07-Str-A4 · December 2014

Question 7 of 9: Slope-deflection with a settling support (22 marks)

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

Notes on this paper

Paper format. 07-Str-A4 Advanced Structural Analysis, National Examination, December 2014 — 3 hours, closed book (approved Sharp or Casio calculator permitted). Nine questions over six pages: Questions 1 and 2 are compulsory (12 and 8 marks), then any two of Questions 3, 4, 5 (18 marks each) and any two of Questions 6, 7, 8, 9 (22 marks each). Six questions constitute a complete paper and total 100 marks; marks are printed in the left margin. All nine questions are solved here, because the set is a study resource rather than a three-hour sitting.

Reference texts.

R. C. Hibbeler, Structural Analysis, 10th ed. — Ch. 6 (influence lines), Ch. 9 (deflections by energy methods), Ch. 10 (force/flexibility method), Ch. 11 (slope-deflection).
A. Kassimali, Structural Analysis, 6th ed. — Ch. 3 (determinacy and stability), Ch. 8 (influence lines for floor-beam systems), Ch. 13 (least work), Ch. 15–16 (slope-deflection with and without sidesway).
A. Ghali, A. M. Neville and T. G. Brown, Structural Analysis: A Unified Classical and Matrix Approach, 7th ed. — Ch. 4 (force method), Ch. 5 (lack of fit and support movement).
W. McGuire, R. H. Gallagher and R. D. Ziemian, Matrix Structural Analysis, 2nd ed. — Ch. 4–5 (member stiffness, assembly, kinematic constraints).
CSA S6:19 Canadian Highway Bridge Design Code and NBCC 2020 — the Canadian codes that these analysis methods feed; no code check is required by this paper.

Sign convention used throughout. End moments are counter-clockwise positive, the convention that matches the standard six-degree-of-freedom stiffness element. In it the slope-deflection equation is

$$M_{ij}=\frac{2EI}{L}\bigl(2\theta_i+\theta_j-3\psi_{ij}\bigr)+\mathrm{FEM}_{ij}$$

the chord rotation is $\psi_{ij}=\bigl[(\mathbf{d}_j-\mathbf{d}_i)\cdot\mathbf{e}_2\bigr]/L$ with $\mathbf{e}_2$ the member axis turned $+90^\circ$, the fixed-end moment of a downward UDL is $\mathrm{FEM}_{i}=+wL^{2}/12$ and $\mathrm{FEM}_{j}=-wL^{2}/12$, and the ordinary sagging bending moment is recovered as $M_{\text{sag}}(i)=-M_{ij}$ and $M_{\text{sag}}(j)=+M_{ji}$. Every translation degree of freedom is closed with the virtual-work statement

$$\sum_{\text{members}}\bigl(M_{ij}+M_{ji}\bigr)\,\psi^{*}_{ij}\;+\;\sum \mathbf{F}\cdot\mathbf{d}^{*}=0$$

in which $\mathbf{F}$ are the equivalent nodal loads and $\mathbf{d}^{*}$, $\psi^{*}$ belong to the unit sway pattern. Mixing this with the clockwise-positive convention of some textbooks produces clean-looking but wrong answers, which is the single most common failure on this subject.

All work below uses the drawing.

Question 7: Slope-deflection with a settling support (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. A 12 m horizontal member on a pin at joint (1) and a 4 m column at joint (2), uniformly loaded, with joint (1) settling 10 mm.

Given data
Quantity	Value
Overhang left of joint (1)	2 m
Span (1)–(2)	8 m
Overhang right of joint (2)	2 m
Uniform load, whole 12 m	3.6 kN/m downward
Column (2)–(3)	4 m, pinned at the base (3)
Settlement of the support at joint (1)	0.01 m downward
Flexural rigidity	$EI=2.4\times 10^{4}\ \text{kN}\cdot\text{m}^{2}$

Find. Shear force and bending moment diagrams for the horizontal member and the column, with the extreme ordinates labelled.

Question 7: pins at joints (1) and (3) give four reactions, so the structure is one degree indeterminate; the settlement enters through the chord rotation of span (1)–(2).

Approach. Reduce the two overhangs to known end moments, express the settlement as a chord rotation of the 8 m span, and solve the two joint rotations by slope-deflection with the column's pinned base condensed out.

Reduce the overhangs. Each 2 m cantilever carries $3.6\ \text{kN/m}$, so it hands its joint a known moment and shear $$M_{oh}=\frac{3.6(2)^{2}}{2}=7.2\ \text{kN}\cdot\text{m (hogging)},\qquad V_{oh}=3.6(2)=7.2\ \text{kN}$$ At joint (1) there is no third member, so this fixes $M_{12}=+7.2$ outright; at joint (2) it is one of the three moments that must balance.
Kinematics of the settlement. The column is inextensible and pinned at (3), so $v_2=0$; the horizontal member is inextensible and pinned at (1), so $u_1=u_2=0$ and the column does not sway. Only span (1)–(2) sees the settlement: $$\psi_{12}=\frac{v_2-v_1}{8}=\frac{0-(-0.01)}{8}=+1.25\times10^{-3}$$
Fixed-end moments and member equations. With $wL^{2}/12=3.6(64)/12=19.2\ \text{kN}\cdot\text{m}$ and $2EI/L=6000$, $$M_{12}=6000\bigl(2\theta_1+\theta_2-3.75\times10^{-3}\bigr)+19.2,\qquad M_{21}=6000\bigl(2\theta_2+\theta_1-3.75\times10^{-3}\bigr)-19.2$$ The column's base is a pin, so it uses the modified stiffness $M_{23}=\tfrac{3EI}{4}\theta_2=18000\,\theta_2$.
Two equations for two rotations. Continuity of moment at joint (1) requires $M_{12}=+7.2$; equilibrium at joint (2) requires $M_{21}+M_{23}+7.2=0$. These give $$2\theta_1+\theta_2=1.75\times10^{-3},\qquad 6000\,\theta_1+30000\,\theta_2=34.5$$ $$\boxed{\theta_1=3.333\times10^{-4}\ \text{rad},\qquad \theta_2=1.0833\times10^{-3}\ \text{rad}}$$
End moments. Back-substitution gives $$\boxed{M_{(1)}=-7.2\ \text{kN}\cdot\text{m},\qquad M_{(2),\text{beam}}=-26.7\ \text{kN}\cdot\text{m},\qquad M_{(2),\text{column}}=+19.5\ \text{kN}\cdot\text{m}}$$ Joint (2) balances as a step in the moment diagram, $26.7-7.2=19.5\ \text{kN}\cdot\text{m}$, which is the column's contribution — joint equilibrium, not an error.
Shears in the horizontal member. From the end moments and the UDL, $$V(0^{+})=+11.9625\ \text{kN},\qquad V(8^{-})=11.9625-3.6(8)=-16.8375\ \text{kN}$$ with $-7.2$ kN just left of joint (1) and $+7.2$ kN just right of joint (2) on the overhangs.
Maximum sagging moment. Zero shear occurs at $$x=\frac{11.9625}{3.6}=3.3229\ \text{m},\qquad M_{\max}=-7.2+\frac{11.9625^{2}}{2(3.6)}=+12.675\ \text{kN}\cdot\text{m}$$
Column and reactions. The column has 19.5 kN·m at the top and zero at the pin, so its shear is constant: $$V_{\text{col}}=\frac{19.5}{4}=4.875\ \text{kN}$$ and the vertical reactions follow from the shear jumps, $$\boxed{R_{(1)}=19.1625\ \text{kN},\qquad R_{(3)}=24.0375\ \text{kN},\qquad H=4.875\ \text{kN}}$$ $$\textstyle\sum V:\;19.1625+24.0375=43.2=3.6(12)\quad\checkmark$$
Isolate the effect of the settlement. Repeating the analysis with no settlement gives a hogging moment of 19.2 kN·m at joint (2); the 10 mm settlement therefore adds 7.5 kN·m, a 39 % increase, and pushes the maximum sagging moment out towards midspan.

The settlement makes the structure worse, not better, because the support that drops is the one that was propping the loaded span. That is the general rule for this configuration and it is worth stating explicitly in an examination answer.

Question 7: shear force in the horizontal member, including both overhangs. The discontinuities are the two support reactions.

Question 7: bending moment in the horizontal member, sagging positive. The step at joint (2) is the column moment.

Question 7: the column carries a linear moment from 19.5 kN·m at the head to zero at the pinned base.

Question 7 — complete results
Member / quantity	Shear (max / min)	Bending moment (max / min)
Left overhang, 2 m	0 to −7.2 kN	0 to −7.2 kN·m
Span (1)–(2), 8 m	+11.9625 to −16.8375 kN	max +12.675 kN·m at $x=3.323$ m; min −26.7 kN·m at (2)
Right overhang, 2 m	+7.2 to 0 kN	−7.2 to 0 kN·m
Column (2)–(3), 4 m	4.875 kN constant	+19.5 kN·m at (2) to 0 at (3)
Reactions	$R_{(1)}=19.1625$ kN, $R_{(3)}=24.0375$ kN, $H=4.875$ kN; joint rotations $\theta_1=3.333\times10^{-4}$, $\theta_2=1.0833\times10^{-3}$ rad

Topic. Slope-deflection analysis of a frame with a prescribed support settlement and cantilever overhangs.

Key relations. $\psi_{ij}=(v_j-v_i)/L$; $\mathrm{FEM}=\pm wL^{2}/12$; modified stiffness $3EI/L$ at a pinned far end; an overhang delivers $M=wL_{oh}^{2}/2$ and $V=wL_{oh}$ to its joint; the moment steps at a joint by the amount carried by the third member.

Why this works. A settlement enters the slope-deflection equation in exactly the same slot as a joint rotation, as the chord rotation $\psi$. Nothing about the method changes; only the numerical value of $\psi$ is non-zero.

Common pitfall. Treating the overhang as an unknown span. It is determinate: its moment and shear at the joint are known before any analysis, and it belongs on the right-hand side of the joint equilibrium equation, not in the stiffness.

Source. Kassimali, Structural Analysis, 6th ed., Ch. 15 (slope-deflection, structures without sidesway, support settlements); Hibbeler, Structural Analysis, 10th ed., §11.3–11.4.

← Question 6 Next: Question 8 →