16-Civ-B1 Advanced Structural Analysis · December 2013
Question 1 of 9: Schematic shear-force and bending-moment diagrams
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Examination, December 2013 — 98-Civ-B1 Advanced Structural Analysis. Closed book, 3 hours, calculator permitted. Nine questions: #1 and #2 are compulsory, two of #3–#5 and two of #6–#9 are selected, and six questions constitute a complete paper (100 marks). Marks are printed in the left margin. All nine questions are solved below, because the set is a study resource rather than an examination script.
Reference texts. R. C. Hibbeler, Structural Analysis, 10th ed. (slope-deflection ch. 11, moment distribution ch. 12, influence lines ch. 6, energy methods ch. 9); A. Kassimali, Structural Analysis, 6th ed. (force/flexibility method ch. 13, matrix stiffness ch. 18); A. Ghali, A. M. Neville & T. G. Brown, Structural Analysis: A Unified Classical and Matrix Approach, 7th ed. (support-movement and lack-of-fit effects, ch. 4 and 6). Canadian design codes (CSA S16, CSA A23.3, NBCC) are not needed here: every question is an elastic-analysis question, not a design question.
Sign convention used throughout (stated once, obeyed everywhere). Global x to the right, y up; moments and rotations are counter-clockwise positive. For a member running from end $i$ to end $j$, let $\mathbf{e}_1$ be the unit vector from $i$ to $j$ and $\mathbf{e}_2$ that vector turned through $+90^\circ$. The chord rotation is $\psi_{ij}=\left[(\boldsymbol{\Delta}_j-\boldsymbol{\Delta}_i)\cdot\mathbf{e}_2\right]/L$, so the slope-deflection equation reads
with $\mathrm{FEM}=+wL^{2}/12$ and $+PL/8$ at the i end of a member carrying a downward uniform load or a central point load (and the negative of those at the j end). Ordinary sagging-positive bending moments follow from $M_{\text{sag}}(i)=-M_{ij}$ and $M_{\text{sag}}(j)=+M_{ji}$. Every number below was computed twice — once by hand from these equations and once by an independent direct-stiffness solution — and the two agree.
Question 1: Schematic shear-force and bending-moment diagrams (12 marks)
Given. Three structures of uniform, unspecified $EI$ whose members are inextensible; all loads are single point loads of magnitude $P$. Structure (a) is a continuous beam: a free overhang of $L/3$ carrying $P$ at its tip, three spans of $L$, and a built-in right-hand end, with two further loads $P$ at the third points of the middle span. Structure (b) is a symmetric trapezoid hinged to the foundation at both feet, loaded by $P$ at mid-length of the top member. Structure (c) is a Z-shaped frame of two equal columns of height $L$ and a horizontal member of length $L$, encastré at both extremities, loaded by $P$ at mid-length of the horizontal member.
Find. The shape of the shear-force and bending-moment diagrams for each structure, together with the ordinates that fix them.
Approach. Each structure is statically indeterminate, so the shapes are obtained by first counting the redundants, then exploiting inextensibility to fix the sway pattern, and finally solving the small set of joint-equilibrium equations; the resulting end moments are converted to diagrams member by member.
1(a) — continuous beam with an overhang
Structure 1(a): free overhang L/3 carrying P, three equal spans L, and a built-in right-hand end. The two interior loads sit at the third points of the middle span.
Count the redundants. Three rollers supply one reaction component each and the encastré end supplies three, so $r=6$ against three equations of statics: the beam is indeterminate to the third degree. Nothing about the diagram can be obtained from statics alone except the overhang.
Fix the one determinate ordinate first. The overhang is a cantilever hanging off the first roller, so the section directly over that support carries $$M_A=-P\left(\tfrac{L}{3}\right)=\boxed{-0.3333\,PL}$$ and the shear throughout the overhang is the constant $-P$. Every diagram must start from these two values; they are the check that the indeterminate solution has been assembled correctly.
Solve the remaining three redundants. Taking the three interior support reactions as the redundants and enforcing zero deflection at each of them (equivalently, running a direct-stiffness solution of the six-element beam) gives the reaction set $$R_A=1.2949P,\quad R_B=0.5641P,\quad R_C=1.4103P,\quad R_D=-0.2692P$$ whose sum is exactly $3P$, and a built-in moment at the far end of $M_D=+0.0897\,PL$. The negative reaction at the fixed end is genuine: the unloaded end span is dragged upward by the hogging carried across support C, so the fixed end must be held down.
Assemble the shear diagram. Marching from the free tip and adding each reaction as it is passed gives the step sequence $-1.000$, $+0.2949$, $+0.8590$ (after $R_B$), $-0.1410$ and $-1.1410$ (after the two interior loads) and finally $+0.2692$ in the end span, which the reaction at D closes to zero.
Assemble the moment diagram. Because every load is a point load the bending moment is piecewise linear; integrating the shears from $M=0$ at the free tip reproduces the ordinates listed below, and the sign changes twice in the middle span and once in the end span.
Shear-force diagram for structure 1(a), in multiples of P. The diagram is a pure step function because all loads are concentrated.
Bending-moment diagram for structure 1(a), in multiples of PL. Sagging is plotted positive.
The characteristic features a marker looks for are: the linear hogging ramp over the overhang, a near-zero moment over support B, sagging peaks under the two interior loads, a large hogging value over support C, and a sagging moment at the built-in end — the last of these being the visible consequence of the uplift reaction found in step 3.
1(b) — symmetric trapezoidal frame on two hinges
Structure 1(b): a symmetric trapezoid hinged to the foundation at both feet. The drawing carries no dimensions, so the ordinates quoted below use the proportions scaled off the paper (leg run 0.75h, top member 1.5h, rise h).
Check: figure 1(b) is dimensionless on the examination paper, so the shapes below are exact but the numerical ordinates depend on the proportions. They were scaled off the printed figure as leg run $0.75h$, top member $1.5h$, rise $h$ (leg length $1.25h$, overall span $3h$); a candidate who assumes a different but self-consistent set of proportions and states the assumption is equally correct.
Classify the frame. Three members, two pinned feet: $r=4$ against three equations, so the frame is indeterminate to the first degree and the redundant is the horizontal thrust $H$. Because the feet are pins, $M=0$ at A and at D for free.
Use symmetry to kill the sway. Inextensibility forces joint B to move perpendicular to leg AB and joint C perpendicular to leg CD, while the top member keeps them the same horizontal distance apart. Under a symmetric load the only displacement field satisfying all three conditions is the trivial one, so B and C do not translate and the single remaining unknown is the joint rotation, with $\theta_C=-\theta_B$ by symmetry.
Write the two member equations. With the pinned foot condensed out, leg AB offers the modified stiffness $3EI/L_{\text{leg}}$, and the top member carries the central-load fixed-end moment $PL_{\text{top}}/8$. Joint equilibrium at B, $M_{BA}+M_{BC}=0$, then yields $$\theta_B=\frac{0.1875\,P h^{2}}{3.7333\,EI}=0.05022\,\frac{Ph^{2}}{EI}$$ and hence the knee moment $$M_B=\boxed{-0.1205\,Ph}\ \ \text{(hogging)}.$$
Recover the thrust from a free body of one leg. Vertical equilibrium and symmetry give $R_{Ay}=R_{Dy}=P/2$ exactly, independent of the proportions. Taking moments about B for member AB, $$H=\frac{0.75h\,(P/2)-|M_B|}{h}=0.375P-0.1205P=\boxed{0.2545P}$$ measured per unit rise, i.e. $H=0.4955P$ for the scaled geometry; the thrust acts inward, so the frame behaves as a shallow arch.
Complete the diagrams. The top member carries a constant shear $\pm P/2$ with a reversal under the load, so its moment diagram is the usual triangle raised on the hogging knee moments, peaking at $$M_{\text{crown}}=-0.1205Ph+\tfrac{P}{2}(0.75h)=\boxed{+0.2545\,Ph}.$$ Each leg carries a constant transverse shear $0.0964P$ and a compressive axial force $0.6973P$, so its moment diagram is a straight line from zero at the hinge to $0.1205Ph$ at the knee.
Bending moment for structure 1(b), developed along A–B–crown–C–D, in multiples of Ph.
Transverse shear in each member of structure 1(b), in multiples of P.
1(c) — stepped (Z-shaped) frame, encastré at both ends
Structure 1(c): lower column fixed at its foot, upper column fixed at its head, horizontal member of length L loaded at mid-length.
Classify and find the degrees of freedom. Two encastré ends give $r=6$, so the frame is indeterminate to the third degree. Kinematically, however, inextensibility of the two vertical columns pins the vertical position of joints B and C, and inextensibility of the horizontal member ties their horizontal movements together: only one sway $\Delta$ and the two joint rotations survive.
Exploit the point symmetry. Rotating the frame through $180^\circ$ about the mid-point of the horizontal member maps it onto itself, so $\theta_C=-\theta_B$. Adding the two joint-equilibrium equations then gives that result directly, and the sway equation $(M_{AB}+M_{BA})=(M_{CD}+M_{DC})$ closes the system.
Solve. Writing $k=EI/L$ and using the fixed-end moment $PL/8$ for the central load, $$\theta_B=-\frac{PL^{2}}{24EI},\qquad \Delta=\frac{PL^{3}}{48EI},$$ from which every end moment follows.
Read off the diagrams. Both knees, the fixed foot and the fixed head all carry the same hogging moment $$|M|=\boxed{\frac{PL}{24}=0.04167\,PL}$$ and the moment under the load is $$M_{\text{load}}=\boxed{+\frac{5PL}{24}=+0.2083\,PL}.$$ Each column therefore carries a constant bending moment and zero shear; the horizontal member carries the constant shear $\pm P/2$ that produces the sagging triangle. Both support reactions are purely vertical, $P/2$ each, with no horizontal thrust at all.
Bending moment for structure 1(c), developed along A–B–load–C–D, in multiples of PL.
Transverse shear for structure 1(c), in multiples of P. The columns are shear-free.
The zero column shear is worth pausing on, because it looks like an error and is not. The load is vertical and the two columns are vertical, so no horizontal force is needed anywhere; the columns nevertheless bend, because the horizontal member drags their heads sideways by $\Delta$ while holding them against rotation, and a member with equal end moments of opposite sense carries constant moment and no shear.
