Mechanical Engineering Principles

72 MECHANICAL ENGINEERING PRINCIPLES

given by:

M=RA×x− 6 (x− 3 )= 2. 4 x− 6 (x− 3 )

= 2. 4 x− 6 x+ 18

i.e. M= 18 − 3. 6 x(clockwise) ( 6. 3 )

The effect of this resultant moment on the left ofx
is shown in the lower diagram of Figure 6.7.

ABC

3 m 2 m

6 kN

2.4 kN 3.6 kN

x

M= 2.4x

−6 (x− 3)

= 18 − 3.6x

M= 3.6 (5 − x)

= 18 − 3.6x

Figure 6.7

Now from Figure 6.7, it can be seen that on the
right side ofx, there is an anti-clockwise moment
of:M=RB×( 5 −x)= 3. 6 ( 5 −x)= 18 − 3. 6 x

i.e. M= 18 − 3. 6 x(anti-clockwise) ( 6. 4 )

The effect of the moment of equation (6.3) and that
of the moment of equation (6.4), is to cause the
beam to sag at this point as shown by the lower
diagram of Figure 6.7, i.e.Mis positive betweenC
andB,and

M=+ 18 − 3. 6 x( 6. 5 )

Shearing force

Consider a distancexbetweenCandB, as shown
in Figure 6.8.

3 m 2 m

6 kN

F= 6 − 2.4

F= 3.6 kN

= 3.6 kN

2.4 kN 3.6 kN

x

C

AB

Figure 6.8

From Figure 6.8, it can be seen that atx, there are

two vertical forces to the left of this section, namely

the 6 kN load acting downwards and the 2.4 kN load

acting upwards, resulting in a net value of 3.6 kN

acting downwards, as shown by the lower diagram

of Figure 6.8. Similarly, by considering the vertical

forces acting on the beam to the right ofx, it can

be seen that there is one vertical force, namely the

3.6 kN load acting upwards, as shown by the lower

diagram of Figure 6.8. Thus, as the right hand of

the section is tending to slide upwards, the shearing

force is said to be negative, i.e. betweenCandB,

F=− 3 .6kN ( 6. 6 )

It should be noted that atC, there is a discontinuity

in the value of the shearing force, where over an

infinitesimal length the shearing force changes from

+ 2 .4kNto− 3 .6 kN, from left to right.

Bending moment and shearing force diagrams

The bending moment and shearing force diagrams

are simply diagrams representing the variation of

bending moment and shearing force, respectively,

along the length of the beam. In the bending moment

and shearing force diagrams, the values of the bend-

ing moments and shearing forces are plotted ver-

tically and the value ofxis plotted horizontally,

wherex=0 at the left end of the beam andx=the

whole length of the beam at its right end.

In the case of the beam of Figure 6.4, bending

moment distribution betweenAandCis given by

equation (6.1), i.e.M= 2. 4 x, where the value ofx

varies betweenAandC.

AtA,x=0, thereforeMA= 2. 4 × 0 = 0

and atC,x =3 m, thereforeMC = 2. 4 × 3 =

7 .2kN.

Additionally, as the equation M = 2. 4 x is

a straight line, the bending moment distribution

betweenAandCwill be as shown by the left side

of Figure 6.9(a).

Similarly, the expression for the variation of

bending moment betweenCandBis given by equa-

tion (6.3), i.e.M= 18 − 3. 6 x, where the value ofx

varies betweenCandB. The equation can be seen

to be a straight line betweenCandB.

AtC,x=3 m, thereforeMC= 18 − 3. 6 × 3 =

18 − 10. 8 = 7 .2kNm

AtB,x=5 m, thereforeMB= 18 − 3. 6 × 5 =

18 − 18 = 0

Therefore, plotting of the equationM= 18 − 3. 6 x

betweenC andB results in the straight line on

the right of Figure 6.9(a), i.e. the bending moment

diagram for this beam has now been drawn.