Question 2 of 9: Schematic Shear-Force and Bending-Moment Diagrams
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. 07-Str-A4 Advanced Structural Analysis, National Examinations, December 2015 — three hours, closed book (an approved Sharp or Casio calculator is the only aid). Nine questions: Questions 1 and 2 are compulsory (8 and 12 marks); the candidate then answers two of Questions 3, 4, 5 (18 marks each) and two of Questions 6, 7, 8, 9 (22 marks each), so six questions constitute a complete 100-mark paper. Marks are printed in the left margin. All nine questions are solved here, because this set is a study resource rather than an examination script.
Reference texts. R. C. Hibbeler, Structural Analysis, 10th ed. (slope-deflection, moment distribution, influence lines, Castigliano); A. Kassimali, Structural Analysis, 6th ed. (force and displacement methods, degrees of freedom); A. Ghali, A. M. Neville & T. G. Brown, Structural Analysis: A Unified Classical and Matrix Approach, 7th ed. (flexibility and stiffness formulations, lack of fit, support movement); W. Weaver Jr. & J. M. Gere, Matrix Analysis of Framed Structures, 3rd ed. (assembly of [K] and {P}); J. C. McCormac, Structural Analysis: Using Classical and Matrix Methods, 4th ed. Canadian design context: CSA S16, CSA A23.3 and the National Building Code of Canada supply the load and resistance rules that these analyses feed, although this paper is purely an analysis paper.
Check: sign convention used throughout. Member end moments \(M_{ij}\) are counter-clockwise positive, which is the convention that matches the standard six-degree-of-freedom stiffness matrix and the vector \(\{\theta_2,\theta_3,\delta\}\) the paper itself defines (Question 9 states “positive counter clockwise”). With that choice the fixed-end moment of a downward uniformly distributed load is \(\mathrm{FEM}_{ij}=+wL^{2}/12\) at the \(i\) end, the chord rotation is \(\psi_{ij}=[(\mathbf{d}_j-\mathbf{d}_i)\cdot\mathbf{e}_\perp]/L\) with \(\mathbf{e}_\perp\) the member axis turned \(+90^{\circ}\), and ordinary sagging moments are recovered as \(M_{\text{sag}}(i)=-M_{ij}\), \(M_{\text{sag}}(j)=+M_{ji}\). Mixing this with Hibbeler’s clockwise-positive convention produces clean-looking integers that are wrong; every diagram below is plotted in the sagging convention.
Check: printed unit in Question 6. The paper prints “EI = 3.2 × 104 kN.mm2”. Taken literally that is a flexural rigidity some ten orders of magnitude too small for an 18 m beam and would make the 36 mm settlement effect vanish. The intended unit is \(\text{kN}\cdot\text{m}^2\): with \(EI=3.2\times10^{4}\ \text{kN}\cdot\text{m}^{2}\) the settlement term \(3EI\Delta/L^{2}=54\ \text{kN}\cdot\text{m}\) is the same order as the load term \(wL^{2}/8=48\ \text{kN}\cdot\text{m}\), which is plainly what the question is testing. The solution below uses \(\text{kN}\cdot\text{m}^{2}\) and flags the typographical error.
Question 2: Schematic Shear-Force and Bending-Moment Diagrams (12 marks)
UDL \(w\) over \(0\le x\le 2L\); point load \(P=wL/4\) at \(x=L/2\)
(b)
Lower beam \(L\), riser \(L/2\), upper beam \(L\)
Built in at joint 1; vertical roller at joint 4
None (all joints rigid)
UDL \(w\) on both horizontal members
Find. The shape and the governing ordinates of the shear-force and bending-moment diagrams for both structures, expressed in the non-dimensional groups \(wL\) and \(wL^{2}\).
Structure 2(a) — three rollers, a built-in end and two internal hinges. Note that the uniformly distributed load stops at the third support.
Approach. Count the releases first: each structure turns out to be indeterminate to only one degree, so a single compatibility statement (or one pass of the displacement method) fixes every ordinate exactly, and the “schematic” diagram the question asks for can be drawn to scale rather than by eye.
Establish the degree of indeterminacy of 2(a). Four supports supply \(1+1+1+3 = 6\) reaction components, plane equilibrium supplies three equations, and each internal hinge supplies one condition equation:
$$r = 6 - 3 - 2 = \boxed{1}$$
So the beam is redundant to one degree only. Better still, the two hinges chop it into a determinate left portion \((0\le x\le 1.25L)\), a determinate suspended span \((1.25L\le x\le 1.75L)\) and a right portion \((1.75L\le x\le 3L)\) that carries the single redundancy.
Analyse the suspended span first. Between the two hinges the beam is a simply supported span of length \(L/2\) carrying \(w\), so it delivers \(wL/4\) to each hinge and its own moment peaks at mid-length:
$$M_{\max}=\frac{w(L/2)^{2}}{8}=\frac{wL^{2}}{32}=+0.031\,wL^{2}$$
That is the entire “middle” of the diagram, and it is the only sagging region between the two interior supports.
Solve the whole beam and record the reactions. Carrying the hinge forces into the outer portions and enforcing continuity across the redundancy gives
$$R_{0}=\tfrac{17}{32}wL=0.531\,wL,\qquad R_{L}=\tfrac{39}{32}wL=1.219\,wL$$
$$R_{2L}=\tfrac{41}{64}wL=0.641\,wL,\qquad R_{3L}=-\tfrac{9}{64}wL=-0.141\,wL$$
with a built-in moment of \(+\tfrac{3}{64}wL^{2}\) (sagging) at \(x=3L\). Vertical equilibrium checks exactly: \(0.531+1.219+0.641-0.141 = 2.250 = 2wL + wL/4\), which is the whole applied load. The negative reaction at the built-in end is a genuine hold-down: that last span carries no load at all and is dragged upward by the hinge force from the left.
Assemble the shear diagram. Shear starts at \(+0.531\,wL\), falls linearly under the UDL to \(+0.031\,wL\) just left of the point load, steps down by \(P=0.25\,wL\), continues to \(-0.719\,wL\) at the second support, jumps to \(+0.500\,wL\), and so on. In the unloaded final span the shear is constant at \(+0.141\,wL\).
Assemble the moment diagram. The moment is zero at the first roller, peaks at the point load, hogs over the interior supports, is identically zero at both hinges, and finishes sagging at the wall:
$$M(L/2)=+\tfrac{9}{64}wL^{2},\quad M(L)=-\tfrac{3}{32}wL^{2},\quad M(1.5L)=+\tfrac{1}{32}wL^{2},$$
$$M(2L)=-\tfrac{3}{32}wL^{2},\quad M(3L)=+\tfrac{3}{64}wL^{2}$$
Structure 2(a) shear force, normalised by \(wL\). The step at \(x=L/2\) is the point load; the vertical jumps at \(x=L\) and \(x=2L\) are the reactions.
Structure 2(a) bending moment, normalised by \(wL^{2}\), sagging positive. The diagram is pinned to zero at both hinges — that is the fastest visual check on any Gerber beam.
Structure 2(b) — stepped frame, built in at joint 1 and resting on a vertical roller at joint 4.
Establish the degree of indeterminacy of 2(b). With \(m=3\), \(j=4\) and \(r_{\text{reac}}=3+1=4\),
$$r = 3(3) + 4 - 3(4) = \boxed{1}$$
The roller supplies no horizontal reaction and there is no horizontal load, so the horizontal reaction at the wall is identically zero — a one-line result that is worth stating before any arithmetic.
Recognise what the riser does. Because no load acts along the riser and no horizontal force can reach it, its two end moments are equal and opposite, so its shear vanishes and it carries a constant bending moment plus a compressive axial force. Solving the frame gives
$$M_{\text{riser}}=\tfrac{4}{19}wL^{2}=0.211\,wL^{2},\qquad N_{\text{riser}}=\tfrac{11}{38}wL=0.289\,wL\ \text{(compression)}$$
Report the reactions and the beam ordinates.
$$R_{1}=\tfrac{49}{38}wL = 1.289\,wL,\qquad R_{4}=\tfrac{27}{38}wL=0.711\,wL,\qquad H_{1}=0$$
The wall hogs at \(-\tfrac{11}{19}wL^{2}=-0.579\,wL^{2}\); the lower beam runs up to \(+0.211\,wL^{2}\) sagging at the corner, crossing zero at \(x=0.579L\). The upper beam starts at that same \(+0.211\,wL^{2}\), peaks at \(+0.252\,wL^{2}\) a distance \(0.289L\) from the corner, and closes at zero on the roller.
Structure 2(b), lower beam: moment normalised by \(wL^{2}\), sagging positive.
Structure 2(b), upper beam: moment normalised by \(wL^{2}\), sagging positive. There is no point of contraflexure in this member.
