mechanics of solid

1. 1
GOVERNMENT ENGINEERING COLLEGE,
PALANPUR
Subject :- Mechanics of Solids

2. Coplanar Non-concurrent Forces.

3. Coplanar Non-concurrent Force System:-
This is the force system in which lines of action of
individual forces lie in the same plane but act at different
points of application.
COPLANAR NON - CONCURRENT
FORCE SYSTEM
F2
F1
F3
F1 F2
F5
F4
F3
3

4. Moment Of A Force :-
The moment of a force is equal to the product of the force
and the perpendicular distance of line of action of force from
the point about which the moment is required.
M = P x X
Here, M = moment
P = Force
X = Perpendicular distance between the line of action of
force and the point about which moment is required.
Unit of moment is N.m. or kN,m.
 Examples of moment are :-
i) To tight the nut by spanner
ii) To open or close the door
 Sign for moment:-
Generally, clockwise moment is taken as positive and anticlockwise
moment is taken as negative.

5. Unit of moment is N.m. Or kN. m.

6. Couple:-
To equal and opposite forces whose lines of action are
different form a couple. Resultant or net force of couple is zero.
Hence, couple acting on a body do not create any translatory
motion of the body. Couple produces only rotational motion of
the body.
• Types of couple:
There are two types of couple.
i ) Clockwise couple
ii) Anticlockwise couple
Clockwise couple:-
 It rotates the body in
clockwise direction.
 It is taken as positive.
Anticlockwise couple:-
 It rotates the body in anticlockwise direction.
 It is taken as negative.

7. Arm of couple :-
The perpendicular distance between the lines of action of
two forces forming couple is know as the arm of couple.
Moment of couple :-
Moment of couple = Forces x arm of couple
M = P x a
unit of couple is N.M. Or KN.M
Example of couple are :-
I. Forces applied to the key of a lock, while locking or
unlocking it.
II. Forces applied on the steering wheel of a car by two
hands to steer the car towards left or right.
III.To open or close wheel valve of a water supply pipe line.

8. CHARACTERISTICS OF A COUPLE :-
I. The algebraic sum of the forces, forming the couple is zero.
II. The algebric sum of the moment of the forces, forming the
couple, about any point is the same and equal to the moment of the
couple itself.
Moment of couple M = P.a
Now, taking moment @ 0,
Mo = P ( a + x ) – P.x
= P.a + P.x – P.x
= P.a
III. A couple cannot be balanced by a single force, but can be balanced
only by a couple, but of opposite nature.
IV. If number of coplanar couples are acting on the body, the they can
be reduced to a single couple, whose magnitude will be equal to the
algebraic sum of moment of all the couples.

9. DIFFRENCE:-
MOMENT COUPLE
 Moment = Force x Perpen-
dicular distance
M = P.x
 It is produced by a single
Force not passing through C.G.
of the body.
 The force move the body in
the direction of force and rotate
the body.
 To balance the force causing
moment, equal and opposite
forces is required
 Two equal and opposite
forces whose lines of action are
different form a couple.
 It is produced by two equal
and opposite parallel, non-
collinear forces.
 Resultant force of couple is
zero. Hence, body does not
move, but rotate only.
 Couple can not be balance by
a single force. It can be balance
by a couple only.

10. Example :-
Replace the couple and force by a single force – couple applied
at a for the lever shown in figure. Also find the distance of a point
C from A where only a single force can replace the force-couple
System.
Solution:- Moment of couple of 250 kN force is
= 250 × 120
= 30000N.mm

11. Single force to replace the force couple system :
Let, 500 N is applied at distance x from A at point C.
Moment of force and couple at b from
= (250×120) + 500 × (200cos 60̇ )
= 30000 + 50000
= 80000 N.mm …..( i )
Moment of force 500 N at C @ B,
= 500× (200 + x) cos60̇
= 250 (200 + x)
= 50000 + 250 x …..( ii )
Equating ( i ) and ( ii )
50000 + 250 x = 80000
250 x = 30000
x = 120 mm.

12. VARIGNON’S PRINCIPLE OF MOMENTS
It states,
‘If number of coplanar forces are acting simultaneously on
a particle, the algebraic sum of the moments of all the forces about
any point is equal to the moment of their resultant force about the
same point.

13. Proof:-
Consider two forces P and Q acting at a point a represented
in magnitude and direction by AB and AC as shown in figure.
let, O be the point, about which
the moments are taken. Through O,
draw a line CD parallel to the
direction of force P, to meet the line
of action of Q at C. Now with AB and AC as two adjacent sides,
complete parallelogram ABCD. Join the diagonal AD of the
parallelogram and OA and OB, diagonal AD represents in
magnitude and direction the resultant of two forces P and Q.
Now, taking moment of forces P,Q and R about O.
Moment of force P about O
= 2 × Area of triangle AOB ….. ( i )
Similarly, moment of force Q, about O,
= 2 × Area of triangle AOC …..( ii )
O C D
R
Q
A P B

14. Moment of resultant R, about O,
= 2 × Area of triangle AOD …..( iii )
Now, from the geometry of the figure,
Area of ∆ AOD = Area of ∆ AOC + Area of ∆ ACD
But area of ∆ACD = Area of ∆ ADB
= Area of ∆ AOB
Therefore, Area of ∆ AOD = Area of ∆ AOC + Area of ∆ AOB
Multiplying both sides by 2,
2 × Area of ∆ AOD = 2 × Area of ∆ AOC + 2 × Area of ∆ AOB
Therefore, moment of force P and O = moment of force P and O
+ moment of force Q and O
( Here, we have considered only two forces. But this principle can
be exerted for any number of forces.)

15. If the body is acted upon by a number of co-planar
non-concurrent forces. It may have one of the following stats:
I. The body may move in any one direction.
II. The body may rotate about itself without moving.
III. The body may move in any one direction, and at the
same time it may also rotate about itself.
IV. The body may be completely at rest.
CONDITION OF EQUILIBRIUM FOR COPLANER
NON – CONCURRENT FORCES:-

16. Example :-
Determine the resultant and locate the same with respect to point
‘A’ of a non-concurrent force system shown in figure.
Solution:-
∑H = 0.5 cos60̇ + 1.2 –
2.1 cos45̇
= 0.25 + 1.2 – 1.485
= - 0.035kN (←)
∑V = 1.1 C 0.5 sin60̇ +
2.1 sin45̇
= 1.1 + 0.433 + 1.485
= 3.02kN (↑)
R = √(∑H)2 + (∑V)2
= √(-0.035)2+ (3.02)2
= 3.02kN

17. tanθ = = = 86.28
∑V
∑H
3.02
0.035
θ = 89.33̇
∑H = -Ve
∑V = +Ve
Therefore, R will be in 2
quadrant.
Location of resultant:
Let R is at perpendicular
distance x from A.
Taking moment @ A.
R.x + (1.2×1.5) + 0.5sin60̇ × 2 = 4.0 +2.1cos45̇ × 1.54
3.02 x + 1.8 + 0.866 = 4 + 2.23
3.02 x = 3.564
x = 1.18 m .