--- title: "Truss: bar forces due to load and temperature" subject: "Theory of Structures II" chapter: 2 key: "semester_5/theory_of_structures_ii:truss-load-and-temperature-9kn-18kn" source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/truss-load-and-temperature-9kn-18kn site: BCT Engineering --- # Truss: bar forces due to load and temperature Theory of Structures II · Chapter 2: Force method - Asked in: 2068 Baishakh, 12+8 marks - Video: [Effect of Temperature Stresses and Lack of Fit on TRUSS analysis.](https://www.youtube.com/watch?v=azRVuTgFGYY) — Ram Kinhikar - 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-load-and-temperature-9kn-18kn - Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-2 ## Question Compute the bar forces in all members due to: (i) given load and (ii) temperature rise by 30°C in the upper chord. Take $E = 2\times10^{5}\ \text{N/mm}^2$, $\alpha = 10.8\times10^{-6}/°C$. Take area of all members to be 30 cm². [Figure: truss with top joints under 9 kN and 18 kN downward loads; three panels of 4 m (total 12 m), height 3 m, diagonals; left support hinged, two roller supports at the lower chord.] ## Short answer Using consistent deformation with the C roller as redundant: load case gives AB 7.10, BC 2.20, CD 2.20, EF -7.10, AE -8.87, BE -3.68, CF -20.03, DF -2.75, BF 6.13 kN; temperature case gives AB 7.49, BC 14.99, CD 14.99, EF -7.49, AE -9.37, BE 5.62, CF 16.86, DF -18.74, BF -9.37 kN. ## Answer **Assumptions (the figure is not clear):** bottom joints A, B, C, D at 0, 4, 8, 12 m; top joints E and F, 3 m above B and C. Members: bottom chord AB, BC, CD; top chord EF; verticals BE, CF; end diagonals AE, DF and one middle diagonal BF (9 members). 9 kN acts downward at E and 18 kN downward at F. A is a hinge; the two rollers are at C and D (vertical reactions only). $EA=2\times10^5\ \text{N/mm}^2\times3000\ \text{mm}^2=6\times10^8$ N $=6\times10^5$ kN for every bar. #### Degree of indeterminacy $m=9$, $r=4$, $j=6$: $D_s=9+4-12=1$ (external). Take the roller reaction at C, $X$ (upward), as the redundant. Primary structure: remove the roller at C, leaving a truss hinged at A and on a roller at D. #### (i) Bar forces due to the loads $S_0$ = forces in the primary truss under the loads; $S_1$ = forces for a unit upward force at C. | Bar | L (m) | $S_0$ (kN) | $S_1$ | $S_0S_1L$ | $S_1^2L$ | Final $S$ (kN) | |---|---|---|---|---|---|---| | AB | 4.000 | 16.00 | -0.444 | -28.44 | 0.790 | 7.10 | | BC | 4.000 | 20.00 | -0.889 | -71.11 | 3.160 | 2.20 | | CD | 4.000 | 20.00 | -0.889 | -71.11 | 3.160 | 2.20 | | EF | 4.000 | -16.00 | 0.444 | -28.44 | 0.790 | -7.10 | | AE | 5.000 | -20.00 | 0.556 | -55.56 | 1.543 | -8.87 | | BE | 3.000 | 3.00 | -0.333 | -3.00 | 0.333 | -3.68 | | CF | 3.000 | 0.00 | -1.000 | 0.00 | 3.000 | -20.03 | | DF | 5.000 | -25.00 | 1.111 | -138.89 | 6.173 | -2.75 | | BF | 5.000 | -5.00 | 0.556 | -13.89 | 1.543 | 6.13 | | **Sum** | | | | **-410.44** | **20.494** | | Vertical displacement of C in the primary truss (positive upward) $\Delta_C=\dfrac{\sum S_0S_1L}{EA}=\dfrac{-410.44}{6\times10^5}=-0.684$ mm, so C would move 0.684 mm downward without the roller. Compatibility (C cannot move): $$ X=-\frac{\sum S_0S_1L}{\sum S_1^2L}=-\frac{-410.44}{20.494}=20.03\ \text{kN (up)} $$ Final bar forces: AB = 7.10, BC = 2.20, CD = 2.20, EF = -7.10, AE = -8.87, BE = -3.68, CF = -20.03, DF = -2.75, BF = 6.13 kN (+ tension, - compression). #### (ii) Temperature rise of 30°C in the upper chord (EF) Free elongation of EF: $e=\alpha\,\Delta T\,L=10.8\times10^{-6}\times30\times4=1.296$ mm. The primary truss expands freely and no forces develop in it ($S_0=0$); only the displacement at C changes. The compatibility equation with the temperature term is $$ \frac{X\sum S_1^2L}{EA}+S_1^{EF}\,e=0\Rightarrow X=-\frac{EA\,S_1^{EF}\,e}{\sum S_1^2L}=-16.86\ \text{kN} $$ (with $S_1^{EF}=0.444$). Bar forces $S=XS_1$: AB = 7.49, BC = 14.99, CD = 14.99, EF = -7.49, AE = -9.37, BE = 5.62, CF = 16.86, DF = -18.74, BF = -9.37 kN (+ tension, - compression). **Answer: (i) AB = 7.10, BC = 2.20, CD = 2.20, EF = -7.10, AE = -8.87, BE = -3.68, CF = -20.03, DF = -2.75, BF = 6.13 kN. (ii) AB = 7.49, BC = 14.99, CD = 14.99, EF = -7.49, AE = -9.37, BE = 5.62, CF = 16.86, DF = -18.74, BF = -9.37 kN.** Both were checked with a full stiffness analysis. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is chosen as the redundant for this truss? A. The roller reaction at C, with the roller removed in the primary truss B. The force in bar EF only C. The hinge at A D. The reaction at D, with the hinge removed Answer: A. m = 9, r = 4, j = 6 gives one degree of external indeterminacy; the roller reaction at C is released. (Key point: Redundant and primary structure) 2. How is X found for the load case? A. X = -sum(S0 S1 L)/sum(S1^2 L), as all bars have the same EA B. X = sum(S0 L)/sum(S1 L) C. X = sum(S1^2 L)/sum(S0 S1 L) D. X = 0 Answer: A. C cannot move, so X f11 + Delta_C = 0 with equal EA for all bars. (Key point: Load case compatibility) 3. How is the temperature effect on the top chord EF entered? A. As a free elongation alpha x dT x L in the compatibility equation, with S0 = 0 B. As an extra external load on EF C. As a change in EA D. It is ignored Answer: A. The primary truss expands freely without forces, so the elongation e = 1.296 mm appears only as a displacement term S1 e. (Key point: Free elongation by temperature) 4. What is the force in bar CF for the load case? A. -20.03 kN (compression) B. +20.03 kN C. 0 kN D. -7.10 kN Answer: A. CF has S0 = 0 and S1 = -1, so S = X S1 = -20.03 kN. 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