Question 9 of 9: Derivation of the stiffness matrix and load vector (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
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
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 9: Derivation of the stiffness matrix and load vector (22 marks)
Given. A symmetric trapezoidal frame, built in at joints (1) and (4), with a horizontal point load at joint (2) and a uniform load on the horizontal member.
Given data
Quantity
Value
Inclined members (1)–(2) and (3)–(4)
6.0 m each, at $60^\circ$ to the horizontal (3 m horizontal projection, 5.196 m rise)
Horizontal member (2)–(3)
6.0 m
Supports at (1) and (4)
fixed — which is why only three unknowns are listed
Horizontal load at joint (2)
46.2 kN to the right
Uniform load on (2)–(3)
6 kN/m downward
$\delta$
translation of joint (3) perpendicular to member (3)–(4), i.e. along $(\cos 30^\circ,\ \sin 30^\circ)$
Find. The translation equation, the two moment equations, and the resulting $[K]$ and $\{P\}$. The equations are not to be solved.
Question 9: fixed bases, $60^\circ$ legs of 6 m, and $\delta$ measured perpendicular to member (3)–(4). The arrow at joint (2) marks the perpendicular direction in which that joint is likewise obliged to move.
Approach. Express the joint displacements in terms of $\delta$ using inextensibility, obtain the three chord rotations, write the six end moments, and assemble the two joint equations plus a virtual-work translation equation.
Kinematics. Joint (3) is constrained to move perpendicular to member (3)–(4), so
$$\mathbf{d}_3=\delta\left(\frac{\sqrt3}{2},\ \frac{1}{2}\right)$$
The horizontal member is inextensible, so $u_2=u_3=\tfrac{\sqrt3}{2}\delta$, and joint (2) must move perpendicular to member (1)–(2), which forces
$$\mathbf{d}_2=\delta\left(\frac{\sqrt3}{2},\ -\frac{1}{2}\right)$$
Joint (2) rises as joint (3) falls, mirror fashion — the frame "scissors".
Chord rotations. Projecting the joint displacements onto each member's transverse axis $\mathbf{e}_2$ with $L=6$ m throughout,
$$\psi_{12}=-\frac{\delta}{6},\qquad \psi_{23}=+\frac{\delta}{6},\qquad \psi_{34}=-\frac{\delta}{6}$$
End moments (part b groundwork). With $2EI/L=EI/3$, $\theta_1=\theta_4=0$ and $\mathrm{FEM}=\pm wL^{2}/12=\pm 18\ \text{kN}\cdot\text{m}$ on member (2)–(3),
$$M_{12}=\tfrac{EI}{3}\Bigl(\theta_2+\tfrac{\delta}{2}\Bigr),\qquad M_{21}=\tfrac{EI}{3}\Bigl(2\theta_2+\tfrac{\delta}{2}\Bigr)$$
$$M_{23}=\tfrac{EI}{3}\Bigl(2\theta_2+\theta_3-\tfrac{\delta}{2}\Bigr)+18,\qquad M_{32}=\tfrac{EI}{3}\Bigl(2\theta_3+\theta_2-\tfrac{\delta}{2}\Bigr)-18$$
$$M_{34}=\tfrac{EI}{3}\Bigl(2\theta_3+\tfrac{\delta}{2}\Bigr),\qquad M_{43}=\tfrac{EI}{3}\Bigl(\theta_3+\tfrac{\delta}{2}\Bigr)$$
Part (b): moment equilibrium at joints (2) and (3). Summing the end moments meeting at each joint,
$$M_{21}+M_{23}=0\;\Longrightarrow\;\boxed{\frac{4EI}{3}\theta_2+\frac{EI}{3}\theta_3=-18}$$
$$M_{32}+M_{34}=0\;\Longrightarrow\;\boxed{\frac{EI}{3}\theta_2+\frac{4EI}{3}\theta_3=+18}$$
The $\delta$ terms cancel identically in both equations because the two members meeting at each joint have equal and opposite chord rotations.
Part (a): the translation equation. Apply a unit virtual translation $\delta^{*}=1$ with $\theta^{*}=0$, giving $\psi^{*}_{12}=\psi^{*}_{34}=-1/6$ and $\psi^{*}_{23}=+1/6$. The end moments produced by a unit real $\delta$ are $M_{12}=M_{21}=EI/6$, $M_{23}=M_{32}=-EI/6$ and $M_{34}=M_{43}=EI/6$, so the internal virtual work is
$$\sum\bigl(M_{ij}+M_{ji}\bigr)\psi^{*}_{ij}=3\left(\frac{EI}{3}\right)\left(-\frac{1}{6}\right)\delta=-\frac{EI}{6}\delta$$
External virtual work. The 46.2 kN load moves $\sqrt3/2$ horizontally, while the uniform load's equivalent nodal forces (18 kN at each end) move $-\tfrac12$ and $+\tfrac12$ and therefore cancel:
$$\sum\mathbf{F}\cdot\mathbf{d}^{*}=46.2\left(\frac{\sqrt3}{2}\right)+18\left(\frac{1}{2}\right)-18\left(\frac{1}{2}\right)=\frac{231\sqrt3}{10}=40.01\ \text{kN}$$
Equating internal and external work gives
$$\boxed{\frac{EI}{6}\,\delta=\frac{231\sqrt3}{10}=40.01}$$
The coefficient can also be read off directly as $36EI/L^{3}=36EI/216=EI/6$.
Part (c): assemble the matrix form. Collecting the three equations in the order $\{\delta,\ \theta_2,\ \theta_3\}$,
$$\boxed{[K]=EI\begin{bmatrix} \dfrac{1}{6} & 0 & 0\\[4pt] 0 & \dfrac{4}{3} & \dfrac{1}{3}\\[4pt] 0 & \dfrac{1}{3} & \dfrac{4}{3}\end{bmatrix}, \qquad \{P\}=\begin{Bmatrix} 231\sqrt3/10\\ -18\\ +18\end{Bmatrix}=\begin{Bmatrix} 40.01\\ -18\\ +18\end{Bmatrix}}$$
with $K_{11}$ in kN/m, the rotational terms in kN·m/rad, $P_1$ in kN and $P_2$, $P_3$ in kN·m.
Check the matrix before stopping. Three tests should be applied to any $[K]$ derived by hand. It is symmetric, as Betti's law requires. Its diagonal is positive. And every term traces to a recognisable stiffness: $EI/6=36EI/L^{3}$ for the sway, $4EI/L$ for a diagonal rotation term and $2EI/L$ for an off-diagonal one, since $4EI/6=2EI/3$ appears here as $\tfrac{4EI}{3}$ only because two members contribute to each joint. The zeros in the first row and column say that this particular geometry decouples the sway from the rotations — a genuine property of the frame, not a slip.
The question says not to solve the equations, and that instruction should be respected: the marks are for the derivation, the sign convention and the correct identification of $\delta$'s direction, not for arithmetic. It is worth noting, however, that the load 46.2 kN was chosen so that $46.2\cos 30^\circ=40.0$ exactly, which is a useful confirmation that the geometry has been read correctly.
