Question 2 of 8: Reactions, shear and bending-moment diagrams for three structures (18 marks)

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) Overhanging beam

Given. Pin at A (x=2 m), roller at B (x=8 m); UDL $$w=6\text{ kN/m}$$ over the left 8 m (0–8 m, i.e. the 2 m left overhang plus the 6 m mid-span); a 12 kN point load at the right tip (x=10 m).

Find. Reactions, and the shear/moment envelope.

Approach. Sum moments about A to get $$R_B$$, then $$\Sigma F_y=0$$ for $$R_A$$; integrate the load to get V and M, locating the zero-shear point for the interior peak.

1. Reactions. UDL resultant $$=6(8)=48\text{ kN}$$ at x=4 m. $$\Sigma M_A=0:\ R_B(6)=48(2)+12(8)\Rightarrow R_B=\dfrac{96+96}{6}=\boxed{32\text{ kN}}$$ $$\Sigma F_y=0:\ R_A=48+12-32=\boxed{28\text{ kN}}$$
2. Shear. $$V(0)=0,\quad V(2^-)=-6(2)=-12,\quad V(2^+)=-12+28=16$$ V decreases linearly (slope $$-6$$) to $$V(8^-)=16-6(6)=-20$$, then $$V(8^+)=-20+32=12$$, constant to the tip where it drops by 12 to zero. Maximum shear magnitude $$=\boxed{20\text{ kN}}$$ just left of B.
3. Moment. Zero-shear point: $$16-6(x-2)=0\Rightarrow x=4.67\text{ m}$$. $$M(2)=-6(2)(1)=\boxed{-12\text{ kN}\cdot\text{m}}\text{ (hogging, overhang)}$$ $$M(4.67)=-6(4.67)^2/2+28(2.67)=\boxed{+9.33\text{ kN}\cdot\text{m}}\text{ (sagging, span peak)}$$ $$M(8)=-12(6)+28(6)-6(6)(3) = \boxed{-24\text{ kN}\cdot\text{m}}\text{ (hogging, at B — governs)}$$ $$M(10)=0$$ (free tip).

(b) Bent cantilever/propped frame

Given. Free tip at x=0 with a 20 kN point load; roller at x=2 m; UDL $$4\text{ kN/m}$$ over 2–12 m; a rigid corner at x=12 m turning into a 10 m vertical leg to a pin base.

Find. Reactions and the shear/moment diagrams for the beam and the column.

Approach. Take moments about the pin base to solve the single roller reaction (this is a determinate L-frame: roller + pin = 3 reactions), then sweep the beam and column for V and M.

1. Reactions. UDL resultant $$=4(10)=40\text{ kN}$$ at x=7 m. Taking moments about the pin base (12,−10), only the horizontal lever arms of the vertical loads matter: $$R_{roll}(2-12) = 20(0-12)+40(7-12)$$ $$-10R_{roll}=-240-200\Rightarrow R_{roll}=\boxed{44\text{ kN}}$$ $$\Sigma F_y=0:\ R_{y,pin}=20+40-44=\boxed{16\text{ kN}},\qquad R_{x,pin}=0$$ (no horizontal load anywhere in the structure).
2. Beam moments. $$M(0)=0,\quad M(2^-)=-20(2)=-40\text{ kN}\cdot\text{m}$$ (hogging at the roller). Past the roller, $$M(x)=-20x+44(x-2)-4\tfrac{(x-2)^2}{2}$$ crosses zero at x=4 m, then sags to a local peak where $$V=0$$: $$V=-20+44-4(x-2)=0\Rightarrow x=8\text{ m},\ \ M(8)=\boxed{+32\text{ kN}\cdot\text{m}}$$ and returns to $$M(12)=0$$ at the corner.
3. Column. Because $$R_{x,pin}=0$$ and the pin sits directly under the corner, the column carries zero bending moment throughout — pure axial compression $$=R_{y,pin}=\boxed{16\text{ kN}}$$. (A quick check: any vertical force acting exactly on the column's own vertical line has zero moment arm about any point on that line.)

(c) Symmetric trapezoidal (gable) frame

Given. Pin base at A; a sloped leg rising 12 m over a 5 m run to the top-left corner; a 16 m top beam (UDL 4.8 kN/m, plus a 24 kN point load at each top corner); a mirrored sloped leg down to a roller base B, 26 m total width.

Find. Reactions and the M-diagram, including the sloped legs.

Approach. The frame is a 3-member open chain (no hinge) with one pin and one roller — exactly 3 reactions for 3 equilibrium equations, so it is determinate despite the sloped legs. Use symmetry to shortcut the vertical reactions.

1. Reactions. Total load $$=24+24+4.8(16)=124.8\text{ kN}$$, symmetric about the centreline, so $$R_A=R_B=\boxed{62.4\text{ kN}}$$. No horizontal load anywhere $$\Rightarrow R_{x,A}=0$$.
2. Left leg (pin to top-left corner). Because $$R_{x,A}=0$$, the reaction is purely vertical; resolving it into components along/normal to the 5:12:13 leg gives an axial force $$=62.4(12/13)=\boxed{57.6\text{ kN (compression)}}$$ and a constant transverse shear $$=62.4(5/13)=\boxed{24.0\text{ kN}}$$. Moment grows linearly from 0 at the pin to $$M=\boxed{312\text{ kN}\cdot\text{m}}$$ at the top corner (taking moments of R_A about points along the leg).
3. Top beam. Moment continues from 312 kN·m (sagging) at the left corner, rises under the UDL to a peak at midspan (x=13 m from A's corner, by symmetry the point of zero shear): $$V=62.4-24-4.8(x-5)=0\Rightarrow x=13\text{ m (from A)}$$ $$M_{peak}=\boxed{465.6\text{ kN}\cdot\text{m}}$$ and falls symmetrically back to 312 kN·m at the right corner.
4. Right leg. Mirrors the left leg: moment decays linearly from 312 kN·m at the top corner to 0 at the roller; axial 57.6 kN compression, transverse shear 24.0 kN.