--- title: "Truss deflection at B: load, temperature, lack of fit" subject: "Theory of Structures I" chapter: 3 key: "semester_4/theory_of_structures_i:truss-defl-b-loading-temp-lackfit-2081" source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/truss-defl-b-loading-temp-lackfit-2081 site: BCT Engineering --- # Truss deflection at B: load, temperature, lack of fit Theory of Structures I · Chapter 3: Analysis by the Virtual Work Method - Asked in: 2081 Chaitra, 10 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_4/theory_of_structures_i/truss-defl-b-loading-temp-lackfit-2081 - Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-3 ## Question For the truss shown in the figure below, calculate the vertical deflection at joint B due to (i) external loading, (ii) temperature of inclined members decreases by 15°C, (iii) lack of fit in vertical members being 8 mm too long. Take $A = 600\ \text{mm}^2$, $E = 200$ GPa and $\alpha = 19\times10^{-6}/^\circ$C. [Figure: truss with bottom joints A (hinge), B (middle) and C (roller), AB = BC = 3 m; top joints D, E and F directly above A, B and C, height 4 m; vertical members AD, BE and CF; inclined members DB and BF; 80 kN downward loads at D and F.] ## Short answer External loading causes 2.67 mm downward deflection; a 15°C temperature drop in inclined members causes 1.78 mm upward; 8 mm excess vertical member length causes 8.00 mm upward. Net vertical deflection at B is 7.11 mm upward. ## Answer #### Given data and method $A = 600$ mm$^2 = 6\times10^{-4}$ m$^2$, $E = 200$ GPa $= 2\times10^8$ kN/m$^2$, $\alpha = 19\times10^{-6}$/$^\circ$C. Joints: A(0,0), B(3,0), C(6,0), D(0,4), E(3,4), F(6,4); members AB, BC, DE, EF, AD, BE, CF, DB, BF. A is a hinge and C a roller; 80 kN acts downward at D and at F. Apply a **unit vertical load at B** (downward) and use the virtual work method. Tension is positive. #### Support reactions By symmetry, $A_y = C_y = 80$ kN. The 80 kN loads act directly over the supports, so they pass straight down the verticals AD and CF ($F = -80$ kN, compression); all other members have zero force. #### Forces in members ($F$ = real loading, $k$ = unit load at B) | Member | L (m) | F (kN) | k | kFL (kN m) | Temp. term (mm) | Fit term (mm) | |---|---|---|---|---|---|---| | AB | 3.000 | 0 | 0 | 0 | 0 | 0 | | BC | 3.000 | 0 | 0 | 0 | 0 | 0 | | DE | 3.000 | 0 | -0.375 | 0 | 0 | 0 | | EF | 3.000 | 0 | -0.375 | 0 | 0 | 0 | | AD | 4.000 | -80.00 | -0.500 | 160.000 | 0 | -4.0000 | | BE | 4.000 | 0 | 0 | 0 | 0 | 0 | | CF | 4.000 | -80.00 | -0.500 | 160.000 | 0 | -4.0000 | | DB | 5.000 | 0 | 0.625 | 0 | -0.8906 | 0 | | BF | 5.000 | 0 | 0.625 | 0 | -0.8906 | 0 | (The real force $F$ is non-zero only in AD and CF.) #### (i) External loading $$ \Delta_{B1} = \sum\frac{kFL}{AE} = \frac{320.00}{6\times10^{-4}\times2\times10^8} = 2.667\ \text{mm}\ (\downarrow) $$ #### (ii) Temperature fall of 15$^\circ$C in the inclined members DB and BF Change in length of each: $\alpha\,\Delta T\,L = 19\times10^{-6}\times(-15)\times5000 = -1.425$ mm. $$ \Delta_{B2} = \sum k\,\alpha\Delta T L = 2\times(0.625)(-1.425) = -1.781\ \text{mm} $$ (Negative: joint B moves up.) #### (iii) Lack of fit: vertical members 8 mm too long $$ \Delta_{B3} = \sum k\,e = (-0.5)(8) + (0)(8) + (-0.5)(8) = -8.000\ \text{mm} $$ (BE has $k = 0$.) #### Total $$ \Delta_B = 2.667 - 1.781 - 8.000 = -7.115\ \text{mm} $$ **Answer:** (i) 2.67 mm down, (ii) 1.78 mm up, (iii) 8.00 mm up. The net vertical deflection of B is 7.11 mm upward. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With 80 kN at D and F directly over the supports, which members carry real force? A. All members B. Only vertical members AD and CF (80 kN compression) C. Only inclined members DB and BF D. Only bottom chord Answer: B. The loads go straight to the supports through AD and CF; every other member has zero force. (Key point: Member forces under load) 2. The deflection at B due to loading is: A. 2.667 mm downward B. 8 mm upward C. 1.78 mm upward D. 26.7 mm downward Answer: A. Sum kFL = 320 and divided by AE = 1.2e5 kN gives 2.667 mm downward. (Key point: Loading deflection) 3. Why does a temperature fall in DB and BF raise joint B? A. The inclined members shorten (negative length change) with positive k, so k alpha dT L is negative B. The inclined members lengthen C. Vertical members shorten D. Loads decrease Answer: A. Each inclined member changes length by -1.425 mm, and with k = 0.625 the sum is -1.781 mm, meaning upward movement. (Key point: Temperature effect) 4. For vertical members 8 mm too long, the deflection at B is: A. +8 mm B. -8 mm (8 mm up), because k = -0.5 for AD and CF and 0 for BE C. -16 mm D. 0, because BE has k = 0 Answer: B. Sum k e = (-0.5)(8) + 0 + (-0.5)(8) = -8 mm, i.e. upward. 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