16-Civ-B1 Advanced Structural Analysis · December 2013

Question 7 of 9: Slope-deflection frame with combined loading and a support settlement

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

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

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 7: Slope-deflection frame with combined loading and a 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. A continuous beam 1–2–3 of two 12 m spans carries 10 kN/m over span 1–2 and an 80 kN point load at the mid-point of span 2–3. Joint 1 is a roller, joint 3 is encastré, and a 12 m column hangs from joint 2 down to a pinned foot at joint 4. Joint 1 settles 12 mm.

Given data
Quantity	Symbol	Value
Span 1–2 and span 2–3	L	12 m each
Column 2–4	h	12 m
Uniform load on span 1–2	w	10 kN/m
Point load at the mid-point of span 2–3	P	80 kN
Settlement of joint 1	Δ	12 mm downward
Flexural rigidity, all members	EI	3.2 × 10⁵ kN·m²
Support at 1 / 3 / 4	—	roller / fixed / pinned

Find. All member end moments, and the shear-force and bending-moment diagrams for every member with their extreme ordinates.

Question 7: two 12 m spans on a 12 m column. The settlement of joint 1 acts alongside the two applied loads, not instead of them.

Approach. Establish first that the frame cannot sway, so the only unknown is the rotation of joint 2; then write the three member equations with the settlement entering as a chord rotation on span 1–2 and solve the single joint equilibrium equation.

Show there is no sway. Span 2–3 is inextensible and joint 3 is fixed, so joint 2 cannot move horizontally; column 2–4 is inextensible and joint 4 is a fixed point, so joint 2 cannot move vertically either. Joint 2 is therefore held in position and the only sway-type action left is the prescribed 12 mm drop of joint 1. This reduces a structure that is indeterminate to the third degree to a single unknown displacement, $\theta_2$, once the pinned ends are condensed out.
Convert the settlement to a chord rotation. For span 1–2 with $\boldsymbol{\Delta}_1=(0,-0.012)$ and $\boldsymbol{\Delta}_2=\mathbf{0}$, $$\psi_{12}=\frac{(0-(-0.012))}{12}=+1.000\times10^{-3}\ \text{rad},$$ while $\psi_{23}=\psi_{24}=0$.
Write the member equations and condense the pins. With $2EI/L=5.333\times10^{4}$ throughout, and the fixed-end moments $+wL^{2}/12=+120$ and $+PL/8=+120$ at the respective i ends, imposing $M_{12}=0$ at the roller and $M_{42}=0$ at the pinned column foot eliminates $\theta_1$ and $\theta_4$ and leaves $$M_{21}=1.5k\theta_2-260,\qquad M_{23}=2k\theta_2+120,\qquad M_{24}=1.5k\theta_2,$$ with $k=2EI/L=5.333\times10^{4}$.
Balance joint 2. Setting $M_{21}+M_{23}+M_{24}=0$ gives $$5k\,\theta_2=140\quad\Longrightarrow\quad \theta_2=+5.250\times10^{-4}\ \text{rad}.$$
Back-substitute. $$M_{21}=\boxed{-218\ \text{kN}\cdot\text{m}},\quad M_{23}=\boxed{+176\ \text{kN}\cdot\text{m}},\quad M_{24}=\boxed{+42\ \text{kN}\cdot\text{m}},\quad M_{32}=\boxed{-92\ \text{kN}\cdot\text{m}}$$ and the three moments meeting at joint 2 sum to zero as required. The 12 mm settlement is worth $3EI\Delta/L^{2}=80$ kN·m on its own, and it relieves the hogging over joint 2 from 298 to 218 kN·m — a 27 % reduction, so it is emphatically not a second-order effect.
Build the diagrams member by member. On span 1–2, $M_{\text{sag}}(x)=V_1x-5x^{2}$ with $V_1=41.83$ kN from $12V_1-720=-218$; the shear falls to $-78.17$ kN at joint 2 and vanishes at $$x=\frac{41.83}{10}=4.183\ \text{m},\qquad M_{\max}=\boxed{+87.50\ \text{kN}\cdot\text{m}}.$$ On span 2–3 the shear is $+47$ kN up to the load and $-33$ kN beyond it, and the moment runs $-176 \to +106 \to -92$ kN·m. The column carries a constant 3.5 kN shear and a moment falling linearly from 42 kN·m at joint 2 to zero at the pin.

Question 7: bending moment along the beam 1–2–3. The step at joint 2 from −218 to −176 kN·m is the 42 kN·m taken by the column, not a plotting error.

Question 7: shear force along the beam 1–2–3.

Question 7: column 2–4 bending moment; the shear is a constant 3.5 kN over the full 12 m.

Question 7 — results
Member	Bending moment (kN·m)	Shear (kN)
Span 1–2	0 at joint 1; max +87.50 at 4.183 m; −218 at joint 2	+41.83 falling to −78.17
Span 2–3	−176 at joint 2; +106 under the load; −92 at joint 3	+47.0 then −33.0
Column 2–4	−42 at joint 2 to 0 at joint 4	3.5 constant
Reactions: joint 1 = 41.83 kN; joint 3 = 33.0 kN vertical, 3.5 kN horizontal, 92 kN·m fixing moment; joint 4 = 125.17 kN vertical, 3.5 kN horizontal. Total vertical = 200 kN = 10(12) + 80. ✓

Topic. Slope-deflection for a non-sway frame carrying loads and a support settlement simultaneously.

Key relations. $M_{ij}=(2EI/L)(2\theta_i+\theta_j-3\psi_{ij})+\mathrm{FEM}_{ij}$; modified end stiffness $3EI/L$ and modified carry-over of zero for a member pinned at its far end; settlement effect on a propped cantilever $=3EI\Delta/L^{2}$; $M_{\text{sag}}(x)=-M_{ij}+V_i x-wx^{2}/2$ within a member.

Why this works. The two inextensibility statements are what turn a three-times-indeterminate frame into a one-equation problem. Once joint 2 is shown to be immovable, the settlement of joint 1 has nowhere to go except into the chord rotation of span 1–2, and everything else follows from a single moment balance. It is also worth noticing which way the settlement acts: dropping the far simple support of a propped cantilever unloads the built-in end, so the settlement here is beneficial for the beam — and would be harmful if joint 1 had been jacked up instead.

Common pitfall. Treating the settlement as an alternative load case and reporting it separately. The question says “in addition to the loads shown”, so the chord rotation and the fixed-end moments must appear in the same set of equations. The other frequent error is forgetting that the beam moment splits at joint 2 — the step of 42 kN·m between −218 and −176 is the column’s share and must be drawn, not averaged away.

Source. Hibbeler, Structural Analysis, 10th ed., §11.3–11.4 (slope-deflection for frames without sidesway) and §12.4 (moment distribution with support settlement).

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