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16-Civ-B1 Advanced Structural Analysis · December 2019

Question 5 of 9: Influence lines for a three-span continuous beam

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. National Examination, December 2019 — 16-Civ-B1 Advanced Structural Analysis, 3 hours, CLOSED BOOK (an approved Casio or Sharp calculator only). Nine questions on seven pages: Questions 1 and 2 are compulsory (8 and 12 marks); the candidate then answers TWO of Questions 3, 4 or 5 (18 marks each) and TWO of Questions 6, 7, 8 or 9 (22 marks each), so six questions make a complete paper of 100 marks. Because this set is a study resource, all nine questions are solved here.

Reference texts. R. C. Hibbeler, Structural Analysis, 10th ed. (Pearson) — Ch. 6 influence lines, Ch. 8–9 deflections and virtual work, Ch. 10 force (flexibility) method, Ch. 11 slope-deflection, Ch. 12 moment distribution, Ch. 15–16 matrix stiffness. A. Kassimali, Structural Analysis, 6th ed. (Cengage) — Ch. 7 Castigliano and least work, Ch. 8 influence lines, Ch. 13 slope-deflection, Ch. 17–18 matrix stiffness. A. Ghali, A. M. Neville and T. G. Brown, Structural Analysis: A Unified Classical and Matrix Approach, 7th ed. (CRC). Canadian context for the settlement question: NBC 2020 Part 4 and CSA S6:19 treat differential support settlement as an imposed deformation to be combined with the permanent loads, so the self-equilibrating moment set computed in Question 6 is a real design action, not a curiosity.

Sign convention used throughout. End moments are written counter-clockwise-positive on the member, which is the convention that matches the standard six-degree-of-freedom element stiffness matrix; the slope-deflection equation is then $M_{ij} = \dfrac{2EI}{L}\left(2\theta_i + \theta_j - 3\psi_{ij}\right) + \mathrm{FEM}_{ij}$ with $\psi$ the counter-clockwise chord rotation and, for a downward UDL, $\mathrm{FEM}_{ij} = +wL^2/12$, $\mathrm{FEM}_{ji} = -wL^2/12$. Ordinary sagging bending moments, which is what every diagram below plots, are recovered as $M_{\text{sag}}(i) = -M_{ij}$ and $M_{\text{sag}}(j) = +M_{ji}$. Mixing this with the clockwise-positive convention of some textbooks produces clean-looking integers that are wrong.

Question 5: Influence lines for a three-span continuous beam (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.

SpanLengthRelative $EI$$L/I$
A–B4 m$EI$4
B–C8 m$2EI$4
C–D4 m$EI$4

All four supports are simple (one is a pin for horizontal restraint), so $M_A = M_D = 0$ and the beam is twice statically indeterminate.

Find. The shape of the influence line for $M_B$ and for $V_{B^{+}}$ over the whole length A to D, and the numerical ordinate of each at the centre of span B–C.

ABCD4 m (EI)8 m (2EI)4 m (EI)
Question 5: three-span continuous beam on four simple supports, spans 4 m (EI), 8 m (2EI) and 4 m (EI).

Approach. By the Mueller-Breslau principle each influence line is the deflected shape produced by releasing the corresponding action, so the shapes follow at once; the ordinate at a particular point is then obtained by placing a unit load there and analysing the beam, which for the required point is a symmetric problem that collapses in two lines of the three-moment equation.

  1. Part (a) — the shape of the influence line for $M_B$. Insert a hinge at B and apply a unit relative rotation. The beam takes up a smooth, continuous deflected shape with a kink at B, negative (downward, hence hogging) through spans A–B and B–C, zero at every support, and a small reversed lobe in the far span C–D. Note the two consequences of continuity that a determinate sketch would miss: the ordinate is non-zero right across span C–D, and the curve is cubic in every span, not made of straight lines.
  2. Part (b) — the shape of the influence line for $V_{B^{+}}$. Cut the beam immediately to the right of B and impose a unit relative transverse displacement across the cut, keeping both faces parallel. The result is a curve that is small and negative over most of A–B, jumps to the value $+1.000$ as the load crosses B, then falls smoothly to zero at C and shows a small reversed lobe in C–D. The jump of exactly one unit at the section is the property that no amount of redundancy can change — a unit load standing on the section is carried wholly by the support on whichever side it stands.
  3. Set up the three-moment equation for a unit load at the required point. Place $P = 1$ at the centre of span B–C, 4 m from each of B and C. Because $L_1/I_1 = L_2/I_2 = L_3/I_3 = 4$ and the load stands on the axis of symmetry, the problem is symmetric and $M_B = M_C$. The simple-beam moment diagram of span B–C has area $A_2 = PL_2^{2}/8 = (1)(8)^2/8 = 8.0$ with its centroid at mid-span, so $\bar a_2 = \bar b_2 = 4.0\ \text{m}$. The three-moment equation at B is $$M_A\frac{L_1}{I_1} + 2M_B\left(\frac{L_1}{I_1}+\frac{L_2}{I_2}\right) + M_C\frac{L_2}{I_2} = -6\left(\frac{A_1\bar a_1}{I_1L_1} + \frac{A_2\bar b_2}{I_2L_2}\right)$$ $$0 + 2M_B(4+4) + 4M_C = -6\left(0 + \frac{8.0\times 4.0}{2\times 8}\right) = -12.0 .$$
  4. Solve. The companion equation at C is the mirror image, $4M_B + 16M_C = -12.0$, so with $M_B = M_C$ $$20M_B = -12.0 \quad\Longrightarrow\quad \boxed{\eta_{M_B} = M_B = -0.600\ \text{m}}$$ The ordinate of a bending-moment influence line has units of length; the minus sign says the support hogs, which is the answer to part (a) at the centre line.
    5. ABCD-0.600 m at mid B-CInfluence line for the bending moment over Bordinate in metres
    Question 5(a): influence line for the bending moment over support B. The ordinate at the centre of span B-C is -0.600 m.
  5. Get the shear ordinate from the same solution. Span A–B carries no load, so its moment varies linearly and $M_B = R_A L_1$, giving $$R_A = \frac{-0.600}{4} = -0.150 \quad(\text{a hold-down}).$$ For span B–C, taking moments about C on the span free body, $$V_{B^{+}} = \frac{M_C - M_B + P L_2/2}{L_2} = \frac{-0.600 + 0.600 + 4.0}{8} = \boxed{\eta_{V_{B^{+}}} = +0.500}$$ The support reaction follows as $R_B = 0.500 - (-0.150) = 0.650$, and the check $R_A + R_B = 0.500 = V_{B^{+}}$ closes because there is no load to the left of the section.
    6. ABCD+0.500 at mid B-CInfluence line for the shear immediately right of Bdimensionless
    Question 5(b): influence line for the shear immediately right of B. It jumps to +1.000 over the support and reads +0.500 at mid B-C.
  6. Read the two answers against intuition. That the shear ordinate comes out at exactly one half is not a coincidence and not the simply supported value by accident: the load stands on the axis of symmetry of a symmetric structure, so it must split equally, and the two hogging end moments of the span are equal and cancel out of the shear expression. The moment ordinate $-0.600\ \text{m}$, by contrast, is genuinely a continuity effect — a determinate two-span reading of the same beam would give zero there.
  7. Where the extremes really are. The largest hogging value of the $M_B$ influence line is $-0.641\ \text{m}$ at about 3.07 m past B (not at the centre line), and the $V_{B^{+}}$ influence line peaks at exactly $+1.000$ with the load standing over B. Both spans A–B and C–D carry reversed lobes, so a design pattern load must be placed span by span, loading only those spans whose ordinates carry the sign being maximised.
QuantityPosition of the unit loadOrdinate
Influence line for $M_B$centre of span B–C −0.600 m (hogging)
Influence line for $V_{B^{+}}$centre of span B–C +0.500
Support moments for that load position— $M_B = M_C = -0.600\ \text{m}$
Reactions for that load position— $R_A = -0.150$, $R_B = +0.650$
Maximum hogging ordinate of the $M_B$ line 3.07 m past B−0.641 m
Maximum ordinate of the $V_{B^{+}}$ lineover B +1.000