Mechanical Engineering Principles

SIMPLY SUPPORTED BEAMS 59

Hence, taking moments aboutPin Figure 5.3,

F 2 ×b=the clockwise moment, and

F 1 ×a=the anticlockwise, or

counter-clockwise, moment

Thus for equilibrium: F 1 a=F 2 b

Problem 3. A system of forces is as shown

in Figure 5.4

F

P

d

5 N 7 N

140 mm

200 mm

Figure 5.4

(a) If the system is in equilibrium find the

distanced.

(b) If the point of application of the 5 N

force is moved to pointP, distance

200 mm from the support, and the 5 N

force is replaced by an unknown force

F, find the value ofF for the system to

be in equilibrium.

(a) From above, the clockwise momentM 1 is due
to a force of 7 N acting at a distancedfrom
the support; the support is called thefulcrum,
i.e.

M 1 =7N×d

The anticlockwise momentM 2 is due to a force

of 5 N acting at a distance of 140 mm from the

fulcrum, i.e.

M 2 =5N×140 mm

Applying the principle of moments, for the

system to be in equilibrium about the fulcrum:

clockwise moment=anticlockwise moment

i.e. 7 N×d= 5 ×140 N mm

Hence, distance,d=

5 ×140 N mm

7N

=100 mm

(b) When the 5 N force is replaced by forceF

at a distance of 200 mm from the fulcrum,

the new value of the anticlockwise moment is

F×200. For the system to be in equilibrium:

clockwise moment = anticlockwise moment

i.e.

( 7 × 100 )Nmm=F×200 mm

Hence,new value of force,

F=

700 N mm

200 mm

= 3 .5N

Problem 4. A beam is supported on its

fulcrum at the pointA, which is at mid-span,

and forces act as shown in Figure 5.5.

Calculate (a) forceFfor the beam to be in

equilibrium, (b) the new position of the 23 N

force whenF is decreased to 21 N, if

equilibrium is to be maintained.

12 N F 23 N

20 mm

80 mm

100 mm

d

A

Figure 5.5

(a) The clockwise moment,M 1 , is due to the 23 N

force acting at a distance of 100 mm from the

fulcrum, i.e.

M 1 = 23 × 100 =2300 N mm

There are two forces giving the anticlockwise

momentM 2. One is the forceF acting at a

distance of 20 mm from the fulcrum and the

other a force of 12 N acting at a distance of

80 mm. Thus

M 2 =(F× 20 )+( 12 × 80 )Nmm

Applying the principle of moments about the

fulcrum:

clockwise moment=anticlockwise moments

i.e. 2300 =(F× 20 )+( 12 × 80 )

Hence F× 20 = 2300 − 960