Department of Civil Engineering, UET Peshawar
CE-117: Engineering Mechanics, Spring 2025
MODULE 4: Moment and Couple
Course Instructor: Prof. Muhammad Javed (mjaved@uetpeshawar.edu.pk)
Lecture Objectives
At the end of this lecture, student should be able to learn
1. How to calculate Moment by using scalar and vector methods.
2. How to calculate Couple Moment
3. How to resolve (Replace) a force into an Equivalent Force-Couple system at
specified point.
4. How to
replace a Force-Couple system with a single Equivalent force
(Resultant force of a Force-Couple system) or another equivalent force-couple
5. How to replace a system of forces to one force and one couple
Computing Moment by Scalar and
Vector Methods.
Moment due to Force(s)
➢The rotational or twisting tendency of a force is known as the moment of the force, or
more simply the moment.
➢Engineers generally use “moment” where physicists use “torque” to describe this concept.
➢Engineers reserve “torque” for moments which are applied about the long axis of a shaft
and produce torsion.
➢ Moment which tends to tends the body is Torsion .whereas, Bending moment tends to
bend the body.
Bending Moment
Torsion
Combined Bending and Torsion
Moment due to Force(s)
Torsion in Bolt due to Force
Moment in wrench due to Force
Torsion , T, acts parallel to the plane of beam
Gravity load of canopy away from beam produces
torsion in addition to Moment
Moment due to Force(s)
➢Moments are vectors, so they have magnitude and
Axis of rotation
(Moment center)
direction and obey all rules of vector addition and
subtraction.
➢Moments have a center of rotation , although it is
more accurate to say that they have an axis of rotation.
➢In two dimensions, the axis of rotation is
perpendicular to the plane of the page and so will
appear as a point of rotation, also called the
moment center.
➢The moment center is defined as any reference axis
about which moment is required to be determined.
Moment arm
Magnitude of Moment due to a Force
The magnitude of the Moment of the force F to rotate the body about the
moment centre O (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 ) using scalar approach is:
𝑀𝐴 = 𝐹 ∗ 𝑑
C
Moment about A vector approach is:
r is position vector
𝐌𝐀 = 𝐫ACx 𝐅
Direction of Moment (Scalar approach)
There is no fixed rule that bounds us to use positive or negative sign
convention for a specific direction. However, while solving a problem one
must be consistent with initially decided sign convention for moment
In these lectures we will plus sign (+) for clockwise moments and a minus sign
(-) for Counter clockwise moments, or vice versa.
Using above sign convention of given Figure,
the moment of F about point A (or about the
z-axis passing through point A) is negative
Direction of Moment (Vector approach)
In a two-dimensional problem the direction of a moment can be determined
easily by inspection as either clockwise or counter-clockwise. A counterclockwise rotation corresponds with a moment vector pointing out of page
and is considered positive.
+z
-z
+z
+x
Point-and-curl right-hand rule technique for moments due to planar forces
Varignon’s Theorem
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.
Varignon’s Theorem
Proof:
θ
β
α θ
Varignon’s Theorem
Proof (contd.):
M
Problem 4.1
In order to raise the lamp post from the
y
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
x
Problem 4.1: Scalar method 1
Slide F to point
Transmissibility)
B
(Principle
of
𝑀𝐴 = −𝐹𝑥 * 𝑑𝑥 + 𝐹𝑦 * 𝑑𝑦
𝑀𝐴 = −200 Cos 𝜃*20 Sin 75𝑜 + 200 Sin𝜃*20 Cos 75𝑜 … . 𝐀
∆𝐴𝐵𝐶
𝐵𝐶 =
20 ft
dx
102 + 202 − 2 ∗ 10 ∗ 20 ∗ 𝐶𝑜𝑠 105𝑜 = 24.57 ft
20
𝐵𝐶
=
Sin 𝜃 Sin 105𝑜
Fx
-
+
𝐵𝐶 2 = 𝐴𝐶 2 + 𝐴𝐵2 − 2 ∗ 𝐴𝐶 ∗ 𝐴𝐵 ∗ 𝐶𝑜𝑠 105𝑜
𝛼=θ
B
B
24.57
ft
75O
A
dy
B′
⇒ 𝜃 = 51.8𝑜
Substituting Value of θ in eqn A
Fy
dx
𝑀𝐴 = −200 Cos 51.8𝑜 *20 Sin 75𝑜 +200 Sin 51.8𝑜 *20 Cos 75𝑜 = −1576 ft . lb
⇒ 𝑀𝐴 = 1576 ft. lb (CCW)
𝜃
A
dy
B′
Why we do not write –ve sign while indicating
direction
Problem 4.1: Scalar method 2 (Easy and preferred approach)
Shift F to intersection point with y-axis (Principle of Transmissibility)
+
Fx
C′
𝑀𝐴 = −𝐹𝑥 * 𝑖𝑦 + 𝐹𝑦 * 0
iy
𝑖𝑦
= Tan 51.8𝑜
10
C′
iy
Fy
A
51.8
o
⇒ 𝑖𝑦 = 10Tan 51.8𝑜
51.8O
C
10 ft
-
A
𝑀𝐴 = −200 Cos 51.8𝑜 *10Tan 51.8𝑜
𝑀𝐴 = −1572 = 1572 ft. lb (CCW)
Problem 4.1: Scalar method 3 (Easy and preferred approach)
Shift F to intersection point with x-axis (Principle of Transmissibility).
+
-
𝑀𝐴 = 𝐹𝑥 * 0- 𝐹𝑦 * 𝑖𝑥
ix
10 ft
𝑀𝐴 =-200 Sin 51.8𝑜 * 𝐴𝐶
C′
𝑀𝐴 = −1572 = 1572 ft. lb (CCW)
Fy
Fx
51.8
o
A
Problem 4.1: Scalar method 4 (Not preferred due to more
computational effort
1. Calculate perpendicular distance of A , from line of action of F
y
A′
180-75= 105o
θ
C
10 ft
𝑑 = 10 ∗ Sin 51.8𝑜 = 7.86 ft
A′
x
θ= 51.8o
A
𝑀𝐴 = 𝐹 ∗ 𝑑 = 200 lb ∗ 7.86 ft
𝑀𝐴 = −1572 ft. lb = 1572 ft. lb(CCW)
Vector method for calculating Moment
Recalling that the moment of F about point A of Figure
may be represented by the cross-product expression
𝐌𝐀 =rAC X F
where rAC is a position vector which runs from the
moment reference point A to any point C on the line of
action of F. The magnitude of this expression is given
by
MA=F*d
C
Position vector
A position vector r : while referring to a force, a
position vector is a vector that defines the direction
and distance from a reference point (often the origin)
to the point where the force is applied. Essentially, it
connects the reference point to the point of action of
෠𝐣
O ෠𝐢
the force.
While calculating moment we can take any point
along the line action of force (by using principle of
Transmissibility) which suits us for calculation
෠𝐣
O ෠𝐢
purpose (usually xi or yi) to moment center (O).
yi
xi
Problem 4.2
Solve Problem 4.1 by Vector approach
Sol:
Step 1: Represent force F in term of unit
vectors i and j
F = Fxi+Fyj = -200 Cos 51.8o i - 200 Sin 51.8o j
F = -123.68 i – 157.17 j
Fx
Fy
Problem 4.2
Step 2: Shift Force F to any convenient point
(say C) along its line of action and determine
position vector of the point , taking moment
centre (point A ) as origin.
rAC = -10 i
Step 3: M =r x F
MA =rACx F = (-10 i)x (-123.68 i – 157.17 j)
MA = 1572 k
According to right hand rules +k indicates C.C. Moment
Hence MA= 1573 ft.lb C.C.W
Which approach should I use: scalar or vector?
➢In simpler problems, such as two-dimensional problems when all forces
are coplanar, the scalar approach for evaluating moments will usually be
easier and faster, and use of clockwise and counterclockwise to
distinguish moment direction is effective.
➢For more complicated problems, such as in three dimensions, a scalar
approach can sometimes be used effectively, but generally the vector
approach is better.
Problem 4.3
The two forces acting on steel beam an be replaced
by an equivalent force R acting at point B
on the beam. Determine the distance b that locates
B.
Solution:
𝜃
= tan−1
3.5
= 54.5°
2.5
𝑀𝐵 = −𝑃 sin 54.5 × 𝑏 + 𝑃 ( 2.5 − 𝑏)
According to the Varignon theorem 𝑀𝑅 = σ 𝑀
i-e σ 𝑀 = 𝑀𝑅 𝐵
R
Since R passes through B, ∴ 𝑀𝑅 𝐵 = 0
𝑀𝑅 𝐵 = 0 = -0.814𝑃𝑏 + 2.5b - 𝑃𝑏
b = 1.37 m
B
Problem 4.4
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
F Cos θ
𝑀𝐴 = 𝐹Cos 𝜃 ∗ 6 + 𝐹Sin 𝜃 ∗ 3
𝑑
𝑀 = −6𝐹Sin 𝜃 + 3𝐹 Cos 𝜃 = 0
𝑑𝜃 𝐴
F
F Sin θ
Tan 𝜃 = 0.5 ⇒ 𝜃 = 26.57o
⇒ (𝑀𝐴 )𝑚𝑎𝑥 = 6Cos26.57𝑜 ∗ 6 + 6Sin26.57o ∗ 3
(𝑀𝐴 )𝑚𝑎𝑥 = 40.24 kN. m
(𝑀𝐴 )𝑚𝑖𝑛 = 𝐹Cos 𝜃 ∗ 6 + 𝐹Sin 𝜃 ∗ 3 = 0
Tan 𝜃 = −2 ⇒ 𝜃 = −63.44 𝑜𝑟 180 − 63.44 = 116.56o
Exercise 4.1
4.1.1 The magnitudes of the forces exerted on
the pillar of a type of suspension bridge at D
by the cables A, B, and C are equal. The
magnitude of the total moment about E due to
the forces exerted by the three cables at D is
135,000 kN-m. What is FA?
Ans: FA = 10,000 kN
4.1.2 The length of bar AB in a simple truss
structure is 350 mm. The moments exerted
about points B and C by the vertical force F are.
1.75 kN.m and 4.20 kN.m CW, respectively
Determine the force F and the length of bar AC.
Ans: F = 10 kN, LAC = 447 mm
Exercise 4.1
4.1.3 The bar AB exerts a force at B that helps
support the vertical retaining wall. The force is
parallel to the bar. The civil engineer wants the bar
to exert a 38 kN-m moment about O. What is the
magnitude of the force the bar must exert?
Ans: F = 10 kN
4.1.4 In order to hold the wheelbarrow in the
position shown, force F must produce a
counterclockwise moment of about the axle at
A. Determine the required magnitude of force F
Ans: F = 114.15 N
Exercise 4.1
4.1.5
Knowing that the two forces shown can be replaced by
an equivalent force acting at O, determine P.
Ans: P = 224 lb
4.1.6
The force F acts on the gripper of the robot
arm. The moments of F about points A and B
are 210 N · m and 90 N · m, respectively—
both counterclockwise. Determine F
and the angle θ.
Ans: F = 333 N and θ = 48.1
Exercise 4.1
4.1.7 Given that T = 28.3 kN and W = 25 kN, determine
the magnitude and sense of the moments about point B
of the following: (a) the force T; (b) the force W; and (c)
forces T and W combined.
Ans: Moment of T = 608 kN m , Moment of W =
608 kN m, Combined moment = 0
4.1.8 The weights of two children sitting at ends
A and B of a seesaw are 84 lb and 64 lb,
respectively. Where should a third child sit so
that the resultant of the weights of the three
children will pass through C if she weighs) 60 lb?
Ans: d = 2.00 ft to the right C
Exercise 4.1
4.1.9
Three forces act on the plate. Determine the
moment due to resultant force of the three
forces about point P.
Ans: MP = 145 N-m (CCW)
4.1.10
Three forces act on the structure. The sum of
the moments due to the forces about A is zero.
Determine the magnitude of the force F.
Ans: F = 2.463 kN
Couple and Associated Concepts
Couple
• A couple is defined to be a system of two forces of
equal magnitude and opposite direction and whose
lines
of
action
are
separated
by
a
distance
• 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
Couple
Magnitude of a Moment Couple
• Scalar Formulation. The moment of a couple is defined
as having a magnitude of
𝑀 = 𝐹∗𝑑
• Vector Formulation. The moment of a couple can also
be expressed by the vector cross product
𝐌 = 𝐫x𝐅
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 (Most important Property, study next
slide)
Characteristics of a Couple
Couple Moment is fixed and it does not depend on Moment Centre
It can be proved that Mo= MA= MB (explanation on
black board).
Thus Couple Moment is a Free vector as its magnitude
is not effected by Moment Centre. This is an important
difference compared to all other vectors we have
encountered.
A
B
Equivalent Couples.
If two couples produce a moment with the same magnitude and direction
then these two couples are equivalent
All couples are equivalent as M in all cases = 2Fd
Replacing (Resolving) a Force by another Force-Couple
System at defined point
Replacing a Force into an equivalent Force–Couple System
Problem 4.5
A 30-lb vertical force P is applied at A to the bracket shown,
which is held by screws at B and C. Replace P with an
equivalent force-couple system at B
Solution
P
P
30 lb
MB=30*5 = 150 in.lb
P= 30 lb
P
Replacing a Force –Couple system by
i. a Single (Resultant) force
ii. Equivalent Force-Couple system on defined
points
Replacing a Force –Couple system by a Single (Resultant) force
M
A
F
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.
The two colinear forces must be added on the side of
line of action of F which cancels the effect of already
acting couple (same magnitude but apposite in
direction)
Problem 4.6
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
4 kN
Solution
4 kN
4 kN
iy
Applying 2 equal and opposite collinear
forces and assuming they their line of
action intersect y-axis above O
4 Sin 15o
4 Cos 15o
40-iy
O′
4 Cos 15o
4 Cos 15o
iy
4 Sin 15o
+
-
𝑀𝑂′ = 4 Cos 15o ∗ 40 − 𝑖𝑦 − 4 Sin 15o ∗ 10 = 300 N. mm
𝑖𝑦 = −40.32 mm
What does –ve sign of distance iy indicate ??
Equivalent Systems
Two or more Coplanar force systems are considered Equivalent Provided that these
fulfil the two conditions σ 𝐹 1 = σ 𝐹 2 = ⋯ … . . and σ 𝑀𝑜 1 = σ 𝑀𝑜 2 = ⋯ … . .
Whereas location of O could be any where
The below two system are equivalent (check using the above two requirements).
Check whether these
two systems are
equivalent or not ?
50 N
25 N
75 N
Problem 4.7
Referring to problem 4.5, find the equivalent force-couple system . Couple
comprises of two horizontal forces at B and C that are equivalent to the couple
obtained in problem 4.5.
30 lb
Solution
MB=30*5 = 150 in.lb
30 lb
F
F
150 in.lb
5 in
F
F
150= F*3 ⇒ F = 50 lb
Replacing a Force System by a
Single (Resultant) Force and Couple
Reduction of coplanar force system to a single force and couple
We already dissed how we can replace a force into a force couple system. Such systems
are called Equivalent systems ?? because………………..
A coplanar force system can be replaced in same manner .
Consider a coplanar force systemin Figure a. In order to
replace the force system by a single force, first reduce the
given system to a force-couple system consisting of the
resultant R and the couple vector 𝐌𝐨 𝐑 (Figure b).
Replacing the force-couple system in
Figure b by a single force has been
discussed and results are produced in
a force diagram in Figure c.
Reduction of coplanar force system to a single force and couple
using rectangular components
Reduction of a system of forces is considerably simplified if the forces are resolved into
rectangular components. The force-couple system at O is then characterized by the components
( Fig. a)
Problem 4.8
Replace the force system acting on the
truss by a resultant force and couple
moment
at
point
C.
Solution
Based on general concept
discussed in previous slide , we can
replace each forces by force-couple
systems at point C
Problem 4.8
500 cos θ
➢ θ= Tan -1 (3/4)= 36.87 o
θ
➢ Mc= ΣMc= 200*2+150*4+100*6+500
500 sin θ
Sin(36.87o)*8+500 Cos (36.87o)*6 = 6400 lb.ft =
6400 lb.ft C.W
➢ 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
=
M = 6400 lb.ft
Rx= 400 lb
Ry= 750 lb
61.9o
R= 850 lb
500 lb
Exercise 4.2
4.2.1 Determine the resultant couple moment
acting on the beam. Solve the problem two ways:
(a) sum moments about point O; and (b) sum
moments about point A.
Ans: MR = 9.69 kN.m (C .W) in both the cases
4.2.2 Determine the magnitude of F so that the
resultant couple moment acting on the beam is 1.5
kN.m clockwise.
Ans: F = 27.6 lb
Exercise 4.2
4.2.3
A 260-lb force is applied at A to the rolled-steel
section shown. Replace that force with an
equivalent force-couple system at the center C of
the section.
Ans: F = 260 lb, MC = ⋅ 200 lb in.
4.2.4 The 80-N horizontal force P acts on a bell
crank as shown. (a) Replace P with an equivalent
force-couple system at B. (b) Find the two vertical
forces at C and D that are equivalent to the couple
found in part a.
Ans: (a) MB =4.00 N m, (b) FC = 100.0 N , FD =
100.000 N
Exercise 4.2
4.2.5 Two systems of forces and moments act on the beam. Are they equivalent?
4.2.6 Two systems of forces act on the beam. Are they equivalent
50 N.m
System 1
System 2
25 N
Exercise 4.2
4.2.7 The bracket, which is fastened to a wall by
anchor bolts at A and B, is loaded by the force P =
120 N and the couple C = 140 N·m. Replace P and
C with (a) an equivalent force-couple system, the
force of which acts at A; and (b) two vertical
forces, one acting at A and the other at B. Ans:
Couple At A= 124 kN.m
4.2.8
Replace the force and couple system acting on the
member in Fig. by an equivalent resultant force and
couple moment acting at point O.
Ans: FR = 461 N , 𝜃= 49.4, (MR)O = 37.5 N.m
Exercise 4.2
4.2.9 Replace the force and couple moment
system acting on the overhang beam by a resultant
force and couple moment at point A
Ans: R= 50.2 kN, θ =84.3o with -ve x-axis
in A.C.W direction, MA= 239 kN.m (C.W)
4.2.10 The shearing forces exerted on the cross section
of a steel channel can be represented by a 900-N vertical
force and two 250-N horizontal forces as shown. Replace
this force and couple with a single force F applied at
Point C, and determine the distance x from C to line BD.
(Point C is defined as the shear center of the section.)
Ans: Mc= = 45 N m , F = 900 N, x = 50.0 mm
Exercise Problems which are part of HA 2
Problems
1. 4.1.6
2. 4.1.8
3. 4.1.10
4. 4.2.1
5. 4.2.8
6. 4.2.10