CHAPTER 8
Thin and Thick Cylinders and Spheres
Problem 1. A shell 3.25 m long and 1 m diameter, is subjected to an internal pressure of 1.2 N/mm2. If the
thickness of the shell is 10 mm find the circumferential and longitudinal stresses. Find also the maximum
shear stress and changes in dimensions of the shell. Take E = 200 kN/mm2 and Poissons ratio = 0.3.
Solution:
L = 3.25 m = 3250 mm
d = 1 m = 1000 mm
t = 10 mm
p = 1.2 N/mm2
2
E = 200 kN/mm = 200 × 103 N/mm2
pd 1.2 × 1000
=
= 60 N/mm2
∴
Hoop stress f1 =
2t
2 × 10
Longitudinal stress f2 =
pd 1.2 × 1000
=
= 30 N/mm2
4t
4 × 10
f1 − f 2 60 − 30
=
= 15 N/mm2
2
2
δd
f
f
Diametral strain =
= e1 = 1 − µ 2
d
E
E
1
=
( f1 − µ f 2 )
E
1
=
(60 − 0.3 × 30)
200 × 103
qmax =
2.55 × 10–4
∴
2.55 × 10–4 × d
2.55 × 10–4 × 1000
0.255 mm.
δL f 2
f
Longitudinal strain =
=
−µ 1
L
E
E
1
=
( f 2 − µ f1 )
E
1
=
(30 − 0.3 × 60)
200 × 103
=
δd =
=
=
Ans.
= 6 × 10–5
δL = 6 × 10–5 × L = 6 × 10–5 × 3250 = 0.195 mm.
Ans.
Change in volume:
δV
V
= 2e1 + e2 = 2 × 2.55 × 10–4 + 6 × 10–5
= 5.7 × 10–4
68
δV = 5.7 × 10–4 × V = 5.7 × 10–4 ×
π
10002 × 3250
4
= 1454950 mm3.
Ans.
Problem 2. A cylindrical shell 2.4 m long 600 mm in diameter with metal thickness 12 mm is completely
filled with water at atmospherical pressure. If an additional 300,000 mm3 water is then pumped in, find the
stresses developed and change in dimensions. Take E = 2 × 105 N/mm2, µ = 0.3.
pd p × 600
Solution:
f1 =
=
= 25 p
2t
2 × 12
f2 =
Diametrical strain e1 =
=
=
Longitudinal strain e2 =
=
=
Volumetric strain =
=
=
∴
p =
=
∴
∴
∴
=
Hoop stress =
δd
=
d
δd =
δL
=
L
pd p × 600
=
= 12.5 p
4t
4 × 12
δd
f
f
= 1 −µ 2
d
E
E
1
[25 p − 0.3 × 12.5 p]
E
21.25 p
E
δL 1
= ( f 2 − µ f1 )
L E
1
[12.5 p − 0.3 × 25 p]
E
5p
E
δV
= 2e1 + e2
V
21.25 p 5 p
2×
+
E
E
p
47.5
E
δV
E
×
V
47.5
300,000
×
2 × 105
47.5
π
× 6002 × 2400
4
1.861413 N/mm2
25 p = 25 × 1.861413 = 46.537 N/mm2
21.25 p
e1 =
E
21.25 p
× 600 = 0.1186 mm
E
5p
e2 =
E
69
5× p
× L = 0.1116 mm
E
f2 = 12.5 p = 23.268 N/mm2
δL =
δV = 300,000 mm3.
Problem 3. The diameter of a riveted boiler is 1.5 m and has to withstand a pressure of 2 N/mm2. Find the
thickness of plates to be used if efficiency is 85% in longitudinal joints and 40% in circumferential joints.
The permissible stress is 150 N/mm2.
Solution: Let ‘t’ be the thickness of plate. Equating bursting force longitudinal joint strength, we get,
pdL = n1 2t L f
t =
pd
2 × 1500
=
= 11.764 mm
2 fn1 2 × 150 × 0.85
Considering longitudinal forces
∴
πd 2
p = n2 πdtf
4
pd
2 × 1500
t =
=
= 12.5 mm
4 f n2 4 × 150 × 0.4
∴ Provide a minimum thickness of 12.5 mm.
Problem 4. A copper cylinder of 100 mm diameter and metal thickness 4 mm is closely wound with steel
wire of 2 mm diameter with tensile stress of 60 N/mm2. Find the stresses in copper cylinder and steel wire
when a fluid is filled at a pressure of 4 N/mm2. Take Es = 2 × 105 N/mm2, Ec = 1.2 × 105 N/mm2 and
µc = 0.28.
Solution: Consider 2 mm length of cylinder
No. of wires = 1
∴ Force exerted by steel wire at diametral section
= fw0 × 2 ×
πd 2
×1
4
π × 22
× 1 = 376.991 N.
4
If initial stress is fc in cylinder, then
= 60 × 2 ×
fc × 2t × 2 = 376.991
fc =
376.991
= 23.561 N/mm2
2×4×2
(comp.)
Let due to fluid pressure alone, stresses developed in steel wire be fw and in cylinder.
Let it be f1 and f2.
Then,
f2 = longitudinal stress
=
pd 4 × 100
=
= 25 N/mm2 (tensile)
4t
4×4
Considering the equilibrium of the cylinder (Ref. Fig. 1) of
2 mm length, we get
π
fw × 2 × × 22 × 1 + f1 × 2 × 4 × 2
4
= 4 × 100 × 2
70
Fig. 1
6.883 fw + 16 f1 = 800
fw + 2.546 f1 = 127.327
... (1)
Equating strain in wire to circumferential strain in cylinder, we get
fw
1
= (f1 – µ × f2)
Es
Ec
fw
5
2 × 10
= (f1 – 0.28 × 25)
1
1.2 × 105
fw = 1.667 f1 – 11.667
... (2)
Substituting it in eqn. (1), we get
1.667 f1 – 11.667 + 2.546 f1 = 127.327
∴
f1 =
138.994
= 32.991 N/mm2
4.213
From eqn. (2),
fw = 1.667 × 32.991 – 11.667
= 43.328 N/mm2
Hence, final stresses are
(a) in steel wire = 60 + 43.328 = 103.328 N/mm2
(b) in cylinder = – 23.561 + 32.991 = 9.43 N/mm2
Problem 5. A thin spherical shell of diameter 1.2 m has metal thickness of 10 mm at atmospheric pressure.
Find the change in diameter and the capacity of the shell if fluid pressure is raised to 2.5 N/mm2.
Take E = 2 × 105 N/mm2, µ = 0.25.
Solution: d = 1.2 m = 1200 mm
p = 2.5 N/mm2
t = 10 mm
pd 2.5 × 1200
Hoop stress = f1 = f2 =
=
= 75 N/mm2
4t
4 × 10
∴
and,
∴
δd
pd
=
= (1 – µ)
d
4tE
δd
2.5 × 1200
=
[1 − 0.25]
1200
4 × 10 × 2 × 105
δd = 0.3375 mm.
δV
pd
= 3
(1 − µ )
V
4 tE
δV =
=
3 × 2.5 × 1200
4 × 10 × 2 × 105
3 × 2.5 × 1200
4 × 10 × 2 × 105
Ans.
(1 − 0.25) ×
× 0.75 ×
π 3
d
6
π
× (1200)3
6
i.e.
δV = 763407 mm3.
Ans.
Problem 6. In a chemical plant a sphere of diameter 900 mm and metal thickness 8 mm is used to store a
gas. If the permissible stress in the metal is 150 N/mm2, find the maximum pressure with which the gas can
be stored if
(i) the sphere is seamless
(ii) the efficiency of joint is 0.65.
71
Solution: (i) When the sphere is seamless :
pd
f =
4t
p × 900
150 =
4×8
∴
p = 5.333 N/mm2.
(ii) When the efficiency of the joint is 0.85,
pd
f =
4t η
Ans.
f 4t η 150 × 4 × 8 × 0.85
=
d
900
Ans.
i.e.
p = 3.467 N/mm2.
Problem 7. A thick cylindrical pipe of outside diameter 300 mm and internal diameter of 200 mm is
subjected to an internal fluid pressure of 20 N/mm2 and external fluid pressure of 5 N/mm2. Determine the
maximum hoop stress developed and draw the variation of hoop stress and radial stress across the
thickness. Indicate values at every 25 mm interval.
200
= 100 mm
Solution:
ri =
2
300
ro =
= 150 mm
2
If the hoop stress and radial stress in the cylinder at a distance x from the centre of cylinder is fx and px
respectively, from Lame’s equation,
b
b
fx =
+ a and px = 2 − a
2
x
x
where a and b are constants.
Now, at
x = 100 mm, px = 20 N/mm2
∴
p =
∴
20 =
At
b
−a
1002
x = 150 mm, px = 5 N/mm2
b
−a
1502
Substituting equation (2) from equation (1), we get
∴
5 =
15 =
b
100
= b
∴
∴ From eqn. (1),
∴
∴
2
−
1 
 1
= b
−

2
150
1502 
 100
b
2
1502 − 1002
1002 × 1502
b = 269978.4
269978.4
20 =
−a
1002
a = 6.9976
269978.4
fx =
+ 6.9976
x2
72
... (1)
... (2)
∴ Maximum hoop stress occurs when x is least i.e. at x = 100 mm
269978.4
∴
fmax =
+ 6.9976
1002
= 34 N/mm2
269978.4
f125 =
+ 6.9976 = 24.4 N/mm2
1252
269978.4
f150 =
+ 6.9976 = 19 N/mm2
1502
269978.4
p100 =
− 6.9976 = 20 N/mm2
1002
269978.4
p125 =
− 6.9976 = 10.481 N/mm2
1252
269978.4
p150 =
− 6.9976 = 5 N/mm2
1502
Hence variation of fx and px are as shown in the Fig. 2:
Fig. 2
Problem 8. A thick cylindrical pipe of internal radius 150 mm and external radius 200 mm is subjected to
an internal fluid pressure of 17.5 N/mm2. Determine the maximum hoop stress in the cross-section. What is
the percentage error if it is determined from thin cylinders theory?
Solution:
r1 = 150 mm r2 = 200 mm
If the radial stress and hoop stress developed in the cylinder at a distance of x from the centre are px and
fx respectively, from Lame’s equation,
px =
b
x2
b
−a
+a
x2
where a and b are arbitrary constants.
At x = 150 mm px = 17.5 N/mm2
fx =
∴
17.5 =
∴
0 =
b
1502
Again when x = 200 mm px = 0
b
2002
−a
... (1)
−a
... (2)
73
From eqn. (1) and (2), we get
17.5 =
b
1502
−
b
2002
= b
2002 − 1502
1502 × 2002
∴
b = 900205.76
b
900205.76
Hence
a =
=
= 22.505
2002
2002
The value of fx is maximum when x is least
i.e. at
x = 150 mm
900205.76
+ 22.505
1502
= 62.5 N/mm2.
Ans.
Internal pressure × Internal diameter
17.5 × 2 × 150
Maximum hoop stress =
=
= 52.5 N/mm2
2 × thickness
2(200 − 150)
∴ Maximum hoop stress =
62.5 − 52.5
× 100
62.5
= 16.
Ans.
Problem 9. The internal and external diameters of a thick cylinder are 300 mm and 500 mm respectively. It
is subjected to an external pressure of 4 N/mm2. Find the internal pressure that can be applied if the
permissible stress in cylinder is limited to 13 N/mm2. Sketch the variation of hoop stress and radial stress
across the thickness of the cylinder.
300
500
= 150 mm r2 =
= 250 mm
Solution:
r1 =
2
2
From Lame’s equations,
b
b
px = 2 − a
fx = 2 + a
x
x
At x = 250 mm px = 4 N/mm2
∴
Percentage error =
b
−a
... (1)
2502
Maximum hoop stress occurs for the least value of x i.e. at inner edge, where x = r1 = 150 mm.
∴
4 =
∴
13 =
b
1502
+a
... (2)
From eqn. (1) and (2), we get
17 =
∴
From eqn. (1),
b
2502
+
b
1502
=b
1502 + 2502
2502 × 1502
b = 281270.68
a =
b
2
−4=
281270.68
250
2502
= 0.5
281270.68
∴
px =
− 0.5
x2
Internal pressure i.e. pressure at x = 150 mm, is
74
−4
pi =
f150 =
f200 =
f250 =
p150 =
p200 =
281270.68
1502
281270.68
2
150
281270.68
2002
281270.68
2502
281270.68
1502
281270.68
2
200
281270.68
− 0.5 = 12 N/mm2
13
12
+ 0.5 = 13 N/mm2
7.53
5.0
+ 0.5 = 7.53
6.53
N/mm2
150
150
4.0
250
+ 0.5 = 5 N/mm2
− 0.5 = 12 N/mm2
Fig. 3
− 0.5 = 6.53 N/mm2
− 0.5 = 4 N/mm2
2502
Problem 10. A compound cylinder of inner radius 100 mm, outer radius 240 mm has common radius at
180 mm. The radial pressure developed at junction is 12 N/mm2. Determine the radial and hoop stresses
developed at inner, common and outer radii when the fluid is admitted at a pressure of 60 N/mm2.
Solution: (a) Inner cylinder : Let Lame’s equation be,
b
b
px = 12 − a1 and fx = 12 + a1
x
x
At x = r1 = 100 mm, px = 0.
and
p250 =
0 =
b1
12 =
b1
1002
At x = r2 = 180 mm, px = 12 N/mm2
∴
1802
− a1
... (1)
− a1
... (2)
From eqns. (1) and (2),
12 =
b1
180
2
−
b1
100
2
∴
b1 = – 173565.99
∴
a1 =
∴
f100 =
b1
1002
+ a1 =
=
f180 =
b1
1802
+ a1 =
=
(b) Outer cylinder : Let Lame’s
px =
At
b1
=
= b1
1002 − 1802
1802 × 1002
− 173565.99
= – 17.357
100
1002
− 173565.99
− 17.357
1002
– 34.717 N/mm2
− 173565.99
− 17.357
1802
– 22.71 N/mm2
equation be,
b2
b
− a2 and fx = 22 + a2
2
x
x
2
x = 180 mm, px = 12
75
∴
12 =
b2
1802
− a2
... (3)
− a2
... (4)
At x = 240 mm, px = 0
∴
0 =
b2
2402
∴ From equations (3) and (4), we get
1 
2402 − 1802
 1
−
= b2
12 = b2 
2
2
240 
1802 × 2402
180
∴
b2 = 888685.71
From equation (4),
a2 =
b2
=
888685.71
= 15.42
2402
888685.71
f180 =
+ a2 =
+ 15.42 = 42.84 N/mm2
1802
1802
888685.71
+ 15.42 = 30.84 N/mm2
f240 =
2402
b1
− 173565.99
− a1 =
− ( − 17.357)
p100 =
2
100
1002
= 0
173565.99
p180 = −
− ( − 17.357) = 12.0 N/mm2
1802
173565.99
− ( − 17.357) = 14.34 N/mm2
p240 =
2402
(c) When fluid is pumped in, let Lame’s equation be,
b
b
px = 32 − a3 and fx = 32 + a3
x
x
At x = 100, px = 60 N/mm2
60 =
240
b2
2
b3
1002
− a3
... (5)
− a3
... (6)
and at x = 240 mm, px = 0
0 =
b3
60 =
b3
2402
From equations (5) and (6), we get
100
2
−
b3
240
∴
b3 = 726050.42
Hence
a3 =
p100
2
= b3
2402 − 1002
1002 × 2402
b3
= 12.605
2402
= 60 N/mm2
p180 =
726050.42
1802
− 12.605 = 9.804 N/mm2
76
p240 = 0
f180
726050.42
+ a3 = 85.209 N/mm2
1002
= 35.014 N/mm2
f100 =
f240 = 25.210 N/mm2
∴ Final stresses are:
f100 = – 34.717 + 85.209 = 50.492 N/mm2
f180, inner = – 22.71 + 35.014 = 12.304 N/mm2
f180, outer = 42.84 + 35.014 = 77.854 N/mm2
f240 = 30.84 + 25.21 = 56.05 N/mm2
Problem 11. A compound cylinder is to made with inner radius of 160 mm and outer radius of 320 mm.
The radius at common junction is to be 240 mm. If the two cylinders with allowance 0.3 mm are used, find
the radial pressure developed at contact surfaces. Also determine the hoop stresses induced at inner edge,
common edge and outer edge of compound cylinder. Take Young’s modulus E = 2 × 105 N/mm2.
Solution:
r1 = 160 mm r2 = 240 mm r0 = 320 mm
Let p be the radial pressure developed at junction. Let Lame’s equations for internal cylinder be
b
b
px = 12 − a1
fx = 12 + a1
x
x
At x = 160 mm px = 0
0 =
b1
1602
− a1
... (1)
At x = 240 mm, px = p
b1
− a1
2402
From eqns. (1) and (2), we get
b1
b
p =
− 12
2
240
160
∴
p =
or
 1602 − 2402 
p = b1 
 1602 × 2402 


∴
b1 = – 46082.94 p.
Hence
a1 =
− 46082.94 p
1602
... (2)
= – 1.8 p.
Hoop stress at junction is
− 46082.94 p
− 1.8 p
2402
= – 2.6 p
Lame’s equation for outer cylinder be,
b
b
px = 22 − a2 and fx = 22 + a2
x
x
At x = 240 mm, px = p
f21 =
77
b2
− a2
... (3)
− a2
3202
From equations (3) and (4), we get
... (4)
p =
2402
At x = 320 mm, px = 0
∴
0 =
b2
p =
b2
2402
−
b2
3202
= b2
(3202 − 2402 )
2402 × 3202
b2 = 131657.14 p.
∴ From eqn. (4),
a2 =
b2
2
=
131657.14 p
320
3202
= 1.285 p.
∴ At junction, hoop stress in outer cylinder is
b2
131657.14 p
=
+ a2 =
+ 1.285 p
2402
2402
= 3.57 p
Considering the circumferential strain, the compatibility equation is,
δr
1
=
(f21 + f20)
r2
E
where f21 is compressive and f20 is tensile.
0.15
1
=
[2.6 p + 3.57 p]
240
2 × 105
p = 20.25 N/mm2
b1
f1 =
+ a1
1602
− 46082.94 × 20.25
− 1.8 × 20.25
=
1602
= – 72.9 N/mm2
b1
− 46082.94 × 20.25
f2, inner =
+ a1 =
− 1.8 × 20.25
2
240
2402
= 52.65 N/mm2
b2
f2, outer =
+ a2 = 3.57 p
2402
= 3.57 × 20.25 = 72.29 N/mm2
b2
131657.14 × 20.25
fouter =
+ a2 =
+ 1.285 × 20.25
3202
3202
= 52 N/mm2.
Problem 12. A spherical shell with internal diameter 320 mm and 640 mm external diameter is subjected
to an internal fluid pressure of 75 N/mm2. Find the hoop stresses developed at 40 mm interval across the
thickness.
Solution:
r0 = 320 mm r1 = 160 mm p = 75 N/mm2
∴
78
The radial pressure and the hoop stress at any radial distance x are given by,
2b
b
px = 3 − a and fx = 3 + a
x
x
Now, at x = 160 mm, px = 75 N/mm2
2b
−a
... (1)
−a
3203
From equations (1) and (2), we get
... (2)
∴
75 =
1603
At x = 320 mm, px = 0
∴
0 =
2b
1 
 1
75 = 2b 
−
3
3203 
 160
∴
From equation (2),
b = 175542857
a =
f =
∴
f160 =
f200 =
f240 =
f280 =
f320 =
2 × 175542857
3203
175542857
x3
175542857
1603
175542857
2003
175542857
2403
175542857
2803
175542857
3203
= 10.714
+ 10.714
+ 10.714 = 53.57 N/mm2
+ 10.714 = 32.656 N/mm2
+ 10.714 = 23.41 N/mm2
+ 10.714 = 18.71 N/mm2
+ 10.714 = 16.071 N/mm2
79
Ans.