Page 275 - Intro to Tensor Calculus

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The stresses are to remain bounded for all values of r and consequently we require c 1 and c 6 to be zero
to avoid infinite stresses for large values of r. The stress σ rr | r=a = 0 requires that

c 3 c 7 c 8
2c 2 + =0 and 2c 5 +6 +4 =0.
a 2 a 4 a 2
The stress σ rθ | r=a = 0 requires that
c 7 c 8
2c 5 − 6 − 2 =0.
a 4 a 2
In the limit as r →∞ we require that the stresses must satisfy the boundary conditions from the previous
F 11 F 11
example 2.4-5. This leads to the equations 2c 2 = and 2c 5 = − . Solving the above system of equations
2 2
produces the Airy stress function
F 11 F 11 2 a 2 F 11 a 2 F 11 2 F 11 a 4
φ = φ(r, θ)= + r − F 11 ln r + c 4 + − r − cos 2θ
4 4 2 2 4 4r 2

and the corresponding stress field is
2 4 2
F 11 a F 11 a a
σ rr = 1 − + 1+3 − 4 cos 2θ
2 r 2 2 r 4 r 2
4 2
F 11 a a
σ rθ = − 1 − 3 +2 sin 2θ
2 r 4 r 2
2 4
F 11 a F 11 a
σ θθ = 1+ − 1+3 cos 2θ.
2 r 2 2 r 4
There is a maximum stress σ θθ =3F 11 at θ = π/2, 3π/2 and a minimum stress σ θθ = −F 11 at θ =0,π.
The effect of the circular hole has been to magnify the applied stress. The factor of 3 is known as a stress
concentration factor. In general, sharp corners and unusually shaped boundaries produce much higher stress
concentration factors than rounded boundaries.

EXAMPLE 2.4-7. Consider an infinite cylindrical tube, with inner radius R 1 and the outer radius R 0 ,
which is subjected to an internal pressure P 1 and an external pressure P 0 as illustrated in the figure 2.4-7.
Find the stress and displacement fields.
Solution: Let u r ,u θ ,u z denote the displacement field. We assume that u θ =0 and u z = 0 since the
cylindrical surface r equal to a constant does not move in the θ or z directions. The displacement u r = u r (r)
is assumed to depend only upon the radial distance r. Under these conditions the Navier equations become

d 1 d
(λ +2µ) (ru r ) =0.
dr r dr

r c 2
This equation has the solution u r = c 1 + and the strain components are found from the relations
2 r
du r u r
e rr = , e θθ = , e zz = e rθ = e rz = e zθ =0.
dr r
The stresses are determined from Hooke’s law (the constitutive equations) and we write

σ ij = λδ ij Θ+2µe ij ,