Shivendra Nandan Galgotias University

1. THIN AND THICK CYLINDERS
They are,
In many engineering applications, cylinders are frequently
used for transporting or storing of liquids, gases or fluids.
Eg: Pipes, Boilers, storage tanks etc.
These cylinders are subjected to fluid pressures. When a
cylinder is subjected to a internal pressure, at any point on the
cylinder wall, three types of stresses are induced on three
mutually perpendicular planes.
INTRODUCTION:

2. 2. Longitudinal Stress (σL ) – This stress is directed along the
length of the cylinder. This is also tensile in nature and tends
to increase the length.
3. Radial pressure ( pr ) – It is compressive in nature.
Its magnitude is equal to fluid pressure on the inside wall and
zero on the outer wall if it is open to atmosphere.
1. Hoop or Circumferential Stress (σC) – This is directed along the
tangent to the circumference and tensile in nature. Thus, there
will be increase in diameter.

3. σ C σ L
1. Hoop Stress (σC) 2. Longitudinal Stress (σL) 3. Radial Stress (pr)
Element on the cylinder
wall subjected to these
three stresses
σ C
σ C
σC
p
σ L
σ L
σ L
p p
pr
σ Lσ L
σ C
σ C
pr
pr

4. INTRODUCTION:
A cylinder or spherical shell is considered to be thin when the
metal thickness is small compared to internal diameter.
i. e., when the wall thickness, ‘t’ is equal to or less than
‘d/20’, where ‘d’ is the internal diameter of the cylinder or shell,
we consider the cylinder or shell to be thin, otherwise thick.
Magnitude of radial pressure is very small compared to
other two stresses in case of thin cylinders and hence neglected.
THIN CYLINDERS

5. Longitudinal
axisLongitudinal stress
Circumferential stress
t
The stress acting along the circumference of the cylinder is called
circumferential stresses whereas the stress acting along the length of
the cylinder (i.e., in the longitudinal direction ) is known as
longitudinal stress

6. The bursting will take place if the force due to internal (fluid)
pressure (acting vertically upwards and downwards) is more than the
resisting force due to circumferential stress set up in the material.
p
σc σc
P - internal pressure (stress)
σc –circumferential stress

7. P - internal pressure (stress)
σc – circumferential stress
dL
σc
p
t

8. EVALUATION OF CIRCUMFERENTIAL or HOOP STRESS (σC):
Consider a thin cylinder closed at both ends and subjected to internal
pressure ‘p’ as shown in the figure.
Let d=Internal diameter, t = Thickness of the wall
L = Length of the cylinder.
p d
t
σcσc
dlt
p
d

9. To determine the Bursting force across the diameter:
Consider a small length ‘dl’ of the cylinder and an elementary
area ‘dA’ as shown in the figure.
rpp dθdldAdF ×××=×=
dθdldFx ××××= θcos
2
d
p
dA
σcσc
dlt
p
d
dθ
θ
Force on the elementary area,
Horizontal component of this force
dθdl ×××=
2
d
p
dθdldFy ××××= θsin
2
d
p
Vertical component of this force

10. The horizontal components cancel out
when integrated over semi-circular
portion as there will be another equal
and opposite horizontal component on
the other side of the vertical axis.
sin
2
d
pforceburstingldiametricaTotal
0
∫ ××××=∴
π
θ dθdl
dA
σcσc
dlt
p
θ
d
dθ
[ ]
surface.curvedtheofareaprojectedp
dpcosdl
2
d
p 0
×=
××=−×××= dlπ
θ

11. dlcc ×××=∴ tσ2)σstressntialcircumfereto(dueforceResisting
dldl ××=××× dptσ2i.e., c
)1....(....................
t2
dp
σstress,ntialCircumfere c
×
×
=∴
dL
σc
p
t
forceBurstingforceResistingum,equillibriUnder =

12. )1....(....................
t2
dp
σstress,ntialCircumfere c
×
×
=∴
Force due to fluid pressure = p × area on which p is acting = p ×(d ×L)
(bursting force)
Force due to circumferential stress = σc × area on which σc is acting
(resisting force) = σc × ( L × t + L ×t ) = σc × 2 L × t
Under equilibrium bursting force = resisting force
p ×(d ×L) = σc × 2 L × t
Assumed as rectangular

13. LONGITUDINAL STRESS (σL):
p
σL
The force, due to pressure of the fluid, acting at the ends of the
thin cylinder, tends to burst the cylinder as shown in figure
P
A
B
The bursting of the cylinder takes
place along the section AB

14. EVALUATION OF LONGITUDINAL STRESS (σL):
d
4
π
pcylinder)ofend(on theforceburstingalLongitudin 2
××=
p
t
σL
tdπσforceResisting L ×××=∴
tdπforcethisresistingsectioncrossofArea ××=
cylinder.theofmaterialtheofstressalLongitudinσLet L =

15. tdπσd
4
π
pi.e., L
2
×××=××
)2.........(..........
t4
dp
σstress,alLongitudin L
×
×
=∴
LC σ2σ(2),&(1)eqsFrom ×=
forceresistingforceburstingum,equillibriUnder =

16. tdπσd
4
π
pi.e.,
forceresistingforceburstingum,equillibriUnder
L
2
×××=××
=
)2.........(..........
t4
dp
σstress,alLongitudin L
×
×
=∴
tdπσ
σσRe
d
4
π
p
actingispon whichareapressurefluidtodueForce
L
LL
2
×××=
×=
××=
×=
actingiswhichonareaforcesisting
p
circumference

17. EVALUATION OF STRAINS
A point on the surface of thin cylinder is subjected to biaxial
stress system, (Hoop stress and Longitudinal stress) mutually
perpendicular to each other, as shown in the figure. The strains due
to these stresses i.e., circumferential and longitudinal are obtained
by applying Hooke’s law and Poisson’s theory for elastic
materials.
σ C=(pd)/(2t)σ C=(pd)/(2t)
σL=(pd)/(4t)
σ L=(pd)/(4t)

18. E
σ
μ
E
σ
ε
:εstrain,ntialCircumfere
LC
C
C
×−=
)3..(..............................μ)2(
Et4
dp
d
δd
εi.e., C −×
××
×
==
σ C=(pd)/(2σC=(pd)/(2t)
σ L=(pd)/(4t)
σ L=(pd)/(4t)
Note: Let δd be the change in diameter. Then
( )
d
d
d
ddd
ncecircumfereoriginal
ncecircumfereoriginalncecircumferefinal
c
δ
π
πδπ
ε
=


 −+
=
−
=
μ)2(
E
σ
E
σ
μ
E
σ
2
L
LL
−×=
×−×=

19. E
σ
μ
E
σ
ε
:εstrain,alLongitudin
CL
L
L
×−=
)4..(..............................μ)21(
Et4
dp
L
δl
εi.e., L ×−×
××
×
==
V
v
STRAIN,VOLUMETRIC
δ
Change in volume = δV = final volume – original volume
original volume = V = area of cylindrical shell × length
L
d
4
2
π
=
μ)21(
E
σ
E
)σ2(
μ
E
σ LLL
×−×=
×
×−=

20. final volume = final area of cross section × final length
[ ] [ ]
[ ] [ ]
[ ]LddLdLdddLLdLd
LLdddd
LLdd
δδδδδδδ
π
δδδ
π
δδ
π
2)(2)(
4
2)(
4
4
2222
22
2
+++++=
+×++=
+×+=
[ ]LdddLLdvolumeFinal
LddandLdLdassuchquantitiessmallertheneglecting
δδ
π
δδδδδ
22
22
2
4
2)(,)(
++=
[ ] [ ]
[ ]LdddLV
LdLdddLLdVvolumeinchange
δδ
π
δ
π
δδ
π
δ
2
222
2
4
4
2
4
+=
−++=

21. [ ]
Ld
4
π
2
4
π
V
dv
2
2
××
+
=
dLdLd δδ
= εL + 2 × εC
V
dV
)5.......(..........μ)45(
Et4
dp
V
dv
i.e., ×−
××
×
=
μ)2(
Et4
dp
2μ)21(
Et4
dp
−
××
×
×+×−
××
×
=
d
d
2
L
L δδ
×+=

22. 2
4t
pd
2t
pd
2
σ-σ
τstress,ShearMaximum
other.eachto
larperpendicuactandnormalarestressesthese
Bothal.longitudinandntialCircumfereviz.,
point,anyatstressesprincipaltwoareThere
:stressShearMaximum
LC
max
−
=
=∴
)5.(....................
8t
pd
τi.e., max =
σ C=(pd)/(2t)
σC=(pd)/(2t)
σ L=(pd)/(4t)
σ L=(pd)/(4t)

23. 2
4t
pd
2t
pd
2
σ-σ
τstress,ShearMaximum
:stressShearMaximum
LC
max
−
=
=∴
)5.(....................
8t
pd
τi.e., max =

24. PROBLEM 1:
A thin cylindrical shell is 3m long and 1m in internal diameter. It is
subjected to internal pressure of 1.2 MPa. If the thickness of the sheet is
12mm, find the circumferential stress, longitudinal stress, changes in
diameter, length and volume . Take E=200 GPa and μ= 0.3.
1. Circumferential stress, σC:
σC= (p×d) / (2×t)
= (1.2×1000) / (2× 12)
= 50 N/mm2
= 50 MPa (Tensile).
SOLUTION:
2. Longitudinal stress, σL:
σL = (p×d) / (4×t)
= σC/2 = 50/2
= 25 N/mm2
= 25 MPa (Tensile).
ILLUSTRATIVE PROBLEMS

25. 3. Circumferential strain, εc:
Change in length = ε L ×L= 5×10-05
×3000 = 0.15 mm (Increase).
E
μ)(2
t)(4
d)(p
εc
−
×
×
×
=
Change in diameter, δd = εc ×d
= 2.125×10-04
×1000 = 0.2125 mm (Increase).
E
μ)2(1
t)(4
d)(p
εL
×−
×
×
×
=
4. Longitudinal strain, εL:
(Increase)102.125
10200
0.3)(2
12)(4
1000)(1.2
04-
3
×=
×
−
×
×
×
=
(Increase)105
10200
0.3)2(1
12)(4
1000)(1.2
05-
3
×=
×
×−
×
×
×
=

26. μ)4(5
Et)(4
d)(p
V
dv
×−×
××
×
=
:
V
dv
strain,Volumetric
V104.75dvin volume,Change -4
××=∴
(Increase)104.75
)3.045(
10200)124(
)10002.1(
4-
3
×=
×−×
×××
×
=
.Litres11919.1
m101.11919mm101.11919
30001000
4
π
104.75
33-36
24-
=
×=×=
××××=

27. A copper tube having 45mm internal diameter and 1.5mm wall
thickness is closed at its ends by plugs which are at 450mm apart. The
tube is subjected to internal pressure of 3 MPa and at the same time
pulled in axial direction with a force of 3 kN. Compute: i) the change
in length between the plugs ii) the change in internal diameter of the
tube. Take ECU = 100 GPa, and μCU = 0.3.
A] Due to Fluid pressure of 3 MPa:
Longitudinal stress, σL = (p×d) / (4×t)
= (3×45) / (4× 1.5) = 22.50 N/mm2
= 22.50 MPa.
SOLUTION:
Change in length, δL= εL × L = 9 × 10-5
×450 = +0.0405 mm (increase)
E
)μ21(
t4
d)(p
εstrain,Long. L
×−
×
×
×
=
5
3
109
10100
)3.021(5.22 −
×=
×
×−×
=

28. Change in diameter, δd= εc × d = 3.825 × 10-4
×45
= + 0.0172 mm (increase)
B] Due to Pull of 3 kN (P=3kN):
Area of cross section of copper tube, Ac = π × d × t
= π × 45 × 1.5 = 212.06 mm2
Longitudinal strain, ε L = direct stress/E = σ/E = P/(Ac × E)
= 3 × 103
/(212.06 × 100 × 103
)
= 1.415 × 10-4
Change in length, δL=εL× L= 1.415 × 10-4
×450= +0.0637mm (increase)
E
)μ2(
t)(4
d)(p
εstrainntialCircumfere C
−
×
×
×
=
Pd/4t = 22.5
4
3
10825.3
10100
)3.02(5.22 −
×=
×
−×
=

29. Lateral strain, εlat= -μ × Longitudinal strain = -μ × εL
= - 0.3× 1.415 × 10-4
= -4.245 × 10-5
Change in diameter, δd = εlat × d = -4.245 × 10-5
×45
= - 1.91 × 10-3
mm (decrease)
C) Changes due to combined effects:
Change in length = 0.0405 + 0.0637 = + 0.1042 mm (increase)
Change in diameter = 0.01721 - 1.91 × 10-3
= + 0.0153 mm (increase)

30. PROBLEM 3:
A cylindrical boiler is 800mm in diameter and 1m length. It is
required to withstand a pressure of 100m of water. If the permissible
tensile stress is 20N/mm2
, permissible shear stress is 8N/mm2
and
permissible change in diameter is 0.2mm, find the minimum thickness
of the metal required. Take E = 200GPa, and μ = 0.3.
Fluid pressure, p = 100m of water = 100×9.81×103
N/m2
= 0.981N/mm2
.
SOLUTION:
1. Thickness from Hoop Stress consideration: (Hoop stress is critical
than long. Stress)
σC = (p×d)/(2×t)
20 = (0.981×800)/(2×t)
t = 19.62 mm

31. 2. Thickness from Shear Stress consideration:
3. Thickness from permissible change in diameter consideration
(δd=0.2mm):
E
μ)(2
t)(4
d)(p
d
δd −
×
×
×
=
Therefore, required thickness, t = 19.62 mm.
t)(8
d)(p
τmax
×
×
=
12.26mm.t
t)(8
800)(0.981
8
=∴
×
×
=
6.67mmt
10200
)3.02(
)t4(
)800981.0(
800
2.0
3
=
×
−
×
×
×
=

32. PROBLEM 4:
A cylindrical boiler has 450mm in internal diameter, 12mm thick and
0.9m long. It is initially filled with water at atmospheric pressure.
Determine the pressure at which an additional water of 0.187 liters
may be pumped into the cylinder by considering water to be
incompressible. Take E = 200 GPa, and μ = 0.3.
Additional volume of water, δV = 0.187 liters = 0.187×10-3
m3
= 187×103
mm3
SOLUTION:
3632
mm10143.14)109.0(450
4
π
V ×=×××=
μ)45(
Et4
dp
V
dV
×−
××
×
=
Solving, p=7.33 N/mm2
)33.045(
10200124
450p
10143.14
10187
36
3
×−
×××
×
=
×
×

33. JOINT EFFICIENCY
Longitudinal
rivets
Circumferential
rivets
Steel plates of only particular lengths and width are available. Hence
whenever larger size cylinders (like boilers) are required, a number
of plates are to be connected. This is achieved by using riveting in
circumferential and longitudinal directions as shown in figure. Due
to the holes for rivets, the net area of cross section decreases and
hence the stresses increase.

34. JOINT EFFICIENCY
The cylindrical shells like boilers are having two types of joints
namely Longitudinal and Circumferential joints. Due to the holes for
rivets, the net area of cross section decreases and hence the stresses
increase. If the efficiencies of these joints are known, the stresses
can be calculated as follows.
Let η L= Efficiency of Longitudinal joint
and η C = Efficiency of Circumferential joint.
...(1)..........
ηt2
dp
σ
L
C
××
×
=
Circumferential stress is given by,

35. ...(2)..........
ηt4
dp
σ
C
L
××
×
=
Note: In longitudinal joint, the circumferential stress is developed
and in circumferential joint, longitudinal stress is developed.
Longitudinal stress is given by,
Circumferential
rivets
Longitudinal
rivets

36. If A is the gross area and Aeff is the effective resisting area then,
Efficiency = Aeff/A
Bursting force = p L d
Resisting force = σc ×Aeff = σc ×ηL ×A = σc ×ηL ×2 t L
Where η L=Efficiency of Longitudinal joint
Bursting force = Resisting force
p L d = σc ×ηL × 2 t L
...(1)..........
ηt2
dp
σ
L
C
××
×
=

37. If η c=Efficiency of circumferential joint
Efficiency = Aeff/A
Bursting force = (π d2
/4)p
Resisting force = σL ×A′eff =σL ×ηc ×A′ = σL ×ηc ×π d t
Where η L=Efficiency of circumferential joint
Bursting force = Resisting force
...(2)..........
ηt4
dp
σ
C
L
××
×
=

38. A cylindrical tank of 750mm internal diameter, 12mm thickness and
1.5m length is completely filled with an oil of specific weight
7.85 kN/m3
at atmospheric pressure. If the efficiency of longitudinal
joints is 75% and that of circumferential joints is 45%, find the
pressure head of oil in the tank. Also calculate the change in volume.
Take permissible tensile stress of tank plate as 120 MPa and E = 200
GPa, and μ = 0.3.
Let p = max permissible pressure in the tank.
Then we have, σL= (p×d)/(4×t) η C
120 = (p×750)/(4×12) 0.45
p = 3.456 MPa.
SOLUTION:
Also, σ C= (p×d)/(2×t) η L
120 = (p×750)/(2×12) 0.75
p = 2.88 MPa.

39. Max permissible pressure in the tank, p = 2.88 MPa.
μ)45(
E)t(4
d)(p
V
dv
Strain,Vol. ×−×
××
×
=
litres.0.567m100.567
.mm10567.01500750
4
π
108.55V108.55dv
108.550.3)4-(5
)1020012(4
750)(2.88
33-
3624-4-
4-
3
=×=
×=××××=××=
×=××
×××
×
=

40. A boiler shell is to be made of 15mm thick plate having a limiting
tensile stress of 120 N/mm2
. If the efficiencies of the longitudinal and
circumferential joints are 70% and 30% respectively determine;
i) The maximum permissible diameter of the shell for an
internal pressure of 2 N/mm2
.
(ii) Permissible intensity of internal pressure when the shell
diameter is 1.5m.
(i) To find the maximum permissible diameter of the shell for an
internal pressure of 2 N/mm2
:
SOLUTION:
ηt2
dp
σe.,i.
L
c
××
×
=
a) Let limiting tensile stress = Circumferential stress = σ c =
120N/mm2
.
d = 1260 mm
7.0512
d2
120
××
×
=

41. ηt4
dp
σe.,i.
C
L
××
×
=
b) Let limiting tensile stress = Longitudinal stress = σ L = 120N/mm2
.
The maximum diameter of the cylinder in order to satisfy both the
conditions = 1080 mm.
d = 1080 mm.
3.0514
d2
120
××
×
=

42. The maximum permissible pressure = 1.44 N/mm2
.
(ii) To find the permissible pressure for an internal diameter of 1.5m:
(d=1.5m=1500mm)
a) Let limiting tensile stress = Circumferential stress = σ c =
120N/mm2
.
ηt2
dp
σe.,i.
L
c
××
×
=
b) Let limiting tensile stress = Longitudinal stress = σ L = 120N/mm2
.
ηt4
dp
σe.,i.
C
L
××
×
=
.N/mm1.68p
7.0512
5001p
120
2
=
××
×
=
.N/mm1.44p
3.0514
5001p
120
2
=
××
×
=

43. PROBLEM 1:
Calculate the circumferential and longitudinal strains for a boiler of
1000mm diameter when it is subjected to an internal pressure of
1MPa. The wall thickness is such that the safe maximum tensile stress
in the boiler material is 35 MPa. Take E=200GPa and μ= 0.25.
(Ans: ε C=0.0001531, ε L=0.00004375)
PROBLEM 2:
A water main 1m in diameter contains water at a pressure head of
120m. Find the thickness of the metal if the working stress in the pipe
metal is 30 MPa. Take unit weight of water = 10 kN/m3
.
(Ans: t=20mm)
PROBLEMS FOR PRACTICE

44. THIN AND THICK
CYLINDERS -33
PROBLEM 3:
A gravity main 2m in diameter and 15mm in thickness. It is subjected
to an internal fluid pressure of 1.5 MPa. Calculate the hoop and
longitudinal stresses induced in the pipe material. If a factor of safety
4 was used in the design, what is the ultimate tensile stress in the pipe
material?
(Ans: σC=100 MPa, σL=50 MPa, σU=400 MPa)
PROBLEM 4:
At a point in a thin cylinder subjected to internal fluid pressure, the
value of hoop strain is 600×10-4
(tensile). Compute hoop and
longitudinal stresses. How much is the percentage change in the
volume of the cylinder? Take E=200GPa and μ= 0.2857.
(Ans: σC=140 MPa, σL=70 MPa, %age change=0.135%.)

45. THIN AND THICK
CYLINDERS -34
PROBLEM 5:
A cylindrical tank of 750mm internal diameter and 1.5m long is to be
filled with an oil of specific weight 7.85 kN/m3 under a pressure head
of 365 m. If the longitudinal joint efficiency is 75% and
circumferential joint efficiency is 40%, find the thickness of the tank
required. Also calculate the error of calculation in the quantity of oil
in the tank if the volumetric strain of the tank is neglected. Take
permissible tensile stress as 120 MPa, E=200GPa and μ= 0.3 for the
tank material. (Ans: t=12 mm, error=0.085%.)

46. THICK CYLINDERS

47. INTRODUCTION:
The thickness of the cylinder is large compared to that of thin
cylinder.
i. e., in case of thick cylinders, the metal thickness ‘t’ is more
than ‘d/20’, where ‘d’ is the internal diameter of the cylinder.
Magnitude of radial stress (pr) is large and hence it cannot be
neglected. The circumferential stress is also not uniform across the
cylinder wall. The radial stress is compressive in nature and
circumferential and longitudinal stresses are tensile in nature.
Radial stress and circumferential stresses are computed by using
‘Lame’s equations’.

48. LAME’S EQUATIONS (Theory) :
4. The material is homogeneous, isotropic and obeys Hooke’s law. (The
stresses are within proportionality limit).
1. Plane sections of the cylinder normal to its axis remain plane and
normal even under pressure.
2. Longitudinal stress (σL) and longitudinal strain (εL) remain constant
throughout the thickness of the wall.
3. Since longitudinal stress (σL) and longitudinal strain (εL) are constant,
it follows that the difference in the magnitude of hoop stress and
radial stress (pr) at any point on the cylinder wall is a constant.
ASSUMPTIONS:

49. LAME’S EQUATIONS FOR RADIAL PRESSURE AND
CIRCUMFERENTIAL STRESS
Consider a thick cylinder of external radius r1 and internal radius
r2, containing a fluid under pressure ‘p’ as shown in the fig.
Let ‘L’ be the length of the cylinder.
p
r2
r1
p

50. Consider an elemental ring of radius ‘r’ and thickness ‘δr’ as shown
in the above figures. Let pr and (pr+ δpr) be the intensities of radial
pressures at inner and outer faces of the ring.
pr
pr+δpr
r2
r1
r
pr
pr+δpr
r2
r1
r
σc σc
r δr
Pr
pr+δpr
External
pressure

51. Consider the longitudinal
section XX of the ring as
shown in the fig.
The bursting force is
evaluated by considering
the projected area,
‘2×r×L’ for the inner face
and ‘2×(r+δr)×L’ for the
outer face .
The net bursting force, P = pr×2×r×L - (pr+δpr)×2×(r+δr)×L
=( -pr× δr - r×δpr - δpr × δr) 2L
Bursting force is resisted by the hoop tensile force developing at the
level of the strip i.e.,
Fr=σc×2 ×δr×L
L
r
pr
pr+δpr
r+δr
X X

52. Thus, for equilibrium, P = Fr
(-pr× δr - r×δpr- δpr × δr) 2L = σ c×2×δr×L
-pr× δr - r×δpr- δpr × δr = σ c×δr
Neglecting products of small quantities, (i.e., δpr × δr)
σ c = - pr – (r× δpr )/ δr ...…………….(1)
Longitudinal strain is constant. Hence we have,
since σL, E and μ are constants (σc – Pr) should be constant . Let it be
equal to 2a. Thus
constant
E
p
μ
E
σ
μ
E
σ rCL
=×+×− Since Pr is compressiveε L =
constant)pσ(
μ
E
σ
rC
L
=−−
E
ε L =

53. σ c- pr = 2a,
i.e., σc = pr + 2a, ………………(2)
From (1), pr+ 2a = - pr – (r× δpr ) / δr
r
r
r
p
-a)p(2
δ
δ
×=+ r
)3.(..........
a)p(
p
2
r
rr
+
=×−
δδ
r
i. e.,
Integrating, (-2 ×loge r) + c = loge (pr + a)
Where c is constant of integration. Let it be taken as loge b, where
‘b’ is another constant.
Thus, loge (pr+a) = -2 ×loge r + loge b = - loge r2
+ loge b = loge
2
r
b

54. .....(4)..........a
r
b
pstress,radialor,
r
b
api.e., 2r2r −==+
The equations (4) & (5) are known as “Lame’s Equations” for radial
pressure and hoop stress at any specified point on the cylinder wall.
Thus, r1≤r ≤r2.
a2a
b
a2pσstress,Hoop 2rc +−=+=
r
.......(5)....................a
r
b
σi.e., 2c +=
Substituting it in equation 2, we get

55. ANALYSIS FOR LONGITUDINAL STRESS
Consider a transverse section near the end wall as shown in the fig.
Bursting force, P =π×r2
2
×p
Resisting force is due to longitudinal stress ‘σ L’.
i.e., FL= σ L× π ×(r1
2
-r2
2
)
For equilibrium, FL= P
σ L× π ×(r1
2
-r2
2
)= π ×r2
2
×p
Therefore, longitudinal stress,
(Tensile)
)r(r
rp
σ 2
2
2
1
2
2
L
−
×
=
p
p
r2
r1
σL
σ LσL
σL
L

56. NOTE:
1. Variations of Hoop stress and Radial stress are parabolic across
the cylinder wall.
2. At the inner edge, the stresses are maximum.
3. The value of ‘Permissible or Maximum Hoop Stress’ is to be
considered on the inner edge.
4. The maximum shear stress (σ max) and Hoop, Longitudinal and
radial strains (εc, εL, εr) are calculated as in thin cylinder but
separately for inner and outer edges.

57. ILLUSTRATIVE PROBLEMS
PROBLEM 1:
A thick cylindrical pipe of external diameter 300mm and internal
diameter 200mm is subjected to an internal fluid pressure of 20N/mm2
and external pressure of 5 N/mm2
. Determine the maximum hoop
stress developed and draw the variation of hoop stress and radial
stress across the thickness. Show at least four points for each case.
SOLUTION:
External diameter = 300mm. External radius, r1=150mm.
Internal diameter = 200mm. Internal radius, r2=100mm.
Lame’s equations:
For Hoop stress, .........(1)
For radial stress, .........(2)
a
r
b
σ 2c +=
a
r
b
p 2r −=

58. Boundary conditions:
At r =100mm (on the inner face), radial pressure = 20N/mm2
i.e.,
Similarly, at r =150mm (on the outer face), radial pressure = 5N/mm2
i.e.,
)........(3..........a
100
b
20 2
−=
)........(4..........a
150
b
5 2
−=
..(5)..........7
r
2,70,000
σ 2c +=
..(6)..........7
r
2,70,000
p 2r −=
Solving equations (3) & (4), we get a = 7, b = 2,70,000.
Lame’s equations are, for Hoop stress,
For radial stress,

59. To draw variations of Hoop stress & Radial stress :
At r =100mm (on the inner face),
(Comp)MPa207
100
2,70,000
pstress,Radial
(Tensile)MPa347
100
2,70,000
σstress,Hoop
2r
2c
=−=
=+=
At r =120mm,
(Comp)MPa11.757
120
2,70,000
pstress,Radial
(Tensile)MPa25.757
120
2,70,000
σstress,Hoop
2r
2c
=−=
=+=
At r =135mm,
(Comp)MPa7.817
135
2,70,000
pstress,Radial
(Tensile)MPa21.817
135
2,70,000
σstress,Hoop
2r
2c
=−=
=+=

60. (Comp)MPa57
150
2,70,000
pstress,Radial
(Tensile)MPa197
150
2,70,000
σstress,Hoop
150mm,rAt
2r
2c
=−=
=+=
=
Variation of Hoop stress & Radial stress
Variation of Hoop
Stress-Tensile
(Parabolic)
Variation of Radial
Stress –Comp
(Parabolic)

61. PROBLEM 2:
Find the thickness of the metal required for a thick cylindrical shell of
internal diameter 160mm to withstand an internal pressure of 8 N/mm2
.
The maximum hoop stress in the section is not to exceed 35 N/mm2
.
SOLUTION:
Internal radius, r2=80mm.
......(2)..........a
r
b
pstress,Radialfor
(1)....................a
r
b
σStress,Hoopfor
are,equationssLame'
2r
2c
−=
+=

62. face)inneronmaxisstressHoop(.N/mm35σstress,Hoopand
,N/mm8pstressradial80mm,rat
are,conditionsBoundary
2
C
2
r
=
==
,600.37,1b13.5,agetwe(4),&(3)equationsSolving ==
)........(4..........a
08
b
35
)........(3..........a
08
b
8i.e.,
2
2
+=
−=
..(6)..........5.13
r
1,37,600
pand
..(5)..........5.13
r
1,37,600
σare,equationssLame'
2r
2c
−=
+=∴

63. .rrat0pi.e.,
0.pressureface,outerOn the
1r ==
=
20.96mm.
r-rmetaltheofThickness 21
=
=∴
100.96mm.r
5.13
r
1,37,600
0
1
2
1
=∴
−=∴

64. PROBLEM 3:
A thick cylindrical pipe of outside diameter 300mm and internal
diameter 200mm is subjected to an internal fluid pressure of 14 N/mm2
.
Determine the maximum hoop stress developed in the cross section.
What is the percentage error if the maximum hoop stress is calculated
by the equations for thin cylinder?
SOLUTION:
Internal radius, r2=100mm. External radius, r1=150mm
.
Lame’s equations:
For Hoop stress, .........(1)
For radial pressure, .........(2)
a
r
b
σ 2c +=
a
r
b
p 2r −=

65. Boundary conditions:
At x =100mm Pr = 14N/mm2
i.e.,
Similarly, at x =150mm Pr = 0
i.e.,
)........(1..........a
100
b
14 2
−=
)........(2..........a
150
b
0 2
−=
..(3)..........2.11
r
22,500
σstress,HoopforequationsLame' 2r +=∴
Solving, equations (1) & (2), we get a =11.2, b = 2,52,000.

66. MPa.36.42.11
100
252000
σ 2max =+=
Max hoop stress on the inner face (where x=100mm):
.23.08%100)
36.4
28-36.4
(errorPercentage =×=
14MPa.pand50mmt200mm,Dewher
t2
dp
σformula,cylinderBy thin max
===
×
×
=
MPa.28
502
20014
σmax =
×
×
=∴

67. PROBLEM 4:
The principal stresses at the inner edge of a cylindrical shell are
81.88 MPa (T) and 40MPa (C). The internal diameter of the
cylinder is 180mm and the length is 1.5m. The longitudinal
stress is 21.93 MPa (T). Find,
(i) Max shear stress at the inner edge.
(ii) Change in internal diameter.
(iii)Change in length.
(iv) Change in volume.
Take E=200 GPa and μ=0.3.
SOLUTION:
2
(-40)-81.88
2
p-σ
τ
:faceinneron thestressshearMaxi)
rC
max ==
= 60.94 MPa

68. σ
E
μ
-p
E
μ
-
E
σ
d
δd
:diameterinnerinChangeii)
Lr
C
××=
σ
E
μ
-p
E
μ
-
E
σ
L
δl
:LengthinChangeiii)
Cr
L
××=
0.078mm.δd
104.365
)40(
10200
0.3
-93.12
10200
0.3
-
10200
81.88
4-
333
+=∴
×=
−×
×
×
××
=
0.070mm.δl
1046.83
81.88
10200
0.3
-)40(
10200
0.3
-
10200
93.12
6-
333
+=∴
×=
×
×
−×
××
=

69. D
δd
2
L
δl
V
δV
:in volumeChangeiv)
×+=
.mm1035.11
)
4
1500180π
(109.198δV
33
2
4-
×=
××
××=∴
= 9.198 ×10-4

70. PROBLEM 5:
Find the max internal pressure that can be allowed into a thick pipe of
outer diameter of 300mm and inner diameter of 200mm so that tensile
tress in the metal does not exceed 16 MPa if, (i) there is no external
luid pressure, (ii) there is a fluid pressure of 4.2 MPa.
SOLUTION:
External radius, r1=150mm.
Internal radius, r2=100mm.
Boundary conditions:
At r=100mm , σc = 16N/mm2
At r=150mm , Pr = 0
Case (i) – When there is no external fluid pressure:

71. )........(2..........a
150
b
0
)........(1..........a
100
b
16i.e.,
2
2
−=
+=
Solving we get, a = 4.92 & b=110.77×103
)........(4..........92.4
10110.77
p
)........(3..........92.4
10110.77
σthatso
2
3
r
2
3
c
−
×
=
+
×
=
r
r
MPa.6.1692.4
001
10110.77
p
100mm,rwherefaceinneron thepressureFluid
2
3
r =−
×
=
=

72. Boundary conditions:
At r=100mm , σc= 16 N/mm2
At r=150mm , pr= 4.2 MPa.
Case (ii) – When there is an external fluid pressure of 4.2 MPa:
)........(2..........a
150
b
4.2
)........(1..........a
100
b
16i.e.,
2
2
−=
+=
Solving we get, a = 2.01 & b=139.85×103
)........(4..........01.2
10139.85
p
)........(3..........01.2
10139.85
σthatso
2
3
r
2
3
r
−
×
=
+
×
=
r
r

73. MPa.11.97501.2
001
10139.85
p
100mm,wherefaceinneron thepressureFluid
2
3
r =−
×
=
=r

74. PROBLEM 1:
A pipe of 150mm internal diameter with the metal thickness of 50mm
transmits water under a pressure of 6 MPa. Calculate the maximum
and minimum intensities of circumferential stresses induced.
(Ans: 12.75 MPa, 6.75 MPa)
PROBLEM 2:
Determine maximum and minimum hoop stresses across the section
of a pipe of 400mm internal diameter and 100mm thick when a fluid
under a pressure of 8N/mm2
is admitted. Sketch also the radial
pressure and hoop stress distributions across the thickness.
(Ans: σmax=20.8 N/mm2
,σmin=12.8 N/mm2
)
PROBLEM 3:
A thick cylinder with external diameter 240mm and internal diameter
‘D’ is subjected to an external pressure of 50 MPa. Determine the
diameter ‘D’ if the maximum hoop stress in the cylinder is not to
exceed 200 MPa. (Ans: 169.7 mm)
PROBLEMS FOR PRACTICE

75. THIN AND THICK
CYLINDERS -63
PROBLEM 4:
A thick cylinder of 1m inside diameter and 7m long is subjected to an
internal fluid pressure of 40 MPa. Determine the thickness of the
cylinder if the maximum shear stress in the cylinder is not to exceed
65 MPa. What will be the increase in the volume of the cylinder?
E=200 GPa, μ=0.3. (Ans: t=306.2mm, δv=5.47×10-3
m3
)
PROBLEM 5:
A thick cylinder is subjected to both internal and external pressure.
The internal diameter of the cylinder is 150mm and the external
diameter is 200mm. If the maximum permissible stress in the
cylinder is 20 N/mm2
and external radial pressure is 4 N/mm2
,
determine the intensity of internal radial pressure. (Ans:
10.72 N/mm2
)

76. THIN AND THICK
CYLINDERS -64