Chapter 11, Problem 11.4.10P

The roof beams of a warehouse are supported by pipe columns (see figure) having an outer diameter d₂= 100 mm and inner diameter d₂, = 90mm. The columns have a length L = 4.0 m, modulus E = 210 GPa, and fixed supports at the base.

Calculate the critical load P_crof one of the columns using the following assumptions: (a) the upper end is pinned and the beam prevents horizontal displacement; (b) the upper end is fixed against rotation and the beam prevents horizontal displacement; (c) the upper end is pinned, but the beam is free to move horizontally; and (d) the upper end is fixed against rotation, but the beam is free to move horizontally.

  

To determine

The critical load when the upper end is pinned and the beam prevents horizontal displacement.

Answer to Problem 11.4.10P

The critical load when the upper end is pinned and the beam prevents horizontal displacement is 447 kN

Explanation of Solution

Given:

E=210 GPa

L= 4 m

d₁= 90 mm

d₂= 100 mm

Concept Used:

  $\begin{array}{l}Momentof\mathrm{int}eriaofthecolumn,I=\frac{\pi \left(d_0{}^4-d_i{}^4\right)}{64}\\ Thecriticalloadofthecolumn,P_{cr}=\frac{{\pi }^{2}EI}{L_{effective}{}^2}\\ Where,\\ I=momentofinertia\\ E=\mathrm{mod}ulusofelasticity\end{array}$

Calculation:

  $\begin{array}{l}Momentof\mathrm{int}eriaofthecolumn,I=\frac{\pi \left(d_0{}^4-d_i{}^4\right)}{64}\\ I=\frac{\pi \left({100}^4-{90}^4\right)}{64}\\ I=1688\times {10}^{3}m{m}^4\\ Thecriticalloadofthecolumn,P_{cr}=\frac{{\pi }^{2}EI}{L_{effective}{}^2}\\ Forpinned-pinnedendcodition\text{}{L}_{effective}=0.699L\\ P_{cr}=\frac{{\pi }^2\times 210\times {10}^3\times 1688\times {10}^3}{{\left(0.699\times 4000\right)}^2}\\ P_{cr}=447kN\end{array}$

Conclusion:

The critical load for the pinned-pinned condition is 447 kN

To determine

The critical load when the upper end is fixed for rotation and the beam prevents horizontal displacement.

Answer to Problem 11.4.10P

The critical load when the upper end is fixed for rotation and the beam prevents horizontal displacement is 875 kN

Explanation of Solution

Given:

E=210 GPa

L= 4 m

d₁= 90 mm

d₂= 100 mm

Concept Used:

  $\begin{array}{l}Momentof\mathrm{int}eriaofthecolumn,I=\frac{\pi \left(d_0{}^4-d_i{}^4\right)}{64}\\ Thecriticalloadofthecolumn,P_{cr}=\frac{{\pi }^{2}EI}{L_{effective}{}^2}\\ Where,\\ I=momentofinertia\\ E=\mathrm{mod}ulusofelasticity\end{array}$

Calculation:

  $\begin{array}{l}Momentof\mathrm{int}eriaofthecolumn,I=\frac{\pi \left(d_0{}^4-d_i{}^4\right)}{64}\\ I=\frac{\pi \left({100}^4-{90}^4\right)}{64}\\ I=1688\times {10}^{3}m{m}^4\\ Thecriticalloadofthecolumn,P_{cr}=\frac{{\pi }^{2}EI}{L_{effective}{}^2}\\ Forfixed-freeendcodition\text{}{L}_{effective}=\frac{L}{2}\\ P_{cr}=\frac{{\pi }^2\times 210\times {10}^3\times 1688\times {10}^3\times 4}{{\left(4000\right)}^2}\\ P_{cr}=875kN\end{array}$

Conclusion:

The critical load when the upper end is fixed for rotation and the beam prevents horizontal displacement is 875 kN

To determine

The critical load when the upper end is pinned but beam free to move horizontally.

Answer to Problem 11.4.10P

The critical load when the upper end is pinned but beam free to move horizontally is 54.7 kN

Explanation of Solution

Given:

E=210 GPa

L= 4 m

d₁= 90 mm

d₂= 100 mm

Concept Used:

  $\begin{array}{l}Momentof\mathrm{int}eriaofthecolumn,I=\frac{\pi \left(d_0{}^4-d_i{}^4\right)}{64}\\ Thecriticalloadofthecolumn,P_{cr}=\frac{{\pi }^{2}EI}{L_{effective}{}^2}\\ Where,\\ I=momentofinertia\\ E=\mathrm{mod}ulusofelasticity\end{array}$

Calculation:

  $\begin{array}{l}Momentof\mathrm{int}eriaofthecolumn,I=\frac{\pi \left(d_0{}^4-d_i{}^4\right)}{64}\\ I=\frac{\pi \left({100}^4-{90}^4\right)}{64}\\ I=1688\times {10}^{3}m{m}^4\\ Thecriticalloadofthecolumn,P_{cr}=\frac{{\pi }^{2}EI}{L_{effective}{}^2}\\ Forfixed-pinnedendcodition\text{}{L}_{effective}=2L\\ P_{cr}=\frac{{\pi }^2\times 210\times {10}^3\times 1688\times {10}^3}{{\left(2\times 4000\right)}^2}\\ P_{cr}=54.7kN\end{array}$

Conclusion:

The critical load when the upper end is pinned but beam free to move horizontally is 54.7 kN .

To determine

The critical load when the upper end is fixed under rotation but the beam is free to move horizontally.

Answer to Problem 11.4.10P

The critical load when the upper end is fixed under rotation but the beam is free to move horizontally is 219 kN

Explanation of Solution

Given:

E=210 GPa

L= 4 m

d₁= 90 mm

d₂= 100 mm

Concept Used:

  $\begin{array}{l}Momentof\mathrm{int}eriaofthecolumn,I=\frac{\pi \left(d_0{}^4-d_i{}^4\right)}{64}\\ Thecriticalloadofthecolumn,P_{cr}=\frac{{\pi }^{2}EI}{L_{effective}{}^2}\\ Where,\\ I=momentofinertia\\ E=\mathrm{mod}ulusofelasticity\end{array}$

Calculation:

  $\begin{array}{l}Momentof\mathrm{int}eriaofthecolumn,I=\frac{\pi \left(d_0{}^4-d_i{}^4\right)}{64}\\ I=\frac{\pi \left({100}^4-{90}^4\right)}{64}\\ I=1688\times {10}^{3}m{m}^4\\ Thecriticalloadofthecolumn,P_{cr}=\frac{{\pi }^{2}EI}{4\times L_{effective}{}^2}\\ Forpinned-pinnedendcodition\text{}{L}_{effective}=\frac{L}{2}\\ P_{cr}=\frac{{\pi }^2\times 210\times {10}^3\times 1688\times {10}^3}{{\left(4000\right)}^2}\\ P_{cr}=219kN\end{array}$

Conclusion:

The critical load when the upper end is fixed under rotation but the beam is free to move horizontally is 219 kN.