07-Str-A4 · December 2015

Question 6 of 9: Three-Span Beam with Support Settlement

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 6: Three-Span Beam with Support Settlement (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.

Given.

Item	Value
Overhang, joint 1 (free) to joint 2	2.0 m
Span 2–3	8.0 m
Span 3–4	8.0 m
Uniformly distributed load	$w=6\ \text{kN/m}$ over the whole 18.0 m
Settlement of support 3	36 mm below supports 2 and 4
Flexural rigidity	$EI = 3.2\times10^{4}\ \text{kN}\cdot\text{m}^{2}$ (see the unit note above)
Supports	Pin at joint 2, rollers at joints 3 and 4

Find. Support moments, reactions, and the complete shear and bending moment diagrams with the maximum and minimum ordinate in each member.

Question 6 — two equal spans with a loaded overhang, the interior support 36 mm low.

Approach. Fix the overhang moment from statics, treat the two 8 m spans as a beam with simple ends, superpose the load effect and the settlement effect at the interior support, and recover shears and reactions span by span.

Take the overhang out of the problem. The 2 m cantilever is determinate, so the moment it delivers to support 2 is known immediately: $$M_{2} = -\frac{w a^{2}}{2} = -\frac{6(2.0)^{2}}{2} = \boxed{-12.00\ \text{kN}\cdot\text{m}}$$ (hogging). That is a known applied moment at joint 2, not an unknown — it does not add a degree of freedom.
Identify what the settlement does on its own. For two equal spans $L$ with simple ends and the interior support settling $\Delta$, the classical result is that the interior support moment changes by $$\Delta M_{3} = \frac{3EI\Delta}{L^{2}} = \frac{3\left(3.2\times10^{4}\right)(0.036)}{8.0^{2}} = \frac{3456}{64} = 54.00\ \text{kN}\cdot\text{m}$$ in the relieving (sagging) sense. Because the middle support drops away, it stops pushing the beam up so hard, so the hogging over it is reduced.
Superpose with the load effect. Solving the beam with the settlement suppressed gives $M_3 = -45.00\ \text{kN}\cdot\text{m}$ (hogging). Adding the settlement contribution, $$M_{3} = -45.00 + 54.00 = \boxed{+9.00\ \text{kN}\cdot\text{m}\ \text{(sagging)}}$$ This is the striking result of the question: a settlement of only 36 mm has reversed the sign of the interior support moment. The threshold is $\Delta M_3 = 45$, i.e. $\Delta = 45(64)/(3\times3.2\times10^{4}) = 30.0\ \text{mm}$; anything beyond 30 mm turns the hog into a sag, and the design question really being asked is whether the top steel over that support is still needed and whether bottom steel now is.
Recover the end shears in span 2–3. With sagging end moments $-12.00$ and $+9.00$ over an 8 m span carrying 6 kN/m, $$V_{2} = \frac{wL}{2} + \frac{M_{3}-M_{2}}{L} = 24.00 + \frac{9.00-(-12.00)}{8.0} = 24.00+2.625 = 26.625\ \text{kN}$$ $$V_{3}^{\,\text{left}} = 26.625 - 6(8.0) = -21.375\ \text{kN}$$
Recover the end shears in span 3–4. The far end is a roller with $M_4=0$: $$V_{3}^{\,\text{right}} = 24.00 + \frac{0-9.00}{8.0} = 24.00-1.125 = 22.875\ \text{kN}$$ $$V_{4} = 22.875 - 48.00 = -25.125\ \text{kN}$$
Assemble the reactions and check global equilibrium. The overhang delivers $6(2.0)=12.00\ \text{kN}$ to joint 2 on top of the span shear: $$R_{2}=12.00+26.625=38.625\ \text{kN},\quad R_{3}=21.375+22.875=44.250\ \text{kN},\quad R_{4}=25.125\ \text{kN}$$ $$\sum R = 38.625+44.250+25.125 = 108.00\ \text{kN} = 6(18.0)\quad\checkmark$$
Locate the span maxima. Zero shear in span 2–3 occurs at $x = 26.625/6 = 4.4375\ \text{m}$ from support 2: $$M_{\max} = -12.00 + 26.625(4.4375) - 3.0(4.4375)^{2} = \boxed{+47.07\ \text{kN}\cdot\text{m}}$$ and in span 3–4 at $x = 22.875/6 = 3.8125\ \text{m}$ from support 3: $$M_{\max} = 9.00 + 22.875(3.8125) - 3.0(3.8125)^{2} = \boxed{+52.61\ \text{kN}\cdot\text{m}}$$ There is one point of contraflexure only, at $x=0.476\ \text{m}$ into span 2–3; span 3–4 sags over its entire length because the settlement has removed the hogging that would normally sit over support 3.

Question 6 shear force. The step at each support is the reaction; the overhang shear runs from zero at the free tip to $-12$ kN at support 2.

Question 6 bending moment, sagging positive. Note that the ordinate over support 3 is above the axis — the 36 mm settlement has flipped it from $-45$ to $+9$ kN·m.

Member	$V$ at left end	$V$ at right end	$M$ maximum	$M$ minimum
Overhang 1–2 (2 m)	0	$-12.00$ kN	0 (free tip)	$-12.00$ kN·m at support 2
Span 2–3 (8 m)	$+26.625$ kN	$-21.375$ kN	$+47.07$ kN·m at 4.44 m	$-12.00$ kN·m at support 2
Span 3–4 (8 m)	$+22.875$ kN	$-25.125$ kN	$+52.61$ kN·m at 3.81 m	0 at support 4
Support moments: $M_2 = -12.00$ kN·m; $M_3 = +9.00$ kN·m (would be $-45.00$ with no settlement); $M_4=0$
Reactions: $R_2 = 38.625$ kN, $R_3 = 44.250$ kN, $R_4 = 25.125$ kN (sum 108.00 kN = $wL_{\text{total}}$)

Topic. Continuous beams under combined load and differential support settlement.

Key relations. Slope-deflection $M_{ij}=\frac{2EI}{L}(2\theta_i+\theta_j-3\psi_{ij})+\mathrm{FEM}_{ij}$ with $\psi = \Delta/L$; for two equal spans with simple ends, $M_{\text{int}} = -\frac{wL^{2}}{8}+\frac{3EI\Delta}{L^{2}}$; end shear $V_i = wL/2 + (M_j-M_i)/L$ in the sagging convention.

Why this works. Settlement enters the slope-deflection equations exactly like a load, through the chord rotation $\psi$, and superposes linearly with the load effect because the structure is elastic. The magnitude of the settlement effect scales with $EI/L^{2}$, so it grows with stiffness and shrinks rapidly with span — the reason short, stiff, continuous members are the ones that suffer most from foundation movement. Quoting the threshold settlement (30 mm here) turns the answer from a number into a design statement.

Common pitfall. Treating the overhang as an unknown span. It is a determinate cantilever: it supplies a known moment and a known shear to support 2 and contributes no stiffness and no degree of freedom. The second trap is disbelieving the positive $M_3$; a sagging moment over an interior support looks wrong but is exactly what a large enough settlement produces, and the diagram then shows only one point of contraflexure in the entire 18 m beam.

Source. Hibbeler, Structural Analysis, 10th ed., §11.2–11.3 (slope-deflection with settlement) and §12.4; Ghali, Neville & Brown, 7th ed., §4.5.

← Question 5 Next: Question 7 →

Member	\(V\) at left end	\(V\) at right end	\(M\) maximum	\(M\) minimum
Overhang 1–2 (2 m)	0	\(-12.00\) kN	0 (free tip)	\(-12.00\) kN·m at support 2
Span 2–3 (8 m)	\(+26.625\) kN	\(-21.375\) kN	\(+47.07\) kN·m at 4.44 m	\(-12.00\) kN·m at support 2
Span 3–4 (8 m)	\(+22.875\) kN	\(-25.125\) kN	\(+52.61\) kN·m at 3.81 m	0 at support 4
Support moments: \(M_2 = -12.00\) kN·m; \(M_3 = +9.00\) kN·m (would be \(-45.00\) with no settlement); \(M_4=0\)
Reactions: \(R_2 = 38.625\) kN, \(R_3 = 44.250\) kN, \(R_4 = 25.125\) kN (sum 108.00 kN = \(wL_{\text{total}}\))