1. The document discusses stresses in thick cylinders subjected to internal and external pressures. 2. It derives Lame's equations which relate radial stress, hoop stress, and constants A and B to the internal and external radii and pressures. 3. Specific cases are examined, such as a cylinder with only internal pressure, only external pressure, and a solid shaft with only external pressure. Expressions are given for the radial and hoop stresses in each case.

1. Class Notes
By
Dr. Sewa Singh
Professor, CTIEMT, Jalandhar

2.  Consider a thick cylinder
 Let length of cylinder = l
 Internal radius = r1
 External radius = r2
 Uniformly distributed internal pressure intensity = p1
 Uniformly distributed external pressure intensity = p2
 Assumptions:
 The cylinder material is linear, homogeneous and isotropic.
 Plane transverse section remaining plain under the pressure. (As a result of
this assumption the longitudinal strain is constant at all the points in the
cylinder wall, i.e. independent of radius)

3.  Let:
 𝑅𝑎𝑑𝑖𝑎𝑙 𝑆𝑡𝑟𝑒𝑠𝑠 𝑎𝑡 𝑎𝑛𝑦 𝑟𝑎𝑑𝑖𝑢𝑠, 𝑟 = 𝞼𝑟
 𝐻𝑜𝑜𝑝 𝑆𝑡𝑟𝑒𝑠𝑠 𝑎𝑡 𝑎𝑛𝑦 𝑟𝑎𝑑𝑖𝑢𝑠, 𝑟 = 𝞼θ
 𝑈𝑛𝑖𝑓𝑜𝑟𝑚 𝑙𝑜𝑛𝑔𝑖𝑡𝑢𝑑𝑖𝑛𝑎𝑙 𝑠𝑡𝑟𝑒𝑠𝑠 = 𝞼𝑧
(all assumed tensile)
Therefore, longitudinal Strain:
ε𝑧 =
1
𝐸
[ 𝞼𝑧 − 𝞄 (𝞼θ+𝞼θ)]
ε𝑧, E, 𝞼𝑧, 𝞄 are all constant
Therefore,
𝞼θ+𝞼θ = 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡 = 2𝐴 (𝑠𝑎𝑦)………Eq. 1

4. Consider an annular ring of
cylinder between radii r and
r+δr as shown in fig. (a)
Let:
𝞼𝑟= Internal Radial Stress
𝞼𝑟 + δ𝞼𝑟 = External Radial
Stress (as shown in fig. (b))
Considering the equilibrium of half the ring:
𝐵𝑢𝑟𝑠𝑡𝑖𝑛𝑔 𝐹𝑜𝑟𝑐𝑒 = 𝞼𝑟 + δ𝞼𝑟 . 2. (r+δr).𝑙−𝞼𝑟. 2𝑟𝑙
Simplifying and neglecting small terms, we get:
𝐵𝑢𝑟𝑠𝑡𝑖𝑛𝑔 𝐹𝑜𝑟𝑐𝑒 = 2𝞼𝑟. δr 𝑙+2δ𝞼𝑟.r 𝑙

5. 𝑅𝑒𝑠𝑖𝑠𝑖𝑡𝑖𝑛𝑔 𝐹𝑜𝑟𝑐𝑒 = 2𝞼θ. 𝑙δr
For equilibrium of the ring
Resisting Force = Bursting Force
i.e.
2𝞼θ. 𝑙δr = 2𝞼𝑟. δr 𝑙+2δ𝞼𝑟.r 𝑙
𝞼θ = 𝞼𝑟 + 𝑟
δ𝞼𝒓
δr
In limiting case, we get
𝞼θ = 𝞼𝑟 + 𝑟
d𝞼𝒓
dr
=
𝑑
𝑑𝑟
𝑟𝞼𝑟 … … . . Eq. 2
Substituting in eq. 1 we get
𝑑
𝑑𝑟
𝑟𝞼𝑟 + 𝞼𝑟 = 2𝐴

6. 𝑟
𝑑𝞼𝒓
𝒅r
= 𝟐(𝑨 − 𝞼𝑟)
𝑑𝞼𝒓
𝒅r
=
𝟐(𝑨 − 𝞼𝑟)
𝒓
Or
𝑑𝞼𝒓
(𝑨 − 𝞼𝑟)
=
𝟐𝒅r
𝒓
Integrating, we get
𝑑𝞼𝒓
(𝑨 − 𝞼𝑟)
=
𝟐𝒅r
𝒓

7. −ln 𝑨 − 𝞼𝑟 = 2 ln 𝑟 − ln 𝐵
ln 𝑨 − 𝞼𝑟 = −2 ln 𝑟 + ln 𝐵
ln 𝑨 − 𝞼𝑟 = ln
𝐵
𝑟2
Taking Anti log on both sides
𝑨 − 𝞼𝑟 =
𝐵
𝑟2
𝞼𝑟 = 𝐴 −
𝐵
𝑟2
… … . . 𝐸𝑞. 3
(Where B is another constant)

8. Substituting Eq.3 in Eq.1
𝞼θ + 𝐴 −
𝐵
𝑟2
= 2𝐴
𝞼θ = 𝐴 +
𝐵
𝑟2
… … … 𝐸𝑞. 4
Eq. 3 & Eq4 are known as Lame’s Equations
Value of the constants A and B can be calculated from the boundary conditions
at 𝑟 = 𝑟1 and 𝑟 = 𝑟2
 At 𝑟 = 𝑟1, 𝞼𝑟 = −𝑝1
 At 𝑟 = 𝑟2, 𝞼𝑟 = −𝑝2

9. Substituting in Eq.3 we get
−𝑝1 = 𝐴 −
𝐵
𝑟1
2
and
−𝑝2 = 𝐴 −
𝐵
𝑟2
2
Therefore
𝑝1 − 𝑝2 = 𝐵
1
𝑟1
2
−
1
𝑟2
2
= 𝐵
𝑟2
2
− 𝑟1
2
𝑟1
2. 𝑟2
2
so
𝐵 =
𝑟1
2
𝑟2
2
(𝑝1 − 𝑝2)
𝑟2
2 − 𝑟1
2

10. And
𝐴 =
𝐵
𝑟1
2
− 𝑝1 =
𝑟2
2
𝑝1 − 𝑝2
𝑟2
2 − 𝑟1
2
− 𝑝1
=
𝑟2
2
𝑝1 − 𝑟2
2
𝑝2 − 𝑟2
2
𝑝1 + 𝑟1
2
𝑝1
𝑟2
2 − 𝑟1
2
=
𝑝1𝑟1
2 − 𝑝2𝑟2
2
𝑟2
2 − 𝑟1
2

11. Therefore
𝞼𝑟 =
𝑝1𝑟1
2 − 𝑝2𝑟2
2
𝑟2
2 − 𝑟1
2
−
𝑟1
2𝑟2
2 𝑝1 − 𝑝2
𝑟2(𝑟2
2
− 𝑟1
2)
… … … … … 𝐸𝑞. 5
And
𝞼θ =
𝑝1𝑟1
2 − 𝑝2𝑟2
2
𝑟2
2 − 𝑟1
2
+
𝑟1
2𝑟2
2 𝑝1 − 𝑝2
𝑟2(𝑟2
2
− 𝑟1
2)
… … … … … 𝐸𝑞. 6

12. 1. Internal Pressure only ,
i.e. 𝑝2 = 0, Therefore
𝞼𝑟 =
𝑝1𝑟1
2
𝑟2
2 − 𝑟1
2
−
𝑟1
2
𝑟2
2
𝑝1
𝑟2(𝑟2
2
− 𝑟1
2)
𝞼𝑟 =
𝑝1𝑟1
2
𝑟2
2 − 𝑟1
2
1 −
𝑟2
2
𝑟2
… … … . 𝐸𝑞. 7
And
𝞼θ =
𝑝1𝑟1
2
𝑟2
2 − 𝑟1
2
1 +
𝑟2
2
𝑟2
… … … . 𝐸𝑞. 8

13. 2. External Pressure only ,
i.e. 𝑝1 = 0, Therefore
𝞼𝑟 =
−𝑝2𝑟2
2
𝑟2
2 − 𝑟1
2
+
𝑟1
2
𝑟2
2
𝑝2
𝑟2(𝑟2
2
− 𝑟1
2)
𝞼𝑟 =
𝑝2𝑟2
2
𝑟2
2 − 𝑟1
2
𝑟1
2
𝑟2
− 1 … … … . 𝐸𝑞. 9
And
𝞼θ =
−𝑝2𝑟2
2
𝑟2
2 − 𝑟1
2
𝑟1
2
𝑟2
+ 1 … … … . 𝐸𝑞. 10

14. 3. Solid Circular Shaft having external radial pressure e only ,
i.e. 𝑟1 = 0 and 𝑝1 = 0 Therefore
𝞼𝑟 = −𝑝2 … … … . . 𝐸𝑞. 11
𝞼θ = −𝑝2 … … … . . 𝐸𝑞. 12
Thus the radial and hoop stresses are equal and constant throughout the shaft
and are of same nature.

15. 4. Longitudinal Stress
Consider the cross-section of the thick cylinder with closed ends subjected to
internal and external pressures (usual notations).
Therefore, for horizontal equilibrium:
𝑝1𝑋 𝜋𝑟1
2
− 𝑝2𝑋 𝜋𝑟2
2
= 𝞼𝑧𝑋 𝜋(𝑟2
2
− 𝑟1
2
)
i.e.
𝞼𝑧 =
𝑝1𝑟1
2 − 𝑝2𝑟2
2
(𝑟2
2 − 𝑟1
2)
… … . . 𝐸𝑞. 13
Note that eq.13 gives the value same as that of constant A, i.e.
Longitudinal stress is constant throughout the length of cylinder

16. 5. Maximum Shear Stress
𝞽𝑚𝑎𝑥 =
1
2
{𝞼1 − 𝞼2}
=
1
2
{𝞼θ − 𝞼𝑟}
=
1
2
𝐴 +
𝐵
𝑟2
− 𝐴 +
𝐵
𝑟2
=
𝐵
𝑟2
Shear stress will be maximum at, 𝑟 = 𝑟1
Therefore,
𝞽𝑚𝑎𝑥 =
𝑟2
2 𝑝1 − 𝑝2
(𝑟2
2
− 𝑟1
2)
… … … 𝐸𝑞. 14