Mechanical Engineering Principles

80 MECHANICAL ENGINEERING PRINCIPLES

it is evident that:

RA=RB ( 6. 30 )

Taking moments aboutBgives:
Clockwise moments about B = anti-clockwise
moments aboutB

i.e. RA×6m= 5

kN

m

×6m×3m ( 6. 31 )

=90 kN m

from which, RA=

90

6

=15 kN=RB

On the right hand side of equation (6.31), the term
5kN/m×6 m is the weight of the UDL, and the
length of 3 m is the distance of the centre of gravity
of the UDL fromB.

Bending moment (BM)

At any distancexin Figure 6.39,

M=RA×x− 5

kN

m

×x×

x

2

( 6. 32 )

i.e. M= 15 x− 2. 5 x^2 (a parabola) ( 6. 33 )

w= 5 kN/m

A

RA= 15 kN

x

Figure 6.39

On the right hand side of equation (6.32), the term
(RA×x) is the bending moment (sagging) caused

by the reaction, and the term

(

5

kN

m

×x×

x

2

)

,

which is hogging, is caused by the UDL,

where

(

5

kN

m

×x

)

is the weight of the UDL up

to the pointx,and

x

2

is the distance of the centre

of gravity of the UDL from the right side of
Figure 6.39.

Atx=0,M= 0

Ax=3m,M= 15 × 3 − 2. 5 × 32 = 22 .5kNm

Atx=6m,M= 15 × 6 − 2. 5 × 62 = 0

Shearing force (SF)

At any distancexin Figure 6.39,

F=RA− 5

kN

m

×x( 6. 34 )

i.e. F= 15 −5x(a straight line) ( 6. 35 )

On the right hand side of equation (6.34), the

term

(

5

kN

m

×x

)

is the weight of the UDL up to the

pointx; this causes a negative configuration to the

shearing force as it is causing the left side to slide

downwards.

Atx=0,F=15 kN

Atx=3m,F= 15 − 5 × 3 = 0

Atx=6m,F= 15 − 5 × 6 =−15 kN

Plotting of equations (6.33) and (6.35) results in the

bending moment and shearing force diagrams of

Figure 6.40.

M= 22.5 kN m

M

M= 15 x− 2.5x^2

0 0

x= 0

x

x= 3 m x= 6 m

(a) BM diagram

00

F +15 kN

−15 kN

F= 15 − 5 x

(b) SF diagram

Figure 6.40

Now try the following exercise

Exercise 32 Further problems on bending

moment and shearing force

diagrams

Determine expressions for the bending mo-

ment and shearing force distributions for each