Mechanical Engineering Principles

BENDING MOMENT AND SHEAR FORCE DIAGRAMS 73

M

M= 2.4x M=^18 − 3.6x

00

0

x

A

x= 0 x= 5 m

F

x

F= 2.4 kN

F=−3.6 kN

+ 6 kN

−

C B

(a) Bending moment diagram

(b) Shearing force diagram

M= 7.2 kN m

0

Figure 6.9 Bending moment and shearing force
diagrams

In the case of the beam of Figure 6.4, the shearing
force distribution along its length fromAtoCis
given by equation (6.2), i.e.F = 2 .4 kN, i.e.F is
constant betweenAandC. Thus the shearing force
diagram betweenAandCis given by the horizontal
line shown on the left ofCin Figure 6.9(b).
Similarly, the shearing force distribution to the
right ofCis given by equation (6.6), i.e.
F=− 3 .6 kN, i.e.F is a constant betweenCand
B, as shown by the horizontal line to the right of
C in Figure 6.9(b). At the pointC, the shearing
force is indeterminate and changes from+ 2 .4kN
to− 3 .6 kN over an infinitesimal length.

Problem 2. Determine expressions for the

distributions of bending moment and

shearing force for the horizontal beam of

Figure 6.10. Hence, sketch the bending and

shearing force diagrams.

5 kN

2 m 2 m

RA RB

3 m 1 m

6 kN

AD BE

C

10 kN

Figure 6.10

Firstly, it will be necessary to calculate the unknown
reactionsRAandRB.

Taking moments aboutBgives:

RA×5m+10 kN×1m=5kN×7m

+6kN×3m

i.e. 5 RA+ 10 = 35 + 18

5 RA= 35 + 18 − 10 = 43

from which, RA=

43

5

= 8 .6kN

Resolving forces vertically gives:

RA+RB=5kN+6kN+10 kN

i.e. 8. 6 +RB= 21

from which, RB= 21 − 8. 6 = 12 .4kN

For the rangeCtoA, see Figure 6.11.

5 kN

C

x

Figure 6.11

To calculate the bending moment distribution (M),

only the resultant of the moments to the left of the

section atxwill be considered, as the resultant of

the moments on the right of the section ofxwill be

exactly equal and opposite.

Bending moment (BM)

From Figure 6.11, at any distancex,

M=− 5 ×x(hogging) =− 5 x ( 6. 7 )

Equation (6.7) is a straight line betweenCandA.

AtC,x=0, thereforeMC=− 5 × 0 =0kNm

AtA,x=2 m, thereforeMA=− 5 × 2

=−10 kN m

Shearing force (SF)

To calculate the shearing force distribution (F)at

any distancex, only the resultant of the vertical

forces to the left ofxwill be considered, as the

resultant of the vertical forces to the right ofxwill

be exactly equal and opposite.

From Figure 6.11, at any distancex,

F=−5kN ( 6. 8 )