Lecture 04BY

Engr Muhammad UsmanMechanical Engineering

DepartmentCECOS University

COURSE OUTLINE

Fundamental concept and principle of mechanics, important vector quantityForce system i.e concurrentNon concurrent and parallel force system

• Resultant of forces (Remaining)• Moment and couple (Start)• Equilibrium of forces (law & type) concept of

free body diagram

3

COUPLE

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.

4

F

F

F

F

d

d

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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COUPLEThe moment produced by two equal, opposite, and non collinear forces is called a couple.

Consider the action of two equal and opposite forces F and -F a distance d apart, as shown in Fig. a. These two forces cannot be combined into a single force because their sum in every direction is zero.

Their only effect is to produce a tendency of rotation.

The combined moment of the two forces about an axis normal to their plane and passing through any point such as O in their plane is the couple M.

This couple has a magnitude

Its direction is counterclockwise when viewed from above for the case illustrated.

Note especially that the magnitude of the couple is independent of the distance a which locates the forces with respect to the moment center O.

It follows that the moment of a couple has the same value for all moment centers.

We may also express the moment of a couple by using vector algebra.

The direction and sense of the couple moment are determined by the right-hand rule, where the thumb indicates this direction when the fingers are curled with the sense of rotation caused by the couple forces.

In all cases, M will act perpendicular to the plane containing these forces.

Magnitude of a CoupleConsider a light bar acted upon by a couple as shown :-

10

The moment of the couple about A is

MA = + ( F x d2 ) - ( F x d1 )  MA = F x ( d2 - d1 )

MA = F x d

d1 d

d2

F

A

F

B

Sample Problem # 01A light bracket ABC is subjected to two forces and two couples as shown. Determine the moment at (a) point A and (b) point B.

11

120 Nm

80 Nm 60o

2 KN

1.5 m

3 KN

2.5 mB

A

Solution:

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a) MA = (2000 cos 60 x 1.5) – (2000 sin 60 x 2.5) + ( 3000 x 0) + 120 – 80 = 1500 – 4330 + 0 +120 - 80 = - 2790 Nm = 2790 Nm

(b)    MB = (2000 cos 60 x 1.5) – (2000 sin 60 x 0) + (3000 x 0) + 120 – 80 = -2558 Nm = 2558 Nm

120 Nm

80 Nm60o

2 KN

1.5 m

3 KN

2.5 mB

A

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?

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A B

dF

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

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B

F

A

d

A B

F F

F

d

A B

F

MA = F x d

Force Couple Equivalent

Equivalent Couples

Changing the values of F and d does not change a given couple as long as the product Fd remains the same.

Likewise, a couple is not affected if the forces act in a different but parallel plane. Figure 2/11 shows four different configurations of the same couple M.

In each of the four cases, the couples are equivalent and are described by the same free vector which represents the identical tendencies to rotate the bodies.

Sample Problem # 02(Single Force System)

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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

solution

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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

Sample Problem # 03(Multiple force system)

18

Another 30 kN horizontal force is added to Sample Problem # 02 at point B. Determine the force-couple equivalent at point A.

40o

20 KN

1.2 m

2.3 mB

A

C

30 KN

Solution

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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

40o

20 KN

1.2 m

2.3 mB

A

C

Fy

Fx

30 KN

Sample Problem # 04Solution

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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

40o

20 KN

1.2 m

2.3 mB

A

C

Fy

Fx

30 KN

Force–Couple SystemsForce acting on an object produces translation and rotation of the object.

The translation effects are represented by the magnitude and direction of the force whereas the rotational effects are represented by the moment of force.

Hence a force can be replaced by a force and couple as shown in the figure.

Consider a force F acting at point A as shown in figure.

Add two forces of magnitude F and –F at point B as shown in figures which are parallel to the force F at A.

The force F at A and the force –F at B forms a couple with moment Fd about any point in the plane.

Thus a force F at A is replaced by a force F at B and a couple of moment Fd as shown in the figure.

Thus, we have replaced the original force at A by the same force acting at a different point B and a couple, without altering the external effects of the original force on the body.

The combination of the force and couple in the right-hand part of Fig. 2/12 is referred to as a force–couple system.

By reversing this process, we can combine a given couple and a force which lies in the plane of the couple (normal to the couple vector) to produce a single, equivalent force.

Replacement of a force by an equivalent force–couple system, and the reverse procedure, have many applications in mechanics and should be mastered.