Chapter 10.3, Problem 80P

A centric load P must be supported by the steel bar AB. Using allowable stress design, determine the smallest dimension d of the cross section that can be used when (a) P = 108 kN, (b) P = 166 kN. Use σ_Y = 250 MPa and E = 200 GPa.

Fig. P10.80

To determine

Find the smallest dimension d of the cross section.

Answer to Problem 80P

The smallest dimension d of the cross section is $\underset{\_}{30.1\text{mm}}$.

Explanation of Solution

Given information:

The length of the column is $L=1.4\text{m}$.

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

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

The centric load acting in the column is $P=108\text{kN}$.

Calculation:

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

$L_e=L=1.4\text{m}$

Find the cross sectional area (A) using the equation.

$A=bd$

Here, the width of the column is b and the depth of the column is d.

Substitute 3d for b.

$\begin{array}{c}A=\left(3d\right)d\\ =3d^2\end{array}$

Find the moment of inertia (I) of the cross section using the equation.

$I=\frac{b{d}^3}{12}$

Substitute 3d for b.

$\begin{array}{c}I=\frac{\left(3d\right)d^3}{12}\\ =\frac{d^4}{4}\end{array}$

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

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

Substitute $\frac{d^4}{4}$ for I and $3d^2$ for A.

$\begin{array}{c}r=\sqrt{\frac{\left(\frac{d^4}{4}\right)}{3d^2}}\\ =\frac{d}{\sqrt{12}}\end{array}$

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 effective length $\left(L_e\right)$ and the minimum radius of gyration (r) as follows;

$\begin{array}{c}\frac{L_e}{r}=\frac{1.4\text{m}\times \frac{\text{1,000}\text{mm}}{\text{1}\text{m}}}{\frac{d}{\sqrt{12}}}\\ =\frac{1,400\sqrt{12}}{d}\end{array}$ (1)

Consider $\frac{1,400\sqrt{12}}{d}\ge \frac{L}{r}=133.22$

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

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

Substitute 200 GPa for E and $\frac{1,400\sqrt{12}}{d}$ for $L_e/r_y$.

$\begin{array}{c}{\sigma }_e=\frac{{\pi }^2\times 200\text{GPa}\times \frac{\text{1,000}\text{MPa}}{\text{1}\text{GPa}}}{{\left(\frac{1,400\sqrt{12}}{d}\right)}^2}\\ =\frac{1,973,920.88d^2}{23,520,000}\end{array}$

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

${\sigma }_{\text{cr}}=0.877{\sigma }_e$

Substitute $\frac{1,973,920.88d^2}{23,520,000}$ for ${\sigma }_e$.

$\begin{array}{c}{\sigma }_{\text{cr}}=0.877\times \frac{1,973,920.88d^2}{23,520,000}\\ =\frac{1,731,128.612d^2}{23,520,000}\end{array}$

Calculate the allowable stress $\left({\sigma }_{all}\right)$ using the relation.

${\sigma }_{all}=\frac{{\sigma }_{\text{cr}}}{1.67}$

Substitute $\frac{1,731,128.612d^2}{23,520,000}$ for ${\sigma }_{\text{cr}}$.

$\begin{array}{c}{\sigma }_{all}=\frac{\left(\frac{1,731,128.612d^2}{23,520,000}\right)}{1.67}\\ =\frac{1,731,128.612d^2}{39,278,400}\end{array}$

Calculate the allowable load $\left(P_{\text{all}}\right)$ using the equation.

$P_{\text{all}}={\sigma }_{\text{all}}A$

Substitute $\frac{1,731,128.612d^2}{39,278,400}$ for ${\sigma }_{\text{all}}$ and $3d^2$ for A.

$\begin{array}{c}{P}_{\text{all}}=\frac{1,731,128.612d^2}{39,278,400}\times 3d^2\\ =\frac{5,193,385.836d^4}{39,278,400}\end{array}$

Consider the allowable load is equal to the centric load.

Substitute 108 kN for $P_{\text{all}}$.

$\begin{array}{l}108\text{kN}\times \frac{\text{1,000}\text{N}}{\text{1}\text{kN}}=\frac{5,193,385.836d^4}{39,278,400}\\ d=30.1\text{mm}\end{array}$

Check:

Substitute 30.1 mm for d in Equation (1).

$\begin{array}{c}\frac{L_e}{r}=\frac{1,400\sqrt{12}}{30.1}\\ =161.12>\frac{L}{r}=133.22\end{array}$

The assumed condition is correct.

Therefore, the smallest dimension d of the cross section is $\underset{\_}{30.1\text{mm}}$.

To determine

Find the smallest dimension d of the cross section.

Answer to Problem 80P

The smallest dimension d of the cross section is $\underset{\_}{33.5\text{mm}}$.

Explanation of Solution

Given information:

The length of the column is $L=1.4\text{m}$.

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

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

The centric load acting in the column is $P=166\text{kN}$.

Calculation:

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

$L_e=L=1.4\text{m}$

Find the cross sectional area (A) using the equation.

$A=bd$

Substitute 3d for b.

$\begin{array}{c}A=\left(3d\right)d\\ =3d^2\end{array}$

Find the moment of inertia (I) of the cross section using the equation.

$I=\frac{b{d}^3}{12}$

Substitute 3d for b.

$\begin{array}{c}I=\frac{\left(3d\right)d^3}{12}\\ =\frac{d^4}{4}\end{array}$

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

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

Substitute $\frac{d^4}{4}$ for I and $3d^2$ for A.

$\begin{array}{c}r=\sqrt{\frac{\left(\frac{d^4}{4}\right)}{3d^2}}\\ =\frac{d}{\sqrt{12}}\end{array}$

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 effective length $\left(L_e\right)$ and the minimum radius of gyration (r) as follows;

$\begin{array}{c}\frac{L_e}{r}=\frac{1.4\text{m}\times \frac{\text{1,000}\text{mm}}{\text{1}\text{m}}}{\frac{d}{\sqrt{12}}}\\ =\frac{1,400\sqrt{12}}{d}\end{array}$ (2)

Consider $\frac{1,400\sqrt{12}}{d}\ge \frac{L}{r}=133.22$

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

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

Substitute 200 GPa for E and $\frac{1,400\sqrt{12}}{d}$ for $L_e/r_y$.

$\begin{array}{c}{\sigma }_e=\frac{{\pi }^2\times 200\text{GPa}\times \frac{\text{1,000}\text{MPa}}{\text{1}\text{GPa}}}{{\left(\frac{1,400\sqrt{12}}{d}\right)}^2}\\ =\frac{1,973,920.88d^2}{23,520,000}\end{array}$

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

${\sigma }_{\text{cr}}=0.877{\sigma }_e$

Substitute $\frac{1,973,920.88d^2}{23,520,000}$ for ${\sigma }_e$.

$\begin{array}{c}{\sigma }_{\text{cr}}=0.877\times \frac{1,973,920.88d^2}{23,520,000}\\ =\frac{1,731,128.612d^2}{23,520,000}\end{array}$

Calculate the allowable stress $\left({\sigma }_{all}\right)$ using the relation.

${\sigma }_{all}=\frac{{\sigma }_{\text{cr}}}{1.67}$

Substitute $\frac{1,731,128.612d^2}{23,520,000}$ for ${\sigma }_{\text{cr}}$.

$\begin{array}{c}{\sigma }_{all}=\frac{\left(\frac{1,731,128.612d^2}{23,520,000}\right)}{1.67}\\ =\frac{1,731,128.612d^2}{39,278,400}\end{array}$

Calculate the allowable load $\left(P_{\text{all}}\right)$ using the equation.

$P_{\text{all}}={\sigma }_{\text{all}}A$

Substitute $\frac{1,731,128.612d^2}{39,278,400}$ for ${\sigma }_{\text{all}}$ and $3d^2$ for A.

$\begin{array}{c}{P}_{\text{all}}=\frac{1,731,128.612d^2}{39,278,400}\times 3d^2\\ =\frac{5,193,385.836d^4}{39,278,400}\end{array}$

Consider the allowable load is equal to the centric load.

Substitute 166 kN for $P_{\text{all}}$.

$\begin{array}{l}166\text{kN}\times \frac{\text{1,000}\text{N}}{\text{1}\text{kN}}=\frac{5,193,385.836d^4}{39,278,400}\\ d=33.5\text{mm}\end{array}$

Check:

Substitute 33.5 mm for d in Equation (2).

$\begin{array}{c}\frac{L_e}{r}=\frac{1,400\sqrt{12}}{33.5}\\ =144.77>\frac{L}{r}=133.22\end{array}$

Therefore, the smallest dimension d of the cross section is $\underset{\_}{33.5\text{mm}}$.