1

Engineering Mechanics

Chapter 3 :

Moments and Couples

2

ObjectivesUnderstand the Principle of Transmissibility of a Force;

Determine the Moment of a Force System from Basic Definition and/or with the application of Varignon’s Theorem;

Understand the Properties of a Couple;

Determine a Force-Couple Equivalent to Replace a System of Forces.

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Introduction

A force will cause motion along its direction.

A force may also cause rotation about a fixed point some distance away.

This rotation or turning effect of a force is called MOMENT.

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Principle of TransmissibilityIf we move (TRANSMIT) the Force P from point A to point B which lies on the Line of Action of Force P, the same effect would be expected.

Principle of transmissibility states that a force acting on a rigid body at different points along the force’s line of action will produce the same effect on the body.

Point of Application

Line of action

P

A

B

Application

P

5

Moment of A Force The moment M of a force F about a fixed point A is defined as the product of the magnitude of force F and the perpendicular distance d from point A to the line of action of force F.

MA = F x d Where force F is in newtons, NAnd distance d is in meters, m

Thus moment MA is in newton-meter, Nm.

A

d F

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Moment Sign Convention

Anti-clockwise : + VE(Counter-clockwise)

Clockwise : - VE

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Example 3.1

Calculate the moment of the 500 N force about the point A as shown in the diagram.

A

1.5 m

500 N

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A

Example 3.1Solution 1.5 m

500 N

Since the perpendicular distance from the force to the axis point A is 1.5 m, from

MA = F x d

  MA = - 500 x 1.5

= - 750 Nm = 750 Nm

Application

9

Example 3.2

Calculate the moment about point A caused by the 500 N force as shown in the diagram.

A 1.5 m

60o

500 N

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Example 3.2Solution:

MA = F x d

= -500

x AB

= -500 x 1.5 sin 60

= -

649.5 Nm

= 649.5 Nm

1.5 m500 N

Line of action

600

600

d

B

A

A 1.5 m

60o

500 N

Application

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Addition of Moments of Coplanar Forces

Coplanar Forces refers to forces acting on the Same Plane ( i.e. 2-D ).For a system of several coplanar forces, the combined turning effect of the forces can be determined by adding algebraically the moment caused by each individual force, taking into account their sense of rotation,

i.e +ve for anti-clockwise and –ve for clockwise rotation.

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Example 3.3 Determine the resulting moment about point A of the system of forces on bar ABC as shown in the diagram.

A B C

60 0

500 N 800 N

1.5 m 0.5m

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60 0

Example 3.3 Solution:

AD = (1.5 + 0.5) sin 60= 2.0 sin 60= 1.732 m

MA = (-500 x 1.5) + (-800 x

1.732) = - 750 – 1385.6 = - 2135.6 Nm

= 2135.6 Nm

A B C

500 800 N1.5 m 0.5 m

A1.5 m 0.5 m

Line of Action

600

D

C

800 N

500 N

Application

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Varignon’s Theorem Varignon’s Theorem states that “ the moment of a force about any point is equal to the sum of the moments of its components about the same point”. To calculate the moment of any force with a slope or at an angle to the x or y-axis, resolve the force into the Fx and the Fy components, and calculate the sum of the moment of these two force components about the same point.

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Example 3.4 Re-Calculate Example 3.3 Above Employing Varignon’s Theorem.

A B C

60 0

500 N 800 N

1.5 m 0.5m

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A

Example 3.4Solution

MA

= (-500 x 1.5) + (- 800 sin 60 x 2)

= -2135.6 Nm = 2135.6 Nm

* Same answer as in example 3.3

B C

60 0

500 N 800 N

1.5 m 0.5m

Application

1.5 m 0.5m 600

A

800 N

B C

Fx

Fy = 800 sin 600

500 N

600

Fx

Fy

= 800 cos 600

Example 3.5 Determine the resulting moment about point A for the system of forces acting on the plate ABCD as shown in the diagram.

15 cm

8cm

10.38 cm 45 N

20 N

35 N

80o

45o

30o

A B

C

D

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Example 3.5Solution

45 N force: Fx = 45 cos 80 = 7.814 N

Fy = 45 sin 80 = 44.32 N

15 cm

8cm

10.38 cm 45 N

20 N35 N

80o

45o

30o

A B

CD

35 N force: Fx = 35 cos 60 = 17.5 N

Fy = 35 sin 60 = 30.31 N

FxFy

45 N800

FxFy

35 N300

Application

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Example 3.5Solution

20 N force: Fx = 20 cos 45 = 14.14 N

Fy = 20 sin 45 = 14.14 N

0.15 m

0.08m

0.1038 m 45 N

20 N

35 N80o

45o60o

A B

CD

MA = (- 45 sin 80 x 0.1038) + (- 35 cos 60 x 0.08)

+ (20 cos 45 x 0.08) + (20 sin 45 x 0.15) = - 4.591 – 1.4 + 1.131 + 2.121 = - 2.74 Nm = 2.74 Nm

FyFx

20 N

450

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3.6 Couples

A couple consists of a pair of 2 forces which has the following properties :- - Equal magnitude and opposite in direction

   - Act along parallel lines of action    - Separated by a perpendicular distance d.

F

F

F

F

d

d

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What a couple does?

A couple causes a body to rotate only without translational motion since the two forces ‘cancels’ out each other giving zero resultant.

 

A couple acting in a system of forces will only contribute to the resulting moment but not to the resulting force.

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Magnitude of a Couple

Consider a light bar acted upon by a couple as shown :-

The moment of the couple about A is

MA = + ( F x d2 ) - ( F x d1 ) 

MA = F x ( d2 - d1 )

MA = F x dWhat is the total moment of the couple about point B ?

d1 d

d2

F

A

F

B

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From above, we see that :

The couple’s moment about any pivot point is equal to F x d

A couple has the same moment about all points on a body.

F

F

F

F

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Example 3.6 A light bracket ABC is subjected to two forces and two couples as shown. Determine the moment at (a) point A and (b) point B.

120 Nm

80 Nm 30o

2 KN

1.5 m

3 KN2.5 m

BA

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Example 3.6Solution:

a) MA = (2000 cos 60 x 2.5) – (2000 sin 60 x 1.5) + ( 3000 x 0)

+ 120 – 80 = 2500 – 2598 + 0 +120 - 80 = - 58 Nm = 58 Nm

(b)    MB = (2000 cos 60 x 0) – (2000 sin 60 x 1.5) + (3000 x 0)

+ 120 – 80 = -2558 Nm = 2558 Nm

120 Nm

80 Nm30o

2 KN

1.5 m

3 KN2.5 m

BA

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3.7 Force Couple Equivalent The Force-Couple Equivalent concept will enable us to transfer a force to another location outside its line of action. Consider a force F acting at a point B on a rigid body as shown in diagram (a) below.

  How do we transfer the force F from point B to point A?

A B

d F

Application

27

The above shows how a force can be replaced by a force-couple equivalent.

B

F

A d

A B

F F

F

d

A B

F

MA = F x d

Force Couple Equivalent

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Example 3.7 (Single Force System)

Determine the force-couple equivalent at point A for the single force of 20 kN acting at point C on the bracket ABC.

40o

20 KN

1.2 m

2.3 mB

A

C

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Example 3.7solution

MA = (20 sin 40 x 2.3) – (20 cos 40 x 1.2)

= 11.18 kNm

 Answer: 11.18 kNm 20 kN 40 0

  A

40o

20 KN

1.2 m

2.3 mB

A

C

Fy

Fx

30

Example 3.8 (Multiple force system)

Another 30 kN horizontal force is added to Example 3.7 at point B. Determine the force-couple equivalent at point A.

40o

20 KN

1.2 m

2.3 mB

A

C

30 KN

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Example 3.8Solution

Rx = Fx = 20 cos 40 + 30

= 45.32 kN Ry = Fy =

20 sin 40 = 12.86 kN

Fy

Fx

400

Therefore R = (45.322 + 12.862) = 47.11 kN

And tan = Ry = 12.86 = 15.84 0 Rx 45.32

Application

40o

20 KN

1.2 m

2.3 mB

A

C

Fy

Fx

30 KN

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Example 3.8Solution

MA = (20 sin 40 x 2.3) – (20 cos 40 x 1.2) + (30 x 0)

= 11.18 kNm  Answer: 47.11 kN

11.18 kNm 15.84 0

Application

40o

20 KN

1.2 m

2.3 mB

A

C

Fy

Fx

30 KN

End of Chapter 3