07-Str-A1 · May 2018

Question 2 of 8: Reactions, Shear and Bending Moment Diagrams

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

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.

Given. Three determinate structures. (a) A 12 m beam with a roller 2 m from the left end and a pin at the right end, carrying 10 kN/m over the 2 m overhang only and a 36 kN point load at mid-span of the 10 m span. (b) A bent frame: a horizontal member from a free end A carrying 12 kN, a pin 3 m along at B, 8 kN/m over the next 9 m to the rigid corner D, then 3 m down to E, 3 m back to an internal hinge at F, and a 3 m vertical bar from F to a pin at G. (c) A 12 m compound beam fixed at the left end, with internal hinges 2 m and 6 m along, rollers at 8 m and 12 m, 5 kN/m over the first 6 m and a 50 kN point load at 10 m.

Find. All reaction components, and complete shear-force and bending-moment diagrams for each structure with the sign and magnitude of every maximum and minimum ordinate.

Approach. Each structure is determinate, so global equilibrium plus one condition equation per internal hinge fixes the reactions; the diagrams then follow from $\mathrm{d}V/\mathrm{d}x = -w$ and $\mathrm{d}M/\mathrm{d}x = V$, with the peak sagging moment always at the point of zero shear.

(a) Beam with an overhang carrying the distributed load

1. Replace the distributed load by its resultant. The 10 kN/m ruling covers only the 2 m overhang, so $$W = w\,L_o = 10 \times 2 = 20\ \text{kN}\quad\text{acting at } x = 1\ \text{m},$$ that is, 1 m to the left of the roller.
2. Take moments about the roller to find the pin reaction. With the roller at $x = 2$ m and the pin at $x = 12$ m, $$R_C = \frac{36(7-2) - 20(2-1)}{10} = \frac{180 - 20}{10} = 16.0\ \text{kN}\ (\uparrow).$$
3. Complete vertical equilibrium. Summing vertical forces, $$R_B = W + P - R_C = 20 + 36 - 16 = \boxed{40.0\ \text{kN}\ (\uparrow)},\qquad R_C = \boxed{16.0\ \text{kN}\ (\uparrow)}.$$ There is no horizontal load, so the horizontal component at the pin is zero.
4. Build the shear diagram. Over the overhang the shear falls linearly from zero at the free tip to $-w L_o = -20$ kN just left of the roller; the roller reaction then lifts it to $-20 + 40 = +20$ kN, where it stays until the 36 kN load drops it to $20 - 36 = -16$ kN for the rest of the span. The extreme ordinates are therefore $V_{\max} = +20$ kN and $V_{\min} = -20$ kN.
5. Integrate for the moment diagram. On the overhang $M = -5x^2$, giving the hogging peak $$M(2) = -\tfrac{1}{2}(10)(2)^2 = -20.0\ \text{kN}\cdot\text{m}$$ over the roller. From there the constant $+20$ kN shear raises the moment at 20 kN·m per metre, so it crosses zero at $x = 3$ m and reaches $$M(7) = -20 + 20(7-2) = \boxed{+80.0\ \text{kN}\cdot\text{m}}$$ directly under the point load, after which the $-16$ kN shear returns it linearly to zero at the pin. The diagram is negative (hogging) for $0 \le x < 3$ m and positive (sagging) for $3 < x \le 12$ m.

(b) Bent frame with an internal hinge and a vertical link

The key observation here is that the 3 m bar between the hinge at F and the pin at G is pinned at both ends and carries no load along its length. It is therefore a two-force member, its force acts along its own axis, and because that axis is vertical the reaction it delivers at F has no horizontal component. That single remark reduces the whole problem to three unknowns and three equations.

1. Resolve the distributed load and note the link direction. The 8 kN/m ruling begins at the pin B and ends at the corner D, a length of 9 m, so $$W = 8 \times 9 = 72\ \text{kN at } x = 7.5\ \text{m}.$$ The link F–G is vertical, so its force $S$ at F is vertical and acts 6 m to the right of B.
2. Take moments about the pin at B for the whole frame. The 12 kN tip load sits 3 m to the left of B and the 72 kN resultant 4.5 m to its right: $$6S = 72(4.5) - 12(3.0) = 324 - 36 = 288 \;\Longrightarrow\; S = \boxed{48.0\ \text{kN}\ (\uparrow)}.$$ The pin at G therefore delivers 48.0 kN upward and the link carries 48.0 kN in compression.
3. Finish the reactions. Vertical equilibrium gives $$B_y = 12 + 72 - 48 = \boxed{36.0\ \text{kN}\ (\uparrow)},$$ and since no horizontal force acts anywhere, $B_x = 0$.
4. Shear in the horizontal member. Between A and B the shear is the constant $-12$ kN carried by the tip load; the pin reaction lifts it to $36 - 12 = +24$ kN, after which the distributed load reduces it linearly to $$V(D) = 24 - 8(9) = -48\ \text{kN}.$$ Zero shear occurs at $x = 3 + 24/8 = 6.0$ m.
5. Moments in the horizontal member. Integrating, $$M(B) = -12(3) = -36.0\ \text{kN}\cdot\text{m},\qquad M(x) = 24x - 144 - 4(x-3)^2 \quad (3 \le x \le 12),$$ so that $M(6) = 144 - 144 - 36 + 36 = 0$ and $$M(D) = 24(12) - 144 - 4(9)^2 = \boxed{-144.0\ \text{kN}\cdot\text{m}}.$$ The moment is hogging everywhere along A–D; it merely touches zero at the point of zero shear, $x = 6$ m, so $M_{\max} = 0$ and $M_{\min} = -144$ kN·m.
6. The rest of the frame. Cutting anywhere in the vertical member D–E and taking the piece below the cut, which is loaded only by the 48 kN reaction at G acting 3 m to its left, gives a constant moment of $$M_{DE} = 48 \times 3 = 144\ \text{kN}\cdot\text{m}$$ with zero shear and 48 kN of axial compression. In the horizontal member E–F the shear is a constant 48 kN and the moment rises linearly from zero at the hinge F to 144 kN·m at E, which is the continuity check on the corner.

(c) Compound beam with two internal hinges

With hinges at 2 m and 6 m the beam separates into three free bodies. The middle piece is pinned at both ends and so behaves as a simply supported span, and once its end reactions are known the outer pieces are elementary.

1. Detach the piece between the hinges. The 4 m length from 2 m to 6 m carries 5 kN/m and is pinned at both ends, so each end delivers $$H = \frac{wL}{2} = \frac{5 \times 4}{2} = 10.0\ \text{kN}.$$
2. Fixed end. The 2 m cantilever carries its own 10 kN of distributed load plus the 10 kN hinge reaction at its tip: $$V_{\text{fixed}} = 10 + 10 = \boxed{20.0\ \text{kN}\ (\uparrow)},\qquad M_{\text{fixed}} = 10(1.0) + 10(2.0) = \boxed{30.0\ \text{kN}\cdot\text{m (hogging)}}.$$
3. Right-hand piece. It carries the 10 kN hinge force at 6 m and the 50 kN load at 10 m on rollers at 8 m and 12 m. Taking moments about the 12 m roller, $$4R_8 = 10(12-6) + 50(12-10) = 60 + 100 = 160 \;\Longrightarrow\; R_8 = \boxed{40.0\ \text{kN}\ (\uparrow)},$$ and vertical equilibrium then gives $R_{12} = 10 + 50 - 40 = \boxed{20.0\ \text{kN}\ (\uparrow)}$.
4. Shear diagram. Over the loaded first 6 m the shear falls linearly from $+20$ kN at the fixed end to $-10$ kN, crossing zero at $x = 4$ m; it holds at $-10$ kN to the first roller, jumps to $+30$ kN there, falls to $-20$ kN under the 50 kN load and is closed by the 20 kN reaction at the far end. Hence $V_{\max} = +30$ kN and $V_{\min} = -20$ kN.
5. Moment diagram. Over the first 6 m $$M(x) = -30 + 20x - 2.5x^2,$$ which is zero at $x = 2$ and $x = 6$ m — exactly the two hinges, as it must be — and peaks at $M(4) = +10.0$ kN·m. Beyond the second hinge the diagram is piecewise linear: $M(8) = -20.0$ kN·m over the first roller, then $$M(10) = -20 + 30(2) = \boxed{+40.0\ \text{kN}\cdot\text{m}}$$ under the point load, returning to zero at the far roller. The moment is negative for $0 \le x < 2$ m and for $6 < x < 8.67$ m, and positive elsewhere; $M_{\min} = -30.0$ kN·m at the fixed end.