CHAPTER 9 BENDING OF BEAMS
EXERCISE 51, Page 124
1. A cantilever of solid circular cross-section is subjected to a concentrated load of 30 N at its free
end, as shown below. If the diameter of the cantilever is 10 mm, determine the maximum stress
in the cantilever.
Second moment of area, I =
d 4 104

 490.9 mm 4
64
64
(1)
The maximum bending moment, namely M , will occur at the built-in end of the beam, i.e. on the
extreme right of the beam.
Maximum bending moment, M = W  L = 30 N  1.2  1000 mm = 36000 N mm
(2)
The maximum stress occurs at the outermost fibre of the beam’s cross-section from NA, at y
By inspection, y =
10 mm
=5m
2
(3)
M

My
=
from which,  =
I
I
y
Hence, from equations (1) to (3),
 =
M y 36000 N mm  5 mm
 367 N / mm 2  367 106 N / m 2
=
4
I
490.9 mm
i.e. the maximum bending stress,  = 367 MPa
2. If the cantilever of problem 1 above were replaced with a tube of the same external diameter, but
of wall thickness 2 mm, what would be the maximum stress due to the load shown.
149
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I=
  D 2 4  D14 
64
 104  64 
=
64
= 427.3 mm 4
y = 5 mm and M = 36000 N mm from Problem 1.
 =
M y 36000 N mm  5 mm
 421N / mm 2  421106 N / m 2
=
I
427.3mm 4
i.e. the maximum bending stress,  = 421 MPa
3. A uniform section beam, simply supported at its ends, is subjected to a centrally placed
concentrated load of 5 kN. The beam’s length is 1 m and its cross-section is a solid circular one.
If the maximum stress in the beam is limited to 30 MPa, determine the minimum permissible
diameter of the beam’s cross-section.
M = W  L = 2.5 kN  0.5m = 2500 N  0.5 1000 mm = 1.25  10 6 N mm
ˆ  30 MPa  30 N / mm2
y=
d
2
My
 =
=
I
d
1.25  106 N mm  mm
2
 30 N / mm 2
d 4
mm 4
64
i.e.
30 
1.25 106  64
2 d 3
from which,
d3 
1.25 106  64
 424413
2 30
Hence,
d  3 424413
i.e.
diameter, d = 75 mm
150
© John Bird & Carl Ross Published by Taylor and Francis
4. If the cross-section of the beam of Problem 3 were of rectangular shape, as shown below,
determine its dimensions. Bending can be assumed to take place about the xx axis.
From Problem 3,  = 30 MPa, M = 1.25 106 N mm
D
 D3 D 4
D
bD
 2

Now y =
and I =
2
12
12
24
3
From above,  =
My
I
1.25  106 
i.e. 30 =
i.e.
D3 
from which,
D=
and
b=
D4
24
D
6
2  1.25 10 12
D3
1.25  106  12
 500000
30
3
500000 = 79.4 mm
D 79.3

= 39.7 mm
2
2
Hence, the dimensions of the beam are 79.4 mm by 39.7 mm
5. If the cross-section of the beam of Problem 3 is a circular tube of external diameter d and
internal diameter d/2, determine the value of d.
4

d 
  d4    

 2  
d
If d is the external diameter, I = 
and y 
64
2
From problem 3, M = 1.25  10 6 N mm
151
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ˆ yˆ
M
ˆ 
I
d
d
1.25 106 
2
2
30 
 4 d4 
 15 4 
 d 
 d  
 16 
16 

64
64
1.25 106 
hence,
  15 4  2 1.25 106
 d  
64  16  d
30
Hence,
i.e.
0.0920d3  41666.67
and
d3 
Hence,
41666.67
= 452898.6
0.0920
external diameter, d =
3
452898.6 = 76.8 mm
6. A cantilever of length 2 m, carries a uniformly distributed load of 30 MN/m, as shown below.
Determine the maximum stress in the cantilever.
l wl2 30N / m  22 m 2

 60 N m  60 103 N mm
M = w l 
2
2
2
d = 25 mm; ŷ 
ˆ 
d 25

= 12.5 mm;
2 2
I=
d 4  254

 19175 mm 4
64
64
ˆ yˆ 60 103 N mm 12.5 mm
M

I
19175 mm 4
i.e. maximum stress = 39.1 MPa
7. If the cantilever of Problem 6 were replaced by a uniform section beam, simply supported at its
ends and carrying the same uniformly distributed load, determine the maximum stress in the
beam. The cross-section of the beam may be assumed to be the same as that of Problem 6.
152
© John Bird & Carl Ross Published by Taylor and Francis
M̂ = 30 N × 1 m – 30 N × 1 m × 0.5 = 15 N m = 15  103 N mm
Using the values from Problem 6 gives:
ˆ 
M̂ y 15 103 N mm 12.5 mm

I
19175 mm 4
i.e. maximum stress = 9.78 MPa
8. If the load in Problem 7 were replaced by a single concentrated load of 120 N, placed at a
distance of 0.75 m from the left support, what would be the maximum stress in the beam due to
this concentrated load.
Taking moments about B gives:
R1  2  120   2  0.75
i.e.
R1 =
150
= 75 N
2
M̂ occurs at point C, and M̂ = R 1 × 0.75 = 75 × 0.75 = 56.25 N m = 56.25 103 N mm
ˆ 
M̂ y 56.25 103 N mm 12.5 mm

I
19175 mm 4
i.e. maximum stress = 36.7 MPa
153
© John Bird & Carl Ross Published by Taylor and Francis
9. If the beam shown in problem 6 were replaced by another beam of the same length, but which
had a cross-section of tee form, as shown below, determine the maximum stress in the beam.
Section
a
1
2
75
175
250

y
y
37.5
17.5
ay
2812.5
3062.5
5875
ay2
105469
53594
159063
i
156.3
17865
18021
 ay  5875 = 23.5 mm
 a 250
IXX   ay2   i = 159063 + 18021 = 177084 mm 4
I NA  IXX  y2  a = 177084 – 23.5 2 × 250 = I = 39022 mm 4
ˆ 
M̂ y 56.25 103 N mm  23.5 mm

I
39022 mm 4
i.e. maximum stress = 33.9 MPa
154
© John Bird & Carl Ross Published by Taylor and Francis
EXERCISE 52, Page 125
Answers found from within the text of the chapter, pages 120 to 125.
EXERCISE 53, Page 125
1. (b) 2. (b) 3. (c)
155
© John Bird & Carl Ross Published by Taylor and Francis