Question 2 of 8: Reactions, shear- and bending-moment diagrams

Question 2: Reactions, shear- and bending-moment diagrams (18 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.

2(a) — Compound (Gerber) beam with an overhang

Given. Pin at $x{=}0$, rollers at $x{=}8$ and $x{=}14$, internal hinge at $x{=}10$, free tip at $x{=}16$; UDL $6\text{ kN/m}$ over the whole $16\text{ m}$; point load $20\text{ kN}\downarrow$ at the free tip. Find. Reactions and the SFD/BMD extremes.

1. Isolate the dropped-in span first (hinge–tip). The right body $[10,16]$ hangs on hinge $D$ and roller $R_{14}$, carrying its own UDL $36\text{ kN}$ at $x{=}13$ and the $20\text{ kN}$ at $x{=}16$. Taking $\sum M_{D}=0$: $R_{14}(4)=36(3)+20(6)$, so $\boxed{R_{14}=57\text{ kN}}$; the hinge then transmits only $1\text{ kN}$ ($36+20-57=-1$, i.e.\ the left body lifts the right body by $1\text{ kN}$).
2. Whole-beam equilibrium for the other two. With total load $6(16)+20=116\text{ kN}$ and $\sum M_{A}=0:\;R_{8}(8)+R_{14}(14)=96(8)+20(16)$ gives $\boxed{R_{8}=36.25\text{ kN}}$, then $\boxed{R_{A}=116-36.25-57=22.75\text{ kN}}$.
3. Shear (SFD). Starting at $+22.75$ and shedding $6\text{ kN/m}$: just left of the roller at $8$, $V=22.75-48=-25.25$ (the minimum); it jumps to $+11$, falls to $-1$ at the hinge, then $-25$ just left of the roller at $14$; that roller lifts it to $\boxed{V_{\max}=+32\text{ kN}}$, decaying to $+20$ at the tip where the $20\text{ kN}$ closes the diagram.
4. Moment (BMD). The sag peak is where $V=0$ in the first span, $x=22.75/6=3.79\text{ m}$: $\boxed{M_{\max}=+43.1\text{kN}\cdot\text{m}}$. Over the roller at $14$ the beam hogs $\boxed{M_{\min}=-52\text{kN}\cdot\text{m}}$ (from the tip cantilever $6\cdot2^2/2+20\cdot2$). At the internal hinge $x=10$, $M=0$ — the check that fixes the analysis.

2(b) — Three-hinged portal frame

Given. Trapezoidal-height rectangular portal: pins at both feet $BL(0,0)$ and $BR(8,0)$, columns $4\text{ m}$, beam $8\text{ m}$, and an internal hinge at the top-right corner $TR(8,4)$. Loads: UDL $6\text{ kN/m}\downarrow$ on the beam and UDL $6\text{ kN/m}$ acting horizontally to the left on the right column. Find. Reactions and BMD extremes.

1. Global equilibrium (4 unknowns, 3 equations + hinge). Beam load $48\text{ kN}\downarrow$ at $x{=}4$; right-column load $24\text{ kN}\leftarrow$ at $(8,2)$. $\sum M_{BL}=0:\;V_{BR}(8)-48(4)+24(2)=0\Rightarrow \boxed{V_{BR}=18\text{ kN}}$, so $\boxed{V_{BL}=30\text{ kN}}$.
2. Use the hinge to split the horizontals. Taking the right column $TR\!-\!BR$ alone, $\sum M_{TR}=0:\;H_{BR}(4)=24(2)\Rightarrow \boxed{H_{BR}=12\text{ kN}}$ ($\rightarrow$), and horizontal equilibrium gives $\boxed{H_{BL}=12\text{ kN}}$ ($\rightarrow$) — the two pins push back against the leftward wind.
3. Member diagrams. The unloaded left column carries only the base thrust, so its moment runs linearly $0\to\boxed{48\text{kN}\cdot\text{m}}$ (hog) at the rigid corner $TL$. The beam, with a $-48$ end moment at $TL$ and $0$ at the hinge $TR$ plus its UDL, sags to $\boxed{+27\text{kN}\cdot\text{m}}$ at $x{=}5$. The right column behaves as a pin-ended strut under its lateral UDL, peaking at $wL^2/8=\boxed{12\text{kN}\cdot\text{m}}$ mid-height.

2(c) — Inclined simply-supported beam (5–12–13)

Given. Straight beam from pin $A(0,0)$ to roller $B(12,5)$ — horizontal span $12\text{ m}$, rise $5\text{ m}$ (length $13\text{ m}$); the roller reacts perpendicular to the beam. Vertical point loads $39\text{ kN}$ at $x{=}4$ and $26\text{ kN}$ at $x{=}8$. Find. Reactions and SFD/BMD extremes.

1. Reactions. $\sum M_{A}=0$ with the roller line $\perp$ to the beam gives moment arm $=13\text{ m}$: $R_{B}(13)=39(4)+26(8)=364\Rightarrow \boxed{R_{B}=28\text{ kN}}$. Resolving, $\boxed{H_{A}=140/13=10.77\text{ kN}}$ and $\boxed{V_{A}=65-336/13=39.15\text{ kN}}$.
2. Shear (measured $\perp$ to the axis). Projecting the left-hand resultant onto the beam normal: $\boxed{V=+32\text{ kN}}$ over $A\!-\!P_1$, $-4\text{ kN}$ over $P_1\!-\!P_2$, and $\boxed{-28\text{ kN}}$ over $P_2\!-\!B$; the axial thrust runs $25\to10\to0\text{ kN}$ (compression).
3. Moment. Integrating the shear along the $13\text{ m}$ axis, $M$ rises to $\boxed{M_{\max}=138.7\text{kN}\cdot\text{m}}$ under the $39\text{ kN}$ load, eases to $121.3\text{kN}\cdot\text{m}$ under the $26\text{ kN}$ load, and returns to $0$ at $B$ — identical to the equivalent horizontal beam, because the perpendicular roller carries the thrust axially.