Question 8 of 8: Overhang Beam Deflection by the Method of Integration

Question 8: Overhang Beam Deflection by the Method of 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 the overhang tip C, by direct double integration of the moment-curvature relation (no superposition).

Approach. Find the reactions, write a single Macaulay-bracket moment expression valid over the whole beam, integrate $EI y''=M(x)$ twice, fix the two constants from the zero-deflection conditions at the two supports, then evaluate the resulting slope and deflection expressions at x = 6 m (point C).

1. Reactions. Summing moments about A (CCW+, load at 6 m): $$R_B(4) - 20(6) = 0 \ \Rightarrow\ R_B = 30\ \text{kN}, \qquad R_A = 20-30 = -10\ \text{kN}\ \text{(i.e. 10 kN downward — the overhang load pulls the near support down)}$$
2. Moment function (Macaulay form, x from A, in mm). $$M(x) = R_A x + R_B\langle x-4000\rangle = -10{,}000\,x + 30{,}000\langle x-4000\rangle \quad [\text{N}\cdot\text{mm}]$$ Check at the tip: $M(6000)=-10{,}000(6000)+30{,}000(2000) = -60\times10^6+60\times10^6=0$ (free end, as required), and at B: $M(4000^-)=-40\times10^6\ \text{N}\cdot\text{mm} = -40$ kN·m, matching $-P\times(\text{overhang length})=-20(2)$ from the free-end side — consistent hogging over the whole overhang.
3. Section property. $$I = 17.1\times10^6\ \text{mm}^4, \qquad EI = 200{,}000(17.1\times10^6) = 3.42\times10^{12}\ \text{N}\cdot\text{mm}^2$$
4. First integration (slope). $$EI\,y' = -5000x^2+15{,}000\langle x-4000\rangle^2 + C_1$$
5. Second integration (deflection) and boundary conditions. $$EI\,y = -\tfrac{5000}{3}x^3+5000\langle x-4000\rangle^3+C_1x+C_2$$ $y(0)=0\Rightarrow C_2=0$. $y(4000)=0$ gives $$-\tfrac{5000}{3}(4000)^3 + C_1(4000) = 0 \ \Rightarrow\ \boxed{C_1 = 2.667\times10^{10}\ \text{N}\cdot\text{mm}^2}$$
6. Evaluate slope and deflection at C (x = 6000 mm). $$EI\,y'(6000) = -5000(6000)^2+15{,}000(2000)^2+2.667\times10^{10} = -9.333\times10^{10}\ \text{N}\cdot\text{mm}^2$$ $$\theta_C = \frac{-9.333\times10^{10}}{3.42\times10^{12}} = -0.02729\ \text{rad} = \boxed{-1.564^\circ}\ \text{(rotating clockwise, tip dipping down)}$$ $$EI\,y(6000) = -\tfrac{5000}{3}(6000)^3+5000(2000)^3+2.667\times10^{10}(6000) = -1.600\times10^{14}\ \text{N}\cdot\text{mm}^3$$ $$\boxed{\delta_C = \frac{-1.600\times10^{14}}{3.42\times10^{12}} = -46.8\ \text{mm}\ (46.8\ \text{mm downward})}$$

Final Results.