Solid Mechanics2020/2021

Class 19

Thick-Walled Cylinders

Examples

December 3, 2020

Recap: Stress in Cylindrical Coordinates

Consider cylindrical coordinates r, θ and z, related to cartesian coordinates by:

x1 = r cosθ, θ = arc tan x2/x1

x2 = r senθ, r2 = x12 + x2

2

x3 = z, z = x3

Stress tensor and components

[σij] = σ

rrσ

rtσ

rz

σtr

σtt

σtz

σzr

σzt

σzz

Stress components incylindrical coordinates

Recap: Cylindrical coordinates

With the body forces b in components (br, bθ, bz) in directions r, θ, z, the equations of equilibrium in cylindrical coordinates are

∂σrr / ∂r + 1/r ∂σrθ / ∂θ + ∂σrz / ∂z + (σrr - σθθ) / r + br = 0

∂σrθ / ∂r + 1/r ∂σθθ / ∂θ + ∂σθz/∂z + 2 σrθ / r + bθ = 0

∂σrz/∂r + 1/r ∂σθz/∂θ + ∂σzz/∂z + σrz / r + bz = 0.

Stress components incylindrical coordinates

Recap: Thin-Walled Cylinders

A cylindrical pressure vessel is thin-walled if t/R < 0.1, where t is the wall thickness and R = (Rext + Rint) / 2 the average radius.

Consider a cylindrical vessel with length L, subjected to internal pressure p.

Because of symmetry, only σrr, σθθ

and σzz ≠ 0.

We simplify the problem assuming that σzz and σθθ are constant across the thickness.

Stresses in a thin-walled cylindrical pressure vessel

Recap: Thin-Walled Cylinders

We determine σθθ by cutting the cylinder in two halves along two longitudinal lines and equilibrating vertical forces.

The upward force due to the internal pressure is balanced by the downward force due to tangential tensile σθθ

stresses.

σθθ (2 L t) = p (2 R L)

σθθ = p R / t

Projected area 2 R L

Stresses in a thin-walled cylindrical pressure vessel

Recap: Thin-Walled Cylinders

The assumption of constant stress across the thickness is not correct for σrr: exterior radius, σrr = 0; interior radius, σrr = – p.

But |σrr| < 0.1 σθθ for a thin-walled cylinder (σθθ = p R / t).

In conclusion:

σθθ = 2 σzz >> |σrr|,

which explains why pressure cylinders fail along longitudinal lines.

Stresses in a thin-walled cylindrical pressure vessel

Recap: Infinitesimal Strain Tensors in Cylindrical Coordinates

The infinitesimal strain tensor in cylindrical coordinates is given by

[e] = err

erθ

erz

eθreθθ eθz

ezr

ezθ

ezz

=

=

𝜕ur

/𝜕r1

2(𝜕u

rr𝜕θ

+𝜕u

θ𝜕r

− uθr)

1

2(𝜕u

r𝜕z

+𝜕u

z𝜕r)

1

2(𝜕u

rr𝜕θ

+𝜕u

θ𝜕r

− uθr) u

r

/r + 𝜕uθ/r𝜕θ1

2(𝜕uθ𝜕z

+𝜕u

zr𝜕θ

)

1

2(𝜕u

z𝜕r

+𝜕u

r𝜕z)

1

2(𝜕u

zr𝜕θ

+𝜕uθ𝜕z

) 𝜕uz

/𝜕z

Recap: Elastic Properties of Isotropic Materials

The isotropic Hookean constitutive relationship is written

eij = 1 / E [(1 + ν) σij – ν σkk δij],

σij = E / (1 + ν) [eij + ν / (1 – 2 ν) ekk δij].

Thick-Walled Cylinders with Internal and External Pressure

Geometry:External radius re

Internal radius ri

Thickness t or length L

Material:Hookean isotropicYoung’s modulus EPoisson’s ratio ν

Loads:Internal pressure pi

External pressure pe

Thick-Walled Cylinders with Internal and External Pressure

The problem is axisymmetric. It does not depend on the coordinate θ.The stresses and the strains do not depend on the coordinate z.

Because of symmetry, only σrr, σθθ and σzz are different from zero.

The shear stress components are zeroσrθ = σrz = σθz = 0.

The shear strain components are zeroerθ = erz = eθz = 0.

Thick-Walled Cylinders with Internal and External Pressure

The strain-displacement relations are:err = ∂ur /∂reθθ = ur / r

ezz = ∂uz / ∂z

The stress-strain relations are:

err = 1 / E [σrr – ν (σθθ + σzz)]eθθ = 1 / E [σθθ – ν (σrr + σzz)]ezz = 1 / E [σzz – ν (σθθ + σrr)]

σrr = E / (1 + ν) [err + ν / (1 – 2 ν) (err + eθθ+ ezz)]σθθ = E / (1 + ν) [eθθ + ν / (1 – 2 ν) (err + eθθ+ ezz)]σzz = E / (1 + ν) [ezz + ν / (1 – 2 ν) (err + eθθ+ ezz)]

Thick-Walled Cylinders with Internal and External Pressure

We consider two geometries for the thick-walled cylinder:the disk the tube, in plane strain, ezz = 0.

For the disk, the stress stateis plane stress, σzz = 0.

Thick-Walled Cylinders with Internal and External Pressure

The stress-strain relations are:

err = 1 / E [σrr – ν (σθθ + σzz)]eθθ = 1 / E [σθθ – ν (σrr + σzz)]ezz = 1 / E [σzz – ν (σθθ + σrr)]

For plane stress, σzz = 0,

ezz = – ν (σθθ + σrr) / E.

For plane strain, ezz = 0,

σzz = ν (σθθ + σrr).

Thick-Walled Cylinders with Internal and External Pressure

The equilibrium equations without body forces,

∂σrr/∂r + 1/r ∂σrθ/∂θ + ∂σrz/∂z + (σrr - σθθ) / r = 0∂σrθ/∂r + 1/r ∂σθθ/∂θ + ∂σθz/∂z + 2 σrθ / r = 0∂σrz/∂r + 1/r ∂σθz/∂θ + ∂σzz/∂z + σrz / r = 0,

reduce to

∂σrr / ∂r + (σrr - σθθ) / r = 00 = 0

∂σzz / ∂z = 0 → σzz = σzz (r)

Thick-Walled Cylinders with Internal and External Pressure

The equilibrium equation without body forces is∂σrr / ∂r + (σrr - σθθ) / r = 0.

Substituting the stress-strain and strain-displacement relations,

E / (1 – ν2) ∂ (∂ur /∂r + ν ur /r) / ∂r + (E / (1 – ν2) (∂ur /∂r + ν ur /r) –E / (1 – ν2) (ur /r + ν ∂ur /∂r)) / r = 0,

obtain the thick-walled cylinder differential equation for ur

∂2ur /∂r2 + 1/r ∂ur / ∂r – ur /r2 = 0, ri < r < re

whose solution isur(r) = A1 r + A2 / r.

Thick-Walled Cylinders with Internal and External Pressure

The solution of the thick-walled cylinder differential equation for ur

∂2ur /∂r2 + 1/r ∂ur / ∂r – ur /r2 = 0, ri < r < re

isur(r) = A1 r + A2 / r.

The integration constants A1 and A2 arefound from the Boundary Conditions (BC):

Displacement BCur(re) = ue and ur(ri) = ui

Stress BC σrr(re) = – pe and σrr(ri) = – pi

Thick-Walled Cylinders with Internal and External Pressure

The solution of the thick-walled cylinder differential equation for ur isur(r) = A1 r + A2 / r.

For stress BC, σrr(re) = – pe and σrr(ri) = – pi,σrr = C1 – C2 / r2

σθθ = C1 + C2 / r2

C1 = (pi ri2 - pe re

2) / (re2 - ri

2), C2 = (pi - pe) re2 ri

2 / (re2 - ri

2).

Note that σrr + σθθ is constant. Then σzz and ezz are constant with r.

For plane stressA1 = (1 – ν) C1 / E, A2 = (1 + ν) C2 / E,

For plane strainA1 = (1 + ν) (1 – 2 ν) C1 / E, A2 = (1 + ν) C2 / E.

Examples in Plane Stress

External pressure pe only,

σrr = C1 – C2 r2 = - pe re2 / (re

2 - ri2) (1 - ri

2 / r2),σθθ = C1 + C2 r2 = - pe re

2 / (re2 - ri

2) (1 + ri2 / r2).

Case 1 – solid cylinder, ri = 0,σrr = - pe re

2 / (re2 - ri

2) (1 - ri2 / r2) = - pe , 0 ≤ r ≤ re

σθθ = - pe re2 / (re

2 - ri2) (1 + ri

2 / r2) = - pe , 0 ≤ r ≤ re

Case 2 – cylinder with small hole, ε = ri / re << 1,σrr = - pe re

2 / (re2 – ε2 re

2) (1 - ri2 / r2), ri ≤ r ≤ re.

At r = ri, σrr = 0. At r = re, σrr = - pe

σθθ = - pe re2 / (re

2 - ε2 re2) (1 + ri

2 / r2), ri ≤ r ≤ re.At r = ri, σθθ ≈ - 2 pe. At r = re, σθθ ≈ - pe

Examples in Plane Stress

External pressure pe only,

Case 3 – solid cylinder made of two cylinders (1 and 2) of different materials in contact at r = rc

Cylinder 1 solid, E1, ν1, 0 ≤ r ≤ re = rc

Cylinder 2 hollow, E2, ν2, ri = rc ≤ r ≤ re

Boundary conditionsσrr

(2)(re) = - pe. At r = 0, σrr is bounded.

Interface conditionsur

(1)(rc) = ur(2)(rc)

σrr(1)(rc) = σrr

(2)(rc) = - pc

Examples in Plane Stress

External pressure pe only,

Case 3 – solid cylinder made of two cylinders of different materials in contact at r = rc

From the interface conditionsur

(1)(rc) = ur(2)(rc)

σrr(1)(rc) = σrr

(2)(rc) = - pc,compute the contact pressure pc

(1 – ν1) (– pc) rc / E1 = (1 – ν2) (pc rc2 – pe re

2) rc / ((re2 - rc

2) E2) +(1 + ν2) (pc – pe) re

2 rc2 / ((re

2 – rc2) E2 rc).

If E1 = 0, pc = 0.If E1 = E2 and ν1 = ν2, pc = pe.

Examples in Plane Stress

Internal pressure pi only,

σrr = pi ri2 / (re

2 - ri2) - pi re

2 ri2 / (re

2 - ri2) r2 =

pi ri2 / (re

2 - ri2) (1 - re

2 / r2) , ri < r < re

σθθ = (pi ri2 (re

2 - ri2) + pi re

2 ri2 / (re

2 - ri2) r2 =

pi ri2 / (re

2 - ri2) (1 + re

2 / r2) , ri < r < re

Consider 3 cases: re / ri = 5, re / ri = 2, re / ri = 1.1.

Compare to the results for thin-walled cylinders:

σθθ = pi R / t, with R = (re + ri) /2 and t = re – ri e σrr = 0.

Examples in Plane Stress

Internal pressure pi only,

Consider 3 cases: re / ri = 5, re / ri = 2, re / ri = 1.1.

Compare with the results for thin-walled cylinders σθθ = pi R / t.

When ε = ri / re << 1, i.e. re → ∞, σθθ(ri) ≈ pi

The maximum shear stress τmax = (pi – (–pi)) / 2 = pi and occurs at r = ri

re / ri σrr σθθ σθθ thin-walled formula

r = ri r = re r = ri r = re

5 - pi 0 13 pi / 12 pi / 12 3 pi / 4

2 - pi 0 5 pi / 3 2 pi / 3 3 pi / 2

1.1 - pi 0 10.5 pi 9.5 pi 10.5 pi

Examples in Plane Stress

Shrink-fit problems.

Two cylinders are assembled, interior cylinder 1 and exterior cylinder 2.

Exterior radius of cylinder 1 is slightly larger than interior radius of cylinder 2. The interference is re

(1) - ri(2) = δ.

Interior cylinder 1, hollow, E, ν, ri ≤ r ≤ re = rc

Exterior cylinder 2 hollow, E, ν, rc ≤ r ≤ re

Boundary conditions:σrr

(1)(ri) = 0 and σrr(2)(re) = 0

Interface conditions:ur

(2)(rc) - ur(1)(rc) = δ and σrr

(1)(rc) = σrr(2)(rc) = - pc

Examples in Plane Stress

Shrink-fit problems.Interface conditions:

ur(2)(rc) - ur

(1)(rc) = δ and σrr(1)(rc) = σrr

(2)(rc) = - pc

(1 – ν) (pc rc2 - pe re

2) rc / ((re2 – rc

2) E) + (1 + ν) (pc - pe) re2 rc

2 / ((re2 – rc

2) E rc) –(1 – ν) (pi ri

2 – pc rc2) rc / ((rc

2 - ri2) E) + (1 + ν) (pi – pc) rc

2 ri2 / ((rc

2 - ri2) E rc)

= δ.If the modulus E are the same in both cylinders,

(1 – ν) pc rc3 / (re

2 – rc2) + (1 + ν) pc re

2 rc / (re2 – rc

2) + (1 – ν) pc rc3 / (rc

2 -ri

2) + (1 + ν) pc rc ri2 / (rc

2 - ri2) = E δ.

Applying an internal pi or external pressure pe will add stresses from a cylinder with exterior radius re and interior radius ri subjected to pi or pe.