--- title: "Truss joint L2 by virtual work and temperature" subject: "Theory of Structures I" chapter: 3 key: "semester_4/theory_of_structures_i:truss-vw-l2-temperature-2069p" source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/truss-vw-l2-temperature-2069p site: BCT Engineering --- # Truss joint L2 by virtual work and temperature Theory of Structures I · Chapter 3: Analysis by the Virtual Work Method - Asked in: 2069 Poush, 8 marks - 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-vw-l2-temperature-2069p - Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-3 ## Question Determine, using virtual work method, the vertical deflection of joint $L_2$. The L/A values for diagonal and vertical members are 12 mm⁻¹ and for horizontal members are 6 mm⁻¹. Take $E = 200\times10^3\ \text{N/mm}^2$ for all members. (i) Find vertical deflection due to loads as shown in figure. (ii) Find the additional deflection if the top boom is subjected to a temperature rise of 20°C. Take $\alpha = 10.8\times10^{-6}/^\circ$C. [Figure: truss, height 12 m, 4 panels of 10 m; bottom joints $L_1$ to $L_4$ and top joints $U_1$ to $U_5$ as drawn; three 10 kN loads on the bottom chord; $L_1$ hinge, $L_4$ roller.] ## Short answer The vertical deflection of joint L2 due to the three 10 kN loads is 4.69 mm downward. A 20°C temperature rise in the top boom causes an upward deflection of 5.40 mm. The net deflection is 0.71 mm upward. ## Answer #### Given data and assumed geometry Truss of height 12 m, 4 panels of 10 m. The joint labels in the figure could not be read exactly, so the support joints are taken as the two ends of the bottom chord, labelled here L0 (hinge) and L4 (roller), with bottom joints L1, L2, L3 between them; top joints U1 to U5 (U1 above L0). Verticals $U_{i+1}L_i$, diagonals U1L1, U2L2, U4L2, U5L3. 10 kN down at L1, L2, L3. $L/A$ = 12 mm$^{-1}$ for diagonals and verticals, 6 mm$^{-1}$ for horizontal members. $E = 200\times10^3$ N/mm$^2 = 200$ kN/mm$^2$. By symmetry the reactions at L0 and L4 are 15 kN. A unit downward load is applied at $L_2$. #### (i) Deflection due to loads $$ \Delta = \sum F\,k\,\frac{L}{AE} $$ | Member | $F$ (kN) | $k$ | $L/A$ (mm$^{-1}$) | $F\,k\,(L/A)$ (kN/mm) | |---|---|---|---|---| | U1U2 | -12.50 | -0.417 | 6 | 31.25 | | U2U3 | -16.67 | -0.833 | 6 | 83.33 | | U3U4 | -16.67 | -0.833 | 6 | 83.33 | | U4U5 | -12.50 | -0.417 | 6 | 31.25 | | L0L1 | 0.00 | 0.000 | 6 | 0.00 | | L1L2 | 12.50 | 0.417 | 6 | 31.25 | | L2L3 | 12.50 | 0.417 | 6 | 31.25 | | L3L4 | 0.00 | 0.000 | 6 | 0.00 | | U1L0 | -15.00 | -0.500 | 12 | 90.00 | | U2L1 | -5.00 | -0.500 | 12 | 30.00 | | U3L2 | -0.00 | -0.000 | 12 | 0.00 | | U4L3 | -5.00 | -0.500 | 12 | 30.00 | | U5L4 | -15.00 | -0.500 | 12 | 90.00 | | U1L1 | 19.53 | 0.651 | 12 | 152.50 | | U2L2 | 6.51 | 0.651 | 12 | 50.83 | | U4L2 | 6.51 | 0.651 | 12 | 50.83 | | U5L3 | 19.53 | 0.651 | 12 | 152.50 | $$ \Delta_1 = \frac{\sum Fk(L/A)}{E} = \frac{938.33}{200} = 4.692\ \text{mm}\ (\downarrow) $$ #### (ii) Additional deflection due to 20$^\circ$C rise in the top boom The top boom has 4 members, each 10 m long: $\alpha\Delta T L = 10.8\times10^{-6}\times20\times10\,000 = 2.16$ mm each. $$ \Delta_2 = \sum k\,\alpha\,\Delta T\,L = 2.16\,(-0.417 - 0.833 - 0.833 - 0.417) = 2.16\times(-2.5) = -5.40\ \text{mm} $$ (upward, because the top boom lengthens and its $k$ values are negative: the top boom is in compression under a downward unit load). **Answer:** (i) vertical deflection of L2 due to loads $= 4.69$ mm downward; (ii) additional deflection due to temperature $= 5.40$ mm upward; net $= 0.71$ mm upward. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Given L/A values, how is the deflection due to loads found? A. Delta = sum F k (L/A) / E B. Delta = sum F k E / (L/A) C. Delta = sum k (L/A) E D. Delta = sum F (L/A) only Answer: A. Since L/(AE) = (L/A)/E, the sum of F k (L/A) is divided by E. (Key point: Formula with L/A) 2. Where is the unit load applied and what are the end reactions? A. Unit downward load at L2, with reactions 15 kN at L0 and L4 for the real loads by symmetry B. Unit upward load at L4 C. Unit load at U3, reactions 10 kN D. Unit load at L1 Answer: A. The deflection of L2 is required, so the unit load goes at L2; the real loads of 3 x 10 kN give 15 kN at each end. (Key point: Unit load position) 3. Why is the temperature effect on L2 upward? A. The temperature has no effect B. Top boom members shorten C. The bottom boom expands D. Top boom members lengthen while their k values are negative, giving a negative sum Answer: D. Delta_2 = 2.16 x (-2.5) = -5.40 mm; negative means upward. (Key point: Temperature effect) 4. What is the net deflection of L2? A. 0.71 mm upward B. 0.71 mm downward C. 4.69 mm downward D. 10.09 mm downward Answer: A. Net is 4.692 - 5.40 = -0.71 mm, i.e. 0.71 mm upward. 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