Chapter 11, Problem 11.9.24P

To determine

The allowable axial load.

Answer to Problem 11.9.24P

The allowable axial load at $L=0.6\text{m}$ is $96.03\text{kN}$ .

The allowable axial load at $L=0.8\text{m}$ is $83.68\text{kN}$ .

The allowable axial load at $L=1\text{m}$ is $60.77\text{kN}$ .

The allowable axial load at $L=1.2\text{m}$ is $42.19\text{kN}$ .

Explanation of Solution

Given:

The outside diameter of the pipe is $d_2=80\text{mm}$ and inside diameter of the pipe is $d_1=72\text{mm}$ .

Concept Used:

Write the expression for the allowable axial load.

  $P_a={\sigma }_{a}A$(1)

The allowable axial load is $P_a$ , the allowable stress is ${\sigma }_a,$ and the area of cross-section is $A$ .

Write the expression for the area of cross-section.

  $A=\frac{\pi }{4}\left(d_2^2-d_1^2\right)$

Substitute $\frac{\pi }{4}\left(d_2^2-d_1^2\right)$ for $A$ in equation (1).

  $P_a={\sigma }_a\times \frac{\pi }{4}\left(d_2^2-d_1^2\right)$(2)

The equation for the aluminum alloy (6061-T6) column allowable stress as follows:

  ${\sigma }_a=20.2-0.126\left(\frac{kL}{r}\right)\text{ksi}\frac{kL}{r}\le 66$ ....... (3)

  ${\sigma }_a=\frac{51000}{{\left(\frac{kL}{r}\right)}^2}\text{ksi}\frac{kL}{r}\ge 66$(4)

The radius of gyration is $r$ and the constant is $k$ .

Since the column is fixed at the base and free at the top so,

  $k=2$

Write the expression for the radius of gyration.

  $r=\frac{I}{A}$(5)

The area moment of inertia is $I$ .

Write the equation of area moment of inertia as follows:

  $I=\frac{\pi }{64}\left(d_2^4-d_1^4\right)$

Substitute $\frac{\pi }{64}\left(d_2^4-d_1^4\right)$ for $I$ and $\frac{\pi }{4}\left(d_2^2-d_1^2\right)$ for $A$ in equation (5).

  $\begin{array}{l}r=\sqrt{\frac{\frac{\pi }{64}\left(d_2^4-d_1^4\right)}{\frac{\pi }{4}\left(d_2^2-d_1^2\right)}}\\ r=\sqrt{\frac{1}{16}\frac{\left(d_2^2+d_1^2\right)\left(d_2^2-d_1^2\right)}{\left(d_2^2-d_1^2\right)}}\\ r=\frac{1}{4}\sqrt{\left(d_2^2+d_1^2\right)}\end{array}$(6)

Calculation:

As per the given problem

  $d_1=72\text{mm}$ , $d_2=80\text{mm}$ , $L=0.6\text{m}$ , $0.8\text{m}$ , $1\text{m}$ and $1.2\text{m}$ .

Convert the diameter into $\text{inch}$ .

  $\begin{array}{c}{d}_1=\left(72\text{mm}\right)\left(\frac{1\text{in}}{25.4\text{mm}}\right)\\ =2.835\text{in}\\ d_2=\left(80\text{mm}\right)\left(\frac{1\text{in}}{25.4\text{mm}}\right)\\ =3.150\text{in}\end{array}$

Substitute $2.835\text{in}$ for $d_1$ and $3.150\text{in}$ for $d_2$ in equation (6).

  $\begin{array}{c}r=\frac{1}{4}\sqrt{\left({2.835}^2+{3.150}^2\right)}\\ =1.059\text{in}\end{array}$

When $L=0.6\text{m}$ ,

  $\begin{array}{c}\frac{kL}{r}=\frac{2\times \left(0.6\text{m}\right)\left(\frac{39.37\text{in}}{1\text{m}}\right)}{1.059}\\ =44.612\end{array}$

Since $\frac{kL}{r}\le 66$ , so

Substitute $44.612$ for $\frac{kL}{r}$ in equation (3).

  $\begin{array}{c}{\sigma }_a=20.2-\left(0.126\times 44.612\right)\text{ksi}\\ =14.58\text{ksi}\end{array}$

Substitute $14.58\text{ksi}$ for ${\sigma }_a$ , $2.835\text{in}$ for $d_1$ and $3.150\text{in}$ for $d_2$ in equation (2).

  $\begin{array}{c}{P}_a=\text{14}.58\times {\text{10}}^3\times \frac{\pi }{4}\left({3.15}^2-{2.835}^2\right)\\ =\left(21588.48\text{lbf}\right)\left(\frac{0.00444822\text{kN}}{1\text{lbf}}\right)\\ =96.03\text{kN}\end{array}$

When $L=0.8\text{m}$ ,

  $\begin{array}{c}\frac{kL}{r}=\frac{2\times 0.8\times \left(\frac{39.37\text{in}}{1\text{m}}\right)}{1.059}\\ =59.483\end{array}$

Since $\frac{kL}{r}\le 66$ , so

Substitute $59.483$ for $\frac{kL}{r}$ in equation (3).

  $\begin{array}{c}{\sigma }_a=20.2-\left(0.126\times 59.483\right)\text{ksi}\\ =\text{12}\text{.705}\text{ksi}\end{array}$

Substitute $\text{12}\text{.705}\text{ksi}$ for ${\sigma }_a$ , $2.835\text{in}$ for $d_1$ and $3.150\text{in}$ for $d_2$ in equation (2).

  $\begin{array}{c}{P}_a=\text{12}\text{.705}\times {\text{10}}^3\times \frac{\pi }{4}\left({3.15}^2-{2.835}^2\right)\\ =\left(18812.186\text{lbf}\right)\left(\frac{0.00444822\text{kN}}{1\text{lbf}}\right)\\ =83.68\text{kN}\end{array}$

When $L=1\text{m}$ ,

  $\begin{array}{c}\frac{kL}{r}=\frac{2\times 1\times \left(\frac{39.37\text{in}}{1\text{m}}\right)}{1.059}\\ =74.35\end{array}$

Since $\frac{kL}{r}\ge 66$ , so

Substitute $74.35$ for $\frac{kL}{r}$ in equation (4).

  $\begin{array}{c}{\sigma }_a=\frac{51000}{{\left(74.35\right)}^2}\text{ksi}\\ =\text{9}\text{.22589}\text{ksi}\end{array}$

Substitute $\text{9}\text{.22589}\text{ksi}$ for ${\sigma }_a$ , $2.835\text{in}$ for $d_1,$ and $3.150\text{in}$ for $d_2$ in equation (2).

  $\begin{array}{c}{P}_a=\text{9}\text{.22589}\times {\text{10}}^3\times \frac{\pi }{4}\left({3.15}^2-{2.835}^2\right)\\ =\left(13660.7\text{lbf}\right)\left(\frac{0.00444822\text{kN}}{1\text{lbf}}\right)\\ =60.77\text{kN}\end{array}$

When $L=1.2\text{m}$ ,

  $\begin{array}{c}\frac{kL}{r}=\frac{2\times 1.2\times \left(\frac{39.37\text{in}}{1\text{m}}\right)}{1.059}\\ =89.223\end{array}$

Since $\frac{kL}{r}\ge 66$ , so

Substitute $89.223$ for $\frac{kL}{r}$ in equation (4).

  $\begin{array}{c}{\sigma }_a=\frac{51000}{{\left(89.223\right)}^2}\text{ksi}\\ =\text{6}\text{.406}\text{ksi}\end{array}$

Substitute $\text{6}\text{.406}\text{ksi}$ for ${\sigma }_a$ , $2.835\text{in}$ for $d_1,$ and $3.150\text{in}$ for $d_2$ in equation (2).

  $\begin{array}{c}{P}_a=\text{6}\text{.406}\times {\text{10}}^3\times \frac{\pi }{4}\left({3.15}^2-{2.835}^2\right)\\ =\left(9485.309\text{lbf}\right)\left(\frac{0.00444822\text{kN}}{1\text{lbf}}\right)\\ =42.19\text{kN}\end{array}$

Conclusion:

The allowable axial load at $L=0.6\text{m}$ is $96.03\text{kN}$ .

The allowable axial load at $L=0.8\text{m}$ is $83.68\text{kN}$ .

The allowable axial load at $L=1\text{m}$ is $60.77\text{kN}$ .

The allowable axial load at $L=1.2\text{m}$ is $42.19\text{kN}$ .