Question 2 of 8: Reactions, shear and moment diagrams (22 marks)

Question 2 — Reactions, shear and moment diagrams (22 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.

2(a) Compound (Gerber) beam

Given. A 12 m beam carrying a uniformly distributed load of 8 kN/m over its first 8 m and a 36 kN point load at the free right-hand tip. Measuring $x$ from the left tip:

Find. The three reactions, and the shear and bending-moment diagrams with their largest positive and negative ordinates.

Approach. With $r=4$ and one hinge the beam is determinate, so cut it at the hinge and solve the short suspended segment first — its hinge force then becomes a known load on the main span.

1. Isolate the suspended segment (8 m to 12 m). This piece carries only the 36 kN tip load and is held by the hinge at $x=8$ and the roller at $x=11$. Taking moments about C, with $V_h$ the upward force the main span applies at the hinge, $$\sum M_C = V_h(8-11) - 36(12-11) = 0 \;\Rightarrow\; V_h = -12\ \text{kN}$$ The negative sign means the main span in fact pulls this segment down with 12 kN. Vertical equilibrium then gives $R_C = 36 + 12 = \boxed{48\ \text{kN}\ \uparrow}$.
2. Transfer the hinge force to the main span. By Newton’s third law the suspended segment pushes up on the main span with 12 kN at $x=8$. The main span (0 m to 8 m) therefore carries the full UDL resultant $W = 8 \times 8 = 64\ \text{kN}$ acting at $x=4$, plus that 12 kN upward force at its right end.
3. Take moments about A to find $R_B$. $$\sum M_A = R_B(7-1) - 64(4-1) + 12(8-1) = 0$$ $$6R_B = 192 - 84 = 108 \;\Rightarrow\; R_B = \boxed{18\ \text{kN}\ \uparrow}$$
4. Vertical equilibrium gives $R_A$. $$R_A = 64 - 12 - R_B = 64 - 12 - 18 = \boxed{34\ \text{kN}\ \uparrow}$$ Checking the whole beam, the reactions total $34+18+48 = 100\ \text{kN}$ against applied loads of $64+36 = 100\ \text{kN}$, and moments about B close to zero.

Working along the beam, the shear starts at zero, falls to $-8$ kN just left of A, jumps to $+26$ kN across the pin, then decreases at 8 kN/m to $-22$ kN just left of B. It jumps to $-4$ kN across B, reaches $-12$ kN where the UDL stops at the hinge, stays constant to C, and jumps to $+36$ kN, which the tip load removes exactly.

1. Locate the maximum sagging moment. Shear crosses zero inside span A–B where $26 - 8(x-1) = 0$, that is at $x = 4.25\ \text{m}$. Integrating the shear from the tip, $$M_{\max}^{+} = -\tfrac{1}{2}(8)(1)^2 + \tfrac{1}{2}(26)(3.25) = -4 + 42.25 = \boxed{38.25\ \text{kN}\cdot\text{m}}$$
2. Evaluate the hogging peak. The largest negative moment is at support C, where the 1 m tip overhang and its 36 kN load act as a cantilever: $$M_C = -36(1) = \boxed{-36\ \text{kN}\cdot\text{m}}$$ and the moment at the hinge is identically zero, which is the check that confirms the whole analysis.

The bending-moment diagram is negative (hogging) over the short end overhangs and over the stretch from B through C, and positive (sagging) through the main span between roughly $x=1.1$ m and $x=7.4$ m.

2(b) Three-hinged portal frame

Given. A portal of 8 m span and 4 m height. The beam carries 8 kN/m downward over its whole span; the right-hand column carries 8 kN/m acting horizontally (right to left) over its whole height. Both bases (A on the left, B on the right) are pinned, and there is an internal hinge at D, the top-left corner. The top-right corner C is rigid.

Find. The four reaction components and the shear and moment diagrams for each of the three members.

Approach. Two pins plus one internal hinge give four unknowns and four equations — three overall plus zero moment at the hinge — so the frame is determinate. Spotting that the left column is a two-force member collapses most of the algebra.

1. Resolve the applied resultants. The beam load totals $W_b = 8(8) = 64\ \text{kN}$ acting downward at midspan, and the column load totals $W_c = 8(4) = 32\ \text{kN}$ acting leftward at mid-height, 2 m above B.
2. Take moments about A for the whole frame. With $B_y$ upward at $(8,0)$, $$\sum M_A = -64(4) + 32(2) + 8B_y = 0 \;\Rightarrow\; B_y = \boxed{24\ \text{kN}\ \uparrow}$$ and vertical equilibrium gives $A_y = 64 - 24 = \boxed{40\ \text{kN}\ \uparrow}$.
3. Exploit the two-force member. Column A–D is pinned at A, hinged at D and carries no load along its length, so it can transmit force only along its own axis. That axis is vertical, hence $$A_x = \boxed{0}, \qquad B_x = 32 - 0 = \boxed{32\ \text{kN}} \ (\rightarrow)$$ Recognising this at the outset is worth several minutes in an examination; the alternative is to take moments about D for the left free body, which gives $4A_x = 0$ and the same answer.
4. Beam D–C. At the hinge the moment is zero and the left column delivers 40 kN upward, so measuring $s$ from D, $$V(s) = 40 - 8s, \qquad M(s) = 40s - 4s^{2}$$ Shear vanishes at $s = 5\ \text{m}$, where $$M_{\max} = 40(5) - 4(25) = \boxed{100\ \text{kN}\cdot\text{m}}$$ and at the rigid corner, $M_C = 40(8) - 4(64) = 64\ \text{kN}\cdot\text{m}$ with $V_C = -24\ \text{kN}$. The whole beam diagram is sagging (positive).
5. Right column C–B. Measuring $y$ upward from B, the free body below the cut carries $B_x = 32\ \text{kN}$ and the part of the horizontal load already passed, so $$M(y) = 32y - 4y^{2}$$ This is zero at the pin, rises monotonically and reaches $M = 32(4) - 4(16) = 64\ \text{kN}\cdot\text{m}$ at C — matching the beam exactly, as a rigid corner with no applied joint moment demands. The column shear runs linearly from 32 kN at B to 0 at C.

2(c) Three-hinged gable frame

Given. A gable frame pinned at A $(0,0)$ and B $(12.5,0)$ with a crown hinge at C $(8,6)$, all dimensions in metres. Two point loads act on the left rafter: 80 kN at 3 m from A and 40 kN at 5 m from A, both measured horizontally. The right rafter carries no load.

Find. The four reaction components and the shear, axial and moment diagrams for both rafters.

Approach. Overall moment equilibrium fixes the vertical reactions; the zero-moment condition at the crown, applied to the unloaded right half, then fixes the horizontal thrust.

1. Vertical reactions from overall equilibrium. Taking moments about A, $$\sum M_A = -80(3) - 40(5) + 12.5B_y = 0 \;\Rightarrow\; B_y = \frac{440}{12.5} = \boxed{35.2\ \text{kN}\ \uparrow}$$ and $A_y = 120 - 35.2 = \boxed{84.8\ \text{kN}\ \uparrow}$.
2. Horizontal thrust from the crown condition. The right half C–B carries no applied load, so taking moments about the crown for that free body, $$\sum M_C = B_y(12.5-8) - B_x(0-6) = 0 \;\Rightarrow\; B_x = -\frac{35.2(4.5)}{6} = -26.4\ \text{kN}$$ so B pushes 26.4 kN to the left and, from horizontal equilibrium, $A_x = \boxed{26.4\ \text{kN}}$ to the right. This inward-acting pair is the characteristic thrust of an arch-like frame.
3. Left-rafter bending moment. At a section a horizontal distance $x$ from A (so at height $0.75x$), $$M(x) = 84.8x - 26.4(0.75x) - 80\langle x-3\rangle - 40\langle x-5\rangle = 65.0x - 80\langle x-3\rangle - 40\langle x-5\rangle$$ The peak falls at the first load point: $$M_{\max} = 65.0(3) = \boxed{195\ \text{kN}\cdot\text{m}}$$ with $M = 165\ \text{kN}\cdot\text{m}$ at the 40 kN point and $M = 0$ at the crown, as required. The whole left-rafter diagram is sagging.
4. Left-rafter axial force and shear. Resolving the reaction along the rafter (direction cosines $0.8$ and $0.6$) and perpendicular to it, the three load stretches give axial forces of 72 kN, 24 kN and 0 (all compression) and shears of $+52$ kN, $-12$ kN and $-44$ kN respectively. That the axial force falls to zero above the 40 kN load is a genuine feature of this geometry, not an arithmetic slip.
5. Right rafter. With a pin at B, a hinge at C and no load between, C–B is a two-force member: it carries pure axial compression of $$N = \sqrt{26.4^{2} + 35.2^{2}} = \boxed{44.0\ \text{kN (C)}}$$ and its bending moment and shear are identically zero over its whole length.