Mechanical Engineering Principles

SIMPLY SUPPORTED BEAMS 65

of a couple are in: N m, N mm, kN m, etc, unlike
that of a force. The method of calculating reactions
on beams due to couples will now be explained with
the aid of worked problems.

Problem 11. Determine the end reactions

for the simply supported beam of

Figure 5.22, which is subjected to an

anti-clockwise couple of 5 N m applied at

mid-span.

5 kN m

RA 1.5 m 1.5 m RB

B

C

A

Figure 5.22

Taking moments aboutB:
Now the reactionRAexerts a clockwise moment
aboutB given by:RA×3 m. Additionally, the
couple of 5 kN m is anti-clockwise and its moment
is 5 kN m regardless of where it is placed.

Clockwise moments

aboutB

}

= moments aboutanti-clockwiseB

}

i.e. RA×3m=5kNm ( 5. 1 )

from which, RA=

5

3

kN

or RA = 1 .667 kN ( 5. 2 )

Resolving forces vertically gives:

Upward forces=downward forces

i.e. RA+RB= 0 ( 5. 3 )

It should be noted that in equation (5.3) the 5 kN m
couple does not appear; this is because it is a couple
and not a force. From equations (5.2) and (5.3),

RB=−RA=− 1 .667 kN

i.e.RB acts in the opposite direction to RA,so
thatRBandRAalso form a couple that resists the
5 kN m couple.

Problem 12. Determine the end reactions

for the simply supported beam of

Figure 5.23, which is subjected to an

anti-clockwise couple of 5 kN m at the

pointC

5 kN m

2 m 1 m

RA

A B

C

RB

Figure 5.23

Taking moments aboutBgives:

RA×3m=5kNm ( 5. 4 )

from which, RA=

5

3

kN

or RA = 1 .667 kN

Resolving forces vertically gives:

i.e. RA+RB= 0

from which, RB =−RA=− 1 .667 kN

It should be noted that the answers for the reactions

are the same for Problems 11 and 12, thereby

proving by induction that the position of a couple

on a beam, simply supported at its ends, does not

affect the values of the reactions.

Problem 13. Determine the reactions for

the simply supported beam of Figure 5.24.

10 kN m 8 kN m 6 kN m

1 m 1 m 1 m 1 m

AB

CDE

RA RB

Figure 5.24

Taking moments aboutBgives:

RA×4m+8kNm=10 kN m+6kNm

i.e. 4 RA= 10 + 6 − 8 = 8

from which, RA =

8

4

=2kN