Chapter 10.3, Problem 84P

Two 89 × 64-mm angles are bolted together as shown for use as a column of 2.4-m effective length to carry a centric load of 325 kN. Knowing that the angles available have thicknesses of 6.4 mm, 9.5 mm, and 12.7 mm, use allowable stress design to determine the lightest angles that can be used. Use σ_Y = 250 MPa and E = 200 GPa.

Fig. P10.84

To determine

Find the lightest angles that can be used.

Answer to Problem 84P

The lightest angle that can be used for the design is $\underset{\_}{\text{L}89\times 64\times 12.7\text{mm}}$.

Explanation of Solution

Given information:

The effective length of the column is $L_e=2.4\text{m}$.

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 $P=180\text{kN}$.

Calculation:

Consider the thickness of the angle section as 9.5 mm.

Refer to Appendix C “Properties of Rolled-Steel Shapes” in the textbook.

For $\text{L}89\times 64\times 9.5\text{mm}$ angle section;

The cross sectional area of the angle (A) is $A=1,360{\text{mm}}^2$.

The moment of inertia in x-axis is $I=I_x=1.07\times {10}^6{\text{mm}}^4$.

The moment of inertia in y-axis is $I=I_y=0.454\times {10}^6{\text{mm}}^4$.

The centroid distance from the flange in x-axis is $x=16.6\text{mm}$.

The area of the two angle section is $A=2\left(1,360\right){\text{mm}}^2$.

The moment of inertia in x-axis is $I_x=2\left(1.07\times {10}^6\right)=2.14\times {10}^6{\text{mm}}^4$.

Find the moment of inertia in y-axis using the relation.

$I_y=2\left[I_y+A{x}^2\right]$

Substitute $0.454\times {10}^6{\text{mm}}^4$ for $I_y$, $1,360{\text{mm}}^2$ for A, and 16.6 mm for x.

$\begin{array}{c}{I}_y=2\left[0.454\times {10}^6+\left(1,360\right){\left(16.6\right)}^2\right]\\ =1,657,523.2{\text{mm}}^4\end{array}$

The minimum moment of inertia is $I=I_y=1,657,523.2{\text{mm}}^4$.

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

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

Substitute $1,657,523.2{\text{mm}}^4$ for I and $2\left(1,360\right){\text{mm}}^2$ for A.

$\begin{array}{c}r=\sqrt{\frac{1,657,523.2}{2\left(1,360\right)}}\\ =24.686\text{mm}\end{array}$

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

$\begin{array}{c}\frac{L_e}{r}=\frac{2.4\text{m}\times \frac{\text{1,000}\text{mm}}{\text{1}\text{m}}}{24.686}\\ =97.22<\frac{L}{r}=133.22\end{array}$

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 97.22 for $L_e/r_y$.

$\begin{array}{c}{\sigma }_e=\frac{{\pi }^2\times 200\text{GPa}\times \frac{\text{1,000}\text{MPa}}{\text{1}\text{GPa}}}{{\left(97.22\right)}^2}\\ =208.832\text{MPa}\end{array}$

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 208.832 MPa for ${\sigma }_e$.

$\begin{array}{c}{\sigma }_{\text{cr}}=\left[{0.658}^{\frac{250}{208.832}}\right]250\\ =151.472\text{MPa}\end{array}$

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

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

Substitute 151.472 MPa for ${\sigma }_{\text{cr}}$.

$\begin{array}{c}{\sigma }_{all}=\frac{151.472}{1.67}\\ =90.702\text{MPa}\end{array}$

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

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

Substitute 90.702 MPa for ${\sigma }_{\text{all}}$ and $2\left(1,360\right){\text{mm}}^2$ for A.

$\begin{array}{c}{P}_{\text{all}}=90.702\times 2\left(1,360\right)\\ =246.71\times {10}^3\text{N}\times \frac{\text{1}\text{kN}}{\text{1,000}\text{N}}\\ =246.71\text{kN}\end{array}$

The centric load is greater than the allowable load. Hence, the design is unsafe.

$P_{\text{all}}=246\text{kN}<P=\text{325}\text{kN}$

Consider the thickness of the angle section as 12.7 mm.

Refer to Appendix C “Properties of Rolled-Steel Shapes” in the textbook.

For $\text{L}89\times 64\times 12.7\text{mm}$ angle section;

The cross sectional area of the angle (A) is $A=1,770{\text{mm}}^2$.

The moment of inertia in x-axis is $I=I_x=1.35\times {10}^6{\text{mm}}^4$.

The moment of inertia in y-axis is $I=I_y=0.566\times {10}^6{\text{mm}}^4$.

The centroid distance from the flange in x-axis is $x=17.8\text{mm}$.

The area of the two angle section is $A=2\left(1,770\right){\text{mm}}^2$.

The moment of inertia in x-axis is $I_x=2\left(1.35\times {10}^6\right)=2.70\times {10}^6{\text{mm}}^4$.

Find the moment of inertia in y-axis using the relation.

$I_y=2\left[I_y+A{x}^2\right]$

Substitute $0.566\times {10}^6{\text{mm}}^4$ for $I_y$, $1,770{\text{mm}}^2$ for A, and 17.8 mm for x.

$\begin{array}{c}{I}_y=2\left[0.566\times {10}^6+\left(1,770\right){\left(17.8\right)}^2\right]\\ =2,253,613.6{\text{mm}}^4\end{array}$

The minimum moment of inertia is $I=I_y=2,253,613.6{\text{mm}}^4$.

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

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

Substitute $2,253,613.6{\text{mm}}^4$ for I and $2\left(1,770\right){\text{mm}}^2$ for A.

$\begin{array}{c}r=\sqrt{\frac{2,253,613.6}{2\left(1,770\right)}}\\ =25.231\text{mm}\end{array}$

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

$\begin{array}{c}\frac{L_e}{r}=\frac{2.4\text{m}\times \frac{\text{1,000}\text{mm}}{\text{1}\text{m}}}{25.231}\\ =95.12<\frac{L}{r}=133.22\end{array}$

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 95.12 for $L_e/r_y$.

$\begin{array}{c}{\sigma }_e=\frac{{\pi }^2\times 200\text{GPa}\times \frac{\text{1,000}\text{MPa}}{\text{1}\text{GPa}}}{{\left(95.12\right)}^2}\\ =218.164\text{MPa}\end{array}$

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 218.164 MPa for ${\sigma }_e$.

$\begin{array}{c}{\sigma }_{\text{cr}}=\left[{0.658}^{\frac{250}{218.164}}\right]250\\ =154.753\text{MPa}\end{array}$

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

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

Substitute 154.753 MPa for ${\sigma }_{\text{cr}}$.

$\begin{array}{c}{\sigma }_{all}=\frac{154.753}{1.67}\\ =92.667\text{MPa}\end{array}$

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

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

Substitute 92.667 MPa for ${\sigma }_{\text{all}}$ and $2\left(1,770\right){\text{mm}}^2$ for A.

$\begin{array}{c}{P}_{\text{all}}=92.667\times 2\left(1,770\right)\\ =328.04\times {10}^3\text{N}\times \frac{\text{1}\text{kN}}{\text{1,000}\text{N}}\\ =328.04\text{kN}\end{array}$

The centric load is less than the allowable load. Hence, the design is unsafe.

$P_{\text{all}}=328.04\text{kN}>P=325\text{kN}$

Therefore, the lightest angle that can be used for the design is $\underset{\_}{\text{L}89\times 64\times 12.7\text{mm}}$.