Question 6 of 8: Overhang beam — shear/moment functions and diagrams

Question 6: Overhang beam — shear/moment functions and diagrams (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. $V(x)$ and $M(x)$ as explicit piecewise functions; the shear and moment diagrams; the locations/values of the maximum positive and negative moment; and any inflection points.

Approach. Find the reactions from statics (treating the applied couple correctly — it contributes to the moment equation but not to $\Sigma F_y$), then build $V(x)$ and $M(x)$ by cutting the beam in each of the three regions, remembering that $M(x)$ jumps discontinuously by $-M_0$ when crossing a CCW-applied couple (verified by the free-end boundary check).

1. Reactions. UDL resultant $W=1(2)=2$ kN at $\bar x=5$ m. Taking $\Sigma M_A=0$ (CCW$+$): $4B_y+M_0-W(5)=0\Rightarrow B_y=\dfrac{2(5)-4}{4}=1.5$ kN, and $A_y=W-B_y=0.5$ kN. $$\boxed{A_y=0.5\ \text{kN}\qquad B_y=1.5\ \text{kN}}$$
2. Shear $V(x)$. No horizontal loads, and the couple does not change $V$: $$V(x)=\begin{cases}0.5\ \text{kN} & 0\le x<4\\[2pt] 2.0-1.0(x-4)\ \text{kN} & 4\le x\le6\end{cases}$$ Check at the free tip: $V(6)=2.0-2.0=0$ — correct.
3. Moment $M(x)$, segment $0\le x<2$. $M(x)=A_yx=0.5x$, rising from 0 to $M(2^-)=1.0$ kN·m.
4. Jump at the applied couple ($x=2$). A CCW applied couple produces a downward jump in $M$ (verified below by the free-tip check): $M(2^+)=M(2^-)-M_0=1.0-4.0=-3.0$ kN·m.
5. Moment, $2 $M(x)=-3.0+0.5(x-2)$, rising to $M(4^-)=-2.0$ kN·m. The roller reaction adds no jump in $M$ (only a slope kink), so $M(4^+)=-2.0$ kN·m too.
6. Moment, $4\le x\le6$. $$M(x)=-2.0+2.0(x-4)-0.5(x-4)^2$$ Check at the free tip: $M(6)=-2.0+4.0-2.0=0$ — exactly zero, confirming both the couple's jump direction and the reactions are correct.
7. Extreme moments and inflection points. $M(x)$ rises monotonically on each of the three smooth segments (no interior stationary point, since $V\ne0$ except momentarily at $x=6$). Hence: $$M_{max}^{+}=\boxed{+1.0\ \text{kN}\!\cdot\!\text{m at }x=2^-}\qquad M_{max}^{-}=\boxed{-3.0\ \text{kN}\!\cdot\!\text{m at }x=2^+}$$ Because the sign reversal at $x=2$ is a discontinuous jump (caused by the point couple) rather than a continuous zero-crossing, there is no smooth inflection point anywhere on the beam — curvature reverses abruptly at the couple, not gradually.