Mechanical Engineering Principles

BENDING MOMENT AND SHEAR FORCE DIAGRAMS 79

downwards, and spread over the entire length

of the cantilever.

w= 10 kN/m

AB

x

5 m

Figure 6.35

Bending moment (BM)

At any distancexin Figure 6.36,

M=−10 kN×x×

x

2

( 6. 27 )

i.e. M=− 5 x^2 (a parabola) ( 6. 28 )

w= 10 kN/m

A

x

Figure 6.36

In equation (6.27), the weight of the uniformly dis-

tributed beam up to the pointxis (10×x). As the

centre of gravity of the UDL is at a distance of

x

2

from the right end of Figure 6.36,

M=− 10 x×

x

2

The equation is negative because the beam is

hogging.

Atx= 0 ,M= 0

Atx=5m,M=− 10 × 5 ×

5

2

=−125 kN m

Shearing force (SF)

At any distancexin Figure 6.36, the weight of the

UDL is( 10 ×x)and this causes the left side to slide

down, or alternatively the right side to slide up.

Hence, F=− 10 x(a straight line) ( 6. 29 )

Atx= 0 ,F= 0

Atx=5m,F=− 10 × 5 =−50 kN

Plotting of equations (6.28) and (6.29) results in the

distributions for the bending moment and shearing

force diagrams shown in Figure 6.37.

(b) SF diagram

AB

0

F

0

−

−50 kN

(a) BM diagram

AB

M

0

−

−125 kN m

M= − 5 x^2

x= 5 m

x= 0

F = − 10 x

0

Figure 6.37

Problem 7. Determine expressions for the

bending moment and shearing force diagrams

for the simply supported beam of

Figure 6.38. The beam is subjected to a

uniformly distributed load (UDL) of 5 kN/m,

which acts downwards, and it is spread over

the entire length of the beam.

w= 5 kN/m

A

RA RB

B

6 m

Figure 6.38

Firstly, it will be necessary to calculate the reactions

RAandRB. As the beam is symmetrically loaded,