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

Question 2: Reactions, Shear Force 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.

Find. The reactions on each of the three structures, and the shear force and bending moment diagram of every member, with the maximum and minimum ordinate of each labelled and the sign of each segment stated.

Approach. Each structure is statically determinate, so take global equilibrium first, then walk the member from one end writing $V(x)$ and $M(x)$, locating the point of zero shear inside every uniformly loaded reach because that is where the sagging moment peaks.

1. Part (a) — reactions of the overhanging beam. The distributed load resolves to $W = 6(8) = 48$ kN acting at $x = 4$ m. Moments about the pin at $x = 2$ m give $$R_B = \frac{48(4-2) + 12(10-2)}{6} = \frac{96 + 96}{6} = 32\ \text{kN}$$ and vertical equilibrium then gives $R_A = 48 + 12 - 32 = 28$ kN. Both reactions act upward.
2. Shear diagram of (a). On the left overhang $V = -6x$, so $V$ falls linearly to $-12$ kN just left of the pin and jumps by $R_A$ to $+16$ kN just right of it. Between the supports $V(x) = 16 - 6(x-2)$, which vanishes at $x = 2 + 16/6 = 4.667$ m and reaches $-20$ kN just left of the roller; the roller jump of 32 kN carries it to $+12$ kN, and the unloaded right overhang holds that value until the 12 kN tip load closes the diagram. The extreme ordinates are therefore $$V_{\max} = +16\ \text{kN}, \qquad V_{\min} = \boxed{-20\ \text{kN}}$$ the minimum occurring immediately to the left of the roller.
3. Moment diagram of (a). The left overhang gives $M = -3x^2$, hence $M = -12$ kN.m at the pin — a hogging, negative segment. Integrating the shear from there, the peak sagging value at the point of zero shear is $$M_{\max} = -12 + \frac{16^2}{2(6)} = -12 + 21.33 = +9.33\ \text{kN.m}$$ and at the roller $$M_{\min} = -12 + 16(6) - 3(6)^2 = \boxed{-24\ \text{kN.m}}$$ The right overhang closes linearly from $-24$ kN.m to zero at the free end, which checks against $-12(2) = -24$ kN.m taken from the right. The diagram is negative (hogging) over $0 \le x \le 2.83$ m, positive (sagging) over $2.83 \le x \le 6.5$ m, and negative again to the tip.

1. Part (b) — reactions of the L-frame. Set the origin at the beam tip, with the roller at $x = 2$ m, the corner at $x = 12$ m and the pin 10 m below the corner. The uniform load resolves to $W = 4(10) = 40$ kN at $x = 7$ m. There is no horizontal loading anywhere, so the pin carries no horizontal reaction, $D_x = 0$. Moments about the pin give $$B_y = \frac{20(12) + 40(5)}{10} = \frac{240 + 200}{10} = 44\ \text{kN}$$ and vertical equilibrium gives $D_y = 60 - 44 = \boxed{16\ \text{kN}}$, both upward.
2. Consequence for the column. With $D_x = 0$ the column has no shear anywhere along it, and since its base is a pin it carries no moment either; it is a pure strut delivering 16 kN of compression to the ground. Equivalently, the moment at the corner taken from the beam side is $-20(12) - 40(5) + 44(10) = 0$, which is the same statement and is worth writing down as the check that the reactions are right.
3. Shear and moment in the beam of (b). On the 2 m overhang $V = -20$ kN constant and $M = -20x$, reaching $-40$ kN.m at the roller. The roller jump takes the shear to $+24$ kN, after which $V(x) = 24 - 4(x-2)$ vanishes at $x = 8$ m and ends at $-16$ kN at the corner. The peak sagging moment is $$M_{\max} = -40 + \frac{24^2}{2(4)} = -40 + 72 = \boxed{+32\ \text{kN.m}}$$ at $x = 8$ m, while the largest hogging value is $M_{\min} = -40$ kN.m over the roller. The beam moment is negative from the tip to $x = 4.6$ m, positive from there to the corner, and exactly zero at the corner.

1. Part (c) — determinacy and global equilibrium. Both bases are pins, so $r = 4$, and the hinge at D supplies one condition; four unknowns against four equations makes the frame determinate. Take A(0, 0), B(5, 12), D(21, 12) and E(26, 0). The uniform load resolves to $W = 4.8(16) = 76.8$ kN at $x = 13$ m, and moments about A give $$E_y = \frac{24(5) + 76.8(13) + 24(21)}{26} = \frac{1622.4}{26} = 62.4\ \text{kN}$$ so $A_y = 124.8 - 62.4 = 62.4$ kN, as symmetry requires.
2. The hinge condition supplies the horizontal reactions. Take the portion to the right of D as a free body and sum moments about the hinge. Nothing acts on it but the reaction at E, so $$5E_y + 12E_x = 0 \;\Rightarrow\; E_x = -\frac{5(62.4)}{12} = -26\ \text{kN}$$ and horizontal equilibrium gives $A_x = +26$ kN. Written as vectors the reactions are $$\boxed{\mathbf{A} = (26,\ 62.4)\ \text{kN}, \qquad \mathbf{E} = (-26,\ 62.4)\ \text{kN}}$$ each support pushing inward, as a spreading frame demands.
3. Both legs turn out to carry axial force only. The unit vector along AB is $(5, 12)/13$, and $5.2 \times (5,\,12) = (26,\ 62.4)$, so the reaction at A lies exactly along the leg. Its component normal to AB is zero, so leg AB has no shear and no moment anywhere — the rigid knee at B is not called upon — and it carries $13(5.2) = 67.6$ kN of compression. Leg DE is unloaded between a hinge and a pin and is a two-force member by inspection, also at 67.6 kN compression. The frame geometry is the funicular shape for this load, which is exactly why the legs bend nowhere.
4. The beam therefore behaves as a simply supported span. Each leg delivers $(26,\ 62.4)$ kN to its knee, from which the 24 kN knee load is subtracted, leaving $38.4$ kN vertical and $26$ kN horizontal entering the beam at each end. The beam has zero moment at both ends, so over its 16 m $$V = \pm 38.4\ \text{kN at the knees}, \qquad M_{\max} = \frac{wL^2}{8} = \frac{4.8(16)^2}{8} = \boxed{+153.6\ \text{kN.m}}$$ at midspan, with the shear passing through zero there. The whole beam diagram is sagging — positive throughout — and the beam also carries a uniform axial thrust of 26 kN compression. Vertical equilibrium checks: $38.4 + 38.4 = 76.8$ kN, the whole distributed load.