Thermal Stresses in Circular Cylinders
In this chapter various techniques are presented to determine the thermal stresses in solid and hollow cylinders. The one-dimensional problems of cylindrical bodies are treated by the displacement method. Plane problems for infinitely long cylinders and f
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Thermal Stresses in Circular Cylinders
In this chapter various techniques are presented to determine the thermal stresses in solid and hollow cylinders. The one-dimensional problems of cylindrical bodies are treated by the displacement method. Plane problems for infinitely long cylinders and for circular plates are treated by the thermal stress function method. Two-dimensional axisymmetric problems and three-dimensional problems are treated with Goodier’s thermoelastic potential and the Boussinesq harmonic functions or Michell’s biharmonic function. The derivation and the general solution of the basic equations related to thermal stresses in circular cylinders are treated in a number of problems. [See also Chap. 24.]
18.1 One-Dimensional Problems The one-dimensional equilibrium equation of a cylindrical body due to axi-symmetric temperature field is σrr − σθθ dσrr + =0 (18.1) dr r The generalized Hooke’s law for plane problems is 1 (σrr − ν ∗ σθθ ) + α∗ τ E∗ 1 = ∗ (σθθ − ν ∗ σrr ) + α∗ τ E 1 σr θ = 2G
rr = θθ r θ
M. Reza Eslami et al., Theory of Elasticity and Thermal Stresses, Solid Mechanics and Its Applications 197, DOI: 10.1007/978-94-007-6356-2_18, © Springer Science+Business Media Dordrecht 2013
(18.2)
445
446
18 Thermal Stresses in Circular Cylinders
where ⎧ ⎨
E E = 1 − ν2 ⎩ E ν ∗ ν = 1−ν ν (1 + ν)α α∗ = α ∗
for plane strain for plane stress for plane strain
(18.3)
for plane stress for plane strain for plane stress
The strain-displacement relations are rr =
du u , θθ = , r θ = 0 dr r
(18.4)
where u is the radial displacement. The components of stress are E ∗ du ∗u ∗ ∗ + ν − (1 + ν )α τ 1 − ν ∗2 dr r u E ∗ ∗ du ∗ ∗ + − (1 + ν ν = )α τ 1 − ν ∗2 dr r =0
σrr = σθθ σr θ
(18.5)
The equilibrium equation with respect to the displacement is d 1 d(r u) dτ = (1 + ν ∗ )α∗ dr r dr dr
(18.6)
The general solution of Eq. (18.6) is ∗
u = (1 + ν )α
∗1
r
τr dr + C1r +
C2 r
(18.7)
where C1 and C2 are constants which may be determined from the boundary conditions. The stresses are α∗ E ∗ E ∗ C2 E∗ C − τr dr + σrr = − 2 1 r 1 − ν∗ 1 + ν∗ r 2 ∗ ∗ ∗ α E E E ∗ C2 τr dr − α∗ E ∗ τ + σθθ = C1 + (18.8) 2 ∗ r 1−ν 1 + ν∗ r 2 σr θ = 0
18.1 One-Dimensional Problems
447
The displacement and stresses in a solid cylinder of radius a with free traction are a 1 − ν∗ r τr dr + τr dr u = (1 + ν )α r 0 1 + ν ∗ a2 0 a
1 r 1 σrr = α∗ E ∗ − 2 τr dr + 2 τr dr r 0 a 0 a
1 r 1 τr dr + 2 τr dr − τ σθθ = α∗ E ∗ 2 r 0 a 0 ⎧ for plane stress ⎨ 0 αE 2ν a σzz = τr dr − τ for plane strain ⎩ 1 − ν a2 0 ∗
∗
1
r
(18.9)
(18.10)
The displacement and stresses in a hollow cylinder with inner radius a and outer radius b with free traction are b a2 1 τr dr 2 2 r a 1+ν r b −a a b 1 r 2 r − a2 = α∗ E ∗ − 2 τr dr + 2 2 τr dr r a r (b − a 2 ) a b r r 2 + a2 ∗ ∗ 1 =α E τr dr + τr dr − τ r2 a r 2 (b2 − a 2 ) a ⎧ 0 for plane stress ⎪ ⎨
b = αE 2ν ⎪ τr dr − τ for plane strain ⎩ 2 1 − ν b − a2 a
u = (1 + ν ∗ )α∗ σrr σθθ
σzz
1
r
τr dr +
1 − ν∗
r+ ∗
(18.11)
(18.12)
Since there
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