Chapter 11, Problem 11.4.5P

A wide-flange steel column (E = 30 × l0⁶ psi) of W12 × 87 shape (see figure) h as a length L = 28 ft. It is supported only at the ends and may buckle in any direction.

Calculate the allowable load P_allowbased upon the critical load with a factor of safety n = 2.5. Consider the following end conditions: (a) pinned-pinned, (b) fixed-free, (c) fixed-pinned, and (d) fixed-fixed.

  

To determine

The allowable load for the pinned-pinned end condition.

Answer to Problem 11.4.5P

The allowable load for the pinned-pinned condition is 252.842 ksi

Explanation of Solution

Given:

E=30 ×10⁶ psi

L= 28 ft

Factor of of safety = 2.5

Column type: W12×87

Concept Used:

  $\begin{array}{l}Factorofsafety=\frac{P_{cr}}{P_{allowable}}\\ Thecriticalloadofthecolumn,P_{cr}=\frac{{\pi }^{2}EI}{L_{effective}{}^2}\\ Where,\\ I=momentofinertia\\ E=\mathrm{mod}ulusofelasticity\end{array}$

Calculation:

  $\begin{array}{l}Momentof\mathrm{int}eriaofthecolumn:\\ I_1=740i{n}^4\\ I_2=241i{n}^4\\ \mathrm{Sin}ce{I}_{2}isthe\min imumvalueweconsider{I}_2.\\ Factorofsafety=\frac{P_{cr}}{P_{allowable}}\\ P_{allowable}=\frac{P_{cr}}{Factorofsafety}\\ P_{allowable}=\frac{{\pi }^{2}EI}{n\times L_{effective}{}^2}\\ Forpinned-pinnedendcodition\text{}{L}_{effective}=L\\ P_{allowable}=\frac{{\pi }^2\times 30\times {10}^6\times 241}{2.5\times {\left(28\times 12\right)}^2}\\ P_{allowable}=252.824ksi\end{array}$

Conclusion:

The allowable load for the pinned-pinned condition is 252.842 ksi

To determine

The allowable load for the fixed-free end condition.

Answer to Problem 11.4.5P

The allowable load for the fixed-free end condition is 63.206 ksi

Explanation of Solution

Given:

E=30 ×10⁶ psi

L= 28 ft

Factor of of safety = 2.5

Column type: W12×87

Concept Used:

  $\begin{array}{l}Factorofsafety=\frac{P_{cr}}{P_{allowable}}\\ Thecriticalloadofthecolumn,P_{cr}=\frac{{\pi }^{2}EI}{L_{effective}{}^2}\\ Where,\\ I=momentofinertia\\ E=\mathrm{mod}ulusofelasticity\end{array}$

Calculation:

  $\begin{array}{l}Momentof\mathrm{int}eriaofthecolumn:\\ I_1=740i{n}^4\\ I_2=241i{n}^4\\ \mathrm{Sin}ce{I}_{2}isthe\min imumvalueweconsider{I}_2.\\ Factorofsafety=\frac{P_{cr}}{P_{allowable}}\\ P_{allowable}=\frac{P_{cr}}{Factorofsafety}\\ Thecriticalloadofthecolumn,P_{cr}=\frac{{\pi }^{2}EI}{L_{effective}{}^2}\\ Forfixed-freeendcodition\text{}{L}_{effective}=2L\\ P_{allowable}=\frac{{\pi }^2\times 30\times {10}^6\times 241}{2.5\times {\left(2\times 28\times 12\right)}^2}\\ P_{allowable}=63.206ksi\end{array}$

Conclusion:

The allowable load for the fixed-free end condition is 63.206 ksi

To determine

The allowable load for the fixed-pinned end condition.

Answer to Problem 11.4.5P

The allowable load for the fixed-pinned end condition is 517.28 ksi

Explanation of Solution

Given:

E=30 ×10⁶ psi

L= 28 ft

Factor of of safety = 2.5

Column type: W12×87

Concept Used:

  $\begin{array}{l}Factorofsafety=\frac{P_{cr}}{P_{allowable}}\\ Thecriticalloadofthecolumn,P_{cr}=\frac{{\pi }^{2}EI}{L_{effective}{}^2}\\ Where,\\ I=momentofinertia\\ E=\mathrm{mod}ulusofelasticity\end{array}$

Calculation:

  $\begin{array}{l}Momentof\mathrm{int}eriaofthecolumn:\\ I_1=740i{n}^4\\ I_2=241i{n}^4\\ \mathrm{Sin}ce{I}_{2}isthe\min imumvalueweconsider{I}_2.\\ Factorofsafety=\frac{P_{cr}}{P_{allowable}}\\ P_{allowable}=\frac{P_{cr}}{Factorofsafety}\\ Thecriticalloadofthecolumn,P_{cr}=\frac{{\pi }^{2}EI}{L_{effective}{}^2}\\ Forfixed-pinnedendcodition\text{}{L}_{effective}=0.699L\\ P_{allowable}=\frac{{\pi }^2\times 30\times {10}^6\times 241}{2.5\times {\left(0.699\times 28\times 12\right)}^2}\\ P_{allowable}=517.28ksi\end{array}$

Conclusion:

The allowable load for the fixed-pinned end condition is 517.28 ksi

To determine

The allowable load for the fixed-fixed end condition.

Answer to Problem 11.4.5P

The allowable load for the fixed-fixed end condition is 1011.296 ksi

Explanation of Solution

Given:

E=30 ×10⁶ psi

L= 28 ft

Factor of of safety = 2.5

Column type: W12×87

Concept Used:

  $\begin{array}{l}Factorofsafety=\frac{P_{cr}}{P_{allowable}}\\ Thecriticalloadofthecolumn,P_{cr}=\frac{{\pi }^{2}EI}{L_{effective}{}^2}\\ Where,\\ I=momentofinertia\\ E=\mathrm{mod}ulusofelasticity\end{array}$

Calculation:

  $\begin{array}{l}Momentof\mathrm{int}eriaofthecolumn:\\ I_1=740i{n}^4\\ I_2=241i{n}^4\\ \mathrm{Sin}ce{I}_{2}isthe\min imumvalueweconsider{I}_2.\\ Factorofsafety=\frac{P_{cr}}{P_{allowable}}\\ P_{allowable}=\frac{P_{cr}}{Factorofsafety}\\ Thecriticalloadofthecolumn,P_{cr}=\frac{{\pi }^{2}EI}{L_{effective}{}^2}\\ Forpinned-pinnedendcodition\text{}{L}_{effective}=\frac{L}{2}\\ P_{allowable}=\frac{{\pi }^2\times 30\times {10}^6\times 241\times 4}{2.5\times {\left(28\times 12\right)}^2}\\ P_{allowable}=1011.296ksi\end{array}$

Conclusion:

The allowable load for the fixed-fixed end condition is 1011.296 ksi