--- title: "Truss deflection at E by virtual work" subject: "Theory of Structures I" chapter: 3 key: "semester_4/theory_of_structures_i:truss-vw-e-2071o" source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/truss-vw-e-2071o site: BCT Engineering --- # Truss deflection at E by virtual work Theory of Structures I · Chapter 3: Analysis by the Virtual Work Method - Asked in: 2071 Magh (old course), 16 marks - Video: [Deflection of Trusses Virtual Work Method](https://www.youtube.com/watch?v=7AJsvPf_o_w) — Nadim Shbeeb - 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-e-2071o - Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-3 ## Question Determine the deflection at E of the pin-jointed truss shown in figure below by using virtual work method. Take area for all members $A = 400\ \text{mm}^2$, $E = 200\times10^3\ \text{N/mm}^2$. [Figure: truss 8 m high with four bottom panels of 10 m each; three 50 kN downward loads on the bottom chord joints, E being the bottom joint at the middle load; supports at the two ends.] ## Short answer Virtual unit load at E. Sum member kFL = 9550.4 kNm. With AE = 80,000 kN, deflection = 9550.4/80000 = 119.4 mm downward. ## Answer #### Given data and assumed geometry Pin-jointed truss, 4 panels of 10 m, height 8 m. Top joints P0 to P4 and bottom joints Q0 to Q4 (E is the bottom middle joint Q2). Members: top chord (4), bottom chord (4), five verticals, and four diagonals P0Q1, P1Q2, P3Q2, P4Q3. 50 kN down at Q1, Q2 and Q3; Q0 hinge, Q4 roller. $A = 400$ mm$^2 = 4\times10^{-4}$ m$^2$, $E = 200\times10^3$ N/mm$^2 = 2\times10^8$ kN/m$^2$ ($AE = 80\,000$ kN). #### Step 1: Reactions and real forces By symmetry $R_{Q0} = R_{Q4} = 75$ kN. Forces by the method of joints (tension +, compression -), e.g. at Q0: $F_{P0Q0} = -75$ kN, then joint P0 gives $F_{P0Q1} = +120.06$ kN and $F_{P0P1} = -93.75$ kN, and so on. #### Step 2: Virtual forces Apply a **unit downward load at E (Q2)**. By symmetry the reactions are 0.5 each, and the $k$ forces follow in the same way. #### Step 3: Table of $F$, $k$, $L$ | Member | L (m) | F (kN) | k | kFL (kN m) | |---|---|---|---|---| | P0P1 | 10.000 | -93.75 | -0.625 | 585.938 | | P1P2 | 10.000 | -125.00 | -1.250 | 1562.500 | | P2P3 | 10.000 | -125.00 | -1.250 | 1562.500 | | P3P4 | 10.000 | -93.75 | -0.625 | 585.938 | | Q0Q1 | 10.000 | -0.00 | -0.000 | 0.000 | | Q1Q2 | 10.000 | 93.75 | 0.625 | 585.938 | | Q2Q3 | 10.000 | 93.75 | 0.625 | 585.938 | | Q3Q4 | 10.000 | 0.00 | 0.000 | 0.000 | | P0Q0 | 8.000 | -75.00 | -0.500 | 300.000 | | P1Q1 | 8.000 | -25.00 | -0.500 | 100.000 | | P2Q2 | 8.000 | -0.00 | -0.000 | 0.000 | | P3Q3 | 8.000 | -25.00 | -0.500 | 100.000 | | P4Q4 | 8.000 | -75.00 | -0.500 | 300.000 | | P0Q1 | 12.806 | 120.06 | 0.800 | 1230.600 | | P1Q2 | 12.806 | 40.02 | 0.800 | 410.200 | | P3Q2 | 12.806 | 40.02 | 0.800 | 410.200 | | P4Q3 | 12.806 | 120.06 | 0.800 | 1230.600 | #### Step 4: Deflection $$ \Delta_E = \sum\frac{kFL}{AE} = \frac{9550.4}{80\,000} = 119.38\ \text{mm} $$ **Answer: vertical deflection at E = 119.4 mm downward.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What unit load is applied to find the deflection of E? A. A unit load at every bottom joint B. A unit downward load at E (the bottom middle joint Q2) C. A unit horizontal load at Q4 D. A unit upward load at P2 Answer: B. The unit load goes at the joint whose deflection is required, in the direction wanted. (Key point: Virtual load) 2. For the middle vertical P2Q2, F = 0 and k = 0. What follows? A. It doubles the deflection B. It carries the greatest load C. It must be removed D. It contributes nothing to the sum of kFL Answer: D. The contribution kFL is zero when F or k is zero. (Key point: Zero-force members) 3. Which formula gives the deflection of E? A. Delta = sum kF/AE B. Delta = sum k L/AE C. Delta = sum kFL/AE, with AE = 80,000 kN D. Delta = sum FL/AE Answer: C. The virtual work formula for trusses multiplies real force F, virtual force k and length L, divided by AE; AE = 4e-4 x 2e8 = 80,000 kN. (Key point: Formula) 4. Sum kFL = 9550.4 kN m. What is the deflection of E? A. 119.4 mm upward B. 119.4 mm downward C. 11.94 mm downward D. 0.119 mm Answer: B. 9550.4/80000 = 0.11938 m = 119.4 mm; positive means in the direction of the unit load (downward). 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