Chapter 11, Problem 11.9.20P

To determine

The value of the outside diameter d for a steel pipe.

Answer to Problem 11.9.20P

The value of the outside diameter d for a steel pipe is 194mm.

Explanation of Solution

Given Information:

The length of the column, L = 3. 0m

The allowable load, P = 800kN

The value of the young modulus, E = 200GPa

The value of maximum stress, s_y = 300MPa

The thickness of the steel pipe, t = 9 mm

here, d is the outside diameter of the steel pipe.

Concept Used:

  $\begin{array}{l}{d}_i=d-2(t)\\ A=\frac{\pi }{4}\left(d_o{}^2-d_i{}^2\right)\\ I=\frac{\pi }{64}\left(d_o{}^4-d_i{}^4\right)\\ r=\sqrt{\frac{I}{A}}\end{array}$

L_c = critical length of column in mm.

Then we choose different diameters d and calculate till the value of P_allow =P

  $\begin{array}{l}i=1,\mathrm{2...4}\\ n_{1_i}=NAi\text{ if }\frac{K{L}_i}{r}>\frac{K{L}_c}{r}\&\\ n_{1_i}=\frac{5}{3}+\frac{3\left(\frac{K{L}_i}{r}\right)}{8\left(\frac{K{L}_c}{r}\right)}-\frac{{\left(\frac{K{L}_i}{r}\right)}^3}{8{\left(\frac{K{L}_c}{r}\right)}^3}if\frac{K{L}_i}{r}\le \frac{K{L}_c}{r}\\ n_{2_i}=\frac{23}{12}\text{ if }\frac{K{L}_i}{r}>\frac{K{L}_c}{r}\&\\ n_{2_i}=NA\text{ if }\frac{K{L}_i}{r}\le \frac{K{L}_c}{r}\end{array}$

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

Calculation:

  $\begin{array}{l}{d}_i=d-2(t)\\ A=\frac{\pi }{4}\left(d^2-{(d-18)}^2\right)\end{array}$

Here, t= 9mm.

  $\begin{array}{l}I=\frac{\pi }{64}\left(d^4-({(d-(18))}^4\right)\\ r=\sqrt{\frac{\frac{\pi }{64}\left(d^4-({(d-(18))}^4\right)}{\frac{\pi }{4}\left(d^2-{(d-18)}^2\right)}}\\ K{L}_c/r=\sqrt{\frac{2{\pi }^{2}200}{275}}=119.81\\ L_c=(114.7)r\end{array}$

L_c = critical length of column in mm.

Select various values of d until

P_allow = P

  $i=1,\mathrm{2...4}$

  $d193194195$

  $A(m{m}^2)520252315259$

  $I(m{m}^4)20.08\times {10}^{6}22.43\times {10}^{6}22.80\times {10}^6$

  $r(mm)65.1365.4865.84$

  $L_c(mm)747075107550$

  $L/r46.0645.8245.57$

  $n_{1}1.8091.8091.808$

For d₁ = 193mm,

  $\begin{array}{l}{\sigma }_{allow}=300\left[\frac{1}{1.809}\left(1-\frac{{\left(46.06\right)}^2}{2\times {\left(114.7\right)}^2}\right)\right]=152.46MPa\\ P_{allow}=152.46\times 5202=793.12kN\end{array}$

For d₂ = 194mm,

  $\begin{array}{l}{\sigma }_{allow}=300\left[\frac{1}{1.808}\left(1-\frac{{\left(45.82\right)}^2}{2\times {\left(114.7\right)}^2}\right)\right]=152.70MPa\\ P_{allow}=152.7\times 5231=798.8kN\end{array}$

For d₃ = 195mm,

  $\begin{array}{l}{\sigma }_{allow}=300\left[\frac{1}{1.808}\left(1-\frac{{\left(45.57\right)}^2}{2\times {\left(114.7\right)}^2}\right)\right]=152.83MPa\\ P_{allow}=152.83\times 5259=803.75kN\\ ForP=800kN,d=194mm.\end{array}$

Conclusion:

The value of the outside diameter d for a steel pipe is 194mm.