Elasticity_ Barber

22.1 Plane problems 349
22.1 Plane problems
If φ and T are independent of z, the non-zero stress components reduce to
σ xx = − ∂2 φ
∂y 2
; σ xy = ∂2 φ
∂x∂y ; σ yy = − ∂2 φ
∂x 2 (22.15)
σ zz = − ∂2 φ
∂x 2 − ∂2 φ
∂y 2 (22.16)
and the solution corresponds to a state of plane strain. In fact, the relations
(22.15) are identical to the Airy function definitions, apart from a difference
of sign, and similarly, (22.10) reduces to a particular integral of (14.7). However,
an important difference is that the present solution also gives explicit
relations for the displacements and can therefore be used for problems with
displacement boundary conditions, including those arising for multiply connected
bodies.
22.1.1 Axisymmetric problems for the cylinder
We consider the cylinder b<r <a in a state of plane strain with a prescribed
temperature distribution. The example in §14.1 and the treatment of isothermal
problems in Chapters 8, 9 show that this problem could be solved for a
fairly general temperature distribution, but a case of particular importance is
that in which the temperature is an axisymmetric function T (r).
If φ is also taken to depend upon r only, we can write (22.10) in the form
1
r
d
dr r dφ
dr
which has the particular integral
dφ
dr
=
2µ(1 + ν)αT (r)
(1 − ν)
=
2µ(1 + ν)α
(1 − ν)r
∫ r
b
, (22.17)
rT (r)dr , (22.18)
where we have selected the inner radius b as the lower limit of integration,
since the required generality will be introduced later through the superposed
isothermal solution.
The corresponding displacement and stress components are then obtained
as
∫ r
u r = 1 dφ (1 + ν)α
=
2µ dr (1 − ν)r b
σ rr = − 1 ∫
dφ + ν)α r
= −2µ(1
r dr (1 − ν)r 2
σ θθ = − d2 φ 2µ(1 + ν)α
=
dr2 (1 − ν)
rT (r)dr (22.19)
b
( 1
r 2 ∫ r
rT (r)dr (22.20)
b
)
rT (r)dr − T (r)
(22.21)
σ rθ = 0 ; u θ = 0 . (22.22)