Question 4.11: A compressed-air tank is supported by two cradles as shown i...

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,

In spherical cap, t = 5/16 in. and r = 15 – (5/16) = 14.688 in. So, t = 0.0212r.
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 (Figure 4.55)

\sigma _1=\sigma _2=\frac{pr}{2t}=\frac{(180  \textrm{psi})(14.688  \textrm{in.})}{2(0.3125  \textrm{in.})} = 4230 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′ represents the maximum shear stress; it occurs on planes inclined at 45º 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 that experiences maximum normal stress on planes inclined at 45º to the bar axis.) Hence,

\textrm{τ}_{\max}=\frac{1}{2}(4230 \textrm{psi})=   2115 psi.

In the cylindrical body of the tank, we have hoop and longitudinal normal stresses:

\sigma _1=\frac{pr}{t}=\frac{(180  \textrm{psi})(14.625  \textrm{in.})}{0.375  \textrm{in.}} = 7020 psi

\sigma _2=\frac{pr}{2t}=\frac{(180  \textrm{psi})(14.625  \textrm{in.})}{2(0.375  \textrm{in.})} = 3510 psi.

Here, the average normal stress is

\sigma _{ave}=\frac{1}{2}(\sigma _1+\sigma _2) =5265 psi

and the radius of Mohr’s circle is

R=\frac{1}{2}(\sigma _1-\sigma _2) =1755 psi.

We want to rotate our axes from their initial configuration, shown at left in Figure 4.56, so that our element has a face parallel to the weld, as shown at right; the transformed σ \acute{_x} and \textrm{τ} \acute{_x} \acute{_y} , or σ_w and \textrm{τ}_w , will be the requested stresses.

Using the average stress (center) and radius R just found, 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′. This point has the following coordinates:

\sigma _\textrm{w} = \sigma_{\textrm{ave}} – R cos 50˚

= 5265 – 1755 cos 50˚

= 4140 psi (tensile)

\textrm{τ}_{\textrm{w}} = R sin 50˚ = 1755 sin 50˚

= 1344 psi.

Since point X′ is below the horizontal axis, \textrm{τ}_w tends to rotate the element counterclockwise, as assumed in Figure 4.57.