Mechanics full notes

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

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

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

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

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…

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

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

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

Two concurrent forces F1, F2
Rx = Σ Fx; Ry = Σ Fy
DERIVATION
F2
F1
R
i
j

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

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

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:

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

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

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

M
-F +Fd
M
-F
+F
d
M
-F
+F d
-2F
d/2+2F
M
EXAMPLE
All four are equivalent couples

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

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

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)

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 ?

Principle of moments
Summarize the above process: R = ΣF
Mo = ΣM = Σ(Fd)
Mo = Rd
First 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

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

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

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’

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 …

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

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

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

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

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

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

80N
30 deg
60 N
40 N
50 N2m
5m
45
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 deg
237 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
LOA of R with x-axis:

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

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 =

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

Modeling the action of forces
Meriam/Kraige
Imp
Imp

FBD - Examples
Meriam/Kraige
Equilibrium equns. Can be
solved,
• Some forces can be zero
• Assumed sign can be
different

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

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

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

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

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

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

Merial / Kraige; 2/117
Replace the 750N tensile force which the cable exerts on point B by a force-
couple system at point O

F = f λ, where λ 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

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

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

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

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

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

Modeling forces in 3D

Types of 3D equilibrium

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

Meriem / Kraige; 3/64

I.H. Shames
Find forces at A, B, D. Pin connection at C; E has welded connection

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

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

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

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

•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

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

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

•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 elements
B 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

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

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

Example
A
B
C
D
EF
L
ΣFy = 0; ΣFx = 0
Finally sign can be changed if
not applied correctly

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

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
1414
FBD
ΣFx = 0 => -FBD + 1414COS45 = 0 => FBD = 1000 N
ΣFy = 0 => -FBC+1414 COS45 = 0 => FBC = 1000 N
1000
FAC
FAB
1414
FBC
FBD

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

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)

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

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

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

Methodology for method of joints
A
B
C
D
EF
L
ΣFy = 0; ΣFx = 0
Finally sign can be changed if
not applied correctly

5B D
A
C E
30 20
5 5
5 55 5
kN, m

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

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

• 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

• 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

Section AA and BB are
possible
convenient

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

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

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

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

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

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

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

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

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

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

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

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

AFAE
BD
Full truss
K, J are un-necessary
here

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

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

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

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

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

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

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

Center of mass & center of gravity
A B
C
W
G
A
B
C
W
G
W
G
AB
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

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 = (∫ 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

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

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

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

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

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

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 B

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

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

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

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

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

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

c1 cclab9, 1/24/2008

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.

:02 =∑ M ( ) 06480515 =+−+− Mxx
( ) cmN352880 ⋅+= xM
From C to D:
∑ = :0yF 0480515 =−− V
N35=V
:01 =∑ M ( ) 040515 2
1 =+−− Mxxx
2
20515 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

• Evaluate equivalent internal force-couple systems
at sections cut within segments AC, CD, and DB.
From D to B:
∑ = :0yF 0400480515 =−−− V
N365−=V
:02 =∑ M
( ) ( ) 01840016006480515 =+−+−−+− Mxxx
( ) cmN365680,11 ⋅−= xM

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

300 lb
4 ft 4 2 2
100 lb/ftFind the shear force and
bending moment for the
loaded beam

Machine

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.

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

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

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

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

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

• 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

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

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

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 ( ) 0N300-N100 5
3 =− F
N80−=F
:0=∑ yF ( ) 0N300- 5
4 =N
N240=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

• If maximum friction force is less than friction force
required for equilibrium, block will slide. Calculate
kinetic-friction force.
( )N240200
N
.
FF kkactual
=
== µ
N48=actualF
Fm
Fk
F
p
Equilibrium Motion

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

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

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)

(B)
490.5 N
N
P

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+µs
2)
ΣMo = 0 => (6) NB - (2.5) (NA) +(W) (1.25) = 0
6µsNA - 2.5 NA + NA(1+µs
2) 1.25 = 0
µs = -2.4 ± 2.6 = > Min µs = 0.2

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

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°

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

P = 440.6 N

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

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

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

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

α
P = M/r
W
R
φ
α > φ Screw will unwind itself
=> M = rW tan (α-φ)
Moment required to
prevent unwinding

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
mm10
mm22
2
tan
==
===
ss
r
L
µφ
ππ
θ °= 3.7θ
°= 7.16sφ
(double square thread)

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

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

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

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

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

Couple moment required to overcome friction
resistance, M
Equilibrium conditions and moment equations are
necessary to solve problems

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

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

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

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

( )( ) 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

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

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

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

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

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

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

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

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

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)

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

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

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

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
In the absence of friction forces, µ = 0 and hence η = 1 => Ideal m/c
For real m/c, η < 1

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

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

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

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

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

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 +−=→

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)

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

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

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)

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

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

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’)

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

( )( ) ( )( )( )
θ
θ
θ
θ
cos43.296.25
cosm3.0sm81.9kg10m08.0mkN4
cos
22
2
2
2
−=
−=
−= mgbka
d
Vd
at θ = 0: 083.32
2
<−=
θd
Vd
unstable
at θ = 51.7o:
036.72
2
>+=
θd
Vd stable

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

(Sin θ varies from 0 to 1)
(Cos θ varies from 1 to 0)

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

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 linearly with distance ‘y’
momentsecond
momentfirst0
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)

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

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

• 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 x
xxx == 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

Beer/Johnston
( )
h
hh
x
yy
h
h
b
dyyhy
h
b
dy
h
yh
bydAyI
0
43
0
32
0
22
43






−=
−=
−
== ∫∫∫
12
3
bh
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

y
MI 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

y
x
a
b
y = k x5/2
Find MI w.r.t Y axis
Beer/Johnston (9.1)

Triangle: bh3/12 (about base)
Circular area: π/4 r4 (about dia)
Rectangular area: bh3/3 (about base)

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
2
AdII +=
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

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

MI of some common geometric shapes

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

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

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

( )( )
( )
23
2
2
12
2
1
mm1072.12
90
mm81.8a-120b
mm2.38
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
46
mm109.45 ×=xI
xI = 46
mm102.138 × − 46
mm103.92 ×

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

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

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)

For the section shown, the moments of inertia with
respect to the x and y axes are Ix = 10.38 cm4 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

( )
°°=
+=
−
−
−=
−
−=
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

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