Question 5.6: A stiffening ring is placed around a cylinder at a distance ...
A stiffening ring is placed around a cylinder at a distance from the ends as shown in Figure 5.12. The cylinder has a radius of 50.0 in. and a thickness of 0.25 in. and is subjected to an internal pressure of 100 psi.
Assuming E=30 \times 10^{6} psi and \mu=0.3, find
a) The discontinuity stress in the shell with the ring assumed to have infinite rigidity.
b) The discontinuity stress in the shell and ring if the ring has a thickness of 0.375 in. and a depth of 4.0 in.
A free-body diagram of the shell-to-ring junction is shown in Figure 5.13. Because the ring is assumed to have infinite rigidity, the deflection due to pressure must be brought back to zero by a force Q_{0}. Also, because the slope at the shell-to-ring junction is zero (due to symmetry), a moment M_{0} must be applied at the junction to reduce the slope created by force Q_{0} to zero. From Figure 5.13,
deflection due to P − deflection due to Q_{0} + deflection due to M_{0} = 0.
The deflection due to P is obtained from Eq. (1) in Example 5.5, whereas the deflections due to M_{0} and Q_{0} are obtained from Eq. (5.24). Hence,
w=\frac{P r^{2}}{E t}\left(1-\frac{\mu}{2}\right) Eq. (1)
\frac{ d ^{3} w}{ d x^{3}}=\frac{-1}{D}\left(2 \beta M_{0} D_{\beta x}-Q_{0} B_{\beta x}\right) (5.24)
\frac{P r^{2}}{E t}\left(1-\frac{\mu}{2}\right)-\frac{Q_{0}}{2 \beta^{3} D}+\frac{M_{0}}{2 \beta^{2} D}=0From Eq. (5.15),
D=\frac{E t^{3}}{12\left(1-\mu^{2}\right)} (5.15)
D = 0.00143E,
and from Eq. (5.20),
\beta^{4}=\frac{E t}{4 r^{2} D}=\frac{3\left(1-\mu^{2}\right)}{r^{2} t^{2}} (5.20)
\beta=0.3636Hence, the deflection compatibility equation becomes
M_{0}-2.750 Q_{0}=-321.39 (1)
The second compatibility equation gives
rotation due to Q_{0} − rotation due to M_{0} = 0
or
Q_{0}-2 \beta M_{0}=0 (2)
Solving Eqs. (1) and (2) gives
M_{0}=321.4 \text { in. }- lb / \text { in }The maximum longitudinal stress is given by
\sigma_{x}=\frac{Pr}{2 t}+\frac{6 M_{0}}{t^{2}}= 40,900 psi.
The maximum hoop moment is given by Eq. (5.17) as
M_{\theta}=\mu M_{x} (5.17)
M_{\theta}=96.4 \text { in. -lb } / \text { in. }The hoop force N_{\theta} is given by Eq. (5.19) as
N_{\theta}=\frac{-E t w}{r} (5.19)
N_{\theta}=\frac{E t w}{r}But because w=0, N_{\theta} is equal to zero and the maximum hoop stress is
\sigma_{\theta}=\frac{6 M_{\theta}}{t^{2}}= 9300 psi.
Solution (b):
The shell deformations are expressed as follows:
Due to P
w=\frac{Pr^{2}}{E t}\left(1-\frac{\mu}{2}\right)\theta=0
Due to Q_{0}
w=\frac{Q_{0}}{2 \beta^{3} D}
\theta=\frac{-Q_{0}}{2 \beta^{2} D}
Due to M_{0},
w=\frac{M_{0}}{2 \beta^{2} D}\theta=\frac{M_{0}}{\beta D}
The ring deformations are expressed as follows:
Due to P,
w=\frac{Pr(r+d / 2)}{d E}\theta=0
Due to Q_{0},
w=\frac{2 Q_{0} r(r+2)}{b d E}\theta=0
Due to M_{0},
w = 0
\theta=0The deflection compatibility is
\left[w_{ p }-w_{Q_{0}}+w_{M_{0}}\right]_{ cyl }=\left[w_{ p }+w_{2 Q_{0}}\right]_{\text {ring }}\frac{P r^{2}}{E t}\left(1-\frac{\mu}{2}\right)-\frac{Q_{0}}{2 \beta^{3} D}+\frac{M_{0}}{2 \beta^{2} D}=\frac{P_{ r }(r+d / 2)}{d E}
+\frac{2 Q_{0}(r+d / 2)}{b d E}
Or
M_{0}-4.06 Q_{0}=-296.8 (3)
Similarly,
\left[\theta_{ p }+\theta_{M_{0}}-\theta_{Q_{0}}\right]_{ cyl }=\left[\theta_{ p }+\theta_{M_{0}}-\theta_{Q_{0}}\right]_{\text {ring }}Or
\frac{M_{0}}{\beta D}-\frac{Q_{0}}{2 \beta^{2} D}=0And
2 \beta M_{0}-Q_{0}=0 (4)
Solving Eqs. (3) and (4) yields
M_{0} = 152.0 in.-lb / in
Q_{0} = 110.6 lb /in.
The hoop stress in the ring is
\sigma_{ r }=\frac{Pr}{d}+\frac{2 Q_{0} r}{b d}= 1250 + 7370
= 8620 psi.
The maximum longitudinal stress in the cylinder is
\sigma_{x}=\frac{P r}{2 t}+\frac{6 M_{0}}{t^{2}}=\frac{100 \times 50}{2 \times 0.25} \times \frac{6 \times 152}{(0.25)^{2}}
= 24,600 psi.
The hoop force at the discontinuity is
N_{\theta}=\frac{E t w}{r}Where
w=w_{ p }-w_{Q_{0}}+w_{M_{0}}=\frac{P r^{2}}{E t}\left(1-\frac{\mu}{2}\right)-\frac{Q_{0}}{2 \beta^{3} D}+\frac{M_{0}}{2 \beta^{2} D}
=\frac{447,500}{E}
Or
N_{\theta}=\frac{E(0.25)(447,500 / E)}{50}= 2238lb /in.,
and the hoop stress at the discontinuity is
\sigma_{\theta}=\frac{N_{\theta}}{t}+\frac{6 M_{\theta}}{t^{2}}=\frac{2238}{0.25}+\frac{6(0.3 \times 152)}{0.25^{2}}
= 13,300 psi,
whereas away from discontinuity,
\sigma_{\theta}=\frac{P r}{t}=20,000 psi