Question 1 of 8: Overhang-Beam Deflection by Integration

Question 1: Overhang-Beam Deflection by Integration (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.

Find. The deflection and slope at C (the free end of the overhang), using the method of double integration.

Approach. Find the reactions, write M(x) for each of the two regions (A–B and B–C), integrate each twice with its own constants, and apply continuity/support boundary conditions to solve for all four constants.

1. Reactions. Taking moments about A: $$R_B(5) = w(3)(5+1.5)\ \Rightarrow\ R_B=39\text{ kN}.$$ Then $$R_A = w(3)-R_B = 30-39=-9\text{ kN}$$ (the pin reaction is actually 9 kN downward — expected, since the whole load sits on the overhang and tries to lift A).
2. Moment equations. Region 1 (0≤x≤5 m, no load): $$M_1(x)=R_A x.$$ Region 2 (5≤x≤8 m, UDL from B to x): $$M_2(x)=R_A x + R_B(x-5) - \dfrac{w(x-5)^2}{2}.$$
3. Integrate region 1 twice. $$EI\,\theta_1=\int M_1\,dx + C_1,\qquad EI\,y_1=\int\!\int M_1\,dx + C_1x+C_2.$$ Boundary conditions $y_1(0)=0$ and $y_1(5\text{ m})=0$ (both supports) fix $C_1,C_2$.
4. Integrate region 2 twice. $$EI\,\theta_2=\int M_2\,dx + C_3,\qquad EI\,y_2=\int\!\int M_2\,dx + C_3x+C_4.$$ Continuity at $x=5$ m requires $y_2(5)=0$ (support B again) and $\theta_2(5)=\theta_1(5)$ (no kink in the elastic curve), which fix $C_3,C_4$.
5. Evaluate at C (x = 8 m). With $EI = (200{,}000)(20.0\times10^6)=4.0\times10^{12}\text{ N}\cdot\text{mm}^2$, substituting x = 8000 mm into $y_2$ and $\theta_2$ gives $$\boxed{y_C = 81.56\ \text{mm (downward)}}\qquad \boxed{\theta_C = -0.03000\ \text{rad} = -1.719^{\circ}}$$