The stresses in thin cylinders and shells subjected to internal pressure or rotational forces are summarized. For thin cylinders under internal pressure, the circumferential (hoop) stress is given by σH=Pd/2t and the longitudinal stress is given by σL=Pd/4t, where P is the internal pressure, d is the internal diameter, and t is the wall thickness. The change in internal volume of the cylinder is given by ΔV=-(5-4v)PV/4tE, where V is the original internal volume, E is Young's modulus, and v is Poisson's ratio. For thin rotating cylinders, the hoop stress is given by σH=ω2R2,

1. CHAPTER 9
THIN CYLINDERS AND SHELLS
Summary
The stresses set up in the walls of a thin cylinder owing to an internal pressure p are:
circumferential or h m p stress aH=
Pd
Pdlongitudinal or axial stress aL= -
4t
where d is the internal diameter and t is the wall thickness of the cylinder.
longitudinal strain cL= -[aL-V a H ]
1
E
1
E
Then:
hoop strain cH = -[aH-vaL]
Fd
4tE
PVchange of volume of contained liquid under pressure = -
K
change of internal volume of cylinder under pressure = -[5 -4v] V
where K is the bulk modulus of the liquid.
For thin rotating cylinders of mean radius R the tensile hoop stress set up when rotating at
For thin spheres:
w rad/s is given by GH = po2R2.
Pd
circumferential or hoop stress aH= -
4t
3Pdchange of volume under pressure = -[1-v] V
4tE
Eflects of end plates and joints-add “joint efficiency factor” ‘1to denominator of stress
equations above.
9.1. Thin cylinders under internal pressure
When a thin-walledcylinder issubjected to internal pressure,three mutually perpendicular
principal stresses will be set up in the cylinder material, namely the circumferential or hoop
198

2. 59.1 Thin Cylinders and Shells 199
stress, the radial stress and the longitudinal stress. Provided that the ratio of thickness to
inside diameter of the cylinder is less than 1/20, it is reasonably accurate to assume that the
hoop and longitudinalstressesareconstant acrossthe wall thicknessand that the magnitude
of the radial stress set up is so small in comparison with the hoop and longitudinal stresses
that it can be neglected.Thisisobviouslyan approximation since,in practice,it will varyfrom
zero at the outside surface to a valueequal to the internal pressure at the inside surface. For
the purpose of the initial derivationof stressformulaeit is also assumed that the ends of the
cylinder and any riveted joints present have no effect on the stresses produced; in practice
they will have an effect and this will be discussed later ( 59.6).
9.1.1. Hoop or circumferential stress
Thisis the stresswhich is set up in resisting the bursting effectof the applied pressure and
can be most conveniently treated by consideringthe equilibrium of half of the cylinder as
shown in Fig. 9.1.
QU Qn
Fig. 9.1. Half of a thin cylinder subjected to internal pressure showing the hoop and
longitudinal stresses acting on any element in the cylinder surface.
Total force on half-cylinder owing to internal pressure = p x projected area = p x dL
Total resisting force owing to hoop stress on set up in the cylinder walls
= 2oH x Lt
. . 2aHLt = pdL
Pd. . circumferential or hoop stress uH= -
2t
9.1.2. Longitudinal stress
Consider now the cylinder shown in Fig. 9.2.
Total force on the end of the cylinder owing to internal pressure
nd2
= pressure x area = p x ~
4

3. 200 Mechanics of Materials 09.1
Fig. 9.2. Cross-section of a thin cylinder.
Area of metal resisting this force = ltdt(approximate1y)
..
i.e.
force d2/4 pd
stress set up = -= p x -= -
area ndt 4t
pdlongitudinal stress uL= -
4t
9.1.3. Changes in dinrensions
(a) Change in length
The change in length of the cylindermay be determined from the longitudinalstrain, i.e.
neglecting the radial stress.
1
E
Longitudinal strain = -[uL-vuH]
and change in length = longitudinal strain x original length
1
E
= -[uL-vuH]L
pd= - [ 1 - 2 v ] L
4tE
(b)Change in diameter
(9.3)
Asabove,thechangein diametermay bedeterminedfrom thestrain on a diameter,i.e. the
diametral strain.
change in diameter
original diameter
Diametral strain =
Now the change in diameter may be found from a consideration of the cipcumferential
change. The stress acting around a circumference is the hoop or circumferential stress on
giving rise to the circumferential strain cH.
Change in circumference= strain x original circumference
= E H X n d

4. $9.2 Thin Cylindersand Sheh 201
New circumference = xd +7cd~H
= d ( 1+EH)
But this is the circumference of a circle of diameter d (1+E,,)
. .
. .
New diameter = d (1+E ~ )
Change in diameter = dEH
d&H
Diametral strain E,, = -= eH
the diametral strain equals the hoop or circumferential strain
d
i.e. (9.4)
d
E
change in diameter = deH = -[aH-voL]Thus
Pd’=---[2-v]
4tE
(c) Change in internal volume
Change in volume = volumetric strain x original volume
From the work of $14.5, page 364.
volumetric strain = sum of three mutually perpendicular direct strains
= E L + 2ED
1
E
= -[UL+2aH-v(aH+2aL)J
= -[Pd 1 +4-v(2+2) J
4tE
Pd
= -[5-4v]
4tE
Therefore with original internal volume V
Pdcbange in internal volume = -[5 -4v] Y
4tE
9.2. Thin rotating ring or cylinder
(9.5)
Considera thin ring or cylinderas shownin Fig. 9.3subjectedto a radialpressurep caused
by the centrifugaleffectof its own masswhen rotating. Thecentrifugaleffectona unit length

5. 202 Mechanics of Materials $9.3
F F
Fig. 9.3. Rotating thin ring or cylinder.
of the circumference is:
p = m o 2 r
Thus, considering the equilibrium of half the ring shown in the figure:
2 F = p x 2 r
F = pr
where F is the hoop tension set up owing to rotation.
constant across the wall thickness.
. .
This tension is transmitted through the complete circumferenceand therefore is restricted by
the complete cross-sectional area.
The cylinder wall is assumed to be so thin that the centrifugal effect can be taken to be
F = pr = mo2r2
where A is the cross-sectional area of the ring.
density p.
Now with unit length assumed, m / Aisthe massof the ring material per unit volume,i.e. the
. . hoop stress = po2r2 (9.7)
9.3. Thin spherical shell under internal pressure
Becauseof the symmetry of the sphere the stressesset up owing to internal pressure will be
two mutually perpendicular hoop or circumferential stresses of equal value and a radial
stress. As with thin cylinders having thickness to diameter ratios less than 1:20, the radial
stress is assumed negligible in comparison with the values of hoop stress set up. The stress
system is therefore one of equal biaxial hoop stresses.
Consider, therefore, the equilibrium of the half-sphere shown in Fig. 9.4.
Force on half-sphere owing to internal pressure
= pressure x projected area
nd2
= p x 4
Resisting force = oHx ltdt (approximately)

6. 59.4 Thin Cylindersand Shells 203
. .
or
Fig. 9.4. Half of a thin sphere subjected to internal pressure showing uniform hoop stresses
acting on a surface element.
nd
4
p x -= CTH x ndt
Pdb H = -
4t
Pdcircumferential or hoop stress = -
4t
i.e.
9.3.1. Change in internal volume
As for the cylinder,
change in volume = original volume x volumetric strain
but
volumetric strain = sum of three mutually perpendicular strains (in this case all equal)
= 3 E D = 3 E H
. . 3Pdchange in internal volume = -[1-v] Y
4tE
(9.9)
9.4. Vessels subjected to fluid pressure
If a fluid is used as the pressurisation medium the fluid itself will change in volume as
pressure is increased and this must be taken into account when calculating the amount of
fluid which must be pumped into the cylinder in order to raise the pressure by a specified
amount, the cylinder being initially full of fluid at atmospheric pressure.
Now the bulk modulus of a fluid is defined as follows:
volumetric stress
volumetric strain
bulk modulus K =

7. 204 Mechanics of Materials 59.5
where, in this case,volumetric stress = pressure p
and volumetric strain = -change in volume 6V
original volume Y
- _
(9.10)
PVi.e. change in volume of fluid under pressure = -
K
Theextra fluid requiredto raise the pressure must, therefore,take up this volumetogether
with the increase in internal volume of the cylinder itself.
.. extra fluid required to raise cylinder pressure by p
= -[5-4v]Pd v+-PV
4tE K
Similarly,for spheres, the extra fluid required is
= ~ 3Pd [l -v]v+ PV-
4tE K
(9.11)
(9.12)
9.5. Cylindrical vessel with hemispherical ends
Considernow thevesselshownin Fig.9.5 in whichthe wallthicknessof thecylindricaland
hemisphericalportions may be different(thisissometimesnecessarysincethe hoop stressin
thecylinderistwicethat inasphereofthe sameradiusand wallthickness).For thepurpose of
the calculation the internal diameter of both portions is assumed equal.
From the preceding sections the following formulae are known to apply:
I c 1
I t t t
I I I
I
1 I
Fig. 9.5. Cross-sectionof a thin cylinder with hemisphericalends.
(a) For the cylindrical portion
Pd
2tc
hoop or circumferential stress = bHc= -

8. $9.6 Thin Cylinders and Shells 205
..
Pdlongitudinal stress = aLc= -
44
1
E
hoop or circumferential strain = -[ g H c-vak]
Pd= -[2-v]
44E
(b) For the hemispherical ends
Pd
4ts
hoop stress = oHs= -
1
E
hoop strain = -[aHs-voHs]
Pd= -[1 -v]
4t,E
Thus equating the two strains in order that there shall be no distortion of the junction,
-[2-v]Pd = -[1Pd -v]
4t,E 4t,E
i.e. (9.13)
With the normally accepted value of Poisson’s ratio for general steel work of 0.3, the
t, 0.7
t, 1.7
thickness ratio becomes
- =-
i.e. the thickness of the cylinder walls must be approximately 2.4 times that of the
hemisphericalendsfor no distortion of thejunction to occur.In thesecircumstances,because
of thereducedwallthicknessof the ends,the maximumstresswill occurin theends.For equal
maximumstresses in the two portions the thicknessof the cylinderwallsmust be twicethat in
the ends but some distortion at the junction will then occur.
9.6. Effects of end plates and joints
The preceding sections have all assumed uniform material properties throughout the
components and have neglected the effects of endplates and joints which are necessary
requirementsfor their production. In general,the strength of thecomponentswill be reduced
by the presenceof, for example,rivetedjoints, and this should be taken into account by the
introduction of ajoint eficiency factor tf into the equations previously derived.

9. 206 Mechanics of Materials 59.7
Thus, for thin cylinders:
Pd
2tqL
hoop stress = ~
where q is the efficiency of the longitudinal joints,
Pdlongitudinal stress = -
4tqc
where qc is the efficiency of the circumferential joints.
For thin spheres:
Pdhoop stress = -
4tV
Normally the joint efficiencyis stated in percentageform and this must be convertedinto
equivalent decimal form before substitution into the above equations.
9.7. Wire-wound thin cylinders
In order to increase the ability of thin cylinders to withstand high internal pressures
without excessiveincreasesin wall thickness,and hence weight and associated materialcost,
theyaresometimeswound with hightensilesteeltapeor wireunder tension.Thissubjectsthe
cylinderto an initialhoop, compressive,stresswhich must be overcomeby the stressesowing
to internal pressure before the material is subjected to tension. There then remains at this
stagethe normal pressurecapacityof thecylinderbeforethe maximumallowablestressin the
cylinder is exceeded.
It is normally required to determine the tension necessary in the tape during winding in
order to ensure that the maximum hoop stressin the cylinderwill not exceed a certainvalue
when the internal pressure is applied.
Consider,therefore,the half-cylinderof Fig. 9.6, where oHdenotes the hoop stressin the
cylinder walls and o,the stress in the rectangular-sectioned tape. Let conditions before
pressure is applied be denoted by suffix 1and after pressure is applied by suffix 2.
Fig. 9.6. Section ofa
Tope
thin cylinder with an external layer of tape wound on with a tension.

10. $9.7 Thin Cylinders and Shells 207
Now force owing to tape = or]x area
= ut,x 2Lt,
ut,x 2Lt, = OH,x 2Ltc
bt x t, = OH]x t,
resistive force in the cylinder material = oH,x 2Lt,
i.e. for equilibrium
or
so that the compressive hoop stress set up in the cylinder walls after winding and before
pressurisation is given by
t
o,,, = crlx 2 (compressive) (9.14)
t c
This equation will be modifiedif wire of circular cross-sectionis used for the windingprocess
in preference to rectangular-sectioned tape. The area carrying the stress ctlwill then beans
where a is the cross-sectional area of the wire and nis the number of turns along the cylinder
length.
After pressure has been applied another force is introduced
= pressure x projected area = pdL
Again, equating forces for equilibrium of the halfcylinder,
pdL = (oH,x 2Ltc)+(or,x 2Lt,)
where o,,,isthe hoop stress in the cylinderafter pressurisation and otlisthe finalstress in the
tape after pressurisation.
Since the limiting value of (iH, is known for any given internal pressure p, this equation
yields the value of or,.
Now the change in strain on the outside surface of the cylinder must equal that on the
inside surface of the tape if they are to remain in contact.
(9.15)
or, -or1
Change in strain in the tape = ~
Et
where E, is Young’s modulus of the tape.
In the absence of any internal pressure originally there will be no longitudinal stress or
strain so that the original strain in the cylinder walls is given by oHl/Ec,where E, is Young’s
modulus of the cylinder material. When pressurised, however,the cylinder will be subjected
to a longitudinal strain so that the final strain in the cylinder walls is given by
.. change in strain on the cylinder =
Thus with b H ,obtained in terms of err, from eqn. (9.14),p and bH,known, and or, found
from eqn. (9.15)the only unknown or, can be determined.

11. 208 Mechanics of Materials
Examples
Example 9.1
A thin cylinder 75 mm internal diameter, 250mm long with walls 2.5 mm thick issubjected
to an internal pressure of 7 MN/mZ.Determine the change in internal diameter and the
change in length.
If,in addition to the internal pressure,the cylinder is subjected to a torque of200 N m,find
the magnitude and nature of the principal stresses set up in the cylinder. E = 200 GN/m2.
v = 0.3.
Solution
Pd2(a) From eqn. (93,change in diameter = -(2-v)
4tE
7 x 106 x 752 x 10-6
(2-0.3)--
4x 2.5 x 10-3x 200 109
= 33.4x m
= 33.4 pm
PdL(b) From eqn. (9.3),change in length = -(1 -2v)
4tE
7 io6 75 10-3 250 x 10-3
(1 -0.6)--
4 2.5x 10-3 x 200 x 109
= 26.2pm
pd 7 x lo6 x 75 x
Hoop stress oH = - =
2t 2 x 2.5 x 10-3
= 105MN/m2
pd 7 x lo6 x 75 x
Longitudinal stress oL = 4t=
2 x 2.5 10-3
= 52.5MN/m2
In addition to these stresses a shear stress 5 is set up.
From the torsion theory,
T T . T R
J R J
- _ -- .. T = -
Now
Then
x (804-754) x (41-31.6)
J = - = - = 0.92 x m4
32 1OI2 32 lo6
200 x 20 x 10-3
shear stress T = -6 = 4.34MN/m2
0.92 x 10-

12. Thin Cylidrs and Shells 209
tOr* On
@T
0, = crL
Enlorgod vmw oi
dement an wrfoce
of cylinder subjected to
torque ond internol pressure
Fig. 9.7. Enlarged view of the stresses acting on an element in the surface of a thin cylinder
subjected to torque and internal pressure.
Thestresssystemthen actingon any element ofthecylinder surfaceisasshownin Fig. 9.7.
The principal stresses are then given by eqn. (13.1l),.
u1and a’ = *(a, +o,,)i-*J[(a,-a,)’ +42,,’]
= $(lo5+52.5)ffJ[(lO5 -52.5)’ +4(4.34)’]
= 3; x 157.5fiJ(2760 +75.3)
= 78.75 f26.6
Then o1 = 105.35MN/m2 and a2 = 52.15MN/m2
The principal stresses are
105.4 MN/mZ and 52.2 MN/m2 both tensile.
Example 9.2
A cylinder has an internal diameter of 230mm, has walls 5 mm thick and is 1m long.It is
m3whenfilledwithaliquidat a pressurep.foundtochangeininternalvolumeby 12.0 x
If E = 200GN/m2 and v = 0.25, and assuming rigid end plates, determine:
(a) the values of hoop and longitudinal stresses;
(b) the modifications to these values if joint efficiencies of 45% (hoop) and 85%
(c) the necessarychangein pressure p to produce a further increase in internal volumeof
(longitudinal)are assumed;
15%. The liquid may be assumed incompressible.
Solution
(a) From eqn..(9.6)
change in internal volume = -pd (5-4v)V
4tE

13. 210 Mechanics of Materials
original volume V = f x 2302x x 1 = 41.6 x m3
p x 230 x x 41.6 x
(5 - 1)4 x 5 x 10-3 x 200 x 109
Then change in volume = 12x =
Thus
Hence,
12 10-6 x 4 x 5 x 10-3 x 200 x 109
= 230 x 10-3 41.6 x 10-3 x 4
= 1.25MN/m2
pd 1.25x lo6x 230 x
hoop stress = - =
2t
Pd
2 x 5 x 10-3
= 28.8MN/m2
longitudinal stress = - = 14.4MN/m2
4t
(b) Hoop stress, acting on the longitudinal joints ($9.6)
pd 1.25x lo6 x 230 x
=-- -
2tVL 2 5 x 10-3 x 0.85
= 33.9 MN/mZ
Longitudinal stress (acting on the circumferential joints)
pd 1.25x lo6x 230 x 1O-j
--
4tv, 4 x 5 x 10-3 x 0.45
= 32MN/m2
(c) Sincethe change in volume is directly proportional to the pressure, the necessary 15%
increase in volume is achieved by increasing the pressure also by 15%.
Necessary increase in p = 0.15 x 1.25x lo6
= 1.86 MN/mZ
Example 9.3
(a) A sphere, 1m internal diameter and 6mm wall thickness, is to be pressure-tested for
safetypurposes with water as the pressure medium. Assuming that the sphere isinitiallyfilled
with water at atmospheric pressure, what extra volume of water is required to be pumped in
to produce a pressure of 3MN/m2 gauge? For water, K = 2.1GN/m2.
(6) The sphere is now placed in serviceand filledwith gas until there isa volumechange of
72 x
(c) To what value can the gas pressure be increased before failure occurs according to the
maximum principal stress theory of elastic failure?
For the material of the sphere E = 200GN/mZ,v = 0.3 and the yield stress 0, in simple
tension = 280MN/m2.
m3. Determine the pressure exerted by the gas on the walls of the sphere.

14. Thin Cylindersand Shells
Solution
21 1
volumetric stress
volumetric strain
(a) Bulk modulus K =
Now
and
volumetric stress = pressure p = 3 MN/mZ
volumetric strain = change in volume +- original volume
i.e.
PK = -
6 V / V
p v 3 X 1 O 6 48
K 2.1x 109 3
change in volume of water = -= x -. .
= 0.748 x m3
(b) From eqn. (9.9)the change in volume is given by
3Pd
6 v = -(1 -v) v4tE
. .
..
3p x 1 x $ ~ ( 0 . 5 ) ~ ( 1-0.3)
72 x =
4 x 6 x
72 x
x 200 x lo9
x 4 x 6 x 200 x lo6x 3
3 x 4n(0.5)3x 0.7
P =
= 314 x lo3N/m2 = 314 kN/mZ
(c) The maximum stress set up in the sphere will be the hoop stress,
i.e. PdaI= aH= -
4t
Now, according to the maximum principal stress theory (see 915.2) failure will occur when
the maximum principal stress equals the value of the yield stress of a specimensubjected to
simple tension,
i.e. when o1 = ay= 280MN/m2
Thus
d
4t
280 x lo6 =
280 x lo6 x 4 x 6 x
1
P'
= 6.72 x lo6N/m2 = 6.7 MN/mZ
The sphere would therefore yield at a pressure of 6.7MN/mZ.
Example 9.4
A closed thin copper cylinder of 150mm internal diameter having a wall thickness of 4 mm
is closely wound with a singlelayer of steeltape having a thickness of 1.5 mm, the tape being

15. 212 Mechanics of Materials
wound on when the cylinder has no internal pressure. Estimate the tensile stressin the steel
tape when it is being wound to ensure that when the cylinder is subjected to an internal
pressureof 3.5 MN/m2the tensile hoop stressin thecylinderwill not exceed 35 MN/m2.For
copper, Poisson’s ratio v = 0.3 and E = 100GN/m2;for steel, E = 200GN/m2.
Solution
Let 6, be thestress in the tape and let conditionsbefore pressure isapplied be denoted by
Consider the halfglider shown (beforepressure is applied)in Fig. 9.6 (see page 206):
s u f b 1and after pressure is applied by s& 2.
force owing to tension in tape = ut1x area
= x 1.5 x 10-3 x L 2
resistive force in the material of cylinder wall = on, x 4 x lo-’ x L x 2
.. 2oH, x 4 x 10-3 x L = 2ot1x 1.5 x 10-3x L
1.5
4
. . oH,= -or,= 0.375otI(compressive)
After pressure is applied another force is introduced
= pressure x projected area
= P W )
Equating forces now acting on the half-cylinder,
pdL = (aH2x 2 x 4 x 10-
but p = 3.5 x lo6N/mZ and oH,= 35 x lo6N/m2
x L)+(ot, x 2 x 1.5 x 10- x L)
:. 3.5 x io6 x 150x 1 0 - 3 ~= (35x io6 x 2 x 4 x 10-3L)+ (ut, x 2 x 1.5 x 10-3 x L)
. . 525 x lo6 = 280 x lo6+3ut2
lo6
(525-280)
3ot, =. .
or,= 82MN/m2
The change in strain on the outside of the cylinder and on the inside of the tape must be
equal:
or2-or1
change in strain in tape = ___
E,
CHI
original strain in cylinder walls = -
E,
(Since there is no pressure in the cylinder in the original condition there will be no
longitudinal stress.)

16. Thh Cylin&rs and Sheh 213
Final strain in cylinder (after pressurising)
ISH vu= I - L
E, E,
Then change in strain in cylinder
Then
Substitutingfor uHlfrom eqn. (1)
10.3 x 3.5 x io6 x SI x 10-3
-0.375 ut,
[35n106- 4 x 4 x - ~ o - ~io0 x 109
8 2 ~ 1 0 ~ - ~ , ~--
200 x 109
82x lo6-ot1= 2(35 x lo6- 10.1x lo6-0.375 otl)
= 49.8 x lo6-0.75or,
Then 1.75t ~ ~ ,= (82.0-49.8)106
32.2 x lo6
1.75
Utl =
= 18.4MN/mZ
Problems
9.1 (A).Determine the hoop and longitudinal stressesset up in a thin boikr shell of circularcroesection,5m
longandof 1.3m internal diameter when the internal pressure reachesavalue of 2.4 bar (240kN/m2).What willthen
be its change in diameter? The wall thickness of the boiler is 25mm. E = 210GN/m2; v = 0.3.
C6.24, 3.12MN/m2;0.033mm.]
9.2 (A). Determine the change in volumeof a thin cylinderof originalvolume65.5 x 10- m3and length 1.3m if
its wall thickness is 6mm and the internal pressure 14 bar (1.4MN/m2).For the cylindermaterial E = 210GN/mZ;
v = 0.3. C17.5x 10-6m3.]
9.3 (A). What must bc the wall thickness of a thin sphericalvessel of diameter 1m if it is to withstand aninternal
9.4 (A/B). A steel cylinder 1m long, of 150mm internal diameter and plate thickness 5mm, is subjected to an
internal pressure of 70bar (7MN/m2); the increase in volume owing to the pressure is 16.8x m3. Find the
values of Poisson's ratio and the modulus of rigidity. Assume E = 210GN/mZ. [U.L.] c0.299; 80.8GN/m2.]
9.5 (B). Definebulk modulus K, and showthat thedecreasein volumeof a fluid under pressure p isp V / K .Hence
derivea formulato find the extra fluidwhichmust bepumped into a thin cylinderto raiseits pressureby an amountp.
How much fluid is required to raise the pressure in a thin cylinderof length 3m, internal diameter 0.7m,and wall
thickness 12mm by 0.7bar (70kN/m2)? E = 210GN/m2 and v = 0.3 for the material of the cylinder and
K = 2.1 GN/m2 for the fluid. C5.981 x m3.]
9.6 (B). A spherical vessel of 1.7m diameter is made from 12mm thick plate, and it is to be subject4 to a
hydraulic test. Determine the additional volume of water which it is necessary to pump into the vessel, whcn the
vessel isinitiallyjust filledwith water, in order to raise the pressure to the proof pressure of 116bar (11.6MN/m2).
The bulk modulus of water is 2.9 GN/m2. For the material of the v e l , E = 200GN/m2, v = 0.3.
C26.14 x m3.]
pressure of 70 bar (7MN/m2)and the hoop stresses are limited to 270 MN/m2? [12.%mm.]

17. 214 Mechanics of Materials
9.7 (B). A thin-walled steelcylinderis subjected to an internal fluid pressure of 21bar (2.1 MN/m’). The boiler is
of 1m insidediameter and 3m long and has a wall thickness of 33mm.Calculate the hoop and longitudinal stresses
present in the cylinderand determine what torque may be applied to the cylinderif the principal stress is limited to
150MN/m2. [35, 17.5MN/m’; 6MNm.l
9.8 (B). A thin cylinder of 300mm internal diameter and 12mm thickness is subjected to an internal pressure
p whiletheendsare subjectedtoan external pressureoft p . Determine the valueof p at whichelasticfailurewilloccur
according to (a)the maximum shear stress theory, and (b)the maximum shear strain energy theory,if the limit of
proportionality ofthe material in simpletension is 270MN/m’. What will be the volumetric strain atthis pressure?
E = 210GN/m2; v = 0.3
9.9 (C). A brass pipe has an internal diameter of 400mm and a metal thickness of 6mm. A singlelayer of high-
tensilewireof diameter 3mm iswound closelyround it at a tension of 500N. Find (a)the stressin the pipewhen there
is no internal pressure; (b)the maximum permissibleinternal pressure in the pipe if the working tensilestressin the
brass is 60MN/m’; (c)the stress in the steel wire under condition (b).Treat the pipe as a thin cylinderand neglect
longitudinal stresses and strains. Es = 200GN/m2; EB= 100GN/m2.
[U.L.] C27.8, 3.04 MN/mZ; 104.8 MNIm’.]
9.10 (B). A cylindricalvessel of 1m diameter and 3m long is made of steel 12mm thick and filled with water at
16°C.The temperature is then raised to 50°C.Find the stressesinduced in the material of the vessel given that over
this range of temperature water increases 0.006per unit volume. (Bulk modulus of water = 2.9GN/m2; E for
steel = 210GN/m2 and v = 0.3.) Neglect the expansion of the steel owing to temperature rise.
[663, 331.5MNjm’.]
9.1 1 (C). A 3m long aluminium-alloy tube, of 150mm outside diameter and 5mm wall thickness, is closely
wound with a single layer of 2.5mm diameter steel wire at a tension of 400N. It is then subjected to an internal
pressure of 70 bar (7MN/m’).
C21.6, 23.6MN/mZ, 2.289 x 2.5 x
(a) Find the stress in the tube before the pressure is applied.
(b) Find the final stress in the tube.
E , = 70GN/m’; vA = 0.28; Es = 200 GN/mZ [-32, 20.5 MN/m’.]
9.12 (B). (a) Derive the equations for the circumferential and longitudinal stresses in a thin cylindrical shell.
(b) A thin cylinder of 300mm internal diameter, 3m long and made from 3mm thick metal, has its ends blanked
off. Working from first principles,except that you may use the equations derivedabove,find the change in capacity
of this cylinder when an internal fluid bressure of 20 bar is applied. E =200GN/m2; v = 0.3. [201 x 10-6m3.]
9.13 (A/B). Show that the tensile hoop stress set up in a thin rotating ring or cylinder is given by:
aH = pw’r’.
Hence determine the maximum angular velocity at which the disc can be rotated if the hoop stress is limited to
20MN/m’. The ring has a mean diameter of 260mm. [3800 rev/min.]