Mechanical Engineering Principles

SIMPLY SUPPORTED BEAMS 61

40 N

15 mm

35 mm

25 N

60 N

d

RA

Figure 5.9

5.3 Simply supported beams having

point loads

Asimply supported beamis said to be one that
rests on two knife-edge supports and is free to move
horizontally.
Two typical simply supported beams having loads
acting at given points on the beam, calledpoint
loading, are shown in Figure 5.10.

abc

F 1 F 2

A

B

C

RA RB

(a)

(b)

RA RB

AB

C

ab

F

Figure 5.10

A man whose mass exerts a forceF vertically
downwards, standing on a wooden plank which is
simply supported at its ends, may, for example, be
represented by the beam diagram of Figure 5.10(a)
if the mass of the plank is neglected. The forces
exerted by the supports on the plank,RAandRB,
act vertically upwards, and are calledreactions.
When the forces acting are all in one plane, the
algebraic sum of the moments can be taken about
anypoint.
For the beam in Figure 5.10(a) at equilibrium:

(i) RA+RB=F,and

(ii) taking moments aboutA,F×a=RB(a+b)

(Alternatively, taking moments aboutC,

RAa=RBb)

For the beam in Figure 5.10(b), at equilibrium

(i) RA+RB=F 1 +F 2 ,and

(ii) taking moments aboutB,

RA(a+b)+F 2 c=F 1 b

Typicalpractical applicationsof simply supported

beams with point loadings include bridges, beams

in buildings, and beds of machine tools.

Problem 6. A beam is loaded as shown in

Figure 5.11. Determine (a) the force acting

on the beam support atB, (b) the force

acting on the beam support atA, neglecting

the mass of the beam.

2 kN 7 kN 3 kN

7 kN 3 kN

0.5 m

1.0 m

0.8 m

0.2 m

2 kN

A B

RA RB

(a)

(b)

Figure 5.11

A beam supported as shown in Figure 5.11 is called

a simply supported beam.

(a) Taking moments about pointAand applying

the principle of moments gives:

clockwise moments = anticlockwise moments

( 2 × 0. 2 )+( 7 × 0. 5 )+( 3 × 0. 8 )kN m

=RB× 1 .0m,

whereRBis the force supporting the beam at

B, as shown in Figure 5.11(b).

Thus ( 0. 4 + 3. 5 + 2. 4 )kN m=RB× 1 .0m

i.e. RB=

6 .3kNm

1 .0m

= 6 .3kN