Engineering Mechanice Lecture 04

Lecture 04
BY
Engr Muhammad Usman
Mechanical Engineering
Department
CECOS University

COURSE OUTLINE
Fundamental concept and principle of
mechanics, important vector quantity
Force system i.e concurrent
Non concurrent and parallel force system
• Resultant of forces (Remaining)
• Moment and couple (Start)
• Equilibrium of forces (law & type) concept of
free body diagram

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.
5

COUPLE
The 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 Couple
Consider 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 # 01
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.
11
120 Nm
80 Nm
60o
2 KN
1.5 m
3 KN
2.5 m
B
A

Solution:
12
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 Nm
60o
2 KN
1.5 m
3 KN
2.5 m
B
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?
13
A B
d
F

The above shows how a force
can be replaced by a force-
couple equivalent.
14
B
F
A
d
A B
FF
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)
16
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 m
B
A
C

solution
17
 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 m
B
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 m
B
A
C
30 KN

Solution
19
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 m
B
A
C
Fy
Fx
30 KN

Sample Problem # 04
Solution
20
 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 m
B
A
C
Fy
Fx
30 KN

Force–Couple Systems
Force 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.

Engineering Mechanice Lecture 04

Engineering Mechanice Lecture 04

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Engineering Mechanice Lecture 04