16-Civ-B1 Advanced Structural Analysis · December 2013

Question 9 of 9: Derivation of the stiffness matrix and load vector

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 9: Derivation of the stiffness matrix and load vector (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 frame with encastré feet at joints 1 and 4. Joint 1 sits at the origin and joint 2 is 3 m to its right and 4 m above it; joint 3 is 5 m to the right of joint 2 at the same level; joint 4 is directly below joint 3 by 4 m and 3 m to its left, so both inclined members are 5 m long and parallel. Member 2–3 is horizontal, 5 m long, and carries 2 kN/m downward. The rigidity $EI$ is the same for all three members.

Given data
Quantity	Symbol	Value
Length of each inclined member	L₁	5 m
Horizontal projection of each inclined member	a	3 m
Vertical projection of each inclined member	b	4 m
Length of the horizontal member 2–3	L₂	5 m
Uniform load on member 2–3	w	2 kN/m
Supports at joints 1 and 4	—	encastré

Find. The translation equation, the two joint-moment equations, and the resulting stiffness matrix and load vector. The equations are not to be solved.

Question 9: the parallelogram frame. Both inclined members have the same 3–4–5 geometry, so joints 2 and 3 translate identically.

Approach. Establish the sway pattern from inextensibility first, which shows that the horizontal member has zero chord rotation; then write the six slope-deflection expressions and assemble the three equilibrium equations, scaling the translation equation so that the matrix comes out symmetric.

Establish the sway pattern. Joint 1 is fixed and member 1–2 is inextensible, so joint 2 moves perpendicular to that member, along the direction $(0.8,\,-0.6)$ shown by the arrow on the figure; likewise joint 3 moves perpendicular to member 3–4. Because the two inclined members are parallel, those two perpendicular directions coincide, and inextensibility of the horizontal member (which equates their horizontal components) then forces the two displacements to be identical: $$\boldsymbol{\Delta}_2=\boldsymbol{\Delta}_3=\delta\,(0.8,\,-0.6).$$ One translation therefore describes the whole sway, as the question anticipates.
Write the three chord rotations. Applying the definition of $\psi$ member by member, $$\psi_{12}=\psi_{34}=-\frac{\delta}{L_1}=-\frac{\delta}{5},\qquad \boxed{\psi_{23}=0}.$$ The horizontal member merely rides along as a rigid body, so it contributes nothing to the translation equation — a result worth getting before any algebra starts, because it removes a third of the work.
Write the six end moments. With $\theta_1=\theta_4=0$ and $\mathrm{FEM}_{23}=+wL_2^{2}/12$, $$M_{12}=\frac{2EI}{L_1}\left(\theta_2+\frac{3\delta}{L_1}\right),\qquad M_{21}=\frac{2EI}{L_1}\left(2\theta_2+\frac{3\delta}{L_1}\right),$$ $$M_{23}=\frac{2EI}{L_2}\left(2\theta_2+\theta_3\right)+\frac{wL_2^{2}}{12},\qquad M_{32}=\frac{2EI}{L_2}\left(2\theta_3+\theta_2\right)-\frac{wL_2^{2}}{12},$$ $$M_{34}=\frac{2EI}{L_1}\left(2\theta_3+\frac{3\delta}{L_1}\right),\qquad M_{43}=\frac{2EI}{L_1}\left(\theta_3+\frac{3\delta}{L_1}\right).$$
(a) Derive the translation equation by virtual work. Give the frame a virtual sway $\delta^{*}=1$ with the joints held against rotation. Every member end moment then works through its own chord rotation, and the only external load that moves is the uniform load, which descends by the vertical component $a/L_1=3/5$ of the unit sway: $$\frac{\left(M_{12}+M_{21}\right)+\left(M_{34}+M_{43}\right)}{L_1}=\frac{a}{L_1}\,wL_2 .$$ Substituting the expressions above gives the first equilibrium equation, $$\boxed{\frac{24EI}{L_1^{3}}\,\delta+\frac{6EI}{L_1^{2}}\,\theta_2+\frac{6EI}{L_1^{2}}\,\theta_3=\frac{a}{L_1}\,wL_2}$$ Free-body statics gives the same result, but virtual work is safer here because the inclined geometry makes the storey-shear free body awkward to draw.
(b) Derive the two joint-moment equations. Joint 2 requires $M_{21}+M_{23}=0$ and joint 3 requires $M_{32}+M_{34}=0$, which reduce to $$\boxed{\frac{6EI}{L_1^{2}}\,\delta+\left(\frac{4EI}{L_1}+\frac{4EI}{L_2}\right)\theta_2+\frac{2EI}{L_2}\,\theta_3=-\frac{wL_2^{2}}{12}}$$ $$\boxed{\frac{6EI}{L_1^{2}}\,\delta+\frac{2EI}{L_2}\,\theta_2+\left(\frac{4EI}{L_1}+\frac{4EI}{L_2}\right)\theta_3=+\frac{wL_2^{2}}{12}}$$ The fixed-end moments appear on the right-hand side with reversed sign, as load terms; they are equal and opposite because the load is symmetric on the member.
(c) Assemble the matrix. Collecting the three boxed equations in the order $\{\delta,\ \theta_2,\ \theta_3\}$ gives $$[K]=EI\begin{bmatrix} \dfrac{24}{L_1^{3}} & \dfrac{6}{L_1^{2}} & \dfrac{6}{L_1^{2}} \\[6pt] \dfrac{6}{L_1^{2}} & \dfrac{4}{L_1}+\dfrac{4}{L_2} & \dfrac{2}{L_2} \\[6pt] \dfrac{6}{L_1^{2}} & \dfrac{2}{L_2} & \dfrac{4}{L_1}+\dfrac{4}{L_2}\end{bmatrix},\qquad \{P\}=\begin{Bmatrix} \dfrac{a}{L_1}wL_2 \\[6pt] -\dfrac{wL_2^{2}}{12} \\[6pt] +\dfrac{wL_2^{2}}{12}\end{Bmatrix}.$$ Substituting $L_1=L_2=5$ m, $a=3$ m and $w=2$ kN/m, $$[K]=EI\begin{bmatrix}0.192 & 0.240 & 0.240\\ 0.240 & 1.600 & 0.400\\ 0.240 & 0.400 & 1.600\end{bmatrix},\qquad \{P\}=\begin{Bmatrix}6.000\\ -4.167\\ +4.167\end{Bmatrix}$$ in kN and kN·m. As instructed, the equations are not solved.

Two properties confirm that the assembly is right without solving anything. First, $[K]$ is symmetric, which is guaranteed by Betti’s theorem for any correctly scaled set of equilibrium equations expressed in energy-conjugate coordinates; if the translation equation had been written as a bare sum of moments rather than divided by $L_1$, the off-diagonal terms would not have matched. Second, $[K]$ is positive definite (leading minors 0.192, 0.250 and 0.323 × EI³, all positive), as any stable elastic structure requires.

Question 9 — results
Item	Expression	Value for the given data
K₁₁ (sway)	24EI/L₁³	0.192 EI
K₁₂ = K₁₃	6EI/L₁²	0.240 EI
K₂₂ = K₃₃	4EI/L₁ + 4EI/L₂	1.600 EI
K₂₃	2EI/L₂	0.400 EI
P₁	(a/L₁)wL₂	6.000 kN
P₂, P₃	∓wL₂²/12	−4.167, +4.167 kN·m
Chord rotations	ψ₁₂ = ψ₃₄ = −δ/L₁, ψ₂₃ = 0	—

Topic. Formulating the displacement-method equations of a sway frame in matrix form.

Key relations. Slope-deflection for each member end; the virtual-work sway equation $\sum\left(M_{ij}+M_{ji}\right)\psi^{*}_{ij}$ balanced against the external virtual work; symmetry of $[K]$ from Betti’s reciprocal theorem; the conjugate pairing of $\delta$ with a force and $\theta$ with a moment.

Why this works. The whole question turns on one geometric observation: the two inclined members are parallel, so joints 2 and 3 translate by the same vector and the horizontal member has no chord rotation at all. That is why a frame with three members and two encastré feet — nominally indeterminate to the third degree — needs only three unknowns. Writing the translation equation by virtual work rather than by free bodies also has a practical payoff: the natural scaling that comes out (dividing by $L_1$) is exactly the one that makes $[K]$ symmetric, so symmetry becomes a free check on the whole derivation.

Common pitfall. Forgetting the external virtual work of the distributed load. The sway mode moves member 2–3 downward by $0.6\delta$, so the 10 kN resultant does real work and $P_1$ is 6 kN, not zero. A student who assumes that a vertical load cannot drive a sway will write $P_1=0$ and predict no sidesway at all. The second pitfall is presenting an unsymmetric $[K]$; if the matrix is unsymmetric the equations are still correct but improperly scaled, and the free check has been thrown away.

Source. Kassimali, Structural Analysis, 6th ed., §18.1–18.4 (matrix stiffness formulation for plane frames); Hibbeler, 10th ed., §11.5; Ghali, Neville & Brown, 7th ed., ch. 6.

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