Question 3.5: The cylindrical pressure as shown in Figure 3.11 is fabricat...
The cylindrical pressure as shown in Figure 3.11 is fabricated from 6 mm thick plate. It is made of number of metal plates that are welded along a helix forming a 30° angle with the horizontal. If the allowable stress normal to the weld line is 75 MPa, find out the largest pressure the tank can withstand. Consider the outer diameter of the tank to be 0.5 m.
From Figure 3.11(b), we find a typical biaxial stress element. Both \sigma_1 \text { as well } \sigma_2 (acting in x and y directions, respectively) contribute to the stress across the welded joint. So, by rotating the element by 30°, we have identified the possible stress components. In Figure 3.11(c), \sigma_2^{\prime} is the normal stress across the welded joint. From the consideration of biaxial stress (refer to Chapter 4 for further details), we can write the following stress transformation equations as:
\sigma_2^{\prime}=\frac{1}{2}\left(\sigma_{x x}+\sigma_{y y}\right)+\frac{1}{2}\left(\sigma_{x x}-\sigma_{y y}\right) \cos 2 \theta-\tau_{x y} \sin 2 \theta
or \sigma_2^{\prime}=\frac{1}{2}\left(\sigma_1+\sigma_2\right)+\frac{1}{2}\left(\sigma_1-\sigma_2\right) \cos 2 \theta-\tau_{x y} \sin 2 \theta
Here \tau_{x y}=0, \sigma_1=p r / t, \sigma_2=p r / 2 t \text { and } \theta=30^{\circ} . Therefore,
\begin{aligned} \sigma_2^{\prime} & =\frac{1}{2}\left\lgroup \frac{p r}{t}+\frac{p r}{2 t} \right\rgroup +\frac{1}{2}\left\lgroup \frac{p r}{t}-\frac{p r}{2 t} \right\rgroup \cos 2\left(90^{\circ}+30^{\circ}\right)-0 \times \sin 2\left(90^{\circ}+30^{\circ}\right) \\ & =\frac{1}{2}\left\lgroup \frac{3 p r}{2 t} \right\rgroup +\frac{1}{2}\left\lgroup \frac{p r}{2 t} \right\rgroup \times \frac{1}{2} \times \frac{p r}{2 t} \times \cos 240^{\circ} \\ & =\frac{3 p r}{4 t}-\frac{p r}{8 t} \\ & =\frac{5}{8} p\left\lgroup \frac{r}{t} \right\rgroup \end{aligned}
where r is internal radius ^4 and t is the thickness. Now, in the present problem, r = R – t, where R is the external radius.
or
\begin{aligned} \frac{r}{t} & =\left\lgroup \frac{R}{t}-1 \right\rgroup \\ & =\frac{250}{6}-1 \\ & =40.67 \end{aligned}
Putting back this value in the expression of \sigma_2^{\prime} , we get
\sigma_2^{\prime}=40.67 \times \frac{5}{8} p
Equating \sigma_2^{\prime} with allowable stress, 75 MPa,
75=40.67 \times \frac{5}{8} p
or p=\frac{75 \times 8}{5 \times 40.67}=2.95 MPa
Therefore, the maximum pressure the tank can withstand is 2.95 MPa.
^4 In the theory of membrane stress equations, all radii are, however, considered as the mean radii. Accordingly, r=\frac{r_{ o }+r_{ i }}{2}=\frac{2 R-t}{2}=R-\frac{t}{2}