Question 2 of 8: Cantilever Deflection by the Method of Integration

Question 2: Cantilever 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. (a) Slope and deflection at the free end by direct double integration (no superposition). (b) Whether the deflection satisfies the L/120 limit.

Approach. Write M(x) piecewise from statics (free body to the right of the cut), integrate $EI\,y''=M(x)$ twice in each region, use the two fixed-end conditions $y(0)=y'(0)=0$ plus continuity of slope and deflection at x = 4 m to fix all constants, then evaluate at the free end.

1. Moment function, by region. For $0\le x\le4$ (m), the free body to the right of the cut carries the remaining UDL plus the end couple: $$M(x) = -\!\left[15(4-x)\!\left(\frac{4-x}{2}\right)+50\right] = -7.5x^2+60x-170\ \ [\text{kN}\cdot\text{m}]$$ For $4\le x\le6$, only the end couple remains: $$M(x)=-50\ \text{kN}\cdot\text{m (constant)}$$ Check at the fixed end: $M(0)=-170\ \text{kN}\cdot\text{m}$, matching the reaction moment by statics ($15(4)(2)+50=120+50=170$ kN·m, hogging) — confirms the sign and magnitude before integrating.
2. Section stiffness. $$EI = 200{,}000\ \text{MPa}\times308\times10^6\ \text{mm}^4 = 6.16\times10^{13}\ \text{N}\cdot\text{mm}^2 = 61{,}600\ \text{kN}\cdot\text{m}^2$$
3. Integrate region 1 (0≤x&le4), fixed-end BCs. $$EI\,y' = -2.5x^3+30x^2-170x+C_1, \qquad EI\,y=-0.625x^4+10x^3-85x^2+C_1x+C_2$$ $y'(0)=0$ and $y(0)=0$ (fixed support) give directly $$\boxed{C_1=0,\ C_2=0}$$
4. Integrate region 2 (4≤x&le6), match at x=4. $$EI\,y'=-50x+D_1, \qquad EI\,y=-25x^2+D_1x+D_2$$ Matching $EI\,y'(4^-)=-2.5(64)+30(16)-170(4)=-360$ to $EI\,y'(4^+)=-200+D_1$ gives $D_1=-160$. Matching $EI\,y(4^-)=-0.625(256)+10(64)-85(16)=-880$ to $EI\,y(4^+)=-400-640+D_2$ gives $D_2=160$: $$EI\,y'(x)=-50x-160, \qquad EI\,y(x)=-25x^2-160x+160 \quad [4\le x\le6]$$
5. (a) Evaluate at the free end, x = 6 m. $$EI\,y'(6)=-50(6)-160=-460\ \text{kN}\cdot\text{m}^2 \ \Rightarrow\ \theta_{free}=\frac{-460}{61{,}600}=\boxed{-7.47\times10^{-3}\ \text{rad}\ (-0.428^\circ, \text{ tip rotating downward})}$$ $$EI\,y(6)=-25(36)-160(6)+160=-1700\ \text{kN}\cdot\text{m}^3\ \Rightarrow\ \delta_{free}=\frac{-1700}{61{,}600}=-0.02760\ \text{m}=\boxed{27.6\ \text{mm, downward}}$$
6. (b) Allowable-deflection check. $$\delta_{allow}=\frac{L}{120}=\frac{6000}{120}=50.0\ \text{mm}$$ $$\boxed{27.6\ \text{mm} < 50.0\ \text{mm}\ \Rightarrow\ \text{the beam SATISFIES the allowable deflection limit}}$$

Final Results.