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

Question 2: Reactions, shear force 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.

Given. Three determinate structures. Measuring $x$ from the left-hand end of each:

Find. Every reaction component, and complete shear force and bending moment diagrams with the maximum and minimum ordinate of every segment labelled.

Approach. Take moments about one support to get the other reaction, then use vertical equilibrium; build the shear diagram by walking along the member accumulating loads, and integrate it for the moment diagram, locating each sagging peak where the shear crosses zero.

2(a) — overhang with two point loads

1. Reactions. Taking moments about the pin at $x = 2$ m, with the roller reaction $R_{B}$ at $x = 12$ m and lever arms measured from the pin, $$-24(2) + 32(4) - 10\,R_{B} = 0$$ $$R_{B} = \frac{-48 + 128}{10} = \boxed{8.0\ \text{kN} \uparrow}$$ and vertical equilibrium gives $R_{A} = 24 + 32 - 8 = 48.0$ kN upward.
2. Shear. Starting from the free left end the shear is $-24$ kN across the overhang, jumps to $-24 + 48 = +24$ kN at the pin, drops to $24 - 32 = -8$ kN at the 32 kN load and is closed by the 8 kN roller reaction at the right-hand end. There is no distributed load, so each segment is constant.
3. Moment. Integrating the shear, $$M(2) = -24(2) = -48.0\ \text{kN}\cdot\text{m},\qquad M(6) = -48 + 24(4) = +48.0\ \text{kN}\cdot\text{m}$$ and $M(12) = 48 - 8(6) = 0$, which closes the diagram at the roller. The extremes are therefore $\boxed{M_{\max} = +48.0\ \text{kN}\cdot\text{m}}$ under the 32 kN load and $-48.0$ kN·m over the pin.

2(b) — double overhang under a uniform load

1. Reactions. The total load is $W = 22(20) = 440$ kN acting at $x = 10$ m. Moments about the pin at $x = 3$ m give $$440(10 - 3) = 11\,R_{B} \;\Rightarrow\; R_{B} = \frac{3080}{11} = \boxed{280\ \text{kN} \uparrow}$$ and $R_{A} = 440 - 280 = 160$ kN upward at $x = 3$ m.
2. Shear. $V = -22x$ over the left overhang, so $V(3^{-}) = -66$ kN; the pin lifts it to $V(3^{+}) = +94$ kN. Between the supports $V = 94 - 22(x - 3)$, which reaches $V(14^{-}) = -148$ kN, and the roller lifts it to $+132$ kN before the right overhang closes it at zero.
3. Where the sagging peak sits. Setting the interior shear to zero, $$x_{0} = 3 + \frac{94}{22} = 7.273\ \text{m}$$ which is the station of maximum sagging moment.
4. Moment. With $M(x) = 160(x - 3) - 11x^{2}$ between the supports, $$M(3) = -\tfrac{1}{2}(22)(3)^{2} = -99.0\ \text{kN}\cdot\text{m}$$ $$M(7.273) = 160(4.273) - 11(52.89) = \boxed{+101.8\ \text{kN}\cdot\text{m}}$$ $$M(14) = -\tfrac{1}{2}(22)(6)^{2} = \boxed{-396\ \text{kN}\cdot\text{m}}$$ The right overhang is the longer one, so it is the right-hand support and not mid-span that governs: the hogging moment there is almost four times the largest sagging moment.

2(c) — continuous bent

1. Read the load path first. The pin at the end of the top arm and the roller under the horizontal member are the only supports, and nothing pushes horizontally, so $\sum F_{x} = 0$ gives $H = 0$ at the pin immediately. That single observation makes the rest of the question a beam problem.
2. Reactions. The load is $W = 8(10) = 80$ kN at $x = 5$ m. Taking moments about the roller at $x = 2$ m, the pin acts 6 m away horizontally (its height does not matter because its horizontal component is zero): $$80(5 - 2) = 6\,V_{A} \;\Rightarrow\; V_{A} = \boxed{40.0\ \text{kN} \uparrow}, \qquad R_{\text{roller}} = 40.0\ \text{kN} \uparrow$$
3. Top arm and column. The 2 m arm carries the 40 kN pin reaction with a constant shear of 40 kN, so it delivers $$M = 40(2) = \boxed{80.0\ \text{kN}\cdot\text{m}}$$ to the corner. The column has no transverse load and zero shear, so that 80 kN·m runs unchanged down its 2 m length while it carries 40 kN in compression.
4. Horizontal member. The shear is $V = -8x$ over the 2 m overhang, so $V(2^{-}) = -16$ kN, jumps to $+24$ kN at the roller, and falls linearly to $-40$ kN at the corner, crossing zero at $$x_{0} = \frac{40}{8} = 5.0\ \text{m}.$$
5. Moments in the horizontal member. With $M(x) = -4x^{2} + 40(x - 2)$ beyond the roller, $$M(2) = -16.0\ \text{kN}\cdot\text{m},\qquad M(5) = \boxed{+20.0\ \text{kN}\cdot\text{m}},\qquad M(10) = -80.0\ \text{kN}\cdot\text{m}$$ and that $-80$ kN·m at the corner is precisely the 80 kN·m carried down the column, which is the check that the three members really were treated as continuous.