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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. 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.
σ C σ L
1. Hoop Stress (C) 2. Longitudinal Stress (L) 3. Radial Stress (pr)
Element on the cylinder wall subjected to these three stresses
σ Cσ C
σC
p
σ L
σ L
σ L
p ppr
σ Lσ L
σ C
σ C
pr
pr
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
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
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
P - internal pressure (stress)
σc – circumferential stress
dL
σc
p
t
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
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 θcos2
dp
dA
σcσc
dlt
p
d
dθ
θ
Force on the elementary area,
Horizontal component of this force
dθdl 2
dp
dθdldFy θsin2
dp
Vertical component of this force
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.
sin2
dpforce bursting ldiametrica Total
0
dθdl
dA
σcσc
dlt
pθ
d
dθ
surface. curved theof area projectedp
dp cosdl2
dp 0
dl
dlcc tσ 2)σ stress ntialcircumfere to(due force Resisting
dldl dptσ2 i.e., c
)1....(....................t2
dpσ stress, ntialCircumfere c
dL
σc
p
t
force Burstingforce Resisting um,equillibriUnder
)1....(....................t2
dpσ stress, 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
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
EVALUATION OF LONGITUDINAL STRESS (σL):
d4
πp cylinder) of end (on the force bursting alLongitudin 2
p
t
σL
tdπσ force Resisting L
td π force thisresistingsection cross of Area
cylinder. theof material theof stress alLongitudinσLet L
tdπσd4
πp i.e., L
2
)2 .........(..........t4
dpσ stress, alLongitudin L
LC σ2σ (2), & (1) eqs From
force resisting force bursting um,equillibriUnder
tdπσd4
πp i.e.,
force resisting force bursting um,equillibriUnder
L2
)2 .........(..........t4
dpσ stress, alLongitudin L
tdπσ
σσRe
d4
πp
acting is pon which area pressure fluid todue Force
L
LL
2
actingiswhichonareaforcesisting
p
circumference
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)
E
σμ
E
σε
:ε strain, ntialCircumfere
LCC
C
)3..(..............................μ)2(Et4
dp
d
δdε i.e., C
σ C=(pd)/(2t)σC=(pd)/(2t)
σ L=(pd)/(4t)
σ L=(pd)/(4t)
Note: Let δd be the change in diameter. Then
d
d
d
ddd
ncecircumfereoriginal
ncecircumfereoriginalncecircumferefinalc
μ)2(E
σ
E
σμ
E
σ2
L
LL
E
σμ
E
σε
:ε strain, alLongitudin
CLL
L
)4..(..............................μ)21(Et4
dp
L
δl ε i.e., L
V
v STRAIN, VOLUMETRIC
Change in volume = δV = final volume – original volume
original volume = V = area of cylindrical shell × length
Ld
4
2
μ)21(E
σ
E
)σ2(μ
E
σ LLL
final volume = final area of cross section × final length
LddLdLdddLLdLd
LLdddd
LLdd
2)(2)(4
2)(4
4
2222
22
2
LdddLLdvolumeFinal
LddandLdLdassuchquantitiessmallertheneglecting
22
22
24
2)(,)(
LdddLV
LdLdddLLdVvolumeinchange
2
222
24
42
4
Ld4π
24π
V
dv
2
2
dLdLd
= εL + 2 × εCV
dV
)5.......(..........μ)45(Et4
dp
V
dv i.e.,
μ)2(Et4
dp2μ)21(
Et4
dp
d
d2
L
L
24tpd
2tpd
2
σ-σ τstress,Shear Maximum
other.each to
lar perpendicuact and normal are stresses these
Both al.longitudin and ntialCircumfere viz.,
point,any at stresses principal twoare There
:stressShear Maximum
LCmax
)5.(....................8t
pdτ i.e., max
σ C=(pd)/(2t)σC=(pd)/(2t)
σ L=(pd)/(4t)
σ L=(pd)/(4t)
24tpd
2tpd
2
σ-σ τstress,Shear Maximum
:stressShear Maximum
LCmax
)5.(....................8t
pdτ i.e., max
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 is12mm, 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
3. Circumferential strain, εc:
Change in length = ε L ×L= 5×10-05×3000 = 0.15 mm (Increase).
E
μ)(2
t)(4
d)(pεc
Change in diameter, δd = εc ×d = 2.125×10-04×1000 = 0.2125 mm (Increase).
E
μ)2(1
t)(4
d)(pεL
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
μ)4(5Et)(4
d)(p
V
dv
:
V
dv strain, Volumetric
V10 4.75 dv in volume, Change -4
(Increase) 10 4.75
)3.045(10200)124(
)10002.1(
4-
3
.Litres 11919.1
m101.11919mm101.11919
300010004
π10 4.75
3 3-3 6
24-
A copper tube having 45mm internal diameter and 1.5mm wall thickness is closed at its ends by plugs which are at 450mm apart. Thetube 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 changein 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)(pε strain, Long. L
53
10910100
)3.021(5.22
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)(pεstrain ntialCircumfere C
Pd/4t = 22.5
43
10825.310100
)3.02(5.22
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)
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
2. Thickness from Shear Stress consideration:
3. Thickness from permissible change in diameter consideration
(δd=0.2mm):
Therefore, required thickness, t = 19.62 mm.
t)(8
d)(pτmax
12.26mm. t
t)(8
800)(0.9818
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 mm 10143.14 )109.0(4504
πV
μ)45(Et4
dp
V
dV
Solving, p=7.33 N/mm2
)33.045(10200124
450p
10143.14
1018736
3
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.
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
dpσ
LC
Circumferential stress is given by,
...(2).......... ηt4
dpσ
CL
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
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
dpσ
LC
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
dpσ
CL
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) 0.45p = 3.456 MPa.
SOLUTION:
Also, σ C= (p×d)/(2×t) η L
120 = (p×750)/(2×12) 0.75 p = 2.88 MPa.
Max permissible pressure in the tank, p = 2.88 MPa.
μ)45(E)t(4
d)(p
V
dv Strain, Vol.
litres. 0.567 m 100.567
.mm 10567.015007504
π108.55 V108.55 dv
108.55 0.3)4-(5)1020012(4
750)(2.88
33-
3624-4-
4-3
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
dpσ e., i.
Lc
a) Let limiting tensile stress = Circumferential stress = σ c = 120N/mm2.
d = 1260 mm
7.0512
d2 120
ηt4
dpσ e., i.
CL
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
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
dpσ e., i.
Lc
b) Let limiting tensile stress = Longitudinal stress = σ L = 120N/mm2.
ηt4
dpσ e., i.
CL
.N/mm 1.68p
7.0512
5001p 120
2
.N/mm 1.44p
3.0514
5001p 120
2
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
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: C=100 MPa, L=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: C=140 MPa, L=70 MPa, %age change=0.135%.)
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%.)
THICK CYLINDERS
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’.
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:
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.
pr2
r1
p
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
rpr
pr+δpr
r2
r1
r
σc σc
r δr
Pr
pr+δprExternal pressure
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
rpr
pr+δpr
r+δrX X
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
constantE
pμ
E
σμ
E
σ rCL Since Pr is compressiveε L =
constant)pσ(μ
E
σrC
L E
ε L =
σ c- pr = 2a,
i.e., σc = pr + 2a, ………………(2)
From (1), pr+ 2a = - pr – (r× δpr ) / δr
r
rr
p-a)p(2
r
)3.(..........a)p(
p2
r
rr
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 = loge2r
b
.....(4).......... ar
bp stress, radial or,
r
bap i.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.
a 2 a b
a 2 pσ stress, Hoop2rc
r
.......(5).................... ar
bσ i.e.,
2c
Substituting it in equation 2, we get
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 ×(r12-r2
2)
For equilibrium, FL= P
σ L× π ×(r12-r2
2)= π ×r22×p
Therefore, longitudinal stress,
(Tensile) )r(r
rpσ 2
22
1
22
L
pp
r2r1
σL
σ LσL
σL
L
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.
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)
ar
bσ
2c
ar
bp
2r
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.......... a100
b20
2
)........(4.......... a150
b5
2
..(5).......... 7r
2,70,000σ
2c
..(6).......... 7r
2,70,000p
2r
Solving equations (3) & (4), we get a = 7, b = 2,70,000.
Lame’s equations are, for Hoop stress,
For radial stress,
To draw variations of Hoop stress & Radial stress :
At r =100mm (on the inner face),
(Comp) MPa 20 7100
2,70,000p stress, Radial
(Tensile) MPa 34 7100
2,70,000σ stress, Hoop
2r
2c
At r =120mm,
(Comp) MPa 11.75 7120
2,70,000p stress, Radial
(Tensile) MPa 25.75 7120
2,70,000σ stress, Hoop
2r
2c
At r =135mm,
(Comp) MPa 7.81 7135
2,70,000p stress, Radial
(Tensile) MPa 21.81 7135
2,70,000σ stress, Hoop
2r
2c
(Comp) MPa 5 7150
2,70,000p stress, Radial
(Tensile) MPa 19 7150
2,70,000σ stress, Hoop
150mm,rAt
2r
2c
Variation of Hoop stress & Radial stress
Variation of Hoop Stress-Tensile (Parabolic)
Variation of Radial Stress –Comp (Parabolic)
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).......... ar
bp stress, Radialfor
(1).................... ar
bσ Stress, Hoopfor
are, equations sLame'
2r
2c
face)inner on max is stress Hoop ( .N/mm 35σ stress, Hoop and
,N/mm 8p stress radial 80mm,rat
are, conditionsBoundary
2C
2r
,600.37,1b 13.5,aget we(4), & (3) equations Solving
)........(4.......... a08
b35
)........(3.......... a08
b8 i.e.,
2
2
..(6).......... 5.13r
1,37,600p and
..(5).......... 5.13r
1,37,600σ are, equations sLame'
2r
2c
.rrat 0p i.e.,
0.pressure face,outer On the
1r
20.96mm.
r-r metal theof Thickness 21
100.96mm.r
5.13r
1,37,6000
1
21
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)
ar
bσ
2c
ar
bp
2r
Boundary conditions:At x =100mm Pr = 14N/mm2
i.e.,
Similarly, at x =150mm Pr = 0
i.e.,
)........(1.......... a100
b14
2
)........(2.......... a150
b0
2
..(3).......... 2.11r
22,500σ stress, Hoopfor equation sLame'
2r
Solving, equations (1) & (2), we get a =11.2, b = 2,52,000.
MPa. 36.42.11100
252000σ
2max
Max hoop stress on the inner face (where x=100mm):
.23.08%100)36.4
28-36.4( error Percentage
14MPa.p and 50mm t200mm,D e whert2
dpσ formula,cylinder By thin max
MPa.28502
20014σ max
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
:faceinner on the stressshear Max i)
rCmax
= 60.94 MPa
σE
μ -p
E
μ -
E
σ
d
δd
:diameterinner in Change ii)
LrC
σE
μ -p
E
μ -
E
σ
L
δl
:Lengthin Change iii)
CrL
0.078mm. δd
104.365
)40(10200
0.3 -93.12
10200
0.3 -
10200
81.88
4-
333
0.070mm. δl
1046.83
81.8810200
0.3 -)40(
10200
0.3 -
10200
93.12
6-
333
D
δd 2
L
δl
V
δV
:in volume Change iv)
.mm 1035.11
)4
1500180π(109.198 δV
33
24-
= 9.198 ×10-4
PROBLEM 5:
Find the max internal pressure that can be allowed into a thick pipe ofouter diameter of 300mm and inner diameter of 200mm so that tensilestress in the metal does not exceed 16 MPa if, (i) there is no external fluid 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:
)........(2.......... a150
b0
)........(1.......... a100
b16 i.e.,
2
2
Solving we get, a = 4.92 & b=110.77×103
)........(4.......... 92.410110.77
p
)........(3.......... 92.410110.77
σ that so
2
3
r
2
3
c
r
r
MPa. 6.16 92.4001
10110.77p
100mm,r wherefaceinner on the pressure Fluid
2
3
r
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.......... a150
b4.2
)........(1.......... a100
b16 i.e.,
2
2
Solving we get, a = 2.01 & b=139.85×103
)........(4.......... 01.210139.85
p
)........(3.......... 01.210139.85
σ that so
2
3
r
2
3
r
r
r
MPa. 11.975 01.2001
10139.85p
100mm, wherefaceinner on the pressure Fluid
2
3
r
r
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: max=20.8 N/mm2, min=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
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=5.47×10-3m3)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)
THIN AND THICK CYLINDERS -64
