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6 Two-Dimensional Problems in Elasticity …
Resultant stresses are
σ C = (−0.013W + 0.00836W ) = −0.00464 W (Compression)
σ D = (0.0088W − 0.00398W ) = 0.00482 W (Tension)
In order to limit the tensile stress to 80 N/mm
2 in the ring, the maximum value of
the force in the pulls is given by.
0.00482 W = 80
∴ W = 16597.51 N or 16.598 kN
6.16 Exercises
1. Is the following function a stress function?
φ = −
P
π
r θ sin θ
If so, find the corresponding stress. What is the problem solved by this function?
2. Investigate what problem of plane stress is solved by the following stress
functions
(a) φ =
P
K
r θ sin θ
(b) φ = −
P
π
r θ sin θ
3. Starting from the stress function ϕ = A log r + Br
2 log r + Cr
2 D, obtain
the stress components σ r and σ θ in a pipe subjected to internal pressure p i
and external pressure p o . Obtain the maximum value of σ θ when p o = 0 and
indicate where it occurs.
4. Check whether the following is a stress function
ϕ = c
r
2
(α − θ ) + r
2
− r
2 cos
2
θ tan α
where α is a constant.
5. Starting frsom the stress function ϕ = C r +
C 1
r
−
(3+μ)
8
+C
2
r r
3 , derive expressions
for σ r and σ θ in case of a rotating disc of inner radius “a” and outer radius “b”.
Obtain the maximum values of σ r and σ θ .
6. Show that the stress function ϕ = A log r + Br
2 log r + Cr
2
+ D solves the
problem of axisymmetric stress distribution. Obtain expressions for σ r and σ θ
in case of a pipe subjected to internal pressure p i and external pressure p 0 .
7. Explain axisymmetric problems with examples.
8. Derive the general expression for the stress function in the case of axisymmetric
stress distribution.
9. Derive the expression for radial and tangential stress in a thick cylinder subjected
to internal and external fluid pressure.