Question 8 of 8: Overhanging Beam – Force for a Deflection Limit

Question 8: Overhanging Beam – Force for a Deflection Limit (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. A beam pinned at C and roller-supported at B (span 3 m), overhanging 3 m to the free end A. A downward force $P$ acts 2 m right of B (1 m left of A); a clockwise couple $M_A=37000$ N·m acts at A. $E=210$ GPa, $I=995\times10^6$ mm$^4$.

Find. The magnitude and sense of $P$ so that the tip deflection at A equals the 10 mm downward limit.

Approach. Superpose the tip deflections from the couple and from $P$ using the unit-load (virtual-work) method, $\delta_A=\int M\,m\,dx/EI$, then solve the linear equation $\delta_A=10$ mm for $P$.

1. Flexural rigidity. $EI=210\times10^9\cdot995\times10^{-6}=2.090\times10^{5}\ \text{N}\cdot\text{m}^2$.
2. Deflection from the couple alone. The clockwise $M_A$ rotates the tip so that A moves down; unit-load integration gives $\delta_{A,M}=+1.33$ mm (downward).
3. Deflection per unit $P$. A downward $P$ on the overhang deflects A downward by $5.10\times10^{-8}\,\text{m}$ per newton, i.e. $0.0510$ mm/kN.
4. Solve for $P$. Require $\delta_{A,M}+P\cdot(0.0510\,\text{mm/kN})=10$ mm: $P=\dfrac{10-1.33}{0.0510}=170\ \text{kN}$, hence $\boxed{P\approx1.70\times10^{5}\ \text{N acting downward}}$.
5. Sense check. Because the clockwise couple already pushes A down by 1.33 mm, $P$ must be downward (not up) and only large enough to add the remaining 8.67 mm.