Chapter 8, Problem 8.5.7P

A cylindrical pressure vessel having a radius r = 14 in. and wall thickness t = 0,5 in, is subjected to internal pressure p = 375 psi, In addition, a torque T = 90 kip-ft acts at each end of the cylinder (see figure),

(a) Determine the maximum tensile stress c_tniXand the maximum in-plane shear stress T_mjv in the wall of the cylinder.

(b) If the allowable in-plane shear stress is 4.5 ksi, what is the maximum allowable torque T\

(c) If 7 = 150 kip-ft and allowable in-plane shear and allowable normal stresses are 4.5 ksi and 11.5 ksi, respectively, what is the minimum required wall thickness

To determine

The maximum tensile stress ${\sigma }_{\max }$ and the maximum in plane shear stress ${\tau }_{\max }$ in the wall of the cylinder.

Answer to Problem 8.5.7P

The maximum tensile stress is $11.09ksi$ and the maximum shear stress is $3.21ksi$

Explanation of Solution

Given Information:

Type of vessel = cylindrical

Radius = $14in$

Internal pressure = $375psi$

Wall thickness = $0.5in$

Torque = $90kip-ft$ at each end of cylinder

Concept used:

  $\text{Longitudinal stress }{\sigma }_x=\frac{pr}{2t}$

  $\text{hoop stress = }\frac{pr}{t}$

Where $p:$ Pressure, $r=\text{radius}$ , $t=\text{thickness}$

Calculation:

Use the following relation to find the longitudinal stress.

  ${\sigma }_x=\frac{pr}{2t}$

Here, wall thickness is $t$ , radius is $r$ , and internal pressure is $\rho$ .

Substitute $375\text{psi for }\rho ,14$ in for $r$ and $0.5$ in. for $t$ .

  $\begin{array}{c}{\sigma }_x=\frac{375\times 14}{2\times 0.5}\\ =5,250\text{psi}\end{array}$

Use the following relation to find the hoop stress.

  ${\sigma }_y=\frac{\mathrm{Pr}}{t}$

Substitute $375\text{psi}\text{for }\rho ,14$

  $r,\text{and 0}\text{.5}$ in. for $t$

  $\begin{array}{c}{\sigma }_y=\frac{375\times 14}{0.5}\\ =10,500\text{psi}\end{array}$

Use the following relation to find the shear stress.

  ${\tau }_{xy}=\frac{-Tr}{I_p}\mathrm{.............}\left(1\right)$

Here, torque is $T$ and polar moment of inertia is $I_P$ .

Express the value of $I_P$ .

  $I_P=\frac{\pi }{2}\left[r^4-{\left(r-t\right)}^4\right]$

Substitute $14\text{ in}\text{. for r, and 0}\text{.5}$ in. for $t$

  $\begin{array}{c}{I}_p=\frac{\pi }{2}\left[{14}^4-{\left(13.5\right)}^4\right]\\ =8,169.63{\text{ in}}^4\end{array}$

From equation $\left(1\right)$ , substitute $8,169.63{\text{ in}}^4$ for $I_P$ , $14$ in. for $r$ , and $90$ k-ft for $T$ .

  $\begin{array}{c}{\tau }_{xy}=\frac{-90\left(\text{k}-\text{ft}\right)\times \text{14in}}{8,169.63{\text{ in}}^4}\\ =\frac{-90\left(k-\text{ft}\right)\times \frac{{10}^3}{k}\times \frac{12\text{in}}{1\text{ft}}\times 14\text{in}}{8,169.63\text{in}}\\ \simeq -1,851\text{psi}\end{array}$

Find the maximum stress $\left({\sigma }_{\max }\right)$ and the maximum in plane shear stress $\left({\tau }_{\max }\right)$ in the wall of the cylinder.

Use the following relation to find the principal stresses.

  ${\sigma }_{12}=\frac{{\sigma }_x+{\sigma }_y}{2}\pm \sqrt{{\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)}^2+{\tau }_{xy}^2}$

Substitute $-1,851psi\text{for }{\tau }_{xy},10,500psi\text{for}{\sigma }_y$ , and $5,250psi\text{for }{\sigma }_x$

  $\begin{array}{c}{\sigma }_{1,2}=\left(\frac{5,250+10,500}{2}\right)\pm \sqrt{{\left(\frac{5,250-10,500}{2}\right)}^2+{\left(-1,851\right)}^2}\\ =\left(\frac{5,250+10,500}{2}\right)\pm \sqrt{6,890,625+3,426,201}\\ =7,875\pm 3,212\end{array}$

Calculate the principal stress, ${\sigma }_2$

  $\begin{array}{c}{\sigma }_2=7,875-3,212\\ =4,663psi\times \frac{ksi}{1000ksi}\\ =4,663ksi\end{array}$

As ${\sigma }_1$ is greater than ${\sigma }_2$ , the maximum stress ${\sigma }_{\max }$ is given by ${\sigma }_1\text{ is 11}\text{.09ksi}$ .

Use the following relation to find the maximum in plane shear stress

  ${\tau }_{\max }=\left(\frac{{\sigma }_1-{\sigma }_2}{2}\right)$

Substitute $11,087psi\text{for }{\sigma }_1\text{and 4,663}psi$ for ${\sigma }_2$

  $\begin{array}{c}{\tau }_{\max }=\frac{11,087-4,663}{2}\\ =3,212psi\times \frac{ksi}{1000psi}\\ =3.21ksi\end{array}$

Hence, the maximum shear stress is $3.21ksi$ .

Conclusion

The values are found by the concept of longitudinal stress and hoop stress.

To determine

Maximum torque.

Answer to Problem 8.5.7P

The maximum torque is $178\text{ kip-ft}$

Explanation of Solution

Given Information:

  $\begin{array}{l}{\tau }_{\max }=4.5ksi\\ {\sigma }_x=5250psi\\ {\sigma }_y=10500psi\\ I_p=8169.63i{n}^4\\ r=14in\end{array}$

Concept used:

  ${\tau }_{\max }=\sqrt{{\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)}^2+{\left(\frac{T_{\max }\times r}{I_p}\right)}^2}$

Calculation:

Find the maximum torque $\left(T_{\max }\right)$ if maximum in plane shear stress is ${\tau }_{\max }=4.5ksi$ .

  ${\tau }_{\max }=\sqrt{{\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)}^2+{\left(\frac{T_{\max }\times r}{I_p}\right)}^2}$

Substitute $10,500psi\text{for }{\sigma }_y,5,250psi\text{for }{\sigma }_x,8,169.63{\text{in}}^4$ for $I_p,4.5ksi\text{for }{\tau }_{\max }$ . Are $14$ in. for $r$ .

  $\begin{array}{c}4.5ksi=\sqrt{{\left(\frac{5,250-10,500}{2}\right)}^2+\left(\frac{{\left(T_{\max }\right)}^2\times {\left(14\right)}^2}{{\left(8,169.63\right)}^2}\right)}\\ {\left(4.5ksi\times {10}^3\frac{psi}{ksi}\right)}^2={\left(\frac{5,250-10,500}{2}\right)}^2+\left(\frac{{\left(T_{\max }\right)}^2\times {\left(14\right)}^2}{{\left(8,169.63\right)}^2}\right)\\ 20.25\times {10}^6=\left(6.89\times {10}^6\right)+\left(T_{\max }^2\times 2.937\times {10}^{-6}\right)\\ T_{\max }^2=\frac{\left(20.25-6.89\right)\times {10}^6}{2.937\times {10}^{-6}}\\ =4.549\times {10}^{12}\\ T_{\max }=2.133\times {10}^{6}lb-in\end{array}$

Conclusion:

The maximum torque is calculated by equating 1p , Tmax, r, etc.

To determine

What is the minimum required wall thickness.

Answer to Problem 8.5.7P

The minimum required wall thickness is $0.519psi$

Explanation of Solution

Given Information:

  $\begin{array}{l}{\tau }_{all}=4.5ksi\\ {\sigma }_{all}=11.5ksi\\ hoop\text{ stress = }\frac{pr}{t}\\ Longitudinal\text{stress=}\frac{pr}{2t}\end{array}$

Concept used:

  ${\sigma }_a=\frac{\frac{pr}{2t}+\frac{pr}{t}}{2}+\sqrt{{\left(\frac{\frac{pr}{2t}+\frac{pr}{t}}{2}\right)}^2+{\left[\frac{T.r}{\frac{\pi }{2}\left[r^4-{\left(r-t\right)}^4\right]}\right]}^2}$

Calculation:

  ${\sigma }_a=\frac{\frac{pr}{2t}+\frac{pr}{t}}{2}+\sqrt{{\left(\frac{\frac{pr}{2t}+\frac{pr}{t}}{2}\right)}^2+{\left[\frac{T.r}{\frac{\pi }{2}\left[r^4-{\left(r-t\right)}^4\right]}\right]}^2}$

Here, allowable in plane shear stress is ${\tau }_a$ and allowable normal stress is ${\sigma }_a$ .

Substitute $11.5ksi\text{for }{\sigma }_a,14$ in. for $r,0.5$ in. for $t,375psi\text{for }p$ , and $150kip-\text{ft}$ for $T$ .

  $\begin{array}{c}{\sigma }_a=\frac{\frac{375\times 14}{2t}+\frac{375\times 14}{t}}{2}+\sqrt{\left(\frac{\frac{375\times 14}{2t}+\frac{375\times 14}{t}}{2}\right)+{\left(\frac{1.8\times {10}^6\times 14\times 2}{\pi \left[\left({14}^4-{\left(14-t\right)}^4\right)\right]}\right)}^2}\\ =\frac{\frac{2,625}{t}+\frac{5,250}{t}}{2}+\sqrt{{\left(\frac{2,625-5,250}{2t}\right)}^2+{\left(A\right)}^2}\end{array}$

Express the value of $A$ form equation $\left(2\right)$ .

  $A=\frac{B}{C}\mathrm{..............}\left(3\right)$

Here, assumed

Variable are $A,B\text{ and }C$ respectively.

Evaluate the value of $B$ from equation $\left(2\right)$ .

  $\begin{array}{c}B=1.8\times {10}^6\times 14\times \frac{2}{\pi }\\ =16.0428\times {10}^6\end{array}$

Evaluate the value of $C$ from Equation $\left(2\right)$

  $\begin{array}{c}C={14}^4-{\left(14-t\right)}^4\\ ={\left({14}^2\right)}^2-{\left({\left[14-t\right]}^2\right)}^2\\ =\left({14}^2+{\left(14-t\right)}^2\right)\left({14}^2-{\left(14-t\right)}^2\right)\\ =\left(196+196+t^2-28t\right)\left(196-196-t^2+28t\right)\\ =\left(392+t^2-28t\right)\left(28t-t^2\right)\\ =10,976t-392t^2+28t^3-t^4-784t^2+28t^3\\ =-t^4+56t^3-1,176t^2+1,0976t\end{array}$

Substitute $-t^4+56t^3-1,176t^2+1,0976t$ for $C$ and $16.0428\times {10}^6\text{ for }B$ .

  $A=\frac{16.0428\times {10}^6}{\left(-t^4+56t^3-1,176t^2+1,0976t\right)}$

Substitute the value of $A\text{ and }{\sigma }_a$ in equation $\left(2\right)$ .

  $\begin{array}{c}11.5\times {10}^3=\frac{\frac{2,625}{t}+\frac{5,250}{t}}{2}+\sqrt{{\left(\frac{2,625-5,250}{2t}\right)}^2+{\left(\frac{16.0428\times {10}^6}{\left(-t^4+56t^3-1,176t^2+1,0976t\right)}\right)}^2}\\ =\frac{3,937.5}{t}+\sqrt{{\left(\frac{-3,312.5}{t}\right)}^2+{\left[\frac{16.0428\times {10}^6}{t\left(-t^3+56t^2-1,176t+10,976\right)}\right]}^2}\end{array}$

Use the trial and error method to find the thickness $t$ value.

Trial $1$

Substitute $t=0.50\text{in}$ . In Equation $\left(2\right)$ .

  $\begin{array}{c}{\sigma }_{allow}=7,875+\sqrt{6,890,625+9,514,743.15}\\ =7,875+4,050.35\\ =11,925.35\text{psi}\end{array}$

Trial $\text{ll}$

Substitute $t=\text{0}\text{.55in}$ . In Equation $\left(2\right)$

  $\begin{array}{c}{\sigma }_{allow}=7,159.09+\sqrt{5,694,731.405+7,948,628.98}\\ =7,159.09+3,693.69\\ =10,852.78\text{psi}\end{array}$

Trial $\text{lll}$

Substitute $t=0.515\text{in}$ . In equation $\left(2\right)$

  $\begin{array}{c}{\sigma }_{allow}=7,645.63+\sqrt{6,495,074.94+8,997,596.67}\\ =7,645.63+3,936.07\\ =11,581.7psi\end{array}$

Substitute $t=0.519\text{in}$ . In Equation $\left(2\right)$ .

  $\begin{array}{c}{\sigma }_{allow}=7,586.71+\sqrt{6,395,343.981+8,867,080.985}\\ =7,586.71+3,906.71\\ =1,1493.4psi\end{array}$

Express the allowable in plane shear stress.

  ${\tau }_{allow}=\sqrt{{\left(\frac{\frac{pr}{2t}-\frac{pr}{t}}{2}\right)}^2+{\left[\frac{T.r}{\frac{\pi }{2}\left[r^4-{\left(r-t\right)}^4\right]}\right]}^2}$

Substitute $14\text{ in for }r,0.5$ . For $t,375\text{psi}\text{for }p$ , and $150kip-ft\text{ for }T$ .

  $\begin{array}{c}{\tau }_{allow}=\sqrt{\left(\frac{\frac{375\times 14}{2t}-\frac{375\times 14}{t}}{2}\right)+{\left(\frac{1.8\times {10}^6\times 14\times 2}{\pi \left[\left({14}^2-{\left(14-t\right)}^4\right)\right]}\right)}^2}\\ 4,500=\sqrt{{\left[\frac{-1312.5}{t}\right]}^2+{\left[\frac{16.0428\times {10}^6}{-t^4+56t^3-117t^2+10,976t}\right]}^2}\end{array}$

Substitute $0.519in$ . For $t$ in equation $\left(4\right)$ .

  $\begin{array}{c}4,500=\sqrt{{\left[\frac{-1312.5}{0.519in}\right]}^2+\left[\frac{16.0428\times {10}^6}{-{\left(0.519in\right)}^4+56{\left(0.519in\right)}^3-1176{\left(0.519in\right)}^2+10,976\left(0.519in\right)}\right]}\\ {\tau }_{allow}=3,906.71psi\end{array}$

As ${\tau }_{allow}\propto \frac{1}{t}$ the thickness value is to be increased to get ${\tau }_{allow}=4,500psi$ .

Hence, from the trial and error method the minimum wall thickness $t_{\min }\text{ is 0}\text{.519in}$

Conclusion:

The minimum thickness is calculated by trial and error method.