Mechanical Engineering Principles

74 MECHANICAL ENGINEERING PRINCIPLES

It is negative, because as the left of the section tends
to slide downwards the right of the section tends
to slide upwards. (Remember, right hand down is
positive).
For the rangeAtoD, see Figure 6.12.

5 kN

A

RA= 8.6 kN

C

2 m

x

Figure 6.12

Bending moment (BM)

At any distancexbetweenAandD

M=− 5 ×x+RA×(x− 2 )

=− 5 x+ 8. 6 (x− 2 )

=− 5 x+ 8. 6 x− 17. 2

i.e. M= 3. 6 x− 17. 2 ( 6. 9 )

(a straight line betweenAandD)

At A,x=2m,MA= 3. 6 × 2 − 17. 2

= 7. 2 − 17. 2 =−10 kN m

At D,x=4m,MD= 3. 6 × 4 − 17. 2

= 14. 4 − 17. 2 =− 2 .8kNm

Shearing force (SF)

At any distancexbetweenAandD,

F=−5kN+ 8 .6kN= 3 .6kN(constant)

( 6. 10 )

For the rangeDtoB, see Figure 6.13.

5 kN

RA= 8.6 kN

AD

6 kN

2 m 2 m

x

Figure 6.13

Bending moment (BM)

At x,M=− 5 ×x+ 8. 6 ×(x− 2 )− 6 ×(x− 4 )

=− 5 x+ 8. 6 x− 17. 2 − 6 x+ 24

i.e. M=− 2. 4 x+ 6. 8 ( 6. 11 )

(a straight line betweenDandB)

At D,x=4m,thereforeMD=− 2. 4 × 4 + 6. 8

=− 9. 6 + 6. 8

i.e. MD=− 2 .8kNm

At B,x=7m,thereforeMB=− 2. 4 × 7 + 6. 8

=− 16. 8 + 6. 8

i.e. MB=−10 kN m

Shearing force (SF)

Atx, F=− 5 + 8. 6 − 6

=− 2 .4kN(constant) ( 6. 12 )

For the rangeBtoE, see Figure 6.14.

C E

5 kN 10 kN

x

8 m

Figure 6.14

Bending moment (BM)

In this case it will be convenient to consider only

the resultant of the couples to the right ofx (−

remember that only one side need be considered,

and in this case, there is only one load to the right

ofx).

At x,M=− 10 ×( 8 −x)=− 80 + 10 x ( 6. 13 )

Equation (6.13) can be seen to be a straight line

betweenBandE.

AtB,x=7 m, thereforeMB=− 80 + 10 × 7

=−10 kN m

AtE,x=8 m, thereforeME=− 80 + 10 × 8

=0kNm

Shearing force (SF)

Atx,F=+ 10 kN(constant) ( 6. 14 )