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. Deflection and slope at the free end of the overhang ($x=9$ m) by direct integration of $EI\,v''=M(x)$; and whether the beam deflects up or down between the supports.

Approach. Find the reactions from statics, build $M(x)$ piecewise for $0\le x\le6$ and $6\le x\le9$, integrate twice per segment, and fix the four constants with $v(0)=0$, $v(6\,\text{m})=0$, and continuity of $v,v'$ at $x=6$ m.

1. Section properties. Two 200×10 mm flanges plus a 380×10 mm web about the centroidal axis: $$I=2\left[\frac{200(10)^3}{12}+200(10)(195)^2\right]+\frac{10(380)^3}{12}=1.9786\times10^{8}\ \text{mm}^4$$ so $EI=(200\,000)(1.9786\times10^8)=3.9572\times10^{13}\ \text{N}\cdot\text{mm}^2$.
2. Reactions. The triangular load's resultant is $W=\tfrac12(20)(9)=90$ kN, acting at its centroid $\bar{x}=\tfrac23(9)=6$ m from A — exactly at the roller B. Taking moments about A: $R_B(6)=W(6)\Rightarrow R_B=90$ kN, and $R_A=W-R_B=\boxed{0}$. (The pin reaction is exactly zero — a direct consequence of the load centroid coinciding with support B.)
3. Bending moment, $0\le x\le6$ m. With $R_A=0$, only the distributed load acts to the left of the cut: $$M_1(x)=-\int_0^x \frac{20\xi}{9}(x-\xi)\,d\xi=-\frac{10}{27}x^3\ \ \text{kN}\!\cdot\!\text{m}\quad(x\text{ in m})$$ giving $M_1(6)=-80\ \text{kN}\!\cdot\!\text{m}$ (hogging).
4. Bending moment, $6\le x\le9$ m. Add the moment of $R_B$: $$M_2(x)=-\frac{10}{27}x^3+90(x-6)\ \ \text{kN}\!\cdot\!\text{m}$$ Check: $M_2(9)=-270+270=0$ — correctly zero at the free tip. $M(x)<0$ (hogging) over the entire span 0–9 m, a signature of the overhang's heavy triangular load dominating the main span.
5. Integrate twice per segment. Working in N·mm ($X$ in mm), $EI\,v_1''=M_1(X)=-3.7037\times10^{-4}X^3$ and $EI\,v_2''=M_2(X)$. Integrating twice gives four constants $C_1,C_2$ (segment 1) and $C_3,C_4$ (segment 2), fixed by $v_1(0)=0$, $v_1(6000)=0$, $v_2(6000)=v_1(6000)$, $\theta_2(6000)=\theta_1(6000)$ (continuity). Solving this 4×4 linear system gives $C_1=6.065\times10^{-4}$, $C_2=0$, $C_3=4.1545\times10^{-2}$, $C_4=-81.876$.
6. Evaluate at the free tip, $X=9000$ mm. $$v(9\,\text{m})=\boxed{-11.94\ \text{mm}}\qquad \theta(9\,\text{m})=\boxed{-4.511\times10^{-3}\ \text{rad}=-0.258^\circ}$$ Both negative: the overhang tip deflects downward 11.94 mm and rotates so that the tip drops further as $x$ increases — consistent with the overhang carrying the largest share of the triangular load.
7. Part (b): sign of the deflection between the supports. Since $M(x)<0$ everywhere on $0above the chord in between (the same way a concave-down parabola with two zero roots is positive between them) — confirmed: $v(1\,\text{m})=+0.606$ mm, $v(3\,\text{m})=+1.706$ mm, $v(5\,\text{m})=+1.570$ mm, with a maximum upward deflection of $+1.947$ mm at $x\approx4.01$ m. The beam deflects upward (never downward) between the two supports — the overhang's hogging moment "lifts" the main span even though the load itself pushes down everywhere, because the entire span is in negative (hogging) curvature.