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

Question 2: Reactions, shear and bending-moment diagrams (24 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.

All three parts use the same sagging-positive convention: a bending moment is plotted positive when it puts the lower fibre of the member in tension, and the shear is the sum of the upward forces to the left of the section, so that $V = \mathrm{d}M / \mathrm{d}x$ everywhere. For the bent member in part (c) the diagrams are developed: the abscissa is the distance measured along the unrolled member chain, not a horizontal projection, which keeps the moment continuous across the rigid corners and makes the shear the true slope of the moment curve.

(a) Overhung beam with two point loads

Given. A straight beam 15 m long. A 72 kN downward load acts at the free left tip; the pin support A sits 3 m from that tip; a 90 kN downward load acts 6 m beyond the pin, at 9 m from the tip; and the roller support B closes the beam a further 6 m along, at 15 m.

Find. The two reactions, and the shear and bending-moment diagrams with every extreme ordinate labelled.

Approach. Take moments about the pin to get the roller reaction, close vertical equilibrium for the pin reaction, then walk the beam from the free tip accumulating shear and moment.

1. Take moments about the pin A. The 72 kN load sits on the overhang, 3 m to the left of A, so it opposes the 90 kN load: $$\sum M_A = 0:\qquad 90(6) - 72(3) - V_B(12) = 0$$ $$V_B = \frac{540 - 216}{12} = \boxed{27\ \text{kN (up)}}$$
2. Close vertical equilibrium. With the roller settled, $$\sum F_y = 0:\qquad V_A = 72 + 90 - 27 = \boxed{135\ \text{kN (up)}}$$ The pin therefore carries five times the roller, which is the signature of a long overhang loaded near its tip.
3. Build the shear diagram from the tip. Walking rightwards and summing upward forces: $V = -72\ \text{kN}$ all along the overhang; the pin adds $+135$, giving $V = +63\ \text{kN}$ from A to the 90 kN load; that load subtracts 90, leaving $V = -27\ \text{kN}$ to the roller, where the final $+27$ closes the diagram. The extreme ordinates are therefore $$V_{\max}^{+} = 63\ \text{kN},\qquad V_{\max}^{-} = -72\ \text{kN}.$$
4. Integrate for the bending moment. Because the loading is a set of point forces, the moment diagram is a chain of straight lines whose slopes are the shear ordinates. At the pin, $$M_A = -72(3) = \boxed{-216\ \text{kN}\cdot\text{m}}$$ which is hogging — the overhang lifts the beam over its support. Continuing at slope $+63$ for 6 m, $$M(9\ \text{m}) = -216 + 63(6) = \boxed{+162\ \text{kN}\cdot\text{m}}$$ sagging, under the 90 kN load. The last reach falls at slope $-27$ for 6 m and lands on $162 - 162 = 0$ at the roller, which is the arithmetic check that both reactions are right.
5. Locate the point of contraflexure. The moment changes sign once, between A and the 90 kN load, where $$-216 + 63(x - 3) = 0 \quad\Rightarrow\quad x = 3 + \frac{216}{63} = 6.43\ \text{m}$$ measured from the free tip. To the left of that point the diagram is negative (hogging, tension on top); to the right it is positive (sagging, tension on the bottom), and the question asks for exactly that statement.

(b) Gerber beam with two internal hinges

Given. A 20 m beam. Reading the dimension string from the left: pin at 0, internal hinge at 5 m, roller at 7 m, a 40 kN downward point load at 10 m, roller at 13 m, internal hinge at 15 m and roller at 20 m. A uniform 4 kN/m runs from the left end to the first roller (0–7 m) and again from the second-last roller to the right end (13–20 m); the ends of the hatched load blocks, not the dimension string, fix those extents. The beam and its loading are symmetric about mid-span.

Find. The four reactions, and the shear and bending-moment diagrams with the extreme ordinates labelled.

Approach. Check determinacy, then decompose at the hinges: the two end reaches behave as simple spans that hand a known force to the central span, which is then an ordinary beam on two supports with overhangs.

1. Confirm the beam is determinate before decomposing it. There are $r = 2 + 1 + 1 + 1 = 5$ restraint components against three equations of statics plus one condition equation per hinge, so $$i = r - (3 + c) = 5 - (3 + 2) = 0,$$ and the beam can be solved by statics alone. Horizontal equilibrium is trivial — the pin is the only horizontal restraint and there is no horizontal load, so $H = 0$ throughout.
2. Detach the left reach, 0 to 5 m. Cut at the hinge. The piece carries the pin at one end and the hinge force at the other, and it is loaded by $4(5) = 20$ kN of uniform load, so it is simply a 5 m simple span: $$V_0 = \tfrac{1}{2}(4)(5) = \boxed{10\ \text{kN (up)}}$$ and an equal 10 kN passes down through the hinge onto the central reach. By the symmetry of the beam the right reach, 15 to 20 m, gives the identical pair: $V_{20} = 10$ kN up and 10 kN delivered down at the second hinge.
3. Solve the central reach, 5 to 15 m. It is supported by the rollers at 7 and 13 m and carries, from left to right: 10 kN at 5 m from the hinge, $4(2) = 8$ kN of uniform load centred at 6 m, the 40 kN load at 10 m, 8 kN centred at 14 m and 10 kN at 15 m. Taking moments about the roller at 7 m, $$\sum M_{7} = 0:\quad -10(2) - 8(1) + 40(3) + 8(7) + 10(8) = 6\,V_{13}$$ $$V_{13} = \frac{228}{6} = \boxed{38\ \text{kN}} ,\qquad V_{7} = 76 - 38 = \boxed{38\ \text{kN}}$$ the two being equal, as the symmetry demands. The four reactions sum to $10 + 38 + 38 + 10 = 96$ kN, which matches the applied total $4(7) + 40 + 4(7) = 96$ kN exactly.
4. Assemble the shear diagram. Over 0 to 7 m, $V = 10 - 4x$, so it falls linearly from $+10$ at the pin through zero at $x = 2.5$ m to $-18$ kN just left of the roller; the roller then lifts it to $+20$ kN, which holds to the 40 kN load. That load drops it to $-20$ kN, held to the second roller, which lifts it to $+18$ kN; from there the uniform load takes it linearly down through zero at 17.5 m to $-10$ kN at the right end. Hence $$V_{\max}^{+} = 20\ \text{kN},\qquad V_{\max}^{-} = -20\ \text{kN}.$$
5. Assemble the bending-moment diagram. On the uniformly loaded reaches the curve is parabolic, elsewhere straight. Working from the left, $$M(x) = 10x - 2x^{2} \quad (0 \le x \le 7),$$ which peaks at $x = 2.5$ m with $M = +12.5$ kN·m, returns to zero at the hinge ($x = 5$, the check that the decomposition was right) and reaches $\boxed{-28\ \text{kN}\cdot\text{m}}$ over the roller at 7 m. The straight reach to the load climbs at $+20$ kN for 3 m to $$M(10) = -28 + 20(3) = \boxed{+32\ \text{kN}\cdot\text{m}},$$ then falls at $-20$ kN back to $-28$ kN·m over the roller at 13 m. The mirror image completes the diagram: zero at the second hinge and $+12.5$ kN·m at 17.5 m.
6. Name the positive and negative regions. The diagram is sagging (positive) on 0–5 m, on roughly 8.6–11.4 m either side of the point load, and on 15–20 m; it is hogging (negative) over each interior roller, from 5 m to about 8.6 m and from about 11.4 m to 15 m. Every zero of the moment diagram either sits at an end support or coincides exactly with a hinge, which is what a correctly decomposed Gerber beam must produce.

(c) Bent member carrying a pair of equal, opposite horizontal loads

Given. A member that runs 4 m horizontally from a pin at A to a bend at B, drops 1 m vertically to a second bend at C, then runs 2 m horizontally to a roller at D on a horizontal bearing. A 20 kN horizontal force acts leftwards at B and a second 20 kN horizontal force acts rightwards at C. Both corners are rigid.

Find. The reactions at A and D, and the developed shear and bending-moment diagrams along the chain A–B–C–D.

Approach. Recognise the applied loading as a pure couple, balance it with a couple formed by the two vertical reactions, then plot the diagrams against distance measured along the unrolled member.

1. Reduce the loading. The two 20 kN forces are equal, opposite and separated by the 1 m height of the riser, so their resultant force is zero and they are statically equivalent to a couple of magnitude $$M_{\text{applied}} = 20(1) = 20\ \text{kN}\cdot\text{m}$$ acting anticlockwise on the member. Horizontal equilibrium follows immediately: $$\sum F_x = 0:\qquad A_x = 20 - 20 = \boxed{0}$$ so the pin carries no horizontal thrust at all — a result worth stating, because it kills the axial force in the upper reach.
2. Balance the couple with the reactions. A couple can only be resisted by a couple. The roller at D delivers a vertical force at 6 m horizontally from the pin, so $$\sum M_A = 0:\qquad 20 + 6\,D_y = 0 \quad\Rightarrow\quad D_y = -\tfrac{20}{6} = \boxed{-3.33\ \text{kN}}$$ $$\sum F_y = 0:\qquad A_y = -D_y = \boxed{+3.33\ \text{kN (up)}}$$ The negative sign at D is real and is the interesting feature of this sub-part: the roller must hold the member down with 3.33 kN. A bearing that can only push would let the member lift, so the idealisation assumes a two-way (tied) roller.
3. Reach A–B (4 m of horizontal member). With $A_x = 0$ the only action carried in is the 3.33 kN vertical, so the shear is constant at $V = +3.33$ kN and the moment grows linearly: $$M_B = 3.33(4) = \boxed{+13.33\ \text{kN}\cdot\text{m}}$$ sagging. The axial force in this reach is zero.
4. Riser B–C (1 m). At B the 20 kN leftward force enters. It is transverse to the vertical riser, so it becomes the riser's shear: $V = -20$ kN, constant over the 1 m. The moment therefore changes by $-20(1) = -20$ kN·m across the riser, $$M_C = 13.33 - 20 = \boxed{-6.67\ \text{kN}\cdot\text{m}},$$ and passes through zero $13.33 / 20 = 0.667$ m below B. Note that the difference between the two corner moments is exactly the applied couple, which is the cleanest available check on the reactions.
5. Reach C–D (2 m). The 20 kN rightward force applied at C cancels the shear that the riser delivers, so the lower horizontal reach carries no axial force; it carries only the 3.33 kN vertical, giving a constant $V = +3.33$ kN and a moment that climbs linearly from $-6.67$ kN·m at C to $$M_D = -6.67 + 3.33(2) = 0$$ at the roller, as a roller must. The developed diagrams therefore close on themselves.
6. Report the extremes and their signs. The largest shear is the 20 kN in the riser; the largest moment is $+13.33$ kN·m at the upper bend, sagging, and the largest moment of the other sign is $-6.67$ kN·m at the lower bend, hogging. The moment diagram is positive from A to a point 0.667 m below B and negative from there to D.