Chapter 7.6, Problem 124P

The compressed-air tank AB has a 250-rnm outside diameter and an 8-mm wall thickness. It is fitted with a collar by which a 40-kN force P is applied at B in the horizontal direction. Knowing that the gage pressure inside the tank is 5 MPa, determine the maximum normal stress and the maximum shearing stress at point K.

Fig. P7.124

To determine

Find the maximum normal stress and maximum shear stress at point K.

Answer to Problem 124P

The maximum normal stress and maximum shear stress at point K are $\underset{\_}{77.4\text{MPa}}$ and $\underset{\_}{38.7\text{MPa}}$.

Explanation of Solution

Given information:

The outer diameter (d) of the tank is $250\text{mm}$.

The wall thickness (t) of the tank is $8\text{mm}$.

The magnitude of the load P is $40\text{kN}$.

The gage pressure (p) inside the tank is $5\text{MPa}$.

Calculation:

Consider an element at point K.

Show the stress acting on the point K as shown in Figure 1.

Calculate the inner radius of the vessel $\left(r\right)$ as shown below.

$r=\frac{d}{2}-t$ (1)

Substitute $250\text{mm}$ for $d$ and $8\text{mm}$ for t in Equation (1).

$\begin{array}{c}r=\frac{250}{2}-8\\ =117\text{mm}\end{array}$

Show the expression for hoop stress acting on the tank as shown below.

${\sigma }_x=\frac{pr}{t}$

Substitute $5\text{MPa}$ for $p$ and $117\text{mm}$ for r, and $8\text{mm}$ for t.

$\begin{array}{c}{\sigma }_x=\frac{5\times 117}{8}\\ =73.125\text{MPa}\end{array}$

Show the expression for longitudinal stress acting on the vessel as shown below.

${\sigma }_y=\frac{pr}{2t}$

Substitute $5\text{MPa}$ for $p$ and $117\text{mm}$ for r, and $8\text{mm}$ for t.

$\begin{array}{c}{\sigma }_y=\frac{5\times 117}{2\times 8}\\ =36.563\text{MPa}\end{array}$

Calculate the stress due to bending at point K as follows:

The point K lies in the neutral axis, then

The stress due to bending at point K, ${\sigma }_y=0$.

Calculate the stress due to transverse shear as follows:

Consider the shear force acting on the tank is denoted by V. Then,

$\begin{array}{c}V=P\\ =40\text{kN}\end{array}$

Consider the distance between the centre of the circular cross-section to the inner and outer surface of the wall are denoted by $c_1$ and $c_2$. Then,

$\begin{array}{c}{c}_2=\frac{d}{2}\\ =\frac{250}{2}\\ =125\text{mm}\end{array}$

$\begin{array}{c}{c}_1=c_2-t\\ =125-8\\ =117\text{mm}\end{array}$

Calculate the shear flow Q using the relation:

$Q=\frac{2}{3}\left(c_2^3-c_1^3\right)$

Substitute $117\text{mm}$ for $c_1$ and $125\text{mm}$ for $c_2$.

$\begin{array}{c}Q=\frac{2}{3}\left({125}^3-{117}^3\right)\\ =\frac{2}{3}\times 0.3515\times {10}^6\\ =0.2343\times {10}^6{\text{mm}}^3\end{array}$

Calculate the moment of inertia I of the cross-section using the relation:

$I=\frac{\pi }{4}\left(c_2^4-c_1^4\right)$

Substitute $117\text{mm}$ for $c_1$ and $125\text{mm}$ for $c_2$.

$\begin{array}{c}I=\frac{\pi }{4}\left({125}^4-{117}^4\right)\\ =\frac{\pi }{4}\times 56.751\times {10}^6\\ =44.573\times {10}^6{\text{mm}}^4\end{array}$

Calculate the shear stress using the relation:

${\tau }_{xy}=\frac{VQ}{It}$

Substitute $40\text{kN}$ for V, $0.2343\times {10}^6{\text{mm}}^3$ for Q, $44.573\times {10}^6{\text{mm}}^4$ for I, and $2\times 8\text{mm}$ for t.

$\begin{array}{c}{\tau }_{xy}=\frac{40\text{kN}\times \left(\frac{{10}^3\text{N}}{1\text{kN}}\right)\times 0.2343\times {10}^6{\text{mm}}^3}{44.573\times {10}^6{\text{mm}}^4\times 2\times 8\text{mm}}\\ =\frac{9372\times {10}^6}{2\times 356.584\times {10}^6}\\ =13.14\text{MPa}\end{array}$

Show the total hoop stress ${\sigma }_x$, total longitudinal stress ${\sigma }_y$, and the total shear stress ${\tau }_{xy}$ as follows:

$\begin{array}{l}{\sigma }_x=73.125\text{MPa}\\ {\sigma }_y=36.563\text{MPa}\\ {\tau }_{xy}=13.14\text{MPa}\end{array}$

Calculate the radius of the Mohr circle using the relation:

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

Substitute $73.125\text{MPa}$ for ${\sigma }_x$, $36.563\text{MPa}$ for ${\sigma }_y$, and $13.14\text{MPa}$ for ${\tau }_{xy}$.

$\begin{array}{c}R=\sqrt{{\left(\frac{73.125-36.563}{2}\right)}^2+{13.14}^2}\\ =\sqrt{506.8545}\\ =22.513\text{MPa}\end{array}$

Show the expression for average stress acting on the vessel as shown below.

${\sigma }_{\text{ave}}=\frac{1}{2}\left({\sigma }_x+{\sigma }_y\right)$

Substitute $73.125\text{MPa}$ for ${\sigma }_x$ and $36.563\text{MPa}$ for ${\sigma }_y$.

$\begin{array}{c}{\sigma }_{\text{ave}}=\frac{1}{2}\left(73.125+36.563\right)\\ =\frac{109.688}{2}\\ =54.844\text{MPa}\end{array}$

Consider the principal stress are denoted by ${\sigma }_a,{\sigma }_b,\text{and }{\sigma }_c$.

Calculate the value of the ${\sigma }_a$ as follows:

${\sigma }_a={\sigma }_{\text{ave}}+R$

Substitute $22.513\text{MPa}$ for R, and $54.844\text{MPa}$ for ${\sigma }_{\text{ave}}$.

$\begin{array}{c}{\sigma }_a=22.513+54.844\\ =77.4\text{MPa}\end{array}$

Calculate the value of the ${\sigma }_b$ as follows:

${\sigma }_b={\sigma }_{\text{ave}}-R$

Substitute $22.513\text{MPa}$ for R, and $54.844\text{MPa}$ for ${\sigma }_{\text{ave}}$.

$\begin{array}{c}{\sigma }_b=22.513-54.844\\ =-32.3\text{MPa}\end{array}$

The stress at point C is 0. Then,

${\sigma }_c=0\text{MPa}$

Compare the value of stress ${\sigma }_a$, ${\sigma }_b$, and ${\sigma }_c$.

Get the maximum and minimum value of the stress as follows:

$\begin{array}{l}{\sigma }_{\max }=77.4\text{MPa}\\ {\sigma }_{\min }=0\text{MPa}\end{array}$

Thus, the maximum normal stress in the tank is $\underset{\_}{77.4\text{MPa}}$.

Calculate the maximum shear stress in the tank using the relation:

${\tau }_{\max }=\frac{1}{2}\left({\sigma }_{\max }-{\sigma }_{\min }\right)$

Substitute $77.4\text{MPa}$ for ${\sigma }_{\max }$ and 0 for ${\sigma }_{\min }$.

$\begin{array}{c}{\tau }_{\max }=\frac{1}{2}\left(77.4-0\right)\\ =38.7\text{MPa}\end{array}$

Thus, the maximum shear stress in the tank is $\underset{\_}{38.7\text{MPa}}$.