Module 4

University of Engineering & Technology Peshawar
CE-117: Engineering Mechanics
MODULE 4:
Moment and Couple
Prof. Dr. Mohammad Javed & Engr. Mudassir Iqbal
mjaved@uetpeshawar.edu.pk mudassiriqbal@uetpeshawar.edu.pk
1

Lecture Objectives
• How to calculate Moment by using scalar and vector methods.
• How to calculate Couple.
• How to resolve Couple into Force-Couple system or calculate the
resultant of Force-Couple system.

Moment of a force
Moment is also referred to as Torque and is the turning effect of a
force.

Moment about a point
• The magnitude of the Moment of the force F to
rotate the body about the axis O-O perpendicular
to the plane of the body is equal to the product of
the magnitude of the force F and to the moment
arm, d (perpendicular distance from the axis to the
line of action of the force).
Thus, the magnitude of the moment about A
(moment centre ) is:

• The right-hand rule is used to identify
the sense of Moment M (vector
quantity) .
• We represent the moment of F about O-
O as a vector pointing in the direction of
the thumb, with the fingers curled in the
direction of the rotational tendency.
Direction of Moment

• Moment directions may be accounted for by
using a stated sign convention, such as a plus
sign (+) for counter clockwise moments and
a minus sign (-) for clockwise moments, or
vice versa.
• For the sign convention of given Figure, the
moment of F about point A (or about the z-
axis passing through point A) is positive.
Sign convention for Moment

Varignon’sTheorem
One of the most useful principles of mechanics is
Varignon’s theorem (sometime known as
Principle of Moment), which states that The
sum of moment of all forces about a point is equal to
the moment of their resultant about the same point.

Proof:
Varignon’sTheorem
θ
θα
β

Proof (contd.):
Varignon’sTheorem
M

Problem 4.1
In order to raise the lamp post from the
position shown, force F is applied to the
cable. If F= 200 lb determine the
moment produced by F about point A.
Solve problem by 4 scalar methods
Ans: 1573 ft.lb (CCW)
x
y

Problem 4.1: Method 1
1. Calculate perpendicular distance
of A from line of action of F
x
y
90o
d = ?

x
y
Fx
Fy
dy
dx
1. Shift F to point B (Principle of
Transmissibility)
2. Calculate perpendicular distances
of point A from line of actions of
Fx and Fy
3. Calculate Moment by using
Varignon’s theorem

x
y
Fx
Fy
iy
1. Shift F to intersection point with
y-axis (Principle ofTransmissibility)
2. Calculate perpendicular distance of
A from line of action of Fx

1. Shift F to intersection point with
x-axis (Principle ofTransmissibility)
2. Calculate perpendicular distance of
A from line of action of Fy
ix = 10 ft
c

• The moment of F about point A of Figure may be
represented by the cross-product expression
where r is a position vector which runs from the
moment reference point A to any point on the line of
action of F. The magnitude of this expression is given by
Vector method of determining Moment

Problem 4.2
• Solve Problem 4.1 byVector approach
Soln:
1. Express Force F in term of unit
vectors i and j
2. Express the coordinates of Point B
(point of application of force) in term
of i and j , taking point A (moment
centre) as origin.
3. MA = rA x F
x
y

• Determine the angle ,θ, of the force F so that it
produces a maximum moment and a minimum
moment about point A. Also, what are the
magnitudes of these maximum and minimum
moments?
• Ans:
θ = 116.57o for minimum moment about A
θ = 26.57o for maximum moment 40.2 kN.m
about A
Problem 4.3

Ans: θ =33.6o Ans: 118 in.lb and 140 in.lb
both in C.W. directions
11 12
Exercise 4.1

Home Assignment 4.1
• SECTION B: Problems 1,7,9
• SECTION D: Problems 2,6,11

Couple
• A Couple is defined as two parallel forces that
have the same magnitude but opposite
directions and are separated by a perpendicular
distance d
• Since the resultant force is zero, the only effect
of a couple is to produce a rotation or
tendency of rotation in a specified direction
• The moment produced by a couple is called a
Couple Moment

Characteristics of a Couple
1. The algebraic sum of magnitude of the member forces forming a
Couple is always zero.
2. The Couple can be balanced by a Couple only, which has same moment
and opposite sense.
3. Moment of a Couple (i.e. Couple Moment) is fixed and it does not
depend on Moment Centre

A
B
It can be shown that Mo= MA= MB
(explanation on black board). Thus Couple
Moment is a Free vector as its magnitude
is not effected by Moment Centre
Characteristics of a Couple

• Scalar Formulation. The moment of a
couple is defined as having a magnitude of
• Vector Formulation. The moment of
a couple can also be expressed by the
vector cross product
Magnitude of a Moment Couple

Equivalent Couples.
If two couples produce a moment with the same magnitude and direction
then these two couples are equivalent
Both couples are equivalent as M1= 30*0.4 =12 N.m , M2 = 40*0.3= 12 N.m

Resolving a Force into an equivalent Force–Couple System
single Force, F Equivalent Force- Couple
system (i.e. F and M)

Resolving a Force into an equivalent Force–Couple System
single Force, F
Equivalent Force- Couple
system (i.e. F and m)

Problem 4.4
Replace the force system acting
on the truss by a resultant force
and couple moment at point C.

Problem 4.4
 θ=Tan -1 (3/4)= 36.87 o
 Rx = ΣFx= 500 Cos 36.87o = 400 lb= 400 lb →
 Ry= ΣFy= -500 Sin 36.87o-200-150-100
= - 750 lb = 750 lb ↓
 R= √(400)2+(750)2 = 850 lb
 It can be calculated that angle of resultant
= 61.9o with +ve x-axis in C.W direction
 Mc= ΣMc= 200*2+150*4+100*6+500
Sin(36.87o)*8+500 Cos (36.87o)*6
= 6400 lb.ft = 6400 lb.ft C.W
500 lb
500 cos θ
500 sin θ
500 sin θ
200
150
100
500 sin θ
150
200
100
500 cos θ
θ
500 cos θ
M = 6400 lb.ft
Rx= 400 lb
Ry= 750 lb
61.9o
R= 850 lb

Ans: MR = 9.69 kN.m (C .W) in both
the cases
Ans: F = 167 lb. Resultant
couple act any where
Exercise 4.2
1
2

Ans: R= 29.9 lb, θ =78.4o with x-axis in A.C.W
direction, Mo= 214 lb.in (A.C.W)
Ans: R= 50.2 kN, θ =84.3o with -ve x-axis in A.C.W
direction, MA= 239 kN.m (C.W)
5
6
Exercise 4.2

Home Assignment 4.3
• SECTION B: Problems 1,4,6
• SECTION D: Problems 2,3,5

By reversing the process, employed
in resolving a force into force-couple
system, we can find the resultant
force of a given force-couple system
Resultant force of Force–Couple System
A
FM
A
FM
F
F
d= M/F
B
M
B
MA
d= M/F
B
A
d= M/F
Step 1 Step 2 Step 3
F F

The device shown is a part of an automobile seat back-release
mechanism. The part is subjected to the 4-N force exerted at A and
a restoring moment exerted by a hidden torsional spring. Determine
the y-intercept of the line of action of the single equivalent force
Problem 4.5
Ans: iy = 40.36 mm along –ve y-xis

Module 4

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