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) — Beam with left overhang

Given. Roller at $x{=}2\text{ m}$, pin at $x{=}12\text{ m}$; UDL $10\text{ kN/m}$ over the $2\text{ m}$ left overhang; point load $36\text{ kN}$ at $x{=}7\text{ m}$. Find. Reactions and the SFD/BMD extremes.

1. Reactions. $\Sigma M_{@2}=0:\;10R_{12}-36(5)-\big(10\cdot2\big)(-1)=0\Rightarrow R_{12}=16\text{ kN}$; then $R_{2}=56-16=40\text{ kN}$. $$\boxed{R_{@2}=40\text{ kN}\uparrow,\qquad R_{@12}=16\text{ kN}\uparrow}$$
2. Shear. Overhang: $V$ falls to $-20\text{ kN}$ just left of the roller; the $40\text{ kN}$ reaction lifts it to $+20\text{ kN}$; the $36\text{ kN}$ load drops it to $-16\text{ kN}$, which the pin returns to zero. $$V_{\max}=+20\text{ kN},\quad V_{\min}=-20\text{ kN}$$
3. Moment. Over the roller the overhang gives $M=-20\text{ kN}\cdot\text{m}$ (hogging). Under the load $M=-20+20(5)=+80\text{ kN}\cdot\text{m}$ (sagging); it returns to zero at the pin. $$\boxed{M_{\max}=+80\text{ kN}\cdot\text{m at }x{=}7,\qquad M_{\min}=-20\text{ kN}\cdot\text{m at }x{=}2}$$

2(b) — Bent frame with an internal hinge

Given. Top beam $T(0,6)\!-\!P(3,6)\!-\!B(12,6)$: $12\text{ kN}$ at the free tip $T$, UDL $8\text{ kN/m}$ over $P\!-\!B$ ($9\text{ m}$), pin at $P$. From $B$ the member drops to $C(12,3)$, runs left to a hinge $H(9,3)$, then drops to a pin $D(9,0)$. Find. Reactions and BMD extremes.

1. Isolate below the hinge. Segment $H\!-\!D$ carries no load and is pinned at $D$; taking moments about $H$ for that free body gives $D_x(3)=0\Rightarrow D_x=0$, so no horizontal force crosses the hinge and (no other horizontal load) $P_x=0$.
2. Reactions. $\Sigma M_{P}=0:\;12(3)-\big(8\cdot9\big)(4.5)+D_y(6)=0\Rightarrow D_y=48\text{ kN}$; $\Sigma F_y:\;P_y=12+72-48=36\text{ kN}$. $$\boxed{P=(0,\,36\text{ kN}),\qquad D=(0,\,48\text{ kN})}$$
3. Bending moment. Along the top beam the overhang gives $-36\text{ kN}\cdot\text{m}$ at $P$, rising to $0$ at $x{=}6$ then falling to the corner value; the column $B\!-\!C$ carries this constant into the frame. The peak magnitude is at the corner $B/C$: $$\boxed{|M|_{\max}=144\text{ kN}\cdot\text{m (hogging, corner }B)}$$ The moment then decays linearly along $C\!-\!H$ to zero at the hinge, and $H\!-\!D$ is momentless.

2(c) — Compound (Gerber) beam

Given. Fixed end at $x{=}0$; internal hinges at $x{=}2$ and $x{=}6$; rollers at $x{=}8$ and $x{=}12$; UDL $5\text{ kN/m}$ over $0\!-\!6\text{ m}$; point load $50\text{ kN}$ at $x{=}10$. Find. Reactions and SFD/BMD extremes.

1. Suspended span first. The span between the hinges ($x{=}2$ to $6$, $L{=}4\text{ m}$, UDL $5$) is simply supported on the two hinges: each hinge reaction $=5(4)/2=10\text{ kN}$, mid-span sag $=wL^2/8=+10\text{ kN}\cdot\text{m}$.
2. Left cantilever. The fixed segment $0\!-\!2$ carries its own UDL plus the $10\text{ kN}$ delivered at the hinge: $R_F=5(2)+10=20\text{ kN}$, $M_F=-\big[5(2)(1)+10(2)\big]=-30\text{ kN}\cdot\text{m}$. $$\boxed{R_F=20\text{ kN}\uparrow,\;\; M_F=30\text{ kN}\cdot\text{m (hogging)}}$$
3. Right span on rollers. Beam $6\!-\!12$ carries the $10\text{ kN}$ hinge load at its left tip and the $50\text{ kN}$ at $x{=}10$: $\Sigma M_{@8}=0\Rightarrow R_{12}=20\text{ kN}$, $R_{8}=40\text{ kN}$. $$\boxed{R_{@8}=40\text{ kN},\qquad R_{@12}=20\text{ kN}}$$
4. Diagram extremes. Hogging $-20\text{ kN}\cdot\text{m}$ over the $x{=}8$ roller and sagging $+40\text{ kN}\cdot\text{m}$ under the $50\text{ kN}$ load; the global extremes are $$M_{\max}=+40\text{ kN}\cdot\text{m},\quad M_{\min}=-30\text{ kN}\cdot\text{m (fixed end)};\quad V_{\max}=+30,\;V_{\min}=-20\text{ kN}.$$