07-Str-A4 · December 2015

Question 1 of 9: Statical Indeterminacy and Slope-Deflection Degrees of Freedom

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

Notes on this paper

Paper format. 07-Str-A4 Advanced Structural Analysis, National Examinations, December 2015 — three hours, closed book (an approved Sharp or Casio calculator is the only aid). Nine questions: Questions 1 and 2 are compulsory (8 and 12 marks); the candidate then answers two of Questions 3, 4, 5 (18 marks each) and two of Questions 6, 7, 8, 9 (22 marks each), so six questions constitute a complete 100-mark paper. Marks are printed in the left margin. All nine questions are solved here, because this set is a study resource rather than an examination script.

Reference texts. R. C. Hibbeler, Structural Analysis, 10th ed. (slope-deflection, moment distribution, influence lines, Castigliano); A. Kassimali, Structural Analysis, 6th ed. (force and displacement methods, degrees of freedom); A. Ghali, A. M. Neville & T. G. Brown, Structural Analysis: A Unified Classical and Matrix Approach, 7th ed. (flexibility and stiffness formulations, lack of fit, support movement); W. Weaver Jr. & J. M. Gere, Matrix Analysis of Framed Structures, 3rd ed. (assembly of [K] and {P}); J. C. McCormac, Structural Analysis: Using Classical and Matrix Methods, 4th ed. Canadian design context: CSA S16, CSA A23.3 and the National Building Code of Canada supply the load and resistance rules that these analyses feed, although this paper is purely an analysis paper.

Check: sign convention used throughout. Member end moments $M_{ij}$ are counter-clockwise positive, which is the convention that matches the standard six-degree-of-freedom stiffness matrix and the vector $\{\theta_2,\theta_3,\delta\}$ the paper itself defines (Question 9 states “positive counter clockwise”). With that choice the fixed-end moment of a downward uniformly distributed load is $\mathrm{FEM}_{ij}=+wL^{2}/12$ at the $i$ end, the chord rotation is $\psi_{ij}=[(\mathbf{d}_j-\mathbf{d}_i)\cdot\mathbf{e}_\perp]/L$ with $\mathbf{e}_\perp$ the member axis turned $+90^{\circ}$, and ordinary sagging moments are recovered as $M_{\text{sag}}(i)=-M_{ij}$, $M_{\text{sag}}(j)=+M_{ji}$. Mixing this with Hibbeler’s clockwise-positive convention produces clean-looking integers that are wrong; every diagram below is plotted in the sagging convention.

Check: printed unit in Question 6. The paper prints “EI = 3.2 × 10⁴ kN.mm²”. Taken literally that is a flexural rigidity some ten orders of magnitude too small for an 18 m beam and would make the 36 mm settlement effect vanish. The intended unit is $\text{kN}\cdot\text{m}^2$: with $EI=3.2\times10^{4}\ \text{kN}\cdot\text{m}^{2}$ the settlement term $3EI\Delta/L^{2}=54\ \text{kN}\cdot\text{m}$ is the same order as the load term $wL^{2}/8=48\ \text{kN}\cdot\text{m}$, which is plainly what the question is testing. The solution below uses $\text{kN}\cdot\text{m}^{2}$ and flags the typographical error.

Question 1: Statical Indeterminacy and Slope-Deflection Degrees of Freedom (8 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. Three plane structures, all members rigidly connected unless a support says otherwise, all members inextensible. (a) A four-column, three-level rigid frame with every column base fixed and a uniformly distributed load on each of the four beams. (b) A beam built into a wall at its left end, propped at mid-length by a vertical hanger running down to a pin support, carrying a roller at its right end and two equal point loads $P$. (c) A stepped frame: a lower beam built into a wall on the left, a vertical riser whose foot sits on a roller, and an upper beam built into a wall on the right, with a uniformly distributed load on each beam.

Find. For each structure, the degree of statical indeterminacy $r$ and the minimum number of structural degrees of freedom $k$ needed for a slope-deflection analysis.

Structure (a) — four fixed-base columns, three framing levels, gravity load on every beam.

Structure (b) — built-in end, interior hanger down to a pin, roller at the far end.

Structure (c) — stepped frame, walls at both extremities, roller under the foot of the riser.

Approach. Count $r$ from the member/joint/reaction census, then count $k$ as the number of independent joint rotations plus the number of independent joint translations (sways) that survive after inextensibility, and discard any rotation at a joint whose moment is known to be zero.

State the two counting rules. For a plane frame with $m$ members, $j$ joints (supports included) and $r_{\text{reac}}$ reaction components, $$r = 3m + r_{\text{reac}} - 3j$$ Equivalently, $r = 3\times(\text{number of closed loops formed with the foundation}) - (\text{number of releases})$; computing both is the cheapest possible check. The kinematic count is $$k = n_{\theta} + n_{\text{sway}}$$ where $n_\theta$ counts joints whose rotation is a genuine unknown and $n_{\text{sway}}$ counts independent joint translations. A pin or roller support contributes no rotation unknown because the moment there is known to be zero — the modified stiffness $3EI/L$ absorbs it.
Census structure (a). The three framing levels sit at the lower beam level, the upper-left beam level and the top beam level. Columns are divided by every beam they meet, giving $2+3+2+1=8$ column segments, and there are four beams, so $m=12$. There are eight free joints plus four fixed bases, $j=12$, and $r_{\text{reac}}=4\times3=12$. Hence $$r = 3(12) + 12 - 3(12) = \boxed{12}$$ Loop check: the graph closes four independent circuits through the ground (left bay lower, left bay upper, centre bay, right bay), and $3\times4=12$ with no releases anywhere. The two counts agree.
Kinematics of structure (a). Every free joint lies on a vertical column line that runs to a fixed base, so with inextensible columns every vertical translation vanishes. Horizontally, the inextensible beams tie joints together level by level: {lower-left pair}, {lower-right pair}, {upper-left pair} and {top pair} each translate as a unit, and no beam joins the lower-left group to the lower-right group. That gives four independent sways. The frame is neither symmetric nor anti-symmetrically loaded, and no joint has a known zero moment, so all eight joint rotations remain: $$k = 8 + 4 = \boxed{12}$$
Census and kinematics of structure (b). Members: the two beam segments either side of the hanger joint and the hanger itself, $m=3$; joints: the wall, the hanger joint, the roller end and the pin at the hanger foot, $j=4$; reactions: $3$ (fixed) $+\,1$ (roller) $+\,2$ (pin) $=6$. Therefore $$r = 3(3) + 6 - 3(4) = \boxed{3}$$ Kinematically the beam is horizontal and inextensible and its left end is built in, so every joint on it has $u=0$; the hanger is vertical and inextensible with a pin at its foot, so $v=0$ at the hanger joint as well. There is no sway. The roller end and the pin foot both carry zero moment, so only the interior rigid joint contributes a rotation: $$k = 1 + 0 = \boxed{1}$$
Census and kinematics of structure (c). Members: lower beam, riser, upper beam, $m=3$; joints: two walls and the two junctions, $j=4$; reactions: $3+1+3=7$. Therefore $$r = 3(3) + 7 - 3(4) = \boxed{4}$$ The roller at the foot of the riser fixes $v$ there; the inextensible riser carries that up to the upper junction; the inextensible lower beam ties the lower junction to the left wall so $u=0$; the inextensible upper beam ties the upper junction to the right wall so $u=0$ there too. No translation survives, and both junctions are rigid moment-carrying joints: $$k = 2 + 0 = \boxed{2}$$

The pattern worth carrying away is that the two counts are almost independent of each other: structure (a) is highly redundant and kinematically expensive, whereas structure (c) is redundant to degree four yet needs only two unknowns, because inextensibility plus a single well-placed roller annihilates every translation.

Structure	Members $m$	Joints $j$	Reactions	Statical indeterminacy $r$	Slope-deflection d.o.f. $k$
(a) four-column rigid frame	12	12	12	12	12 (8 rotations + 4 sways)
(b) propped beam with hanger	3	4	6	3	1 (one rotation, no sway)
(c) stepped frame	3	4	7	4	2 (two rotations, no sway)

Topic. Statical versus kinematic indeterminacy of plane frames.

Key relations. $r = 3m + r_{\text{reac}} - 3j$; $r = 3\,n_{\text{loops}} - n_{\text{releases}}$; $k = n_\theta + n_{\text{sway}}$; inextensibility constraint $(\mathbf{d}_j-\mathbf{d}_i)\cdot\mathbf{e}_{ij}=0$ for every member.

Why this works. Each plane member carries three independent internal force parameters once its own equilibrium is imposed, and each joint supplies three equilibrium equations; the surplus of unknowns over equations is precisely $r$. The kinematic count is the mirror image: the displacement field of a frame of inextensible members is completely described by the joint rotations plus the small number of translation patterns that the axial constraints permit, and the slope-deflection method writes one equilibrium equation for each of those unknowns. Symmetry, anti-symmetry and known-zero-moment joints shrink $k$ without changing $r$, which is exactly why the question asks for both.

Common pitfall. Counting a rotation unknown at every pin and roller support. The moment there is identically zero, so the modified stiffness $3EI/L$ (paired with $wL^{2}/8$ rather than $wL^{2}/12$) removes that unknown at no cost in accuracy — on structure (b) the naive count gives $k=3$ instead of the minimum $k=1$ the question demands. The second trap is forgetting that a beam with no beam continuing to the next bay does not tie those two bays into one sway group; structure (a) has four sways, not three.

Source. Kassimali, Structural Analysis, 6th ed., §3.4 (static determinacy) and §16.1–16.2 (degrees of freedom); Hibbeler, Structural Analysis, 10th ed., §11.1 and §12.1.

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