Mechanical Engineering Principles

BENDING MOMENT AND SHEAR FORCE DIAGRAMS 75

0

0

(a) BM diagram (kN m)

(b) SF diagram (kN)

0 x

x

M

F

A

AD B

B

0 E

E

D

− 10

− 10

+ 10 + 10

+3.6 +3.6

− 5 −2.4 −2.4

−2.8

C

Figure 6.15

Equation (6.14) is positive because the shearing
force is causing the right side to slide downwards.
The bending moment and shearing force diagrams
are plotted in Figure 6.15 with the aid of equa-
tions (6.7) to (6.14) and the associated calculations
atC,A,D,BandE.

Problem 3. Determine expressions for the

bending moment and shearing force

distributions for the beam of Figure 6.16.

Hence, sketch the bending moment and

shearing force diagrams.

15 kN m

2 m 2 m 3 m 1 m

30 kN m

ABD

CE

10 kN

RA RB

Figure 6.16

Firstly, it will be necessary to calculate the reactions
RAandRB.
Taking moments aboutBgives:

15 kN m+RA×5m+10 kN×1m=30 kN m

i.e. 5 RA= 30 − 10 − 15 = 5

from which, RA=

5

5

=1kN

Resolving forces vertically gives:

RA+RB=10 kN

i.e. 1 +RB= 10

from which, RB= 10 − 1 =9kN

15 kN m

C

x

Figure 6.17

For the spanCtoA, see Figure 6.17.

Bending moment (BM)

At x, M=15 kN m(constant) ( 6. 15 )

Shearing force (SF)

At x=0, F=0kN ( 6. 16 )

For the spanAtoD, see Figure 6.18.

2 m

15 kN m

RA= 1 kN

A

C

x

Figure 6.18

Bending moment (BM)

At x, M=15 kN m +RA×(x− 2 )

= 15 + 1 (x− 2 )

= 15 +x− 2

i.e. M= 13 +x(a straight line) (6.17)

AtA,x=2 m, thereforeMA= 13 + 2 =15 kN m

AtD,x=4 m, thereforeMD(−)= 13 + 4

=17 kN m

Note thatMD(−)means thatMis calculated to the

left ofD

Shearing force (SF)

At x, F=1kN(constant) ( 6. 18 )

For the spanDtoB, see Figure 6.19.

C

15 kN m 30 kN m

AD

1 kN

2 m 2 m

x

Figure 6.19