07-Str-A4 · December 2014

Question 5 of 9: Slope-deflection with a lack of fit (18 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 5: Slope-deflection with a lack of fit (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. A symmetric structure: a continuous 16 m top member on two splayed pin-based legs, propped at midspan by a hanger that is pinned at both ends and was made 8 mm too long.

Given data
Quantity	Value
Top member, joints (2)–(3)–(4)	two spans of 8 m, continuous through (3)
Leg (1)–(2)	1.8 m across, 2.4 m up ⇒ length 3.0 m; pinned at (1), rigid at (2)
Leg (4)–(5)	the mirror image
Member (3)–(6)	vertical, 2.4 m, pinned at both ends
Fabrication error	8 mm too long
Flexural rigidity	$EI=8.0\times 10^{5}\ \text{kN}\cdot\text{m}^{2}$, all members
Applied load	none

Find. Shear and bending moment diagrams for every member, with the maximum and minimum ordinates labelled.

Question 5: the hanger (3)–(6) carries an open circle at each end, so it is a two-force member; forcing an over-length link into place therefore imposes a displacement, not a load.

Approach. Convert the fabrication error into a prescribed 8 mm lift of joint (3), prove from inextensibility and symmetry that joints (2) and (4) do not move, and solve the resulting two-span beam on rotational springs.

Interpret the lack of fit. Member (3)–(6) has a pin at each end and no load along it, so it is a two-force member, and it is inextensible. A link 8 mm longer than the gap can only be installed by pushing joint (3) up by the same amount: $$\boxed{v_3=+0.008\ \text{m (upward, prescribed)}}$$ The force it then carries is an unknown to be recovered, not an input.
Show that joints (2) and (4) do not translate. The top member is inextensible, so $u_2=u_4$. The structure and the imposed displacement are symmetric about joint (3), and a symmetric response requires $u_2=-u_4$; therefore $u_2=u_4=0$. Each leg is inextensible along its own axis $(0.6,\,0.8)$, so $$0.6u_2+0.8v_2=0\;\Longrightarrow\;v_2=v_4=0$$ The whole problem collapses to a two-span beam with a jacked centre support.
Replace each leg by a rotational spring. A member of length 3.0 m whose far end is pinned and whose near end cannot translate has the modified stiffness $$k_{\theta}=\frac{3EI}{L}=\frac{3EI}{3}=EI$$ so joint (2) is restrained by a spring of stiffness $EI$, and joint (4) likewise.
Chord rotations of the two spans. With $v_2=v_4=0$ and $v_3=+0.008$ m, $$\psi_{23}=\frac{v_3-v_2}{8}=+1.00\times10^{-3},\qquad \psi_{34}=\frac{v_4-v_3}{8}=-1.00\times10^{-3}$$ Symmetry gives $\theta_3=0$ and $\theta_4=-\theta_2$.
Solve joint (2). The slope-deflection equations with no fixed-end moments give $M_{21}=EI\theta_2$ and $M_{23}=\tfrac{EI}{4}(2\theta_2-3\psi_{23})$, so $$EI\theta_2+\frac{EI}{4}\bigl(2\theta_2-0.003\bigr)=0\;\Longrightarrow\;1.5\,\theta_2=7.5\times10^{-4}$$ $$\boxed{\theta_2=+5.00\times 10^{-4}\ \text{rad},\qquad \theta_4=-5.00\times 10^{-4}\ \text{rad},\qquad \theta_3=0}$$
End moments. Back-substituting with $EI=8.0\times 10^{5}$, $$M_{21}=+400\ \text{kN}\cdot\text{m},\quad M_{23}=-400,\quad M_{32}=-500,\quad M_{34}=+500,\quad M_{43}=+400,\quad M_{45}=-400$$ so in sagging terms the top member reads $$\boxed{M_{(2)}=+400\ \text{kN}\cdot\text{m},\qquad M_{(3)}=-500\ \text{kN}\cdot\text{m},\qquad M_{(4)}=+400\ \text{kN}\cdot\text{m}}$$ Each joint balances: $400-400=0$ at (2), $-500+500=0$ at (3).
Shears and the link force. The spans carry no load, so the moment is linear and the shear constant: $$V_{\text{beam}}=\frac{-500-400}{8}=\mp112.5\ \text{kN},\qquad V_{\text{leg}}=\frac{400}{3.0}=133.33\ \text{kN}$$ The jump in beam shear at joint (3) is the hanger force: $$\boxed{N_{(3)-(6)}=112.5+112.5=225\ \text{kN COMPRESSION}}$$ A member made too long and forced into place must finish in compression; had the arithmetic produced tension, the sign of $\psi$ would have been inverted.
Axial forces and reactions. Resolving at joint (2), the leg delivers a horizontal thrust of 251.0 kN into the beam, so the top member carries 251.0 kN of axial compression, and $$R_{(1)}=R_{(5)}=112.5\ \text{kN acting DOWNWARD},\qquad H_{(1)}=-H_{(5)}=251.0\ \text{kN}$$ $$\textstyle\sum V:\;-112.5-112.5+225=0\quad\checkmark$$

Check: both outer supports are pulled down by 112.5 kN. A self-strained structure of this kind needs holding-down anchors at (1) and (5) capable of that uplift-reversal; a nominal bearing detail would simply lift off and release the whole 225 kN prop force.

Question 5: bending moment in the continuous top member. Straight lines because there is no applied load — every ordinate comes from the 8 mm lack of fit.

Question 5: shear in the top member, constant at 112.5 kN in each span and reversing across the hanger.

Question 5 — complete results
Member	Bending moment (max / min)	Shear	Axial
Leg (1)–(2), 3.0 m	0 at the pin to 400 kN·m at (2)	133.33 kN constant	240.6 kN compression
Span (2)–(3), 8 m	max +400, min −500 kN·m	112.5 kN constant	251.0 kN compression
Span (3)–(4), 8 m	max +400, min −500 kN·m	112.5 kN constant	251.0 kN compression
Leg (4)–(5), 3.0 m	400 kN·m at (4) to 0 at the pin	133.33 kN constant	240.6 kN compression
Hanger (3)–(6), 2.4 m	0 (two-force member)	0	225 kN compression
Joint rotations	$\theta_2=+5.00\times10^{-4}$, $\theta_3=0$, $\theta_4=-5.00\times10^{-4}$ rad

Topic. Self-straining of an indeterminate structure by a fabrication error (lack of fit), analysed by slope-deflection.

Key relations. A pin-ended, inextensible link that is $\delta$ too long imposes a displacement $\delta$ at the joint it props; modified stiffness $3EI/L$ for a member whose far end is pinned; $\psi_{ij}=(v_j-v_i)/L$; $M_{ij}=\tfrac{2EI}{L}(2\theta_i+\theta_j-3\psi_{ij})$.

Why this works. A lack of fit is a kinematic input, not a load. Because the structure is indeterminate it cannot accommodate the misfit as a rigid-body motion, so the misfit is absorbed as strain and the internal forces follow — which is why a completely unloaded frame can develop 500 kN·m and 225 kN.

Common pitfall. Applying the misfit as a force, or getting its sign backwards. The physical check is unambiguous: over-length plus forced into place equals compression in the link, and every other quantity flows from that.

Source. Ghali, Neville and Brown, Structural Analysis: A Unified Classical and Matrix Approach, 7th ed., Ch. 5 (lack of fit, temperature and support movement); Kassimali, Structural Analysis, 6th ed., §16.4.

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