Chapter 3, Problem 105P

Repeat Prob. 3-104 with the tank being pressurized to 50 psig.

To determine

The maximum state of stress in the tank.

The principle normal stress.

The principle shear stress.

Answer to Problem 105P

The tangential stress on the cylinder is $17681.4018\text{psi}$, radial stress is $-73.826\text{psi}$ and longitudinal stress is $5725.5\text{psi}$.

The principle normal stresses are $17681.4018\text{psi}$, $5725.5\text{psi}$, and $-73.826\text{psi}$ respectively in decreasing order.

The principle shear stresses are $8877.614\text{psi}$, $5977\text{psi}$ and $2899.6\text{psi}$ respectively in decreasing order.

Explanation of Solution

Write the expression for tangential stress in pressurized cylinders due to water present in the tank.

${\sigma }_{t_1}=\frac{p_{i_w}{r}_i^2-p_o{r}_o^2-r_i^2{r}_o^2\frac{\left(p_o-p_{i_w}\right)}{r^2}}{r_o^2-r_i^2}$ (I)

Here, the tangential stress is ${\sigma }_{t_1}$, the internal pressure due to weight of water is $p_{i_w}$, the external pressure is $p_o$, the inside radius of cylinder is $r_i$, outside radius of cylinder is $r_o$ and the radius under consideration is $r$.

Write the expression for tangential stress in pressurized cylinders due to internal pressure.

${\sigma }_{t_2}=\frac{p_i{r}_i^2-p_o{r}_o^2-r_i^2{r}_o^2\frac{\left(p_o-p_i\right)}{r^2}}{r_o^2-r_i^2}$ (II)

Here, the tangential stress is ${\sigma }_{t_2}$, the internal pressure is $p_i$.

Write the expression for total tangential stress.

${\sigma }_t={\sigma }_{t_1}+{\sigma }_{t_2}$ (III)

Here, the total tangential stress is ${\sigma }_{t_1}$.

Write the expression for inner radius of cylinder.

$r_i=r_o-t$ (IV)

Here, the thickness of cylinder is $t$.

Write the expression for radial stress in pressurized cylinders due to weight of water.

${\sigma }_{r_1}=\frac{p_i{r}_i^2-p_o{r}_o^2+r_i^2{r}_o^2\frac{\left(p_o-p_i\right)}{r^2}}{r_o^2-r_i^2}$ (V)

Here, the radial stress is ${\sigma }_{r_1}$.

Write the expression for radial stress in pressurized cylinders due to internal pressure.

${\sigma }_{r_2}=\frac{p_i{r}_i^2-p_o{r}_o^2+r_i^2{r}_o^2\frac{\left(p_o-p_i\right)}{r^2}}{r_o^2-r_i^2}$ (VI)

Here, the radial stress is ${\sigma }_{r_2}$.

Write the expression for the total radial stress.

${\sigma }_r={\sigma }_{r_1}+{\sigma }_{r_2}$ (VII)

Here, the total radial stress is ${\sigma }_r$.

Write the expression for the area of the wall.

$A_w=\frac{\pi }{4}\left(d_0^2-d_i^2\right)$ (VIII)

Here, the area of the wall is $A_w$, the outer diameter is $d_o$ and the inner diameter is $d_i$.

Write the expression for the volume of the wall of the cylinder.

$V_w=A_w\times h$ (IX)

Here, the volume of the cylinder wall is $V_w$ and the height of the wall is $h$.

Substitute $\frac{\pi }{4}\left(d_0^2-d_i^2\right)$ for $A_w$ in Equation (IX).

$V_w=\frac{\pi }{4}\left(d_0^2-d_i^2\right)\times h$ (X)

Write the expression for the volume of dome.

$V_{dome}=\frac{2\pi }{3}\left(r_o^3-r_i^3\right)$ (XI)

Here, the volume of dome is $V_{dome}$.

Write the expression for the weight of the complete structure.

$W=w\left(V_w+V_{dome}\right)$ . (XII)

Here, the weight of the structure is $W$, the unit weight is $w$

Write the expression for the stress at the wall.

${\sigma }_w=-\frac{W}{A_w}$ (XIII)

Here, the stress at the wall is ${\sigma }_w$.

Write the expression for the internal pressure on cylinder due to water.

$p_i=\omega \times \delta$ (XIV)

Here, the weight density of water is $\omega$ and height of water in tank is $\delta$.

Write the expression for principle shear stress.

${\tau }_{1/2}=\frac{{\sigma }_1-{\sigma }_2}{2}$ (XI)

Here, the principle shear stress is ${\tau }_{1/2}$, the principle normal stress is ${\sigma }_1$ and principle normal stress along the plane perpendicular to ${\sigma }_1$ is ${\sigma }_2$.

Write the expression for principle shear stress.

${\tau }_{2/3}=\frac{{\sigma }_2-{\sigma }_3}{2}$ (XII)

Here, the principle shear stress is ${\tau }_{2/3}$ and the principle normal stress along the plane perpendicular to ${\sigma }_1$ and ${\sigma }_2$ is ${\sigma }_3$.

Write the expression for principle shear stress.

${\tau }_{1/3}=\frac{{\sigma }_1-{\sigma }_3}{2}$ (XIII)

Here, the principle shear stress is ${\tau }_{1/3}$ and the principle stresses on the planes are arranged respectively with ${\sigma }_1>{\sigma }_2>{\sigma }_3$.

Write the expression for longitudinal stress on the cylindrical vessel.

${\sigma }_{l_1}=\frac{p_i{r}_i^2}{r_o^2-r_i^2}$ (XVIII)

Here, the longitudinal stress is ${\sigma }_{l_1}$.

Write the expression for the total longitudinal stress.

${\sigma }_l={\sigma }_{l_1}+{\sigma }_w$ (XIX)

Here, the total longitudinal stress is ${\sigma }_l$.

Conclusion:

Convert the outer diameter from feet to inch.

$\begin{array}{c}{d}_o=30\text{ft}\\ =\left(\text{30}\text{ft}\right)\left(\frac{\text{12}\text{in}}{1\text{ft}}\right)\\ =\text{360}\text{in}\end{array}$

The outer radius of the cylinder is $180\text{in}$.

Substitute $180\text{in}$ for $r_o$ and $0.75\text{in}$ for $t$ in Equation (IV).

$\begin{array}{c}{r}_i=180\text{in}-0.75\text{in}\\ =179.25\text{in}\end{array}$

The inner diameter of the cylinder is $358.5\text{in}$.

Substitute $\text{360}\text{in}$ for $d_o$ and $358.5\text{in}$ for $d_i$ in Equation (VIII).

$\begin{array}{c}{A}_w=\frac{\pi }{4}\left({\left(360\text{in}\right)}^2-{\left(358.5\text{in}\right)}^2\right)\\ =\frac{\pi }{4}\left(1077.75{\text{in}}^2\right)\\ \approx 846.5{\text{in}}^{\text{2}}\end{array}$

Substitute $\text{360}\text{in}$ for $d_o$, $358.5\text{in}$ for $d_i$ and $60\text{ft}$ for $h$ in Equation (X).

$\begin{array}{c}{V}_w=\frac{\pi }{4}\left({\left(360\text{in}\right)}^2-{\left(358.5\text{in}\right)}^2\right)\times \left(\frac{60\text{ft}\times 12\text{in}}{1\text{ft}}\right)\\ =\frac{\pi }{4}\left(1077.75{\text{in}}^2\right)\left(60\text{ft}\right)\left(\frac{12\text{in}}{1\text{ft}}\right)\\ \approx 609453.27{\text{in}}^{\text{3}}\end{array}$

Substitute $\text{180}\text{in}$ for $d_o$ and $179.25\text{in}$ for $d_i$ in Equation (XI).

$\begin{array}{c}{V}_{dome}=\frac{2\pi }{3}\left[{\left(180\text{in}\right)}^3-{\left(179.25\text{in}\right)}^3\right]\\ =\frac{2\pi }{3}\left[72596.67{\text{in}}^{\text{3}}\right]\\ \approx 152046.11{\text{in}}^{\text{3}}\end{array}$

Refer to Table A-5 “Physical Constants of Materials” to obtain the properties of unit weight of carbon steel as,

$w=0.282\text{lbf}/{\text{in}}^{\text{3}}$

Substitute $0.282\text{lbf}/{\text{in}}^{\text{3}}$ for $w$, $609453.27{\text{in}}^{\text{3}}$ for $V_w$ and $152046.11{\text{in}}^{\text{3}}$ for $V_{dome}$ in Equation (XII).

$\begin{array}{c}W=\left(0.282\text{lbf}/{\text{in}}^{\text{3}}\right)\left(609453.27{\text{in}}^3+152046.11{\text{in}}^3\right)\\ =\left(0.282\text{lbf}/{\text{in}}^{\text{3}}\right)\left(761499.38{\text{in}}^3\right)\\ \approx 214742.83\text{lbf}\end{array}$

Substitute $214742.83\text{lbf}$ for $W$ and $846.5{\text{in}}^{\text{2}}$ for $A$ in Equation (XIII).

$\begin{array}{c}{\sigma }_w=-\frac{214742.83\text{lbf}}{846.5{\text{in}}^2}\\ =\left(-253.68{\text{lbf/in}}^2\right)\left(\frac{1\text{psi}}{1{\text{lbf/in}}^2}\right)\\ =-253.68\text{psi}\end{array}$

Substitute $62.4\text{lbf}/{\text{ft}}^{\text{3}}$ for $\omega$ and $55\text{ft}$ for $\delta$ in Equation (XIV).

$\begin{array}{c}{p}_i=\left(62.4\text{lbf}/{\text{ft}}^{\text{3}}\right)\left(55\text{ft}\right)\\ =\left(62.4\text{lbf}/{\text{ft}}^{\text{3}}\right)\left(\frac{{\left(1\text{ft}\right)}^{\text{3}}}{{\left(12\text{in}\right)}^{\text{3}}}\right)\left(55\text{ft}\right)\left(\frac{\text{12}\text{in}}{1\text{ft}}\right)\\ =\left(0.0361{\text{lbf/in}}^3\right)\left(660\text{in}\right)\\ \approx 23.826\text{psi}\end{array}$

Since the external pressure is zero, the maximum tangential stress will occur at inside radius.

$r=r_i$

Substitute $0\text{psi}$ for $p_o$, $23.826\text{psi}$ for $p_i$, $180\text{in}$ for $r_o$, $179.25\text{in}$ for $r_i$ and $179.25\text{in}$ for $r$ in Equation (I).

$\begin{array}{c}{\sigma }_{t_1}=\frac{\left(23.826\text{psi}\right){\left(179.25\text{in}\right)}^2-\left(0\text{psi}\right){\left(180\text{in}\right)}^2-{\left(179.25\text{in}\right)}^2{\left(180\text{in}\right)}^2\frac{\left(0\text{psi}-23.826\text{psi}\right)}{{\left(179.25\text{in}\right)}^2}}{{\left(180\text{in}\right)}^2-{\left(179.25\text{in}\right)}^2}\\ =\frac{\left(23.826\text{psi}\right){\left(179.25\text{in}\right)}^2-{\left(180\text{in}\right)}^2\left(-23.826\text{psi}\right)}{269.4375{\text{in}}^2}\\ \approx 5706.3518\text{psi}\end{array}$

Substitute $0\text{psi}$ for $p_o$, $50\text{psi}$ for $p_i$, $180\text{in}$ for $r_o$, $179.25\text{in}$ for $r_i$ and $179.25\text{in}$ for $r$ in Equation (II).

$\begin{array}{c}{\sigma }_{t_2}=\frac{\left(50\text{psi}\right){\left(179.25\text{in}\right)}^2-\left(0\text{psi}\right){\left(180\text{in}\right)}^2-{\left(179.25\text{in}\right)}^2{\left(180\text{in}\right)}^2\frac{\left(0\text{psi}-50\text{psi}\right)}{{\left(179.25\text{in}\right)}^2}}{{\left(180\text{in}\right)}^2-{\left(179.25\text{in}\right)}^2}\\ =\frac{\left(50\text{psi}\right){\left(179.25\text{in}\right)}^2-{\left(180\text{in}\right)}^2\left(-50\text{psi}\right)}{269.4375{\text{in}}^2}\\ \approx 11975.05\text{psi}\end{array}$

Substitute $5706.3518\text{psi}$ for ${\sigma }_{t_1}$ and $11975.05\text{psi}$ for ${\sigma }_{t_2}$ in Equation (III).

$\begin{array}{c}{\sigma }_t=5706.3518\text{psi}+11975.05\text{psi}\\ =17681.4018\text{psi}\end{array}$

Thus, the tangential stress on the cylinder is $17681.4018\text{psi}$.

Substitute $0\text{psi}$ for $p_o$, $23.826\text{psi}$ for $p_i$, $180\text{in}$ for $r_o$, $179.25\text{in}$ for $r_i$ and $179.25\text{in}$ for $r$ in Equation (V).

$\begin{array}{c}{\sigma }_{r_1}=\frac{\left(23.826\text{psi}\right){\left(179.25\text{in}\right)}^2-\left(0\text{psi}\right){\left(180\text{in}\right)}^2+{\left(179.25\text{in}\right)}^2{\left(180\text{in}\right)}^2\frac{\left(0\text{psi}-23.826\text{psi}\right)}{{\left(179.25\text{in}\right)}^2}}{{\left(180\text{in}\right)}^2-{\left(179.25\text{in}\right)}^2}\\ =\frac{\left(23.826\text{psi}\right){\left(179.25\text{in}\right)}^2+{\left(180\text{in}\right)}^2\left(-23.826\text{psi}\right)}{269.4375{\text{in}}^2}\\ \approx -23.826\text{psi}\end{array}$

Substitute $0\text{psi}$ for $p_o$, $50\text{psi}$ for $p_i$, $180\text{in}$ for $r_o$, $179.25\text{in}$ for $r_i$ and $179.25\text{in}$ for $r$ in Equation (VI).

$\begin{array}{c}{\sigma }_{t_2}=\frac{\left(50\text{psi}\right){\left(179.25\text{in}\right)}^2-\left(0\text{psi}\right){\left(180\text{in}\right)}^2+{\left(179.25\text{in}\right)}^2{\left(180\text{in}\right)}^2\frac{\left(0\text{psi}-50\text{psi}\right)}{{\left(179.25\text{in}\right)}^2}}{{\left(180\text{in}\right)}^2-{\left(179.25\text{in}\right)}^2}\\ =\frac{\left(50\text{psi}\right){\left(179.25\text{in}\right)}^2+{\left(180\text{in}\right)}^2\left(-50\text{psi}\right)}{269.4375{\text{in}}^2}\\ \approx -50\text{psi}\end{array}$

Substitute $-23.826\text{psi}$ for ${\sigma }_{r_1}$ and $-50\text{psi}$ for ${\sigma }_{r_2}$ in Equation (VII).

$\begin{array}{c}{\sigma }_r=-23.826-50\\ =-73.826\text{psi}\end{array}$

Thus, the radial stress on the cylinder is $-73.826\text{psi}$.

Substitute $50\text{psi}$ for $p_i$, $180\text{in}$ for $r_o$, $179.25\text{in}$ for $r_i$ and $179.25\text{in}$ for $r$ in Equation (XVIII).

$\begin{array}{c}{\sigma }_{l_1}=\frac{\left(50\text{psi}\right){\left(179.5\text{in}\right)}^2}{{\left(180\text{in}\right)}^2-{\left(179.25\text{in}\right)}^2}\\ =\frac{\left(1611012.5\text{psi}\cdot {\text{in}}^2\right)}{{\left(180\text{in}\right)}^2-{\left(179.25\text{in}\right)}^2}\\ \approx 5979.2\text{psi}\end{array}$

Substitute $5979.2\text{psi}$ for ${\sigma }_{l_1}$ and $-253.68\text{psi}$ for ${\sigma }_w$ in Equation (XIX).

$\begin{array}{c}{\sigma }_l=5979.2\text{psi}-253.68\text{psi}\\ =5725.5\text{psi}\end{array}$

Thus, the longitudinal stress is $5725.5\text{psi}$.

Since the maximum principle normal stresses are assumed as ${\sigma }_1>{\sigma }_2>{\sigma }_3$, the magnitude of principle stresses are:

$\begin{array}{l}{\sigma }_1={\sigma }_t=17681.4018\text{psi}\\ {\sigma }_2={\sigma }_w=5725.5\text{psi}\\ {\sigma }_3={\sigma }_r=-73.826\text{psi}\end{array}$

Thus, the principle normal stresses are $17681.41\text{psi}$, $5725.5\text{psi}$, and $-73.826\text{psi}$ respectively in decreasing order.

Substitute $17681.4018\text{psi}$ for ${\sigma }_1$ and $5725.5\text{psi}$ for ${\sigma }_2$ in Equation (XI).

$\begin{array}{c}{\tau }_{1/2}=\frac{17681.41\text{psi}-5725.5\text{psi}}{2}\\ =\frac{11955.91\text{psi}}{2}\\ \approx 5977\text{psi}\end{array}$

Substitute $17681.4018\text{psi}$ for ${\sigma }_1$ and $-73.826\text{psi}$ for ${\sigma }_3$ in Equation (XI).

$\begin{array}{c}{\tau }_{1/3}=\frac{17681.4018\text{psi}-\left(-73.826\text{psi}\right)}{2}\\ =\frac{17755.2278\text{psi}}{2}\\ \approx 8877.614\text{psi}\end{array}$

Substitute $5725.5\text{psi}$ for ${\sigma }_2$ and $-73.826\text{psi}$ for ${\sigma }_3$ in Equation (XI).

$\begin{array}{c}{\tau }_{2/3}=\frac{5725.5\text{psi}-\left(-73.826\text{psi}\right)}{2}\\ =\frac{5799.326\text{psi}}{2}\\ \approx 2899.6\text{psi}\end{array}$

Write the decreasing arrangement of principle shear stress.

${\tau }_{1/3}>{\tau }_{1/2}>{\tau }_{2/3}$

Thus, The principle shear stresses are $8877.614\text{psi}$, $5977\text{psi}$ and $2899.6\text{psi}$ respectively in decreasing order.