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07-Str-A4 · December 2017

Question 4 of 9: Least work — forces in a three-bar truss (16 marks)

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

Notes on this paper

Paper format. National Examinations, December 2017 — 07-Str-A4 Advanced Structural Analysis. Three hours, CLOSED BOOK (an approved Sharp or Casio calculator is the only aid). Nine questions: answer BOTH #1 and #2, then ONLY TWO of #3, #4 or #5, and ONLY TWO 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 worked below, because the complete set is the 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 4: Least work — forces in a three-bar truss (16 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 bars pinned to a rigid ceiling at U1, U2 and U3 and meeting at a single free joint L1, 6 m below U2, which carries a 30 kN vertical load.

Question 4 — bar geometry and stiffness
BarHorizontal offset of the support from L1LengthRelative $EA$Flexibility $L/EA$
U1–L14.5 m to the left7.5 m$1.5EA$$5/EA$
U2–L10, vertical6.0 m$1.2EA$$5/EA$
U3–L18.0 m to the right10.0 m$2EA$$5/EA$

Find. The axial force in each of the three bars.

U1 1.5EA U2 1.2EA U3 2EA L1 30 kN 4.5 m 8 m 6 m
Q4: three bars pinned to a rigid ceiling at U1, U2 and U3 and meeting at the loaded joint L1; the relative axial stiffness EA is marked on each bar.

Approach. Three bar forces meet at one joint that supplies only two equilibrium equations, so the truss is once redundant; carry one bar force as the redundant and minimise the strain energy with respect to it.

  1. Confirm the degree of redundancy. With $m=3$ bars, $j=4$ joints and $r=6$ reaction components from the three pinned ceiling supports, $$m+r-2j=3+6-8=1$$ so the truss is statically indeterminate to the first degree — exactly the case least work is written for.
  2. Write joint equilibrium at L1. Taking tension positive with unit vectors from L1 towards each support, $\mathbf{u}_{1}=(-0.6,\,0.8)$, $\mathbf{u}_{2}=(0,\,1)$, $\mathbf{u}_{3}=(0.8,\,0.6)$: $$-0.6N_{1}+0.8N_{3}=0\;\Rightarrow\;N_{3}=0.75N_{1}$$ $$0.8N_{1}+N_{2}+0.6N_{3}=30\;\Rightarrow\;N_{2}=30-1.25N_{1}$$
  3. Note the calibration built into the data. Every bar has the same flexibility, $L/EA=7.5/1.5=6.0/1.2=10/2=5$ in units of $1/EA$. Confirming that before starting is the cheapest check that the geometry and the relative stiffnesses have been read correctly, and it is what makes the algebra below collapse.
  4. Apply the least work condition. With $U=\sum N_{i}^{2}L_{i}/(2EA_{i})$, compatibility at the redundant requires $$\frac{\partial U}{\partial N_{1}}=\sum N_{i}\frac{L_{i}}{EA_{i}}\frac{\partial N_{i}}{\partial N_{1}}=0,\qquad \frac{\partial N_{2}}{\partial N_{1}}=-1.25,\quad \frac{\partial N_{3}}{\partial N_{1}}=0.75$$
  5. Solve. The common factor $5/EA$ cancels, leaving $$N_{1}\left(1+1.25^{2}+0.75^{2}\right)=1.25\times 30\;\Longrightarrow\;N_{1}=\frac{37.5}{3.125}=\boxed{12.0\ \text{kN}}$$
  6. Back-substitute and check. $$N_{2}=30-1.25(12)=\boxed{15.0\ \text{kN}},\qquad N_{3}=0.75(12)=\boxed{9.0\ \text{kN}}$$ Horizontal equilibrium: $-0.6(12)+0.8(9)=-7.2+7.2=0$. Vertical: $0.8(12)+15+0.6(9)=9.6+15+5.4=30$ kN. Both close exactly.

All three forces come out positive, so every bar is in tension — which is what a hanging joint under gravity must produce, and a useful sanity check before the arithmetic is trusted. The vertical bar takes exactly half the applied load and the two inclined bars share the remainder in proportion to their direction cosines rather than to their stiffnesses, because the flexibilities were made equal by design. Change any one of the three relative $EA$ values and that neat split disappears, which is the point the question is making.

Question 4 — bar forces
BarForceSenseVertical share of the 30 kN
U1–L1 ($1.5EA$, 7.5 m)12.0 kNtension9.6 kN
U2–L1 ($1.2EA$, 6.0 m)15.0 kNtension15.0 kN
U3–L1 ($2EA$, 10.0 m)9.0 kNtension5.4 kN
Horizontal check at L1$-0.6(12)+0.8(9)=0$
Check: the lettering on the bars. The question text says “the relative EA value is shown on the diagram”, while the hand lettering on the bars reads more like EI. A pin-jointed bar has no flexural stiffness, so only axial stiffness can be meant, and 1.5, 1.2 and 2 are treated as relative $EA$ throughout.