Chapter 10.4, Problem 106P

To determine

Find the thickness of the lightest tube.

Answer to Problem 106P

The thickness of the lightest steel tube is $\underset{\_}{15\text{mm}}$.

Explanation of Solution

Given information:

The length of the steel tube is $L=2.2\text{m}$.

The outer diameter of the steel tube is $d_o=80\text{mm}$.

The magnitude of the axial load is $P=165\text{kN}$.

The eccentricity of the load in steel tube is $e=15\text{mm}$.

The allowable yield stress of the steel tube is ${\sigma }_Y=250\text{MPa}$.

The modulus of elasticity of the steel tube is $E=200\text{GPa}$.

The allowable stress in bending is ${\left({\sigma }_{\text{all}}\right)}_{\text{bending}}=150\text{MPa}$

Calculation:

The effective length of the column $\left(L_e\right)$ is equal to the length of the column (L).

$L_e=L=2.2\text{m}$

Find the inner diameter of the steel tube $\left(d_i\right)$ using the relation.

$d_i=d_o-2t$

Here, the thickness of the steel tube is t.

Substitute 80 mm for $d_o$.

$d_i=80-2t$

Find the cross sectional area of the steel tube (A) using the equation.

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

Substitute 80 mm for $d_o$ and $\left(80-2t\right)$ for $d_i$.

$A=\frac{\pi \times \left({80}^2-{\left(80-2t\right)}^2\right)}{4}$

Find the moment of inertia of the steel tube (I) using the equation.

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

Substitute 80 mm for $d_o$ and $\left(80-2t\right)$ for $d_i$.

$I=\frac{\pi \left({80}^4-{\left(80-2t\right)}^4\right)}{64}$

Find the minimum radius of gyration (r) using the relation.

$r=\sqrt{\frac{I}{A}}$

Substitute $\frac{\pi \left({80}^4-{\left(80-2t\right)}^4\right)}{64}$ for I and $\frac{\pi \times \left({80}^2-{\left(80-2t\right)}^2\right)}{4}$ for A.

$\begin{array}{c}r=\sqrt{\frac{\left(\frac{\pi \left({80}^4-{\left(80-2t\right)}^4\right)}{64}\right)}{\left(\frac{\pi \times \left({80}^2-{\left(80-2t\right)}^2\right)}{4}\right)}}\\ =\sqrt{\frac{\left({80}^4-{\left(80-2t\right)}^4\right)}{16\left({80}^2-{\left(80-2t\right)}^2\right)}}\end{array}$

Find the distance between the neutral axis to the extreme fibre (c) using the relation.

$c=\frac{d_o}{2}$

Substitute 80 mm for $d_o$.

$\begin{array}{c}c=\frac{80}{2}\\ =40\text{mm}\end{array}$

Find the slenderness ratio $\frac{L}{r}$ using the equation.

$\frac{L}{r}=4.71\sqrt{\frac{E}{{\sigma }_Y}}$

Here, the modulus of elasticity of the material is E and the allowable yield strength is ${\sigma }_Y$

Substitute 200 GPa for E and 250 MPa for ${\sigma }_Y$:

$\begin{array}{c}\frac{L}{r}=4.71\sqrt{\frac{200\text{GPa}\times \frac{\text{1,000}\text{MPa}}{\text{1}\text{GPa}}}{250}}\\ =133.22\end{array}$

Find the ratio of the effective length to the minimum radius of gyration.

$\begin{array}{c}\frac{L_e}{r}=\frac{2.2\text{m}\times \frac{\text{1,000}\text{mm}}{\text{1}\text{m}}}{\sqrt{\frac{\left({80}^4-{\left(80-2t\right)}^4\right)}{16\left({80}^2-{\left(80-2t\right)}^2\right)}}}\\ =\frac{8,800}{\sqrt{\frac{\left({80}^4-{\left(80-2t\right)}^4\right)}{\left({80}^2-{\left(80-2t\right)}^2\right)}}}\end{array}$ (1)

Consider $\frac{8,800}{\sqrt{\frac{\left({80}^4-{\left(80-2t\right)}^4\right)}{\left({80}^2-{\left(80-2t\right)}^2\right)}}}<\frac{L}{r}=133.22$.

Find the effective stress $\left({\sigma }_e\right)$ using the equation.

${\sigma }_e=\frac{{\pi }^{2}E}{{\left(L_e/r\right)}^2}$

Substitute 200 GPa for E and $\frac{8,800}{\sqrt{\frac{\left({80}^4-{\left(80-2t\right)}^4\right)}{\left({80}^2-{\left(80-2t\right)}^2\right)}}}$ for $L_e/r$.

$\begin{array}{c}{\sigma }_e=\frac{{\pi }^2\times 200\text{GPa}\times \frac{\text{1,000 MPa}}{\text{1}\text{GPa}}}{{\left(\frac{8,800}{\sqrt{\frac{\left({80}^4-{\left(80-2t\right)}^4\right)}{\left({80}^2-{\left(80-2t\right)}^2\right)}}}\right)}^2}\\ =\frac{{\pi }^2\times 200\text{GPa}\times \frac{\text{1,000 MPa}}{\text{1}\text{GPa}}\times \frac{\left({80}^4-{\left(80-2t\right)}^4\right)}{\left({80}^2-{\left(80-2t\right)}^2\right)}}{{\left(8,800\right)}^2}\\ =\frac{0.02549\left({80}^4-{\left(80-2t\right)}^4\right)}{\left({80}^2-{\left(80-2t\right)}^2\right)}\end{array}$

Find the critical stress $\left({\sigma }_{\text{cr}}\right)$ using the relation.

${\sigma }_{\text{cr}}=\left[{0.658}^{\frac{{\sigma }_Y}{{\sigma }_e}}\right]{\sigma }_Y$

Substitute 250 MPa for ${\sigma }_Y$ and $\frac{0.02549\left({80}^4-{\left(80-2t\right)}^4\right)}{\left({80}^2-{\left(80-2t\right)}^2\right)}$ for ${\sigma }_e$.

$\begin{array}{c}{\sigma }_{\text{cr}}=\left[{0.658}^{\frac{250}{\left[\frac{0.02549\left({80}^4-{\left(80-2t\right)}^4\right)}{\left({80}^2-{\left(80-2t\right)}^2\right)}\right]}}\right]250\\ =\left[{0.658}^{\frac{9,807.891\left({80}^2-{\left(80-2t\right)}^2\right)}{\left({80}^4-{\left(80-2t\right)}^4\right)}}\right]250\end{array}$

Find the allowable stress ${\left({\sigma }_{\text{all}}\right)}_{bending}$ due to centric load using the relation.

${\left({\sigma }_{\text{all}}\right)}_{\text{centric}}=\frac{{\sigma }_{\text{cr}}}{1.67}$

Substitute $\left[{0.658}^{\frac{9,807.891\left({80}^2-{\left(80-2t\right)}^2\right)}{\left({80}^4-{\left(80-2t\right)}^4\right)}}\right]250$ for ${\sigma }_{\text{cr}}$.

$\begin{array}{c}{\left({\sigma }_{\text{all}}\right)}_{centric}=\frac{\left[{0.658}^{\frac{9,807.891\left({80}^2-{\left(80-2t\right)}^2\right)}{\left({80}^4-{\left(80-2t\right)}^4\right)}}\right]250}{1.67}\\ =149.701\left[{0.658}^{\frac{9,807.891\left({80}^2-{\left(80-2t\right)}^2\right)}{\left({80}^4-{\left(80-2t\right)}^4\right)}}\right]\end{array}$

Find the maximum moment (M) using the relation.

$M=Pe$

Here, the allowable load is P and the eccentricity of the load is e.

Substitute 165 kN for P and 15 mm for e.

$\begin{array}{c}M=165\text{kN}\times \frac{\text{1,000}\text{N}}{\text{1}\text{kN}}\times 15\\ =2,475,000\text{N-mm}\end{array}$

Find the thickness of the lightest tube (t) using the centric and bending equation.

$\frac{P/A}{{\left({\sigma }_{\text{all}}\right)}_{\text{centric}}}+\frac{Mc/I}{{\left({\sigma }_{\text{all}}\right)}_{\text{bending}}}\le 1$

Substitute 165 kN for P, $\frac{\pi \times \left({80}^2-{\left(80-2t\right)}^2\right)}{4}$ for A, $149.701\left[{0.658}^{\frac{9,807.891\left({80}^2-{\left(80-2t\right)}^2\right)}{\left({80}^4-{\left(80-2t\right)}^4\right)}}\right]$ for ${\left({\sigma }_{\text{all}}\right)}_{centric}$, 2,475,000 N-mm for M, 40 mm for c, $\frac{\pi \left({80}^4-{\left(80-2t\right)}^4\right)}{64}$ for I, and 150 MPa for ${\left({\sigma }_{\text{all}}\right)}_{,\text{bending}}$.

$\frac{\left(\frac{165\text{kN}\times \frac{\text{1,000}\text{N}}{\text{1}\text{kN}}}{\frac{\pi \times \left({80}^2-{\left(80-2t\right)}^2\right)}{4}}\right)}{149.701\left[{0.658}^{\frac{9,807.891\left({80}^2-{\left(80-2t\right)}^2\right)}{\left({80}^4-{\left(80-2t\right)}^4\right)}}\right]}+\frac{\left(\frac{2,475,000\times 40}{\frac{\pi \left({80}^4-{\left(80-2t\right)}^4\right)}{64}}\right)}{150}=1$

Solve the equation;

The thickness is $t=13.901\text{mm}$.

The nearest 3 mm increment of the thickness is 15 mm.

Check:

Substitute 15 mm for t in Equation (1).

$\begin{array}{c}\frac{L_e}{r}=\frac{8,800}{\sqrt{\frac{\left({80}^4-{\left(80-2\left(15\right)\right)}^4\right)}{\left({80}^2-{\left(80-2\left(15\right)\right)}^2\right)}}}\\ =\frac{8,800}{\sqrt{\frac{34,710,000}{3,900}}}\\ =93.28<\frac{L}{r}=133.22\end{array}$

Therefore, the thickness of the lightest steel tube is $\underset{\_}{15\text{mm}}$.