Mechanical Engineering Principles

BENDING MOMENT AND SHEAR FORCE DIAGRAMS 77

Problem 5. Determine the bending moment

and shearing force diagram for the cantilever

shown in Figure 6.24, which is rigidly

constrained at the endB.

A

5 kN 10 kN

B

C

2 m 1 m

Figure 6.24

For the spanAtoC, see Figure 6.25.

5 kN

x

A

Figure 6.25

Bending moment (BM)

At x, M=−5kN×x

i.e. M=− 5 x(a straight line) ( 6. 23 )

Shearing force (SF)

At x, F=−5kN(constant) ( 6. 24 )

For the spanCtoB, see Figure 6.26.

5 kN 10 kN

C

A

2m

x

Figure 6.26

Bending moment (BM)

At x, M=−5kN×x−10 kN×(x− 2 )

=− 5 x− 10 x+ 20

i.e. M= 20 − 15 x(a straight line) ( 6. 25 )

At C, x=2m,thereforeMC= 20 − 15 × 2

= 20 − 30

i.e. MC=−10 kN m

(b) SF diagram (kN)

A

F

− 5 − 5

− 15 − 15

0 CB 0 x

(a) BM diagram (kN m)

A

M

B

C

0 x

M=− 5 x − 10

− 25

M= 20 − 15 x

x= 0

Figure 6.27

AtB, x=3m,thereforeMB= 20 − 15 × 3

= 20 − 45

i.e. MB=−25 kNm

Shearing force (SF)

Atxin Figure 6.26,

F=−5kN−10 kN

i.e. F=−15 kN(constant) ( 6. 26 )

From equations (6.23) to (6.26) and the associated

calculations, the bending moment and shearing force

diagrams can be plotted, as shown in Figure 6.27.

Now try the following exercise

Exercise 31 Further problems on bending

moment and shearing force

diagrams

Determine expressions for the bending mo-

ment and shearing force distributions for each

of the following simply supported beams;

hence, or otherwise, plot the bending moment

and shearing force diagrams.

1. Figure 6.28
   [see Figure 6.43(a) on page 82)]