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Batch: Jan - May 2008 R. Ganesh Narayanan 1
Engineering Mechanics – Statics
Instructor
R. Ganesh Narayanan
Department of Mechanical Engineering
IIT Guwahati
Batch: Jan - May 2008 R. Ganesh Narayanan 2
-These lecture slides were prepared and used by me to conduct lectures for 1st year B. Tech. 
students as part of ME 101 – Engineering Mechanics course at IITG.
- Theories, Figures, Problems, Concepts used in the slides to fulfill the course requirements are 
taken from the following textbooks 
- Kindly assume that the referencing of the following books have been done in this slide 
- I take responsibility for any mistakes in solving the problems. Readers are requested to rectify 
when using the same 
- I thank the following authors for making their books available for reference
R. Ganesh Narayanan
1. Vector Mechanics for Engineers – Statics & Dynamics, Beer & Johnston; 7th edition
2. Engineering Mechanics Statics & Dynamics, Shames; 4th edition
3. Engineering Mechanics Statics Vol. 1, Engineering Mechanics Dynamics Vol. 2, Meriam & 
Kraige; 5th edition
4. Schaum’s solved problems series Vol. 1: Statics; Vol. 2: Dynamics, Joseph F. Shelley
R. Ganesh Narayanan 3
Engineering mechanics
- Deals with effect of forces on objects
Mechanics principles used in vibration, spacecraft 
design, fluid flow, electrical, mechanical m/c design 
etc.
Statics: deals with effect of force on bodies which 
are not moving
Dynamics: deals with force effect on moving bodies
We consider RIGID BODIES – Non deformable
R. Ganesh Narayanan 4
Scalar quantity: Only magnitude; time, volume, speed, 
density, mass…
Vector quantity: Both direction and magnitude; Force, 
displacement, velocity, acceleration, moment…
V = IvI n, where IvI = magnitude, n = unit vector
n = V / IvI
n - dimensionless and in direction of vector ‘V’
In our course:
y
x
z
j
i
k
i, j, k – unit vectors
R. Ganesh Narayanan 5
Dot product of vectors: A.B = AB cos θ; A.B = B.A (commutative)
A.(B+C) = A.B+A.C (distributive operation)
A.B = (Axi+Ayj+Azk).(Bxi+Byj+Bzk) = AxBx+AyBy+AzBz
Cross product of vectors: A x B = C; ICI = IAI IBI Sin θ; AxB = -(BxA)
C x (A+B) = C x A + C x B
i j k
A
B
θ
i . i = 1
i . j = 0
k i
j
k x j = -i; 
i x i = 0
AxB = (Axi+Ayj+Azk)x(Bxi+Byj+Bzk) = (AyBz- AzBy)i+( )j+( )k
i j k
Ax AY AZ
BX BY BZ
R. Ganesh Narayanan 6
Force:
- action of one body on another
- required force can move a body in the direction of action, 
otherwise no effect
- some times plastic deformation, failure is possible
- Magnitude, direction, point of application; VECTOR
Force 
< P kN
Force, 
P kN
Direction of motion
Body moves
Body does 
not move
P, kN
bulging
R. Ganesh Narayanan 7
Force system:
θ
P
WIREBracket
Magnitude, direction and point of application 
is important
External effect: Forces applied (applied force); Forces exerted by 
bracket, bolts, foundation….. (reactive force)
Internal effect: Deformation, strain pattern – permanent strain; 
depends on material properties of bracket, bolts…
R. Ganesh Narayanan 8
Transmissibility principle:
A force may be applied at any point on a line of action 
without changing the resultant effects of the force 
applied external to rigid body on which it acts
Magnitude, direction and line of action is important; not 
point of application
PP
Line of 
action
R. Ganesh Narayanan 9
Concurrent force:
Forces are said to be concurrent at a point if their lines of 
action intersect at that point
A
F1
F2
R
F1, F2 are concurrent forces
R will be on same plane
R = F1+F2
Plane
Parallelogram law of forces
Polygon law of forces
A
F1
F2
R
F2
F1
A
F1
F2
R
Use triangle law
A F1
R
F2
R does not 
pass through ‘A’
R = F1+F2 R = F1+F2
R. Ganesh Narayanan 10
Two dimensional force system
Rectangular components:
Fx
Fy
j
i
F θ
F = Fx + Fy; both are vector components in x, y direction
Fx = fx i ; Fy = fy j; fx, fy are scalar quantities
Therefore, F = fx i + fy j
Fx = F cos θ; Fy = F sin θ
F = fx2 + fy2 ; θ = tan -1 (fy/fx)
+ ve
+ ve
- ve
- ve
R. Ganesh Narayanan 11
Two concurrent forces F1, F2
Rx = Σ Fx; Ry = Σ Fy
DERIVATION
F2
F1
R
i
j
R. Ganesh Narayanan 12
Moment: Tendency to rotate; torque
Moment about a point: M = Fd
Magnitude of moment is 
proportional to the force ‘F’ and
moment arm ‘d’ i.e, perpendicular
distance from the axis of rotation
to the LOA of force
UNIT : N-m
Moment is perpendicular to plane about axis O-O
Counter CW = + ve; CW = -ve
B
A
F
d
r
O
O
M
α
R. Ganesh Narayanan 13
Cross product:
M = r x F; where ‘r’ is the position vector which runs from 
the moment reference point ‘A’ to any point on the 
LOA of ‘F’
M = Fr sin α; M = Fd
M = r x F = -(F x r): sense is important
B
A
d
r α
Sin α = d / r
R. Ganesh Narayanan 14
Varignon’s theorem:
The moment of a force about any point is equal to the 
sum of the moments of the components of the forces 
about the same point
o
Q
P R
r
B Mo = r x R = r x (P+Q) = r x P + r x Q
Moment of ‘P’
Moment of ‘Q’
Resultant ‘R’ – moment arm ‘d’
Force ‘P’ – moment arm ‘p’; Force ‘Q’ – moment arm ‘q’
Mo= Rd = -pP + qQ
Concurrent forces – P, Q
Usefulness:
R. Ganesh Narayanan 15
Pb:2/5 (Meriam / Kraige):
Calculate the magnitude of the moment
about ‘O’ of the force 600 N
1) Mo = 600 cos 40 (4) + 600 sin 40 (2)
= 2610 Nm (app.)
2) Mo = r x F = (2i + 4j) x (600cos40i-600sin40j)
= -771.34-1839 = 2609.85 Nm (CW);
mag = 2610 Nm
o
600N4
2
A
in mm
40 deg
r
i
j
R. Ganesh Narayanan 16
Couple: Moment produced by two equal, opposite and 
non-collinear forces
-F
+F
a
d
o
=>-F and F produces rotation
=>Mo = F (a+d) – Fa = Fd; 
Perpendicular to plane
⇒Independent of distance from ‘o’, 
depends on ‘d’ only
⇒ moment is same for all moment 
centers
M
R. Ganesh Narayanan 17
Vector algebra method
-F
+F
o
rb
ra
r M = ra x F + rb x (-F) = (ra-rb) x F = r x F
CCW 
Couple
CW 
Couple
Equivalent couples
•Changing the F and d values does not change a given couple 
as long as the product (Fd) remains same
•Changing the plane will not alter couple as long as it is parallel
R. Ganesh Narayanan 18
M
-F +Fd
M
-F
+F
d
M
-F
+F d
-2F
d/2+2F
M
EXAMPLE
All four are equivalent couples
R. Ganesh Narayanan 19
Force-couple system
=>Effect of force is two fold – 1) to push or pull, 2) 
rotate the body about any axis
⇒Dual effect can be represented by a force-couple 
syatem
⇒ a force can be replaced by a force and couple
F
A
B
F
A
B F
-F
B F
M = Fd
R. Ganesh Narayanan 20
o
80N
o
80N
80 N80 N o80 N
Mo = Y N m
60deg
9 m
Mo = 80 (9 sin 60) = 624 N m; CCW
EXAMPLE
9
60 deg
R. Ganesh Narayanan 21
Resultants
To describe the resultant action of a group or system of forces
Resultant: simplest force combination which replace the original 
forces without altering the external effect on the body to which
the forces are applied
R
R = F1+F2+F3+….. = Σ F
Rx = Σ Fx; Ry = Σ Fy; R = (Σ Fx)2 + (Σ Fy)2
Θ = tan -1 (Ry/Rx)
R. Ganesh Narayanan 22
F1 F2 
F3 
F1 – D1; F2 – D2; F3 – D3
F1 F2 
F3 
M1 = F1d1; 
M2 = F2d2; 
M3 = F3d3
R= ΣF
Mo= ΣFd
NON-CONCURRENT FORCES
R
d
Mo=Rd
How to obtain resultant force ?
R. Ganesh Narayanan 23
Principle of moments
Summarize the above process: R = ΣF
Mo = ΣM = Σ(Fd)
Mo = RdFirst two equations: reduce the system of forces to a force-couple 
system at some point ‘O’
Third equation: distance ‘d’ from point ‘O’ to the line of action ‘R’
=> VARIGNON’S THEOREM IS EXTENDED HERE FOR NON-
CONCURENT FORCES
R= ΣF
Mo= ΣFd
R
d
Mo=Rd
R. Ganesh Narayanan 24
STATICS – MID SEMESTER – DYNAMICS
Tutorial: Monday 8 am to 8.55 am
1. Vector Mechanics for Engineers – Statics & Dynamics, Beer & Johnston; 7th edition
2. Engineering Mechanics Statics & Dynamics, Shames; 4th edition
3. Engineering Mechanics Statics Vol. 1, Engineering Mechanics Dynamics Vol. 2, 
Meriam & Kraige; 5th edition
4. Schaum’s solved problems series Vol. 1: Statics; Vol. 2: Dynamics, Joseph F. Shelley
Reference books
R. Ganesh Narayanan 25
ENGINEERING MECHANICS
TUTORIAL CLASS: Monday 8 AM TO 8.55 AM
07010605 (5 Students)07010601
Dr. Saravana Kumar120507010449 (36 Students)07010414TG5
07010413 (13 Students)07010401
Dr. M. Pandey120207010353 (28 Students)07010326TG4
07010325 (25 Students)07010301
R. Ganesh Narayanan1G207010249 (16 Students)07010234TG3
07010233 (33 Students)07010201
Dr. senthilvelan1G107010149 (8 Students)07010142TG2
Prof. R. TiwariL207010141 (41 Students)07010101TG1
ToFrom
TutorsClass RoomRoll NumbersTutorial Groups
LECTURE CLASSES: LT2 (one will be optional):
Monday 3 pm to 3.55 pm
Tuesday 2 pm to 2.55 pm
Thursday 5 pm to 5.55 pm
Friday 4 pm to 4.55 pm
R. Ganesh Narayanan 26
Three dimensional force system
Rectangular components
Fx = F cos θx; Fy = F cos θy; Fz = F cos θz
F = Fx i + Fy j + Fz k 
= F (i cos θx + j cos θy + k cos θz) = F (l i + m j + n k)
F = F nf
o
Fx i
Fy j
Fz k
F
θz
θx
θy
l, m, n are directional cosines of ‘F’
R. Ganesh Narayanan 27
F
r
Mo
d
αA
A - a plane in 3D structure
Mo = F d (TEDIOUS to find d)
or Mo = r x F = – (F x r) (BETTER)
Evaluating the cross product
Described in determinant form: i j k
rx rY rZ
FX FY FZ
Moment in 3D
Expanding …
R. Ganesh Narayanan 28
Mo = (ryFz - rzFy) i + (rzFx – rxFz) j + (rxFy – ryFx) k
Mx = ryFz – rzFy; My = rzFx – rxFz; Mz = rxFy – ryFx
Moment about any arbitrary axis λ:
F
r
Mo n
o
λ
Magnitude of the moment Mλ of F about λ
= Mo . n (scalar reprn.)
Similarly, Mλ = (r x F.n) n (vector reprn.)
Scalar triple product
rx ry rz
Fx FY FZ
α β γ
α, β, γ – DCs of n
R. Ganesh Narayanan 29
Varignon’s theorem in 3D
o F1
F3
F2
r
B
Mo = rxF1 + rxF2 + rx F3 +…= Σ(r x F)
= r x (F1+F2+F3+…)
= r x (ΣF) = r x R
Couples in 3D
B
M
A
r
ra
rb
d
-F
+F
M = ra x F + rb x –F = (ra-
rb) x F = rxF
R. Ganesh Narayanan 30
Beer-Johnston; 2.3
F1 = 150N
30
F4 = 100N
15
F3 = 110N
F2 = 80N
20
• Evaluate components of F1, F2, F3, F4
• Rx = ΣFx; Ry = ΣFy
• R = Rx i + Ry j
• α = tan -1 (Ry/Rx)
Ry
Rx
R
α
• R = 199i + 14.3j; α = 4.1 deg
2D force system; equ. Force-couple; principle of 
moments
R. Ganesh Narayanan 31
F1
F2 R =3000 N
30 DEG
45 DEG
15 DEG
Find F1 and F2
3000 (cos15i – sin 15j) = F1 (cos 30i – Sin 30j)+ F2 (cos45i – sin 45j)
EQUATING THE COMPONENTS OF VECTOR,
F1 = 2690 N; F2 = 804 N
R = F1 + F2
Boat
R. Ganesh Narayanan 32
o
A
B
20 DEG
C
OC – FLAG POLE
OAB – LIGHT FRAME
D – POWER WINCH
D
780 N
Find the moment Mo of 780 N 
about the hinge point
10m
10
10
T = -780 COS20 i – 780 sin20 j
= -732.9 i – 266.8 j
r = OA = 10 cos 60 i + 10 sin 60 j = 5 i + 8.6 j
Mo = r x F = 5014 k ; Mag = 5014 Nm
Meriam / kraige; 2/37
R. Ganesh Narayanan 33
Meriam / kraige; 2/6
Replace couple 1 by eq. couple p, -p; find Θ
M = 100 (0.1) = 10 Nm (CCW)
M = 400 (0.04) cos θ
10 = 400 (0.04) cos θ
=> Θ = 51.3 deg
M
P
-P
40
100
100
100
100N 100N
60
θ
θ
1
2
1
2
R. Ganesh Narayanan 34
80N
30 deg
60 N
40 N
50 N2m 5m45
2m
2m
1m
o
140Nm
Find the resultant of four forces and one 
couple which act on the plate
Rx = 40+80cos30-60cos45 = 66.9 N
Ry = 50+80sin 30+60cos45 = 132.4 N
R = 148.3 N; Θ = tan-1 (132.4/66.9) = 63.2 deg
Mo = 140-50(5)+60cos45(4)-60sin45(7) = -237 Nm
o
R = 148.3N
63.2 deg237 Nm
o
R = 148.3N
63.2 deg
148.3 d = 237; d = 1.6 mFinal LOA of R:
o
R = 148.3N
b
x
y
(Xi + yj) x (66.9i+132.4j) = -237k
(132.4 x – 66.9 y)k = -237k
132.4 x -66.9 y = -237
Y = 0 => x = b = -1.792 m
Meriam / kraige; 2/8
LOA of R with x-axis:
R. Ganesh Narayanan 35
Couples in 3D
B
M
A
r
ra
rb
d
-F
+F
M = ra x F + rb x –F = (ra-
rb) x F = rxF
F
AB
F
M = Fd
F
AB
F
-F
r
B
Equivalent couples
R. Ganesh Narayanan 36
How to find resultant ?
R = ΣF = F1+F2+F3+…
Mo = ΣM = M1+M2+M3+… = Σ(rxF)
M = Mx2 + My2 + Mz2; R = ΣFx2 + ΣFy2 + ΣFz2
Mx = ; My = ; Mz = 
R. Ganesh Narayanan 37
Equilibrium
Body in equilibrium - necessary & sufficient condition:
R = ΣF = 0; M = ΣM = 0
Equilibrium in 2D
Mechanical system: body or group of bodies which can be conceptually 
isolated from all other bodies
System: single body, combination of bodies; rigid or non-rigid; 
combination of fluids and solids
Free body diagram - FBD:
=> Body to be analyzed is isolated; Forces acting on the body are 
represented – action of one body on other, gravity attraction, 
magnetic force etc.
=> After FBD, equilibrium equns. can be formed
R. Ganesh Narayanan 38
Modeling the action of forces
Meriam/Kraige
Imp
Imp
R. Ganesh Narayanan 39
FBD - Examples
Meriam/Kraige
Equilibrium equns. Can be 
solved, 
• Some forces can be zero
• Assumed sign can be 
different
R. Ganesh Narayanan 40
Types of 2D equilibrium
x
F1
F2
F3
Collinear: ΣFx = 0 F1 F2
F3
F4
Parallel: ΣFx = 0; ΣMz = 0
F1
F2
F3
F4
X
Y
Concurrent at a point: ΣFx = 0; ΣFy = 0
X
Y
M
General: ΣFx = 0; ΣFy = 0; ΣMz = 0
R. Ganesh Narayanan 41
General equilibrium conditions
ΣFx = 0; ΣFy = 0; ΣFz = 0
ΣMx = 0; ΣMy = 0; ΣMz = 0
⇒These equations can be used to solve unknown forces, 
reactions applied to rigid body
⇒For a rigid body in equilibrium, the system of external forces will 
impart no translational, rotational motion to the body
⇒Necessary and sufficient equilibrium conditions
R. Ganesh Narayanan 42
Written in three alternate ways,
A
B
C
D
PY
Px
QY
Qx
RY
Rx
W
BY
AX
AY
ΣMB = 0 => will not provide new information; used to check the 
solution; To find only three unknowns
A B
C D
P Q R
RollerPin
ΣFx = 0; ΣFy = 0; ΣMA = 0 I
R. Ganesh Narayanan 43
ΣFx = 0; ΣMA = 0; ΣMB = 0 II
• Point B can not lie on the line that passes through point A
• First two equ. indicate that the ext. forces reduced to a single vertical force at A
• Third eqn. (ΣMB = 0) says this force must be zero
Rigid body in equilibrium =>
ΣMA = 0; ΣMB = 0; ΣMc = 0; III
Body is statically indeterminate: more unknown reactions than 
independent equilibrium equations
R. Ganesh Narayanan 44
Meriam / Kraige; 2/10
z
x
12 m
9
B
O
A
15 T = 10kN
Y
Find the moment Mz of T about the z-axis passing 
thro the base O
3D force system
R. Ganesh Narayanan 45
F = T = ITI nAB = 10 [12i-15j+9k/21.21] = 10(0.566i-0.707j+0.424k) k N
Mo = rxF = 15j x 10(0.566i-0.707j+0.424k) = 150 (-0.566k+0.424i) k Nm
Mz = Mo.k= 150 (-0.566k+0.424i).k = -84.9 kN. m
R. Ganesh Narayanan 46
Merial / Kraige; 2/117
Replace the 750N tensile force which the cable exerts on point B by a force-
couple system at point O
R. Ganesh Narayanan 47
F = f λ, whereλ is unit vector along BC
= (750) BC/IBCI = 750 (-1.6i+1.1j+0.5k/2.005)
F = -599i+412j+188.5k
rob = OB = 1.6i-0.4j+0.8k
Mo = rob x F
= (1.6i-0.4j+0.7k) x (-599i+412j+188.5k)
Mo = - 363i-720j+419.2k
R. Ganesh Narayanan 48
Meriem / Kraige; 3/4
ΣMA = (T cos 25) (0.25) + (T sin 25) (5-0.12) –
10(5-1.5-0.12) – 4.66 (2.5-0.12) = 0
T = 19.6 kN T
25 deg
y
Ax
Ay
10 kN
0.5 m
4.66 kN
5m
1.5m
0.12 m
ΣFx = Ax – 19.6 cos 25 = 0
Ax = 17.7 kN
ΣFy = Ay+19.61 sin 25-4.66-10 = 0
Ay = 6.37 kN
A = Ax2 + Ay2 = 18.88kN
2D equilibrium
Find T and force at A; I-beam with mass of 95 
kg/meter of length
95 kg/meter => 95(10-3)(5)(9.81) = 4.66kN
R. Ganesh Narayanan 49
A
B
40 50 30 10
60
60 a 80
mm, N
Beer/Johnston; 4.5
Find reactions at A, B if (a) a = 100 mm; (b) 
a=70 mm
40 50 30 10
Bx
By
Ay
a = 100 mm
ΣMa = 0 => (-40x60)+(-50x120)+(-30x220)+
(-10x300)+(-Byx120) = 0
By = 150 N
ΣFy = 0 => By-Ay-40-50-30-10 = 0
= 150-Ay-130 = 0 => Ay = 20 N
a = 70 mm
By = 140 N Ay = 10 N
R. Ganesh Narayanan 50
A B
20 20 20 20
C
2.25
3.75
E
D
4.5
F
1.8
A B
20 20 20 20
C
2.25
3.75
E
D
4.5
F
1.8
150 kN
Ex
Ey
Find the reaction at the fixed end 
‘E’
DF = 7.5 m
ΣFx = Ex + 150 (4.5/7.5) = 0 => Ex = - 90 kN (sign change)
ΣFy = Ey – 4(20)-150 (6/7.5) = 0 => Ey = 200 kN
ΣME= 20 (7.2) + 20 (5.4) + 20 (3.6) +20 (1.8) – (6/7.5) (150) (4.5) + 
ME= 0
ME= +180 kN.m => ccw
ME
Beer/Johnston; 4.4
R. Ganesh Narayanan 51
Instructions for TUTORIAL
• Bring pen, pencil, tagged A4 sheets, calculator, text books
• Submitted in same tutorial class
• Solve div II tutorial problems also
• Solve more problems as home work
• Tutorial : 10 % contribution in grading
• Do not miss any tutorial class
QUIZ 1 – FEB, 11TH, 2008
R. Ganesh Narayanan 52
3D equilibrium
3D equilibrium equns. can be written in scalar and vector form 
ΣF = 0 (or) ΣFX = 0; ΣFY = 0; ΣFZ = 0
ΣM = 0 (or) ΣMX = 0; ΣMY = 0; ΣMZ = 0
ΣF = 0 => Only if the coefficients of i, j, k are zero; ΣFX = 0 
ΣM = 0 => Only if the coefficients of i, j, k are zero; ΣMX = 0
R. Ganesh Narayanan 53
Modeling forces in 3D
R. Ganesh Narayanan 54
Types of 3D equilibrium
R. Ganesh Narayanan 55
Meriem / Kraige
B
A
7 m
6 m
2 m
y
x
z
By
Bx
G
W=mg=200 x 9.81
W = 1962 N
Ay
AzAx
h
3.5
3.5
7 = 22 + 62 + h2 => h = 3 m
rAG = -1i-3j+1.5k m; rAB = -2i-6j+3k m
ΣMA = 0 => rAB x (Bx+By) + rAG x W = 0
(-2i-6j+3k) x (Bx i + By j) + (-i-3j+1.5k) x (-1962k) = 0
(-3By+5886)i + (3Bx-1962)j + (-2By+6Bx)k = 0
=> By = 1962 N; Bx = 654 N
ΣF = 0 => (654-Ax) i + (1962-Ay) j + (-1962+Az)k = 0
=> Ax = 654 N; Ay = 1963 N; Az = 1962 N; find A
R. Ganesh Narayanan 56
Meriem / Kraige; 3/64
R. Ganesh Narayanan 57
I.H. Shames
Find forces at A, B, D. Pin connection at C; E has welded connection
R. Ganesh Narayanan 58
F.B.D. - 1
F.B.D. - 2
ΣMc = 0 => (Dy) (15) – 200 (15) (15/2) –
(1/2)(15)(300)[2/3 (15)] = 0
Dy = 3000 N
F.B.D. - 2
R. Ganesh Narayanan 59
F.B.D. - 1
ΣMB = 0 => -Ay (13) +(3000) (21) – 200 
(34) (34/2-13) – ½ (300) (15) [6+2/3(15)] 
= 0
Ay = -15.4 N
ΣFy = 0 => Ay+By+3000-200(34)-
(1/2)(300)(15) = 0
Sub. ‘Ay’ here,
=> By = 6065 N
R. Ganesh Narayanan 60
2D, 3D force system
• Rectangular components
• Moment
• Varignon’s theorem
• Couple
• Force-couple system
• Resultant
• Principle of moment
Equilibrium equations
ΣFx = 0; ΣFy = 0; ΣMA = 0
ΣFx = 0; ΣMA = 0; ΣMB = 0
ΣMA = 0; ΣMB = 0; ΣMc = 0 
2D
ΣF = 0 (or) ΣFX = 0; ΣFY = 0; ΣFZ = 0
ΣM = 0 (or) ΣMX = 0; ΣMY = 0; ΣMZ = 0
3D
R. Ganesh Narayanan 61
Structures
Truss: Framework composed of members joined at their ends to form a rigid 
structures
Plane truss: Members of truss lie in same plane
Bridge truss
Roof truss
R. Ganesh Narayanan 62
•Three bars joined with pins at end
• Rigid bars and non-collapsible
• Deformation due to induced internal strains is negligible
B D
CA
Non-rigid rigid
E
Non rigid body can be made rigid by 
adding BC, DE, CE elements
B
C
D
A
A
B
c
R. Ganesh Narayanan 63
Instructions for TUTORIAL
• Bring pen, pencil, tagged A4 sheets, calculator, text books
• Submitted in same tutorial class
• Solve div II tutorial problems also
• Solve more problems as home work
• Tutorial : 10 % contribution in grading
• Do not miss any tutorial class
QUIZ 1 – FEB, 11TH, 2008
R. Ganesh Narayanan 64
Structures
Truss: Framework composed of members joined at their ends to form a rigid 
structures
Plane truss: Members of truss lie in same plane
Bridge truss
Roof truss
R. Ganesh Narayanan 65
•Three bars joined with pins at end
• Rigid bars and non-collapsible
• Deformation due to induced internal strains is negligible
Non rigid body can be made rigid by 
adding BC, DE, CE elementsB D
CA
Non-rigid rigid
E
B
C
D
A
A
B
c
Simple truss: structures built from basic triangle
More members are present to prevent collapsing => statically indeterminate truss; 
they can not be analyzed by equilibrium equations
Additional members not necessary for maintaining equilibrium - redundant
R. Ganesh Narayanan 66
In designing simples truss or truss => assumptions are followed
1. Two force members – equilibrium only in two forces; either tension or compression
2. Each member is a straight link joining two points of application of force
3. Two forces are applied at the end; they are equal, opposite and collinear for 
equilibrium
4. Newton’s third law is followed for each joint
5. Weight can be included; effect of bending is not accepted
6. External forces are applied only in pin connections
7. Roller or rocker is also provided at joints to allow expansion and contraction due to 
temperature changes and deformation for applied loads
T
T c
c weight
TWO FORCE MEMBERS
R. Ganesh Narayanan 67
Method of joints
•This method consists of satisfying the conditions of equilibrium for the 
forces acting on the connecting pin of each joint
•This method deals with equilibrium of concurrent forces and only two 
independent equilibrium equations are solved
• Newton’s third law is followed
Two methods to analyze force in simple truss
R. Ganesh Narayanan 68
Example
A
B
C
D
EF
L
ΣFy = 0; ΣFx = 0
Finally sign can be changed if 
not applied correctly
R. Ganesh Narayanan 69
Internal and external redundancy
external redundancy: If a plane truss has more supports than are necessary to 
ensure a stable equilibrium, the extra supports constitute external redundancy
Internal redundancy: More internal members than are necessary to prevent collapse, 
the extra members constitute internal redundancy
Condition for statically determinate truss: m + 3 = 2j
- Equilibrium of each joint can be specified by two scalar force equations, then ‘2j’
equations are present for a truss with ‘j’ joints
-The entire truss composed of ‘m’ two force members and having the maximum of 
three unknown support reactions, there are (m + 3) unknowns
j – no. of joints; m – no. of members
m + 3 > 2 j =>more members than independent equations; statically 
indeterminate
m + 3 < 2 j => deficiency of internal members; truss is unstable
R. Ganesh Narayanan 70
A
C E
F
DB
1000
10
10
10 10
1000
I. H. Shames
Determine the force transmitted by each member;
A, F = 1000 N
Pin A
FAB
FAC
1000
A
ΣFx = 0 =>FAC – 0.707FAB = 0
ΣFy = 0 => -0.707FAB+1000 = 0
FAB = 1414 N; FAC = 1000 N
Pin B
B
FBC
1414FBD
ΣFx = 0 => -FBD + 1414COS45 = 0 => FBD = 1000 N
ΣFy = 0 => -FBC+1414 COS45 = 0 => FBC = 1000 N
1000
FAC
FAB
1414
FBC
FBD
R. Ganesh Narayanan 71
Pin C
B
1000
1000
FCE
FDC
1000
10001000
1000
FDC
FCE
ΣFx = 0 => -1000 + FCE + FDC COS 45 = 0 => FCE = 1000 N
ΣFy = 0 => -1000+1000+ FDC COS 45 = 0 => FDC = 0
SIMILARLY D, E, F pins are solved
R. Ganesh Narayanan 72
B D
A
C E
30 20
5 5
5
5 55 5
kN, m
Find the force in each member of the loaded 
cantilever truss by method of joints
Meriem / Kraige (similar pbm. 6.1 in Beer/Johnston)
R. Ganesh Narayanan 73
ΣME = 0 => 5T-20(5)-30 (10) = 0; T = 80 kN
ΣFx = 0 => 80 cos 30 – Ex = 0; Ex = 69.28 kN
ΣFy = 0 => Ey +80sin30-20-30 = 0 => Ey = 10kN
ΣFx = 0; ΣFy = 0
Find AB, AC forces
ΣFx = 0; ΣFy = 0
Find BC, BD forces
ΣFx = 0; ΣFy = 0
Find CD, CE forces
ΣFy = 0
Find DE forces
ΣFx = 0 can be checked
FBD of entire truss
FBD of joints
R. Ganesh Narayanan 74
Q = 100 N; smooth surfaces; Find 
reactions at A, B, C Q
Q
A
B
30°
c
roller
100
100
RA
Rc
roller
RB
ΣF = 0 => (-RA cos 60 - RB cos 60 + Rc) i + (-2 x 100 + RB
sin 60 + RA sin 60) j = 0
RC = (RA + RB)/2
RB + RA = 230.94
RC = 115.5 N
RB
100
Rc
RAB
30°
ΣF = 0 => (-RAB cos 30 - RB cos 60 + Rc) i + (RB Sin 60 – 100 - RAB
sin 30) j = 0
0.866 RAB + 0.5 RB = 115.5; -0.5 RAB + 0.866 RB = 100
RAB = 50 N (app.); RB = 144.4 N; RA = 230.94-144.4 = 86.5 N
R. Ganesh Narayanan 75
Method of joints
•This method consists of satisfying the conditions of equilibrium for the 
forces acting on the connecting pin of each joint
•This method deals with equilibrium of concurrent forces and only two 
independent equilibrium equations are solved
• Newton’s third law is followed
Two methods to analyze force in plane truss
Method of sections
R. Ganesh Narayanan 76
Methodology for method of joints
A
B
C
D
EF
L
ΣFy = 0; ΣFx = 0
Finally sign can be changed if 
not applied correctly
R. Ganesh Narayanan 77
5B D
A
C E
30 20
5 5
5 55 5
kN, m
R. Ganesh Narayanan 78
Method of sections
• In method of joints, we need only two equilibrium equations, as we 
deal with concurrent force system
• In method of sections, we will consider three equilibrium 
equations, including one moment equilibrium eqn.
• force in almost any desired member can be obtained directly from 
an analysis of a section which has cut the member
• Not necessary to proceed from joint to joint
•Not more than three members whose forces are unknown should 
be cut. Only three independent equilibrium eqns. are present
•Efficiently find limited information
R. Ganesh Narayanan 79
A
B
C
D
EF
L
•The external forces are obtained initially from method of joints, by 
considering truss as a whole
• Assume we need to find force in BE, then entire truss has to be 
sectioned across FE, BE, BC as shown in figure; we have only 3 
equilibrium equns.
• AA – section across FE, BE, BC; Forces in these members are 
initially unknown
A
B
C
D
EF
LR1 R2
A
A
Methodology for method of sections
R. Ganesh Narayanan 80
• Now each section will apply opposite forces on each other
• The LHS is in equilibrium with R1, L, three forces exerted on the cut 
members (EF, BE, BC) by the RHS which has been removed
• IN this method the initial direction of forces is decided by moment about 
any point where known forces are present
• For eg., take moment about point B for the LHS, this will give BE, BC to 
be zero; Then moment by EF should be opposite to moment by R1; 
Hence EF should be towards left hand side - compressive
Section 1 Section 2
R. Ganesh Narayanan 81
• Now take moment about ‘F’ => BE should be opposite to R1 
moment; Hence BE must be up and to the right; So BE is tensile
• Now depending on the magnitudes of known forces, BC direction 
has to be decided, which in this case is outwards i.e., tensile
Σ MB = 0 => FORCE IN EF; BE, BC = 0
Σ Fy = 0 => FORCE IN BE; BC, EF = 0
Σ ME = 0 => FORCE IN BC; EF, BE = 0
Section 1 Section 2
R. Ganesh Narayanan 82
Section AA and BB are 
possible
convenient
R. Ganesh Narayanan 83
Important points
• IN method of sections, an entire portion of the truss is considered a 
single body in equilibrium
• Force in members internal to the section are not involved in the 
analysis of the section as a whole
• The cutting section is preferably passed through members and not 
through joints
• Either portion of the truss can be used, but the one with smaller 
number of forces will yield a simpler solution
• Method sections and method of joints can be combined
• Moment center can be selected through which many unknown forces
pass through
• Positive force value will sense the initial assumption of force direction
R. Ganesh Narayanan 84
Meriem/Kraige
Find the forces included in members KL, 
CL, CB by the 20 ton load on the cantilever 
truss
Section 1 Section 2
x
Σ Moment abt. ‘L’ => CB is compressive => creates CW moment
Σ Moment abt. C => KL is tensile => creates CW moment 
CL is assumed to be compressive
y
KL
20 T
C CB
CL
G
P
L
K
R. Ganesh Narayanan 85
y
KL
20 T
C CB
CL
G
P
L
K
x
Section 1 Section 2
Σ ML = 0 => 20 (5) (12)- CB (21) = 0 => CB = 57.1 t (C)
Σ Mc = 0 => 20 (4)(12) – 12/13 (KL) (16) = 0; KL = 65 t (T)
Σ Mp = 0 => find PC distance and find CL; CL = 5.76 t (C)
BL = 16 + (26-16)/2 = 12 ft
Θ = tan -1 (5/12) => cos Θ = 12/13
θ
R. Ganesh Narayanan 86
Meriem/Kraige
Find the force in member DJ of the truss 
shown. Neglect the horizontal force in 
supports
Section 2 cuts four members, but we have only 
3 equi. Equns
Hence consider section 1 which cuts only 3 
members – CD, CJ, KJ
Consider FBD for whole truss and find 
reaction at A
ΣMG = -Ay (24) +(10) (20) + 10(16) + 10 
(8) = 0
Ay = 18. 3 kN => creates CW moment
Force direction
Σ Moment abt. A => CD, JK – Eliminated; CJ will be upwards creating CCW moment
Σ Moment abt. C => JK must be towards right creating CCW moment
ASSUME CD TO HAVE TENSILE FORCE
R. Ganesh Narayanan 87
ΣMA = 0 => CJ (12) (0.707) – 10 (4) -10( 8) =0; CJ = 14.14 Kn
ΣMJ = 0 => 0.894 (CD) (6) +18.33 (12)-10(4)-10(8) = 0; CD = -18.7 kN
CD direction is changed
From section 1 FBD
From section 2 FBD
ΣMG = 0 => 12 DJ +10(16)+10(20)-18.3 (24)-
14.14 (0.707)(12) = 0
DJ = 16.7 kN
R. Ganesh Narayanan 88
I.H. Shames FBD - 1
FBD - 2
From FBD-2
ΣMB = 0 => -(10)(500)+30 (789)- FAC Sin 30 (30) = 0
FAC = 1244.67 N
From FBD -1
ΣFx = 0 => FDA Cos 30 – (1244.67) cos 30 – 1000 sin 30 = 0 ; 
FDA = 1822 N
ΣFy = 0 => (1822)Sin 30 + (1244.67) sin 30 +FAB – 1000 Cos 30 = 0; FAB = -667 N
R. Ganesh Narayanan 89
Frames and machines
Multi force members: Members on which three or more forces acting 
on it (or) one with two or more forces and one or more couples acting 
on it
Frame or machine: At least one of its member is multi force member
Frame: Structures which are designed to support applied loads and 
are fixed in position
Machine: Structure which contain moving parts and are designed to 
transmit input forces or couples to output forces or couples
Frames and machines contain multi force members, the forces in 
these members will not be in directions of members
Method of joints and sections are not applicable
R. Ganesh Narayanan 90
Inter-connected rigid bodies with multi force members
• Previously we have seen equilibrium of single rigid bodies
• Now we have equilibrium of inter-connected members which 
involves multi force members
• Isolatemembers with FBD and applying the equilibrium equations
• Principle of action and reaction should be remembered
• Statically determinate structures will be studied
R. Ganesh Narayanan 91
Force representation and FBD
• Representing force by rectangular components
• Calculation of moment arms will be simplified
• Proper sense of force is necessary; Some times arbitrary assignment 
is done; Final force answer will yield correct force direction
• Force direction should be consistently followed
R. Ganesh Narayanan 92
Frames and machines
Multi force members: Members on which three or more forces acting 
on it (or) one with two or more forces and one or more couples acting 
on it
Frame or machine: At least one of its member is multi force member
Frame: Structures which are designed to support applied loads and 
are fixed in position
Machine: Structure which contain moving parts and are designed to 
transmit input forces or couples to output forces or couples
Frames and machines contain multi force members, the forces in 
these members will not be in directions of members
Method of joints and sections are not applicable
R. Ganesh Narayanan 93
Inter-connected rigid bodies with multi force members
• Previously we have seen equilibrium of single rigid bodies
• Now we have equilibrium of inter-connected members which 
involves multi force members
• Isolate members with FBD and applying the equilibrium equations
• Principle of action and reaction should be remembered
• Statically determinate structures will be studied
R. Ganesh Narayanan 94
Force representation and FBD
• Representing force by rectangular components
• Calculation of moment arms will be simplified
• Proper sense of force is necessary; Some times arbitrary assignment 
is done; Final force answer will yield correct force direction
• Force direction should be consistently followed
R. Ganesh Narayanan 95
AFAE
BD
Full truss
K, J are un-necessary 
here
R. Ganesh Narayanan 96
Meriem/Kraige
A
B
C
D
E
F
30 lb
50 lb
12
12
30 ft
20 ft
20 ft
Find the forces in all the frames; 
neglect weight of each member
Ax
Ay
50 lb
30 lb
Cx
Cy
Σ Mc = 0 => 50 (12) +30(40)-30 (Ay) = 0; Ay = 60 lb
Σ Fy = 0 => Cy – 50 (4/5) – 60 = 0 => Cy = 100 lb
FBD of full frame
R. Ganesh Narayanan 97
ED:
ΣMD = 0 => 50(12)-12E = 0 => E = 50 lb
ΣF = 0 => D-50-50 = 0 => D= 100 lb
(components will be eliminated)
EF: Two force member; E, F are 
compressive
EF: F = 50 lb (opposite and equal to E)
AB:
ΣMA = 0 => 50(3/5)(20)-Bx (40) = 0 => Bx = 15 lb
ΣFx = 0 => Ax+15-50(3/5) = 0 => Ax = 15 lb
ΣFy = 0 => 50 (4/5)-60-By = 0 =>By = -20 lb
BC: ΣFx = 0 => 30 +100 (3/5)-15-Cx = 0 => Cx = 75 lb
FBD of individual members
D
Σ Fx = -50 (cos 53.1)+15+15 = -30+15+15 = 0
F
Fx
Fy
53.1 deg
E
R. Ganesh Narayanan 98
A
B
C
E
D
60
100 150
480 N
160
60
80
Find the force in link DE and components of 
forces exerted at C on member BCD
A
B
C
E
D
100 150
480 N
160
Ax
Ay
Bx
80
θ
Σ Fy = 0 => Ay-480 = 0 =>Ay = 480 N
Σ MA = 0 => Bx (160)-480 (100) = 0 => Bx = 300 N
Σ Fx = 0 => 300+Ax = 0 => Ax = -300 N
FBD of full frame
Θ = tan -1 (80/150) = 28.07 deg
R. Ganesh Narayanan 99
FBD of BCD
B
C
D
480 N
300 Cx
Cy
FDE
Σ Mc = 0 => -FDE sin 28.07 (250) – 300(80)-480 (100) = 0; FDE = -561 N
Σ Fx = 0 => Cx – (-561) cos 28.07 +300 = 0 => Cx = -795 N
Σ Fy = 0 => Cy – (-561) sin 28.07 – 480 = 0 => Cy = 216 N
θ
E
D
FDE
FDE
DE: Two force member
D
FDE
A
E
Ax
Ay
FDE
Cy
Cx
FBD of AE
FBD of DE
R. Ganesh Narayanan 100
A
B
C
D
E F
400 kg
3m 2m
1.5m
0.5m
1.5m
1.5m
R =0.5 m
Find the horizontal and vertical 
components of all the forces; neglect 
weight of each member
Meriem/Kraige
FBD of full frame
Ay
Ax
0.4 x 9.81 = 3.92
Σ MA = 0 => 5.5 (-0.4) (9.81) + 5Dx = 0 => Dx = 4.32 kN
Σ Fx = 0 => -Ax + 4.32 = 0 => Ax = 4.32 kN
Σ Fy = 0 => Ay – 3.92 = 0 => Ay = 3.92 kN
Dx
R. Ganesh Narayanan 101
FBD of individual members
3.92
4.32
3.92
Bx
By
4.32
Cx
Cy
A
D
3.92
3.92
3.92
3.92
F
C
E
Cx
Cy
ExEy
E
EyEx
Bx
By
B
3.92
3.92
A
B
C
D
E F
400 kg
3m 2m
1.5m
0.5m
1.5m
1.5m
R =0.5 m
Apply equilibrium equn. And solve for 
forces
R. Ganesh Narayanan 102
Machines
• Machines are structures designed to transmit and modify forces. Their main purpose 
is to transform input forces into output forces.
• Given the magnitude of P, determine the 
magnitude of Q. 
Taking moments about A, 
P
b
a
QbQaPM A =−==∑ 0
R. Ganesh Narayanan 103
Center of mass & center of gravity
A B
C
W
G
A
B
C
W
G
W
GA
B
C
G
•Body of mass ‘m’
•Body at equilibrium w.r.t. forces in the cord and resultant of gravitational 
forces at all particles ‘W’
•W is collinear with point A
•Changing the point of hanging to B, C – Same effect
•All practical purposes, LOA coincides with G; G – center of gravity
BODY
R. Ganesh Narayanan 104
dw
G
w
z Y
X
Moment abt. Y axis = dw (x)
Sum of moments for small regions through out the 
body: ∫ x dw
Moment of ‘w’ force with Y axis = w x
∫ x dw = w x
Sum of moments Moment of the sum
W = mg
r
r
X = (∫ x dm) / m
X = (∫ x dw) / w Y = (∫ y dw) / w Z = (∫ z dw) / w
Y = (∫ y dm) / m Z = (∫ z dm) / m
1
2
R. Ganesh Narayanan 105
r = (∫ r dm) / mIn vector form,
ρ = m/V; dm = ρ dv
X = (∫ x ρ dv) / ∫ ρ dv
Y = (∫ y ρ dv) / ∫ ρ dv
Z = (∫ z ρ dv) / ∫ρ dv
ρ = not constant through out 
body
3
4
Equns 2, 3, 4 are independent of ‘g’; They depend only on mass distribution; 
This define a co-ordinate point – center of mass
This is same as center of gravity as long as gravitational field is uniform and parallel
R. Ganesh Narayanan 106
Centroids of lines, areas, volumes
X = (∫ xc dv) / v Y = (∫ yc dv) / v Z = (∫ zc dv) / v
Suppose if density is constant, then the expression define a purely 
geometrical property of the body; It is called as centroid
Centroid of volume
X = (∫ x dA) / A Y = (∫ y dA) / A Z = (∫ z dA) / A
Centroid of area
X = (∫ x dL) / L Y = (∫ y dL) / L Z = (∫ z dL) / L
Centroid of line
R. Ganesh Narayanan 107
x
y
h
xy
dy
Find the y-coordinate of centroid of the triangular area
AY = ∫ y dA
½ b h (y) = ∫ y (x dy) = ∫ y [b (h-y) / h] dy = b h2 / 6
b
X / (h-y) = b/h
0
h
0
h
Y = h / 3
R. Ganesh Narayanan 108
Beams
Structural members which offer resistance to bending due to 
applied loads
• Reactions at beam supports are determinate if they involve only three 
unknowns. Otherwise, they are statically indeterminate
R. Ganesh Narayanan 109
External effects in beams
Reaction due to supports, distributed load, concentrated loads
Internal effects in beams
Shear, bending, torsion of beams
v
v
M M
SHEAR BENDING TORSION
R. Ganesh Narayanan 110
Cx
W
D
E
B
C
F
G A
SECTION - J
F
D
J
T
V
M
V – SHEAR FORCE
F – AXIAL FORCE
M – BENDING MOMENT AT J
T
D
Cy
FBE
AX
AY
J
A
A
J
F
V
M
Internal forces in beam
compression
Tension
R. Ganesh Narayanan 111
Shear force and bending moment in beam
To determine bending moment and shearing 
force at any point in a beam subjected to 
concentrated and distributed loads.
1. Determine reactions at supports by 
treating whole beam as free-body
FINDING REACTION FORCES AT A AND BR. Ganesh Narayanan 112
2. SECTION beam at C and draw free-body 
diagrams for AC and CB. By definition, 
positive sense for internal force-couple 
systems are as shown.
DIRECTION OF V AND M
SECTION C
SECTION C SECTION C
+ VE SHEAR FORCE
+VE BENDING MOMENT
V
M
V
M
R. Ganesh Narayanan 113
EVALUATING V AND M
Apply vertical force equilibrium eqn. to AC, shear force at ‘C’, i.e., 
‘V’ can be determined
Apply moment equilibrium eqn. at C, bending moment at ‘C’, i.e., 
‘M’ can be determined; Couple if any should be included
+ ve value of ‘V’ => assigned shear force direction is correct
+ ve value of ‘M’ => assigned bending moment is correct
R. Ganesh Narayanan 114
Evaluate the Variation of shear and bending 
moment along beam
Beer/Johnston
ΣMB= 0 =>RA (-L)+P (L/2) = 0; RA= +P/2
RB = +P/2
SECTION AT C
Between A & D
SECTION AT E
Between D & B
R. Ganesh Narayanan 115
SECTION AT C; C is at ‘x’ distance from A
Member AC: ΣFy = 0 => P/2-V = 0; V = +P/2
ΣMc = 0 => (- P/2) (X) + M = 0; M = +PX/2
Any section between A and D will 
yield same result
V = +P/2 is valid from A to D
V = +P/2 yields straight line from A 
to D (or beam length : 0 to L/2)
M = +PX/2 yield a linear straight line 
fit for beam length from 0 to L/2
R. Ganesh Narayanan 116
SECTION AT E; E is at ‘x’ distance from A
CONSIDER AE:
ΣFy = 0 => P/2-P-V = 0; V = -P/2
ΣME = 0 => (- P/2) (X) +P(X-L/2)+ M = 0; M = +P(L-X)/2
EB CAN ALSO BE CONSIDERED
R. Ganesh Narayanan 117
V = V0 + (NEGATIVE OF THE AREA UNDER THE LOADING 
CURVE FROM X0 TO X) = V0 - ∫w dx
M = M0 + (AREA UNDER SHEAR DIAGRAM FROM X0 TO X) = M0
+ ∫V dx
c1
Slide 117
c1 cclab9, 1/24/2008
R. Ganesh Narayanan 118
Beer/Johnston
:0=∑ AM
( ) ( )( ) ( )( ) 0cm22N400cm6N480cm32 =−−yB
N365=yB:0=∑ BM
( )( ) ( )( ) ( ) 0cm32cm10N400cm26N480 =−+ A
N515=A
:0=∑ xF 0=xB
• The 400 N load at E may be replaced by a 400 N force and 1600 N-cm couple at 
D.
• Taking entire beam as free-body, calculate 
reactions at A and B.
• Determine equivalent internal force-couple 
systems at sections cut within segments AC, 
CD, and DB.
R. Ganesh Narayanan 119
:02 =∑M ( ) 06480515 =+−+− Mxx
( ) cmN 352880 ⋅+= xM
From C to D:
∑ = :0yF 0480515 =−− V
N 35=V
:01 =∑M ( ) 040515 21 =+−− Mxxx
220515 xxM −=
From A to C:
∑ = :0yF 040515 =−− Vx
xV 40515 −=
V = 515 + (-40 X) = 515-40X = 515 - ∫40 dx
M = ∫515-40x dx = 515x-20 x2
0
x
0
x
R. Ganesh Narayanan 120
• Evaluate equivalent internal force-couple systems 
at sections cut within segments AC, CD, and DB.
From D to B:
∑ = :0yF 0400480515 =−−− V
N 365−=V
:02 =∑M
( ) ( ) 01840016006480515 =+−+−−+− Mxxx
( ) cmN 365680,11 ⋅−= xM
R. Ganesh Narayanan 121
Shear force & Bending moment plot
AC: (35X12) + (1/2 x 12 x 480) = 3300
0 to 3300
CD: 3300 +(35X6) = 3510
3300 to 3510
DB: 365 x 14 = 5110
5110 to 0
AREA UNDER SHEAR FORCE DIAGRAM GIVES BM DIAGRAM
R. Ganesh Narayanan 122
300 lb
4 ft 4 2 2
100 lb/ftFind the shear force and 
bending moment for the 
loaded beam
R. Ganesh Narayanan 123
Machine
R. Ganesh Narayanan 124
R. Ganesh Narayanan 125
Friction
•Earlier we assumed action and reaction forces at contacting surfaces 
are normal
• Seen as smooth surface – not practically true
• Normal & tangential forces are important
• Tangential forces generated near contacting surfaces are 
FRICTIONAL FORCES
• Sliding of one contact surface to other – friction occurs and it is 
opposite to the applied force
• Reduce friction in bearings, power screws, gears, aircraft propulsion, 
missiles through the atmosphere, fluid flow etc.
• Maximize friction in brakes, clutches, belt drives etc.
• Friction – dissipated as heat – loss of energy, wear of parts etc.
R. Ganesh Narayanan 126
Friction
Dry friction
(coulomb friction)
Fluid friction
• Occurs when un-lubricated surfaces are 
in contact during sliding
• friction force always oppose the sliding 
motion
• Occurs when the adjacent layers in a 
fluid (liquid, gas) are moving at different 
velocities
• This motion causes friction between 
fluid elements
• Depends on the relative velocity 
between layers
• No relative velocity – no fluid friction
• depends on the viscosity of fluid –
measure of resistance to shearing action 
between the fluid layers
R. Ganesh Narayanan 127
Dry friction: Laws of dry friction
W
N
• W – weight; N – Reaction of the surface
• Only vertical component
• P – applied load
• F – static friction force : resultant of many forces acting over 
the entire contact area
• Because of irregularities in surface & molecular attraction
A
W
P
N
F
A
R. Ganesh Narayanan 128
P
W
N
F
A B
•‘P’ is increased; ‘F’ is also increased and continue to oppose ‘P’
• This happens till maximum ‘Fm’ is reached – Body tend to move till Fm is reached
• After this point, block is in motion
• Block in motion: ‘Fm’ reduced to ‘Fk – lower value – kinetic friction force’ and it 
remains same – related to irregularities interaction
•‘N’ reaches ‘B’ from ‘A’ – Then tipping occurs abt. ‘B’
Fm
Fk
F
p
Equilibrium Motion
More irregularities 
interaction
Less irregularities 
interaction
R. Ganesh Narayanan 129
EXPERIMENTAL EVIDENCE: 
Fm proportional to N
Fm = µs N; µs – static friction co-efficient
Similarly, Fk = µk N; µk – kinetic friction co-efficient
µs and µk depends on the nature of 
surface; not on contact area of 
surface 
µk = 0.75 µs
R. Ganesh Narayanan 130
Four situations can occur when a rigid body is in contact with a
horizontal surface:
We have horizontal and vertical force equilibrium equns. and 
F = µ N
• No motion,
(Px < Fm)
Fm
Fk
F
p
Equilibrium Motion
R. Ganesh Narayanan 131
• No motion • Motion• No friction • Motion impending
It is sometimes convenient to replace normal force N and friction force 
F by their resultant R:
ss
sm
s
N
N
N
F
µφ
µ
φ
=
==
tan
tan
kk
kk
k
N
N
N
F
µφ
µ
φ
=
==
tan
tan
Φs – angle of static 
friction – maximum angle 
(like Fm)
Φk – angle of kinetic 
friction; Φk < Φs
R. Ganesh Narayanan 132
Consider block of weight W resting on board with variable inclination 
angle θ.
Angle of inclination = 
angle of repose; θ= φs
R – Not vertical
ANGLE OF INCLINATION IS INCREASING
R. Ganesh Narayanan 133
Three categories of problems
• All applied forces are given, co-effts. of friction are known
• Find whether the body will remain at rest or slide
• Friction force ‘F’ required to maintain equilibrium is unknown 
(magnitude not equal to µs N)
• Determine ‘F’ required for equilibrium, by solving equilibrium equns; Also find 
‘N’
• Compare ‘F’ obtained with maximum value ‘Fm’ i.e., from Fm = µs N
• ‘F’ is smaller or equal to ‘Fm’, then body is at rest
• Otherwise body starts moving
• Actual friction force magnitude = Fk = µk N
Solution
First category: to know a body slips or not 
R. Ganesh Narayanan 134
A 100 N force acts as shown on a 300 N block 
placed on an inclined plane. The coefficients of 
friction between the block and plane are µs = 0.25 
and µk = 0.20. Determine whether the block is in 
equilibrium and find the value of the friction force.
Beer/Johnston
:0=∑ xF ( ) 0N 300 - N 100 5
3 =− F
N 80−=F
:0=∑ yF ( ) 0N 300 - 5
4 =N
N 240=N
The block will slide down the plane. 
Fm < F
Fm = µs N = 0.25 (240) = 60 N
Θ = 36.9 DEG
Θ = 36.9 
DEG
R. Ganesh Narayanan 135
• If maximum friction force is less than frictionforce 
required for equilibrium, block will slide. Calculate 
kinetic-friction force.
( )N 240200
N
.
FF kkactual
=
== µ
N 48=actualF
Fm
Fk
F
p
Equilibrium Motion
R. Ganesh Narayanan 136
Meriam/Kraige; 6/8
M
30°
Cylinder weight: 30 kg; Dia: 400 mm
Static friction co-efft: 0.30 between cylinder and surface
Calculate the applied CW couple M which cause the cylinder 
to slip
30 x 9.81
NA
FA = 0.3 NANB
FB = 0.3 NB
M
C
ΣFx = 0 = -NA+0.3NB Cos 30-NB Sin 30 = 0
ΣFy = 0 =>-294.3+0.3NA+NBCos 30-0.3NB Sin 30 = 0
Find NA & NB by solving these two equns.
ΣMC = 0 = > 0.3 NA (0.2)+0.3 NB (0.2) - M = 0
Put NA & NB; Find ‘M’
NA = 237 N & NB = 312 N; M = 33 Nm
R. Ganesh Narayanan 137
Meriam/Kraige; 6/5
Wooden block: 1.2 kg; Paint: 9 kg
Determine the magnitude and direction of (1) the friction 
force exerted by roof surface on the wooden block, (2) 
total force exerted by roof surface on the wooden block
Θ = tan-1 (4/12) = 18.43°
Paint
Wooden 
block
12
4
Roof 
surface
Θ
(2) Total force = 10.2 x 9.81 = 100.06 N UP
N
F
10.2x 9.81
X
Y
(1)ΣFx = 0 => -F+100.06 sin 18.43 => F = 31.6 N
ΣFy = 0 => N = 95 N
Second category: Impending relative motion when two or 
three bodies in contact with each other
R. Ganesh Narayanan 138
Beer/Johnston
20 x 9.81 = 196.2 N
N1
F1
T
30 x 9.81 = 294.3 N
F2
For 20 kg block For 30 kg block
F1
P
N1
N2
(a)
R. Ganesh Narayanan 139
(B)
490.5 N
N
P
R. Ganesh Narayanan 140
Beer/Johnston
A
B
6 m
2.5 m
A 6.5-m ladder AB of mass 10 kg leans against a wall as shown. 
Assuming that the coefficient of static friction on µs is the same at 
both surfaces of contact, determine the smallest value of µs for which 
equilibrium can be maintained.
A
B
FB
NB
FA
NA
W
1.25 1.25
O
Slip impends at both A and B, FA= µsNA, FB= µsNB
ΣFx=0=> FA−NB=0, NB=FA=µsNA
ΣFy=0=> NA−W+FB=0, NA+FB=W
NA+µsNB=W; W = NA(1+µs2)
ΣMo = 0 => (6) NB - (2.5) (NA) +(W) (1.25) = 0
6µsNA - 2.5 NA + NA(1+µs2) 1.25 = 0
µs = -2.4 ± 2.6 = > Min µs = 0.2
R. Ganesh Narayanan 141
Wedges
Wedges - simple machines used to raise heavy 
loads like wooden block, stone etc.
Loads can be raised by applying force ‘P’ to 
wedge
Force required to lift block is significantly less 
than block weight
Friction at AC & CD prevents wedge from sliding 
out
Want to find minimum force P to raise block
A – wooden block
C, D – Wedges
R. Ganesh Narayanan 142
0
:0
0
:0
21
21
=+−−
=
=+−
=
∑
∑
NNW
F
NN
F
s
y
s
x
µ
µ
FBD of block
( )
( ) 06sin6cos
:0
0
6sin6cos
:0
32
32
=°−°+−
=
=+
°−°−−
=
∑
∑
s
y
ss
x
NN
F
P
NN
F
µ
µµ
FBD of wedge
N3
6°
F3
6°
R. Ganesh Narayanan 143
Two 8° wedges of negligible weight are used to move 
and position a 530-N block. Knowing that the 
coefficient of static friction is 0.40 at all surfaces of 
contact, determine the magnitude of the force P for 
which motion of the block is impending
Beer/Johnston
Φs = tan−1 µs = tan−1 (0.4) = 21.801°
21.8°
R1
FBD of block
20°
21.8°
530
R2
530
R2
R141.8°
91.8° 46.4° (R2/Sin 41.8) = (530/sin 46.4)
R2 = 487.84 N
Using sine law,
slip impends at wedge/block 
wedge/wedge and block/incline
R. Ganesh Narayanan 144
P = 440.6 N
R. Ganesh Narayanan 145
Beer/Johnston
A 6° steel wedge is driven into the end of an ax handle 
to lock the handle to the ax head. The coefficient of 
static friction between the wedge and the handle is 
0.35. Knowing that a force P of magnitude 60 N was 
required to insert the wedge to the equilibrium position 
shown, determine the magnitude of the forces exerted 
on the handle by the wedge after force P is removed.
P = 60 N Φs = tan −1 µs= tan −1 (0.35 ) = 19.29°
19.29°
3°
19.29°
3°
6°
By symmetry R1= R2; in EQUILIBRIUM
ΣFy = 0: 2R1 sin 22.29° −60 N =0
R1 = R2 = 79.094 N
WHAT WILL HAPPEN IF ‘P’ IS REMOVED ?
R1R2
R. Ganesh Narayanan 146
Vertical component of R1, R2 will be eliminated
Hence, H1 = H2 = 79.094 N cos22.29° = 73.184 N
Final force = 73.184 N
Since included angle is 3°(< φs) from the normal, the 
wedge is self-locking and will remain in place.
• No motion
R. Ganesh Narayanan 147
Screws
Used for fastening, transmitting power or motion, lifting body
Square threaded jack - screw jack V-thread is also 
possible
W- AXIAL LOAD
M – APPLIED MOMENT ABOUT AXIS OF SCREW
M = P X r
L – LEAD DISTANCE – Advancement per revolution
α – HELIX ANGLE
M
Upward 
motion
R. Ganesh Narayanan 148
α
2πr
L
W
φ
α
RP = M/r
One full thread 
of screw
To raise load
M
F
Φ – angle of friction
R
P
w Φ+α
tan (Φ+α) = P/W = M/rW
=> M = rW tan (Φ+α)
α = tan-1 (L/2πr)
To lower load – unwinding condition
φ
α
P = M/r
W
R
α < φ
Screw will remain in place –
self locking
=> M = rW tan (Φ-α)
α = φ In verge of un-winding
Moment required to 
lower the screw
R. Ganesh Narayanan 149
α
P = M/r
W
R
φ
α > φ Screw will unwind itself
=> M = rW tan (α-φ)
Moment required to 
prevent unwinding
R. Ganesh Narayanan 150
Beer/Johnston A clamp is used to hold two pieces of wood together 
as shown. The clamp has a double square thread of 
mean diameter equal to 10 mm with a pitch of 2 mm.
The coefficient of friction between threads is µs = 
0.30. 
If a maximum torque of 40 Nm is applied in 
tightening the clamp, determine (a) the force exerted 
on the pieces of wood, and (b) the torque required to 
loosen the clamp.
Lead distance = 2 x pitch = 2 x 2 = 4 mm
r = 5 mm
( )
30.0tan
1273.0
mm 10
mm22
2
tan
==
===
ss
r
L
µφ
ππ
θ °= 3.7θ
°= 7.16sφ
(double square thread)
R. Ganesh Narayanan 151
a) Forces exerted on the wooded pieces
M/r tan (Φ+α) = W
W = 40 / (0.005) tan (24) = 17.96 kN
b) the torque required to loosen the clamp
M = rW tan (Φ-α) = 0.005 (17.96) tan (9.4)
M = 14.87 Nm
R. Ganesh Narayanan 152
The position of the automobile jack shown is 
controlled by a screw ABC that is single-
threaded at each end (right-handed thread at A, 
left-handed thread at C). Each thread has a pitch 
of 2 mm and a mean diameter of 7.5 mm. If the 
coefficient of static friction is 0.15, determine the 
magnitude of the couple M that must be applied 
to raise the automobile.
Beer/Johnston
FBD joint D:
ΣFy = 0 => 2FADsin25°−4 kN=0
FAD = FCD = 4.73 kN
By symmetry:
4 kN
FAD FCD
25° 25°D
R. Ganesh Narayanan 153
FBD joint A:
4.73 kN
FAC
FAE = 4.73
25°
25°A
ΣFx = 0 => FAC−2(4.73) cos25°=0
FAC = 8.57 kN
L = Pitch = 2 mm
W = FAC = 8.57
φ
θ
R
Joint A
P = M/r
Π (7.5)
θ
Here θ is used instead of α used earlier
MA = rW tan (Φ+α) = (7.5/2) (8.57) tan (13.38) = 7.63 Nm
Similarly, at ‘C’, Mc = 7.63 Nm (by symmetry); Total moment = 7.63 (2) = 15.27 Nm
R. Ganesh Narayanan 154
Journal & Thrust bearing
Journal bearings provide lateral support to rotating shafts
Thrust bearings provide axial support
Journal bearing - Axle friction Thrust bearing - Disc friction
shaft
bearing
shaft
bearing
R. Ganesh Narayanan 155
Friction between two 
ring shaped areas
Friction in full circular area
- DISK FRICTION (Eg., Disc clutch)
Consider Hollow shaft (R1, R2)
M – Moment required for shaft 
rotation at constant speed
P – axial force which maintains 
shaft in contact with bearing
R. Ganesh Narayanan 156
Couple moment required to overcomefriction 
resistance, M
Equilibrium conditions and moment equations are 
necessary to solve problems
R. Ganesh Narayanan 157
A .178 m-diameter buffer weighs 10.1 N. The 
coefficient of kinetic friction between the buffing pad 
and the surface being polished is 0.60. Assuming 
that the normal force per unit area between the pad 
and the surface is uniformly distributed, determine 
the magnitude Q of the horizontal forces required to 
prevent motion of the buffer.
Beer/Johnston
O
M
Q - Q
0.2 m
Σ Mo = 0 => (0.2) Q – M = 0; Q = M / 0.2
M = 2/3 (0.6) (10.1) (0.178/2) = 0.36 Nm 
Q = M / 0.2 = 0.36/0.2 = 1.8 N
R. Ganesh Narayanan 158
Belt friction
Draw free-body diagram for PP’ element of belt
( ) 0
2
cos
2
cos:0 =∆−
∆
−
∆
∆+=∑ NTTTF sx µ
θθ
( ) 0
2
sin
2
sin:0 =
∆
−
∆
∆+−∆=∑
θθ
TTTNFy
dT / T = µS dθ
∫ dT / T = µS ∫ dθ
T1
T2
0
β
ln (T2/T1) = µS β; T2/T1 = e µS β
Consider flat belt, cylindrical 
drum
β – angle of 
contact
R. Ganesh Narayanan 159
α
V- Belt
T2/T1 = e µS β/sin (α/2)
ln (T2/T1) = µS β; T2/T1 = e µS β
Applicable to belts passing over fixed drums; ropes wrapped around a post; belt 
drives
T2 > T1
This formula can be used only if belt, rope are about to slip;
Angle of contact is radians; rope is wrapped ‘n’ times - 2πn rad
In belt drives, pulley with lesser β value slips first, with ‘µS’ remaining same
R. Ganesh Narayanan 160
Beer/Johnston
A flat belt connects pulley A to pulley B. The 
coefficients of friction are µs= 0.25 and µk= 0.20 
between both pulleys and the belt. 
Knowing that the maximum allowable tension in the 
belt is 600 N, determine the largest torque which can 
be exerted by the belt on pulley A.
Since angle of contact is smaller, slippage will occur on pulley B first. Determine 
belt tensions based on pulley B; β = 120 deg = 2π/3 rad
( )
N4.355
1.688
N600
688.1
N600
1
3225.0
11
2 s
==
===
T
e
T
e
T
T πβµ
R. Ganesh Narayanan 161
( )( ) 0N600N4.355mc8:0 =−+=∑ AA MM
mcN8.1956 ⋅=AM
Check for belt not sliping at pulley A:
ln (600/355.4) = µ x 4π/3 => µ = 0.125 < 0.25
R. Ganesh Narayanan 162
A 120-kg block is supported by a rope which is 
wrapped 1.5 - times around a horizontal rod. Knowing 
that the coefficient of static friction between the rope 
and the rod is 0.15, determine the range of values of P 
for which equilibrium is maintained.
Beer/Johnston
P
W = 9.81 X 120 = 
1177.2 N
β= 1.5 turns = 3π rad For impending motion of W up
P = W e µsβ = (1177.2 N) e (0.15)3π
= 4839.7 N
For impending motion of W down
P = W e−µsβ = (1177.2 N) e−(0.15)3π
= 286.3 N
For equilibrium: 286 N ≤ P ≤ 4.84 kN
R. Ganesh Narayanan 163
In the pivoted motor mount shown, the weight W of the 
175-N motor is used to maintain tension in the drive 
belt. Knowing that the coefficient of static friction 
between the flat belt and drums A and B is 0.40, and 
neglecting the weight of platform CD, determine the 
largest couple which can be transmitted to drum B when 
the drive drum A is rotating clockwise.
Beer/Johnston
For impending belt slip: CW rotation β = π radians
Obtain FBD of motor and mount; ΣMD = 0 => find T1 and T2
Obtain FBD of drum at B; MB in CCW; ΣMB = 0; Find MB
T1 = 54.5 N, T2 = 191.5 N
MB=10.27 N.m
R. Ganesh Narayanan 164
Virtual work
We have analyzed equilibrium of a body by isolating it with a FBD 
and equilibrium equations
Class of problems where interconnected members move relative to 
each other; equilibrium equations are not the direct and 
conventional method
Concept of work done by force is more direct => Method of virtual 
work
R. Ganesh Narayanan 165
Work of a force
U = +(F cos α) ∆S (+ ve)
F
A
α
A’
∆S
F
A
α
A’
∆S
U = +F (cos α ∆S)
Work done ‘U’ by the force F on the body during 
displacement is the compt. Of force in the 
displacement direction times the displacement
Work is a scalar quantity as we get same result regardless 
of direction in which we resolve vectors
F
A
α
A’
∆S U = -(F cos α) ∆S 
U = 0 if ∆S = 0 and α = 90 deg
R. Ganesh Narayanan 166
F
A
A’
drA1
A2
Work done by force F during 
displacement dr is given by, dU = F.dr
dU = (Fx i + Fy j + Fz k).(dx i + dy j + dz k)
= Fx dx + Fy dy + Fz dz
U = ∫ F.dr = ∫ Fx dx + Fy dy + Fz dz
We should know relation between the force and their coordinates
Work of a couple
dθ
M
dU = M dθ
U = ∫M d θ
F
-F
Moment can be taken 
instead of forces
R. Ganesh Narayanan 167
Forces which do no work
• ds = 0; cos α = 0
• reaction at a frictionless pin due to rotation of a body around the 
pin 
• reaction at a frictionless surface due to motion of a body along the 
surface
• weight of a body with cg moving horizontally
• friction force on a wheel moving without slipping
Only work done by applied forces, loads, friction forces need to
be considered
R. Ganesh Narayanan 168
Sum of work done by several forces may be zero
• bodies connected by a frictionless pin
=> W.D by F and –F is opposite and will cancel
• bodies connected by an inextensible cord
• internal forces holding together particles of a rigid 
body
Rigid body
A, B – particles
F, -F are acting as shown
Though dr, dr’ are different, components of these 
displacements along AB must be equal, otherwise 
distance between the particles will change and this is 
not a rigid body; so U done by F and –F cancel each 
other, i.e, U of internal forces = 0
R. Ganesh Narayanan 169
Principle of virtual work
Imagine the small virtual displacement of particle which is 
acted upon by several forces F1, F2, ….. Fn
Imagine the small displacement A to A’
This is possible displacement, but will not occur
AA’ ---- VIRTUAL DISPLACEMENT, δr (not dr)
Work done by these forces F1, F2, ….Fn during virtual 
displacement δr is called VIRTUAL WORK, δU
δU = F1. δr + F2. δr + …..+ Fn. δr = R . δr 
Total virtual work of the 
forces
Virtual work of 
the resultant
R. Ganesh Narayanan 170
Principle of virtual work for particle
Principle of virtual work for rigid body
Principle of virtual work for system of interconnected rigid bodies
Work of internal forces is zero (proved earlier)
R. Ganesh Narayanan 171
Applications of Principle of virtual work
Mainly applicable to the solutions of problems involving machines or mechanisms 
consisting of several interconnected rigid bodies
Wish to determine the force of the vice on the block for a given force 
P assuming no friction
Virtual displacement ‘δθ’ is given; This results in δxB and –δyc. 
Here no work is done by Ax, Ay at A and N at B
TOGGLE VISE
R. Ganesh Narayanan 172
δUQ = -Q δxB ; δUP = -P δyc
In this problem, we have eliminated all un-known reactions, while 
ΣMA = 0 would have eliminated only TWO unknowns
The same problem can be used to find ‘θ’ for which the 
linkage is in equilibrium under two forces P and Q
Output work = Input work
R. Ganesh Narayanan 173
Real machines
For an ideal machine without friction, the output work is equal to the input 
work; 2Ql cos θ δθ = Pl sin θ δθ
In real machine, output work < input work => because of presence of 
friction forces
( )µθ
θδθµθδθθδθ
δδδδ
−=
−+−=
=−−−=
tan
cossincos20
0
2
1 PQ
PlPlQl
xFyPxQU BCB
Output work = Input work –
friction force work
Q = 0 if tan θ = µ => θ = φ, angle of friction
R. Ganesh Narayanan 174
Mechanical efficiency
Mechanical efficiency of m/c, η = Output work / Input work
For toggle vise, η = 2Ql cos θ δθ / Pl sin θ δθ
Substituting Q = ½ P (tan θ – µ) here
η = 1 – µ cot θ
Inthe absence of friction forces, µ = 0 and hence η = 1 => Ideal m/c
For real m/c, η < 1
R. Ganesh Narayanan 175
Beer/Johnston
Determine the magnitude of the couple M 
required to maintain the equilibrium of the 
mechanism.
Virtual displacement = δθ, δxD, Work done by Ex, 
Ey, A is zero
By virtual work principle,
δU = δUM + δUp = 0
M δθ + P δxD = 0
δxD = -3l sin θ δθ can be obtained from 
geometry
M = 3Pl sin θ
R. Ganesh Narayanan 176
Beer/Johnston
A hydraulic lift table consisting of two identical 
linkages and hydraulic cylinders is used to raise a 
1000-kg crate. Members EDB and CG are each 
of length 2a and member AD is pinned to the 
midpoint of EDB. 
Determine the force exerted by each cylinder in 
raising the crate for θ = 60o, a = 0.70 m, and L = 
3.20 m.
Work done is zero for Ex, Ey, Fcg; Work done by W, FDH will be considered
R. Ganesh Narayanan 177
W, δy are in opposite 
direction, (-)sign will come
FDH, δs are in same direction, 
(+) sign will come 
1)
2) Express δy, δs in terms of δθ
δy = 2a cos δθ; δs = (aL sinθ/s) δθ
---- (1)
Substituting δy & δs in (1) gives,
-(1/2) W (2a cos δθ) + (FDH) (aL sinθ/s) δθ = 0
FDH = W (s/L) cot θ
3) Apply numerical data
FDH = (1000 X 9.81) (2.91/3.2) cot 60 = 5.15 kN ‘S’ is obtained from this triangle
R. Ganesh Narayanan 178
The mechanism shown is acted upon by the force P. 
Derive an expression for the magnitude of the force Q 
required for equilibrium.
Beer/Johnston
YF
W.D. by Ay, Bx, By will be zero
δ U = 0 => +Q (δXA) - P (δYF)
Find δXA and δYF in terms of δθ
(Check calculation of δXA and δYF)
δ U = Q(2l cosθ δθ) - P(3l sinθ δθ) = 0
Q Bx
By
P
Ay
δYf
XAδXA
δθ
x
y
R. Ganesh Narayanan 179
Work of force using finite displacement
Work of force F corresponding to infinitesimal displacement, 
dr = dU = F. dr
Work of F corresponding to a finite displacement of particle 
from A1 to A2 and covering distances S1, S2,
U1-2 = ∫ F . dr or U1-2 = ∫ (F cosα) ds = F (S2-S1)
A2
A1
S2
S1
S1, S2 – distance along the path traveled by the particle
Area under curve = U1-2
Similarly, work of a couple of moment M, dU = M dθ
U1-2 = ∫M dθ = M (θ2-θ1)
θ2
θ1
R. Ganesh Narayanan 180
Work of a weight
yW
WyWy
WdyU
WdydU
y
y
∆−=
−=
−=
−=
∫→
21
21
2
1
Work is equal to product of W and 
vertical displacement of CG of body; 
Body moves upwards; Body moving 
downwards will have +ve work done
( )
2
22
12
12
1
21
2
1
kxkx
dxkxU
dxkxFdxdU
x
x
−=
−=
−=−=
∫→
Work of a spring
F = k x
k – spring 
constant, N/m
+ve work done is expected if x2 < x1, i.e., 
when spring is returning to its un-deformed 
position
( ) xFFU ∆212
1
21 +−=→
R. Ganesh Narayanan 181
Potential Energy
Work of a weight: 2121 WyWyU −=→
The work is independent of path and depends only on 
positions (A1, A2) or Wy
potential energy of the body with 
respect to the force of gravity W
r== gVWy
( ) ( )
2121 gg
VVU −=→
Vg1 < Vg2 => PE is increasing with displacement in this 
case, work done is negative
Work is positive, if PE decreases
Unit of PE – Joule (J)
R. Ganesh Narayanan 182
Work of a spring
( ) ( )
=
−=
−=→
e
ee
V
VV
kxkxU
21
2
22
12
12
1
21
potential energy of the body with 
respect to the elastic force F
r
Here PE increases, work done is (-ve)
Now in general, it is possible to find a function V, called potential energy, such 
that, dU = -dV
U1-2 = V1 – V2 => A force which satisfies this eqn. is conservative force
Work is independent of path & negative of change in PE for the 
cases seen
R. Ganesh Narayanan 183
Potential energy & equilibrium
Considering virtual displacement, δU = -δV = 0
=> (dV / dθ) = 0 => position of the variable defined by single independent 
variable, θ
In terms of potential energy, the virtual work principle states that if a system is in 
equilibrium, the derivative of its total potential energy is zero
(δV/δθ) = 0
Example:
Initial spring length = AD
Work is done only by W, F
For equilibrium, δU = (δVe + δVg)
1
2
R. Ganesh Narayanan 184
Total potential energy of the system, V = Vg + Ve
For ‘W’ For ‘F’
dv/dθ = 4kl2sinθ cosθ – Wl sinθ = 0
Θ = 0 and θ = cos-1 (W/4kl)
R. Ganesh Narayanan 185
PO
C
k
B
A
θ
b
b
b
b
Two uniform links of mass, m are connected as 
shown. As the angle θ increases with P applied in 
the direction shown, the light rod connected at A, 
passes thro’ pivoted collar B, compresses the 
spring (k). If the uncompressed position of the 
spring is at θ = 0, find the force which will 
produce equilibrium at the angle θ
Compression distance of spring, x = movement of ‘A’ away from B; X = 2b sin θ/2
δU = (δVe + δVg)
Find Ve, δVe; Vg, δVg; δU (of P)
Vg = 0
Ve = ½ k x2; Vg = mgh
δU = P δ (4b sin θ/2)
2Pb cos θ/2 δθ = 2kb2sin θ/2 cos θ/2 δθ + mgb sin θ/2 δθ
P = kb sin θ/2 + ½ mg tan θ/2
4bsinθ/2
R. Ganesh Narayanan 186
Meriam/Kraige, 7/39
For the device shown the spring would be un-stretched 
in the position θ=0. Specify the stiffness k of the spring 
which will establish an equilibrium position θ in the 
vertical plane. The mass of links are negligible.
Spring stretch distance, x = 2b-2b cos θ
Ve = ½ k [(2b)(1-cos θ)]2 = 2kb2 (1-cos θ)2
Vg = -mgy = -mg (2bsinθ) = -2mgbsin θ
V = 2kb2 (1-cos θ)2 - 2mgbsin θ
For equilibrium, dv / dθ = 4kb2(1-cos θ) sin θ - 2mgb cosθ = 0
=> K = (mg/2b) (cot θ/1-cos θ)
k
b b
b
θ
m
y
Vg = 0
R. Ganesh Narayanan 187
0=
θd
dV
0=
θd
dV
d2V / dθ2 < 0d2V / dθ2 > 0
Must examine higher 
order derivatives and are 
zero
AB AB
Stability of equilibrium (one DOF ‘θ’)
R. Ganesh Narayanan 188
Knowing that the spring BC is un-stretched when θ = 0, 
determine the position or positions of equilibrium and state 
whether the equilibrium is stable, unstable, or neutral.
Beer/Johnston
( ) ( )θθ cos2
2
1
2
2
1
bmgak
mgyks
VVV ge
+=
+=
+=
( )( )
( )( )( )
θ
θθθ
θθ
θ
8699.0
m3.0sm81.9kg10
m08.0mkN4
sin
sin0
2
22
2
=
==
−==
mgb
ka
mgbka
d
dV
°=== 7.51rad902.00 θθ
R. Ganesh Narayanan 189
( )( ) ( )( )( )
θ
θ
θ
θ
cos43.296.25
cosm3.0sm81.9kg10m08.0mkN4
cos
22
2
2
2
−=
−=
−= mgbka
d
Vd
at θ = 0: 083.3
2
2
<−=
θd
Vd
unstable
at θ = 51.7o:
036.7
2
2
>+=
θd
Vd stable
R. Ganesh Narayanan 190
Spring AB of constant 2 kN/m is attached to two 
identical drums as shown.
Knowing that the spring is un-stretched when θ = 
0, determine (a) the range of values of the mass 
m of the block for which a position of equilibrium 
exists, (b) the range of values of θ for which the 
equilibrium is stable.
Beer/Johnston (10.81)
A
B
A
B
R. Ganesh Narayanan 191
(Sin θ varies from 0 to 1)
(Cos θ varies from 1 to 0)
R. Ganesh Narayanan 192
XV = (∫ xc dv) YV = (∫ yc dv) ZV = (∫ zc dv)
Centroid of volume:
XA = (∫ xc dA) YA = (∫ yc dA) ZA = (∫ zc dA)
Centroid of area:
Moments of inertia : The moment of inertia of an object about a given axis 
describes how difficult it is to change its angular motion about that axis. 
First moment of volume
w.r.t. ‘yz’ plane
Symmetry plane
Centroid of 
volume ∫ xc dv = 0
R. Ganesh Narayanan 193
Consider a beam subjected to pure bending. 
Internal forces vary linearly with distance from 
the neutral axis which passes through the section 
centroid.
X-axis => neutral axis => centroid of section 
passes
∆F = k y ∆A vary linearlywith distance ‘y’
moment second 
momentfirst 0
22 ==
====
∆=∆
∫∫
∫∫
dAydAykM
QdAydAykR
AkyF
x
r
∆MX = y ∆F = k y2 ∆A;
Moment of inertia of beam 
section w.r.t x-axis, IX (+VE)
R. Ganesh Narayanan 194
Second moments or moments of inertia of an area with 
respect to the x and y axes,
∫∫ == dAxIdAyI yx
22
For a rectangular area,
3
3
1
0
22 bhbdyydAyI
h
x === ∫∫
IY = ∫ x2 dA = ∫ x2 h dx = 1/3 b3h
0
b
Rectangular moment of inertia
R. Ganesh Narayanan 195
The polar moment of inertia is an important parameter in problems involving 
torsion of cylindrical shafts and rotations of slabs.
∫= dArJ
2
0
The polar moment of inertia is related to the rectangular moments of 
inertia,
( )
xy II
dAydAxdAyxdArJ
+=
+=+== ∫∫∫∫
22222
0
Polar moment of inertia
R. Ganesh Narayanan 196
• Consider area A with moment of inertia Ix. Imagine that 
the area is concentrated in a thin strip parallel to the x axis 
with equivalent Ix.
A
I
kAkI xxxx ==
2
kx = radius of gyration with respect to the x axis
A
J
kAkJ
A
I
kAkI
O
OOO
y
yyy
==
==
2
2
222
yxO kkk +=
Radius of gyration
R. Ganesh Narayanan 197
Beer/Johnston
( )
h
hh
x
yy
h
h
b
dyyhy
h
b
dy
h
yh
bydAyI
0
43
0
32
0
22
43






−=
−=
−
== ∫∫∫
12
3bh
I x=
Determine the moment of inertia of a 
triangle with respect to its base.
For similar triangles,
dy
h
yh
bdA
h
yh
bl
h
yh
b
l −
=
−
=
−
=
dA = l dy
Determination of MI by area of integration
R. Ganesh Narayanan 198
yMI of rectangular area:
dA = bdy
x
b
h
y
dy
Ix = ∫ y2 dA = ∫ y2 bdy = 1/3 bh3; Iy = 1/3 hb3
0
h
MI - Ix and Iy for elemental strip:
y dIx = 1/3 dx (y3) = 1/3 y3 dx
dIy = x2dA = x2y dx or 1/3 x3dy
x
Y
X
dA = Ydx From this, MI of whole area can be 
calculated by integration
dx
dy
About centroidal axis (X, Y): Ix = 1/12 bh3; Iy = 1/12 b3h
X
Y
R. Ganesh Narayanan 199
y
x
a
b
y = k x5/2
Find MI w.r.t Y axis
Beer/Johnston (9.1)
R. Ganesh Narayanan 200
Triangle: bh3/12 (about base)
Circular area: π/4 r4 (about dia)
Rectangular area: bh3/3 (about base)
R. Ganesh Narayanan 201
Parallel axis theorem
• Consider moment of inertia I of an area A with respect 
to the axis AA’
∫= dAyI
2
• The axis BB’ passes through the area centroid and 
is called a centroidal axis.
( )
∫∫∫
∫∫
+′+′=
+′==
dAddAyddAy
dAdydAyI
22
22
2
2AdII +=
dA
A A’
y
CB B’
d
y’
C – Centroid
BB’ – Centroidal axis
MI of area with 
centroidal axis
0
Parallel axis theorem
Jo = Jc + Ad2
First moment of 
area
R. Ganesh Narayanan 202
Moments of Inertia of Composite Areas
The moment of inertia of a composite area A about a given axis is 
obtained by adding the moments of inertia of the component areas A1, 
A2, A3, ... , with respect to the same axis.
x
y
It should be noted that the radius of gyration of a composite area is not 
equal to sum of radii of gyration of the component areas
R. Ganesh Narayanan 203
MI of some common geometric shapes
R. Ganesh Narayanan 204
Moment of inertia IT of a circular area with respect to a 
tangent to the circle T,
( )
4
4
5
224
4
12
r
rrrAdIIT
π
ππ
=
+=+=
Application 1:
Application 2: Moment of inertia of a triangle with respect to a 
centroidal axis,
( )
3
36
1
2
3
1
2
13
12
12
2
bh
hbhbhAdII
AdII
AABB
BBAA
=
−=−=
+=
′′
′′
IDD’ = IBB
’ + ad’2 = 1/36 bh3 + 1/2bh (2/3h)2 = ¼ bh3
R. Ganesh Narayanan 205
Find the centroid of the area of the un-equal Z section. Find the 
moment of inertia of area about the centroidal axes
shames
Ai xi yi Aixi Aiyi
2x1=2 1 7.5 2 15
8x1=8 2.5 4 20 32
4x1=4 5 0.5 20 2
ΣAi = 14 ΣAixi = 42 ΣAiyi = 49
1
2
3
Xc = 42/14 = 3 in.; Yc = 49/14 = 3.5in
y
x
1
6
2 1 4
1
1
2
3
Xc, Yc
Ixcxc = [(1/12)(2)(13)+(2)(42)] + [(1/12)(1)(83)+(8)(1/2)2] + 
[(1/12)(4)(13)+(4)(32)] = 113.16 in4
Similarly, Iycyc = 32.67 in4
R. Ganesh Narayanan 206
Beer/Johnston:
Determine the moment of inertia of the shaded area 
with respect to the x axis.
Rectangle:
( )( ) 46
3
13
3
1 mm102.138120240 ×=== bhIx
3
Half-circle:
moment of inertia with respect to AA’,
( ) 464
8
14
8
1 mm1076.2590 ×===′ ππrI AA
R. Ganesh Narayanan 207
( )( )
( )
23
2
2
12
2
1
mm1072.12
90
mm 81.8a-120b
mm 2.38
3
904
3
4
×=
==
==
===
ππ
ππ
rA
r
a
moment of inertia with respect to x’,
( )( )
46
362
mm1020.7
1072.121076.25
×=
××=−= ′′ AaII AAx =25.76x106 – (12.72x103) (38.2)2
moment of inertia with respect to x,
( )( )
46
2362
mm103.92
8.811072.121020.7
×=
×+×=+= ′ AbII xx
46mm109.45 ×=xI
xI = 46mm102.138 × − 46mm103.92 ×
R. Ganesh Narayanan 208
Product of inertia, Ixy
∫= dAxyI xy
[Similar to Ixx (or Ix), Iyy (or Iy)]
When the x axis, the y axis, or both are an axis of symmetry, 
the product of inertia is zero.
The contributions to Ixy of dA and dA’ will cancel out
Parallel axis theorem for products of inertia:
AyxII xyxy +=
Centroid ‘C’ is defined by x, y
R. Ganesh Narayanan 209
Moment of inertia, Product of inertia about rotated axes
Given
∫
∫∫
=
==
dAxyI
dAxIdAyI
xy
yx
22
we wish to determine moments and product of 
inertia with respect to new axes x’ and y’
x, y rotated to x’, y’
θθ
θθ
θθ
2cos2sin
2
2sin2cos
22
2sin2cos
22
xy
yx
yx
xy
yxyx
y
xy
yxyx
x
I
II
I
I
IIII
I
I
IIII
I
+
−
=
+
−
−
+
=
−
−
+
+
=
′′
′
′
The change of axes yields
Ix’+Iy’ = Ix+Iy
R. Ganesh Narayanan 210
y
xa
θ
Imax
Imin
•Assume Ixx, Iyy, Ixy are known for the reference axes x, y
•At what angle of θ, we have maximum and minimum ‘I’
•Minimum angle will be at right angles to maximum angle
•These axes are called Principal axes & MI are Principal MI
Principal axes & Principal MI
Imax, min = (Ix+Iy/2) ± (Ix-Iy/2)2 + Ixy2
tan 2θ = 2Ixy / (Iy-Ix)
R. Ganesh Narayanan 211
For the section shown, the moments of inertia with 
respect to the x and y axes are Ix = 10.38 cm
4 and Iy = 
6.97 cm4.
Determine (a) the orientation of the principal axes of the 
section about O, and (b) the values of the principal 
moments of inertia about O.
Apply the parallel axis theorem to each rectangle,
( )∑ += ′′ AyxII yxxy
Note that the product of inertia with respect to centroidal axes parallel to the xy
axes is zero for each rectangle.
56.6
28.375.125.15.1
0005.1
28.375.125.15.1
cm,cm,cm,cmArea,Rectangle
42
−=
−−+
−+−
∑ Ayx
III
II
I
Ayxyx
R. Ganesh Narayanan 212
( )
°°=
+=
−
−
−=
−
−=
255.4and4.752
85.3
97.638.10
56.622
2tan
m
yx
xy
m
II
I
θ
θ
°=°= 7.127and7.37 mm θθ
( )2
2
2
2
minmax,
56.6
2
97.638.10
2
97.638.10
22
−+




 −±
+
=
+




 −
±
+
= xy
yxyx
I
IIII
I
4
min
4
max
cm897.1
cm45.15
==
==
II
II
b
a