--- title: "Truss: force method, 60 kN at D" subject: "Theory of Structures II" chapter: 2 key: "semester_5/theory_of_structures_ii:truss-force-method-60kn-3m-4m" source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/truss-force-method-60kn-3m-4m site: BCT Engineering --- # Truss: force method, 60 kN at D Theory of Structures II · Chapter 2: Force method - Asked in: 2074 Asoj, 12 marks - Video: [SA26: Force Method (Truss Analysis)](https://www.youtube.com/watch?v=AuBcV1oehD8) — Dr. Structure - 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-force-method-60kn-3m-4m - Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-2 ## Question Analyse the truss shown in figure below using the "Force Method". Take the cross-sectional stiffness EA of the members to be constant. [Figure: truss ABCDE; bottom chord A-D-E (4 m + 4 m), top joints B and C at 3 m height, with diagonals; A hinged, E roller; 60 kN downward at D.] ## Short answer Using force method with redundants in diagonals AC and BE, by symmetry VA=VE=30kN. Final member forces (kN, + tension): AB & CE -30.80, BC -28.33, AD & DE +25.83, BD & CD +36.06, AC & BE -9.78. ## Answer Reading of the figure: A (0, 0) hinged, E (8, 0) roller; D (4, 0) on the bottom chord; top joints B (2, 3) and C (6, 3); members AB, BC, CE (top and sides), AD, DE (bottom chord), BD, CD and the two crossing diagonals AC and BE. The 60 kN load acts downward at D. $EA$ is constant. (The member arrangement is read from the description; the method is the same for any arrangement.) $m = 9$, $j = 5$, $r = 3$, so $D_s = m + r - 2j = 2$. Choose $X_1$ = force in AC and $X_2$ = force in BE (tension +). #### Reactions By symmetry $V_A = V_E = 30$ kN and $H_A = 0$. #### Force tables (tension +) | Member | $L$ (m) | $S_0$ | $n_1$ | $n_2$ | Final $S$ | |---|---|---|---|---|---| | AB | 3.606 | -36.056 | -0.537 | 0.000 | -30.796 | | BC | 4.000 | -40.000 | -0.596 | -0.596 | -28.331 | | CE | 3.606 | -36.056 | 0.000 | -0.537 | -30.796 | | AD | 4.000 | +20.000 | -0.596 | 0.000 | +25.834 | | DE | 4.000 | +20.000 | 0.000 | -0.596 | +25.834 | | BD | 3.606 | +36.056 | +0.537 | -0.537 | +36.056 | | CD | 3.606 | +36.056 | -0.537 | +0.537 | +36.056 | | AC | 6.708 | 0.000 | +1.000 | 0.000 | -9.785 | | BE | 6.708 | 0.000 | 0.000 | +1.000 | -9.785 | #### Flexibility coefficients (in $1/AE$ units) $$ \begin{aligned} f_{11} &= f_{22} = 12.677, \qquad f_{12} = f_{21} = -0.661 \\ \Delta_{10} &= \sum S_0n_1L = 117.576, \qquad \Delta_{20} = \sum S_0n_2L = 117.576 \end{aligned} $$ #### Compatibility $$ \begin{aligned} 12.677\,X_1 + (-0.661)\,X_2 &= -117.576 \\ -0.661\,X_1 + 12.677\,X_2 &= -117.576 \end{aligned} $$ $$ X_1 = S_{AC} = -9.78\ \text{kN},\qquad X_2 = S_{BE} = -9.78\ \text{kN} $$ #### Final member forces $S = S_0 + n_1X_1 + n_2X_2$ | Member | Force (kN) | Nature | |---|---|---| | AB | -30.80 | compression | | BC | -28.33 | compression | | CE | -30.80 | compression | | AD | +25.83 | tension | | DE | +25.83 | tension | | BD | +36.06 | tension | | CD | +36.06 | tension | | AC | -9.78 | compression | | BE | -9.78 | compression | **Answer: AB $= CE = -30.80$ kN, BC $= -28.33$ kN, AD $= DE = +25.83$ kN, BD $= CD = +36.06$ kN, AC $= BE = -9.78$ kN (positive = tension).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With 9 members, 5 joints and 3 reactions, what is D_s? A. 1 B. 3 C. 0 D. 2 Answer: D. D_s = m + r - 2j = 9 + 3 - 10 = 2, so two redundants X1 (AC) and X2 (BE) are chosen. (Key point: Degree of indeterminacy) 2. What are the support reactions? A. VA = VE = 30 kN, HA = 0 B. VA = 60 kN, VE = 0 C. VA = VE = 60 kN D. VA = 30 kN, HA = 30 kN Answer: A. Symmetric loading of 60 kN at mid-span gives 30 kN each and no horizontal reaction. (Key point: Reactions) 3. What form do the two compatibility equations take? A. f11X1 + f12X2 = -Δ10 and f21X1 + f22X2 = -Δ20 B. X1 = X2 = 0 C. f11X1 = f22X2 D. ΣS = 60 Answer: A. Relative displacements at the cuts in AC and BE vanish: [f]{X} + {Δ0} = 0. (Key point: Compatibility equations) 4. How is f12 computed? A. Σ n1^2 L/AE B. Σ S0 n1 L/AE C. Σ n1 n2 L/AE D. Σ S0 L/AE Answer: C. f12 = Σ n1 n2 L/AE, using member forces for unit loads in AC and BE. (Key point: Coefficients) 5. What are the forces in AC and BE? A. +9.78 kN each (tension) B. -9.78 kN each (compression) C. -30.8 kN each D. 0 Answer: B. Solving symmetric equations: 12.677 X - 0.661 X = -117.576 gives X = -9.78 kN (compression). 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