--- title: "Truss: loads, temperature rise, lack of fit" subject: "Theory of Structures II" chapter: 2 key: "semester_5/theory_of_structures_ii:truss-loads-temp-lack-of-fit" source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/truss-loads-temp-lack-of-fit site: BCT Engineering --- # Truss: loads, temperature rise, lack of fit Theory of Structures II · Chapter 2: Force method - Asked in: 2076 Chaitra, 10 marks - Video: [Lack of Fit in Trusses | Structural Analysis](https://www.youtube.com/watch?v=wazo1ndN_1Y) — GATE ACADEMY by Umesh Dhande - 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-loads-temp-lack-of-fit - Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-2 ## Question Find the member forces of the given loaded truss for the given external loadings and due to rise in temperature of all diagonal members by 20°C. Take $AE = 5000$ kN for all members and coefficient of thermal expansion $2.06\times10^{-6}/°C$. Additionally, vertical members are 5 mm too long. [Figure: rectangular truss ABCD with both diagonals, 4 m wide and 3 m high; 30 kN horizontal at B and 50 kN downward at C; A hinged, D roller.] ## Short answer Member forces: AB = +13.40 kN, BC = -12.13 kN, CD = -59.10 kN, AD = +17.87 kN, AC = +15.16 kN, BD = -22.34 kN (tension positive). Reactions: H_A = 30 kN left, V_A = 22.50 kN downward, V_D = 72.50 kN upward. ## Answer Reading of the figure: A (0, 0) hinged and D (4, 0) roller at the base, B (0, 3) and C (4, 3) at the top, members AB, CD (verticals, 3 m), BC, AD (horizontals, 4 m) and diagonals AC, BD (5 m). Loads: 30 kN horizontal at B (towards C) and 50 kN downward at C. The truss has $D_s = 6 + 3 - 8 = 1$ (internal). Take the force in AC as the redundant $X$ (tension +). #### Effects to be included - Diagonals AC and BD: temperature rise 20$^\circ$C, so $e = \alpha\,\Delta t\,L = 2.06\times10^{-6}\times20\times5 = 2.06\times10^{-4}$ m (elongation). - Verticals AB and CD: 5 mm too long, so $e = +0.005$ m. - $AE = 5000$ kN for all members. #### Reactions (statics) $\Sigma M_A = 0$: $V_D(4) = 50(4) + 30(3)$, so $V_D = 72.50$ kN ($\uparrow$). $\Sigma F_y$: $V_A = 50 - 72.50 = -22.50$ kN, i.e. 22.50 kN downward. $\Sigma F_x$: $H_A = 30$ kN acting towards the left. #### Primary structure (AC removed) and unit force system $S_0$ = force due to loads; $n$ = force due to $X = 1$ in AC. Tension is positive. | Member | $L$ (m) | $S_0$ (kN) | $n$ | $S_0 nL$ | $n^2L$ | $n\,e$ (mm) | |---|---|---|---|---|---|---| | AB | 3 | +22.500 | -0.60 | -40.500 | 1.080 | -3.0000 | | BC | 4 | 0.000 | -0.80 | 0.000 | 2.560 | 0.0000 | | CD | 3 | -50.000 | -0.60 | +90.000 | 1.080 | -3.0000 | | AD | 4 | +30.000 | -0.80 | -96.000 | 2.560 | 0.0000 | | AC | 5 | 0.000 | +1.00 | 0.000 | 5.000 | +0.2060 | | BD | 5 | -37.500 | +1.00 | -187.500 | 5.000 | +0.2060 | | Total | | | | -234.000 | 17.280 | -5.5880 | #### Compatibility $$ \frac{\sum S_0 nL}{AE} + X\frac{\sum n^2L}{AE} + \sum n\,e = 0 $$ $$ \frac{-234.000}{5000} + X\,\frac{17.280}{5000} - 0.005588 = 0 $$ $$ X = \frac{-(-234.000 + 5000 \times -0.005588)}{17.280} = 15.159\ \text{kN} $$ (The thermal and lack-of-fit terms enter only through $\sum n\,e$, which is $-0.005588$ m.) #### Final forces $S = S_0 + nX$ (kN) | Member | Force | Nature | |---|---|---| | AB | +13.40 | tension | | BC | -12.13 | compression | | CD | -59.10 | compression | | AD | +17.87 | tension | | AC | +15.16 | tension | | BD | -22.34 | compression | Check at joint C (vertical): $-S_{CD} - 0.6\,S_{AC} = 50.00$ kN, equal to the 50 kN load. **Answer: AB $= +13.40$ kN, BC $= -12.13$ kN, CD $= -59.10$ kN, AD $= +17.87$ kN, AC $= +15.16$ kN, BD $= -22.34$ kN (positive = tension). Reactions: $H_A = 30$ kN (left), $V_A = 22.50$ kN (downward), $V_D = 72.50$ kN (upward).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the degree of indeterminacy and the redundant? A. Ds = 3; the reactions are redundant B. Ds = 1 (internal); the force in diagonal AC is taken as the redundant C. Ds = 0; no redundant D. Ds = 2; both diagonals are redundants Answer: B. m + r - 2j = 6 + 3 - 8 = 1; the supports are determinate, so one member is redundant. (Key point: Redundant) 2. What is the elongation of a diagonal (5 m) due to a 20°C rise? A. 2.06 x 10^-6 x 20 x 5 = 2.06 x 10^-4 m B. 2.06 x 10^-6 x 20 = 4.12 x 10^-5 m C. 2.06 x 10^-6 x 5 = 1.03 x 10^-5 m D. 0.005 m Answer: A. e = alpha Delta t L. The 5 mm figure applies to the verticals as lack of fit. (Key point: Thermal elongation) 3. How do the temperature and lack-of-fit effects enter the compatibility equation? A. They are added to S0 as extra loads B. They change AE C. Through the sum of n x e, added to sum S0 n L/AE + X sum n^2 L/AE D. They are ignored because they are small Answer: C. Each member's e is multiplied by its unit-force value n and summed: sum n e = -0.005588 m. (Key point: Compatibility terms) 4. How are the final member forces obtained? A. S = S0 x nX B. S = S0 + nX, e.g. AB = 22.5 + (-0.6)(15.159) = 13.40 kN tension C. S = X for every member D. S = S0 - X Answer: B. Superposition of the load forces S0 and the redundant effect nX gives the final forces. 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