Mechanical Engineering Principles

76 MECHANICAL ENGINEERING PRINCIPLES

Bending moment (BM)

Atx, M=15 kN m+1kNm×(x− 2 )

−30 kN m

= 15 +x− 2 − 30

i.e. M=x− 17 (a straight line) (6.19)

AtD,x=4 m, thereforeMD(+)= 4 − 17
=−13 kN m

Note thatMD(+)means thatMis calculated just to
the right ofD.

AtB,x=7 m, thereforeMB(−)= 7 − 17
=−10 kN m

Shearing force (SF)

At x, F=−1kN(constant) ( 6. 20 )

For the spanBtoE, see Figure 6.20.

15 kN m 10 kN

C

E

x

8 m

Figure 6.20

In this case we will consider the right of the beam
as there is only one force to the right of the section
atx.

M=− 10 ×( 8 −x)=− 80 + 10 x(a straight line)

M

F

C

15

0

(b) SF diagram (kN)

0

0

AD B^0 E x

x= 0

x= 0

x= 8 m

x= 8 m

− 13

+ 10 + 10

+ 1 + 1

− 10

(a) BM diagram (kN m)

+ 17

x

Figure 6.21

Atx, F =10 kN (positive as the right hand is

going down, and constant)

Plotting the above equations for the various spans,

results in the bending moment and shearing force

diagrams of Figure 6.21.

Problem 4. Calculate and plot the bending

moment and shearing force distributions for

the cantilever of Figure 6.22.

5 kN

A

B

x

2 m

Figure 6.22

In the cantilever of Figure 6.22, the left hand end is

free and the right hand end is firmly fixed; the right

hand end is called theconstrained end.

Bending moment (BM)

Atxin Figure 6.22 , M=−5kN×x

i.e. M=− 5 x ( 6. 21 )

(a straight line)

Shearing force (SF)

Atxin Figure 6.22, F=−5kN(a constant)

( 6. 22 )

For equations (6.21) and (6.22), it can be seen that

the bending moment and shearing force diagrams

are as shown in Figure 6.23.

M=− 5 x

x

x

−5 kN −5 kN

−10 kN m

(a) BM diagram

(b) SF diagram

F

M

0

0 0

0

Figure 6.23