Mechanical Engineering Principles

106 MECHANICAL ENGINEERING PRINCIPLES

The total weight on the beam=wL,andasthe
beam is symmetrically loaded, the values of the end

reactions,R=wL
2

, as shown in Figure 8.6.

L

w

R=

wL

2

R=

wL

2

Figure 8.6

Now the maximum bending moment,Mˆ, occurs at
the mid-span, where

Mˆ=R×L

2

−

wL

2

×

L

4

but R=

wL

2

,

hence Mˆ=

wL

2

×

L

2

−

wL

2

×

L

4

=

wL^2

4

−

wL^2

8

=wL^2

(

2 − 1

8

)

=

wL^2

8

=

2

tonnes

m

×(3m)^2

8

= 2 .25 tonnes m

= 2 .25 tonnes m×

1000 kg

tonne

×

9 .81 N

kg

i.e.maximum bending moment,Mˆ =22073 N m

Second moment of area,

I=

0. 1 × 0. 23

12

−

0. 06 × 0. 163

12

= 6. 667 × 10 −^5 − 2. 048 × 10 −^5

i.e. I= 4. 619 × 10 −^5 m^4

The maximum stressσˆ occurs in the fibre of the
beam’s cross-section, which is the furthest distance
fromNA, namelyyˆ.

By inspection,yˆ=

0. 2

2

=0.1m.

From

σˆ

yˆ

=

M

I

σˆ=

Myˆ

I

=

22073 N m× 0 .1m

4. 619 × 10 −^5 m^4

i.e. maximum stress,σˆ= 47. 79 × 106 N/m^2

= 47 .79 MPa

Problem 3. A cantilever beam, whose

cross-section is a tube of external diameter

0.2 m and wall thickness of 0.02 m, is

subjected to a point load, at its free end, of

3 kN, as shown in Figure 8.7. Determine the

maximum bending stress in this cantilever.

1.5 m

3 kN

Figure 8.7

From problem 10, page 92,I=

π

(

D 24 −D 14

)

64

where D 2 =the external diameter of the tube,

and D 1 =the internal diameter of the tube.

Hence I=

π( 0. 24 − 0. 164 )

64

i.e. I= 4. 637 × 10 −^5 m^4

The maximum bending moment, namely Mˆ, will

occur at the built-in end of the beam, i.e. on the

extreme right of the beam of Figure 8.7.

Maximum bending moment,

Mˆ =W×L

=3kN× 1 .5m× 1000

N

kN

=4500 N m