors.
R12x R12 cos θ12( )⋅:= R12x 9.334− in⋅= R12y R12 sin θ12( )⋅:= R12y 9.334 in⋅=
R14x R14 cos θ14( )⋅:= R14x 76.151− in⋅= R14y R14 sin θ14( )⋅:= R14y 21.836− in⋅=
R32x R32 cos θ32( )⋅:= R32x 0.566 in⋅= R32y R32 sin θ32( )⋅:= R32y 0.566 in⋅=
R34x R34 cos θ34( )⋅:= R34x 31.412− in⋅= R34y R34 sin θ34( )⋅:= R34y 6.106 in⋅=
RPx RP cos θP( )⋅:= RPx 123.966− in⋅= RPy RP sin θP( )⋅:= RPy 10.846− in⋅=
3. Write equations 3(b) for link 4, the walking beam.
ΣΣΣΣ F
x
: F14x F34x+ 0= (1)
ΣΣΣΣ F
y
: Fcable− F14y+ F34y+ W4− 0= (2)
ΣΣΣΣ M
z
: Rpx Fcable⋅ R14x F14y⋅ R14y F14x⋅−( )+ R34x F34y⋅ R34y F34x⋅−( )+ 0= (3)
4. The direction (but not the sense) of F34 is known so write the equation that relates the x- and y-components of
this force.
F34y F34x tan θ3( )⋅− 0= (4)
5. There are four unknowns in the four equations above. Solving for F14x, F14y, F34x, and F34y,
A
1
0
R14y−
in
0
0
1
R14x
in
0
1
0
R34y−
in
tan θ3( )−
0
1
R34x
in
1
�
�
�
�
�
�
�
�
�
�
�
�
�
�
:=
B
0
Fcable W4+
lbf
RPx− Fcable⋅
in lbf⋅
0
��
�
�
�
�
�
��
��
�
�
�
�
�
��
:=
F14x
F14y
F34x
F34y
�
�
�
�
�
�
�
�
�
�
�
�
A
1−
B⋅ lbf⋅:=
F14x 2446 lbf⋅= F14y 10687− lbf⋅= F34x 2446− lbf⋅= F34y 16363 lbf⋅=
6. From Newton's thrid law and, since the connecting rod (3) is a two-force member
F43x F34x−:= F43x 2446 lbf⋅= F43y F34y−:= F43y 16363− lbf⋅=
F23x F43x−:= F23x 2446− lbf⋅= F23y F43y−:= F23y 16363 lbf⋅=
7. Write equations 3(b) for link 2, the crank.
ΣΣΣΣ F
x
: F12x F32x+ 0= (5)
ΣΣΣΣ F
y
: F12y F32y+ W2− 0= (6)
ΣΣΣΣ M
z
: T2 R12x F12y⋅ R12y F12x⋅−( )+ R32x F32y⋅ R32y F32x⋅−( )+ 0= (7)
8. There are three unknowns in the three equations above. Solving for F12x, F12y, and T2, since
F32x F23x−:= F32x 2446 lbf⋅= F32y F23y−:= F32y 16363− lbf⋅=
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-44-3
A
1
0
R12y−
in
0
1
R12x
in
0
0
1
��
�
�
�
�
��
�
�
�
�
:= B
F32x−
lbf
W2 F32y−
lbf
R32x F32y⋅ R32y F32x⋅−( )−
in lbf⋅
��
�
�
�
�
�
�
�
�
�
�
�
�
�
�
:=
F12x
F12y
T2
�
�
�
�
�
�
�
�
�
�
A
1−
B⋅:=
F12x 2446−= lbf F12y 16961= lbf
T2 146128= in-lbf
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-45-1
PROBLEM 3-45
Statement: Figure P3-20 shows an aircraft overhead bin mechanism in end view. For the position shown, draw
free-body diagrams of links 2 and 4 and the door (3) using variable names similar to those used in
Case Studies 1A and 2A. There are stops that prevent further clockwise motion of link 2 (and the
identical link behind it at the other end of the door) resulting in horizontal forces being applied to
the door at points A. Assume that the mechanism is symmetrical so that each set of links 2 and 4
carry one half of the door weight. Ignore the weight of links 2 and 4 as they are negligible.
Assumptions: 1. A two-dimensional model is adequate.
2. Inertia forces may be ignored as the mechanism is at rest against stops.
Solution: See Mathcad file P0345.
1. Isolate each of the elements to be analyzed, starting with the door. Place the force, Fstop, at the point A and the
known weight at the CG. Assume the forces in links 2 and 4 to be positive (tensile). The position vectors R43
and R23 will be known as will the angles θ2 and θ4 that links 2 and 4 make with the horizontal axis.
Fstop 23RA
θ2
23F
F
θ
W3
2
3
y
R
43
43
x
B
4
R
32
F
A
32
y
θ
R12
x
2
O2
2
F12
2. Links 2 and 4 are two-force members with the
forces acting at the pinned ends along the line
joining the pin centers. The assumption made in
step 1 is that these are tensile forces on links 2
and 4.
θ
4O
F14
4
4 R34
R14
B
x
F34
y
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-46-1
PROBLEM 3-46
Statement: For the overhead bin mechanism of Problem 3-45 and the data of Table P3-4, determine the pin
forces on the door (3), and links 2 & 4 and the reaction force on each of the two stops.
Given: R23 180.0 mm⋅:= θ23 160.345 deg⋅:= R43 180.0 mm⋅:= θ43 27.862 deg⋅:=
W3 45 N⋅:= θ2 85.879 deg⋅:= θ4 172.352 deg⋅:=
Solution: See Mathcad files P0345 and P0346.
1. Draw free-body diagrams of each element (see Problem 3-45).
R
32
F
A
32
y
θ
R12
x
2
O2
2
F12
θ
4O
F14
4
4 R34
R14
B
x
F34
y
Fstop 23RA
θ2
23F
F
θ
W3
2
3
y
R
43
43
x
B
4
2. Calculate the x- and y-components of the position vectors on the door (3).
R23x R23 cos θ23( )⋅:= R23x 169.512− mm⋅= R23y R23 sin θ23( )⋅:= R23y 60.544 mm⋅=
R43x R43 cos θ43( )⋅:= R43x 159.134 mm⋅= R43y R43 sin θ43( )⋅:= R43y 84.122 mm⋅=
3. Write equations 3(b) for link 3, the door.
ΣΣΣΣ F
x
: Fstop F23x+ F43x+ 0= (1)
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-46-2
ΣΣΣΣ F
y
: F23y F43y+ 0.5 W3⋅− 0= (2)
ΣΣΣΣ M
z
: R23x− Fstop⋅ R23x F23y⋅ R23y F23x⋅−( )+ R43x F43y⋅ R43y F43x⋅−( )+ 0= (3)
4. The directions (but not the sense) of F23 and F43 are known so write the equations that relates the x- and
y-components of these forces.
F23y F23x tan θ2( )⋅− 0= (4)
F43y F43x tan θ4( )⋅− 0= (5)
5. There are five unknowns in the five equations above. Solving for F23x, F23y, F43x, F43y, and Fstop:
A
1
0
R23y−
mm
tan θ2( )−
0
0
1
R23x
mm
1
0
1
0
R43y−
mm
0
tan θ4( )−
0
1
R43x
mm
0
1
1
0
R23x−
mm
0
0
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
:= B
0
0.5 W3⋅
N
0
0
0
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
:=
F23x
F23y
F43x
F43y
Fstop
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
A
1−
B⋅ N⋅:=
F23x 1.49 N⋅= F23y 20.63 N⋅= F43x 13.96− N⋅= F43y 1.87 N⋅=
The pin forces at A and B are:
F23 F23x
2
F23y
2
+:= F23 20.68 N⋅= F43 F43x
2
F43y
2
+:= F43 14.08 N⋅=
The force on each stop is: Fstop 12.47 N⋅=
6. From Newton's thrid law and, since links 2 and 4 are two-force members
F34x F43x−:= F34x 13.96 N⋅= F34y F43y−:= F34y 1.87− N⋅=
F32x F23x−:= F32x 1.49− N⋅= F32y F23y−:= F32y 20.63− N⋅=
The pin forces at O2 and O4 are numerically equal to those at A and B, respectively.
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-47-1
PROBLEM 3-47
Statement: A particular automobile front suspensionconsists of two A-arms , the wheel (with tire), a coil
spring, and a shock absorber (damper). The effective stiffness of the suspension (called the "ride
rate") is a function of the coil spring stiffness and the tire stiffness. The A-arms are designed to
give the wheel a nearly vertical displacement as the tire rides over bumps in the road. The entire
assembly can be modeled as a spring-mass-damper system as shown in Figure 3-15(b). If the
sprung mass (mass of the portion of the vehicle supported by the suspension system) weighs 675
lb, determine the ride rate that is required to achieve an undamped natural frequency of 1.4 Hz.
What is the static deflection of the suspension for the calculated ride rate?
Units: Hz 2 π⋅ rad⋅ sec
1−
⋅:=
Given: Sprung mass Ws 675 lbf⋅:= Natural frequency ωn 1.4 Hz⋅:=
Solution: See Figure 3-15(b) and Mathcad file P0347.
1. Calculate the sprung mass Ms
Ws
g
:= Ms 1.748 lbf sec
2
⋅ in
1−
⋅=
2. Using equation 3.4, calculate the required ride rate
Ride rate k ωn
2
Ms⋅:= k 135.28
lbf
in
=
3. Calculate the static deflection using equation 3.5
Static deflection δ
Ws
k
:= δ 4.99in=
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-48-1
PROBLEM 3-48
Statement: The independent suspension system of Problem 3-47 has an unsprung weight (the weight of the
axle, wheel, A-arms, etc.) of 106 lb. Calculate the natural frequency (hop resonance) of the
unsprung mass if the sum of the tire and coil spring stiffnesses is 1100 lb/in.
Units: Hz 2 π⋅ rad⋅ sec
1−
⋅:=
Given: Unsprung mass Wu 106 lbf⋅:= Stiffness k 1100
lbf
in
⋅:=
Solution: See Figure 3-15(b) and Mathcad file P0348.
1. Calculate the unsprung mass Mu
Wu
g
:= Mu 0.275 lbf sec
2
⋅ in
1−
⋅=
2. Using equation 3.4, calculate the natural frequency
Natural frequency ωn
k
Mu
:= ωn 10.1 Hz=
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-49-1
PROBLEM 3-49
Statement: The independent suspension system of Problem 3-47 has a sprung weight of 675 lb and a ride rate
of 135 lb/in. Calculate the damped natural frequency of the sprung mass if the damping coefficient
of the shock absorber is a constant 12 lb-sec/in.
Units: Hz 2 π⋅ rad⋅ sec
1−
⋅:=
Given: Sprung mass Ws 675 lbf⋅:= Ride rate k 135
lbf
in
⋅:=
Damping coefficient d 12
lbf sec⋅
in
⋅:=
Solution: See Figure 3-15(b) and Mathcad file P0349.
1. Calculate the sprung mass Ms
Ws
g
:= Ms 1.748lbf sec
2
⋅ in
1−
⋅=
2. Using equation 3.7, calculate the damped natural frequency
Damped natural frequency ωd
k
Ms
d
2 Ms⋅
�
�
�
�
�
�
2
−:= ωd 1.29Hz=
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-50-1
PROBLEM 3-50_______________________________________________________
Statement: Figure P3-22 shows a powder compaction mechanism. For the position shown, draw free-body
diagrams of the input arm (2), connecting rod (3) and compacting ram (4) using variable names
similar to those used in Case Studies 1A and 2A. Assume that the input arm turns slowly
enough that accelerations can be ignored. Ignore the weights of the arm, connecting rod, and
compacting ram. Neglect friction.
Assumptions: 1. A two-dimensional model is adequate.
2. Inertia forces may be ignored.
3. The reactions at slider bearings E and F can be modeled as concentrated forces acting
horizontally at the center of each bearing.
D
E
F
F
com
P
x
y
F
14F
F
14E
F
34
R
14E
R
14F
R
P
3
R
34
Solution: See Mathcad file P0350.
1. Isolate each of the elements to be analyzed, starting with the
compacting rod, since the external force on it is known. Place the
known force, Fcom, at the point P. The position vectors R14E,
R14F, and Rp will be known as will the angle, θ3,that the
compacting ram makes with the vertical axis.
2. The connecting rod is a two-force member with the forces
acting at the interfaces B and D along the line joining points B
and D. The assumption made in step 1 is that these are tensile
forces on link 3.
3. The input arm is acted on by forces at A, B, and C. Assume that the
unknown reaction force at A is positive.
B
D
x
y
F
43
F
23
R
43
R
23
Compacting Ram (4)
C
F
in
B
A
F
32
x
y
F
12y
F
12x
R
in
R
32
R
12
Connecting Rod (3)
Input Arm (2)
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-51-1
PROBLEM 3-51______________________________________________________
Statement: For the compaction mechanism of Problem 3-50 and the data of Table P3-5, determine the pin forc
on the compacting ram, connecting rod, and input arm.
Given: R12 148.4 mm⋅:= θ12 45− deg⋅:= R14E 57.0 mm⋅:= θ14E 90 deg⋅:=
R14F 62.9 mm⋅:= θ14F 270 deg⋅:= R34 32.00 in⋅:= θ34 105.64− deg⋅:=
R23 87.6 mm⋅:= θ23 254.36 deg⋅:= R43 87.6 mm⋅:= θ43 74.36 deg⋅:=
R32 42.9 mm⋅:= θ32 74.36 deg⋅:= R34 15.0 mm⋅:= θ34 90 deg⋅:=
Rin 152.6 mm⋅:= θin 225 deg⋅:= RP 105.0 mm⋅:= θP 270 deg⋅:=
Fcom 100 N⋅:= θ3 254.36 deg⋅:=
Solution: See Mathcad files P0350 and P0351.
1. Draw free-body diagrams of each element (see Problem
3-50).
D
E
F
F
com
P
x
y
F
14F
F
14E
F
34
R
14E
R
14F
R
P
θ
3
R
34
B
D
x
y
F
43
F
23
R
43
R
23
Compacting Ram (4)
C
F
in
B
A
F
32
x
y
F
12y
F
12x
R
in
R
32
R
12
Connecting Rod (3)
Input Arm (2)
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-51-2
2. Calculate the x- and y-components of the position vectors.
R12x R12 cos θ12( )⋅:= R12x 104.935 mm⋅= R12y R12 sin θ12( )⋅:= R12y 104.935− mm⋅=
R14Ex R14E cos θ14E( )⋅:= R14Ex 0 mm⋅= R14Ey R14E sin θ14E( )⋅:= R14Ey 57.000 mm⋅=
R14Fx R14F cos θ14F( )⋅:= R14Fx 0.000− mm⋅= R14Fy R14F sin θ14F( )⋅:= R14Fy 62.900− mm⋅=
R23x R23 cos θ23( )⋅:= R23x 23.616− mm⋅= R23y R23 sin θ23( )⋅:= R23y 84.357− mm⋅=
R32x R32 cos θ32( )⋅:= R32x 11.566 mm⋅= R32y R32 sin θ32( )⋅:= R32y 41.312 mm⋅=
R34x R34 cos θ34( )⋅:= R34x 0.000 mm⋅= R34y R34 sin θ34( )⋅:= R34y 15.000 mm⋅=
R43x R43 cos θ43( )⋅:= R43x 23.616 mm⋅= R43y R43 sin θ43( )⋅:= R43y 84.357 mm⋅=
RPx RP cos θP( )⋅:= RPx 0.000− mm⋅= RPy RP sin θP( )⋅:= RPy 105.000− mm⋅=
Rinx Rin cos θin( )⋅:= Rinx 107.904− mm⋅= Riny Rin sin θin( )⋅:= Riny 107.904− mm⋅=
3. Write equations 3(b) for link 4, the compacting ram.
ΣΣΣΣ F
x
: F14E F14F+ F34x+ 0= (1)
ΣΣΣΣ F
y
: Fcom F34y+ 0= (2)
ΣΣΣΣ M
z
: R14Ey− F14E⋅( ) R14Fy− F14F⋅( )+ R34x F34y⋅ R34y F34x⋅−( )+ 0= (3)
4. The direction (but not the sense) of F34 is known so write the equation that relates the x- and y-components of
this force.
F34y F34x tan θ3( )⋅− 0= (4)
5. There are four unknowns in the four equations above. Solving for F14x, F14y, F34x, and F34y,
A
1
0
R14Ey−
mm
0
1
0
R14Fy−
mm
0
1
0
R34y−
mm
tan θ3( )−
0
1
R34x
mm
1
�
�
�
�
�
�
�
�
�
�
�
�
�
�
:=
B
0
Fcom
N
−
0
0
�
�
�
�
�
�
�
�
�
�
�
�
�
�
:=
F14E
F14F
F34x
F34y
�
�
�
�
�
�
�
�
�
�
�
�
A
1−
B⋅ N⋅:=
F14E 18.2 N⋅= F14F 9.8 N⋅= F34x 28.0− N⋅= F34y 100.0− N⋅=
6. From Newton's thrid law and, since the connecting rod (3) is a two-force member
F43x F34x−:= F43x 28.0 N⋅= F43y F34y−:= F43y 100.0 N⋅=
F23x F43x−:= F23x 28.0− N⋅= F23y F43y−:= F23y 100.0− N⋅=7. Write equations 3(b) for link 2, the input arm.
ΣΣΣΣ F
x
: F12x F32x+ Finx+ 0= (5)
ΣΣΣΣ F
y
: F12y F32y+ Finy+ 0= (6)
ΣΣΣΣ M
z
: R12x F12y⋅ R12y F12x⋅−( ) R32x F32y⋅ R32y F32x⋅−( )+ Rinx Finy⋅ Riny Finx⋅−( )+ 0= (7)
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-51-3
8. The direction (but not the sense) of F
in
is known so write the equation that relates the x- and
y-components of this force.
Finy Finx tan θin( )⋅− 0= (8)
9. There are four unknowns in the four equations above. Solving for F12x, F12y, Finx, and Finy, since
F32x F23x−:= F32x 28 N⋅= F32y F23y−:= F32y 100 N⋅=
A
1
0
R12y−
mm
0
0
1
R12x
mm
0
1
0
Riny−
mm
tan θin( )−
0
1
Rinx
mm
1
�
�
�
�
�
�
�
�
�
�
�
�
�
�
:=
B
F32x−
N
F32y−
N
R32x F32y⋅ R32y F32x⋅−( )−
N mm⋅
0
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
:=
F12x
F12y
Finx
Finy
�
�
�
�
�
�
�
�
�
�
�
�
A
1−
B⋅ N⋅:=
F12x 36.0 N⋅= F12y 36.0− N⋅= Finx 64.0− N⋅= Finx 64.0− N⋅=
F12 F12x
2
F12y
2
+��
�
�:= Fin Finx
2
Finy
2
+��
�
�:=
F12 51 N⋅= Fin 91 N⋅=
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-52-1
PROBLEM 3-52
Statement: Figure P3-23 shows a drag link slider crank mechanism. For the position shown, draw free-body
diagrams of links 2 through 6 using variable names similar to those used in Case Studies 1A and
2A. Assume that the crank turns slowly enough that accelerations can be ignored. Ignore the
weights of the links and any friction forces or torques.
Assumptions: 1. A two-dimensional model is adequate.
2. Inertia forces may be ignored.
3. Links 4 and 6 are three-force bodies.
Solution: See Figure P3-23 and Mathcad file P0352.
D
P
F
P
F
56
F
16
y
x
θ5
1. Isolate each of the elements to be analyzed,
starting with the slider, link 6, since the external
forces on it are known. Place the known force,
FP, at the point P. This is a three-force member
so the forces are coincident at point D and there
is no turning moiment on the link. The angle,
θ5,that link 5 makes with the horizontal axis is
known.
2. Link 5 is a two-force member with the forces
acting at the interfaces C and D along the line
joining points C and D. The assumption made in
step 1 is that these are compressive forces on link
5.
Slider block 6
D
C
x
y
F
45
F
65
θ5
R
45
R
65
Link 5
3. Link 4 is a three-force body with the three forces meeting at a point. The position vectors R14, R34, and
R54 will be known as will the angles,� 3 and �5,that links 3 and 5, respectively, make with the horizontal
axis.
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-52-2
B
O4
C
E
x
y
F
34
F
54
F
14x
F
14y
R
54
R
34
R
14
A
B
F
23
F
43
x
y
R
43
R
23
Link 4
4. Link 3 is a two-force member with the forces acting at the
interfaces A and B along the line joining points A and B.
5. The crank is acted on by forces at A and O2, and a torque
which we will assume to be positive (CCW). As in step 1,
assume that the unknown reaction force at O2 is positive.
A
O2
F
12x
12y
F
x
y
T
F
32
R
12
R
32
Link 3
Link 2
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-53-1
PROBLEM 3-53
Statement: For the drag link slider crank mechanism of Problem 3-52 and the data of Table P3-6, determine the
pin forces on the slider, connecting rods, and crank and the reaction torque on the crank.
Given: R12 63.5 mm⋅:= θ12 45.38 deg⋅:= R14 93.6 mm⋅:= θ14 55.89− deg⋅:=
R23 63.5 mm⋅:= θ23 267.8 deg⋅:= R32 63.5 mm⋅:= θ32 225.38 deg⋅:=
R34 103.5 mm⋅:= θ34 202.68 deg⋅:=
R43 63.5 mm⋅:= θ43 87.80 deg⋅:= R45 190.5 mm⋅:= θ45 156.65 deg⋅:=
R54 103.5 mm⋅:= θ54 45.34 deg⋅:= R65 190.5 mm⋅:= θ65 23.35− deg⋅:=
FP 85 N⋅:= θ5 156.65deg:= θ3 87.80 deg⋅:=
Solution: See Mathcad files P0352 and P0353.
D
P
F
P
F
56
F
16
y
x
θ5
1. Draw free-body diagrams of each element (see
Problem 3-52).
Slider block 6
D
C
x
y
F
45
F
65
θ5
R
45
R
65
Link 5
2. Calculate the x- and y-components of the position vectors.
R12x R12 cos θ12( )⋅:= R12x 44.602 mm⋅= R12y R12 sin θ12( )⋅:= R12y 45.198 mm⋅=
R14x R14 cos θ14( )⋅:= R14x 52.489 mm⋅= R14y R14 sin θ14( )⋅:= R14y 77.497− mm⋅=
R23x R23 cos θ23( )⋅:= R23x 2.438− mm⋅= R23y R23 sin θ23( )⋅:= R23y 63.453− mm⋅=
R32x R32 cos θ32( )⋅:= R32x 44.602− mm⋅= R32y R32 sin θ32( )⋅:= R32y 45.198− mm⋅=
R34x R34 cos θ34( )⋅:= R34x 95.497− mm⋅= R34y R34 sin θ34( )⋅:= R34y 39.908− mm⋅=
R43x R43 cos θ43( )⋅:= R43x 2.438 mm⋅= R43y R43 sin θ43( )⋅:= R43y 63.453 mm⋅=
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-53-2
R45x R45 cos θ45( )⋅:= R45x 174.898− mm⋅= R45y R45 sin θ45( )⋅:= R45y 75.504 mm⋅=
R54x R54 cos θ54( )⋅:= R54x 72.75 mm⋅= R54y R54 sin θ54( )⋅:= R54y 73.619 mm⋅=
R65x R65 cos θ65( )⋅:= R65x 174.898 mm⋅= R65y R65 sin θ65( )⋅:= R65y 75.504− mm⋅=
A
B
F
23
F
43
x
y
R
43
R
23
B
O4
C
E
x
y
F
34
F
54
F
14x
F
14y
R
54
R
34
R
14
Link 4
A
O2
F
12x
12y
F
x
y
T
F
32
R
12
R
32
Link 3
Link 2
3. Write equations 3(b) for link 5, the slider.
ΣΣΣΣ F
x
: F56x FP− 0= (1)
ΣΣΣΣ F
y
: F16 F56y+ 0= (2)
4. The direction (but not the sense) of F56 is known so write the equation that relates the x- and y-components of
this force.
F56y F56x tan θ5( )⋅− 0= (3)
5. There are three unknowns in the three equations above. Solving for F56x, F56y, and F16,
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-53-3
A
1
0
tan θ5( )−
0
1
1
0
1
0
�
�
�
�
�
�
�
�
:=
B
FP
N
0
0
��
�
�
�
�
��
�
�
�
�
:=
F56x
F56y
F16
�
�
�
�
�
�
�
�
�
�
A
1−
B⋅ N⋅:=
F56x 85.0 N⋅= F56y 36.7− N⋅= F16 36.7 N⋅=
6. From Newton's thrid law and, since the connecting rod (5) is a two-force member
F65x F56x−:= F65x 85− N⋅= F65y F56y−:= F65y 36.7 N⋅=
F45x F65x−:= F45x 85 N⋅= F45y F65y−:= F45y 36.7− N⋅=
and, for link 4
F54x F45x−:= F54x 85− N⋅= F54y F45y−:= F54y 36.7 N⋅=
7. Write equations 3(b) for link 4, the rocker.
ΣΣΣΣ F
x
: F34x F54x+ F14x+ 0= (4)
ΣΣΣΣ F
y
: F34y F54y+ F14y+ 0= (5)
ΣΣΣΣ M
z
: R14x F14y⋅ R14y F14x⋅−( ) R34x F34y⋅ R34y F34x⋅−( )+ R54x F54y⋅ R54y F54x⋅−( )+ 0=
8. The direction (but not the sense) of F34 is known so write the equation that relates the x- and y-components of
this force.
F34y F34x tan θ3( )⋅− 0= (7)
9. There are four unknowns in the four equations above. Solving for F34x, F34y, F14x, and F14y,
A
1
0
R34y−
mm
tan θ3( )−
0
1
R34x
mm
1
1
0
R14y−
mm
0
0
1
R14x
mm
0
�
�
�
�
�
�
�
�
�
�
�
�
�
�
:= B
F54x−
N
F54y−
N
R54x F54y⋅ R54y F54x⋅−( )−
N mm⋅
0
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
:=
F34x
F34y
F14x
F14y
�
�
�
�
�
�
�
�
�
�
�
�
A
1−
B⋅ N⋅:=
F34x 3.5 N⋅= F34y 90.9 N⋅= F14x 81.5 N⋅= F14y 127.6− N⋅=
10. From Newton's thrid law and, since the connecting rod (3) is a two-force member
F43x F34x−:= F43x 3.5− N⋅= F43y F34y−:= F43y 90.9− N⋅=
F23x F43x−:= F23x 3 N⋅= F23y F43y−:= F23y 90.9 N⋅=
and, for link 2
F32x F23x−:= F32x 3.5− N⋅=F32y F23y−:= F32y 90.9− N⋅=
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-53-4
11. Write equations 3(b) for link 2, the crank.
ΣΣΣΣ F
x
: F12x F32x+ 0= (8)
ΣΣΣΣ F
y
: F12y F32y+ 0= (9)
ΣΣΣΣ M
z
: T2 R12x F12y⋅ R12y F12x⋅−( )+ R32x F32y⋅ R32y F32x⋅−( )+ 0= (10)
12. There are three unknowns in the three equations above. Solving for F12x, F12y, and T2
A
1
0
R12y−
mm
0
1
R12x
mm
0
0
1
��
�
�
�
�
��
�
�
�
�
:= B
F32x−
N
F32y−
N
R32x F32y⋅ R32y F32x⋅−( )−
N mm⋅
��
�
�
�
�
�
�
�
�
�
�
�
�
�
�
:=
F12x
F12y
T2
�
�
�
�
�
�
�
�
�
�
A
1−
B⋅:=
F12x 3.5= N F12y 90.9= N T2 7796−= N*mm
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-1a-1
PROBLEM 4-1a
Statement: A differential stress element has a set of applied stresses on it as indicated in each row of Table
P4-1. For row a, draw the stress element showing the applied stresses, find the principal stresses
and maximum shear stress using Mohr's circle diagram, and draw the rotated stress element
showing the principal stresses.
Given: σx 1000:= σy 0:= σz 0:=
τxy 500:= τyz 0:= τzx 0:=
1000
500
x
y
Solution: See Figure 4-1a and Mathcad file P0401a.
1. Draw the stress element, indicating the x and y axes.
2. Draw the Mohr's circle axes, indicating the τ and σ
axes with CW up and CCW down.
3. Plot the positive x-face point, which is (+1000, -500),
and label it with an "x."
FIGURE 4-1aA
4. Plot the positive y-face point, which is (0, +500), and
label it with a "y."
Stress Element for Problem 4-1a
5. Draw a straight line from point x to point y. Using the point where this line intersects the σ-axis as the center of
the Mohr circle, draw a circle that goes through points x and y.
6. The center of the circle will be at σc
σx σy+
2
:= σc 500=
7. The circle will intersect the σ-axis at two of the principal stresses. In this case, we see that one is positive and
the other is negative so they will be σ1 and σ3. The third principal stress is σ2 = 0.
8. Calculate the radius of the circle R
σx σy−
2
�
�
�
�
�
�
2
τxy
2
+:= R 707.1=
τ CCW τ CCW
1500-500
y
1000 1500500
0
500 x
2
σ
σσ
φ
13
500
τ CW
-500 1000500
0
500
σσ 13 σ2
τ2-3
500
τ CW
τ
τ
1-3
1-2
σ
FIGURE 4-1aB
2D and 3D Mohr's Circle Diagrams for Problem 4-1a
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-1a-2
9. Calculate the principal stresses σ1 σc R+:= σ1 1207= σ2 0:=
σ3 σc R−:= σ3 207−=
10. Draw the three Mohr's circles to represent the complete 3D stress state.
207
22.5°
x
1207
y11. Calculate the principal shear stresses
τ12 0.5 σ1 σ2−( )⋅:= τ12 603.6=
τ23 0.5 σ2 σ3−( )⋅:= τ23 103.6=
τ13 0.5 σ1 σ3−( )⋅:= τ13 707.1=
The maximum principal stress is always τ13.
12. Determine the orientation of the principal normal
stress (σ1) with respect to the x-axis. From the 2D
Mohr's circle diagram, we see that the angle 2φ from x
to σ1 is CCW and is given by
FIGURE 4-1aC
Rotated Stress Element for Problem 4-1a
ϕ
1
2
acos
σx σc−
R
�
�
�
�
�
�
⋅:= ϕ 22.5 deg=
13. Draw the rotated 2D stress element showing the two nonzero principal stresses.
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-1h-1
PROBLEM 4-1 h
Statement: A differential stress element has a set of applied stresses on it as indicated in each row of Table
P4-1. For row h, draw the stress element showing the applied stresses, find the principal stresses
and maximum shear stress and draw the Mohr's circle diagram.
Given: σx 750:= σy 500:= σz 250:=
τxy 500:= τyz 0:= τzx 0:=
x
750
y
500 500 500
z
250
Solution: See Figures 4-1h and Mathcad file P0401h.
1. Calculate the coefficients (stress invariants) of equation (4.4c).
C2 σx σy+ σz+:= C2 1.500 10
3
×=
C1
σx
τxy
τxy
σy
�
�
�
�
�
�
σx
τzx
τzx
σz
�
�
�
�
�
�
+
σy
τyz
τyz
σz
�
�
�
�
�
�
+:=
C1 4.375 10
5
×=
C0
σx
τxy
τzx
τxy
σy
τyz
τzx
τyz
σz
�
�
�
�
�
�
�
�
�
�
:= C0 3.125 10
7
×= FIGURE 4-1hA
Stress Element for Problem 4-1h
2. Find the roots of the triaxial stress equation: σ
3
C2 σ
2
⋅− C1 σ⋅+ C0− 0=
v
C0−
C1
C2−
1
�
�
�
�
�
�
�
�
�
�
�
�
:= r polyroots v( ):= r
110
250
1140
�
�
�
�
�
�
�
�
=
τ CCW
1500
τ1-2
-500 1000500
0
500
σ σ 1σ
τ2-3
23
500
τ CW
τ1-3
σ
3. Extract the principal stresses from
the vector r by inspection.
σ1 r3
:= σ1 1140=
σ2 r2
:= σ2 250=
σ3 r1
:= σ3 110=
4. Using equations (4.5), evaluate
the principal shear stresses.
τ13
σ1 σ3−
2
:= τ13 515=
τ12
σ1 σ2−
2
:= τ12 445=
τ23
σ2 σ3−
2
:= τ23 70=
FIGURE 4-1hB
5. Draw the three-circle Mohr diagram. The Three Mohr's Circles for Problem 4-1h
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-2-1
PROBLEM 4-2
Statement: A 400-lb chandelier is to be hung from two 10-ft-long solid steel cables in tension. Choose a
suitable diameter for the cable such that the stress will not exceed 5000 psi. What will be the
deflection of the cables? State all assumptions.
Given: Weight of chandelier W 400 lbf⋅:=
Length of cable L 10 ft⋅:= L 120 in=
Allowable stress σallow 5000 psi⋅:=
Number of cables N 2:=
Young's modulus E 30 10
6
⋅ psi⋅:=
Assumptions: The cables share the load equally.
Solution: See Mathcad file P0402.
1. Determine the load on each cable P
W
N
:= P 200 lbf=
2. The stress in each cable will be equal to the load on the cable divided by its cross-sectional area. Using
equation (4.7), and setting the stress equal to the allowable stress, we have
σallow
4 P⋅
π d
2
⋅
=
3. Solve this equation for the unknown cable diameter.
d
4 P⋅
π σallow⋅
:= d 0.226 in=
4. Round this up to the next higher decimal equivalent of a common fractional size: d 0.250 in⋅:=
5. Using equation (4.8), determine the deflection in each cable.
Cross-section area A
π d
2
⋅
4
:= A 0.049 in
2
=
Cable deflection ∆s
P L⋅
A E⋅
:= ∆s 0.016 in=
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-3-1
PROBLEM 4-3
Statement: For the bicycle pedal-arm assembly in Figure 4-1 with rider-applied force of 1500 N at the pedal,
determine the maximum principal stress in the pedal arm if its cross-section is 15 mm in dia. The
pedal attaches to the pedal arm with a 12-mm screw thread. What is the stress in the pedal screw?
Given: Distances (see figure) a 170 mm b 60 mm
Rider-applied force Frider 1.5 kN
Pedal arm diameter dpa 15 mmScrew thread diameter dsc 12 mm
Pedal
x
b Arm
rider
F
a
c
F
Mc
y
TcC
z
Solution: See Figure 4-3 and Mathcad file P0403.
1. From the FBD in Figure 4-3A (and on
the solution for Problem 3-3), we see that
the force from the rider is reacted in the
pedal arm internally by a moment, a
torque, and a vertical shear force. There
are two points at section C (Figure 4-3B)
that we should investigate, one at z = 0.5
dpa (point A), and one at y = 0.5 dpa (point
B).
2. Refering to the FBD resulting from
taking a section through the arm at C, the
maximum bending moment Mc is found by
summing moments about the y-axis, and
the maximum torque Tc is found by
summing moments about the x-axis.
FIGURE 4-3A
Free Body Diagram for Problem 4-3
Arm
x
B
y
Section C
A
z
M
y
: Frider a Mc 0=
M
x
: Frider b Tc 0=
Maximum bending moment:
Mc Frider a Mc 255 N m
Maximum torque:
Tc Frider b Tc 90 N m
Vertical shear:
FIGURE 4-3B
Fc Frider Fc 1.500 kN Points A and B at Section C
3. Determine the stress components at point A where we have the effects of bending and torsion, but where the
transverse shear due to bending is zero because A is at the outer fiber. Looking down the z-axis at a stress
element on the surface at A,
Distance to neutral axis cpa 0.5 dpa cpa 7.5 mm
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-3-2
Moment of inertia of
pedal-arm Ipa
π dpa
4
64
Ipa 2.485 10
3
mm
4
Bending stress
(x-direction)
σx
Mc cpa
Ipa
σx 769.6 MPa
Stress in y-direction σy 0 MPa
Torsional stress
due to Tc
τxy
Tc cpa
2 Ipa
τxy 135.8 MPa CW
Principal stresses at A,
equation (4.6a) σ1A
σx σy
2
σx σy
2
2
τxy
2
σ3A
σx σy
2
σx σy
2
2
τxy
2
σ1A 793 MPa σ2A 0 MPa σ3A 23 MPa
4. Determine the stress components at point B where we have the effects of transverse shear and torsion, but
where the bending stress is zero because B is on the neutral plane. Looking down the y-axis at a stress
element at B,
Cross-section area
of pedal-arm Apa
π dpa
2
4
Apa 176.7 mm
2
Torsional stress
due to Tc and shear
stress due to Fc
τzx
4
3
Fc
Apa
τxy τzx 124.5 MPa CW
Normal stresses σx 0 MPa σz 0 MPa
Principal stresses at B σ1B
σx σz
2
σx σz
2
2
τzx
2
σ3B
σx σz
2
σx σz
2
2
τzx
2
σ1B 124 MPa σ2B 0 MPa σ3B 124 MPa
5. The maximum principal stress is at point A and is σ1A 793 MPa
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-3-3
6. Determine the stress in the pedal screw.
Bending moment Msc Frider b Msc 90 N m
Distance to neutral axis csc 0.5 dsc csc 6 mm
Moment of inertia of
pedal screw Isc
π dsc
4
64
Isc 1.018 10
3
mm
4
Bending stress
(y-direction)
σy
Msc csc
Isc
σy 530.5 MPa
Stress in z-direction σz 0 MPa
Torsional stress τxy 0 MPa
Since there is no shear stress present at the top of the screw where the bending stress is a maximum, the
maximum principal stress in the pedal screw is
σ1 σy σ1 530.5 MPa
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-4-1
PROBLEM 4-4
Statement: The trailer hitch shown in Figure P4-2 and Figure 1-1 (p. 12) has loads applied as defined in
Problem 3-4. The tongue weight of 100 kg acts downward and the pull force of 4905 N acts
horizontally. Using the dimensions of the ball bracket shown in Figure 1-5 (p. 15), determine:
(a) The principal stresses in the shank of the ball where it joins the ball bracket.
(b) The bearing stress in the ball bracket hole.
(c) The tearout stress in the ball bracket.
(d) The normal and shear stresses in the 19-mm diameter attachment holes.
(e) The principal stresses in the ball bracket as a cantilever.
Given: a 40 mm b 31 mm c 70 mm d 20 mm
Mtongue 100 kg Fpull 4.905 kN dsh 26 mm t 19 mm
Assumptions: 1. The nuts are just snug-tight (no pre-load), which is the worst case.
2. All reactions will be concentrated loads rather than distributed loads or pressures.
Solution: See Figure 4-4 and Mathcad file P0404.
1. The weight on the tongue is Wtongue Mtongue g Wtongue 0.981 kN
2. Solving first for the reactions on the ball by summing the horizontal and vertical forces and the moments
about A.
40 = a
19 = t
31 = b
70 = c
20 = d
2
2
C
D
A
B
F
F
F
F
F
F
F
C
D
A
B
F
a1x
b1
a1y
a2x
b2
c2x
d2
c2y
a2yF
1 1
W
Fpull
tongue
FIGURE 4-4A
Dimensions and Free Body Diagram for Problem 4-4
F
x
: Fpull Fa1x Fb1 0= (1)
F
y
: Fa1y Wtongue 0= (2)
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-4-2
M
A
: Fb1 t Fpull a 0= (3)
3. Solving equation (3) for Fb1 Fb1
Fpull a
t
Fb1 10.326 kN
4. Substituting into (1) and solving for Fa1x Fa1x Fpull Fb1 Fa1x 15.231 kN
5. Solving (2) for Fa1y Fa1y Wtongue Fa1y 0.981 kN
6. Now, refering to the FBD of the bracket, we can apply the equations of equilibrium to determine the reactions
at C and D on the bracket.
F
x
: Fa2x Fb2 Fc2x Fd2 0= (4)
F
y
: Fc2y Fa2y 0= (5)
M
C
: Fd2 d Fb2 b Fa2x b t( ) Fa2y c 0= (6)
7. Note also that the interface forces between part 1 (ball) and part 2 (bracket) have been drawn on their
respective FBDs in opposite senses. Therefore,
Fa2x Fa1x Fa2y Fa1y Fb2 Fb1
8. Solving equation (6) for Fd2 Fd2
Fa2x b t( ) Fa2y c Fb2 b
d
Fd2 25.505 kN
9. Substituting into (4) and solving for Fc2x Fc2x Fa2x Fb2 Fd2 Fc2x 30.41 kN
10. Solving (5) for Fa1y Fc2y Fa2y Fc2y 0.981 kN
11. Determine the principal stresses in the shank of the ball where it joins the ball bracket.
The internal bending moment at A on the FBD of the ball is
M Fpull a M 196.2 N m
Distance to neutral axis csh 0.5 dsh csh 13 mm
Moment of inertia of shank Ish
π dsh 4
64
Ish 2.243 10
4
mm
4
Bending stress (x-direction) σx
M csh
Ish
σx 113.7 MPa
Stress in y-direction σy 0 MPa
Shear stress at A τxy 0 MPa
Since the shear stress is zero,
x
is the maximum principal stress, thus
σ1 σx σ1 114 MPa σ2 0 MPa σ3 0 MPa
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-4-3
12. Determine the bearing stress in the ball bracket hole.
Bearing area Abearing dsh t Abearing 494 mm
2
Bearing stress σbearing
Fpull
Abearing
σbearing 9.93 MPa
d R
Tearout length
13. Determine the tearout stress in the ball bracket.
Shear area (see Figure 4-4B)
Atear 2 t R
2
0.5 d( )
2
=
Atear 2 t 32 mm( )
2
0.5 dsh 2
Atear 1111 mm
2
Stress
τtear
Fpull
Atear
FIGURE 4-4B
Tearout Diagram for Problem 4-4
τtear 4.41 MPa
14. Determine the normal and shear stresses in the attachment bolts if they are 19-mm dia.dbolt 19 mm
Bolt cross-section area (2 bolts)
Abolt 2
π dbolt
2
4
Abolt 567.1 mm
2
Normal stress (tension) σbolt
Fc2x
Abolt
σbolt 53.6 MPa
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-4-4
c
R
M
F
2
pull
a
1
W tongue
Shear stress
τbolt
Fc2y
Abolt
τbolt 1.7 MPa
15. Determine the principal stresses in the ball bracket as a
cantilever (see Figure 4-4C).
Bending moment
M Fpull a Wtongue c M 264.8 N m
Width of bracket w 64 mm
Moment of inertia I
w t
3
12
I 36581 mm
4
FIGURE 4-4C
Cantilever FBD for Problem 4-4
Total tensile stress σ
M t
2 I
Fpull
w t
σ 72.8 MPa
Since there are no shear stress at the top and bottom of the bracket where the bending stresses are maximum,
they are also the principal stresses, thus
σ1 σ σ1 72.8 MPa σ2 0 MPa σ3 0 MPa
τmax
σ
2
τmax 36.4 MPa
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-5-1
PROBLEM 4-5
Statement: Repeat Problem 4-4 for the loading conditions of Problem 3-5, i.e., determine the stresses due to a
horizontal force that will result on the ball from accelerating a 2000-kg trailer to 60 m/sec in 20 se
Assume a constant acceleration. From Problem 3-5, the pull force is 6000 N. Determine:
(a) The principal stresses in the shank of the ball where it joins the ball bracket.
(b) The bearing stress in the ball bracket hole.
(c) The tearout stress in the ball bracket.
(d) The normal and shear stresses in the 19-mm diameter attachment holes.
(e) The principal stresses in the ball bracket as a cantilever.
Given: a 40 mm b 31 mm c 70 mm d 20 mm
Mtongue 100 kg Fpull 6 kN dsh 26 mm t 19 mm
Assumptions: 1. The nuts are just snug-tight (no pre-load), which is the worst case.
2. All reactions will be concentrated loads rather than distributed loads or pressures.
Solution: See Figure 4-5 and Mathcad file P0405.
1. The weight on the tongue is Wtongue Mtongue g Wtongue 0.981 kN
2. Solving first for the reactions on the ball by summing the horizontal and vertical forces and the moments about
A.
40 = a
19 = t
31 = b
70 = c
20 = d
2
2
C
D
A
B
F
F
F
F
F
F
F
C
D
A
B
F
a1x
b1
a1y
a2x
b2
c2x
d2
c2y
a2yF
1 1
W
Fpull
tongue
FIGURE 4-5A
Dimensions and Free Body Diagram for Problem 4-5
F
x
: Fpull Fa1x Fb1 0= (1)
F
y
: Fa1y Wtongue 0= (2)
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-5-2
M
A
: Fb1 t Fpull a 0= (3)
3. Solving equation (3) for Fb1 Fb1
Fpull a
t
Fb1 12.632 kN
4. Substituting into (1) and solving for Fa1x Fa1x Fpull Fb1 Fa1x 18.632 kN
5. Solving (2) for Fa1y Fa1y Wtongue Fa1y 0.981 kN
6. Now, refering to the FBD of the bracket, we can apply the equations of equilibrium to determine the reactions
at C and D on the bracket.
F
x
: Fa2x Fb2 Fc2x Fd2 0= (4)
F
y
: Fc2y Fa2y 0= (5)
M
C
: Fd2 d Fb2 b Fa2x b t( ) Fa2y c 0= (6)
7. Note also that the interface forces between part 1 (ball) and part 2 (bracket) have been drawn on their respective
FBDs in opposite senses. Therefore,
Fa2x Fa1x Fa2y Fa1y Fb2 Fb1
8. Solving equation (6) for Fd2 Fd2
Fa2x b t( ) Fa2y c Fb2 b
d
Fd2 30.432 kN
9. Substituting into (4) and solving for Fc2x Fc2x Fa2x Fb2 Fd2 Fc2x 36.432 kN
10. Solving (5) for Fa1y Fc2y Fa2y Fc2y 0.981 kN
11. Determine the principal stresses in the shank of the ball where it joins the ball bracket.
The internal bending moment at A on the FBD of the ball is
M Fpull a M 240 N m
Distance to neutral axis csh 0.5 dsh csh 13 mm
Moment of inertia of shank Ish
π dsh 4
64
Ish 2.243 10
4
mm
4
Bending stress (x-direction) σx
M csh
Ish
σx 139.1 MPa
Stress in y-direction σy 0 MPa
Shear stress at A τxy 0 MPa
Since the shear stress is zero,
x
is the maximum principal stress, thus
σ1 σx σ1 139 MPa σ2 0 MPa σ3 0 MPa
12. Determine the bearing stress in the ball bracket hole.
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-5-3
Bearing area Abearing dsh t Abearing 494 mm
2
Bearing stress σbearing
Fpull
Abearing
σbearing 12.15 MPa
d R
Tearout length
13. Determine the tearout stress in the ball bracket.
Shear area (see Figure 4-4B)
Atear 2 t R
2
0.5 d( )
2
=
Atear 2 t 32 mm( )
2
0.5 dsh 2
Atear 1111 mm
2
Stress
τtear
Fpull
Atear
FIGURE 4-5B
τtear 5.4 MPa Tearout Diagram for Problem 4-5
14. Determine the normal and shear stresses in the attachment bolts if they are 19-mm dia. dbolt 19 mm
Bolt cross-section area (2 bolts)
Abolt 2
π dbolt
2
4
Abolt 567.1 mm
2
Normal stress (tension) σbolt
Fc2x
Abolt
σbolt 64.2 MPa
c
R
M
F
2
pull
a
1
W tongue
Shear stress
τbolt
Fc2y
Abolt
τbolt 1.7 MPa
15. Determine the principal stresses in the ball bracket as a
cantilever (see Figure 4-4C).
Bending moment
M Fpull a Wtongue c M 308.6 N m
Width of bracket w 64 mm
Moment of inertia I
w t
3
12
I 36581 mm
4
FIGURE 4-5C
Cantilever FBD for Problem 4-5
Total tensile stress σ
M t
2 I
Fpull
w t
σ 85.1 MPa
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-5-4
Since there are no shear stress at the top and bottom of the bracket where the bending stresses are maximum,
they are also the principal stresses, thus
σ1 σ σ1 85.1 MPa σ2 0 MPa σ3 0 MPa
τmax
σ
2
τmax 42.5 MPa
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-6-1
PROBLEM 4-6
Statement: Repeat Problem 4-4 for the loading conditions of Problem 3-6, i.e., determine the stresses due to a
horizontal force that will results from an impact between the ball and the tongue of the 2000-kg
trailer if the hitch deflects 2.8 mm dynamically on impact. The tractor weighs 1000 kg and the
velocity at impact is 0.3 m/sec. Determine:
(a) The principal stresses in the shank of the ball where it joins the ball bracket.
(b) The bearing stress in the ball bracket hole.
(c) The tearout stress in the ball bracket.
(d) The normal and shear stresses in the 19-mm diameter attachment holes.
(e) The principal stresses in the ball bracket as a cantilever.
Given: a 40 mm b 31 mm c 70 mm d 20 mm
Mtongue 100 kg Fpull 55.1 kN dsh 26 mm t 19 mm
Assumptions: 1. The nuts are just snug-tight (no pre-load),which is the worst case.
2. All reactions will be concentrated loads rather than distributed loads or pressures.
Solution: See Figure 4-6 and Mathcad file P0406.
1. The weight on the tongue is Wtongue Mtongue g Wtongue 0.981 kN
2. Solving first for the reactions on the ball by summing the horizontal and vertical forces and the moments
about A.
40 = a
19 = t
31 = b
70 = c
20 = d
2
2
C
D
A
B
F
F
F
F
F
F
F
C
D
A
B
F
a1x
b1
a1y
a2x
b2
c2x
d2
c2y
a2yF
1 1
W
Fpull
tongue
FIGURE 4-6A
Dimensions and Free Body Diagram for Problem 4-6
F
x
: Fpull Fa1x Fb1 0= (1)
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-6-2
F
y
: Fa1y Wtongue 0= (2)
M
A
: Fb1 t Fpull a 0= (3)
3. Solving equation (3) for Fb1 Fb1
Fpull a
t
Fb1 116 kN
4. Substituting into (1) and solving for Fa1x Fa1x Fpull Fb1 Fa1x 171.1 kN
5. Solving (2) for Fa1y Fa1y Wtongue Fa1y 0.981 kN
6. Now, refering to the FBD of the bracket, we can apply the equations of equilibrium to determine the reactions
at C and D on the bracket.
F
x
: Fa2x Fb2 Fc2x Fd2 0= (4)
F
y
: Fc2y Fa2y 0= (5)
M
C
: Fd2 d Fb2 b Fa2x b t( ) Fa2y c 0= (6)
7. Note also that the interface forces between part 1 (ball) and part 2 (bracket) have been drawn on their respective
FBDs in opposite senses. Therefore,
Fa2x Fa1x Fa2y Fa1y Fb2 Fb1
8. Solving equation (6) for Fd2 Fd2
Fa2x b t( ) Fa2y c Fb2 b
d
Fd2 251.382 kN
9. Substituting into (4) and solving for Fc2x Fc2x Fa2x Fb2 Fd2 Fc2x 306.482 kN
10. Solving (5) for Fa1y Fc2y Fa2y Fc2y 0.981 kN
11. Determine the principal stresses in the shank of the ball where it joins the ball bracket.
The internal bending moment at A on the FBD of the ball is
M Fpull a M 2.204 10
3
N m
Distance to neutral axis csh 0.5 dsh csh 13 mm
Moment of inertia of shank Ish
π dsh 4
64
Ish 2.243 10
4
mm
4
Bending stress (x-direction) σx
M csh
Ish
σx 1277 MPa
Stress in y-direction σy 0 MPa
Shear stress at A τxy 0 MPa
Since the shear stress is zero,
x
is the maximum principal stress, thus
σ1 σx σ1 1277 MPa σ2 0 MPa σ3 0 MPa
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-6-3
12. Determine the bearing stress in the ball bracket hole.
Bearing area Abearing dsh t Abearing 494 mm
2
Bearing stress σbearing
Fpull
Abearing
σbearing 111.54 MPa
d R
Tearout length
13. Determine the tearout stress in the ball bracket.
Shear area (see Figure 4-4B)
Atear 2 t R
2
0.5 d( )
2
=
Atear 2 t 32 mm( )
2
0.5 dsh 2
Atear 1111 mm
2
Stress
τtear
Fpull
Atear
FIGURE 4-6B
Tearout Diagram for Problem 4-6
τtear 49.59 MPa
14. Determine the normal and shear stresses in the attachment bolts if they are 19-mm dia. dbolt 19 mm
Bolt cross-section area (2 bolts)
Abolt 2
π dbolt
2
4
Abolt 567.1 mm
2
Normal stress (tension) σbolt
Fc2x
Abolt
σbolt 540 MPa
c
R
M
F
2
pull
a
1
W tongue
Shear stress
τbolt
Fc2y
Abolt
τbolt 1.7 MPa
15. Determine the principal stresses in the ball bracket as a
cantilever (see Figure 4-4C).
Bending moment
M Fpull a Wtongue c M 2.3 10
3
N m
Width of bracket w 64 mm
Moment of inertia I
w t
3
12
I 36581 mm
4
FIGURE 4-6C
Cantilever FBD for Problem 4-6
Total tensile stress σ
M t
2 I
Fpull
w t
σ 635.5 MPa
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-6-4
Since there are no shear stress at the top and bottom of the bracket where the bending stresses are maximum,
they are also the principal stresses, thus
σ1 σ σ1 636 MPa σ2 0 MPa σ3 0 MPa
τmax
σ
2
τmax 318 MPa
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-7-1
PROBLEM 4-7
Statement: Design the wrist pin of Problem 3-7 for a maximum allowable principal stress of 20 ksi if the pin is
hollow and loaded in double shear.
Given: Force on wrist pin Fwristpin 12.258 kN Fwristpin 2756 lbf
Allowable stress σallow 20 ksi
Assumptions: Choose a suitable outside diameter, say od 0.375 in
Solution: See Figure 4-12 in the text and Mathcad file P0407.
1. The force at each shear plane is F
Fwristpin
2
F 1378 lbf
2. With only the direct shear acting on the plane, the Mohr diagram will be a circle with center at the origin and
radius equal to the shear stress. Thus, the principal normal stress is numerically equal to the shear stress, which
in this case is also the principal shear stress, so we have = 1 = allow.
3. The shear stress at each shear plane is τ
F
A
=
4 F
π od
2
id
2
= σallow=
4. Solving for the inside diameter, id od
2 4 F
π σallow
id 0.230 in
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-8-1
PROBLEM 4-8
Statement: A paper mill processes rolls of paper having a density of 984 kg/m3. The paper roll is 1.50 m
outside dia (OD) by 0.22 m inside dia (ID) by 3.23 m long and is on a simply supported, hollow,
steel shaft. Find the shaft ID needed to obtain a maximum deflection at the center of 3 mm if the
shaft OD is 22 cm.
Given: Paper density ρ 984
kg
m
3
Roll dimensions Shaft outside dia od 220 mm
Outside diameter OD 1.50 m Young's modulus E 207 GPa
Inside diameter ID 0.22 m Allowable deflection δ 3 mm
Lemgth L 3.23 m
Assumptions: The shaft (beam) supporting the paper roll is simply-supported at the ends and is the same
length as the paper roll. The paper acts as a distributed load over the length of the shaft.
Solution: See Mathcad file P0408.
1. The weight of the paper roll is equal to its volume times the paper density times g.
Wroll
π
4
OD
2
ID
2
L ρ g Wroll 53.89 kN
2. The intensity of the distributed load is w
Wroll
L
w 16.686
N
mm
3. Using Figure B-2(b) in Appendix B with a = 0, the maximum deflection is at the midspan and is
y
w x
24 E I
2 L x
2
x
3
L
3
=
For x = L/2, this reduces to y
5 w L
4
384 E I
=
Letting = -y and solving for I, we have I
5 w L
4
384 E δ
I 3.808 10
7
mm
4
4. The area moment of inertia for a hollow circular cross-section is I
π
64
od
4
id
4
=
Solving this for the id yields id od
4 64 I
π
1
4
id 198.954 mm
Round this down (for slightly less deflection) to id 198 mm
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-9-1
PROBLEM 4-9
Statement: A ViseGrip plier-wrench is drawn to scale in Figure P4-3, and for which the forces were analyzedin Problem 3-9, find the stresses in each pin for an assumed clamping force of P = 4000 N in the
position shown. The pins are 8-mm dia and are all in double shear.
Given: Pin forces as calculated in Problem 3-9:
Member 1 F21 7.5 kN F41 5.1 kN
Member 2 F12 7.5 kN F32 5.1 kN
Member 3 F23 5.1 kN F43 5.1 kN
Member 4 F14 5.1 kN F34 5.1 kN
Pin diameter d 8 mm
Assumptions: Links 3 and 4 are in a toggle position, i.e., the pin that joins links 3 and 4 is in line with the pins
that join 1 with 4 and 2 with 3.
Solution: See Figure 4-9 and Mathcad file P0409.
1. The FBDs of the assembly and each individual link are shown in Figure 4-9. The dimensions, as scaled from
Figure P4-3 in the text, are shown on the link FBDs.
50.0 = a
22.0 = d
21.2 = h
F
26.9 = f
32
2.8 = g
28.0 = e
2
F12
P
14F
4
34F
F 4
2
1
P
P
F
55.0 = b
F
39.5 = c
3
F41
43F
23
F
129.2°
F21
1
F
3
P
FIGURE 4-9
Free Body Diagrams for Problem 4-9
2. The cross-sectional area for all pins is the same and is A
π d
2
4
A 50.265 mm
2
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-9-2
3. The pin that joins members 1 and 2 is the most highly stressed while the stress on each of the remaining pins
is the same. Since the pins are in double shear, we will divide the pin load by 2 in each case.
Pin joining 1 and 2 τ12
F12
2 A
τ12 74.6 MPa
All other pins τ14
F14
2 A
τ14 50.7 MPa
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-10-1
PROBLEM 4-10
Statement: The over-hung diving board of problem 3-10 is shown in Figure P4-4a. Assume cross-section
dimensions of 305 mm x 32 mm. The material has E = 10.3 GPa. Find the largest principal stress
at any location in the board when a 100-kg person is standing at the free end.
700 = a
2000 = L
R
R
1
2
P
Given: Weight of person W 100 kgf
Board dimensions
Distance to support a 0.7 m
Length of board L 2 m
Cross-section w 305 mm
t 32 mm
FIGURE 4-10
Assumptions: The weight of the beam is negligible
compared to the applied load and so can
be ignored.
Free Body Diagram for Problem 4-10
Solution: See Figure 4-10 and Mathcad file P0410.
1. From the FBD of the diving board and Figure B-3 (Appendix B), the reactions at the supports are
R1 W 1
L
a
R1 1821 N
R2 W
L
a
R2 2802 N
2. Also from Figure D-3, the maximum bending moment occurs at the right-hand support where, in the FBD
above, x = a.
Mmax R1 a Mmax 1275 N m
3. The maximum bending stress will occur on the top and bottom surfaces of the board at the section where the
maximum bending moment occurs which, in this case, is at x = a. The only stress present on the top or
bottom surface of the board is the bending stress x. Therefore, on the top surface where the stress is
tensile, x. is the principal stress 1 . Thus,
Distance to extreme fiber c
t
2
c 16 mm
Moment of inertia I
w t
3
12
I 8.329 10
5
mm
4
Bending stress σx
Mmax c
I
σx 24.492 MPa
Maximum principal stress σ1 σx σ1 24.5 MPa
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-11-1
PROBLEM 4-11
Statement: Repeat Problem 4-10 using the loading conditions of Problem 3-11. Assume the board weighs 29
kg and deflects 13.1 cm statically when the person stands on it. Find the largest principal stress
at any location in the board when the 100-kg person in Problem 4-10 jumps up 25 cm and lands
back on the board. Find the maximum deflection.
700 = a
2000 = L
R
R
1
2
F i
Given: Beam length L 2000 mm
Distance to support a 700 mm
Mass of person mpers 100 kg
Mass of board mboard 29 kg
Static deflection δst 131 mm
Height of jump h 250 mm
Cross-section w 305 mm
FIGURE 4-11
t 32 mm Free Body Diagram for Problem 4-11
Assumptions: The apparent Young's modulus for fiberglas is
E 1.03 10
4
MPa
Solution: See Figure 4-11 and Mathcad file P0411.
1. From Problem 3-11, the dynamic load resulting from the impact of the person with the board isFi 3.056 kN
2. From the FBD of the diving board and Figure B-3(a) (Appendix B), the reactions at the supports are
R1 Fi 1
L
a
R1 5.675 kN
R2 Fi
L
a
R2 8.731 kN
3. Also from Figure D-3(a), the maximum bending moment occurs at the right-hand support where, in the
FBD above, x = a.
Mmax R1 a Mmax 3.973 kN m
4. The maximum bending stress will occur on the top and bottom surfaces of the board at the section where the
maximum bending moment occurs which, in this case, is at x = a. The only stress present on the top or bottom
surface of the board is the bending stress x. Therefore, on the top surface where the stress is tensile, x is the
principal stress 1 . Thus,
Distance to extreme fiber c
t
2
c 16 mm
Moment of inertia I
w t
3
12
I 8.329 10
5
mm
4
Bending stress σx
Mmax c
I
σx 76.322 MPa
Maximum principal stress σ1 σx σ1 76.3 MPa
5. Calculate the maximum deflection from the equation given in Figure D-3(a) at x = L. Let b in the figure be a in
our problem and let a in the figure be equal to L, then
ymax
Fi
6 a E I
a L( ) L
3
L L a( )
3
a
2
L a( ) L ymax 401.4 mm
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-12-1
PROBLEM 4-12
Statement: Repeat Problem 4-10 using the cantilevered diving board design in Figure P4-4b.
P
R 1
700
2000
1300 = L
M1
Given: Beam length L 1300 mm
Weight at free end P 100 kgf
Cross-section w 305 mm
t 32 mm
Assumptions: The apparent Young's modulus for
fiberglas is
E 1.03 10
4
MPa
Solution: See Figure 4-12 and Mathcad file P0412. FIGURE 4-12
Free Body Diagram for Problem 4-12
1. From the FBD of the diving board and Figure B-1(a) (Appendix B), the reactions at the supports are
R1 P R1 981 N
M1 P L M1 1275 N m
2. Also from Figure D-1, the maximum bending moment occurs at the support where, in the FBD above, x = 0.
Mmax M1 Mmax 1275 N m
3. The maximum bending stress will occur on the top and bottom surfaces of the board at the section where the
maximum bending moment occurs which, in this case, is at x = 0. The only stress present on the top or bottom
surface of the board is the bending stress x. Therefore, on the top surface where the stress is tensile, x is the
principal stress 1 . Thus,
Distance to extreme fiber c
t
2
c 16 mm
Moment of inertia I
w t
3
12
I 8.329 10
5
mm
4
Bending stress σx
Mmax c
I
σx 24.492 MPa
Maximum principal stress σ1 σx σ1 24.5 MPa
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-13-1
PROBLEM 4-13
Statement: Repeat Problem 4-11 using the diving board design shown in Figure P4-4b. Assume the board
weighs 19 kg and deflects 8.5 cm statically when the person stands on it.
Given: Unsupported length L 1300 mm
2000
1300 = L
R 1
700
M1
F i
Mass of board mboard 19 kg
Static board deflection δstat 85 mm
Mass of person mperson 100 kg
Height of jump h 250 mm
Cross-section w 305 mm
t 32 mm
Assumptions: The apparent Young's modulus for fiberglas is
FIGURE 4-13
E 1.03 10
4
MPa
Free Body Diagram for Problem 4-13
Solution: See Figure 4-13 and Mathcad file P0413.
1. From Problem 3-13, the dynamic load resulting from the impact of the person with the board isFi 3.487 kN
2. From the FBD of the diving board and Figure B-1(a) (Appendix B), the reactions at the supports are
R1 Fi R1 3487 N
M1 Fi L M1 4533 N m
3. Also from Figure D-1, the maximum bending moment occurs at the support where, in the FBD above, x = 0.
Mmax M1 Mmax 4533 N m
4. The maximum bending stress will occur on the top and bottom surfaces of the board at the section where the
maximum bending moment occurs which, in this case, is at x = 0. The only stress present on the top or bottom
surface of the board is the bending stress x. Therefore, on the top surface where the stress is tensile, x is the
principal stress 1 . Thus,
Distance to extreme fiber c
t
2
c 16 mm
Moment of inertia I
w t
3
12
I 8.329 10
5
mm
4
Bending stress σx
Mmax c
I
σx 87.086 MPa
Maximum principal stress σ1 σx σ1 87.1 MPa
5. Calculate the maximum deflection from the equation given in Figure D-3(a) at x = L. Let b in the figure be a in
our problem and let a in the figure be equal to L, then
ymax
Fi L
3
3 E I
ymax 297.7 mm
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-14-1
PROBLEM 4-14
Statement: Figure P4-5 shows a child's toy called a pogo stick. The child stands on the pads, applying
half her weight on each side. She jumps off the ground, holding the pads up against her
feet, and bounces along with the spring cushioning the impact and storing energy to help
each rebound. Design the aluminum cantilever beam sections on which she stands to
survive jumping 2 in off the ground. Assume an allowable stress of 20 ksi. Define and size
the beam shape.
P
F /2i F /2i
Given: Allowable stress σallow 20 ksi
Young's modulus E 10.3 10
6
psi
Assumptions: The beam will have a rectangular
cross-section with the load applied at a
distance of 5 in from the central support.
L 5 in
Solution: See Figure 4-14 and Mathcad file P0414.
1. From Problem 3-14, the total dynamic force on both
foot supports is
Fi 224 lbf
Therefore, the load on each support is
P
Fi
2
P 112 lbf
2. To give adequate support to the childs foot, let the
width of the support beam be
w 1.5 in
3. From Figure B-1(a) in Appendix B, the maximum
bending moment at x = 0 is
M P L M 560 in lbf
FIGURE 4-14
4. We can now calculate the minimum required section modulus, Z = I/c.
Free Body Diagram for Problem 4-14
Bending stress σ
M
Z
= σallow=
Solving for Z, Z
M
σallow
Z 458.8 mm
3
5. For a rectangular cross-section, I
w t
3
12
= and c
t
2
= so Z
w t
2
6
=
Solving for t, t
6 Z
w
t 0.335 in
Round this up to the next higher decimal equivalent of a common fraction, t 0.375 in
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-15-1
PROBLEM 4-15
Statement: Design a shear pin for the propeller shaft of an outboard motor if the shaft through which the
pin is placed is 25-mm diameter, the propeller is 20-cm diameter, and the pin must fail when a
force > 400 N is applied to the propeller tip. Assume an ultimate shear strength for the pin
material of 100 MPa.
d
F
pin Propeller Shaft
T Propeller Hub
Shear Pin
Fpin
Given: Propeller shaft dia d 25 mm
Propeller dia D 200 mm
Max propeller tip force Fmax 400 N
Ultimate shear strength Sus 100 MPa
Assumptions: A shear pin is in direct, double shear.
Solution: See Figure 4-15 and Mathcad file P0415.
1. Calculate the torque on the propeller shaft that will
result from a tip force on the propeller of Fmax.
FIGURE 4-15
Free Body Diagram for Problem 4-15
T Fmax
D
2
T 40000 N mm
2. This will be reacted by the shear pin's couple on the shaft. Determine the magnitude of the direct shear force.
Fpin
T
d
Fpin 1600 N
3. Determine the maximum pin diameter that will shear at this force.
Direct shear stress τ
Fpin
A
=
4 Fpin
π dpin
2
= Sus=
Solving for the pin diameter dpin
4 Fpin
π Sus
dpin 4.514 mm
Round this to dpin 4.5 mm
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-16-1
PROBLEM 4-16
Statement: A track to guide bowling balls is designed with two round rods as shown in Figure P4-6. The
rods are not parallel to one another but have a small angle between them. The balls roll on the
rods until they fall between them and drop onto another track. The angle between the rods is
varied to cause the ball to drop at different locations. Find the maximum stress and deflection
in the rods assuming that they are
(a) simply supported at each end, and
(b) fixed at each end.
Given: Rod length L 30 in
2RR 1 L
a Fball
Rod diameter d 1.00 in
Distance to load a 23.15 in
Young's modulus E 30 10
6
psi
Assumptions: The analysis of Problem 3-16 yielded
the following for a simply supported
beam:
FIGURE 4-16A
Free Body Diagram for Problem 4-16(a), taken
on a plane through the rod axis and ball center
Max ball load Fball 13.89 lbf
Max moment Mmax 73.4 in lbf
Reactions R1 3.17 lbf
R2 10.72 lbf
Solution: See Figure 4-16 and Mathcad file P0416.
1. The maximum bending stress will occur at the outer fibers of the rod at the section where the maximum
bending moment occurs which, in this case, is at x = a. The only stress present on the top or bottom surface of
the rod is the bending stress x. Therefore, on the bottom surface where the stress is tensile, x is the principal
stress 1 . Thus, for a simply supported rod,
Distance to extreme fiber c
d
2
c 0.5 in
Moment of inertia I
π d
4
64
I 0.0491 in
4
Bending stress σx
Mmax c
I
σx 748 psi
Maximum principal stress σ1 σx σ1 748 psi
2. Calculate the maximum deflection for the simply supported case from the equation given in Figure D-2(a),
ymax
Fball
6 E I
2 a
3
a
4
L
L a
2
ymax 0.0013 in
L
a
R
M1
1 2R
M
2
Fball
3. For the case where the rod is built in at each end,
the beam is statically indeterminate. As seen in
Figure 4-16B, there are four unknown reactions and
only two equilibrium equationscan be written using
statics. We will find the reactions using Example 4-7
as a model. FIGURE 4-16B
Free Body Diagram for Problem 4-16(b), taken on a
plane through the rod axis and ball center
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-16-2
4. Write an equation for the load function in terms of equations 3.17 and integrate the resulting function four
times using equations 3.18 to obtain the shear and moment functions. Note use of the unit doublet function to
represent the moment at the wall. For the beam in Figure 4-16B,
q(x) = -M
1
<x - 0>-2 + R
1
<x - 0>-1 - F<x - a>-1 + R
2
<x - L>-1 + M2<x - L>
-2
V(x) = -M
1
<x - 0>-1 + R
1
<x - 0>0 - F<x - a>0 + R
2
<x - L>0 + M2<x - L>
-1 + C1
M(x) = -M
1
<x - 0>0 + R
1
<x - 0>1 - F<x - a>1 + R
2
<x - L>1 + M2<x - L>
0 + C1x+ C2
(x) = ( -M
1
<x - 0>1 + R
1
<x - 0>2/2 - F<x - a>2/2 + R
2
<x - L>2/2 + M2<x - L>
1 + C1x
2/2 + C2x + C3) / EI
y(x) = ( -M
1
<x - 0>2/2 + R
1
<x - 0>3/6 - F<x - a>3/6 + R
2
<x - L>3/6 + M2<x - L>
2 /2+ C1x
3/6 + C2x
2/2 + C3x + C4) / EI
5. Because the reactions have been included in the loading function, the shear and moment diagrams both
close to zero at each end of the beam, making C1 = C2 = 0. This leaves six unknowns; the four reactions and the
constants of integration, C3 and C4. There are four boundary conditions that we can use and two equilibrium
equations. The boundary conditions are: at x = 0, = 0 and y = 0; and at x = L, = 0 and y = 0. Applying the
boundary conditions at x = 0 results in C3 = C4 = 0. Applying the BCs at x = L results in the following two
equations, which are solved for R1 and M1.
At x = L, θ 0= 0
R1
2
L
2
M1 L
F
2
L a( )
2
=
y 0= 0
R1
6
L
3
M1
2
L
2
F
6
L a( )
3
=
Solving these two equations simultaneously for R1 and M1,
M1
Fball
L
L a( )
2 L a( )
3
L
M1 16.765 in lbf
R1 2
M1
L
Fball
L a( )
2
L
2
R1 1.842 lbf
6. The remaing two reactions can be found by using the equations of equilibrium.
F
y
= 0: R1 Fball R2 0=
M = 0: M1 Fball a R2 L M2 0=
Solving these two equations simultaneously for R2 and M2,
R2 Fball R1 R2 12.048 lbf
M2 M1 Fball a R2 L M2 56.657 in lbf
7. Define the range for x, x 0 in 0.005 L L
8. For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than z,
and a value of one when it is greater than or equal to z.
S x z( ) if x z 1 0( )
9. Write the shear, moment, slope, and deflection equations in Mathcad form, using the function S as a multiplying
factor to get the effect of the singularity functions.
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-16-3
V x( ) R1 S x 0 mm( ) Fball S x a( ) R2 S x L( )
M x( ) M1 S x 0 mm( ) R1 S x 0 mm( ) x 0 mm( ) Fball S x a( ) x a( )
M2 S x L( ) R2 S x L( ) x L( )
θ x( )
1
E I
M1 S x 0 mm( ) x
R1
2
S x 0 mm( ) x 0 mm( )
2
Fball
2
S x a( ) x a( )
2
M2 S x L( ) x L( )
R2
2
S x L( ) x L( )
2
y x( )
1
E I
M1
2
S x 0 mm( ) x
2
R1
6
S x 0 mm( ) x 0 mm( )
3
Fball
6
S x a( ) x a( )
3
M2
2
S x L( ) x L( )
2
R2
6
S x L( ) x L( )
3
10. Plot the shear, moment, slope, and deflection diagrams.
(a) Shear Diagram (b) Moment Diagram
0 10 20 30
15
10
5
0
5
Distance along beam, x - in
S
h
ea
r,
V
-
l
b
0 10 20 30
60
40
20
0
20
40
Distance along beam, x - in
M
o
m
en
t,
M
-
l
b
i
n
(c) Slope Diagram (d) Deflection Diagram
0 10 20 30
0.1
0
0.1
Distance along beam, x - in
S
lo
p
e
-
T
h
o
u
sa
n
d
s
o
f
R
a
d
0 10 20 30
0.8
0.6
0.4
0.2
0
Distance along beam, x - in
D
ef
le
ct
io
n
-
t
h
o
u
sa
n
d
th
s
o
f
in
FIGURE 4-16C
Shear, Moment, Slope, and Deflection Diagrams for Problem 4-16(b)
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-16-4
11 The maximum moment occurs at x = L and is
Mmax M2 Mmax 56.7 in lbf
12 Calculate the maximum bending and principal stresses.
Bending stress σx
Mmax c
I
σx 577 psi
Maximum principal stress σ1 σx σ1 577 psi
13. To find the maximum deflection, first determine at what point on the beam the slope is zero. Let this be at x = e.
From the slope diagram, we see that e < a. Using the slope equation and setting it equal to zero, we have
For = 0 0 M1 e
R1
2
e
2
=
Solving for e e
2 M1
R1
e 18.204 in
Maximum deflection ymax y e( ) ymax 0.00063 in
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-17-1
PROBLEM 4-17
Statement: A pair of ice tongs is shown in Figure P4-7. The ice weighs 50 lb and is 10 in wide across the
tongs. The distance between the handles is 4 in, and the mean radius r of the tong is 6 in. The
rectangular cross-sectional dimensions are 0.75 x 0.312 in. Find the stress in the tongs.
F
W/2
B
FB
A
O
C
F
F
C
O
3.5 = cy
12.0 = by
2.0 = cx
5.0 = bx
11.0 = ax
Given: Mean radius of tong rc 6.00 in
Tong width w 0.312 in
Tong depth h 0.75 in
Assumptions: The tong can be analyzed as a curved
beam.
Solution: See Problem 3-17, Figure 4-17,
and Mathcad file P0417.
1. The maximum bending moment and axial force in the
tong were found in Problem 3-17 at point A. They are
Maximum moment MA 237.5 in lbf
Axial force at D FAn 25 lbf FIGURE 4-17
Free Body Diagram for Problem 4-17
2. Calculate the section area, inside radius and outside radus.
Area of section A h w A 0.234 in
2
Inside and outside radii
of section
ri rc 0.5 h ri 5.625 in
ro rc 0.5 h ro 6.375 in
3. Use the equation in the footnote on page 195 of the text to calculate the radius of the neutral axis.
Radius of neutral axis rn
ro ri
ln
ro
ri
rn 5.992 in
4. Calculate the eccentricty and the distances from the neutral axis to the extreme fibers.
Eccentricity e rc rn e 0.007821 in
Distances from neutral
axis to extreme fibers
ci rn ri ci 0.3672 in
co ro rn co 0.3828 in
Stresses at inner and
outer radii
σi
MA
e A
ci
ri
FAn
A
σi 8.58 ksi
σo
MA
e A
co
ro
FAn
A
σo 7.69 ksi
5. The shear stress is zero at the outer fibers. Therefore, these are the principal stresses. At the inner surface
σ1 σi σ1 8.58 ksi σ2 0 ksi σ3 0 ksi
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-18-1
PROBLEM 4-18
Statement: A set of steel reinforcing rods is to be stretched axially in tension to createa tensile stress of 30
ksi prior to being cast in concrete to form a beam. Determine how much force will be required to
stretch them the required amount and how much deflection is required. There are 10 rods; each
is 0.75-in diameter and 30 ft long.
Given: Desired stress σ 30 ksi Rod diameter d 0.75 in
Number of rods Nrods 10 Young's modulus E 30 10
6
psi
Rod length L 30 ft
Assumptions: The rods share the load equally.
Solution: See Mathcad file P0418.
1. Calculate the cross-sectional area of one rod. A
π d
2
4
A 0.442 in
2
2. Determine the force required to achieve the desired stress level in one rod.
σ
F
A
= F σ A F 13.254 kip
3. Determine the total force required to achieve the desired stress level in all rods.
Ftotal Nrods F Ftotal 132.5 kip
4. Determine the amount the rods will deflect under the applied load.
δ
F L
A E
δ 0.360 in
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-19-1
PROBLEM 4-19
Statement: The clamping fixture used to pull the rods in Problem 4-18 is conected to the hydraulic ram by a
clevis like that shown in Figure P4-8. Determine the size of the clevis pin needed to withstand
the applied force. Assume an allowable shear stress of 20 000 psi and an allowable normal
stress of 40 000 psi. Determine the required outside radius of the clevis end to not exceed the
above allowable stresses in either tear out or bearing if the clevis flanges are each 0.8 in thick.
Given: Desired rod stress σrod 30 ksi Rod diameter d 0.75 in
Number of rods Nrods 10 Young's modulus E 30 10
6
psi
Rod length L 30 ft Clevis flange thickness t 0.8 in
Clevis strength Ssallow 20 ksi
Sballow 40 ksi
Assumptions: The rods share the load equally, and there is one clevis for all ten rods.
Solution: See Figures 4-12 and 4-13 in the text, Figure 4-19, and Mathcad file P0419.
1. Calculate the cross-sectional area of one rod. A
π d
2
4
A 0.442 in
2
2. Determine the force required to achieve the desired stress level in one rod.
σrod
F
A
= F σrod A F 13.254 kip
3. Determine the total force required to achieve the desired stress level in all rods.
Ftotal Nrods F Ftotal 132.5 kip
This force is transmitted through the clevis pin, which is in double shear.
4. Calculate the minimum required clevis pin diameter for the allowable shear stress. Divide the load by 2
because of the double shear loading.
τpin
Ftotal
2 Apin
=
2 Ftotal
π d
2
= Ssallow=
Solving for the pin diameter d
2 Ftotal
π Ssallow
d 2.054 in
Round this up to the next higher decimal equivalent of a common fraction ( 2 1/8) d 2.125 in
5. Check the bearing stress in the clevis due to the pin on one side of the clevis.
Bearing stress area Ab d t Ab 1.700 in
2
Bearing force Fb
Ftotal
2
Fb 66.268 kip
Bearing stress σb
Fb
Ab
σb 39.0 ksi
Since this is less than Sballow, this pin diameter is acceptable.
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-19-2
6. Determine the tearout stress in the clevis.
d R
Tearout length
Shear area (see Figure 4-19) Atear 2 t R
2
0.5 d( )
2
=
Shear force
Ftear
Ftotal
2
Ftear 66.268 kip
Shear stress and strength
FIGURE 4-19
τ
Ftear
Atear
=
Ftear
2 t R
2
0.5 d( )
2
= Ssallow= Tearout Diagram for Problem 4-19
Solving for the clevis radius, R R
Ftear
2 t Ssallow
2
0.5 d( )
2
R 2.328 in
Round this up to the next higher decimal equivalent of a common fraction ( 2 3/8) R 2.375 in
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-20-1
PROBLEM 4-20
Statement: Repeat Problem 4-19 for 12 rods, each 1 cm in diameter and 10 m long. The desired rod stress is 20
MPa. The allowable normal stress in the clevis and pin is 280 MPa and their allowable shear stress
is 140 MPa. Each clevis flange is 2 cm wide.
Units: kN 10
3
newton MPa 10
6
Pa GPa 10
9
Pa
Given: Desired rod stress σrod 200 MPa Rod diameter d 10 mm
Number of rods Nrods 12 Young's modulus E 207 GPa
Rod length L 10 m Clevis flange thickness t 20 mm
Clevis strength Ssallow 140 MPa
Sballow 280 MPa
Assumptions: The rods share the load equally, and there is one clevis for all twelve rods.
Solution: See Figures 4-12 and 4-13 in the text, Figure 4-20, and Mathcad file P0420.
1. Calculate the cross-sectional area of one rod. A
π d
2
4
A 78.54 mm
2
2. Determine the force required to achieve the desired stress level in one rod.
σrod
F
A
= F σrod A F 15.708 kN
3. Determine the total force required to achieve the desired stress level in all rods.
Ftotal Nrods F Ftotal 188.5 kN
This force is transmitted through the clevis pin, which is in double shear.
4. Calculate the minimum required clevis pin diameter for the allowable shear stress. Divide the load by 2 because
of the double shear loading.
τpin
Ftotal
2 Apin
=
2 Ftotal
π d
2
= Ssallow=
Solving for the pin diameter d
2 Ftotal
π Ssallow
d 29.277 mm
Round this up to the next higher even mm d 30 mm
5. Check the bearing stress in the clevis due to the pin on one side of the clevis.
Bearing stress area Ab d t Ab 600 mm
2
Bearing force Fb
Ftotal
2
Fb 94.248 kN
Bearing stress σb
Fb
Ab
σb 157.1 MPa
Since this is less than Sballow, this pin diameter is acceptable.
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-20-2
6. Determine the tearout stress in the clevis.
d R
Tearout length
Shear area (see Figure 4-19) Atear 2 t R
2
0.5 d( )
2
=
Shear force
Ftear
Ftotal
2
Ftear 94.248 kN
Shear stress and strength
τ
Ftear
Atear
=
Ftear
2 t R
2
0.5 d( )
2
= Ssallow= FIGURE 4-20
Tearout Diagram for Problem 4-20
Solving for the clevis radius, R R
Ftear
2 t Ssallow
2
0.5 d( )
2
R 22.544 mm
Round this up to the next higher even mm R 24 mm
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-21-1
PROBLEM 4-21
Statement: Figure P4-9 shows an automobile wheel with two common styles of lug wrench being used to
tighten the wheel nuts, a single-ended wrench in (a), and a double-ended wrench in (b). In each
case two hands are required to provide forces respectively at A and B as shown. The distance
between points A and B is 1 ft in both cases and the handle diameter is 0.625 in. The wheel nuts
require a torque of 70 ft-lb. Find the maximum principle stress and maximum deflection in each
wrench design.
Given: Distance between A and B dAB 1 ft
Tightening torque T 70 ft lbf
Wrench diameter d 0.625 in
Assumptions: 1. The forces exerted by the user's hands lie in a plane through the wrench that is also parallel
to the plane of the wheel.
2. The applied torque is perpendicular to the plane of the forces.
3.By virtue of 1 and 2 above, this is a planar problem that can be described in a 2D FBD.
Solution: See Figure 4-21 and Mathcad file P0421.
12" = dAB
12" = dAB
6"
F
(b) Double-ended Wrench
T
F
T
(a) Single-ended Wrench
F
F1. In Problem 3-21 we found that for both cases
F 70 lbf
2. From examination of the FBDs, we see that, in
both cases, the arms are in bending and the stub
that holds the socket wrench is in pure torsion.
The maximum bending stress in the arm will occur
near the point where the arm transitions to the
stub. The stress state at this transition is very
complicated, but we can find the nominal bending
stress there by treating the arm as a cantilever
beam, fixed at the transition point. For both cases
the torque in the stub is the same.
Case (a)
2. The bending moment at the transition is
Ma F dAB Ma 840 lbf in
FIGURE 4-21
3. The tensile stress at this point is found from Free Body Diagrams for Problem 4-21
Moment of inertia I
π d
4
64
I 0.00749 in
4
Dist to extreme fibre c 0.5 d c 0.313 in
Stress σx
Ma c
I
σx 35.05 ksi
4. There are no other stress components present at this point, so x is the maximum principle stress here and
σ1 σx σ1 35.0 ksi σ2 0 psi σ3 0 psi
5. The torque in the stub is T 840 in lbf
6. The shear stress at any point on the outside surface of the stub is found from
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-21-2
Polar moment of inertia J 2 I J 0.0150 in
4
Shear stress τxy
T c
J
τxy 17.52 ksi
7. There are no other stress components present along the outside surface of the stub, so
σ1 τxy σ1 17.5 ksi σ2 0 psi σ3 σ1
8. Thus, the maximum principle stress for case (a) is on the upper surface of the handle (arm) near the point
where it transitions to the stub.
There will be two deflection components that we can calculate separately and then add (superposition).
One will come from the bending of the arm and one will come from the twisting of the stub, projected out
to the end of the arm.
9. Deflection of the arm due to bending only for a stub length of stub 3 in :
Assuming that the wrenches are made from steel E 30 10
6
psi G 11.7 10
6
psi
From Figure B-1(a), Appendix B, yarm
F dAB
3
3 E I
yarm 0.179 in
From equation (4.24), the
angular twist of the stub is θstub
T stub
J G
θstub 0.014 rad
The deflection at the end of
the arm due to the stub twist
is
ystub dAB θstub ystub 0.173 in
So, the total deflection is ya yarm ystub ya 0.352 in
Case (b)
10. The bending moment at the transition is Mb
F dAB
2
Mb 420 lbf in
11. The tensile stress at this point is found from
Stress σx
Mb c
I
σx 17.52 ksi
12. There are no other stress components present at this point, so x is the maximum principle stress here and
σ1 σx σ1 17.5 ksi σ2 0 psi σ3 0 psi
13. The torque in the stub is T 840 in lbf
14. The shear stress at any point on the outside surface of the stub is found from
Shear stress τxy
T c
J
τxy 17.52 ksi
15. There are no other stress components present along the outside surface of the stub, so
σ1 τxy σ1 17.5 ksi σ2 0 psi σ3 σ1
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-21-3
16. Thus, the maximum principle stress for case (b) is the same on the upper surface of the handle (arm) near the
point where it transitions to the stub, and on the outside surface of the stub.
There will be two deflection components that we can calculate separately and then add (superposition).
One will come from the bending of the arm and one will come from the twisting of the stub, projected out to
the end of the arm.
Deflection of the arm due to bending only:
From Figure B-1(a), Appendix B, yarm
F 0.5 dAB 3
3 E I
yarm 0.022 in
From equation (4.24), the
angular twist of the stub is θstub
T stub
J G
θstub 0.014 rad
The deflection at the end of
the arm due to the stub twist
is
ystub
dAB
2
θstub ystub 0.086 in
So, the total deflection is yb yarm ystub yb 0.109 in
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-22-1
PROBLEM 4-22
Statement: A roller-blade skate is shown in Figure P4-10. The polyurethane wheels are 72 mm dia. The
skate-boot-foot combination weighs 2 kg. The effective "spring rate" of the person-skate
subsystem is 6000 N/m. The axles are 10-mm-dia steel pins in double shear. Find the stress in
the pins for a 100-kg person landing a 0.5-m jump on one foot. (a) Assume all 4 wheels land
simultaneously. (b) Assume that one wheel absorbs all the landing force.
Given: Axle pin diameter d 10 mm
Solution: See Figure P4-10 and Mathcad file P0422.
1. From Problem 3-22, we have the forces for cases (a) and (b): Fa 897 N Fb 3.59 kN
2. In both cases, this is the force on one axle. The shear force will be one half of these forces because the pins
are in double shear.
Shear area As
π d
2
4
As 78.54 mm
2
Shear stress
Case (a) all wheels landing τa
Fa
2 As
τa 5.71 MPa
Case (b) one wheel landing τb
Fb
2 As
τb 22.9 MPa
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-23a-1
PROBLEM 4-23a
Statement: A beam is supported and loaded as shown in Figure P4-11a. Find the reactions, maximum shear,
maximum moment, maximum slope, maximum bending stress, and maximum deflection for the
data given in row a from Table P4-2.
Given: Beam length L 1 m
R 2
F
R 1
a
b
L
w
Distance to distributed load a 0.4 m
Distance to concentrated load b 0.6 m
Distributed load magnitude w 200 N m
1
Concentrated load F 500 N
Moment of inertia I 2.85 10
8
m
4
Distance to extreme fiber c 2.00 10
2
m FIGURE 4-23A
Free Body Diagram for Problem 4-23
Solution: See Figures 4-23 and Mathcad file P0423a.
1. The reactions, maximum shear and maximum moment were all found in Problem 3-23a. Those results are
summarized here.
Load function q(x) = R
1
<x - 0>-1 - w<x - 0>0 + w<x - a>0 - F<x - b>-1 + R2<x - L>
-1
Shear function V(x) = R
1
<x - 0>0 - w<x - 0>1 + w<x - a>1 - F<x - b>0 + R2<x - L>
0
Moment function M(x) = R
1
<x - 0>1 - w<x - 0>2/2 + w<x - a>2/2 - F<x - b>1 + R2<x - L>
1
Modulus of elasticity E 207 GPa
Reactions R1 264.0 N R2 316.0 N
Maximum shear Vmax 316 N (negative, from x = b to x =L)
Maximum moment Mmax 126.4 N m (at x = b)
2. Integrate the moment function, multiplying by 1/EI, to get the slope.
(x) = [R
1
<x>2/2 - w<x>3/6 + w<x - a>3/6 - F<x - b>2/2 + R
2
<x - L>2/2 + C
3
]/EI
3. Integrate again to get the deflection.
y(x) = [R
1
<x>3/6 - w<x>4/24 + w<x - a>4/24 - F<x - b>3/6 + R
2
<x-L>3/6 + C
3
x +C
4
]/EI
4. Evaluate C3 and C4
At x = 0 and x = L, y = 0, therefore, C4 = 0.
0
R1
6
L
3
w
24
L
4
w
24
L a( )
4
F
6
L b( )
3
C3 L=
C3
1
L
R1
6
L
3
w
24
L
4
w
24
L a( )
4
F
6
L b( )
3
C3 31.413 N m
2
5. Define the range for x x 0 m 0.005 L L
6. For a Mathcad solution, definea step function S. This function will have a value of zero when x is less than z,
and a value of one when it is greater than or equal to z.
S x z( ) if x z 1 0( )
7. Write the slope and deflection equations in Mathcad form, using the function S as a multiplying factor to
get the effect of the singularity functions. See Figure 4-23aB where these functions are plotted.
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-23a-2
θ x( )
1
E I
R1
2
S x 0 in( ) x
2
w
6
S x 0 in( ) x
3
w
6
S x a( ) x a( )
3
R2
2
S x L( ) x L( )
2
F
2
S x b( ) x b( )
2
C3
y x( )
1
E I
R1
6
S x 0 in( ) x
3
w
24
S x 0 in( ) x
4
w
24
S x a( ) x a( )
4
R2
6
S x L( ) x L( )
3
F
6
S x b( ) x b( )
3
C3 x
8. Maximum slope occurs at x = L θmax θ L( ) θmax 0.335 deg
9. Maximum deflection occurs at x = c, where = 0 and c < b.
θ0
1
E I
R1
2
c
2
w
6
c
3
w
6
c a( )
3
C3
= 0=
Solving for c,
A
R1
2
3
w
6
a B 3
w
6
a
2
C C3
w
6
a
3
A 92.000 N B 16.000 N m C 33.547 N m
2
c
B B
2
4 A C
2 A
c 0.523 m
Substituting c into the deflection equation, ymax y c( ) ymax 1.82 mm
0 0.2 0.4 0.6 0.8 1
0.01
0.005
0
0.005
0.01
SLOPE, radians
θ x( )
x
m
0 0.2 0.4 0.6 0.8 1
2
1.5
1
0.5
0
DEFECTION, mm
y x( )
mm
x
m
FIGURE 4-23aB
Slope and Deflection Diagrams for Problem 4-23a
10. The maximum bending stress occurs at x = b, where the moment is a maximum. For
c 2.00 10
2
m c 20 mm
σmax
Mmax c
I
σmax 88.7 MPa
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-24a-1
PROBLEM 4-24a
Statement: A beam is supported and loaded as shown in Figure P4-11b. Find the reactions, maximum
shear, maximum moment, maximum slope, maximum bending stress, and maximum deflection for
the data given in row a from Table P4-2.
F
w
R 1
M1
L
a
Given: Beam length L 1 m
Distance to distributed load a 0.4 m
Distance to concentrated load b 0.6 m
Distributed load magnitude w 200 N m
1
Concentrated load F 500 N
Moment of inertia I 2.85 10
8
m
4
Distance to extreme fiber c 2.00 10
2
m
FIGURE 4-24A
Free Body Diagram for Problem 4-24
Solution: See Figures 4-24 and Mathcad file P0424a.
1. The reactions, maximum shear and maximum moment were all found in Problem 3-24a. Those results are
summarized here.
Load function q(x) = -M1<x - 0>
-2 + R
1
<x - 0>-1 - w<x - a>0 - F<x - L>-1
Shear function V(x) = -M1<x - 0>
-1 + R
1
<x - 0>0 - w<x - a>1 - F<x - L>0
Moment function M(x) = -M1<x - 0>
0 + R
1
<x - 0>1 - w<x - a>2/2 - F<x - L>1
Modulus of elasticity E 207 GPa
Reactions R1 620.0 N M1 584.0 N m
Maximum shear Vmax 620 N (positive, at x = 0)
Maximum moment Mmax 584 N m (negative, at x = 0)
2. Integrate the moment function, multiplying by 1/EI, to get the slope.
(x) = [-M
1
<x-0>1 + R
1
<x - 0>2/2 - w<x - a>3/6 - F<x - L>2/2 + C
3
]/EI
3. Integrate again to get the deflection.
y(x) = [-M
1
<x-0>2/2 + R
1
<x - 0>3/6 - w<x - a>4/24 - F<x - L>3/6 + C
3
x +C
4
]/EI
4. Evaluate C
3
and C
4
. At x = 0, = 0 and y = 0, therefore, C
3
= 0 and C
4
= 0.
5. Define the range for x x 0 m 0.005 L L
6. For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than z,
and a value of one when it is greater than or equal to z.
S x z( ) if x z 1 0( )
7. Write the slope and deflection equations in Mathcad form, using the function S as a multiplying factor to get
the effect of the singularity functions. See Figure 4-24aB where these functions are plotted.
θ x( )
1
E I
M1 S x 0 in( ) x
R1
2
S x 0 in( ) x
2
w
6
S x a( ) x a( )
3
F
2
S x L( ) x L( )
2
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-24a-2
y x( )
1
E I
M1
2
S x 0 in( ) x
2
R1
6
S x 0 in( ) x
3
w
24
S x a( ) x a( )
4
F
6
S x L( ) x L( )
3
8. Maximum slope occurs at x = L θmax θ L( ) θmax 2.73 deg
9. Maximum deflection occurs at x = L ymax y L( ) ymax 32.2 mm
10. The maximum bending stress occurs at x = 0, where the moment is a maximum. For c 20 mm
σmax
M1 c
I
σmax 410 MPa
SLOPE, radians DEFLECTION, mm
0 0.2 0.4 0.6 0.8 1
0.05
0.04
0.03
0.02
0.01
0
θ x( )
x
m
0 0.2 0.4 0.6 0.8 1
40
30
20
10
0
y x( )
mm
x
m
FIGURE 4-24aB
Slope and Deflection Diagrams for Problem 4-24a
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-25a-1
PROBLEM 4-25a
Statement: A beam is supported and loaded as shown in Figure P4-11c. Find the reactions, maximum
shear, maximum moment, maximum slope, maximum bending stress, and maximum deflection for
the data given in row a from Table P4-2.
b
L
a
1R 2R
F
w
Given: Beam length L 1 m
Distance to distributed load a 0.4 m
Distance to reaction load b 0.6 m
Distributed load magnitude w 200 N m
1
Concentrated load F 500 N
Moment of inertia I 2.85 10
8
m
4
Distance to extreme fiber c 2.00 10
2
m FIGURE 4-25A
Free Body Diagram for Problem 4-25
Solution: See Figures 4-25 and Mathcad file P0425a.
1. The reactions, maximum shear and maximum moment were all found in Problem 3-25a. Those results are
summarized here.
Load function q(x) = R
1
<x - 0>-1 - w<x - a>0 + R2<x - b>
-1 - F<x - L>-1
Shear function V(x) = R
1
<x - 0>0 - w<x - a>1 + R2<x - b>
0 - F<x - L>0
Moment function M(x) = R
1
<x - 0>1 - w<x - a>2/2 + R2<x - b>
1 - F<x - L>1
Modulus of elasticity E 207 GPa
Reactions R1 353.3 N R2 973.3 N
Maximum shear Vmax 580 N (positive, at x = b)
Maximum moment Mmax 216 N m (negative, at x = b)
2. Integrate the moment function, multiplying by 1/EI, to get the slope.
(x) = [R
1
<x - 0>2/2 - w<x - a>3/6 + R
2
<x - b>2/2 - F<x - L>2/2 + C
3
]/EI
3. Integrate again to get the deflection.
y(x) = [R
1
<x - 0>3/6 - w<x - a>4/24 + R
2
<x-b>3/6 - F<x - L>3/6 + C
3
x +C
4
]/EI
4. Evaluate C3 and C4
At x = 0 and x = b, y = 0, therefore, C4 = 0.
0
R1
6
b
3
w
24
b a( )
4
C3 b=
C3
1
b
R1
6
b
3
w
24
b a( )
4
C3 21.22 N m
2
5. Define the range for x x 0 m 0.005 L L
6. For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than z,
and a value of one when it is greater than or equal to z.
S x z( ) if x z 1 0( )
7. Write the slope and deflection equations in Mathcad form, using the function S as a multiplying factor to get
the effect of the singularityfunctions. See Figure 4-25aB where these functions are plotted.
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-25a-2
θ x( )
1
E I
R1
2
S x 0 in( ) x
2
w
6
S x a( ) x a( )
3
R2
2
S x b( ) x b( )
2
F
2
S x L( ) x L( )
2
C3
y x( )
1
E I
R1
6
S x 0 in( ) x
3
w
24
S x a( ) x a( )
4
R2
6
S x b( ) x b( )
3
F
6
S x L( ) x L( )
3
C3 x
8. Maximum slope occurs at x = L θmax θ L( ) θmax 0.823 deg
9. Maximum deflection occurs at x = L. ymax y L( ) ymax 4.81 mm
10. The maximum bending stress occurs at x = b, where the moment is a maximum. For c 20 mm
σmax
Mmax c
I
σmax 152 MPa
0 0.2 0.4 0.6 0.8 1
0.015
0.01
0.005
0
0.005
SLOPE, radians
θ x( )
x
m
0 0.2 0.4 0.6 0.8 1
6
4
2
0
2
DEFLECTION, mm
y x( )
mm
x
m
FIGURE 4-25aB
Slope and Deflection Diagrams for Problem 4-25a
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-26a-1
PROBLEM 4-26a
Statement: A beam is supported and loaded as shown in Figure P4-11d. Find the reactions, maximum
shear, maximum moment, maximum slope, maximum bending stress, and maximum deflection for
the data given in row a from Table P4-2.
Given: Beam length L 1 m
b
L
a
1R
F
3R 2 R
w
Distance to distributed load a 0.4 m
Distance to R2 b 0.6 m
Distributed load magnitude w 200 N m
1
Concentrated load F 500 N
Moment of inertia I 2.85 10
8
m
4
Distance to extreme fiber c 2.00 10
2
m FIGURE 4-26A
Free Body Diagram for Problem 4-26
Modulus of elasticity E 207 GPa
Solution: See Figures 4-26 and Mathcad file P0426a.
1. From inspection of Figure P4-11d, write the load function equation
q(x) = R
1
<x>-1 - F<x - a>-1 - w<x - a>0 + R
2
<x - b>-1 - R3<x - L>
-1
2. Integrate this equation from - to x to obtain shear, V(x)
V(x) = R
1
<x>0 - F<x - a>0 - w<x - a>1 + R
2
<x - b>0 - R3<x - L>
0
3. Integrate this equation from - to x to obtain moment, M(x)
M(x) = R
1
<x>1 - F<x - a>1 - w<x - a>2/2 + R
2
<x - b>1 - R3<x - L>
1
4. Integrate the moment function, multiplying by 1/EI, to get the slope.
(x) = [R
1
<x>2/2 - F<x - a>2/2 - w<x - a>3/6 + R
2
<x - b>2/2 + R
3
<x - L>2/2 + C
3
]/EI
5. Integrate again to get the deflection.
y(x) = [R
1
<x>3/6 - F<x - a>3/6 - w<x - a>4/24 + R
2
<x - b>3/6 + R
3
<x - L>3/6 + C
3
x + C
4
]/EI
6. Evaluate R1, R2, R3, C3 and C4
At x = 0, x = b, and x = L; y = 0, therefore, C4 = 0.
At x = L+, V = M = 0
Guess R1 100 N R2 100 N R3 100 N C3 5 N m
2
Given
R1
6
b
3
F
6
b a( )
3
w
24
b a( )
4
C3 b 0 N m
3
=
R1
6
L
3
F
6
L a( )
3
w
24
L a( )
4
R2
6
L b( )
3
C3 L 0 N m
3
=
R1 F w L a( ) R2 R3 0 N=
R1 L F L a( )
w
2
L a( )
2
R2 L b( ) 0 N m=
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-26a-2
R1
R2
R3
C3
Find R1 R2 R3 C3
R1 112.33 N R2 559.17 N R3 51.50 N C3 5.607 N m
2
7. Define the range for x x 0 in 0.002 L L
8. For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than z,
and a value of one when it is greater than or equal to z.
S x z( ) if x z 1 0( )
9. Write the shear and moment equations in Mathcad form, using the function S as a multiplying factor to get
the effect of the singularity functions.
V x( ) R1 S x 0 in( ) F S x a( ) w S x a( ) x a( ) R2 S x b( ) R3 S x L( )
M x( ) R1 S x 0 in( ) x F S x a( ) x a( )
w
2
S x a( ) x a( )
2
R2 S x b( ) x b( )
10. Plot the shear and moment diagrams.
0 200 400 600 800 1 10
3
600
400
200
0
200
SHEAR, N
V x( )
N
x
mm
0 200 400 600 800 1 10
3
40
15
10
35
60
MOMENT, N-m
M x( )
N m
x
mm
FIGURE 4-26aB
Shear and Moment Diagrams for Problem 4-26a
11. From the diagram, we see that maximum shear occurs at x = b -,
Vmax V b 0.001 mm( ) Vmax 428 N
12. The maximum moment occurs at x = a,
Mmax M a( ) Mmax 44.9 N m
13. Write the slope and deflection equations in Mathcad form, using the function S as a multiplying factor to get t
effect of the singularity functions. See Figure 4-26aB where these functions are plotted.
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-26a-3
θ x( )
1
E I
R1
2
S x 0 in( ) x
2
F
2
S x a( ) x a( )
2
w
6
S x a( ) x a( )
3
R2
2
S x b( ) x b( )
2
R3
2
S x L( ) x L( )
2
C3
y x( )
1
E I
R1
6
S x 0 in( ) x
3
F
6
S x a( ) x a( )
3
w
24
S x a( ) x a( )
4
R2
6
S x b( ) x b( )
3
R3
6
S x L( ) x L( )
3
C3 x
14. Maximum slope occurs between x = a and x = b θmax 0.0576 deg
15. Maximum deflection occurs between x = 0 and x = a ymax 0.200 mm
16. The maximum bending stress occurs at x = a, where the moment is a maximum. For c 20 mm
σmax
Mmax c
I
σmax 31.5 MPa
0 0.2 0.4 0.6 0.8 1
0.1
0.05
0
0.05
0.1
SLOPE, deg.
θ x( )
deg
x
m
0 0.2 0.4 0.6 0.8 1
0.3
0.2
0.1
0
0.1
DEFLECTION, mm
y x( )
mm
x
m
FIGURE 4-26aC
Slope and Deflection Diagrams for Problem 4-26a
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-27-1
PROBLEM 4-27
Statement: A storage rack is to be designed to hold the paper roll of Problem 4-8 as shown in Figure P4-12.
Determine suitable values for dimensions a and b in the figure. Consider bending, shear, and
bearing stresses. Assume an allowable tensile/compressive stress of 100 MPa and an allowable
shear stress of 50 MPa for both stanchion and mandrel, which are steel. The mandrel is solid
and inserts halfway into the paper roll. Balance the design to use all of the material strength.
Calculate the deflection at the end of the roll.
Given: Paper roll dimensions OD 1.50 m Material properties Sy 100 MPa
ID 0.22 m Sys 50 MPa
Lroll 3.23 m E 207 GPa
Roll density ρ 984 kg m
3
Assumptions: 1. The paper roll's weight creates a concentrated load acting at the tip of the mandrel.
2. The mandrel's root in the stanchion experiences a distributed load over its length of
engagement
LM1
R1
y
m
x
WSolution: See Figures 4-27 and Mathcad file P0427.
1. In Problem 3-27, we were concerned only with the
portion of the mandrel outside of the stanchion.
Therefore, we modeled it as a cantilever beam with a
shear andmoment reaction at the stanchion.
Unfortunately, this tells us nothing about the stress or
force distributions in the portion of the mandrel that is
inside the stanchion. To do this we need to modify the
model by replacing the concentrated moment (and
possibly the concentrated shear force) with a force
system that will yield information about the stress
distribution in the mandrel on that portion that is inside
the stanchion. Figure 4-27A shows the FBD used in
Problem 3-27. Figure 4-27B is a simple model, but is not
representative of a built-in condition. It would be
appropriate if the hole in the stanchion did not fit tightly
around the mandrel. Figure 4-27C is an improvement
that will do for our analysis.
FIGURE 4-27A
Free Body Diagram for Problem 3-27
L
R2
1R
y
m
x
W
2. Determine the weight of the roll and the length of the
mandrel.
W
π
4
OD
2
ID
2
Lroll ρ g W 53.9 kN FIGURE 4-27B
Simplified Free Body Diagram, not used
Lm 0.5 Lroll Lm 1.615 m
Lm
R
x
a
b
w
y W3. From inspection of Figure 4-27C, write the
load function equation
q(x) = -w<x>0 + w<x - b>0 + R<x - b>-1 - W<x - b -L
m
>-1
4. Integrate this equation from - to x to obtain shear,
V(x)
V(x) = -w<x>1 + w<x - b>1 + R<x - b>0 - W<x - b -L
m
>0
5. Integrate this equation from - to x to obtain
moment, M(x)
FIGURE 4-27C
Free Body Diagram used in Problem 4-27
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-27-2
M(x) = -(w/2)<x>2 + (w/2)<x - b>2 + R<x - b>1 - W<x - b -L
m
>1
6. Solve for the reactions by evaluating the shear and moment equations at a point just to the right of x = b + Lm,
where both are zero.
At x = (b + L
m
)+ , V = M = 0
0 w b Lm w Lm R W= R W w b=
0
w
2
b Lm 2 w
2
Lm
2
R Lm=
w
2
b Lm 2 w
2
Lm
2
W w b( ) Lm= w
2 W Lm
b
2
=
Note that R is inversely proportional to b and w is inversly proportional to b2.
7. To see the value of x at which the shear and moment are maximum, let
b 400 mm then w
2 W Lm
b
2
and R W w b L b Lm
8. Define the range for x x 0 mm 0.002 L L
9. For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than z,
and a value of one when it is greater than or equal to z.
S x z( ) if x z 1 0( )
10. Write the shear and moment equations in Mathcad form, using the function S as a multiplying factor to get
the effect of the singularity functions.
V x( ) w S x 0 mm( ) x w S x b( ) x b( ) R S x b( ) W S x L( )
M x( )
w
2
S x 0 mm( ) x
2
w
2
S x b( ) x b( )
2
R S x b( ) x b( ) W S x L( ) x L( )
11. Plot the shear and moment diagrams.
0 400 800 1200 1600 2000
800
600
400
200
0
200
Shear Diagram
V x( )
kN
x
mm
0 400 800 1200 1600 2000
100
50
0
50
Moment Diagram
M x( )
kN m
x
mm
FIGURE 4-27D
Shear and Moment Diagram Shapes for Problem 4-27
12. From Figure 4-27D, the maximum internal shear and moment occur at x = b and are
Vmax
2 W Lm
b
= Mmax W Lm Mmax 87.04 kN m
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-27-3
13. The bending stress will be a maximum at the top or bottom of the mandrel at a section through x = b.
σmax
Mmax a
2 I
= where I
π a
4
64
= so, σmax
32 Mmax
π a
3
= Sy=
Solving for a, a
32 W Lm
π Sy
1
3
a 206.97 mm
Round this to a 210 mm
14. Using this value of a and equation 4.15c, solve for the shear stress on the neutral axis at x = b.
τmax
4 Vmax
3 A
=
8 W Lm
3
π a
2
4
b
= Sys=
Solving for b b
8 W Lm
3
π a
2
4
Sys
b 134.026 mm
Round this to b 134 mm
15. These are minimum values for a and b. Using them, check the bearing stress.
Magnitude of distributed load w
2 W Lm
b
2
w 9695
N
mm
Bearing stress σbear
w b
a b
σbear 46.2 MPa
Since this is less than Sy, the design is acceptable for a 210 mm and b 134 mm
16. Assume a cantilever beam loaded at the tip with load W and a mandrel diameter equal to a calculated above.
Moment of inertia I
π a
4
64
I 9.547 10
7
mm
4
Deflection at tip (Appendix B) ymax
W Lm
3
3 E I
ymax 3.83 mm
This can be accomodated by the 220-mm inside diameter of the paper roll.
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-28-1
PROBLEM 4-28
Statement: Figure P4-13 shows a forklift truck negotiating a 15 deg ramp to to drive onto a 4-ft-high loading
platform. The truck weighs 5 000 lb and has a 42-in wheelbase. Design two (one for each side)
1-ft-wide ramps of steel to have no more than 1-in deflection in the worst case of loading as the
truck travels up them. Minimize the weight of the ramps by using a sensible cross-sectional
geometry.
Given: Ramp angle θ 15 deg Ramp width w 12 in
Platform height h 4 ft Allowable deflection δmax 1.0 in
Truck weight W 5000 lbf Young's modulus E 30 10
6
psi
Truck wheelbase Lt 42 in
Assumptions: 1. The worst case is when the truck CG is located at the center of the beam's span.
2. Use a coordinate frame that has the x-axis along the long axis of the beam.
3. Ignore traction forces and the weight components along the x-axis of the beam.
4. There are two ramps, one for each side of the forklift.
Solution: See Figure 4-28 and Mathcad file P0428.
a
b
L
W
R1
Fa
y
xF
R 2
b
a
Wb
CGa
CGb
FIGURE 4-28A
Dimensions and Free Body Diagram for Problem 4-28
1. Determine the length of the beam between supports and the distances a and b for the worst-case loading.
Length of beam L
h
sin θ( )
L 15.455 ft
From Problem 3-28, a 5.061 ft b 8.561 ft
2. The load distribution of the wheels on a single ramp is given in Problem 3-28 as
Fa 575.0 lbf Fb 1839.9 lbf
3. From inspection of Figure 4-28A, write the load function equation
q(x) = R
1
<x - 0>-1 - Fa<x - a>
-1 - Fb<x - b>
-1 + R2<x - L>
-1
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-28-2
4. Integrate this equation from - to x to obtain shear, V(x)
V(x) = R
1
<x - 0>0 - Fa<x - a>
0 - Fb<x - b>
0 + R2<x - L>
0
5. Integrate this equation from - to x to obtain moment, M(x)
M(x) = R
1
<x - 0>1 - Fa<x - a>
1 - Fb<x - b>
1 + R2<x - L>
1
6. The reactions are given in Problem 3-28 as R1 1207.4 lbf R2 1207.4 lbf
7. Integrate the moment function, multiplying by 1/EI, to get the slope.
(x) = [R
1
<x>2/2 - Fa<x - a>
2/2 - Fb<x - b>
2/2 + R
2
<x - L>2/2 + C
3
]/EI
8. Integrate again to get the deflection.
y(x) = [R
1
<x>3/6 - Fa<x - a>
3/6 - Fb<x - b>
3/6 + R
2
<x-L>3/6 + C
3
x +C
4
]/EI
9. Evaluate C3 and C4
At x = 0 and x = L, y = 0, therefore, C4 = 0.
0 R1 L
3
Fa L a( )
3
Fb L b( )
3
6 C3 L=
C3
1
6 L
R1 L
3
Fa L a( )
3
Fb L b( )
3
C34.983 10
6
lbf in
2
8. Define the range for x x 0 m 0.005 L L
9. For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than z,
and a value of one when it is greater than or equal to z.
S x z( ) if x z 1 0( )
10. Write the slope and deflection equations in Mathcad form, using the function S as a multiplying factor to get
the effect of the singularity functions. Use an assumed value of I so that the value of x that corresponds to ymax
can be found. Let I 10 in
4
θ x( )
1
E I
R1
2
S x 0 m( ) x
2
Fa
2
S x a( ) x a( )
2
Fb
2
S x b( ) x b( )
2
R2
2
S x L( ) x L( )
2
C3
y x( )
1
E I
R1
6
S x 0 m( ) x
3
Fa
6
S x a( ) x a( )
3
Fb
6
S x b( ) x b( )
3
R2
6
S x L( ) x L( )
3
C3 x
11. Plot the shear and moment diagrams using the assumed value of I.
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-28-3
0 4 8 12 16
0.02
0.01
0
0.01
0.02
SLOPE, radians
θ x( )
x
ft
0 4 8 12 16
1.5
1
0.5
0
DEFLECTION, in
y x( )
in
x
ft
FIGURE 4-28B
Slope and Deflection Diagrams for Problem 4-28, Using an Assumed Value for I
12. Maximum deflection occurs at x = c, where = 0 and c < b.
θ0
1
E I
R1
2
c
2
Fa
2
c a( )
2
C3
= 0=
Solving for c,
A
R1
2
Fa
2
B a Fa C C3
a
2
Fa
2
A 316.200 lbf B 3.492 10
4
lbf in C 6.043 10
6
in
2
lbf
c
B B
2
4 A C
2 A
c 7.804 ft
13. The maximum deflection occurs at x = c and is ymax
1
E I
R1 c
3
6
Fa
6
c a( )
3
C3 c
= δmax=
Solving for I
I
1
E δmax
R1 c
3
6
Fa
6
c a( )
3
C3 c
I 10.159 in
4
This is the minimum allowable value of the moment of inertia.
14. Assume a channel section such as that shown in Figure 4-28C. To keep it simple, let the thickness of
the flanges and web be the same. Choose 3/8-in thick plate, which is readily available. Then,
t 0.375 in
15. The cross-sectional area of the ramp is A h( ) w t 2 t h t( )
16. The distance to the CG is cg h( )
1
A h( )
w t
2
2
t h
2
t
2
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-28-4
t
w
h
Flange
Web
17. The moments of inertia of the web and a flange are
Iweb h( )
w t
3
12
w t cg h( )
t
2
2
Ifl h( )
t h t( )
3
12
h t cg h( )
h t
2
2
18. Using the known moment of inertia, solve
for the unknown flange height, h. Guess h 1 in
Given
I Iweb h( ) 2 Ifl h( )=
h Find h( ) h 3.988 in
Round this up to h 4.00 in
FIGURE 4-28C
Channel Section for Problem 4-28
19. Summarizing, the ramp design dimensions are:
Length L 185.5 in Flange height h 4.00 in Shape channel
Width w 12.00 in Thickness t 0.375 in Material steel
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-29a-1
PROBLEM 4-29a
Statement: Find the spring rate of the beam in Problem 4-23 at the applied concentrated load for row a in
Table P4-2.
R 2
F
R 1
a
b
L
w
Given: Beam length L 1 m
Distance to distributed load a 0.4 m
Distance to concentrated load b 0.6 m
Distributed load magnitude w 200 N m
1
Concentrated load Fb 500 N
Moment of inertia I 2.85 10
8
m
4
FIGURE 4-29
Modulus of elasticity E 207 GPa
Free Body Diagram for Problem 4-23
Solution: See Figure 4-29 and Mathcad file P0429a.
1. The deflection equation was found in Problem 4-23. Those results are summarized here.
Load function q(x) = R
1
<x - 0>-1 - w<x - 0>0 + w<x - a>0 - F<x - b>-1 + R2<x - L>
-1
Shear function V(x) = R
1
<x - 0>0 - w<x - 0>1 + w<x - a>1 - F<x - b>0 + R2<x - L>
0
Moment function M(x) = R
1
<x - 0>1 - w<x - 0>2/2 + w<x - a>2/2 - F<x - b>1 + R2<x - L>
1
Slope function (x) = [R
1
<x>2/2 - w<x>3/6 + w<x - a>3/6 - F<x - b>2/2 + R
2
<x - L>2/2 + C
3
]/EI
Deflection function y(x) = [R
1
<x>3/6 - w<x>4/24 + w<x - a>4/24 - F<x - b>3/6 + R
2
<x-L>3/6 + C
3
x +C
4
]/EI
2. To determine the stiffness under the load F we will need to find the incremental beam deflection due to F
alone. The procedure will be to find the deflection at x = b when F = 0, and then find it when Fb 500 N . The
stiffness will then be the force divided by the incremental deflection.
3. For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than z,
and a value of one when it is greater than or equal to z.
S x z( ) if x z 1 0( )
4. Write the reactions (from Problem 3-23), integration constant, and deflection (from problem 4-23) equations in
Mathcad form, using the function S as a multiplying factor to get the effect of the singularity functions.
R1 F( )
w
2
L
F
L
L b( )
w
2 L
L a( )
2
R2 F( ) w a F R1 F( )
C3 F( )
1
L
R1 F( )
6
L
3
w
24
L
4
w
24
L a( )
4
F
6
L b( )
3
y x F( )
1
E I
R1 F( )
6
S x 0 in( ) x
3
w
24
S x 0 in( ) x
4
w
24
S x a( ) x a( )
4
R2 F( )
6
S x L( ) x L( )
3
F
6
S x b( ) x b( )
3
C3 F( ) x
5. The deflection at x = b for F 0 N is y0 y b F( ) y0 0.137 mm
6. The deflection at x = b for F Fb is yF y b F( ) yF 1.765 mm
7. The deflection due to F alone is ∆y yF y0 ∆y 1.627 mm
8. The stiffness of the beam under the load F at x = b is k
F
∆y
k 307
N
mm
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-30a-1
PROBLEM 4-30a
Statement: Find the spring rate of the beam in Problem 4-24 at the applied concentrated load for row a in
Table P4-2.
F
w
R 1
M1
L
a
Given: Beam length L 1 m
Distance to distributed load a 0.4 m
Distributed load magnitude w 200 N m
1
Concentrated load FL 500 N
Moment of inertia I 2.85 10
8
m
4
Modulus of elasticity E 207 GPa
Solution: See Figure 4-30 and Mathcad file P0430a.
FIGURE 4-30
1. The deflection equation was found in Problem 4-24.
Those results are summarized here.
Free Body Diagram for Problem 4-24
Load function q(x) = -M1<x - 0>
-2 + R
1
<x - 0>-1 - w<x - a>0 - F<x - L>-1
Shear function V(x) = -M1<x - 0>
-1 + R
1
<x - 0>0 - w<x - a>1 - F<x - L>0
Moment function M(x) = -M1<x - 0>
0 + R
1
<x - 0>1 - w<x - a>2/2 - F<x - L>1
Slope function (x) = [-M
1
<x-0>1 + R
1
<x - 0>2/2 - w<x - a>3/6 - F<x - L>2/2 + C
3
]/EI
Deflection function y(x) = [-M
1
<x-0>2/2 + R
1
<x - 0>3/6 - w<x - a>4/24 - F<x - L>3/6 + C
3
x +C
4
]/EI
2. To determine the stiffness under the load F we will need to find the incremental beam deflection due to
F alone. The procedure will be to find the deflection at x = L when F = 0, and then find it when FL 500 N
. The stiffnesswill then be the force divided by the incremental deflection.
3. For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than z,
and a value of one when it is greater than or equal to z.
S x z( ) if x z 1 0( )
4. Write the reaction (from Problem 3-24) and deflection (from problem 4-24) equations in Mathcad form, using
the function S as a multiplying factor to get the effect of the singularity functions.
R1 F( ) w L a( ) F
M1 F( )
w
2
L a( )
2
R1 F( ) L
y x F( )
1
E I
M1 F( )
2
S x 0 in( ) x
2
R1 F( )
6
S x 0 in( ) x
3
w
24
S x a( ) x a( )
4
F
6
S x L( ) x L( )
3
5. The deflection at x = L for F 0 N is y0 y L F( ) y0 3.912 mm
6. The deflection at x = L for F FL is yF y L F( ) yF 32.163 mm
7. The deflection due to F alone is ∆y yF y0 ∆y 28.251 mm
8. The stiffness of the beam under the load F at x = L is k
F
∆y
k 17.7
N
mm
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-31a-1
PROBLEM 4-31a
Statement: Find the spring rate of the beam in Problem 4-25 at the applied concentrated load for row a in
Table P4-2.
b
L
a
1R 2R
F
w
Given: Beam length L 1 m
Distance to distributed load a 0.4 m
Distance to concentrated load b 0.6 m
Distributed load magnitude w 200 N m
1
Concentrated load FL 500 N
Moment of inertia I 2.85 10
8
m
4
FIGURE 4-31
Modulus of elasticity E 207 GPa
Free Body Diagram for Problem 4-25
Solution: See Figure 4-31 and Mathcad file P0431a.
1. The deflection equation was found in Problem 4-25. Those results are summarized here.
Load function q(x) = R
1
<x - 0>-1 - w<x - a>0 + R2<x - b>
-1 - F<x - L>-1
Shear function V(x) = R
1
<x - 0>0 - w<x - a>1 + R2<x - b>
0 - F<x - L>0
Moment function M(x) = R
1
<x - 0>1 - w<x - a>2/2 + R2<x - b>
1 - F<x - L>1
Slope function (x) = [R
1
<x - 0>2/2 - w<x - a>3/6 + R
2
<x - b>2/2 - F<x - L>2/2 + C
3
]/EI
Deflection function y(x) = [R
1
<x - 0>3/6 - w<x - a>4/24 + R
2
<x-b>3/6 - F<x - L>3/6 + C
3
x +C
4
]/EI
2. To determine the stiffness under the load F we will need to find the incremental beam deflection due to
F alone. The procedure will be to find the deflection at x = L when F = 0, and then find it when
FL 500 N . The stiffness will then be the force divided by the incremental deflection.
3. For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than z,
and a value of one when it is greater than or equal to z.
S x z( ) if x z 1 0( )
4. Write the reactions (from Problem 3-25), integration constant, and deflection (from problem 4-25) equations in
Mathcad form, using the function S as a multiplying factor to get the effect of the singularity functions.
R1 F( )
1
b
w
2
L a( )
2
F L b( ) w L a( ) L b( )
R2 F( ) w L a( ) F R1 F( )
C3 F( )
1
b
R1 F( )
6
b
3
w
24
b a( )
4
y x F( )
1
E I
R1 F( )
6
S x 0 in( ) x
3
w
24
S x a( ) x a( )
4
R2 F( )
6
S x b( ) x b( )
3
F
6
S x L( ) x L( )
3
C3 F( ) x
5. The deflection at x = L for F 0 N is y0 y L F( ) y0 0.288 mm
6. The deflection at x = L for F FL is yF y L F( ) yF 4.808 mm
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-31a-2
7. The deflection due to F alone is ∆y yF y0 ∆y 4.52 mm
8. The stiffness of the beam under the load F at x = L is k
F
∆y
k 111
N
mm
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-32a-1
PROBLEM 4-32a
Statement: Find the spring rate of the beam in Problem 4-26 at the applied concentrated load for row a in
Table P4-2.
b
L
a
1R
F
3R 2 R
w
Given: Beam length L 1 m
Distance to distributed load a 0.4 m
Distance to concentrated load b 0.6 m
Distributed load magnitude w 200 N m
1
Concentrated load Fa 500 N
Moment of inertia I 2.85 10
8
m
4
FIGURE 4-32
Modulus of elasticity E 207 GPa
Free Body Diagram for Problem 4-26
Solution: See Figure 4-32 and Mathcad file P0432a.
1. The deflection equation was found in Problem 4-26. Those results are summarized here.
Load function q(x) = R
1
<x>-1 - F<x - a>-1 - w<x - a>0 + R
2
<x - b>-1 - R3<x - L>
-1
Shear function V(x) = R
1
<x>0 - F<x - a>0 - w<x - a>1 + R
2
<x - b>0 - R3<x - L>
0
Moment function M(x) = R
1
<x>1 - F<x - a>1 - w<x - a>2/2 + R
2
<x - b>1 - R3<x - L>
1
Slope function (x) = [R
1
<x>2/2 - F<x - a>2/2 - w<x - a>3/6 + R
2
<x - b>2/2 + R
3
<x - L>2/2 + C
3
]/EI
Deflection function y(x) = [R
1
<x>3/6 - F<x - a>3/6 - w<x - a>4/24 + R
2
<x - b>3/6 + R
3
<x - L>3/6 + C
3
x + C
4
]/EI
2. To determine the stiffness under the load F we will need to find the incremental beam deflection due to
F alone. The procedure will be to find the deflection at x = a when F = 0, and then find it when Fa 500 N
. The stiffness will then be the force divided by the incremental deflection.
3. For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than z,
and a value of one when it is greater than or equal to z.
S x z( ) if x z 1 0( )
4. Write the reactions, integration constant, and deflection (from problem 4-26) equations in Mathcad form,
using the function S as a multiplying factor to get the effect of the singularity functions.
Let f1 F( )
F
6
b a( )
3
w
24
b a( )
4
f2 F( )
F
6
L a( )
3
w
24
L a( )
4
f3 F( ) F L a( )
w
2
L a( )
2
then
R1 F( )
3
L b L b( )
L
b
f1 F( ) f2 F( )
L b( )
2
6
f3 F( )
R1 Fa 112.333 N
R2 F( )
1
L b( )
f3 F( ) L R1 F( ) R2 Fa 559.167 N
R3 F( ) F w L a( ) R1 F( ) R2 F( ) R3 Fa 51.500 N
C3 F( )
1
b
f1 F( )
b
2
6
R1 F( )
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-32a-2
y x F( )
1
E I
R1 F( )
6
S x 0 in( ) x
3
F
6
S x a( ) x a( )
3
w
24
S x a( ) x a( )
4
R2 F( )
6
S x b( ) x b( )
3
R3 F( )
6
S x L( ) x L( )
3
C3 F( ) x
5. The deflection at x = a for F 0 N is y0 y a F( ) y0 0.00126 mm
6. The deflection at x = a for F Fa is yF y a F( ) yF 0.177 mm
7. The deflection due to F alone is ∆y yF y0 ∆y 0.176 mm
8. The stiffness of the beam under the load F at x = a is k
F
∆y
k 2844
N
mm
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-33a-1
PROBLEM 4-33a
Statement: For the bracket shown in Figure P4-14 and the data in row a of Table P4-3, determine the
bending stress at point A and the shear stress due to transverseloading at point B. Also the
torsional shear stress at both points. Then determine the principal stresses at points A and B.
F
R
L
M
T
A
B
y
T x
Given: Tube length L 100 mm
Arm length a 400 mm
Arm thickness t 10 mm
Arm depth h 20 mm
Applied force F 50 N
Tube OD OD 20 mm
Tube ID ID 14 mm
Modulus of elasticity E 207 GPa FIGURE 4-33
Free Body Diagram of Tube for Problem 4-33
Solution: See Figure 4-33 and Mathcad file P0433a.
1. Determine the bending stress at point A. From the FBD of the tube in Figure 4-33 we see that
Reaction force R F R 50.0 N
Reaction moment M F L M 5.00 N m
Distance from NA
to outside of tube ct 0.5 OD ct 10.0 mm
Moment of inertia It
π
64
OD
4
ID
4
It 5968 mm4
Bending stress
at point A σxA
M ct
It
σxA 8.38 MPa
2. Determine the shear stress due to transverse loading at B.
Cross-section area A
π
4
OD
2
ID
2
A 160.2 mm2
Maximum shear V R
Maximum shear stress
(Equation 4.15d) τVmax 2
V
A
τVmax 0.624 MPa
3. Determine the torsional shear stress at both points. Using equation 4.23b and the FBD above
Torque on tube T F a T 20.0 N m
Polar moment of
inertia J
π
32
OD
4
ID
4
J 11936 mm4
Maximum torsional
stress at surface τTmax
T ct
J
τTmax 16.76 MPa
4. Determine the principal stress at point A.
Stress components σxA 8.378 MPa σzA 0 MPa
τxz τTmax τxz 16.76 MPa
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-33a-2
Principal stresses
σ1
σxA σzA
2
σxA σzA
2
2
τxz
2
σ1 21.46 MPa
σ2 0 MPa
σ3
σxA σzA
2
σxA σzA
2
2
τxz
2
σ3 13.08 MPa
τ13
σ1 σ3
2
τ13 17.27 MPa
5. Determine the principal stress at point B.
Stress components σxB 0 MPa σyB 0 MPa
τxy τTmax τVmax τxy 16.13 MPa
Principal stresses
σ1
σxB σyB
2
σxB σyB
2
2
τxy
2
σ1 16.13 MPa
σ2 0 MPa
σ3
σxB σyB
2
σxB σyB
2
2
τxy
2
σ3 16.13 MPa
τ13
σ1 σ3
2
τ13 16.13 MPa
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-34a-1
PROBLEM 4-34a
Statement: For the bracket shown in Figure P4-14 and the data in row a of Table P4-3, determine the
deflection at load F.
Given: Tube length L 100 mm Applied force F 50 N
Arm length a 400 mm Tube OD OD 20 mm
Arm thickness t 10 mm Tube ID ID 14 mm
Arm depth h 20 mm Modulus of elasticity E 207 GPa
Modulus of rigidity G 80.8 GPa
Solution: See Figure 4-34 and Mathcad file P0434a.
1. The deflection at load F can be determined by superimposing the rigid-body deflection of the arm due to the
twisting of the tube with the beam deflection of the tube and the arm alone.
2. Determine the rigid-body deflection due to twisting of the tube. Refering to Figure 4-34, the torque in the
tube is
Torque on tube T F a T 20.0 N m
Polar moment of inertia Jt
π
32
OD
4
ID
4
Jt 11936 mm4
Tube angle of twist θ
T L
Jt G
θ 2.07368 10
3
rad
θ 0.119 deg
Deflection at F due to δθ a θ δθ 0.829 mm
3. Determine the rigid-body deflection due to bending of the tube.
Moment of inertia It
Jt
2
It 5968 mm
4
Deflection of tube
end and arm end
(see Appendix B)
δtb
F L
3
3 E It
δtb 0.013 mm
F
R
L
M
F
F
T
a
hT
A
B
y
T x z
y
FIGURE 4-34
Free Body Diagrams of Tube and Arm for Problem 4-34
4. Determine the beam bending of arm alone.
Moment of inertia Ia
t h
3
12
Ia 6667 mm
4
Deflection at F δa
F a
3
3 E Ia
δa 0.773 mm
5. Determine the total deflection by superposition.
δtot δθ δtb δa δtot 1.616 mm downward
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-35a-1
PROBLEM 4-35a
Statement: For the bracket shown in Figure P4-14 and the data in row a of Table P4-3, determine the spring
rate of the tube in bending, the spring rate of the arm in bending, and the spring rate of the tube
in torsion. Combine these into an overall spring rate in terms of the force F and the linear
deflection at F.
Given: Tube length L 100 mm Applied force F 50 N
Arm length a 400 mm Tube OD OD 20 mm
Arm thickness t 10 mm Tube ID ID 14 mm
Arm depth h 20 mm Modulus of elasticity E 207 GPa
Modulus of rigidity G 80.8 GPa
Solution: See Figure 4-35 and Mathcad file P0435a.
1. Determine the spring rate due to bending of the tube.
Moment of inertia It
π
64
OD
4
ID
4
It 5968 mm4
Deflection of tube
end and arm end
(see Appendix B)
δtb
F L
3
3 E It
δtb 0.013 mm
Spring rate due to
bending in tube ktb
F
δtb
ktb 3706
N
mm
2. Determine the spring rate due to beam bending of arm alone.
Moment of inertia Ia
t h
3
12
Ia 6667 mm
4
Deflection at F δa
F a
3
3 E Ia
δa 0.773 mm
Spring rate due to
bending in arm ka
F
δa
ka 64.7
N
mm
F
R
L
M
F
F
T
a
hT
A
B
y
T x z
y
FIGURE 4-35
Free Body Diagrams of Tube and Arm for Problem 4-35
3. Determine the spring rate of the tube in torsion. Refering to Figure 4-35, the torque in the tube is
Torque on tube T F a T 20.0 N m
Polar moment of inertia Jt
π
32
OD
4
ID
4
Jt 11936 mm4
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-35a-2
Tube angle of twist θ
T L
Jt G
θ 2.07368 10
3
rad
θ 0.119 deg
Deflection at F due to q δθ a θ δθ 0.829 mm
Spring rate due to
torsion in tube kθ
F
δθ
kθ 60.28
N
mm
4. Determine the overall spring rate. The springs are in series, thus
1
koa
1
kθ
1
ktb
1
ka
=
koa
kθ ktb ka
ktb ka kθ ka kθ ktb
koa 30.9
N
mm
Checking, δtot
F
koa
δtot 1.616 mm
which is the same total deflection gotten in Problem 4-34.
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-36a-1
PROBLEM 4-36a
Statement: For the bracket shown in Figure P4-14 and the data in row a of Table P4-3, redo Problem 4-33
considering the stress concentration at points A and B. Assume a stress concentration factor
of 2.5 in both bending and torsion.
Given: Tube length L 100 mm
F
R
L
M
T
A
B
y
T x
Arm length a 400 mm
Arm thickness t 10 mm
Arm depth h 20 mm
Applied force F 50 N
Tube OD OD 20 mm
Tube ID ID 14 mm
Modulus of elasticity E 207 GPa
FIGURE 4-36
Stress-concentration
factors
Ktb 2.5 Free Body Diagram of Tube for Problem 4-36
Kts 2.5
Solution: See Figure 4-36 and Mathcad file P0436a.
1. Determine the bending stress at point A. From the FBD of the tube in Figure 4-36 we see thatReaction force R F R 50.0 N
Reaction moment M F L M 5.00 N m
Distance from NA
to outside of tube ct 0.5 OD ct 10.0 mm
Moment of inertia It
π
64
OD
4
ID
4
It 5968 mm4
Bending stress
at point A σxA Ktb
M ct
It
σxA 20.94 MPa
2. Determine the shear stress due to transverse loading at B.
Cross-section area A
π
4
OD
2
ID
2
A 160.2 mm2
Maximum shear V R
Maximum shear stress
(Equation 4.15d) τVmax 2
V
A
τVmax 0.624 MPa
3. Determine the torsional shear stress at both points. Using equation 4.23b and the FBD above
Torque on tube T F a T 20.0 N m
Polar moment of
inertia J
π
32
OD
4
ID
4
J 11936 mm4
Maximum torsional
stress at surface τTmax Kts
T ct
J
τTmax 41.89 MPa
4. Determine the principal stress at point A.
Stress components σxA 20.944 MPa σzA 0 MPa
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-36a-2
τxz τTmax τxz 41.89 MPa
Principal stresses
σ1
σxA σzA
2
σxA σzA
2
2
τxz
2
σ1 53.6 MPa
σ2 0 MPa
σ3
σxA σzA
2
σxA σzA
2
2
τxz
2
σ3 32.71 MPa
τ13
σ1 σ3
2
τ13 43.18 MPa
5. Determine the principal stress at point B.
Stress components σxB 0 MPa σyB 0 MPa
τxy τTmax τVmax τxy 41.26 MPa
Principal stresses
σ1
σxB σyB
2
σxB σyB
2
2
τxy
2
σ1 41.26 MPa
σ2 0 MPa
σ3
σxB σyB
2
σxB σyB
2
2
τxy
2
σ3 41.26 MPa
τ13
σ1 σ3
2
τ13 41.26 MPa
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-37-1
PROBLEM 4-37
Statement: A semicircular, curved beam as shown in Figure 4-37 has the dimensions given below. For the
load pair applied along the diameter and given below, find the eccentricity of the neutral axis
and the stress at the inner and outer fibers.
F
F
rc
(a) Entire Beam
(b) Critical Section
M
idod
F
F
w
Given: Outside diameter od 150 mm
Inside diameter id 100 mm
Width of beam w 25 mm
Load F 14 kN
Solution: See Figure 4-37 and Mathcad file P0437.
1. Calculate the section depth, area, inside radius and
outside radus.
Section depth h
od id
2
h 25 mm
Area of section A h w A 625 mm
2
Centroid radius rc
od id
4
rc 62.5 mm
Inside and outside
radii of section
ri rc 0.5 h ri 50 mm
ro rc 0.5 h ro 75 mm
2. The critical section is the one that is along the horizontal
centerline. There, the bending moment is
FIGURE 4-37
Free Body Diagrams for Problem 4-37
Bending moment M F rc M 0.875 kN m
3. Use the equation in the footnote of the text to calculate the radius of the neutral axis.
Radius of neutral axis rn
ro ri
ln
ro
ri
rn 61.658 mm
4. Calculate the eccentricty and the distances from the neutral axis to the extreme fibers.
Eccentricity e rc rn e 0.8424 mm
Distances from neutral
axis to extreme fibers
ci rn ri ci 11.66 mm
co ro rn co 13.34 mm
Stresses at inner and
outer radii
σi
M
e A
ci
ri
F
A
σi 409.9 MPa
σo
M
e A
co
ro
F
A
σo 273.2 MPa
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-38-1
PROBLEM 4-38
Statement: Design a solid, straight, steel torsion bar to have a spring rate of 10 000 in-lb per radian per foot
of length. Compare designs of solid round and solid square cross-sections. Which is more
efficient in terms of material use?
Given: Length of rod L 12 in Modulus of rigidity G 11.7 10
6
psi
Spring rate k 10000
in lbf
rad
Solution: See Mathcad file P0438.
1. Determine the rod diameter and volume for a round rod.
Spring rate k
J G
L
= J
π d
4
32
=
Rod diameter d
32 L k
π G
1
4
d 0.569 in
Volume of rod V
π d
2
4
L V 3.046 in
3
2. Determine the rod width and volume for a square rod using equation 4.26b and Table 4-6.
Spring rate k
K G
L
= K 2.25 a
4
=
Rod half-width a
L k
2.25 G
1
4
a 0.260 in 2 a 0.520 in
Volume of rod V 2 a( )
2
L V 3.241 in
3
3. Even though the square rod width is less than the round rod diameter, it takes slightly more material when a
square rod is used than when a round rod is used. Thus, the round rod is more efficient.
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-39-1
PROBLEM 4-39
Statement: Design a 1-ft-long steel, end-loaded cantilever spring for a spring rate of 10 000 lb/in. Compare
designs of solid round and solid square cross-sections. Which is more efficient in terms of
material use?
Given: Length of rod L 12 in Modulus of rigidity E 30 10
6
psi
Spring rate k 10000
lbf
in
Solution: See Figure B-1(a) in Appendix B and Mathcad file P0439.
1. Determine the rod diameter and volume for a round rod.
Spring rate k
3 E I
L
3
= I
π d
4
64
=
Rod diameter d
64 L
3
k
3 π E
1
4
d 1.406 in
Volume of rod V
π d
2
4
L V 18.64 in
3
2. Determine the rod width and volume for a square rod using equation 4.26b and Table 4-6.
Spring rate k
3 E I
L
3
= I
a
4
12
=
Rod width a
4 L
3
k
E
1
4
a 1.232 in
Volume of rod V a
2
L V 18.215 in
3
3. It takes slightly more material when a round rod is used than when a square rod is used. Thus, the square rod
is more efficient.
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-40-1
PROBLEM 4-40
Statement: Redesign the roll support of Problem 4-8 to be like that shown in Figure P4-16. The stub
mandrels insert to 10% of the roll length at each end. Choose appropriate dimensions a and
b to fully utilize the mandrel's strength, which is the same as in Problem 4-27. See Problem
4-8 for additional data.
Given: Paper roll dimensions OD 1.50 m Material properties Sy 100 MPa
ID 0.22 m Sys 50 MPa
Lroll 3.23 m E 207 GPa
Roll density ρ 984 kg m
3
Assumptions: 1. The paper roll's weight creates a concentrated load acting at the tip of the mandrel.
2. The mandrel's root in the stanchion experiences a distributed load over its length of
engagement
Solution: See Figures 4-40 and Mathcad file P0440.
1. Model the support in such a way that stresses in the
portion of the mandrel that is inside the stanchion can
be determined. There are several assumptions that can
be made about the loads on this portion of the mandrel.
Figure 4-40A shows the one that will be used for this
design.
x
a
b
R
Lm
w
y F
2. Determine the weight of the roll, the load on each
support, and the length of the mandrel.
W
π
4
OD
2
ID
2
Lroll ρ g W 53.9 kN FIGURE 4-40A
Free Body Diagram used in Problem 4-40
F 0.5 W F 26.95 kN
Lm 0.1 LrollLm 323 mm
3. From inspection of Figure 4-40A, write the load function equation
q(x) = -w<x>0 + w<x - b>0 + R<x - b>-1 - F<x - b -L
m
>-1
4. Integrate this equation from - to x to obtain shear, V(x)
V(x) = -w<x>1 + w<x - b>1 + R<x - b>0 - F<x - b -L
m
>0
5. Integrate this equation from - to x to obtain moment, M(x)
M(x) = -(w/2)<x>2 + (w/2)<x - b>2 + R<x - b>1 - F<x - b -L
m
>1
6. Solve for the reactions by evaluating the shear and moment equations at a point just to the right of x = b + Lm,
where both are zero.
At x = (b + L
m
)+ , V = M = 0
0 w b Lm w Lm R F= R F w b=
0
w
2
b Lm 2 w
2
Lm
2
R Lm=
w
2
b Lm 2 w
2
Lm
2
F w b( ) Lm= w
2 F Lm
b
2
=
Note that R is inversely proportional to b and w is inversly proportional to b2.
7. To see the value of x at which the shear and moment are maximum, let
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-40-2
b 200 mm then w
2 F Lm
b
2
and R F w b L b Lm
8. Define the range for x x 0 mm 0.002 L L
9. For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than z,
and a value of one when it is greater than or equal to z.
S x z( ) if x z 1 0( )
10. Write the shear and moment equations in Mathcad form, using the function S as a multiplying factor to get the
effect of the singularity functions.
V x( ) w S x 0 mm( ) x w S x b( ) x b( ) R S x b( ) F S x L( )
M x( )
w
2
S x 0 mm( ) x
2
w
2
S x b( ) x b( )
2
R S x b( ) x b( ) F S x L( ) x L( )
11. Plot the shear and moment diagrams.
Shear Diagram Moment Diagram
0 100 200 300 400 500 600
200
100
0
100
200
V x( )
kN
x
mm
0 100 200 300 400 500 600
10
7
4
1
2
M x( )
kN m
x
mm
FIGURE 4-40B
Shear and Moment Diagram Shapes for Problem 4-40
12. From Figure 4-40B, the maximum internal shear and moment occur at x = b and are
Vmax
2 F Lm
b
= Mmax F Lm Mmax 8.704 kN m
13. The bending stress will be a maximum at the top or bottom of the mandrel at a section through x = b.
σmax
Mmax a
2 I
= where I
π a
4
64
= so, σmax
32 Mmax
π a
3
= Sy=
Solving for a, a
32 W Lm
π Sy
1
3
a 121.037 mm
Round this to a 125 mm
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-40-3
14. Using this value of a and equation 4.15c, solve for the shear stress on the neutral axis at x = b.
τmax
4 Vmax
3 A
=
8 F Lm
3
π a
2
4
b
= Sys=
Solving for b b
8 F Lm
3
π a
2
4
Sys
b 37.828 mm
Round this to b 38 mm
15. These are minimum values for a and b. Using them, check the bearing stress.
Magnitude of distributed load w
2 F Lm
b
2
w 12055
N
mm
Bearing stress σbear
w b
a b
σbear 96.4 MPa
Since this is less than Sy, the design is acceptable for a 125 mm and
b 38 mm
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-41-1
PROBLEM 4-41
Statement: A 10-mm ID steel tube carries liquid at 7 MPa. Determine the principal stresses in the wall if its
thickness is: a) 1 mm, b) 5 mm.
Given: Tubing ID ID 10 mm Inside pressure pi 7 MPa
Assumption: The tubing is long therefore the axial stress is zero.
Solution: See Mathcad file P0441.
(a) Wall thickness is t 1 mm
1. Check wall thickness to radius ratio to see if this is a thick or thin wall problem.
ratio
t
0.5 ID
ratio 0.2
Since the ratio is greater than 0.1, this is a thick wall problem.
2. Using equations 4.48a and 4.48b, the stresses are maximum at the inside wall where
Inside radius ri 0.5 ID ri 5 mm
Outside radius ro ri t ro 6 mm
Tangential stress σt
ri
2
pi
ro
2
ri
2
1
ro
2
ri
2
σt 38.82 MPa
Radial stress σr
ri
2
pi
ro
2
ri
2
1
ro
2
ri
2
σr 7.00 MPa
3. Determine the principal stresses (since, for this choice of coordinates, the shear stress is zero),
σ1 σt σ1 38.82 MPa
σ2 0 MPa
σ3 σr σ3 7.00 MPa
The maximum shear stress is
τmax
σ1 σ3
2
τmax 22.91 MPa
(b) Wall thickness is t 5 mm
1. Check wall thickness to radius ratio to see if this is a thick or thin wall problem.
ratio
t
0.5 ID
ratio 1
Since the ratio is greater than 0.1, this is a thick wall problem.
2. Using equations 4.48a and 4.48b, the stresses are maximum at the inside wall where
Inside radius ri 0.5 ID ri 5 mm
Outside radius ro ri t ro 10 mm
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-41-2
Tangential stress σt
ri
2
pi
ro
2
ri
2
1
ro
2
ri
2
σt 11.67 MPa
Radial stress σr
ri
2
pi
ro
2
ri
2
1
ro
2
ri
2
σr 7.00 MPa
3. Determine the principal stresses (since, for this choice of coordinates, the shear stress is zero),
σ1 σt σ1 11.67 MPa
σ2 0 MPa
σ3 σr σ3 7.00 MPa
The maximum shear stress is
τmax
σ1 σ3
2
τmax 9.33 MPa
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-42-1
PROBLEM 4-42
Statement: A cylindrical tank with hemispherical ends is required to hold 150 psi of pressurized air at room
temperature. Find the principal stresses in the 1-mm-thick wall if the tank diameter is 0.5 m and
its length is 1 m.
Given: Tank ID ID 500 mm
Wall thickness t 1 mm
Inside pressure pi 150 psi pi 1034 kPa
Solution: See Mathcad file P0442.
1. Check wall thickness to radius ratio to see if this is a thick or thin wall problem.
ratio
t
0.5 ID
ratio 4 10
3
Since the ratio is less than 0.1, this is a thin wall problem.
2. Using equations 4.49a, 4.49b and 4.49c, the stresses are
Radius r 0.5 ID r 250 mm
Tangential stress σt
pi r
t
σt 258.55 MPa
Radial stress σr 0 MPa σr 0.00 MPa
Axial stress σa
pi r
2 t
σa 129.28 MPa
3. Determine the principal stresses (since, for this choice of coordinates, the shear stress is zero),
σ1 σt σ1 259 MPa σ1 37.5 ksi
σ2 σa σ2 129 MPa σ2 18.75 ksi
σ3 0 MPa σ3 0.00 MPa σ3 0.00 MPa
The maximum shear stress is
τmax
σ1 σ3
2
τmax 129 MPa
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-43-1
PROBLEM 4-43
Statement: Figure P4-17 shows an off-loading station at the end of a paper rolling machine. The finished
paper rolls are 0.9-m OD by 0.22-m ID by 3.23-m long and have a density of 984 kg/m3. The
rolls are transfered from the machine conveyor (not shown) to the forklift truck by the
V-linkage of the off-load station, which is rotated through 90 deg by an air cylinder. The paper
thenrolls onto the waiting forks of the truck. The forks are 38-mm thick by 100-mm wide by
1.2-m long and are tipped at a 3-deg angle from the horizontal. Find the stresses in the two
forks on the truck when the paper rolls onto it under two different conditions (state all
assumptions):
(a) The two forks are unsupported at their free end.
(b) The two forks are contacting the table at point A.
Given: Paper roll dimensions OD 0.90 m Fork dimensions t 38 mm
ID 0.22 m w 100 mm
Lroll 3.23 m Lfork 1200 mm
Roll density ρ 984 kg m
3
θfork 3 deg
Assumptions: 1. The greatest bending moment will occur when the paper roll is at the tip of the fork for case (a)
and when it is midway between supports for case (b).
2. Each fork carries 1/2 the weight of a paper roll.
3. For case (a), each fork acts as a cantilever beam (see Appendix B-1(a)).
4. For case (b), each fork acts as a beam that is built-in at one end and simply-supported at the
other.
Solution: See Figure 4-43 and Mathcad file P0443.
F
R1
M1
L fork
t
R
t
F
R1 2
0.5 Lfork
L fork
Case (a), Cantilever Beam
M2
Case (b), Fixed-Simply Supported Beam
1. Determine the weight of the roll and the load on each
fork.
W
π
4
OD
2
ID
2
Lroll ρ g W 18.64 kN
F 0.5 W F 9.32 kN
2. The moment of inertia and the distance to the extreme
fiber for a fork are
I
w t
3
12
I 4.573 10
5
mm
4
c
t
2
c 19 mm
Case (a)
3. From Figure D-1(a), the moment is a maximum at the
support and is FIGURE 4-43A
Free Body Diagrams used in Problem 4-43
Mmax F Lfork Mmax 11.186 kN m
4. The bending stress is maximum at the support and is σa
Mmax c
I
σa 464.8 MPa
Case (b)
5. This beam is statically indeterminate. However, using singularity functions and the method shown in Example
4-7, we can determine the reactions and find the maximum moment.
6. Calculate the distance from the left support to the load and the distance between supports.
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-43-2
a 0.5 Lfork a 600 mm
L Lfork L 1200 mm
7. From inspection of Figure 4-43A, write the load function equation
q(x) = R
1
<x>-1 - F<x - a>-1 + R
2
<x - L>-1 + M2<x - L>
-2
8. Integrate this equation from - to x to obtain shear, V(x)
V(x) = R
1
<x>0 - F<x - a>0 + R
2
<x - L>0 + M2<x - L>
-1
9. Integrate this equation from - to x to obtain moment, M(x)
M(x) = R
1
<x>1 - F<x - a>1 + R
2
<x - L>1 + M2<x - L>
0
10. Integrate the moment function, multiplying by 1/EI, to get the slope.
(x) = [R
1
<x>2/2 - F<x - a>2/2 + R
2
<x - L>2/2 + M
2
<x - L>1 + C
3
]/EI
11. Integrate again to get the deflection.
y(x) = [R
1
<x>3/6 - F<x - a>3/6 + R
2
<x - L>3/6 + M
2
<x - L>2/2 + C
3
x + C
4
]/EI
12. Evaluate R1, R2, M2, C3 and C4
At x = 0 and x = L; y = 0, therefore, C
4
= 0. At x = L, = 0
At x = L+, V = M = 0
Guess R1 1 kN R2 1 kN M2 1 kN m C3 1 kN m
2
Given
R1 L
3
6
F L a( )
3
6
C3 L 0 kN m
3
=
R1 L
2
2
F L a( )
2
2
C3 0 kN m
2
=
R1 R2 F 0 kN=
R1 L F L a( ) M2 0 kN m=
R1
R2
M2
C3
Find R1 R2 M2 C3
R1 2.913 kN R2 6.409 kN M2 2.097 kN m C3 0.419 kN m
2
13. Define the range for x x 0 in 0.002 L L
14. For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than
z, and a value of one when it is greater than or equal to z.
S x z( ) if x z 1 0( )
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-43-3
15. Write the shear and moment equations in Mathcad form, using the function S as a multiplying factor to get the
effect of the singularity functions.
V x( ) R1 S x 0 in( ) F S x a( ) R2 S x L( )
M x( ) R1 S x 0 in( ) x F S x a( ) x a( ) R2 S x L( ) x L( )
16. Plot the shear and moment diagrams.
Shear Diagram Moment Diagram
0 200 400 600 800 1000 1200
10
5
0
5
10
V x( )
kN
x
mm
0 200 400 600 800 1000 1200
3
2
1
0
1
2
M x( )
kN m
x
mm
FIGURE 4-43B
Shear and Moment Diagrams for Problem 4-43
17. The maximum moment occurs at x = L, Mmax M L( ) Mmax 2.097 kN m
18. The bending stress is maximum at the support and is σa
Mmax c
I
σa 87.2 MPa
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-44-1
PROBLEM 4-44
Statement: Determine a suitable thickness for the V-links of the off-loading station of Figure P4-17 to limit
their deflections at the tips to 10-mm in any position during their rotation. Two V-links support
the roll, at the 1/4 and 3/4 points along the roll's length, and each of the V arms is 10-cm wide by
1-m long. The V arms are welded to a steel tube that is rotated by the air cylinder. See Problem
4-43 for more information.
Given: Roll OD OD 0.90 m Arm width wa 100 mm
Roll ID ID 0.22 m Arm length La 1000 mm
Roll length Lroll 3.23 m Max tip deflection δtip 10 mm
Roll density ρ 984 kg m
3
Mod of elasticity E 207 GPa
Assumptions: 1. The maximum deflection on an arm will occur just after it begins the transfer and just before it
completes it, i.e., when the angle is either zero or 90 deg., but after the tip is no longer
supported by the base unit.
2. At that time the roll is in contact with both arms ("seated" in the V) and will remain in that
state throughout the motion. When the roll is in any other position on an arm the tip will be
supported.
3. The arm can be treated as a cantilever beam with nonend load.
4. A single arm will never carry more than half the weight of a roll.
5. The pipe to which the arms are attached has OD = 160 mm.
Solution: See Figure 4-44 and Mathcad file P0444.
370 = a
1000 = L
F
F
M
4501. Determine the weight of the roll and the load on each
V-arm.
W
π
4
OD
2
ID
2
Lroll ρ g W 18.64 kN
F 0.5 W F 9.32 kN
2. From Appendix B, Figure B-1, the tip deflection of a
cantilever beam with a concentrated load located at a
distance a from the support is
ymax
F a
2
6 E I
a 3 L( )=
where L is the beam length and I is the
cross-section moment of inertia. In this case
I
wa ta
3
12
=
FIGURE 4-44
3. Setting ymax δtip= and a 370 mm Free Body Diagram used in Problem 4-44
substituting for I and solving for ta
ta
2 F a
2
3 La a
E δtip wa
1
3
ta 31.889 mm
Let the arm thickness be ta 32 mm
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-45-1
PROBLEM 4-45
Statement: Determine the critical load on the air cylinder rod in Figure P4-17 if the crank arm that rotates it is
0.3 m long and the rod has a maximum extension of 0.5 m. The 25-mm-dia rod is solid steel with a
yield strength of 400 MPa. State all assumptions.
Given: Rod length L 500 mm Young's modulus E 207 GPa
Rod diameter d 25 mm Yield strength Sy 400 MPa
Assumptions: 1. The rod is a fixed-pinnedcolumn.
2. Use a conservative value of 1 for the end factor (see Table 4-7 in text).
Solution: See Mathcad file P0445.
1. Calculate the slenderness ratio that divides the unit load vs slenderness ratio graph into Johnson and Euler
regions.
SrD π
2 E
Sy
SrD 101.07
2. Calculate the cross-section area and the moment of inertia.
Area A
π
4
d
2
A 490.87 mm
2
Moment of inertia I
π
64
d
4
I 1.92 10
4
mm
4
3. Using Table 4-7, calculate the effective column length.
Leff 1 L Leff 500 mm
4. Calculate the slenderness ratio for the column.
Radius of gyration k
I
A
k 6.25 mm
Slenderness ratio Sr
Leff
k
Sr 80.00
Since the Sr for this column is less than SrD, it is a Johnson column.
5. Calculate the critical load using the Johnson equation.
Pcr A Sy
1
E
Sy Sr
2 π
2
Pcr 134.8 kN
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-46-1
PROBLEM 4-46
Statement: The V-links of Figure P4-17 are rotated by the crank arm through a shaft that is 60 mm dia by
3.23 m long. Determine the maximum torque applied to this shaft during motion of the
V-linkage and find the maximum stress and deflection for the shaft. See Problem 4-43 for more
information.
Given: Paper roll dimensions OD 900 mm Shaft dims d 60 mm
ID 220 mm Lshaft 3230 mm
Lroll 3230 mm
Roll density ρ 984 kg m
3
Modulus of rigidity G 79 GPa
Assumptions: The greatest torque will occur when the link is horizontal and the paper roll is located as shown
in Figure P4-17 or Figure 4-46.
Solution: See Figure 4-46 and Mathcad file P0446.
y
450.0
60-mm-dia shaft
yR
W
T
1. Determine the weight of the roll on the V-arms.
W
π
4
OD
2
ID
2
Lroll ρ g
W 18.64 kN
2. Summing moments about the shaft center,
T
OD
2
W T 8.390 kN m
3. Calculate the polar moment of inertia.
J
π d
4
32
J 1.272 10
6
mm
4
4. The maximum torsional stress will be at the outside
diameter of the shaft. The radius of the OD is,
r
d
2
r 30 mm FIGURE 4-46
Free Body Diagram used in Problem 4-46
5. Determine the maximum torsional stress using equation (4.23b).
τmax
T r
J
τmax 197.8 MPa
6. Use equation (4.24) to determine the angular shaft deflection.
θ
T Lshaft
J G
θ 15.447 deg
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-47-1
PROBLEM 4-47
Statement: Determine the maximum forces on the pins at each end of the air cylinder of Figure P4-17.
Determine the stress in these pins if they are 30-mm dia and in single shear.
Given: Paper roll dimensions OD 0.90 m Pin diameter d 30 mm
ID 0.22 m
Lroll 3.23 m
Roll density ρ 984 kg m
3
Assumptions: 1. The maximum force in the cylinder rod occurs when the transfer starts.
2. The cylinder and rod make an angle of 8 deg to the horizontal at the start of transfer.
3. The crank arm is 300 mm long and is 45 deg from vertical at the start of transfer.
4. The cylinder rod is fully retracted at the start of the transfer. At the end of the transfer it will
have extended 500 mm from its initial position.
Solution: See Figure 4-47 and Mathcad file P0447.
212.1
212.1
450.0
8°
y
yR
W
Rx x
A
F
1. Determine the weight of the roll on the
forks.
W
π
4
OD
2
ID
2
Lroll ρ g
W 18.64 kN
2. From the assumptions and
Figure 4-47, the x and y
distances from the origin to
point A are,
Rax 300 cos 45 deg( ) mm
Ray 300 sin 45 deg( ) mm
Rax 212.132 mm
Ray 212.132 mm
3. From Figure 4-47, the x
distance from the origin to
point where W is applied is, FIGURE 4-47
Free Body Diagram at Start of Transfer for V-link of Problem 4-47
Rwx
OD
2
Rwx 450 mm
4. Sum moments about the pivot point and solve for the compressive force in the cylinder rod.
W Rwx F Rax sin 8 deg( ) F Ray cos 8 deg( ) 0=
F
W Rwx
Ray cos 8 deg( ) Rax sin 8 deg( )
F 46.469 kN
This is the shear force in the pins
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-47-2
5. Determine the cross-sectional area of the pins and the direct shear stress.
Shear area A
π d
2
4
A 706.858 mm
2
Shear stress τ
F
A
τ 65.7 MPa
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-48-1
PROBLEM 4-48
Statement: A 100-kg wheelchair marathon racer wants an exerciser that will allow indoor practicing in any
weather. The design shown in Figure P4-18 is proposed. Two free-turning rollers on bearings
support the rear wheels. A platform supports the front wheels. Design the 1-m-long rollers as
hollow tubes of aluminum to minimize the height of the platform and also limit the roller
deflections to 1 mm in the worst case. The wheelchair has 650-mm-dia drive wheels separated
by a 700-mm track width. The flanges shown on the rollers limit the lateral movement of the
chair while exercising and thus the wheels can be anywhere between those flanges. Specify
suitable sized steel axles to support the tubes on bearings. Calculate all significant stresses.
Given: Mass of chair M 100 kg Maximum deflection δ 1 mm
Wheel diameter dw 650 mm Modulus elasticity
Track width T 700 mm Aluminum Ea 71.7 GPa
Roller length Lr 1000 mm Steel Es 207 GPa
Assumptions: 1. The CG of the chair with rider is
sufficiently close to the rear wheel that all of
the weight is taken by the two rear wheels.
2. The small camber angle of the rear wheels
does not significantly affect the magnitude
of the forces on the rollers.
3. Both the aluminum roller and the steel axle
are simply supported. The steel axles that
support the aluminum tube are fixed in the
mounting block and do not rotate. The
aluminum tube is attached to them by two
bearings (one on each end of the tubes, one
for each axle). The bearings' inner race is
fixed, and the outer race rotates with the
aluminum tube. Each steel axle is considered
to be loaded as a simply supported beam.
Their diameter must be less than the inner
diameter of the tubes to fit the roller bearings
between them.
F F
W/2
FIGURE 4-48A
Solution: See Figures 4-48 and Mathcad file P0448.
Free Body Diagram of One Wheel
used in Problem 4-48
1. Calculate the weight of the chair with rider.
Weight of chair W M g W 980.7 N
2. Calculate the forces exerted by the wheels on the rollers (see Figure 4-48A). From the FBD of a wheel,
summing vertical forces
2 F cos θ( )
W
2
0=
Let θ 20 deg then F
W
4 cos θ( )
F 260.9 N
3. The worst condition (highest moment at site of a stress concentration) will occur when the chair is all the
way to the left or right. Looking at the plane through the roller that includes the forces exerted by the wheels
(the FBD is shown in Figure 4-48B) the reactions R1 and R2 come from the bearings, which are inside the
hollow roller and are, themselves, supported by the steel axle.4. Solving for the reactions. Let the distance from R
1
to F be a 15 mm
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-48-2
1000
700
15
R1
F F
2R
M
1 R2 Lr F a T( ) F a 0=
F
y R1 2 F R2 0=
R2
F 2 a T( )
Lr
R2 190.5 N
R1 2 F R2 R1 331.3 N
FIGURE 4-48B
Free Body Diagram of One Tube used in Problem 4-48
5. The maximum bending moment will be at the right-hand load and will be
Mrmax R2 Lr a T( ) Mrmax 54.3 N m
Note, if the chair were centered on the roller the maximum moment would be
Mc F
Lr T
2
Mc 39.1 N m
and this would be constant along the axle between the two loads, F.
6. Note that the bearing positions are fixed regardless of the position of the chair on the roller.
Because of symmetry,
1000
1130
65
R a1
R1
R a2
R 2
Ra1 R1 Ra1 331.3 N
Ra2 R2 Ra2 190.5 N
7. The maximum bending moment
occurs
at R
1
and is for b 65 mm
Mamax Ra1 b FIGURE 4-48C
Free Body Diagram of One Axle used in Problem 4-48
Mamax 21.5 N m
8. Determine a suitable axle diameter. Let the factor of safety against yielding in the axle be
Nsa 3
9. Tentatively choose a low-carbon steel for the axle, say AISI 1020, cold rolled with
Sy 393 MPa
10. At the top of the axle under the load R1 there is only a bending stress. Set this stress equal to the yield
strength divided by the factor of safety.
σx
32 Mamax
π da
3
=
Sy
Nsa
=
Solving for the axle diameter, da da
32 Nsa Mamax
π Sy
1
3
da 11.875 mm
Let the axle diameter be da 15 mm made from cold-rolled AISI 1020 steel.
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-48-3
11. Suppose that bearing 6302 from Chapter
10, Figure 10-23. It has a bore of 15 mm and an
OD of 42 mm. Thus, the inside diameter of the
roller away from the bearings where the
moment is a maximum will be di 40 mm .
This will provide a 1-mm shoulder for axial
location of the bearings.
1000
15
700150
F F
F
F
12. The maximum deflection of the roller will
occur when the chair is in the center of the
roller. For this case the reactions are both equal
to the loads, F (see Figure 4-48D). The
maximum deflection is at the center of the roller.
FIGURE 4-48D
Free Body Diagram of Roller with Chair in the Center.
13. Write the load function and then integrate four times to get the deflection function.
q(x) = F<x>-1 - F<x - a>-1 - F<x - b>-1 + F<x - L>-1
y(x) = F[<x>3 - <x - a>3 - <x - b>3 + <x - L>3 + C3x]/(6EI)
where C3
1
L
L a( )
3
a
3
L
3
=
14. Write the deflection function at x = L/2 for
a 150 mm
ymax
F
6 Ea I
L
2
3
L
2
a
3
1
2
L a( )
3
a
3
L
3
=
15. Set this equation equal to the allowed deflection and solve for the required moment of inertia, I.
I
F
6 Ea δ
Lr
2
3
Lr
2
a
3
1
2
Lr a 3 a3 Lr3
I 6.618 10
4
mm
4
16. Knowing the inside diameter of the tube, solve for the outside diameter.
I
π
64
do
4
di
4
= do
64 I
π
di
4
1
4
do 44.463 mm
Round this up to do 46 mm
DESIGN SUMMARY
Axles Rollers
Material AISI 1020 steel, cold-rolled Material 2024-T4 aluminum
Diameter da 15 mm Outside diameter do 46 mm
Length 1220 mm Inside diameter di 40 mm
Length 1040 mm
Spacing c dw do sin θ( )
c 238 mm
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-49a-1
PROBLEM 4-49a
Statement: A hollow, square column has the dimensions and properties below. Determine if it is a Johnson
or an Euler column and find the critical load:
(a) If its boundary conditions are pinned-pinned.
(b) If its boundary conditions are fixed-pinned.
(c) If its boundary conditions are fixed-fixed.
(d) If its boundary conditions are fixed-free.
Given: Length of column L 100 mm Material Steel
Outside dimension so 4 mm Yield strength Sy 300 MPa
Inside dimension si 3 mm Modulus of elasticity E 207 GPa
Solution: See Mathcad file P0449a.
1. Calculate the slenderness ratio that divides the unit load vs slenderness ratio graph into Johnson and Euler
regions.
SrD π
2 E
Sy
SrD 116.7
2. Calculate the cross-section area and the moment of inertia.
Area A so
2
si
2
A 7.00 mm
2
Moment of inertia I
1
12
so
4
si
4
I 14.58 mm4
(a) pinned-pinned ends
3. Using Table 4-7, calculate the effective column length.
Leff 1 L Leff 100 mm
4. Calculate the slenderness ratio for the column.
Radius of gyration k
I
A
k 1.443 mm
Slenderness ratio Sr
Leff
k
Sr 69.28
Since the Sr for this column is less than SrD, it is a Johnson column.
5. Calculate the critical load using the Johnson equation.
Pcr A Sy
1
E
Sy Sr
2 π
2
Pcr 1.73 kN
(b) fixed-pinned ends
6. Using Table 4-7, calculate the effective column length.
Leff 0.8 L Leff 80 mm
7. Calculate the slenderness ratio for the column.
Radius of gyration k
I
A
k 1.443 mm
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-49a-2
Slenderness ratio Sr
Leff
k
Sr 55.43
Since the Sr for this column is less than SrD, it is a Johnson column.
8. Calculate the critical load using the Johnson equation.
Pcr A Sy
1
E
Sy Sr
2 π
2
Pcr 1.86 kN
(c) fixed-fixed ends
9. Using Table 4-7, calculate the effective column length.
Leff 0.65 L Leff 65 mm
10. Calculate the slenderness ratio for the column.
Radius of gyration k
I
A
k 1.443 mm
Slenderness ratio Sr
Leff
k
Sr 45.03
Since the Sr for this column is less than SrD, it is a Johnson column.
11. Calculate the critical load using the Johnson equation.
Pcr A Sy
1
E
Sy Sr
2 π
2
Pcr 1.94 kN
(d) fixed-free ends
12. Using Table 4-7, calculate the effective column length.
Leff 2.1 L Leff 210 mm
13. Calculate the slenderness ratio for the column.
Radius of gyration k
I
A
k 1.443 mm
Slenderness ratio Sr
Leff
k
Sr 145.49
Since the Sr for this column is greater than SrD, it is an Euler column.
14. Calculate the critical load using the Euler equation.
Pcr A
π
2
E
Sr
2
Pcr 676 N
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-50a-1
PROBLEM 4-50a
Statement: A hollow, round column has the dimensions and properties below. Determine if it is a Johnson
or an Euler column and find the critical load:
(a) If its boundary conditions are pinned-pinned.
(b) If its boundary conditions are fixed-pinned.
(c) If its boundary conditions are fixed-fixed.
(d) If its boundary conditions are fixed-free.
Given: Length of column L 1500 mm Material Steel
Outside diameter od 20 mm Yield strength Sy 300 MPa
Inside diameter id 14 mmModulus of elasticity E 207 GPa
Solution: See Mathcad file P0450a.
1. Calculate the slenderness ratio that divides the unit load vs slenderness ratio graph into Johnson and Euler
regions.
SrD π
2 E
Sy
SrD 116.7
2. Calculate the cross-section area, moment of inertia, and the radius of gyration.
Area A
π
4
od
2
id
2
A 160.22 mm2
Moment of inertia I
π
64
od
4
id
4
I 5968 mm4
Radius of gyration k
I
A
k 6.103 mm
3. Define functions to determine column type and critical load.
Type type Sr "Euler" Sr SrDif
"Johnson" otherwise
Critical load Pcr Sr A π
2
E
Sr
2
return type Sr "Euler"=if
A Sy
1
E
Sy Sr
2 π
2
otherwise
(a) pinned-pinned ends
4. Using Table 4-7, calculate the effective column length.
Leff 1 L Leff 1500 mm
5. Calculate the slenderness ratio for the column.
Slenderness ratio Sr
Leff
k
Sr 245.77
6. Determine the type and critical load using the functions defined above.
type Sr "Euler" Pcr Sr 5.42 kN
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-50a-2
(b) fixed-pinned ends
7. Using Table 4-7, calculate the effective column length.
Leff 0.8 L Leff 1200 mm
8. Calculate the slenderness ratio for the column.
Slenderness ratio Sr
Leff
k
Sr 196.62
9. Determine the type and critical load using the functions defined above.
type Sr "Euler" Pcr Sr 8.47 kN
(c) fixed-fixed ends
10. Using Table 4-7, calculate the effective column length.
Leff 0.65 L Leff 975 mm
11. Calculate the slenderness ratio for the column.
Slenderness ratio Sr
Leff
k
Sr 159.75
12. Determine the type and critical load using the functions defined above.
type Sr "Euler" Pcr Sr 12.8 kN
(d) fixed-free ends
13. Using Table 4-7, calculate the effective column length.
Leff 2.1 L Leff 3150 mm
14. Calculate the slenderness ratio for the column.
Slenderness ratio Sr
Leff
k
Sr 516.12
15. Determine the type and critical load using the functions defined above.
type Sr "Euler" Pcr Sr 1.23 kN
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-51a-1
PROBLEM 4-51a
Statement: A solid, rectangular column has the dimensions and properties below. Determine if it is a
Johnson or an Euler column and find the critical load:
(a) If its boundary conditions are pinned-pinned.
(b) If its boundary conditions are fixed-pinned.
(c) If its boundary conditions are fixed-fixed.
(d) If its boundary conditions are fixed-free.
Given: Length of col. L 100 mm Material Steel
Thickness t 10 mm Yield strength Sy 300 MPa
Height h 20 mm Modulus of elasticity E 207 GPa
Solution: See Mathcad file P0451a.
1. Calculate the slenderness ratio that divides the unit load vs slenderness ratio graph into Johnson and Euler
regions.
SrD π
2 E
Sy
SrD 116.7
2. Calculate the cross-section area, moment of inertia, and the radius of gyration.
Area A h t A 200.00 mm
2
Moment of inertia I
h t
3
12
I 1667 mm
4
Radius of gyration k
I
A
k 2.887 mm
3. Define functions to determine column type and critical load.
Type type Sr "Euler" Sr SrDif
"Johnson" otherwise
Critical load Pcr Sr A π
2
E
Sr
2
return type Sr "Euler"=if
A Sy
1
E
Sy Sr
2 π
2
otherwise
(a) pinned-pinned ends
4. Using Table 4-7, calculate the effective column length.
Leff 1 L Leff 100 mm
5. Calculate the slenderness ratio for the column.
Slenderness ratio Sr
Leff
k
Sr 34.64
6. Determine the type and critical load using the functions defined above.
type Sr "Johnson" Pcr Sr 57.36 kN
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-51a-2
(b) fixed-pinned ends
7. Using Table 4-7, calculate the effective column length.
Leff 0.8 L Leff 80 mm
8. Calculate the slenderness ratio for the column.
Slenderness ratio Sr
Leff
k
Sr 27.71
9. Determine the type and critical load using the functions defined above.
type Sr "Johnson" Pcr Sr 58.31 kN
(c) fixed-fixed ends
10. Using Table 4-7, calculate the effective column length.
Leff 0.65 L Leff 65 mm
11. Calculate the slenderness ratio for the column.
Slenderness ratio Sr
Leff
k
Sr 22.52
12. Determine the type and critical load using the functions defined above.
type Sr "Johnson" Pcr Sr 58.9 kN
(d) fixed-free ends
13. Using Table 4-7, calculate the effective column length.
Leff 2.1 L Leff 210 mm
13. Calculate the slenderness ratio for the column.
Slenderness ratio Sr
Leff
k
Sr 72.75
14. Determine the type and critical load using the functions defined above.
type Sr "Johnson" Pcr Sr 48.34 kN
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-52a-1
PROBLEM 4-52a
Statement: A solid, circular column, loaded eccentrically, has the dimensions and properties below. Find
the critical load:
(a) If its boundary conditions are pinned-pinned.
(b) If its boundary conditions are fixed-pinned.
(c) If its boundary conditions are fixed-fixed.
(d) If its boundary conditions are fixed-free.
Given: Length of column L 100 mm Material Steel
Outside diameter od 20 mm Yield strength Sy 300 MPa
Eccentricity (t) e 10 mm Modulus of elasticity E 207 GPa
Solution: See Mathcad file P0452a.
1. Calculate the cross-section area, distance to extreme fiber, and the moment of inertia.
Area A
π
4
od
2
A 314.16 mm
2
Distance to extreme fiber c 0.5 od c 10 mm
Moment of inertia I
π
64
od
4
I 7854 mm
4
4. Calculate the radius of gyration and eccentricity ratio for the column.
Radius of gyration k
I
A
k 5.00 mm
Eccentricity ratio Er
e c
k
2
Er 4.0
(a) pinned-pinned ends
3. Using Table 4-7, calculate the effective column length.
Leff 1 L Leff 100 mm
4. Calculate the slenderness ratio for the column.
Slenderness ratio Sr
Leff
k
Sr 20.00
5. Calculate the critical load using the Secant equation.
Guess P 1 kN
Given
P
Sy A
1 Er sec Sr
P
4 E A
=
Pcr Find P( ) Pcr 18.63 kN
(b) fixed-pinned ends
6. Using Table 4-7, calculate the effective column length.
Leff 0.8 L Leff 80 mm
7. Calculate the slenderness ratio for the column.
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MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-52a-2
Slenderness ratio Sr
Leff
k
Sr 16.00
8. Calculate the critical load using the Secant equation.
Guess P 1 kN
Given
P
Sy A
1 Er sec Sr
P
4 E A
=
Pcr Find P( ) Pcr 18.71 kN
(c) fixed-fixed ends
9. Using Table 4-7, calculate the effective column length.
Leff 0.65 L Leff 65 mm
10. Calculate the slenderness ratio for the column.
Slenderness ratio Sr
Leff
k
Sr 13.00
11. Calculate the critical load using the Secant equation.
Guess P 1 kN
Given
P
Sy A
1 Er sec Sr
P
4 E A
=
Pcr Find P( ) Pcr 18.76 kN
(d) fixed-free ends
12. Using Table 4-7, calculate the effective column length.
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