--- title: "Castigliano: square truss member forces" subject: "Theory of Structures II" chapter: 2 key: "semester_5/theory_of_structures_ii:castigliano-square-truss-120kn-90kn" source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/castigliano-square-truss-120kn-90kn site: BCT Engineering --- # Castigliano: square truss member forces Theory of Structures II · Chapter 2: Force method - Asked in: 2070 Chaitra, 10 marks - Video: [Castigliano methods eg4](https://www.youtube.com/watch?v=HpMPrRzZL3c) — Anup Patil - 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/castigliano-square-truss-120kn-90kn - Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-2 ## Question Use Castigliano's theorem to determine forces induced in each member of the square truss loaded as shown below. [Figure: square truss ABCD, 4 m x 4 m, with both diagonals, AE constant; 120 kN downward at D and 90 kN horizontal (toward the left) at D; A hinged, B roller.] ## Short answer Use Castigliano with force in AC as redundant, ∂U/∂X=ΣS n L/AE=0. Final forces: AB=-32.57, BC=+57.43, CD=+57.43, DA=-152.57, AC=-81.21 kN compression, and BD=+46.07 kN tension. ## Answer Reading of the figure: A (0, 0) hinged, B (4, 0) roller, C (4, 4) and D (0, 4); members AB, BC, CD, DA (4 m) and diagonals AC, BD (5.657 m); 120 kN downward and 90 kN horizontal (towards the left) at D. $AE$ is constant. $D_s = 6 + 3 - 8 = 1$. Take the force in AC as the redundant $X$. #### Reactions $\Sigma M_A = 0$: $V_B(4) + 90(4) = 0$, hence $V_B = -90.00$ kN (downward 90 kN). $V_A = 120 + 90 = 210.00$ kN ($\uparrow$), $H_A = 90$ kN ($\rightarrow$). #### Member forces in terms of $X$ ($S = S_0 + nX$) | Member | $L$ (m) | $S_0$ | $n_1$ | Final $S$ | |---|---|---|---|---| | AB | 4.000 | -90.000 | -0.707 | -32.574 | | BC | 4.000 | 0.000 | -0.707 | +57.426 | | CD | 4.000 | 0.000 | -0.707 | +57.426 | | DA | 4.000 | -210.000 | -0.707 | -152.574 | | AC | 5.657 | 0.000 | +1.000 | -81.213 | | BD | 5.657 | +127.279 | +1.000 | +46.066 | #### Castigliano's theorem The strain energy of the truss is $U = \sum \dfrac{S^2L}{2AE}$. The relative displacement of the cut ends of AC must be zero, so $$ \frac{\partial U}{\partial X} = \sum \frac{S\,n\,L}{AE} = \frac{1}{AE}\left[\sum S_0nL + X\sum n^2L\right] = 0 $$ $$ \sum S_0nL = 1568.528,\qquad \sum n^2L = 19.314 $$ $$ X = S_{AC} = -\frac{1568.528}{19.314} = -81.21\ \text{kN} $$ #### Forces in all members | Member | Force (kN) | Nature | |---|---|---| | AB | -32.57 | compression | | BC | +57.43 | tension | | CD | +57.43 | tension | | DA | -152.57 | compression | | AC | -81.21 | compression | | BD | +46.07 | tension | **Answer: AB $= -32.57$, BC $= +57.43$, CD $= +57.43$, DA $= -152.57$, AC $= -81.21$, BD $= +46.07$ kN (positive = tension).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For the square truss with both diagonals (6 members, 3 reactions, 4 joints), what is D_s? A. 0 B. 1 C. 2 D. 3 Answer: B. D_s = 6 + 3 - 2(4) = 1, so one redundant (force in AC) is taken. (Key point: Degree of indeterminacy) 2. Which condition determines the redundant X? A. Σ S n L/AE = ∂U/∂X = 0 (no relative displacement of the cut ends) B. U = 0 C. Σ S L = 0 D. ∂U/∂P = 0 for the 120 kN load Answer: A. The cut ends of AC must have zero relative movement, so ∂U/∂X = 0. (Key point: Castigliano condition) 3. With S = S0 + nX, the redundant is: A. X = ΣS0nL × Σn^2L B. X = ΣS0nL/Σn^2L C. X = -ΣS0nL/Σn^2L D. X = -Σn^2L/ΣS0nL Answer: C. Setting ΣS0nL + XΣn^2L = 0 gives X = -ΣS0nL/Σn^2L = -81.21 kN. (Key point: Formula for X) 4. Which member forces are correct? A. AC = +81.21 tension B. DA = -152.57 compression and AC = -81.21 compression C. BD = -46.07 compression D. AB = +32.57 tension Answer: B. X is negative so AC is in compression; DA is -152.57 kN, BD +46.07 kN tension. 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