Question 2 of 8: Shear and Moment Diagrams for an Overhanging Beam

Question 2: Shear and Moment Diagrams for an Overhanging Beam (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. Pin support at A ($x=0$); roller support at B ($x=6\text{ m}$); free tip C ($x=8\text{ m}$). UDL $w=5\text{ kN/m}$ over $AB$; point load $P=100\text{ kN}$ downward at C.

Find. V(x) and M(x) over the full beam, the shear/moment diagrams, and the locations of maximum/negative moment and any inflection points.

Approach. Find the reactions from global equilibrium, then cut the beam in each of the two regions (span AB and overhang BC) and write V(x), M(x) from the free body to the left of the cut.

1. Reactions. $\sum M_A=0$: $R_B(6)=w(6)(3)+P(8)=90+800=890\Rightarrow R_B=148.33\text{ kN}$. $\sum F_y=0$: $R_A=w(6)+P-R_B=130-148.33=\boxed{-18.33\text{ kN}}$ — negative, so the pin at A must actually hold the beam DOWN (the overhang's lever arm dominates the span load).
2. V(x), M(x) on span AB ($0\le x\le6$m). $V(x)=R_A-wx=-18.33-5x$; $M(x)=R_A x-\tfrac{1}{2}wx^2=-18.33x-2.5x^2$. Since both terms are negative for $x\gt0$, $M(x)\lt0$ over the ENTIRE span — the beam is hogging throughout, with no positive (sagging) region and no interior inflection point.
3. V(x), M(x) on overhang BC ($6\le x\le8$m). Just right of B, $V=R_A-w(6)+R_B=-48.33+148.33=100\text{ kN}$ and stays constant to C (no load between B and C except the tip load, which acts exactly at C). $M(x)=M_B+100(x-6)$ with $M_B=M(6)=-18.33(6)-2.5(36)=\boxed{-200\text{ kN}\cdot\text{m}}$ (max hogging moment, at the roller B), rising linearly back to $M(8)=-200+100(2)=0$ at the free tip, as required.
4. Diagram shape. $V(x)$ is a straight line from $-18.33$ to $-48.33$ over AB, jumps up by $R_B=148.33$ to $+100$ at B, then stays flat at $100$ to C. $M(x)$ is a downward parabola from $0$ at A to $-200$ at B (vertex of the hogging curve lies past B since the slope $V$ never reaches zero inside $0\lt x\lt6$), then rises linearly from $-200$ to $0$ over the overhang. There is no positive moment and no inflection point anywhere on this beam.