Question 16.10: Design of a Conical Storage Tank A thin-walled container of ...
Design of a Conical Storage Tank
A thin-walled container of conical shape supported from the top and filled with a heavy-liquid metal of specific weight γ is shown in Figure 16.22 (see Example 6.6). Determine
a. The expressions for the tangential stress \sigma _{\theta} and meridional stress \sigma _{\phi}
b. The factor of safety n against yielding, using the maximum shear stress criterion
Given: The container is made of ASTM A36 structural steel.
\begin{array}{l} h=3 m , \quad \alpha=45^{\circ} \\ t=3 mm , \quad \gamma=80 kN / m ^3 \end{array}
S_{y} = 250 MPa (from Table B.1)
Assumption: The simple support depicted in the figure, free to move as the shell deforms under loading, ensures that no bending is produced in the neighborhood of the edge.
Referring to Figure 16.22, we write
\phi=\frac{\pi}{2}-\alpha, \quad r_0=y \tan \alpha
where \alpha is the half-cone angle. At any arbitrary level y, the pressure is
p=p_z=\gamma(h-y)
a. Substitution of the foregoing expressions into the first of Equation 16.67, after
division by t, results in the hoop stress:
\begin{array}{l} N_\theta=p_r r_\theta=-\frac{p_r r_0}{\sin \theta} \\ N_\phi=N_s=-\frac{F}{2 \pi r_0 \sin \phi} \end{array} (16.67)
\sigma_\theta=\frac{\gamma(h-y) y}{t} \frac{\tan \alpha}{\cos \alpha} (16.71a)
The load equals the weight of the liquid of volume a c O d b . Then,
F=-\pi \gamma y^2\left(h-y+\frac{1}{3} y\right) \tan ^2 \alpha
Carrying this value into the second of Equation 16.67, and dividing the resulting expression by t, gives the meridional stress:
\sigma_\phi=\sigma_s=\frac{\gamma(h-2 y / 3) y}{2 t} \frac{\tan \alpha}{\cos \alpha} (16.72a)
Comments: We see from these expressions that at the bottom (y = 0), the pressure and the membrane stresses vanish. The exact stress distribution at the apex and edge is obtained by application of the bending theory [4].
b. Differentiating with respect to y and equating to 0, it can be verified that the maximum values of the preceding membrane stresses and their location are as follows:
\sigma_{\theta, \max }=\frac{\gamma h^2}{4 t} \frac{\tan \alpha}{\cos \alpha} \quad(\text { at } y=h / 2) (16.71b)
\sigma_{s, \max }=\frac{3 \gamma h^2}{16 t} \frac{\tan \alpha}{\cos \alpha} \quad(\text { at } y=3 h / 4) (16.72b)
Note that the largest stress occurs at midheight and is given by Equation 16.71b. The design equation is given by Equation 6.18 in Example 6.6 as
\frac{S_y}{n}=\frac{\gamma h^2}{4 t} \frac{\tan \alpha}{\cos \alpha} (6.18)
\frac{\gamma h^2}{4 t} \frac{\tan \alpha}{\cos \alpha}=\frac{S_y}{n} (16.73)
Introducing the required values, we have
\frac{80(3)^2}{4(3)} \frac{\tan 45^{\circ}}{\cos 45^{\circ}}=\frac{250}{n}
from which n = 2.95.