Chapter 11, Problem 11.9.7P

To determine

The value of the allowable axial load pressure for the different lengths.

Answer to Problem 11.9.7P

The value of the allowable axial load pressure for the different lengths is:

S.No	Length of column( ft)	Axial load ,Pallow( k )
1	06	58.8
2	12	43.0
3	18	23.1
4	24	13.0

Explanation of Solution

Given Information:

The outside diameter of the steel pipe column, d_o =4.5 in.

The thickness of the steel pipe column, t = .237 in.

The value of the young modulus, E = 29000 ksi

The value of maximum stress, s_y = 36 ksi

Concept Used:

For pinned end, K=1

Here, r = radius of gyration.

  $\begin{array}{l}\frac{L_c}{r}=\sqrt{\frac{2{\pi }^{2}E}{{\sigma }_y}}\\ L_c=\text{critical}\text{length}\text{ of}\text{column}\text{ in}\text{ ft}\end{array}$

  ${\sigma }_{allow\_\max }=\frac{{\sigma }_y}{n_1}$

Then we try different value of ${\sigma }_{allow}$

  $\begin{array}{l}\\ A=\frac{P}{{\sigma }_{allow}}\end{array}$

Then select the trial column

  $\begin{array}{l}{n}_1=NA\text{ if}\left(\frac{KL}{r}\right)>\frac{K{L}_c}{r}\&\\ n_1=\frac{5}{3}+\frac{3\left(\frac{KL}{r}\right)}{8\left(\frac{K{L}_c}{r}\right)}-\frac{{\left(\left(\frac{KL}{r}\right)\right)}^3}{8{\left(\frac{K{L}_c}{r}\right)}^3}if\left(\frac{KL}{r}\right)\le \frac{K{L}_c}{r}\\ n_2=\frac{23}{12}\text{ if}\left(\frac{KL}{r}\right)>\frac{K{L}_c}{r}\&\\ n_2=NA\text{ if}\left(\frac{KL}{r}\right)\le \frac{K{L}_c}{r}\end{array}$

  $\begin{array}{l}{\sigma }_{allow}={\sigma }_y\left[\frac{1}{n_1}\left(1-\frac{{\left(\frac{KL}{r}\right)}^2}{2{\left(\frac{K{L}_c}{r}\right)}^2}\right)\right]if\left(\frac{KL}{r}\right)\le \frac{K{L}_c}{r}\\ {\sigma }_{allow}={\sigma }_y\left[\frac{{\left(\frac{K{L}_c}{r}\right)}^2}{2n_{2_i}{\left(\frac{KL}{r}\right)}^2}\right]if\left(\frac{KL}{r}\right)>\frac{K{L}_c}{r}\\ P_{allow}={\sigma }_{allow}\times A\end{array}$

Calculation:

For pinned end, K = 1.

  $\begin{array}{l}{d}_i=4.5-2(0.237)=4.026in.\\ A=\frac{\pi }{4}\left({4.5}^2-{4.026}^2\right)=3.1740in{.}^2\\ I=\frac{\pi }{64}\left({4.5}^4-{4.026}^4\right)=7.2326in{.}^4\\ r=\sqrt{\frac{7.2326}{3.1740}}=1.5095in.\\ {\left(\frac{L}{r}\right)}_{\max }=200\\ L_{\max }=200\times 1.5095=301.9in.=25.16ft\end{array}$

Here, r = radius of gyration.

  $\begin{array}{l}{L}_c=1.5095\sqrt{\frac{2{\pi }^{2}29000}{36}}=1.5095\times 126.09=190.33in.=15.86ft\\ \text{For}i=1,L_1=6ft\\ \text{as}{L}_1<L_c\end{array}$

  $\begin{array}{l}\text{So,}\\ n_{1_1}=\frac{5}{3}+\frac{3\left(\frac{6}{1.5095}\right)}{8\left(\frac{15.86}{1.5095}\right)}-\frac{{\left(\frac{6}{1.5095}\right)}^3}{8{\left(\frac{15.86}{1.5095}\right)}^3}=1.667+.1419-0.00234=1.806\\ \text{For}i=2,L_2=12ft\end{array}$

  $\begin{array}{l}As{L}_2<L_c\\ \text{So,}\\ n_{1_2}=\frac{5}{3}+\frac{3\left(\frac{12}{1.5095}\right)}{8\left(\frac{15.86}{1.5095}\right)}-\frac{{\left(\frac{12}{1.5095}\right)}^3}{8{\left(\frac{15.86}{1.5095}\right)}^3}=1.667+.2837-0.0541=1.896\\ \text{For}i=3,L_3=18ft\end{array}$

  $\begin{array}{l}\text{ As }{L}_3>L_c\\ n_{2_3}=\frac{23}{12}=1.917\\ \text{For}i=4,L_4=24ft\\ As\\ L_4>L_c\\ n_{2_4}=\frac{23}{12}=1.917\end{array}$

  $\begin{array}{l}{\sigma }_{allow}={\sigma }_y\left[\frac{1}{n_{1_i}}\left(1-\frac{{\left(\frac{K{L}_i}{r}\right)}^2}{2{\left(\frac{K{L}_c}{r}\right)}^2}\right)\right]if\frac{K{L}_i}{r}\le \frac{K{L}_c}{r}\text{this}\text{is}\text{used}\text{for}{L}_1\&L_2\\ {\sigma }_{allow}={\sigma }_y\left[\frac{{\left(\frac{K{L}_c}{r}\right)}^2}{2n_{2_i}{\left(\frac{K{L}_i}{r}\right)}^2}\right]if\frac{K{L}_i}{r}>\frac{K{L}_c}{r}\text{this}\text{is}\text{used}\text{for}{L}_3\&L_4\end{array}$

For L₁ = 6 ft.

  $\begin{array}{l}{\sigma }_{allow}=36\left[\frac{1}{1.805}\left(1-\frac{{\left(\frac{6}{1.5095}\right)}^2}{2{\left(\frac{15.86}{1.5095}\right)}^2}\right)\right]=18.517ksi\\ P_{allow}=18.517\times 3.1740=58.77k=58.8k\end{array}$

For L₂ =12 ft

  $\begin{array}{l}{\sigma }_{allow}=36\left[\frac{1}{1.896}\left(1-\frac{{\left(\frac{12}{1.5095}\right)}^2}{2{\left(\frac{15.86}{1.5095}\right)}^2}\right)\right]=13.55ksi\\ P_{allow}=13.55\times 3.1740=43.0k=43k\end{array}$

For L₃ =18 ft.

  $\begin{array}{l}{\sigma }_{allow}=36\left[\frac{{\left(\frac{15.86}{1.5095}\right)}^2}{2\times 1.917{\left(\frac{18}{1.5095}\right)}^2}\right]=7.289ksi\\ P_{allow}=7.289\times 3.1740=23.14k=23.1k\end{array}$

For L₄ = 24 ft,

  $\begin{array}{l}{\sigma }_{allow}=36\left[\frac{{\left(\frac{15.86}{1.5095}\right)}^2}{2\times 1.917{\left(\frac{24}{1.5095}\right)}^2}\right]=4.10ksi\\ P_{allow}=4.10\times 3.1740=13.01k=13.0k\end{array}$

Conclusion:

The value of the allowable axial load pressure for the different lengths is:

S.No	Length of column( ft)	Axial load ,Pallow( k )
1	06	328
2	12	243
3	18	134
4	24	75.3