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

Question 2: Reactions, shear and bending-moment diagrams for three structures (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.

(a) Beam with a left overhang

Given. A 12 m beam: free tip at x=0; UDL $$w=10\text{ kN/m}$$ over the 2 m overhang (0–2 m); roller at x=2 m; a 36 kN point load at x=7 m; pin at x=12 m.

Find. Reactions, and the shear/moment envelope.

Approach. Sum moments about the roller to get the pin reaction, then $$\Sigma F_y=0$$ for the roller reaction; integrate the load along the length to get V and M, locating the zero-shear point for the interior peak.

1. Reactions. UDL resultant $$=10(2)=20\text{ kN}$$ at x=1 m (1 m to the left of the roller). $$\Sigma M_{roller}=0:\ R_{pin}(10)=36(5)+20(-1)\Rightarrow R_{pin}=\dfrac{180-20}{10}=\boxed{16\text{ kN}}$$ $$\Sigma F_y=0:\ R_{roller}=20+36-16=\boxed{40\text{ kN}}$$
2. Shear. $$V(0)=0,\quad V(2^-)=-10(2)=-20\text{ kN},\quad V(2^+)=-20+40=+20\text{ kN}$$ constant to x=7 m, where the 36 kN load drops it to $$20-36=-16\text{ kN}$$, constant to the pin. Maximum shear magnitude $$=\boxed{20\text{ kN}}$$ (either side of the roller).
3. Moment. $$M(2)=-10(2)(1)=\boxed{-20\text{ kN}\cdot\text{m}}$$ (hogging, at the overhang support — this is the tip cantilever moment). $$M(7)=-20+20(5)=\boxed{+80\text{ kN}\cdot\text{m}}$$ (sagging, under the point load — governs). $$M(12)=80-16(5)=0$$ ✓ (pin).

(b) Bent frame with a cantilevered tip and a hinge

Given. A 12 kN tip load at the free left end; a pin at x=3 m; UDL $$8\text{ kN/m}$$ over the 9 m span from the pin to a rigid corner (x=12 m); the beam then turns down 3 m to an internal hinge, across 3 m, and down another 3 m to a base pin.

Find. Both pin reactions, the hinge force, and the beam moment envelope.

Approach. The hinge transmits shear but no moment: isolate the lower member (hinge–corner–base pin, no load on it) first — with only 2 external forces acting on it (the hinge force and the base reaction), equilibrium forces those two forces to be equal, opposite and collinear along the 3 m×3 m diagonal between the hinge and the base pin. Then close the upper member (which carries the tip load and the UDL) about the pin A.

1. Lower member (hinge to base pin, no load between them). With the base pin 3 m below and 3 m to the side of the hinge, equilibrium requires $$H_x=H_y=H$$ (the hinge force and the base reaction act along the 45° line joining them).
2. Upper member (tip – pin A – corner – hinge). UDL resultant $$=8(9)=72\text{ kN}$$ at mid-span (x=7.5 m from the tip, 4.5 m right of A). Taking moments about A (3,0), with the hinge reaction $$(-H,-H)$$ acting at the hinge (12,−3), i.e. (9,−3) from A: $$\Sigma M_A=0:\ 12(-3)(1)-72(4.5) -\big[9(-H)-(-3)(-H)\big]=0$$ $$36-324-12H=0\Rightarrow H=\boxed{-24\text{ kN}}$$ (24 kN, directed opposite the assumed tension sense).
3. Reactions. $$R_{x,A}=H=\boxed{-24\text{ kN}}\ (\text{24 kN acting left}),\qquad R_{y,A}=12+72+H=\boxed{60\text{ kN}}$$ Base pin: $$S_x=-H=\boxed{24\text{ kN}},\qquad S_y=-H=\boxed{24\text{ kN}}$$ Check: $$\Sigma F_x=-24+24=0,\ \Sigma F_y=60+24-12-72=0$$ ✓
4. Moment along the upper (horizontal) member. Tip to A: $$M(3^-)=-12(3)=\boxed{-36\text{ kN}\cdot\text{m}}$$ (hogging, cantilever). Past A, shear jumps to $$-12+60=48\text{ kN}$$ and falls under the UDL: zero shear at $$x=3+48/8=9\text{ m}$$, giving the span peak $$M(9)=-36+48(6)-8(6)^2/2=\boxed{+108\text{ kN}\cdot\text{m}}$$ (sagging — governs). At the corner (x=12 m): $$M=-36+48(9)-8(9)^2/2=\boxed{+72\text{ kN}\cdot\text{m}}$$, carried unchanged around the rigid corner into the drop, decaying to zero at the hinge (no load on the vertical drop).

(c) Fixed-base beam released to determinacy by two hinges

Given. Fixed at x=0; UDL $$5\text{ kN/m}$$ over 0–6 m; hinges at x=2 m and x=6 m; roller at x=8 m; a 50 kN point load at x=10 m (midspan of the last panel); roller at x=12 m.

Find. The fixed-end moment, both roller reactions, and the moment envelope (a compound/Gerber beam).

Approach. Two hinges exactly cancel the 2 extra restraints of a fixed+roller+roller beam ($$i=5-3-2=0$$), so solve the hinge-to-hinge “suspended” span first (it carries only its own UDL and has no other restraint), then work outward to the cantilever stub and the roller-supported tail.

1. Suspended span (hinge–hinge, x=2 to x=6, UDL only). A simple 4 m span carrying $$5(4)=20\text{ kN}$$ centred at its own mid-span: by symmetry each hinge transmits $$V_h=\boxed{10\text{ kN}}$$ down to its neighbour. Its own mid-span moment (a check): $$M=10(2)-5(2)^2/2=\boxed{+10\text{ kN}\cdot\text{m}}$$.
2. Fixed stub (x=0 to hinge at x=2). Carries its own UDL $$5(2)=10\text{ kN}$$ at x=1 m, plus the 10 kN handed down from the hinge at x=2 m. $$R_{y0}=10+10=\boxed{20\text{ kN}}$$ $$M_0=10(1)+10(2)=\boxed{30\text{ kN}\cdot\text{m}}$$ (hogging — check: $$M(\text{hinge})=-30+20(2)-5(2)^2/2=0$$ ✓).
3. Tail (hinge at x=6 to the free end at x=12, carrying the 10 kN handed down, both rollers, and the 50 kN load at x=10). $$\Sigma M_{(x=8)}=0:\ R_{12}(4)=50(2)+10(-2)\Rightarrow R_{12}=\dfrac{100-20}{4}=\boxed{20\text{ kN}}$$ $$\Sigma F_y=0:\ R_8=10+50-20=\boxed{40\text{ kN}}$$
4. Tail moments. $$M(8)=-10(2)=\boxed{-20\text{ kN}\cdot\text{m}}$$ (hogging, over the interior roller). $$M(10)=-20+40(2)=\boxed{+40\text{ kN}\cdot\text{m}}$$ (sagging, under the 50 kN load — governs). $$M(12)=40-20(2)=0$$ ✓.