A thin-walled spherical tank has a diameter of 0.75 m and an internal pressure of 20 MPa. The yield $20 \mathrm { MPa }$ stress in tension is 920 MPa, the yield stress in shear is 475 MPa, and the factor of safety is 2. $475 \mathrm { MPa }$ of elasticity is 210 $\text { GPa, }$ Poisson's ratio is 0.28, and maximum normal strain is $1220 \times 10 ^ { - 6 }$ The minimum permissible thickness of the tank is approximately:

Question

Quizplus · Accepted Answer

The maximum stress produced in the wall of a thin spherical shell is given by
$$\sigma_{max} = \dfrac{P r}{t}$$
Where $P$ is the internal pressure, $r$ is the radius of the sphere, and $t$ is the thickness of the shell.

From the given data, we have
\begin{align*}
P &= 20 \mathrm {~MPa}\
r &= \dfrac{0.75}{2} = 0.375 \mathrm {~m}\
\sigma_{max} &= \dfrac{P r}{t}
\end{align*}

We need to find the minimum permissible thickness of the tank, so we can rearrange the above equation as
$$t = \dfrac{P r}{\sigma_{max}}$$

Now, we need to calculate the maximum stress that the shell can withstand in tension and in shear, and take the minimum of these two values to ensure that the factor of safety is satisfied.

In tension, the yield stress is 920 MPa, but we need to apply a factor of safety of 2, so the maximum permissible stress in tension is
$$\sigma_{max}^{tension} = \dfrac{920}{2} = 460 \mathrm {~MPa}$$

In shear, the yield stress is 475 MPa, and we need to apply a factor of safety of 2, so the maximum permissible stress in shear is
$$\sigma_{max}^{shear} = \dfrac{475}{2} = 237.5 \mathrm {~MPa}$$

The shell is made of an elastic material with Young's modulus of 210 GPa and Poisson's ratio of 0.28. The maximum normal strain in the shell is given by
$$\epsilon_{max} = \dfrac{\sigma_{max}}{E}\left(1-\nu^2\right)$$

Substituting the values, we get
$$\epsilon_{max} = \dfrac{\sigma_{max}}{E}\left(1-\nu^2\right) = \dfrac{\sigma_{max}}{210 \times 10^9}\left(1-0.28^2\right) = 1220 \times 10^{-6}$$

Solving for the maximum permissible stress in tension, we get
\begin{align*}
\sigma_{max}^{tension} &= \epsilon_{max} E \left(\dfrac{1}{1+\nu}\right)\
460 &= 1220 \times 10^{-6} \times 210 \times 10^9 \times \left(\dfrac{1}{1+0.28}\right)
\end{align*}

Solving for the thickness of the shell, we get
\begin{align*}
t &= \dfrac{P r}{\sigma_{max}^{tension}}\
&= \dfrac{20 \times 10^6 \times 0.375}{460}\
&\approx 10.5 \mathrm{~mm}
\end{align*}

Therefore, the minimum permissible thickness of the tank is approximately 10.5 mm, which corresponds to choice C.

A Thin-Walled Spherical Tank Has a Diameter of 0 20MPa20 \mathrm { MPa }20MPa

A Thin-Walled Spherical Tank Has a Diameter of 0 $20 \mathrm { MPa }$