Question 5 of 8: Rigid Bar Propped by a Buckling-Governed Column

Question 5: Rigid Bar Propped by a Buckling-Governed Column (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Rigid bar pinned at $A$; column at $C$, $2\text{ m}$ from $A$; load $P$ at $B$, $3\text{ m}$ beyond $C$ ($5\text{ m}$ from $A$). Column: length $2.5\text{ m}$, $50\times50\text{ mm}$ square, pinned–pinned, $E=200\text{ GPa}$, $\sigma_Y=350\text{ MPa}$, $FS_{buckling}=3$.

Find. Maximum allowable $P$ at B without buckling the column (in-plane only).

Approach. Take moments of the rigid bar about the pin at A to relate the column's axial force to P, then size P from the column's ALLOWABLE (factored) Euler buckling load.

1. Rigid-bar statics. $\sum M_A=0$: $F_{col}(2)-P(5)=0\Rightarrow F_{col}=2.5P$ (compression in the column).
2. Euler buckling load. $I=\dfrac{50^4}{12}=520{,}833\text{ mm}^4$; pinned–pinned $\Rightarrow K=1$: $P_{cr}=\dfrac{\pi^2EI}{(KL)^2}=\dfrac{\pi^2(200{,}000)(520{,}833)}{2500^2}=\boxed{164.5\text{ kN}}$.
3. Apply the buckling factor of safety. $F_{col,allow}=P_{cr}/FS=164.5/3=\boxed{54.83\text{ kN}}$. Check the stress at this load: $54{,}830/50^2=21.9\text{ MPa}\ll\sigma_Y=350\text{ MPa}$ — the column is slender and buckling governs long before yielding.
4. Solve for P. $F_{col}=2.5P\le54.83\Rightarrow P\le54.83/2.5=\boxed{21.9\text{ kN}}$.