Question 3 of 9: Three-bar truss by the theorem of least work
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Examinations, May 2017 — 07-Str-A4 Advanced Structural Analysis. Three hours, CLOSED BOOK (an approved Sharp or Casio calculator is permitted). Answer BOTH Questions 1 and 2, ONLY TWO of Questions 3, 4 or 5, and ONLY TWO of Questions 6, 7, 8 or 9; six questions constitute a complete paper for 100 marks. Marks are printed in the left margin. All nine questions are worked below, because the complete set is the study resource.
Reference texts. R. C. Hibbeler, Structural Analysis, 10th ed. (Pearson) — Ch. 8 influence lines, 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), Ch. 13 slope-deflection, Ch. 16 sidesway frames; W. McGuire, R. H. Gallagher & R. D. Ziemian, Matrix Structural Analysis, 2nd ed., for the element stiffness matrix used as the independent check. Once the analysis is complete, member design follows CSA A23.3:19, CSA S16:19 and CSA O86:19, with load combinations from NBCC 2020.
Check: sign convention 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 stiffness matrix, so every answer below can be checked against a direct-stiffness solution. Chord rotation is $\psi_{ij}=\left[(\mathbf{D}_{j}-\mathbf{D}_{i})\cdot\mathbf{e}_{2}\right]/L$, where $\mathbf{e}_{2}$ is the member axis turned $+90^\circ$. The fixed-end moment of a downward uniform load is $\mathrm{FEM}_{ij}=+wL^{2}/12$ at the $i$ end and $-wL^{2}/12$ at the $j$ end; for a member released at its far end it becomes $+wL^{2}/8$. Ordinary sagging moments follow as $M_{\text{sag}}(i)=-M_{ij}$ and $M_{\text{sag}}(j)=+M_{ji}$, and all diagrams are plotted sagging positive. Member shear at the $i$ end is $V_{i}=(M_{ij}+M_{ji})/L$ plus the equivalent nodal shear of any span load. Mixing this with Hibbeler's clockwise-positive convention produces clean-looking integers that are wrong, so the convention is stated once and used everywhere.
Question 3: Three-bar truss by the theorem of least work (16 marks)
Given. Three pin-ended bars meeting at the single free joint 3, all with the same axial rigidity $EA$, loaded by a horizontal 40.5 kN force at that joint.
Given data
Member
From joint 3 towards
Direction cosines
Length
2–3
joint 2
$(-1,\,0)$
4 m
3–4
joint 4
$(0,\,+1)$
3 m
1–3
joint 1
$(-0.8,\,-0.6)$
5 m
load
—
$(+1,\,0)$
40.5 kN
Find. The axial force in each of the three members, with its sense.
Three bars meet at the loaded joint 3: two equilibrium equations, three unknowns, so the truss is one degree statically indeterminate and least work supplies the missing equation.
Approach. Joint 3 supplies only two equilibrium equations for three bar forces, so choose the diagonal force as the redundant, write the other two in terms of it, and make the strain energy stationary.
Confirm the degree of indeterminacy. Every bar runs from the single free joint to a pinned support, so the only equilibrium available is that of joint 3: two equations for three unknown bar forces. The truss is therefore one degree statically indeterminate, and exactly one compatibility condition is needed — which is what least work supplies.
Express the bar forces in terms of one redundant. Take tension positive and let $X=N_{13}$. Resolving at joint 3, with each bar pulling the joint towards its far end,$$\sum F_{x}:\;-N_{23}-0.8X+40.5=0,\qquad \sum F_{y}:\;N_{34}-0.6X=0,$$so that $N_{23}=40.5-0.8X$ and $N_{34}=0.6X$. Every internal force is now a linear function of the single unknown $X$.
Write the strain energy. For a pin-jointed bar the stored energy is $N^{2}L/2EA$, so$$U=\frac{1}{2EA}\Big[(40.5-0.8X)^{2}(4)+(0.6X)^{2}(3)+X^{2}(5)\Big].$$Because $EA$ is common to all three bars it will cancel: the answer depends only on the ratios of the bar lengths, which is why the question can withhold a numerical $EA$.
Apply the theorem of least work. The redundant takes the value that makes the strain energy stationary, $\partial U/\partial X=0$:$$-6.4\,(40.5-0.8X)+2.16X+10X=0 \;\Longrightarrow\; 17.28X=259.2 .$$Solving,$$X=N_{13}=\boxed{15.0\ \text{kN (tension)}}.$$The coefficient $17.28=0.64(4)+0.36(3)+5$ is the flexibility of the redundant path and is always positive, so the stationary point is the minimum that gives the theorem its name.
Back-substitute and check. The remaining two forces follow immediately:$$N_{23}=40.5-0.8(15.0)=28.5\ \text{kN},\qquad N_{34}=0.6(15.0)=9.0\ \text{kN},$$both tensile. Substituting all three back into the joint-3 equations closes them exactly ($-28.5-12.0+40.5=0$ and $9.0-9.0=0$), and an independent direct-stiffness solve of the same three-bar assembly returns the identical set, which is the check worth doing because the least-work algebra is where sign slips hide.
All three members are in tension. That is worth a sentence of physical sense: the applied force drags joint 3 to the right, the horizontal bar and the diagonal both resist by pulling it back towards the wall, and the vertical bar is dragged into tension by the downward component the diagonal introduces.
