Question 7.S-P.5: A compressed-air tank is supported by two cradles as shown; ...
A compressed-air tank is supported by two cradles as shown; one of the cradles is designed so that it does not exert any longitudinal force on the tank. The cylindrical body of the tank has a 30-in. outer diameter and is fabricated from a \frac{3}{8}-in. steel plate by butt welding along a helix that forms an angle of 25° with a transverse plane. The end caps are spherical and have a uniform wall thickness of \frac{5}{16} in. For an internal gage pressure of 180 psi, determine (a) the normal stress and the maximum shearing stress in the spherical caps, (b) the stresses in directions perpendicular and parallel to the helical weld.
a. Spherical Cap. Using Eq. (7.36), we write
s _{1}= s _{2}=\frac{p r}{2 t} (7.36)
p = 180 psi, t = \frac{5}{16} in. = 0.3125 in., r = 15 – 0.3125 = 14.688 in.
s _{1}= s _{2}=\frac{p r}{2 t}=\frac{(180 psi )(14.688 in .)}{2(0.3125 in .)} s = 4230 psi
We note that for stresses in a plane tangent to the cap, Mohr’s circle reduces to a point (A, B) on the horizontal axis and that all in-plane shearing stresses are zero. On the surface of the cap the third principal stress is zero and corresponds to point O. On a Mohr’s circle of diameter AO, point D^{\prime} represents the maximum shearing stress; it occurs on planes at 45° to the plane tangent to the cap.
t _{\max }=\frac{1}{2}(4230 psi ) t_{\max }=2115 psi
b. Cylindrical Body of the Tank. We first determine the hoop stress s_{1} and the longitudinal stress s_{2}. Using Eqs. (7.30) and (7.32), we write
s _{1}=\frac{p r}{t} (7.30)
s _{1}=2 s _{2} (7.32)
p = 180 psi, t = \frac{3}{8} in. = 0.375 in., r = 15 – 0.375 = 14.625 in.
s _{1}=\frac{p r}{t}=\frac{(180 psi )(14.625 in. )}{0.375 in .}=7020 psi s _{2}=\frac{1}{2} s _{1}=3510 psi
s _{ ave }=\frac{1}{2}\left( s _{1}+ s _{2}\right)=5265 psi R=\frac{1}{2}\left( s _{1}- s _{2}\right)=1755 psi
Stresses at the Weld. Noting that both the hoop stress and the longitudinal stress are principal stresses, we draw Mohr’s circle as shown.
An element having a face parallel to the weld is obtained by rotating the face perpendicular to the axis Ob counterclockwise through 25°. Therefore, on Mohr’s circle we locate the point X^{\prime} corresponding to the stress components on the weld by rotating radius CB counterclockwise through 2u = 50°.
s _{w}= s _{\text {ave }}-R \cos 50^{\circ}=5265-1755 \cos 50^{\circ} s _{w}=+4140 psi
t _{w}=R \sin 50^{\circ}=1755 \sin 50^{\circ} t _{w}=1344 psi
Since X^{\prime} is below the horizontal axis, t_{w} tends to rotate the element counterclockwise.
