Chapter 11, Problem 11.9.18P

To determine

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

Answer to Problem 11.9.18P

The value of the outside diameter d for a steel pipe is 99 mm.

Explanation of Solution

Given Information:

The length of the column, L = 3.5 m

The allowable load, P = 130 kN

The value of the young modulus, E = 200 GPa

The value of maximum stress, s_y = 275 MPa

The thickness of the steel pipe, t = d/20

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 ft.

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:

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

Here,

  $\begin{array}{l}t=d/20\\ A=.14923d^2\\ I=\frac{\pi }{64}\left(d^4-({(d-2(d/20))}^4\right)=\frac{\pi }{64}(d^4-{(}^{.9})=0.016881d^4\\ r=\sqrt{\frac{0.016881d^4}{0.14932d^2}}=.33623d\\ {\left(\frac{KL}{r}\right)}_{\max }=200,hereK=1\\ L_{\max }=200\times .33623d=67.246d\\ K{L}_c/r=\sqrt{\frac{2{\pi }^{2}200}{275}}=119.81\end{array}$

  $L_c=(119.81)rin.$

L_c= critical length of column in mm.

Select various values of d until P_allow= P.

  $i=1,\mathrm{2...4}$

  $d979899$

  $A(m{m}^2)140414331463$

  $I(m{m}^4)1494\times {10}^{3}1557\times {10}^{3}1463\times {10}^3$

  $r(mm)32.6232.9633.30$

  $L_c(mm)390839503989$

  $L/r107.3106.2105.1$

  $n_{1}1.9131.9121.911$

For d₁=97mm,

  $\begin{array}{l}{\sigma }_{allow}=275\left[\frac{1}{1.913}\left(1-\frac{{\left(107.3\right)}^2}{2\times {\left(119.8\right)}^2}\right)\right]=86.09MPa\\ P_{allow}=86.09\times 1404=120.88kN\end{array}$

For d₂ = 98mm,

  $\begin{array}{l}{\sigma }_{allow}=275\left[\frac{1}{1.912}\left(1-\frac{{\left(106.2\right)}^2}{2\times {\left(119.8\right)}^2}\right)\right]=87.32MPa\\ P_{allow}=87.32\times 1433=125.12kN\end{array}$

For d₃= 99mm.

  $\begin{array}{l}{\sigma }_{allow}=275\left[\frac{1}{1.911}\left(1-\frac{{\left(105.1\right)}^2}{2\times {\left(119.8\right)}^2}\right)\right]=88.52MPa\\ P_{allow}=88.52\times 1463=129.51kN=130kN\\ ForP=130kN,d=99mm.\end{array}$

Conclusion:

The value of the outside diameter d for a steel pipe is 99 mm .