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

Question 2: Reactions, shear force and bending moment diagrams for three determinate 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.

Given. Three determinate structures, all dimensions read from the figures.

Find. The reactions, then the shear force and bending moment diagrams with the maximum positive and maximum negative ordinate of each.

Approach. Take global moments about one support to get the other reaction in parts (a) and (c); in part (b) release the structure at the internal hinge first, because the hinge supplies the fourth equation the four reaction components need. Then walk the shear from the free end, locate every point of zero shear, and evaluate the moment there and at every support and load point.

1. Part (a) — reactions. Measure $x$ from the free left end, so the pin is at $x = 3$ m and the roller at $x = 15$ m. Taking moments about the pin, with the 15 m run of 4 kN/m acting as 60 kN at $x = 7.5$ m, $$R_{15}\,(15 - 3) = 10\,(0 - 3) + 60\,(7.5 - 3) + 25\,(9 - 3)$$ $$R_{15} = \frac{-30 + 270 + 150}{12} = \boxed{32.5\ \text{kN}\ \uparrow}$$ and vertical equilibrium gives $$R_{3} = 10 + 60 + 25 - 32.5 = \boxed{62.5\ \text{kN}\ \uparrow}$$
2. Part (a) — shear. Starting from the free end the shear falls under the 10 kN load and then under the distributed load at 4 kN per metre, reaching $-10 - 4(3) = -22$ kN just left of the pin. The pin adds 62.5 kN, so the shear jumps to $+40.5$ kN, then falls to $40.5 - 4(6) = +16.5$ kN just left of the 25 kN load, drops to $-8.5$ kN just right of it, and reaches $-8.5 - 4(6) = -32.5$ kN at the roller, where the 32.5 kN reaction closes the diagram. The extreme ordinates are $$V_{\max}^{+} = \boxed{+40.5\ \text{kN}}, \quad V_{\max}^{-} = \boxed{-32.5\ \text{kN}}$$
3. Part (a) — bending moment. Over the overhang $M = -10x - 2x^{2}$, which reaches $-30 - 18 = -48$ kN·m at the pin. That is the largest hogging value, because the shear does not pass through zero anywhere in the 3 m to 9 m reach (it runs from $+40.5$ to $+16.5$ kN) nor in the 9 m to 15 m reach (from $-8.5$ to $-32.5$ kN). The moment therefore peaks at the 25 kN load, where the shear changes sign: $$M(9) = -10(9) - 2(9)^{2} + 62.5(6) = -90 - 162 + 375 = \boxed{+123.0\ \text{kN}\cdot\text{m}}$$ Between the two the moment vanishes at $-2x^{2} + 52.5x - 187.5 = 0$, that is at $x = 4.264$ m, which is the point of contraflexure.
4. Part (b) — use the hinge to unlock the reactions. Four reaction components (one at each roller, two at the pin) face three equilibrium equations, and the internal hinge at $x = 4$ m supplies the fourth. Isolate the 4 m piece to the left of the hinge, which carries $4(4) = 16$ kN of distributed load at its own mid-length, and take moments about the hinge: $$R_{0}\,(4) - 16\,(2) = 0 \;\Rightarrow\; R_{0} = \boxed{8.0\ \text{kN}\ \uparrow}$$
5. Part (b) — the remaining two reactions. Carry the hinge force into the right-hand piece. That piece runs from the hinge to the right roller and carries the remaining $4(2) = 8$ kN of distributed load at $x = 5$ m, the 24 kN load at $x = 10$ m, and the 8 kN downward pull that the left piece exerts through the hinge. Moments about the pin at $x = 6$ m give $$R_{12}\,(6) = 24\,(10 - 6) - 8\,(6 - 4) - 8\,(6 - 5) = 96 - 16 - 8 = 72$$ $$\begin{aligned} R_{12} &= \boxed{12.0\ \text{kN}\ \uparrow} \\ R_{6} &= 4(6) + 24 - 8 - 12 = \boxed{28.0\ \text{kN}\ \uparrow} \end{aligned}$$
6. Part (b) — diagrams. The shear starts at $+8$ kN, falls at 4 kN per metre to zero at $x = 2$ m, continues to $-16$ kN just left of the pin, jumps to $+12$ kN, holds until the 24 kN load takes it to $-12$ kN, and holds to the right roller. Hence $V_{\max}^{+} = +12$ kN and $V_{\max}^{-} = -16$ kN. Integrating, $M(2) = 8(2) - 2(2)^{2} = +8$ kN·m, the moment is zero at the hinge as it must be, and $$M(6) = 8(6) - 2(6)^{2} = 48 - 72 = \boxed{-24.0\ \text{kN}\cdot\text{m}}$$ $$M(10) = M(6) + 12(4) = -24 + 48 = \boxed{+24.0\ \text{kN}\cdot\text{m}}$$ The moment is negative from $x = 4$ m to $x = 8$ m and positive elsewhere.
7. Part (c) — geometry and total load. The member is a single bent beam: 3 m level, then 12 m horizontal by 5 m rise, then 2 m level, so the sloping reach is $\sqrt{12^{2} + 5^{2}} = 13$ m and its direction cosines are $12/13$ and $5/13$. The 33.8 kN/m acts on the horizontal projection, so $$W = 33.8\,(17) = 574.6\ \text{kN} \quad\text{at}\quad \bar{x} = 8.5\ \text{m}$$
8. Part (c) — reactions. The roller sits on a plane parallel to the incline, so its reaction $R$ is normal to the incline, with components $-5R/13$ horizontally and $12R/13$ vertically. Moments about the pin at the upper bend $(15,\,5)$, noting that the perpendicular distance from that pin to the line of $R$ is exactly the 13 m sloping length, $$574.6\,(15 - 8.5) = 13R \;\Rightarrow\; R = \frac{3734.9}{13} = \boxed{287.3\ \text{kN}}$$ so the roller pushes 110.5 kN to the left and 265.2 kN upward, and the pin carries $$\begin{aligned} H &= \boxed{110.5\ \text{kN}\rightarrow} \\ V &= 574.6 - 265.2 = \boxed{309.4\ \text{kN}\uparrow} \end{aligned}$$
9. Part (c) — diagrams. On the 3 m cantilever the shear is the accumulated load, $-33.8x$, reaching $-101.4$ kN, and the moment reaches $-33.8(3)^{2}/2 = -152.1$ kN·m (hogging). Just past the roller the shear taken perpendicular to the incline is $42.5 + (265.2 - 101.4)(12/13) = +193.7$ kN, and it falls to $42.5 - 241.8(12/13) = -180.7$ kN just below the pin, vanishing where $265.2 - 33.8x = -46.04$, that is at $x = 9.208$ m. There the sagging moment is greatest: $$M_{\max} = \boxed{499.3\ \text{kN}\cdot\text{m}\ \text{sagging}}$$ The right-hand 2 m cantilever hogs $33.8(2)^{2}/2 = 67.6$ kN·m at the pin and closes to zero at the free end.