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CHAPTER TWO

THICK-WALLED CYLINDERS andSPINNING DISKS

* Important concepts and equations in MECH 202** Calculation of mechanical and thermal stresses in thick-

walled cylinders and disks

1.1 The Theory of Elasticity (MECH 202) Method -------Equilibrium Equations and Boundary Conditions and Strain-Displacement Relations

1.2 Equilibrium Equations, Stress, Strain and DisplacementFormulas for thick-walled cylinders and spinning disks

1.3 Thick-walled cylinder under Pressure, Compound Cylinders1.4 Stresses in spinning disks1.5 Thermal stress in cylinders and disks

Review and Summary

1.1 THE THEORY OF ELASTICITY (MECH 202)METHOD

• The deformation mode does not have to be described in order tosolve a problem

• The solution satisfies(1) condition of equilibrium at every point(2) continuity of displacement field(3) loading and support conditions (boundary condition)

2

0

0

xyxx xy x xy

xy yxy y xy y

tdy tdx dx tdy dy tdxx y

tdy tdx dx tdy dy tdxx y

τσσ τ σ τ

τ στ σ τ σ

∂ ∂ − − + + + + = ∂ ∂ ∂ ∂

− − + + + + = ∂ ∂

• Equilibrium equations and boundary conditions

(1) In general stresses are functions of the coordinates

(2) Consider the equilibrium of a differential element

(3) The equilibrium of forces in X and Y directions requires(t – thickness of the element)

(4) Differential equations of equilibriumx direction:

y direction:

y

x

dx

dy

0

0

xyx

xy y

x y

x y

τσ

τ σ

∂∂+ =

∂ ∂∂ ∂

+ =∂ ∂

3

cos 2 sin 22 2

sin 2 cos 22

x y x yB xy

x yB xy

σ σ σ σσ θ τ θ

σ στ θ τ θ

+ −= + +

−= − +

(5) Boundary conditions

on CD: σB = -P, τB = 0on DEA: σB = 0, τB = 0

(a) displacement boundary conditions

on ABC, u = 0, v = 0

(b) stress boundary conditions (surface tractions are related to theinternal stress by Mohr's circle)

4

• Strain-displacement relations

(1) displacements in X, Y and Z directions,displacement fields ---- dictate the strain fields

FIGURE 2.3.1. Lines OA and OB are drawn on a body in its strain-free state. As a result ofloading, configuration OAB is converted to O9A9B9. Equations 2.3.3 are derived from thissketch.

(2) Normal strain

Similarly,

(3) Shear strain

( ) ( ) ( )1x final original

original

OA OAOA

ε = −

( )1 u udx u dx u dx

dx x x

∂ ∂ = + + − − = ∂ ∂

y

v

yε

∂=

∂

xy

u vu dy u v dx v u vy xdy dx y x

γ α β

∂ ∂+ − + − ∂ ∂∂ ∂= + = + = +∂ ∂

, ,x y xy

u v u v

x y y xε ε γ

∂ ∂ ∂ ∂= = = +

∂ ∂ ∂ ∂

u = u(x,y)v = v(x,y)

εx = εx(x,y)εy = εy(x,y)γxy = γxy(x,y)

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• Equations in polar coordinates

FIGURE 2.6.1. Stresses that act on a plane differential element of unit thickness in polarcoordinates.

(1) Hooke’s law (for plane stress, σz = τrz = τθz = 0)

(2) Equilibrium equations (no body force)

(3) Strain displacement relations

y

x

( ) ( )1 1,r r r

rr

E E

G

θ θ θ

θθ

ε σ υσ ε σ υσ

τγ

= − = −

=

: 0

21: 0

r rr

r r

r directionr r r

directionr r r

θ θ

θ θ θ

τ σ σσθ

σ τ τθ

θ

∂ −∂− + + =

∂ ∂∂ ∂

− + + =∂ ∂

1

1

rr

r

rr

u

ruu

r ru uu

r r r

θθ

θ θθ

ε

εθ

γθ

∂=

∂∂

= +∂

∂∂= + −

∂ ∂

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• Saint-Venant's Principle

If a system of forces acting on a small region of an elastic solid isreplaced by another force system, acting within the same regionand having the same resultant force and moment as the firstsystem, then stresses change appreciably only in theneighborhood of the loaded region.

FIGURE 2.2.3. Illustration of Saint-Venant's principle. When distributed pressure loads (a) arereplaced by statically equivalent concentrated loads (b), stresses change considerably near theloads (shaded regions), but change little elsewhere.

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1.2 EQUILIBRIUM EQUATIONS, STRESS,STRAIN AND DISPLACEMENT FORMULASFOR THICK-WALLED CYLINDERS ANDSPINNING DISKS

• Thin- and thick-walled cylinders

• Equilibrium equations for cylinder and disk

FIGURE 3.2.1. (a) Cross section of a thick-walled cylinder, under internal pressure Pi andexternal pressure Po , or plan view of a disk spinning with constant angular velocity ω. (b) Forcethat act on a differential element of dimensions r dθ by dr by h, where h is the (constant) thicknessof the disk or a typical length along the cylinder. The mass density is ρ.

2 0rrdw r

dr rθσ σσ

ρ−

+ + = * two kinds of loads: Po or(&) Pi ; ω

* only two stress components: σθ, σr, (τrθ = 0), they are

only functions of r, i.e., σθ = σθ(r), σr = σr(r)

0.1t

r<

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• Strain, displacement and elastic stress formulas

(1) strain (due to axial symmetry, there are only 2 straincomponents εθ and εr, and they only depend on r)

( u = u(r) is the only displacement component, uθ = 0 )

(2) stress

(state of plane stress)

(3) equilibrium equation expressed in terms ofdisplacement, displacement and stress solution

( )2 2,

2 r

r u r u du

r r drθ

π πε ε

π+ −

= =

( )

( )

2 2

2 2

1 1

1 1

r r

r

E E du u

dr v

E E u du

r dr

θ

θ θ

σ ε υε υυ υ

σ ε υε υυ υ

= + = + − − = + = + − −

( )

( )

( )

( )

( )

2

2 22

2 2

2 322

1

21 22 2

2 21 22 2

0

1 1,

1

8, ,

1 31

1 8

1 1 31

1 8

rr

r r

r

dr

dr ru

d u du ur u u r

dr r dr r E

C ru C r

r Eu

EC C r

r

EC C r

r

θ

θ θ

θ

σ σσρω

σ ε

υρω

υρω

ε ε σ σ

υ υσ υ ρω

υ

υ υσ υ ρω

υ

−+ + =

→ →

−+ − = − =

−= + −

→ →

− + = + − − − − + = + + − −

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C1, C2 are constants of integration, they are determined byboundary conditions.1.3 THICK-WALLED CYLINDER UNDER

PRESSURE, COMPOUND CYLINDER

• Thick-walled cylinder under pressure (ωω = 0)

(a) boundary conditionsσr = -Pi at r = bσr = -Po at r = bτrθ = 0 at r = a, r = b

(b) stress formulasBy using above boundary conditionsthe constants C1 and C2 are found tobe

Finally, we have stress formulas(Lamé solution)

(a) Internal pressure

( )

2 2

1 2 2

2 2

2 2 2

1

1

i o

i o

b P a PC

E a b

a b P PC

E a b

υ

υ

−−=−

−+=

−

u

(b) External pressure

FIGURE 3.4.1. Stresses σr and σθ in thick-walled cylinders under internal and externalpressure. Stresses are drawn to scale for thecase a = 3b.

2 22 2

2 2 2 2 2 2

2 22 2

2 2 2 2 2 2

1 1

1 1

i or

i o

Pb P aa b

a b r a b r

Pb P aa b

a b r a b rθ

σ

σ

= − − − − −

= + − + − −

internal pressure external pressure

uσ ε→ →

10

(1) If t/b ≤ 0.1 (thin-wall), the stress solution σθ for thin-wall isalmost the same as thick-wall cylinder.

(2) If t/b → ∞, (or > 4), the σθb/Pi → 1.0, i.e., σθb ≈ Pi, σrb = -Pi,σθa → 0. τmax = Pi

(3) In a cylinder with closed ends, axial force is provided bypressure against the end caps, the axial stress due to internaland external pressure load is

(4) The largest shear stress is on the inside surface, at r = b,

FIGURE 3.4.2. Stress σθ at r = b due tointernal pressure only. The upper curve is theLamé solution. The lower curve is the thin-walled equation, σθ = Pib/t. The middle curveuses the mean radius R = (a + b)/2 in place ofb in the thin-walled equation.

FIGURE 3.4.3. Pressurized holes in a flatbody of arbitrary contour. If holes are widelyseparated and not close to an edge, stresses atthe holes due to pressure P are much likestresses in a very thick pressurized cylinder.

2 2

2 2i o

z

Pb P a

a bσ

−=

−

( )2

max 2 2

1

2i

r

Pa

a bθτ σ σ= − =−

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• Compound cylinder

(1) More efficient use of material

(2) Shrink-fitting

• The contact pressure Pc due toshrink-fitting depends on theradius mismatch ∆, i.e.,

Ur(outer-cylinder) (Pc , …)Ur(inner-cylinder) (Pc , …) = ∆

(a)

(b)

FIGURE 3.6.1. (a) Stresses produced byshrink-fit contact pressure Pi at thin interfaceof a compound cylinder. (b) Stressesproduced by the combination of shrink-fit andinternal pressure. Dashed lines representstresses due to Pi alone.

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• Once Pc is known, the residual stress in the 2 cylinders can becalculated by Lamé solution

• Superposing the stresses of internal pressure Pi → moreefficient use of material

1.4 STRESSES IN SPINNING DISKS OFCONSTANT THICKNESS

• Solid Disk

• Disk with a central holeσr = 0 at r = a and r = b → C1 = …, C2 = …

* A small central hole doubles the stress over the case of no hole!

( )

2

22

1

22 2

2

22 2

2

0, 0

, 0

13

8

31

8

3 1 31

8 3

r

r

r C

r a

aC

E

ra

a

ra

aθ

σσ

υυ ρω

υσ ρω

υ υσ ρω

υ

→ ≠ ∞ → == = →

−= +

+= −

+ +

= − + (a)

(b)

2 22 2 2 2

2

2 22 2 2 2

2

3

8

3 1 3

8 3

r

a ba b r

r

a ba b r

rθ

υσ ρω

υ υσ ρω

υ

+= + − −

+ +

= + + − +

FIGURE 3.7.1. Stresses σr and σθ inelastic spinning disks of constantthickness. (a) Solid disk. (b) Disk witha central hole.

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( )oo r

ru r

Eθ θε σ υσ= = −

• Example

A steel disk and a solid shaft are connected by shrink-fitting. Todetermine:(1) What are the stresses at standstill;(2) At what speed the shrink-fit will loosen; and what are the

stresses at this speed;(3) What are the largest σθ and contact pressure at half this speed.

(1) , at r = ro , σθ = -Pc

(in the shaft); σθ = +1.093Pc (in the disk), σr = -Pc

(2)

(a) (b)FIGURE 3.7.2. Two days to connect a disk and a shaft. The second way, witha solid disk and a discontinuous shaft, results in lower stresses. Larger flanges(dashed lines) allow bolts to be placed farther from the axis, where disk stressesare lower.

( ) ( )0.08 0.081.093 0.0001c c c cP P P P

E Eυ υ− − − − − − =

∆

22 2

2

30 | 1

8

518 /

r ror c L

L

bP a

a

rad s

υσ ρω

ω

=

+= = − + −

→ =

14

ocP

(3) at ω = 259 rad/s and r = ro , σθ = σθ(ω) + σθ(fit) = 161Mpa;Pc = + σr(ω) = 89.3 MPa

1.5 THERMAL STRESS IN CYLINDERS ANDDISKS

• Disk (plane stress) (hollow or solid disk)

• Cylinder (plane strain) (with or without central hole)

For a steady state of heat flow, we have

( )

( )

1 22 2 2

1 22 2 2

11

1

11

1

r

r rb

r

rb

E EC C T dr

r r

E EC C T dr ET

r rθ

υ ασ υ

υ

υ ασ υ α

υ

− = + − − − − = + + + − −

∫

∫

( )

( )

2 2

2 2 2

2 22

2 2 2

1

1

a r

r b b

a r

b b

E r bTrdr Trdr

r a b

E r bTrdr Trdr Tr

r a bθ

ασ

υ

ασ

υ

−= − − −

+= + − − −

∫ ∫

∫ ∫

( )0 ln , | , |

lni

i i r b o r a

T T rT T T T T T

a b b = =−

= + = =