Chapter 10.2, Problem 52P

To determine

Select the lightest W200 shape.

Answer to Problem 52P

The lightest W200 shape is $\underset{\_}{\text{W}200\times 35.9}$.

Explanation of Solution

Given information:

The magnitude of axial load is $P=345\text{kN}$.

The eccentricity of the load is $e=6\text{mm}$.

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

The allowable stress in the column is ${\sigma }_{all}=120\text{MPa}$

Calculation:

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

$L_e=2L=3.6\text{m}$

Consider the section $\text{W}200\times 41.7$:

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

For $\text{W}200\times 41.7$ Rolled-steel shape;

The cross sectional area is $A=5,320{\text{mm}}^2$.

The minimum moment of inertia in the y-axis is $I=I_y=9.03\times {10}^6{\text{mm}}^4$.

The minimum radius of gyration in the y-axis is $r=r_y=41.1\text{mm}$.

The width of the flange is $b_f=166\text{mm}$.

Find the distance between the neutral axis and the outermost fibre (c) using the relation.

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

Substitute 166 mm for $b_f$.

$\begin{array}{c}c=\frac{166}{2}\\ =83\text{mm}\end{array}$

Determine the critical load $\left(P_{cr}\right)$ using the equation.

$P_{cr}=\frac{{\pi }^{2}EI}{{\left(L_e\right)}^2}$

Here, the modulus of elasticity is E, the minimum moment of inertia is I, and the effective length is $L_e$.

Substitute 200 GPa for E, $9.03\times {10}^6{\text{mm}}^4$ for I, and 3.6 m for $L_e$.

$\begin{array}{c}{P}_{\text{cr}}=\frac{{\pi }^2\times 200\text{GPa}\times \frac{{\text{10}}^{\text{9}}{\text{N/m}}^{\text{2}}}{\text{1}\text{GPa}}\times 9.03\times {10}^6{\text{mm}}^4\times {\left(\frac{\text{1}\text{m}}{\text{1,000}\text{mm}}\right)}^4}{{3.6}^2}\\ =1,375\times {10}^3\text{N}\times \frac{\text{1}\text{kN}}{\text{1,000}\text{N}}\\ =1,375\text{kN}\end{array}$

Find the maximum stress $\left({\sigma }_m\right)$ using the equation.

${\sigma }_{\max }=\frac{P}{A}\left(1+\frac{ec}{r^2}\sec \left(\frac{\pi }{2}\sqrt{\frac{P}{P_{\text{cr}}}}\right)\right)$

Here, the maximum stress in the column is ${\sigma }_{\max }$, the axial load is P, the cross sectional area of the section is A, the eccentricity of the load is e, and the minimum radius of gyration is r.

Substitute 345 kN for P, 1,375 kN for $P_{\text{cr}}$, $5,320{\text{mm}}^2$ for A, 6 mm for e, 83 mm for c, and 41.1 mm for r.

$\begin{array}{c}{\sigma }_m=\frac{345\text{kN}\times \frac{\text{1,000}\text{N}}{\text{1}\text{kN}}}{5,320{\text{mm}}^2\times {\left(\frac{\text{1}\text{m}}{\text{1,000}\text{mm}}\right)}^2}\left(1+\frac{6\times 83}{{41.1}^2}\sec \left(\frac{\pi }{2}\sqrt{\frac{345}{\text{1,375}}}\right)\right)\\ =91.93\times {10}^6\text{Pa}\times \frac{\text{1}\text{MPa}}{{\text{10}}^{\text{6}}\text{Pa}}\\ =91.93\text{MPa}<{\sigma }_{all}\text{=120}\text{MPa}\end{array}$

Consider the section $\text{W}200\times 26.6$:

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

For $\text{W}200\times 26.6$ Rolled-steel shape;

The cross sectional area is $A=3,390{\text{mm}}^2$.

The minimum moment of inertia in the y-axis is $I=I_y=3.32\times {10}^6{\text{mm}}^4$.

The minimum radius of gyration in the y-axis is $r=r_y=31.2\text{mm}$.

The width of the flange is $b_f=133\text{mm}$.

Find the distance between the neutral axis and the outermost fibre (c) using the relation.

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

Substitute 133 mm for $b_f$.

$\begin{array}{c}c=\frac{133}{2}\\ =66.5\text{mm}\end{array}$

Determine the critical load $\left(P_{cr}\right)$ using the equation.

$P_{cr}=\frac{{\pi }^{2}EI}{{\left(L_e\right)}^2}$

Substitute 200 GPa for E, $3.32\times {10}^6{\text{mm}}^4$ for I, and 3.6 m for $L_e$.

$\begin{array}{c}{P}_{\text{cr}}=\frac{{\pi }^2\times 200\text{GPa}\times \frac{{\text{10}}^{\text{9}}{\text{N/m}}^{\text{2}}}{\text{1}\text{GPa}}\times 3.32\times {10}^6{\text{mm}}^4\times {\left(\frac{\text{1}\text{m}}{\text{1,000}\text{mm}}\right)}^4}{{3.6}^2}\\ =505.66\times {10}^3\text{N}\times \frac{\text{1}\text{kN}}{\text{1,000}\text{N}}\\ =505.66\text{kN}\end{array}$

Find the maximum stress $\left({\sigma }_m\right)$ using the equation.

${\sigma }_{\max }=\frac{P}{A}\left(1+\frac{ec}{r^2}\sec \left(\frac{\pi }{2}\sqrt{\frac{P}{P_{\text{cr}}}}\right)\right)$

Substitute 345 kN for P, 505.66 kN for $P_{\text{cr}}$, $3,390{\text{mm}}^2$ for A, 6 mm for e, 66.5 mm for c, and 31.2 mm for r.

$\begin{array}{c}{\sigma }_m=\frac{345\text{kN}\times \frac{\text{1,000}\text{N}}{\text{1}\text{kN}}}{3,390{\text{mm}}^2\times {\left(\frac{\text{1}\text{m}}{\text{1,000}\text{mm}}\right)}^2}\left(1+\frac{6\times 66.5}{{31.2}^2}\sec \left(\frac{\pi }{2}\sqrt{\frac{345}{\text{505}\text{.66}}}\right)\right)\\ =256.31\times {10}^6\text{Pa}\times \frac{\text{1}\text{MPa}}{{\text{10}}^{\text{6}}\text{Pa}}\\ =256.31\text{MPa}<{\sigma }_{all}\text{=120}\text{MPa}\end{array}$

Consider the section $\text{W}200\times 35.9$:

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

For $\text{W}200\times 35.9$ Rolled-steel shape;

The cross sectional area is $A=4,570{\text{mm}}^2$.

The minimum moment of inertia in the y-axis is $I=I_y=7.62\times {10}^6{\text{mm}}^4$.

The minimum radius of gyration in the y-axis is $r=r_y=40.9\text{mm}$.

The width of the flange is $b_f=165\text{mm}$.

Find the distance between the neutral axis and the outermost fibre (c) using the relation.

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

Substitute 165 mm for $b_f$.

$\begin{array}{c}c=\frac{165}{2}\\ =82.5\text{mm}\end{array}$

Determine the critical load $\left(P_{cr}\right)$ using the equation.

$P_{cr}=\frac{{\pi }^{2}EI}{{\left(L_e\right)}^2}$

Substitute 200 GPa for E, $7.62\times {10}^6{\text{mm}}^4$ for I, and 3.6 m for $L_e$.

$\begin{array}{c}{P}_{\text{cr}}=\frac{{\pi }^2\times 200\text{GPa}\times \frac{{\text{10}}^{\text{9}}{\text{N/m}}^{\text{2}}}{\text{1}\text{GPa}}\times 7.62\times {10}^6{\text{mm}}^4\times {\left(\frac{\text{1}\text{m}}{\text{1,000}\text{mm}}\right)}^4}{{3.6}^2}\\ =1,161\times {10}^3\text{N}\times \frac{\text{1}\text{kN}}{\text{1,000}\text{N}}\\ =1,161\text{kN}\end{array}$

Find the maximum stress $\left({\sigma }_m\right)$ using the equation.

${\sigma }_{\max }=\frac{P}{A}\left(1+\frac{ec}{r^2}\sec \left(\frac{\pi }{2}\sqrt{\frac{P}{P_{\text{cr}}}}\right)\right)$

Substitute 345 kN for P, 1,161 kN for $P_{\text{cr}}$, $4,570{\text{mm}}^2$ for A, 6 mm for e, 82.5 mm for c, and 40.9 mm for r.

$\begin{array}{c}{\sigma }_m=\frac{345\text{kN}\times \frac{\text{1,000}\text{N}}{\text{1}\text{kN}}}{4,570{\text{mm}}^2\times {\left(\frac{\text{1}\text{m}}{\text{1,000}\text{mm}}\right)}^2}\left(1+\frac{6\times 82.5}{{40.9}^2}\sec \left(\frac{\pi }{2}\sqrt{\frac{345}{\text{1,161}}}\right)\right)\\ =109.58\times {10}^6\text{Pa}\times \frac{\text{1}\text{MPa}}{{\text{10}}^{\text{6}}\text{Pa}}\\ =109.58\text{MPa}<{\sigma }_{all}\text{=120}\text{MPa}\end{array}$

The calculated maximum stress is closer to the allowable stress.

The shape $\text{W}200\times 35.9$ satisfies the condition.

Therefore, the lightest W200 shape is $\underset{\_}{\text{W}200\times 35.9}$.