16-Civ-A1 Elementary Structural Analysis · Undated paper

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

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

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

Part (a) — simple span with a loaded overhang

Given. A straight beam 15 m long. A 72 kN downward point load sits at the free left tip; a pin support is 3 m to its right; a 90 kN downward point load acts 6 m beyond the pin; a roller closes the beam a further 6 m along, at the right-hand end.

Find. Both reactions, and the maximum positive and negative ordinates of the shear and bending-moment diagrams.

Approach. The beam is determinate ($r = 3$), so take moments about the pin for the roller reaction, use vertical equilibrium for the pin reaction, then build the diagrams by walking from the left tip.

1. Roller reaction, by moments about the pin at $x = 3\text{ m}$. Taking clockwise moments as positive, the tip load sits 3 m to the left of the pin and therefore acts anticlockwise: $$\sum M_{\text{pin}} = -72(3) + 90(6) - R_C(12) = 0$$ $$R_C = \frac{-216 + 540}{12} = \boxed{27\ \text{kN} \uparrow}$$
2. Pin reaction, by vertical equilibrium. $$R_B = 72 + 90 - 27 = \boxed{135\ \text{kN} \uparrow}$$
3. Shear-force diagram. Walking from the left tip, the shear is $-72\text{ kN}$ across the overhang, jumps by $+135$ at the pin to $+63\text{ kN}$, drops by $90$ under the second load to $-27\text{ kN}$, and is closed to zero by the roller. There is no distributed load, so every segment is constant: $$V_{\max}^{+} = \boxed{+63\ \text{kN}}, \qquad V_{\max}^{-} = \boxed{-72\ \text{kN}}$$
4. Bending-moment diagram. Integrating the shear (or taking the moment of everything to the left of each section) gives a straight line in each segment. At the pin the overhang hogs the beam: $$M(3) = -72(3) = \boxed{-216\ \text{kN}\cdot\text{m}\ \text{(hogging)}}$$ and under the 90 kN load the beam sags: $$M(9) = -72(9) + 135(6) = -648 + 810 = \boxed{+162\ \text{kN}\cdot\text{m}\ \text{(sagging)}}$$ From the roller the moment closes at zero, $M(15) = 162 - 27(6) = 0$, which confirms the reactions.
5. Point of contraflexure. Between the pin and the 90 kN load the moment climbs at the constant rate $V = 63\text{ kN}$, so it crosses zero at $$x = 3 + \frac{216}{63} = \boxed{6.43\ \text{m from the left tip}}$$ The diagram is negative (hogging) from the tip to this point and positive (sagging) from there to the roller.

Part (b) — 20 m compound (Gerber) beam with two internal hinges

Given. Measuring from the left end:

A uniformly distributed load of 4 kN/m covers 0–7 m and, symmetrically, 13–20 m. Total load $= 4(14) + 40 = 96\text{ kN}$. The structure and its loading are symmetric about $x = 10\text{ m}$.

Find. All four reactions and the extreme ordinates of the shear and bending-moment diagrams.

Approach. With $r = 5$ and $c = 2$ the beam is determinate. Decompose it at the hinges into a suspended central span carrying two simply supported end spans, solve each end span first, then carry its hinge reaction onto the central span as an applied load.

1. Left end span, 0 to 5 m. This piece is supported by the pin at $x = 0$ and by the hinge at $x = 5$, and carries $4 \times 5 = 20\text{ kN}$ of UDL. Being symmetric, $$R_A = \tfrac{1}{2}(20) = \boxed{10\ \text{kN} \uparrow}$$ and the hinge delivers 10 kN downward onto the central span. By symmetry the right end span (15 to 20 m) behaves identically, so $R_{20} = 10\text{ kN} \uparrow$ and the hinge at $x = 15$ delivers 10 kN downward.
2. Central span, 5 to 15 m. It carries the two 10 kN hinge forces at its ends, the 40 kN point load at midspan, and the two overhanging strips of UDL (5–7 m and 13–15 m, $4 \times 2 = 8\text{ kN}$ each). The total is $$W = 2(10) + 2(8) + 40 = 76\ \text{kN}$$ and, because everything is symmetric about $x = 10$, $$R_7 = R_{13} = \tfrac{1}{2}(76) = \boxed{38\ \text{kN} \uparrow}$$
3. Check global equilibrium. $$10 + 38 + 38 + 10 = 96\ \text{kN} = 4(14) + 40 \quad\checkmark$$
4. Shear-force diagram. From the pin the shear falls linearly under the UDL from $+10$ at $x=0$ through $-10$ at the hinge to $-18\text{ kN}$ just left of the roller at 7 m; the roller lifts it to $+20\text{ kN}$, and with no load between 7 and 10 m it stays there until the 40 kN load takes it to $-20\text{ kN}$. The right half mirrors the left. Hence $$V_{\max}^{+} = \boxed{+20\ \text{kN}}, \qquad V_{\max}^{-} = \boxed{-20\ \text{kN}}$$
5. Bending-moment diagram. In the end spans the moment is the simple-beam parabola $$M(x) = 10x - 2x^{2} \quad (0 \le x \le 5)$$ which peaks at $x = 2.5\text{ m}$ with $M = +12.5\text{ kN}\cdot\text{m}$ and returns to zero at the hinge, as it must. Continuing past the hinge the same expression gives the hogging peak over the roller, $$M(7) = 10(7) - 2(7)^{2} = \boxed{-28\ \text{kN}\cdot\text{m}}$$ and from there the shear of $+20\text{ kN}$ carries the moment up to midspan: $$M(10) = -28 + 20(3) = \boxed{+32\ \text{kN}\cdot\text{m}}$$
6. Confirm the hinges. A hinge cannot carry moment, so $M(5) = M(15) = 0$ is a necessary condition; both are satisfied exactly, as is $M(20) = 0$ at the free-ended roller. The diagram is sagging in the two end spans and across the middle 8 m, and hogging in the two 3 m zones straddling the interior rollers.

Part (c) — stepped beam loaded by a horizontal couple

Given. A pin support at A carries a horizontal member 4 m long to E; the beam then steps down 1 m through a vertical riser E–F and runs 2 m along the lower level to a roller at G. A 20 kN horizontal force acts to the left at E (the top of the riser) and an equal 20 kN force acts to the right at F (the foot of the riser).

Find. The reactions and the bending-moment diagram with its extreme ordinates.

Approach. Recognise the two horizontal forces as a couple, so the horizontal reaction vanishes and the vertical reactions must form an equal and opposite couple; then walk the moment along the developed axis A–E–F–G.

1. Horizontal equilibrium. The two 20 kN forces cancel, and the roller at G is horizontal-free, so $$A_x = 20 - 20 = \boxed{0}$$
2. The applied couple. The forces are equal, opposite and offset by the 1 m height of the riser: $$M_{\text{couple}} = 20 \times 1 = \boxed{20\ \text{kN}\cdot\text{m}\ \text{(anticlockwise)}}$$
3. Vertical reactions. The only way a couple can be resisted by two vertical forces 6 m apart is by an equal and opposite couple, so taking moments about A, $$20 + 6R_G = 0 \;\Rightarrow\; R_G = -\tfrac{20}{6}$$ $$R_G = \boxed{3.33\ \text{kN} \downarrow}, \qquad A_y = \boxed{3.33\ \text{kN} \uparrow}$$ The roller pulls down: this is a real result, not a sign slip, and it is the reason a step loaded by a couple needs a hold-down detail.
4. Upper member A–E. With no transverse load the shear is the constant $A_y = 3.33\text{ kN}$ and the moment grows linearly: $$M_E = 3.33 \times 4 = \boxed{13.33\ \text{kN}\cdot\text{m}}$$ The axial force in this member is zero, because the 20 kN at E is carried away as shear in the riser.
5. Lower member F–G. Working back from the roller, the same argument gives a constant shear of $3.33\text{ kN}$ and $$M_F = 3.33 \times 2 = \boxed{6.67\ \text{kN}\cdot\text{m}}$$ bending the lower member the opposite way to the upper one.
6. The riser E–F. Cutting the riser and taking the lower portion, the vertical member carries a shear of 20 kN (it alone equilibrates the 20 kN at F) and an axial tension of 3.33 kN. Its moment therefore changes by $20 \times 1 = 20\text{ kN}\cdot\text{m}$ over the 1 m height, which is exactly the jump from $+13.33$ at E to $-6.67$ at F: $$13.33 - (-6.67) = 20\ \text{kN}\cdot\text{m} \quad\checkmark$$ The moment passes through zero $6.67/20 = 0.33\text{ m}$ above F.