This document discusses generalized Hooke's Law, strain relationships, and the mechanics of solids, specifically focusing on shear stress and strain, as well as the behavior of pressure vessels. It covers the properties of isotropic and anisotropic materials, including stress transformation and normal strain calculations for various shapes like cylindrical and spherical vessels. Additionally, the document explores thick-walled cylinders and the mathematical theory of elasticity to understand stress conditions and design considerations.

1. Chapter 3
Generalized Hooke’s Law, Pressure Vessels, and
Thick-Walled Cylinders
Mechanics of Solids

2. Part A- Constitutive Relationships for Shear
Stress-Strain Relationships for Shear
• A body subjected to shear stresses causes shear deformations.
• Shear Strain (𝛾) = The change in the initial right angle between any
two imaginary planes in a body. Unit = rad
• Linear relationship between pure shear stress and angle 𝛾.
𝜏 = 𝐺𝛾
• Where, G = Shear modulus of shear modulus of elasticity or modulus
of rigidity.

3. Fig. 1: Element in pure shear
Fig. 2: Shear stress-strain diagrams: (a) typical and
(b) idealized for a ductile material

4. Elastic Strain Energy for Shear Stresses
• Consider an element in a state of shear as shown in Fig. 3
Fig. 3: (a) an element in a state of shear, (b) The deformed shape of this element
• It is assumed that the bottom plane of the element is fixed in
position.
• The force on the top plane reaches a final value 𝜏 𝑑𝑥 𝑑𝑧.

5. • The total displacement of this force for a small deformation of the
element is 𝛾 𝑑𝑦.
• Actual work done on the element is equal to the internal recoverable
elastic strain energy,
• Where, dV is the volume of the infinitesimal element.
• Strain energy density for shear,
𝑈0 𝑠ℎ𝑒𝑎𝑟 =
𝑑𝑈
𝑑𝑉 𝑠ℎ𝑒𝑎𝑟
=
𝜏𝛾
2
=
𝜏2
2𝐺
∵ 𝜏 = 𝐺𝛾
𝑈𝑠ℎ𝑒𝑎𝑟 =
𝜏2
2𝐺
𝑑𝑉

6. Part B- Generalized Concepts of Strain and Hooke’s
Law
Fig. 4: One and two-dimensional strained elements in initial and final positions

7. • As shown in Fig. 4(a), some points like A and B move to A’ and B’
because of extensional strain in one direction.
• During straining point A experiences displacement u and point B
u+Δu.
• Normal Strain,
𝜀 = lim
∆𝑥 →0
∆𝑢
∆𝑥
=
𝑑𝑢
𝑑𝑥
𝜀 𝑥 = lim
∆𝑥→0
𝐷′ 𝐶′ − 𝐷𝐶
𝐷𝐶
• Where, the vectorial displacements of points C and D are uc = CC’ and
uD = DD’.
• If at a point of a body, u, v, and w are the three displacements
components occurring, respectively in the x, y, and z directions of the
co-ordinate axis, Fig. 4(b)

8. • Double script analogously can be used for strains, where one
designates the direction of the line element and the other, the
direction of the displacements. Positive sign for elongations.
• An element can also experience a shear strain as shown in for
example in the x-y plane in Fig. 4(c)
• v is the displacement in y direction, as one moves in the x direction,
𝑑𝑣 𝑑𝑥 is the slope of the initial horizontal side of the infinitesimal
element.

9. • Similarly the vertical side tilts through an angle 𝑑𝑢 𝑑𝑦.
• Therefore, for small angle changes, the definition of the shear strain
associated with the xy coordinates is
• Here, it is assumed that tangents of small angles are equal to angle
themselves in radian manner.
• The shear strains for the xz and yz planes are as follows:
• Six strain displacement equations depend only on three displacement
components u, v, and w. so, they are dependent. Three independent
equations can be developed showing interrelationships among 𝜀 𝑥𝑥,
𝜀 𝑦𝑦, 𝜀 𝑧𝑧, 𝛾𝑥𝑦, 𝛾𝑦𝑧, and 𝛾𝑧𝑥.

10. Strain Tensor
• The amount of shear deformation 𝛾𝑥𝑦 consist of two 𝛾𝑥𝑦 2’s as
shown in Fig. 5
Fig. 5: Shear deformation
• Since, the element is not rotated as a rigid body, the strain is said to
be pure or irrotational.

11. • The strain tensor in matrix representation can be assembled as
follows:
𝜀 𝑥
𝛾 𝑥𝑦
2
𝛾 𝑥𝑧
2
𝛾 𝑦𝑥
2 𝜀 𝑦
𝛾 𝑦𝑧
2
𝛾 𝑧𝑥
2
𝛾 𝑧𝑦
2
𝜀 𝑧
=
𝜀 𝑥𝑥 𝜀 𝑥𝑦 𝜀 𝑥𝑧
𝜀 𝑦𝑥 𝜀 𝑦𝑦 𝜀 𝑦𝑧
𝜀 𝑧𝑥 𝜀 𝑧𝑦 𝜀 𝑧𝑧
• The strain tensor is symmetric. This notion has wide acceptance in
continuum mechanics (elasticity, plasticity, rheology, etc.). The strain
tensor can also be diagonalized, having 3 diagonal components.
• The strains give only the relative displacement of points; rigid body
displacements do not affect the strains.

12. Generalized Hooke’s Law for Isotropic Materials
• Only applicable to homogeneous isotropic materials.
• For anisotropic materials such as wood has different properties in the
longitudinal, radial and transverse directions. Such ‘Orthotropic’
materials have nine independent material constants, whereas
isotropic materials have only two.
• For fully anisotropic crystalline materials the number of independent
material constants can be as large as 21.
• The extent of lateral deformation is analytically formulated using
poisson’s ratio.

13. Fig. 6: Element deformations caused by normal stresses acting in directions of coordinate axis
• From Fig. 6(b), element is subject to a tensile stress 𝜎𝑥.
𝜀 𝑥
′ = 𝜎𝑥 𝐸 is the strain in the x direction
• The corresponding lateral strains 𝜀 𝑦
′ and 𝜀 𝑧
′ along the y and z axes,
respectively can be found using poisson’s ratio.

14. 𝜀 𝑦
′ = 𝜀 𝑧
′ = −𝜈𝜎𝑥 𝐸
• Similar expressions for 𝜀 𝑥
′′, 𝜀 𝑦
′′, and 𝜀 𝑧
′′ apply when the element is
stressed as shown in the Fig. 6(c), and again for strains 𝜀 𝑥
′′′, 𝜀 𝑦
′′′, and
𝜀 𝑧
′′′, when stressed, as shown in Fig. 6(d).
• By superposing these strains, complete expressions for normal strains
𝜀 𝑥, 𝜀 𝑦, and 𝜀 𝑧 are obtained.
• Since, the three elastic constants, E, 𝜈, and G are dependent to each
other for isotropic materials, there are only two constants.

15. • No changes in the shear strain occurs due to a change in temperature
take place in isotropic materials since such materials have the same
properties in all directions.
• If stress in one or two direction exits, then there may be strain along
z-direction.
• For plane strain, 𝜀 𝑧 = 0. σ 𝑧 should not be zero from generalized
equations.

16. E,G, and 𝜈 Relationships
• In a state of pure shear, such as shown in Fig. 7(a), can be
transformed into an equivalent system of normal stresses.
• Bisect square element ABCD by diagonal AC and isolate a triangular
element, as shown in Fig. 7(b). If this element is dz thick, then each
area associated with sides AB or BC is dA, and that associated with
the diagonal AC is 2 dA.
Fig. 7: Transformation of pure shear stress into equivalent normal stresses

17. • Since the shear stress acting on the areas dA is τ, the forces acting on
these areas are τ dA. The components of these forces acting toward
diagonal BD are in equilibrium.
• While, the components parallel to diagonal BD develop a resultant
τ 2 dA acting normal to AC.
• This gives rise to a force 𝜎1 2 𝑑𝐴 shown in the figure. the shear
stress resultant and this force must be equal, 𝜎1 = τ.
• By isolating an element with a side BD, as shown in Fig. 7(d), and
proceeding in the same manner as before, 𝜎2 = −τ.
• The result of the two analyses are displayed in the Fig. 7(e). This
representation of a stress is completely equivalent to that shown in
Fig. 7(a).
• A pure shear stress at a point can be alternatively represented by the
normal stresses at 45° with the directions of the shear stresses.

18. • Consider the deformed element shown in the Fig. 8, determining the
strain in diagonal DB on two different bases.
Fig. 8: Kinematics of element
Deformation for establishing a
Relationship between shear and’
Axial strains
• Considering only infinitesimal deformations, and letting sin 𝛾 ≈
𝑡𝑎𝑛 𝛾 ≈ 𝛾 and cos 𝛾 ≈ 1, it follows that displacement BB’ due to
shear is 𝑎𝛾. The projection of this displacement onto diagonal DB’, is
equal to the stretch of DB, is 𝑎𝛾 2.

19. • Since the length of DB is 2𝑎, its normal strain 𝜀45° is 𝛾 2.
𝜀45° =
𝜏
2𝐺
• If the x axis is directed along diagonal DB, an alternative expression
for the normal strain in diagonal DB is found.
𝜀45° =
𝜎1
𝐸
− 𝜈
𝜎2
𝐸
=
𝜏
𝐸
1 + 𝜈
• Equating the two alternative relations for the strain along the shear
diagonal and simplifying,
𝐸 = 2𝐺 1 + 𝜈
• Note:
𝜈 > 0 and (shear modulus) G < (elastic modulus) E
For most materials, 𝜈 is in the neighborhood of 1
4
.

20. Dilation and Bulk Modulus
• The sides dx, dy, and dz of an infinitesimal element after straining
become 1 + 𝜀 𝑥 𝑑𝑥, 1 + 𝜀 𝑦 𝑑𝑦, and 1 + 𝜀 𝑧 𝑑𝑧, respectively.
• The change in volume is determined by subtracting the initial volume
from the volume of the strained element.
1 + 𝜀 𝑥 𝑑𝑥 1 + 𝜀 𝑦 𝑑𝑦 1 + 𝜀 𝑧 𝑑𝑧 − 𝑑𝑥 𝑑𝑦 𝑑𝑧 ≈ 𝜀 𝑥 + 𝜀 𝑦 + 𝜀 𝑧 dx dy dz
• Where, the products of strain 𝜀 𝑥 𝜀 𝑦 + 𝜀 𝑦 𝜀 𝑧 + 𝜀 𝑧 𝜀 𝑥 + 𝜀 𝑥 𝜀 𝑦 𝜀 𝑧, being
small are neglected.
• Dilation (e) = In the infinitesimal(small) strain theory, the change in
volume per unit volume
𝑒 = 𝜀 𝑥 + 𝜀 𝑦 + 𝜀 𝑧 =
1 − 2𝜈
𝐸
𝜎𝑥 + 𝜎 𝑦 + 𝜎𝑧

21. • Dilation is proportional to algebraic sum of all normal stresses.
• If an elastic body is subjected to hydrostatic pressure of uniform
intensity p, so that
𝜎𝑥 = 𝜎 𝑦 = 𝜎𝑧 = −𝑝
𝑒 = −
3 1 − 2𝜈
𝐸
𝑝
or −
𝑝
𝐸
= 𝑘 =
𝐸
3 1−2𝜈
• Modulus of Compression/ Bulk Modulus (k) = the ratio of hydrostatic
compressive stress to the decrease in volume.

22. Part C- Thin Walled Pressure Vessels
Cylindrical and Spherical Pressure Vessels
Fig. 9: Diagrams for analysis of thin-walled cylindrical pressure vessels

23. • A sphere is an ideal shape for a closed pressure vessel if the contents
are of negligible weight.
• Considering a cylindrical pressure vessel such as a boiler, as shown in
Fig. 9(a). The segment is isolated from this vessel by passing two
planes perpendicular to the axis of cylinder and one additional
longitudinal plane through the same axis as Fig. 9(b).
• There can be two normal stresses, 𝜎1 and 𝜎2. These stresses
multiplied by the respective areas on which they act, maintain the
element of the cylinder in equilibrium against the internal pressure.
• Let the internal pressure in excess of the internal pressure be p psi or
Pa (gage pressure), and let the internal radius of the cylinder be ri .
• Then the force on the infinitesimal area Lri dθ (where dθ is an
infinitesimal angle) of the cylinder caused by the internal pressure
acting normal thereto is pLri dθ (Fig. 9(c))

24. • The component of this force acting in the horizontal direction is pLri
dθ (cos θ).
• The total resisting force of 2P acting on the cylindrical segment is,
2𝑃 = 2
0
𝜋 2
2𝑝𝐿𝑟𝑖 𝑐𝑜𝑠𝜃 𝑑𝜃 = 2𝑝𝑟𝑖 𝐿
• From symmetry, half of this total force is resisted at the top cut
through the cylinder and the other half is resisted at the bottom.
Alternate: -
• The two forces P resist the force developed by the internal pressure p
which acts perpendicular to the projected area A1 of the cylindrical
segment onto the diametric plane (Fig. 9(d)). This area in Fig. 9(b) is
2riL; hence 2P = A1p = 2riLp.

25. • This force is resisted by the forces developed in the material in the
longitudinal cuts, since the outside radius of the cylinder is ro, the
area of both longitudinal cuts is 2A = 2L(ro - ri).
• If the average normal stress acting on the longitudinal cut is 𝜎1, the
force resisted by the walls of the cylinder is 2L(ro - ri) 𝜎1.
• Thickness of the cylinder, t = (ro - ri) .
𝜎1 =
𝑝𝑟𝑖
𝑡
• The normal stress given by above equation is often referred to as the
‘circumferential’ or ‘hoop’ stress for thin walled cylinder. Here, the
force P in the hoop must be considered acting tangentially to the
cylinder.

26. • The other normal stress 𝜎2 acting in a cylindrical pressure vessel acts
longitudinally.
• By passing a section through the vessel perpendicular to its axis, a
free body as shown in Fig. 9(f) is obtained.
• The force developed by the internal pressure is 𝑝𝜋𝑟𝑖
2
and the force
developed by the longitudinal stress 𝜎2 in the wall is 𝜎2 𝜋𝑟𝑜
2 − 𝜋𝑟𝑖
2
.
Equating these two forces and solving for 𝜎2,
𝜎2 =
𝑝𝑟𝑖
2
𝑟𝑜
2
− 𝑟𝑖
2
• However, 𝑟𝑜 − 𝑟𝑖 ≈ 𝑡 and for thin-walled vessels, 𝑟𝑜 ≈ 𝑟𝑖 ≈ 𝑟; hence,
𝜎2 =
𝑝𝑟
2𝑡

27. Spherical Vessels
Fig. 10: Thin-walled spherical pressure vessel
• The maximum membrane stresses for thin-walled spherical pressure
vessels,
𝜎1 = 𝜎2 =
𝑝𝑟
2𝑡
• The maximum shear stresses associated with these normal stresses
are half as large.

28. • The planes on which these shear stresses act may be identified on
elements viewed toward a section through the wall of a vessel. Such a
section is shown in Fig. 11
• The stress 𝜎2 acts perpendicularly to the plane of the figure.
Fig. 11: In yielded steel pressure vessels shear slip planes at 45° can be observed on etched specimens

29. Remarks on Thin-Walled Pressure Vessels
• The maximum normal stress in a spherical pressure vessel is only
about one-half as large as that in a cylindrical one, as shown in Fig. 12
and Fig. 13
Fig. 12: An element of a thin-walled cylindrical Fig. 13: An element of a thin-walled spherical
pressure vessel pressure vessel
• In a cylindrical pressure vessel, the longitudinal stresses, 𝜎2 parallel to
the vessel’s axis, do not contribute to maintaining the equilibrium of
the internal pressure p acting on the curved surface; whereas in a
spherical vessel, a system of equal stresses resists the applied internal
pressure.

30. • These longitudinal and hoop stresses are treated as biaxial, although
the internal pressure p acting on the wall causes local compressive
stresses on the inside equal to this pressure.
• Such stresses are small in comparison with the membrane stresses 𝜎1
and 𝜎2, and are generally ignored for thin-walled pressure vessels.
• A much more important problem arises at geometrical changes in the
shape of a vessel. This can cause a disturbance in the membrane
action.
• If a cylindrical pressure vessel has hemispherical ends, as shown in
Fig. 14(a), and if initially the cylinder and the heads were independent
of each other, under pressurization they would expand, as shown by
the dashed lines. The cylinder and the ends would expand by
different amounts and would tend to create a discontinuity in the
wall, as shown in Fig. 14

31. Fig. 14: Exaggerated deformations of pressure vessels at discontinuity
• However, physical quantity of the wall must be maintained by local
bending and shear stresses in the neighborhood of the juncture, as
shown in Fig. 14(b)
• If instead of relatively flexible hemi-spherical ends, thick end plates
are used, the local bending and shear stresses increase, as Fig. 14(c).
Flat ends are very undesirable.

32. • A majority of pressure vessels are manufactured from curved sheets
that are joined together by welding. Examples of welds used in
pressure vessels are shown in Fig. 15, with preference given to the
different types of butt joints.
Fig. 15: Examples of welds used in pressure vessels
(a) Double fillet lap-joint,
(b) Double-welded butt joint with V-grooves.
• The formulas derived for thin-walled pressure vessels in the
preceding section should be used only for cases of internal pressure.
• Because, If a vessel is to be designed for external pressure, as in the
case of a vacuum tank or a submarine, instability (bucking) of the
walls may occur.

33. Part D – Thick-Walled Cylinders
• In order to solve the posed problem, a characteristic method of the
‘mathematical theory of elasticity’ is employed.
• This consists of assuring equilibrium for each infinitesimal element,
and the use of geometric relations, allowing only their compatible
(possible) deformations. The equilibrium conditions are related to
those of deformation using the generalized Hooke’s law.
• The governing differential equation established on the preceding
bases is solved subject to the prescribed boundary conditions.
• This solution is also useful for the design of extrusion molds and other
mechanical equipments

34. Solution of the General Problem
• Consider a long cylinder with axially restrained ends whose cross
section has the dimensions shown in Fig. 16
Fig. 16: Thick-walled cylinder
• 𝑟𝑖 = The inside radius of this cylinder; 𝑟𝑜 = the outside radius.
• 𝑝𝑖 = The internal pressure in the cylinder; 𝑝 𝑜 = external pressure.

35. • Since the cylinder is long, every ring of unit thickness measured
perpendicular to the plane is stressed alike.
• A typical infinitesimal element of unit thickness is defined by two
radii, 𝑟 and r + 𝑑𝑟, and an angle 𝑑Φ as shown in Fig. 16(b)
• If the normal radial stress acting on the infinitesimal element at a
distance r from the center of the cylinder is 𝜎𝑟, the variable stress at a
distance 𝑟 + 𝑑𝑟 will be 𝜎𝑟 + 𝑑𝜎𝑟 𝑑𝑟 𝑑𝑟.
• Both normal ‘tangential’ stresses acting on the other two faces are 𝜎𝑟.
• Since, from the condition of symmetry, every element at the same
radial distance from the center must be stressed alike, no shear
stresses act on the element shown.
• The axial stresses 𝜎𝑥 on the two faces of the element are equal and
opposite normal to the plane.

36. Static Equilibrium
• The area on which 𝜎𝑟 acts is 1 × 𝑟 𝑑Φ; that on which 𝜎𝑟 + 𝑑𝜎𝑟 acts is
1 × 𝑟 + 𝑑𝑟 𝑑Φ; and each area on which 𝜎𝑡 acts is 1 × 𝑑𝑟. The
weight of the element itself is neglected.
• Since the angle included between the sides of the element is 𝑑Φ,
both tangential stresses are inclined 1
2
𝑑Φ to the line perpendicular to
OA. Then, summing the force along the radial line,
𝐹𝑟 = 0
• Simplifying, and neglecting the infinitesimals of higher order,

37. • This equation has two unknown stresses, 𝜎𝑡 and 𝜎𝑟. Intermediate
steps are required to express this equation in terms of one unknown
so that it can be solved. This is done by introducing the geometry of
deformations and properties of the material into the problem.
Geometric Compatibility
• The deformation of an element is described by its strains in the radial
and tangential directions. If u represents the radial displacement or
movement of a cylindrical surface of radius r, Fig. 16(a), u +
𝑑𝑢 𝑑𝑟 𝑑𝑟 is the radial displacement of the adjacent surface of
radius 𝑟 + 𝑑𝑟. Strain,

38. • The strain in the tangential direction, 𝜀𝑡 follows by subtracting from
the length of the circumference of the deformed cylindrical surface of
radius 𝑟 + 𝑢 the circumference of the unstrained cylinder of radius 𝑟
and dividing the difference by the latter length.
• One unknown variable is u.

39. Properties of a Material
• The generalized Hooke’s law relating strains to stresses:
• However, in the case of the thick-walled cylinder with axially
restrained deformation, the problem is one of plane strain, i.e., 𝜀 𝑥 =
0
• Relation for the axial stress, 𝜎𝑥 = 𝜈 𝜎𝑟 + 𝜎𝑡
• Stresses 𝜎𝑟 and 𝜎𝑡 in terms of strains:

40. • This equations bring the plane strain condition for elastic material.
Formation of the Differential Equation
• By expressing the strains 𝜀 𝑟 and 𝜀𝑡 in terms of the displacement 𝑢, as
shown earlier:

41. • By substituting these values in the equation of static equilibrium and
simplifying, the desired governing differential equation:
𝑑2
𝑢
𝑑𝑟2
+
1
𝑟
𝑑𝑢
𝑑𝑟
−
𝑢
𝑟2
= 0
Solution of the Differential Equation
• The general solution of above equation, which gives the radial
displacement 𝑢 of any point on the cylinder, is
𝑢 = 𝐴1 𝑟 + 𝐴2 𝑟
where, the constants 𝐴1 and 𝐴2 must be determined from the
conditions at the boundaries of the body.
• The displacement 𝑢 is not known at either the inner or the outer
boundary of the cylinder’s wall. However, the known pressures are
equal to the radial stresses acting on the elements at the respective
radii.

42. • Hence, 𝜎𝑟 𝑟𝑖 = −𝑝𝑖 and 𝜎𝑟 𝑟𝑜 = −𝑝 𝑜
• Where the minus signs are used to indicate compressive stresses.
• Since, 𝑢 = 𝐴1 𝑟 + 𝐴2 𝑟 and 𝑑𝑢 𝑑𝑟 = 𝐴1 − 𝐴2 𝑟2
can be substituted
in the expression for 𝜎𝑟. So, the boundary conditions become
• Solving these equations simultaneously for 𝐴1 and 𝐴2 yields,

43. • These constants, permit the determination of the radial displacement
of any point on the elastic cylinder subjected to the specified
pressures using equation of the radial displacement 𝑢.
• If 𝑢 and 𝑑𝑢 𝑑𝑟 can be found using these constants 𝐴1 and 𝐴2,
general equations for the radial and tangential stresses at any point of
an elastic cylinder are obtained.
𝜎𝑟 = 𝐶1 −
𝐶2
𝑟2 and 𝜎𝑡 = 𝐶1 +
𝐶2
𝑟2
where, 𝐶1 =
𝑝 𝑖 𝑟𝑖
2
−𝑝 𝑜 𝑟 𝑜
2
𝑟 𝑜
2−𝑟𝑖
2 and 𝐶2 =
𝑝 𝑖−𝑝 𝑜 𝑟𝑖
2
𝑟 𝑜
2
𝑟 𝑜
2−𝑟𝑖
2
• We can see that 𝜎𝑟 + 𝜎𝑡 is constant over the whole cross-sectional
area of the cylinder. Implying that 𝜎𝑥 = 𝜈 𝜎𝑟 + 𝜎𝑡 is also constant.

44. Remarks on The Thin-Disk Problem
• If an annular thin-disk were to be considered, the plane stress
condition (i.e., 𝜎𝑥 = 0 and 𝜀 𝑥 = −𝜈 𝜎𝑥 + 𝜎 𝑦 𝐸) governs.
𝜀 𝑟 =
1
𝐸
𝜎𝑟 − 𝜈𝜎𝑡 and 𝜀𝑡 =
1
𝐸
−𝜈𝜎𝑟 + 𝜎𝑡
• And by solving these equations simultaneously,
𝜎𝑟 =
𝐸
1−𝜈2 𝜀 𝑟 + 𝜈𝜀𝑡 and 𝜎𝑡 =
𝐸
1−𝜈2 𝜀𝑡 + 𝜈𝜀 𝑟
• The resulting differential equation remains the same as early. While, a
different constant 𝐴1 must be used in equation of the radial
displacement 𝑢. The constant 𝐴2 remains the same as early.
𝐴1 =
1 − 𝜈
𝐸
𝑝𝑖 𝑟𝑖
2
− 𝑝 𝑜 𝑟𝑜
2
𝑟𝑜
2
− 𝑟𝑖
2

45. Special Cases
• Internal pressure only, i.e., 𝑝𝑖 ≠ 0 and 𝑝0 = 0, Fig. 17
Fig. 17: An element in which 𝜏 𝑚𝑎𝑥 occurs

46. • Since 𝑟𝑜
2
𝑟2
≥ 1, 𝜎𝑟 is always a compressive stress and is maximum
at 𝑟 = 𝑟𝑖.
• 𝜎𝑡 is always a tensile stress and is maximum at 𝑟 = 𝑟𝑖.
• For brittle material,𝜎𝑡 generally governs the design. For ductile
material such as mild steel, it is more appropriate to adopt the
criterion for the initiation of yielding due to shear rather than the
material’s capacity for resisting normal stress.
• For, Thin-walled cylinders, the maximum radial stresses, equal to 𝑝𝑖 is
negligible in comparison with 𝜎1.Also, 𝜏 𝑚𝑎𝑥 = 𝜎1 2
• For, Thick-walled cylinder, the radial stress 𝜎𝑟 may be of the same
order of magnitude as 𝜎𝑡. So, the maximum shear stress must be
found by superposing the effects from both of the large normal
stresses in the manner shown in Fig. 18

47. Fig. 18: Stress transformations for obtaining maximum shear stresses
• Both of these stresses reach their maximum values at the inner
surface of the cylinder.
• If, 𝜎 𝑦𝑝 is the normal yield stress in uniaxial tension.
𝜏 𝑚𝑎𝑥 =
𝜎𝑡 𝑚𝑎𝑥 − 𝜎 𝑟 𝑚𝑎𝑥
2
=
𝑝 𝑖 𝑟 𝑜
2
𝑟 𝑜
2 − 𝑟𝑖
2 =
𝜎 𝑦𝑝
2
and 𝑝𝑖 = 𝑝 𝑦𝑝 =
𝜎 𝑦𝑝 𝑟 𝑜
2 − 𝑟𝑖
2
2𝑟 𝑜
2

48. • External pressure only, i.e., 𝑝𝑖 = 0 and 𝑝0 ≠ 0
• Since 𝑟𝑖
2
𝑟2 ≤ 1, both stresses are always compressive. The
maximum compressive stress is 𝜎𝑡 and it occurs at 𝑟 = 𝑟𝑖.
• Above equations can’t be used for very thin-walled cylinders. Buckling
of walls may occur and strength formula give misleading results.

49. Behavior of Ideally Plastic Thick-Walled Cylinders
• In previous section equation was derived for the onset of yield at the
inner surface of the cylinder due to the maximum shear.
• Upon subsequent increase in the internal pressure, the yielding
progresses toward the outward surface, and an elastic-plastic state
prevails in the cylinder with a limiting radius ‘c’ beyond which the
cross-section remains elastic.
• As the pressure increases, the radius ‘c’ also increases until, the entire
cross-section becomes fully plastic at the ultimate load.
• The maximum shear criterion for ideally plastic material,
𝜏 𝑚𝑎𝑥 =
𝜎𝑡 − 𝜎𝑟
2
=
𝜎 𝑦𝑝
2

50. • The solutions to be obtained in this section will be valid only as long
as 𝑟0 < 5.75𝑟𝑖 with 𝜈 = 0.3
Plastic Behavior of Thick-Walled Cylinders
• Static equilibrium:
• Yield criterion:
• By combining these two equations, the basic differential equation,

51. • The solution of this can be written as,
𝜎𝑟 = 𝜎 𝑦𝑝 ln 𝑟 + 𝐶
Fig. 19: Cylinder with inner and outer radius as a and b respectively
• For a cylinder with inner radius ‘a’ and outer radius ‘b’, the boundary
condition (zero external pressure) can be expressed as

52. 𝜎𝑟 𝑏 = 0 = 𝜎 𝑦𝑝 ln 𝑏 + 𝐶
The integration constant, 𝐶 = −𝜎 𝑦𝑝 ln 𝑏
• The radial and tangential stresses are then obtained using above
solution of differential equation and yield criterion, respectively.
𝜎𝑟 = 𝜎 𝑦𝑝 ln 𝑟 − ln 𝑏 = 𝜎 𝑦𝑝 𝑙𝑛 𝑟 𝑏
𝜎𝑡 = 𝜎 𝑦𝑝 + 𝜎𝑟 = 𝜎 𝑦𝑝 1 + 𝑙𝑛 𝑟 𝑏
• The stress distributions given by above equations are shown in Fig.
19(c), whereas Fig. 19(b) shows the elastic stress distributions.
• The ultimate internal pressure,
𝑝 𝑢𝑙𝑡 = 𝜎𝑟 𝑎 = 𝜎 𝑦𝑝 𝑙𝑛 𝑎 𝑏

53. Elastic-Plastic Behavior of Thick-Walled Cylinders
• For 𝑝 𝑦𝑝 < 𝑝𝑖 < 𝑝 𝑢𝑙𝑡 given by,
𝑝 𝑦𝑝 =
𝜎 𝑦𝑝 𝑟 𝑜
2 − 𝑟𝑖
2
2𝑟 𝑜
2 and 𝑝 𝑢𝑙𝑡 = 𝜎 𝑦𝑝 𝑙𝑛 𝑎 𝑏
The cross-section of the cylinder between the inner radius 𝑎 and an
intermediate radius 𝑐 is fully plastic, whereas that between 𝑐 and the
outer radius 𝑏 is in the elastic domain, Fig. 20
• At the elastic-plastic interface, the yield condition is just satisfied, and
the corresponding radial stress 𝑋 can be computed using equation of
𝑝 𝑦𝑝 with 𝑟𝑖 = 𝑐 and 𝑟𝑜 = 𝑏; hence,
𝑋 =
𝜎 𝑦𝑝
2
𝑏2
− 𝑐2
𝑏2
• This stress becomes the boundary condition for a fully plastic
segment with inner radius 𝑎 and outer radius 𝑐.

54. Fig. 20: Thick-walled cylinder
𝜎𝑟 𝑐 = −𝑋 = −
𝜎 𝑦𝑝
2
𝑏2 − 𝑐2
𝑏2
= 𝜎 𝑦𝑝 ln 𝑐 + 𝐶
and, 𝐶 = −
𝜎 𝑦𝑝
2
𝑏2−𝑐2
𝑏2 − 𝜎 𝑦𝑝 ln 𝑐

55. • The radial stress in the plastic region,
𝜎𝑟 = 𝜎 𝑦𝑝 𝑙𝑛
𝑟
𝑐
−
𝜎 𝑦𝑝
2
𝑏2 − 𝑐2
𝑏2
• The radial stress in the plastic zone,
𝜎𝑡 = 𝜎 𝑦𝑝 + 𝜎𝑟 = 𝜎 𝑦𝑝 1 + ln
𝑟
𝑐
−
𝜎 𝑦𝑝
2
𝑏2 − 𝑐2
𝑏2
• The internal pressure 𝑝𝑖 at which the plastic zone extends from 𝑎 to 𝑐
can be obtained, simply as 𝑝𝑖 = 𝜎𝑟 𝑎
• Equations of 𝜎𝑟 and 𝜎𝑡 for special cases,(slide 45) with 𝑟𝑖 = 𝑐 and
𝑟𝑜 = 𝑏, provide the necessary relations for calculating the stress
distributions in the elastic zone.