Question 2 of 8: Shear and Moment Diagrams for a Stepped-UDL Overhang Beam

Question 2: Shear and Moment Diagrams for a Stepped-UDL Overhang 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. Overhang tip $A$ ($x=0$, concentrated load $30\text{ kips}$ downward); pin support $B$ ($x=5\text{ ft}$); roller support $D$ ($x=25\text{ ft}$). UDL $w_1=2\text{ kips/ft}$ over $A$ to $C$ ($0\le x\le15\text{ ft}$); UDL $w_2=5\text{ kips/ft}$ over $C$ to $D$ ($15\le x\le25\text{ ft}$).

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

Approach. Find the reactions from global equilibrium, then integrate the load directly from the free (overhang) end at A, adding the jump from each reaction as it is reached, to build V(x) and M(x) piecewise over the three regions.

1. Reactions. $\sum M_B=0$ (taking moments of the 30-kip load and each UDL resultant about $B$, plus $R_D$ at $20\text{ ft}$ from B): solving simultaneously with $\sum F_y=0$ gives $R_B=\boxed{76.25\text{ kips}}$, $R_D=\boxed{33.75\text{ kips}}$ (sum $=110\text{ kips}=30+2(15)+5(10)$, checks against the total load).
2. V(x), M(x) on the overhang $0\le x\le5$ (before B). Starting from the free tip, the 30-kip load causes an immediate jump: $V(x)=-30-2x$; $M(x)=-30x-x^2$. At $x=5$: $V=-40\text{ kips}$, $M=\boxed{-175\text{ kip}\cdot\text{ft}}$ (hogging).
3. V(x), M(x) on $5\le x\le15$ (B to C, still under $w_1$). The reaction $R_B$ jumps $V$ up by $76.25$: $V(x)=-40+76.25-2(x-5)=36.25-2(x-5)$; $M(x)=-175+\int_5^xV\,d\xi$. At $x=15$: $V=16.25\text{ kips}$, $M=\boxed{+87.5\text{ kip}\cdot\text{ft}}$ — the moment has crossed zero between B and C, so an inflection point exists there.
4. Locate the inflection point. Setting $M(x)=0$ in the $[5,15]$ segment gives $x=\boxed{10.73\text{ ft}}$ (moment changes sign from hogging to sagging at this point).
5. V(x), M(x) on $15\le x\le25$ (C to D, under $w_2$). $V(x)=16.25-5(x-15)$; setting $V=0$ gives $x=15+16.25/5=\boxed{18.25\text{ ft}}$, where the moment is maximum: $M(18.25)=\boxed{+113.9\text{ kip}\cdot\text{ft}}$ (the maximum POSITIVE/sagging moment on the whole beam). Continuing to $x=25$: $V(25^-)=-33.75\text{ kips}$, jumping to $V(25^+)=0$ once $R_D=33.75\text{ kips}$ is added, and $M(25)=0$ as required at the free/roller end.
6. Diagram summary. $V(x)$ is piecewise-linear: from $-30$ at A down to $-40$ at $B^-$, jumps to $+36.25$ at $B^+$, ramps down to $+16.25$ at C, continues down (steeper) to $-33.75$ at $D^-$, then jumps to $0$ at $D^+$. $M(x)$ is piecewise-quadratic: $0$ at A, down to the maximum HOGGING value $-175\text{ kip}\cdot\text{ft}$ at B (the overall maximum negative moment), crossing zero (inflection) at $x=10.73\text{ ft}$, rising to the maximum SAGGING value $+113.9\text{ kip}\cdot\text{ft}$ at $x=18.25\text{ ft}$ (where $V=0$), then back down to $0$ at D.