--- title: "Truss forces: member AC 10 mm too long" subject: "Theory of Structures II" chapter: 2 key: "semester_5/theory_of_structures_ii:truss-lack-of-fit-ac-10mm" source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/truss-lack-of-fit-ac-10mm site: BCT Engineering --- # Truss forces: member AC 10 mm too long Theory of Structures II · Chapter 2: Force method - Asked in: 2078 Kartik, 6 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/truss-lack-of-fit-ac-10mm - Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-2 ## Question Determine forces in all members of the truss shown in the figure if the member AC is too long by 10 mm. Take AE constant for all members. [Figure: rectangular truss ABCD, 4 m wide and 3 m high, with both diagonals; A hinged, D roller.] ## Short answer Take the force in AC as redundant X. With lack of fit 10 mm, flexibility f11 = 17.28/AE. Compatibility gives X = –0.01 AE/17.28 = –5.787×10⁻⁴ AE. Member forces: AC and BD = –5.787×10⁻⁴ AE (compression), AB and CD = +3.472×10⁻⁴ AE, BC and AD = +4.630×10⁻⁴ AE (tension). ## Answer Reading of the figure: A (0, 0) hinged, D (4, 0) roller at the base, B (0, 3) and C (4, 3) at the top, with both diagonals AC and BD. Bars AB and CD are 3 m, BC and AD are 4 m, and AC and BD are 5 m. The truss is indeterminate to the 1st degree (internal); the supports are determinate. Take the force in AC as the redundant $X$ (tension +). Lack of fit $\lambda = +10$ mm $= 0.01$ m (member too long). No external load acts, so the primary structure carries no force and the fabrication error alone produces the member forces. #### Unit force system ($X = 1$ in AC) | Member | $L$ (m) | $n$ | $n^2L$ | |---|---|---|---| | AB | 3 | $-0.6$ | 1.08 | | BC | 4 | $-0.8$ | 2.56 | | CD | 3 | $-0.6$ | 1.08 | | AD | 4 | $-0.8$ | 2.56 | | BD | 5 | $+1$ | 5.00 | | AC | 5 | $+1$ | 5.00 | | Total | | | 17.28 | $$ f_{11} = \sum \frac{n^2 L}{AE} = \frac{17.28}{AE} $$ #### Compatibility The member is too long, so after it is forced into place it is compressed. Compatibility requires the net relative displacement at the cut in AC, due to $X$ plus the lack of fit, to be zero: $$ f_{11}X + \lambda = 0 \ \Rightarrow\ X = -\frac{0.01\,AE}{17.28} = -0.0005787\,AE $$ #### Member forces $S = n X$ (AE in kN) | Member | Force | Nature | |---|---|---| | AC | $-0.0005787\,AE$ | compression | | BD | $-0.0005787\,AE$ | compression | | AB, CD | $+0.0003472\,AE$ | tension | | BC, AD | $+0.0004630\,AE$ | tension | Example: for $AE = 2\times10^5$ kN, AC = BD $= -115.74$ kN, AB = CD $= 69.44$ kN, BC = AD $= 92.59$ kN. Check at joint A (horizontal): $S_{AD} + 0.8\,S_{AC} = (0.4630 - 0.8\times0.5787)\times10^{-3}AE = 0$. There is no external load, so all reactions are zero. **Answer: AC = BD $= -5.787\times10^{-4}\,AE$ (compression); AB = CD $= +3.472\times10^{-4}\,AE$; BC = AD $= +4.630\times10^{-4}\,AE$ (tension).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. No external load acts, so why do forces develop in the members? A. They do not; all members carry zero force B. AC is forced into a gap that is too short for it, so the fabrication error produces a self-equilibrating set of internal forces C. The supports react to the lack of fit with horizontal reactions D. Only AC is stressed, other members stay unstressed Answer: B. The truss is internally indeterminate, so the lack of fit cannot be absorbed freely and stresses every member, while reactions stay zero. (Key point: Source of forces) 2. With X = 1 in AC, the forces in AB, CD are -0.6 and in BC, AD are -0.8. Why? A. Because the sides are shorter than the diagonals, they carry half the force B. Because AE is constant, all sides carry the same force C. The unit forces in AC and BD have components 0.6 and 0.8 at each joint, balanced by the sides (3-4-5 geometry) D. Because the supports provide these forces Answer: C. Joint equilibrium with a 3-4-5 diagonal gives components 3/5 = 0.6 and 4/5 = 0.8, which the sides must resist in compression. (Key point: Unit-force pattern) 3. Why does the compatibility equation f11 X + lambda = 0 give a negative X for AC? A. A too-long member is squeezed on assembly, so AC is in compression (X = -0.0005787 AE) B. A too-long member is stretched, so AC is in tension C. The sign is arbitrary and has no physical meaning D. X is negative because AE is negative Answer: A. Fitting a long bar into a shorter gap compresses it; AC and BD are compressed while the sides are in tension. 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