Question 5.10: A compressed air tank is supported by two cradles as shown. ......
A compressed air tank is supported by two cradles as shown. Relative to the effects of the air pressure inside the tank, the effects of the cradle supports are negligible. The cylindrical body of the tank has a 30-in outer diameter and is fabricated from a 3/8-in steel plate by welding along a helix that forms an angle of 25° with a transverse (vertical) plane. The end caps are spherical and have a uniform wall thickness of 5/16 in. For an internal gage pressure of 180 psi, determine
a. The normal stresses and maximum shear stresses in the spherical caps
b. The stresses in directions perpendicular and parallel to the helical weld
Given: Dimensions of and pressure on compressed air tank.
Find: Stress states in spherical end caps and along welds in cylindrical body.
Assume: Thin-walled pressure vessel theory applies.
First, we validate our assumption that thin-walled theory will apply in both the spherical end caps and the cylindrical body. We must have t ≤ 0.1r in both sections. So:
a. In spherical cap, t = 5/16 in and r = 15 − (5/16) = 14.688 in. So, t = 0.0212r.
b. In cylindrical body, t = 3/8 in and inner radius r = 14.625 in. So, t = 0.0256r.
In a spherical pressure vessel, we have equal hoop and longitudinal stresses:
\sigma_{\text {sphere }}=\frac{p r}{2 t}=\frac{(180 \mathrm{psi})(14.688 \mathrm{in})}{2(0.3125 \mathrm{in})}=4230 \mathrm{psi}So, in a plane tangent to the cap, Mohr’s circle reduces to a point (A, B) on the horizontal (σ) axis, and all in-plane shear stresses are zero. On the surface of the cap, the third principal stress is zero, corresponding to point O. On a Mohr’s circle of diameter AO, point D^{\prime} represents the maximum shear stress; it occurs on planes inclined at 45° to the plane tangent to the cap. (This is as we would expect for purely normal loading in the reference axes, as for an axially loaded bar which experiences maximum normal stress on planes inclined at 45° to the bar axis.) Hence,
\tau_{\max }=\frac{1}{2}(4230 \mathrm{psi})=2115 \mathrm{psi}In the cylindrical body of the tank, we have hoop and longitudinal normal stresses:
\begin{gathered} \sigma_{\theta \theta}=\frac{p r}{t}=\frac{(180 \mathrm{psi})(14.625 \mathrm{in})}{0.375 \mathrm{in}}=7020 \mathrm{psi}, \\ \sigma_{x x}=\frac{p r}{2 t}=\frac{(180 \mathrm{psi})(14.625 \mathrm{in})}{2(0.375 \mathrm{in})}=3510 \mathrm{psi} . \end{gathered}Here, the average normal stress is σ_{av} = \frac{1}{2} (σ_1 + σ_2) = 5265 psi, and the radius of Mohr’s circle is R = \frac{1}{2} (σ_1 − σ_2) = 1755 psi. We want to rotate our axes from their initial configuration, shown at left below, so that our element has a face parallel to the weld, as shown at right; the transformed \sigma_{x^{\prime} x^{\prime}} and \sigma_{x^{\prime} y^{\prime}} , or σw and τ_w, will be the requested stresses.
Using the average stress (center) and radius R found above, we construct Mohr’s circle and find these transformed stress components.
Since we want to rotate the element by θ = 25°, we rotate around Mohr’s circle by 2θ = 50°, to arrive at point X^{\prime} . This point has coordinates:
\begin{aligned} \sigma_{\mathrm{W}} & =\sigma_{\mathrm{av}}-R \cos 50^{\circ} \\ & =5265-1755 \cos 50^{\circ} \\ & =4140 \text { psi (tensile) } \end{aligned} ,
\begin{aligned} \tau_{\mathrm{W}} & =R \sin 50^{\circ}=1755 \sin 50^{\circ} \\ & =1344 \text { psi. } \end{aligned} .
Since point X^{\prime} is below the horizontal axis, τ_w tends to rotate the element counterclockwise, as assumed in the sketch below.
