Question 3 of 8: Vertical deflection at the overhang tip

Question 3: Vertical deflection at the overhang tip (16 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. Overhang tip $A$ ($x{=}0$), pin $B$ ($x{=}3\text{ m}$), $60\text{ kN}$ at $x{=}6\text{ m}$, roller $C$ ($x{=}12\text{ m}$); UDL $8\text{ kN/m}$ over the $3\text{ m}$ overhang $A$–$B$. Find. $\delta_A$.

Approach. Reactions first; then unit-load (virtual-work) $\delta_A=\frac{1}{EI}\int M\,m\,dx$, with $M$ the real moment and $m$ the moment from a unit vertical load at $A$.

1. Real reactions. $\sum M_B=0:\;R_C(9)=60(3)-8(3)(1.5)\Rightarrow R_C=\boxed{16\text{ kN}}$; $R_B=8(3)+60-16=\boxed{68\text{ kN}}$.
2. Virtual system. Unit downward load at $A$: $R_{B}^{v}=1.25$, $R_{C}^{v}=-0.25$ (from $\sum M_B$: $R_C^{v}(9)=-1(3)$).
3. Integrate. Combining the overhang and span contributions, $\displaystyle \delta_A=\frac{1}{EI}\int_0^{12} M\,m\,dx=\frac{495}{EI}$, and with $EI=2.7\times10^{4}$: $\delta_A=\dfrac{495}{27000}=\boxed{0.01833\text{ m}=18.3\text{ mm}}$.
4. Direction. The result is positive upward: the large near-support $60\text{ kN}$ rotates the span so much at $B$ that the tip lifts, overpowering the overhang’s own downward bending. $\boxed{\delta_A\approx 18.3\text{ mm}\ \uparrow}$.

The two competing effects are worth separating. Acting alone, the $8\text{ kN/m}$ over the $3\text{ m}$ overhang would bend the free tip downward by roughly $wL^4/8EI\approx 3\text{ mm}$. Against that, the $60\text{ kN}$ in the main span drives a clockwise rotation of the elastic curve at $B$; because the overhang projects to the left of $B$, that rotation carries the tip up by about $21\text{ mm}$. The algebraic sum, $-3+21\approx 18\text{ mm}$ upward, is exactly what the virtual-work integral returns, and it is why the sign of the answer — not just its magnitude — must be reported. A quick check on the reactions ($R_B+R_C=84\text{ kN}$ equals the total applied load $24+60$) confirms the equilibrium on which the whole calculation rests.