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THICK CYLINDERS
ATMIYA INSTITUTE OF
TECHNOLOGY & SCIENCE
MECHANICAL
DEPARTMENT
5th SEMESTER
MADE BY:- DHEIRYA JOSHI
DJ996
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’.
DJ996
LAME’S EQUATIONS (Theory) :
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.
4. The material is homogeneous, isotropic and obeys Hooke’s law. (The stresses are
within proportionality limit).
ASSUMPTIONS:
DJ996
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
DJ996
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
DJ996
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
DJ996
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
compressive
ε L =
constant)pσ(
μ
E
σ
rC
L

E
ε L =
DJ996
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


 r
)3.(..........
a)p(
p
2
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 = loge
2
r
b
DJ996
.....(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
DJ996
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
DJ996
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.
IMPORTANT POINTSDJ996
THANK YOU
DJ996

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MCQ Soil mechanics questions (Soil shear strength).pdf
 

Thick cylinders

  • 1. THICK CYLINDERS ATMIYA INSTITUTE OF TECHNOLOGY & SCIENCE MECHANICAL DEPARTMENT 5th SEMESTER MADE BY:- DHEIRYA JOSHI DJ996
  • 2. 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’. DJ996
  • 3. LAME’S EQUATIONS (Theory) : 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. 4. The material is homogeneous, isotropic and obeys Hooke’s law. (The stresses are within proportionality limit). ASSUMPTIONS: DJ996
  • 4. 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 DJ996
  • 5. 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 DJ996
  • 6. 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 DJ996
  • 7. 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 compressive ε L = constant)pσ( μ E σ rC L  E ε L = DJ996
  • 8. 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    r )3.(.......... a)p( p 2 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 = loge 2 r b DJ996
  • 9. .....(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 DJ996
  • 10. 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 DJ996
  • 11. 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. IMPORTANT POINTSDJ996