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. National Exams, May 2017 — 16-Civ-B1 Advanced Structural Analysis. Three hours, CLOSED BOOK (an approved Sharp or Casio calculator is permitted). Answer BOTH Questions 1 and 2, ONLY TWO of Questions 3, 4 or 5, and ONLY TWO of Questions 6, 7, 8 or 9; six questions constitute a complete paper for 100 marks. Marks are printed in the left margin. All nine questions are worked below, because the complete set is the study resource.
Reference texts. R. C. Hibbeler, Structural Analysis, 10th ed. (Pearson) — Ch. 8 influence lines, Ch. 9–10 the force (flexibility) method, Ch. 11 slope-deflection, Ch. 14–16 the stiffness method; A. Kassimali, Structural Analysis, 6th ed. (Cengage) — Ch. 7 deflections by work and energy (Castigliano), Ch. 13 slope-deflection, Ch. 16 sidesway frames; W. McGuire, R. H. Gallagher & R. D. Ziemian, Matrix Structural Analysis, 2nd ed., for the element stiffness matrix used as the independent check. Once the analysis is complete, member design follows CSA A23.3:19, CSA S16:19 and CSA O86:19, with load combinations from NBCC 2020.
Check: sign convention used throughout. End moments $M_{ij}$ are counter-clockwise positive on the member end — the convention that matches the six-degree-of-freedom planar frame element stiffness matrix, so every answer below can be checked against a direct-stiffness solution. Chord rotation is $\psi_{ij}=\left[(\mathbf{D}_{j}-\mathbf{D}_{i})\cdot\mathbf{e}_{2}\right]/L$, where $\mathbf{e}_{2}$ is the member axis turned $+90^\circ$. The fixed-end moment of a downward uniform load is $\mathrm{FEM}_{ij}=+wL^{2}/12$ at the $i$ end and $-wL^{2}/12$ at the $j$ end; for a member released at its far end it becomes $+wL^{2}/8$. Ordinary sagging moments follow as $M_{\text{sag}}(i)=-M_{ij}$ and $M_{\text{sag}}(j)=+M_{ji}$, and all diagrams are plotted sagging positive. Member shear at the $i$ end is $V_{i}=(M_{ij}+M_{ji})/L$ plus the equivalent nodal shear of any span load. Mixing this with Hibbeler's clockwise-positive convention produces clean-looking integers that are wrong, so the convention is stated once and used everywhere.
Question 2: Schematic shear force and bending moment diagrams (12 marks)
Check: the source figures carry no dimensions. The question asks only for the shape of the diagrams, and the shapes below are dimension-independent. So that every ordinate can be stated and checked rather than waved at, representative dimensions in the drawn proportions have been adopted: for (a) $4-1-2-1-4-4$ m between the seven support and hinge points, $w=10$ kN/m from the first roller to the third support, and $P=20$ kN on a 1 m stub; for (b) a 6 m loaded cantilever, a 4 m vertical member, a 6 m arm and $w=10$ kN/m. Changing the dimensions scales the ordinates but moves no zero and changes no sign.
Given. Structure (a): a continuous member built in at A and G, on rollers at B, E and F, with internal hinges at C and D, $w=10$ kN/m from B to E, and a horizontal $P=20$ kN applied 1 m above F on a rigid stub. Structure (b): a 6 m cantilever tip loaded at $w=10$ kN/m, a roller at the knee B, a 4 m vertical member B–C and a 6 m arm C–D built in to a wall at D. All members share the same $EI$ and are inextensible.
Find. The shear force and bending moment diagrams for both structures, with the governing ordinates, the points of zero moment and the discontinuities identified.
Structure (a): the two hinges isolate a simply supported interior length; the stub couple $P\!\times\!h$ makes the moment jump at F.
Approach. Cut the structure at every internal hinge, analyse the resulting determinate and singly indeterminate pieces, then reassemble; treat the stub as an equivalent joint force plus couple.
Isolate the length between the two hinges. The segment C–D has zero moment at both ends, so it is a simply supported span of its own. It therefore delivers$$V_{\text{hinge}}=\frac{wL_{CD}}{2}=\frac{10(2)}{2}=\boxed{10\ \text{kN}}$$to each hinge, and carries a parabolic sagging moment peaking at $wL_{CD}^{2}/8=+5$ kN·m at its mid-point. Nothing else about the structure can change these two numbers, which is what makes the hinges so useful.
Reduce the left-hand piece to a propped cantilever. To the left of hinge C the structure is built in at A and propped at B, with the length B–C acting as a determinate overhang. The overhang delivers a hogging moment at B of$$M_{B}=-\left[\frac{w\,(1)^{2}}{2}+V_{\text{hinge}}(1)\right]=-(5+10)=-15\ \text{kN}\cdot\text{m}.$$A moment applied at the propped end of a propped cantilever carries over to the built-in end at half its size and with the opposite bending sense, so $M_{A}=+7.5$ kN·m; the span A–B carries no load, so its shear is constant at $(M_{B}-M_{A})/4=-5.625$ kN.
Solve the right-hand piece. To the right of hinge D the piece is a two-span beam E–F–G on a roller at E and F and built in at G, loaded by the D–E overhang (which delivers $M_{E}=-15$ kN·m by the same arithmetic as step 2) and by the stub at F. The stub is a determinate cantilever, so replace it by a horizontal force $P$ and a clockwise couple$$M_{\text{stub}}=P\,h=20(1)=20\ \text{kN}\cdot\text{m}$$applied at F. Two degrees of redundancy remain; solving them gives $M_{F}^{-}=-4.29$ kN·m immediately left of F, $M_{F}^{+}=+15.71$ kN·m immediately right of it — the jump is exactly the stub couple — and $M_{G}=-7.86$ kN·m at the built-in end.
Read the shear diagram off the reactions. The support reactions come to $-5.63$, $+25.63$, $+22.68$, $-8.57$ and $+5.89$ kN at A, B, E, F and G, summing to the applied $wL_{BE}=40$ kN. The shear is therefore constant at $-5.63$ kN over A–B, falls linearly from $+20$ kN at B through zero inside C–D to $-20$ kN at E, is constant at $+2.68$ kN over E–F, and constant at $-5.89$ kN over F–G. Note the hold-down at F: the stub couple is strong enough to reverse that reaction.
Structure (b): the loaded cantilever fixes the knee moment, the vertical member carries it unchanged, and the arm halves it into the wall.
Structure (b) is even more transparent, because the horizontal equilibrium of the whole assembly does the work.
The cantilever fixes the knee moment. The length to the left of B is a plain cantilever hanging off the knee, so$$M_{B}=\frac{wa^{2}}{2}=\frac{10(6)^{2}}{2}=\boxed{180\ \text{kN}\cdot\text{m}}$$hogging, with a parabolic moment and a shear rising linearly to $wa=60$ kN at B.
The vertical member carries a constant moment. There is no horizontal load anywhere and the only horizontal restraint is at the wall, so the horizontal reaction vanishes; the arm C–D therefore carries no axial force, and joint C then forces the shear in the vertical member B–C to be zero. A member with zero shear carries a constant moment, so the full 180 kN·m travels up B–C unchanged.
The arm halves it into the wall. Both ends of the arm are held against vertical movement (the knee by the roller through the inextensible column, the far end by the wall) and the wall also fixes the rotation, so the arm behaves as a propped cantilever driven by the 180 kN·m delivered at C. Carry-over gives$$M_{D}=\tfrac{1}{2}M_{C}=90\ \text{kN}\cdot\text{m}$$of the opposite bending sense, a constant arm shear of $(180+90)/6=45$ kN, and a point of zero moment 4 m from C. Vertical equilibrium then gives $R_{B}=wa+45=105$ kN up and a 45 kN hold-down at the wall.
