Chapter 11, Problem 11.4.6P

i.

To determine

The allowable load for the pinned-pinned end condition.

Answer to Problem 11.4.6P

The allowable load for the pinned-pinned condition is 677.94 kN

Explanation of Solution

Given:

E=200 GPa

L= 7.5 m

Factor of of safety = 2.5

Column type: W250×89

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:

  $\begin{array}{l}Momentof\mathrm{int}eriaofthecolumn:\\ I_1=142\times {10}^{6}m{m}^4\\ I_2=48.3\times {10}^{6}m{m}^4\\ \mathrm{Sin}ce{I}_{2}isthe\min imumvalueweconsider{I}_2.\\ Factorofsafety=\frac{P_{cr}}{P_{allowable}}\\ P_{allowable}=\frac{P_{cr}}{Factorofsafety}\\ P_{allowable}=\frac{{\pi }^{2}EI}{n\times L_{effective}{}^2}\\ Forpinned-pinnedendcodition\text{}{L}_{effective}=L\\ P_{allowable}=\frac{{\pi }^2\times 200\times {10}^3\times 48.3\times {10}^6}{2.5\times {\left(7500\right)}^2}\\ P_{allowable}=677.97kN\end{array}$

Conclusion:

The allowable load for the pinned-pinned condition is 677.97 kN

ii.

To determine

The allowable load for the fixed-free end condition.

Answer to Problem 11.4.6P

The allowable load for the fixed-free end condition is 169.49 kN

Explanation of Solution

Given:

E=200 GPa

L= 7.5 m

Factor of of safety = 2.5

Column type: W250×89

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:

  $\begin{array}{l}Momentof\mathrm{int}eriaofthecolumn:\\ I_1=142\times {10}^{6}m{m}^4\\ I_2=48.3\times {10}^{6}m{m}^4\\ \mathrm{Sin}ce{I}_{2}isthe\min imumvalueweconsider{I}_2.\\ Factorofsafety=\frac{P_{cr}}{P_{allowable}}\\ P_{allowable}=\frac{P_{cr}}{Factorofsafety}\\ Thecriticalloadofthecolumn,P_{cr}=\frac{{\pi }^{2}EI}{L_{effective}{}^2}\\ Forfixed-freeendcodition\text{}{L}_{effective}=2L\\ P_{allowable}=\frac{{\pi }^2\times 200\times {10}^3\times 48.3\times {10}^6}{2.5\times {\left(2\times 7500\right)}^2}\\ P_{allowable}=169.49kN\end{array}$

Conclusion:

The allowable load for the fixed-free end condition is 169.49 kN

iii.

To determine

The allowable load for the fixed-pinned end condition.

Answer to Problem 11.4.6P

The allowable load for the fixed-pinned end condition is 1387.58 kN

Explanation of Solution

Given:

E=200 GPa

L= 7.5 m

Factor of of safety = 2.5

Column type: W250×89

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:

  $\begin{array}{l}Momentof\mathrm{int}eriaofthecolumn:\\ I_1=142\times {10}^{6}m{m}^4\\ I_2=48.3\times {10}^{6}m{m}^4\\ \mathrm{Sin}ce{I}_{2}isthe\min imumvalueweconsider{I}_2.\\ Factorofsafety=\frac{P_{cr}}{P_{allowable}}\\ P_{allowable}=\frac{P_{cr}}{Factorofsafety}\\ Thecriticalloadofthecolumn,P_{cr}=\frac{{\pi }^{2}EI}{L_{effective}{}^2}\\ Forfixed-pinnedendcodition\text{}{L}_{effective}=0.699L\\ P_{allowable}=\frac{{\pi }^2\times 200\times {10}^3\times 48.3\times {10}^6}{2.5\times {\left(0.699\times 7500\right)}^2}\\ P_{allowable}=1387.58kN\end{array}$

Conclusion:

The allowable load for the fixed-pinned end condition is 1387.58 kN

iv.

To determine

The allowable load for the fixed-fixed end condition.

Answer to Problem 11.4.6P

The allowable load for the fixed-fixed end condition is 2711.90 kN

Explanation of Solution

Given:

E=200 GPa

L= 7.5 m

Factor of of safety = 2.5

Column type: W250×89

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:

  $\begin{array}{l}Momentof\mathrm{int}eriaofthecolumn:\\ I_1=142\times {10}^{6}m{m}^4\\ I_2=48.3\times {10}^{6}m{m}^4\\ \mathrm{Sin}ce{I}_{2}isthe\min imumvalueweconsider{I}_2.\\ Factorofsafety=\frac{P_{cr}}{P_{allowable}}\\ P_{allowable}=\frac{P_{cr}}{Factorofsafety}\\ Thecriticalloadofthecolumn,P_{cr}=\frac{{\pi }^{2}EI}{L_{effective}{}^2}\\ Forpinned-pinnedendcodition\text{}{L}_{effective}=\frac{L}{2}\\ P_{allowable}=\frac{{\pi }^2\times 200\times {10}^3\times 48.3\times {10}^6\times 4}{2.5\times {\left(7500\right)}^2}\\ P_{allowable}=2711.90kN\end{array}$

Conclusion:

The allowable load for the fixed-fixed end condition is 2711.90 kN