--- title: "Truss: temperature decrease in vertical members" subject: "Theory of Structures II" chapter: 2 key: "semester_5/theory_of_structures_ii:truss-temp-decrease-vertical-ab-ef" source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/truss-temp-decrease-vertical-ab-ef site: BCT Engineering --- # Truss: temperature decrease in vertical members Theory of Structures II · Chapter 2: Force method - Asked in: 2078 Bhadra, 6 marks - Video: [Virtual Work for Temperature Change Causing Deflection in a Truss](https://www.youtube.com/watch?v=Ufn7V5TysB4) — STATICS THE EASY WAY - 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-temp-decrease-vertical-ab-ef - Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-2 ## Question Compute the bar force in the member AB, and EF due to the decrease in temperature of 20°C in all vertical members only. Take $\alpha = 12\times10^{-6}/°C$ and $E = 2\times10^{5}\ \text{N/mm}^2$, cross-sectional area of all members is 35 cm². [Figure: truss with bottom chord A-B-C-D (3 panels of 3 m), top joints F and E, height 3 m, with diagonals; A hinged, D roller.] ## Short answer Take CF as redundant and apply compatibility for the temperature shortening of verticals. Bar force in AB is 0, and in EF is 34.79 kN tension. ## Answer Reading of the figure: bottom chord A-B-C-D with three panels of 3 m (A hinged, D roller); top joints F above B and E above C at 3 m height; members AB, BC, CD, FE, AF, ED, the verticals BF and CE, and the two diagonals BE and CF. The truss has $m = 10$, $j = 6$, $r = 3$, so $D_s = m + r - 2j = 1$ (internal). Take the force in CF as the redundant $X$. #### Data - $E = 2\times10^5$ N/mm$^2 = 2\times10^8$ kN/m$^2$, $A = 35$ cm$^2 = 35\times10^{-4}$ m$^2$, so $AE = 700000$ kN. - Vertical members BF and CE: $\Delta t = -20^\circ$C, so each shortens by $e = \alpha\,\Delta t\,L = 12\times10^{-6}(-20)(3) = -7.20\times10^{-4}$ m $= -0.72$ mm. There is no external load, so the primary structure carries no force. The temperature change alone causes the forces. #### Force $n$ due to $X = 1$ in CF | Member | $L$ (m) | $n$ | |---|---|---| | AB | 3.000 | 0.0000 | | BC | 3.000 | -0.7071 | | CD | 3.000 | 0.0000 | | FE | 3.000 | -0.7071 | | AF | 4.243 | 0.0000 | | ED | 4.243 | 0.0000 | | BF | 3.000 | -0.7071 | | CE | 3.000 | -0.7071 | | BE | 4.243 | +1.0000 | | CF | 4.243 | +1.0000 | The outer members AB, CD, AF and ED carry zero force in this system. #### Compatibility $$ \begin{aligned} f_{11} &= \sum\frac{n^2 L}{AE} = \frac{4(0.5)(3) + 2(1)(4.243)}{AE} = \frac{14.485}{AE} = 2.0693\times10^{-5}\ \text{m/kN} \\ \Delta_{1t} &= \sum n\,e = 2(-0.7071)(-0.00072) = 1.0182\times10^{-3}\ \text{m} \end{aligned} $$ $$ f_{11}X + \Delta_{1t} = 0 \ \Rightarrow\ X = -\frac{1.0182\times10^{-3}}{2.0693\times10^{-5}} = -49.21\ \text{kN} $$ #### Member forces $S = nX$ | Member | Force (kN) | |---|---| | AB | 0.00 | | EF (FE) | 34.79 (tension) | | BC | 34.79 (tension) | | BF = CE | 34.79 (tension) | | BE = CF | -49.21 (compression) | | AF, ED, CD | 0 | **Answer: bar force in AB $= 0$; bar force in EF $= 34.79$ kN (tension).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. By how much does each vertical member BF, CE (3 m) change for a 20°C fall? A. It shortens by 12 x 10^-6 x 20 x 3 = 0.72 mm B. It lengthens by 0.72 mm C. It shortens by 7.2 mm D. It shortens by 0.072 mm Answer: A. e = alpha x Delta t x L = 12e-6 x (-20) x 3 = -7.2e-4 m, a shortening of 0.72 mm. (Key point: Thermal change in length) 2. Why is the force in AB zero? A. Because AB is not a vertical member, so it cannot change length, hence carries the largest force B. n for AB is zero for the unit force in CF, and there is no load, so S = nX = 0 C. Because A is hinged D. Because X is zero Answer: B. Only BC, FE, BF, CE, BE and CF carry n values; the outer members AB, CD, AF, ED have n = 0. (Key point: Why AB is zero) 3. X = -49.21 kN (CF in compression). Why is FE in tension? A. Because FE is a vertical member B. Because temperature fall always puts the top chord in compression C. n for FE is -0.7071, so S = (-0.7071)(-49.21) = +34.79 kN D. Because n for FE is +1, giving +49.21 kN Answer: C. The negative redundant multiplied by the negative unit-force value gives a tensile force in FE, equal to 34.79 kN. 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