16-Civ-B1 Advanced Structural Analysis · December 2017
Question 1 of 9: Minimum number of structural degrees of freedom (8 marks)
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Examination, December 2017 — 16-Civ-B1 Advanced Structural Analysis. Three hours, closed book (an approved Sharp or Casio calculator is the only aid). Nine questions: both #1 and #2 are compulsory, then two only of #3, #4 or #5 and two only of #6, #7, #8 or #9 — six questions constitute a complete paper. Marks are printed in the left margin: 8, 12, 16, 16, 16, 24, 24, 24, 24. All nine questions are solved here, because the set is a study resource rather than a sitting.
Reference texts. R. C. Hibbeler, Structural Analysis, 10th ed. (Pearson) — Ch. 4 shear and moment diagrams, Ch. 9–10 the force (flexibility) method, Ch. 11 slope-deflection, Ch. 14–16 the stiffness method; A. Kassimali, Structural Analysis, 6th ed. (Cengage) — Ch. 7 deflections by work and energy (Castigliano and least work), Ch. 13 slope-deflection, Ch. 16 sidesway frames; W. McGuire, R. H. Gallagher and R. D. Ziemian, Matrix Structural Analysis, 2nd ed., for the planar frame element stiffness matrix used as the independent check on every answer below. Member sizing that would follow such an analysis is governed in Canada by CSA A23.3:19, CSA S16:19 and CSA O86:19, with load combinations from NBCC 2020.
Check: sign convention, stated once and used throughout. End moments $M_{ij}$ are counter-clockwise positive on the member end — the convention that matches the six-degree-of-freedom planar frame element, so every result below can be reproduced by a direct-stiffness solution. The chord rotation is $\psi_{ij}=\left[(\mathbf{D}_{j}-\mathbf{D}_{i})\cdot\mathbf{n}\right]/L$ with $\mathbf{n}$ the member axis turned $+90^\circ$; the fixed-end moment of a downward UDL is $\mathrm{FEM}_{ij}=+wL^{2}/12$ at the $i$ end; and the ordinary sagging moment is $M_{\text{sag}}(i)=-M_{ij}$, $M_{\text{sag}}(j)=+M_{ji}$. Mixing this with the clockwise-positive convention produces clean-looking integers that are wrong.
Question 1: Minimum number of structural degrees of freedom (8 marks)
Given. Four structures, all with inextensible members: (a) a four-span continuous beam under a UDL over its whole length, pinned at the left end and on rollers elsewhere, with internal hinges $3L/4$ in from each end; (b) a two-storey two-bay frame joined to a taller single-bay frame, four fixed bases, UDL on the upper beam of each part; (c) a symmetric hipped frame on pinned bases under a symmetric UDL over the full projection; (d) an unsymmetric bent on two fixed bases under a UDL over the full projection.
Find. The minimum number of structural degrees of freedom $k$ for each structure, together with the arrows that represent them.
Q1(a): four equal spans carrying a UDL over the whole length, pin at the left end, rollers elsewhere, internal hinges 3L/4 in from each end.
Q1(b): four fixed bases; nine rigid joints (blue) and three storey sways, k = 12.
Q1(c): symmetric hipped frame on pinned bases under a symmetric UDL applied over the full horizontal projection.
Q1(d): unsymmetric bent, both bases fixed, UDL over the whole projection.
Approach. Count the independent joint rotations and the independent joint translations separately, then delete every rotation that a zero-moment release makes redundant and every pair that symmetry ties together.
State the counting rule. The slope-deflection unknowns are the joint displacements, so $$k=n_{\theta}+n_{\Delta},\qquad n_{\Delta}=2\,j_{\text{free}}-m_{\text{inext}}-n_{\text{support}}$$ where $j_{\text{free}}$ counts joints free to move, $m_{\text{inext}}$ the inextensible members and $n_{\text{support}}$ the translation components held by supports. A rotation is not counted at a joint where every member framing in is released, because the modified stiffness $3EI/L$ absorbs it.
(a) Strip the determinate end pieces. The left hinge sits $0.75L$ from the pin, so the piece between them carries $M=0$ at both ends and has no interior support: it is a simple span of length $0.75L$ that hands a known force $$V_{\text{hinge}}=\frac{w(0.75L)}{2}=0.375\,wL$$ to the hinge. The mirror-image piece at the right-hand end behaves identically, so neither contributes an unknown.
(a) Count what is left, then use symmetry. The residue is a three-support beam with a $0.25L$ overhang at each end, giving $\theta_{B}$, $\theta_{C}$, $\theta_{D}$. Structure and load are symmetric about the centre support, so $\theta_{C}=0$ and $\theta_{D}=-\theta_{B}$: $$k_{(a)}=3-2=\boxed{1}$$
Structure (b) offers none of those reductions, so it must be counted in full. Every beam is inextensible, which locks all the joints on one floor to a single horizontal movement, and every column is inextensible on a fixed base, which stops any joint moving vertically.
(b) Count joints and storeys. Three rigid joints at the lower floor, four at the intermediate floor and two at the roof of the tall bay give nine rotations; three floors give three sways: $$k_{(b)}=9+3=\boxed{12}$$
(c) Count the unreduced totals. Four free joints give four rotations (the pinned bases contribute none). Their eight displacement components are tied by five inextensible members, and the two pinned bases fix nothing else, so $$n_{\Delta}=8-5=3,\qquad k_{(c),\text{unreduced}}=4+3=7$$
(c) Apply symmetry. A mirror about the centre line maps the frame and the load onto themselves, so the two eave rotations pair off and so do the two ridge-beam rotations. The symmetric translation field requires $u_{3}=-u_{4}$ while the horizontal member joining those joints is inextensible, so $u_{3}=u_{4}=0$ and only the spread of the eaves survives: $$k_{(c)}=(4-2)+(3-2)=\boxed{3}$$
(d) No reductions are available. Three free joints give three rotations; their six components less four inextensible members leave two translations, and nothing is symmetric or released: $$k_{(d)}=3+2=\boxed{5}$$ A convenient independent pair is the horizontal movement of the left knee and the vertical movement of the top of the inclined member; every other component follows from the four member constraints.
The arrows the question asks for are drawn on the four sketches: a single curved arrow at the first interior support in (a); nine curved arrows and three straight storey arrows in (b); two curved arrows on one half of (c) plus one straight arrow showing the eaves spreading; and three curved arrows plus two straight arrows in (d).
Question 1 — minimum structural degrees of freedom