Mechanical Engineering Principles

66 MECHANICAL ENGINEERING PRINCIPLES

Resolving forces vertically gives:

RA+RB= 0

from which, RB=−RA=−2kN

Problem 14. Determine the reactions for

the simply supported beam of Figure 5.25.

1 m1 m1 m1 m

10 kN m 8 kN m 6 kN

RA RB

D

A

CE

B

Figure 5.25

Taking moments aboutBgives:

RA×4m+8kNm+6kN×1m= 10 kN m

i.e. 4 RA= 10 − 8 − 6 =− 4

from which, RA=−

4

4

=−1kN

(acting downwards)

Resolving forces vertically gives:

RA+RB+ 6 = 0

from which, RB=−RA− 6 =−(− 1 )− 6

i.e. RB= 1 − 6 =−5kN
(acting downwards)

Now try the following exercise

Exercise 28 Further problems on simply

supported beams with cou-

ples

For each of the following problems, determine

the reactions acting on the simply supported

beams:

1. Figure 5.26

[RA=−1kN, RB=1kN]

5 kN m

RA RB

A B

(a)

3 m 2 m

Figure 5.26

2. Figure 5.27

[RA=−1kN, RB=1kN]

5 kN m

2.5 m 2.5 m

RA RB

A B

(b)

Figure 5.27

3. Figure 5.28

[RA=1kN, RB=−1kN]

(c)

10 kN m 6 kN m 12 kN m

R 2 m 2 m 2 m 2 m

A RB

Figure 5.28

4. Figure 5.29 [RA= 0 ,RB=0]

(d)

10 kN m 10 kN m

1 m 5 m 2 m

RA RB

Figure 5.29