--- title: "Redundant truss: force in member BF" subject: "Theory of Structures II" chapter: 2 key: "semester_5/theory_of_structures_ii:truss-redundant-bf-member-force" source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/truss-redundant-bf-member-force site: BCT Engineering --- # Redundant truss: force in member BF Theory of Structures II · Chapter 2: Force method - Asked in: 2070 Asar, 5 marks - 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-redundant-bf-member-force - Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-2 ## Question Determine the force in member BF of the redundant truss shown in figure below. Cross section areas of each member in cm² are given in the figure within brackets (values 15, 20, 25 and 30). [Figure: truss ABCDEF with 4 m height; bottom panels 3 m, 4 m and 3 m; loads 30 kN and 50 kN downward at bottom joints; A hinged, D roller; member areas as printed in the figure.] ## Short answer Use force method taking BF as redundant. Compute bar forces in the released truss under given loads and under unit tensile load along BF. Apply compatibility ΣS0S1L/A + XΣS1²L/A = 0, giving X = 15.10 kN. Force in BF is 15.10 kN (tension). ## Answer The truss has one degree of redundancy ($m + r - 2j = 10 + 3 - 12 = 1$). Take the force $X$ in the diagonal BF as the redundant and solve it by the force method (unit-load / strain energy). **Assumed data (the figure is not fully legible):** joints A, B, C, D on the bottom chord at 0, 3, 7, 10 m; E, F on top, 4 m above B and C; loads 30 kN at B and 50 kN at C; A hinged, D roller; both diagonals BF and CE present in the middle panel. Areas: bottom chord 20 cm², top chord 25 cm², verticals 15 cm², all diagonals 30 cm². E is constant, so it cancels. #### Step 1: Released (determinate) truss, loads only Remove BF. Reactions: $R_D = (30\times3 + 50\times7)/10 = 44$ kN, $R_A = 36$ kN, $H_A = 0$. Solving by joints gives the forces $S_0$ in the table. #### Step 2: Unit load on the released truss Apply a pair of unit tensile forces along BF and find the forces $S_1$ (BF itself has $S_1 = +1$). #### Step 3: Compatibility $$ \sum S_0 S_1 \frac{L}{A} + X\sum S_1^2\frac{L}{A} = 0 \quad\Rightarrow\quad X = -\frac{\sum S_0S_1L/A}{\sum S_1^2L/A} $$ | Member | L (m) | A (cm²) | $S_0$ (kN) | $S_1$ | $S_0S_1L/A$ | $S_1^2L/A$ | |---|---|---|---|---|---|---| | AB | 3.000 | 20 | 27.00 | 0.000 | 0.000 | 0.0000 | | BC | 4.000 | 20 | 27.00 | -0.707 | -3.818 | 0.1000 | | CD | 3.000 | 20 | 33.00 | 0.000 | 0.000 | 0.0000 | | EF | 4.000 | 25 | -33.00 | -0.707 | 3.734 | 0.0800 | | AE | 5.000 | 30 | -45.00 | 0.000 | 0.000 | 0.0000 | | BE | 4.000 | 15 | 30.00 | -0.707 | -5.657 | 0.1333 | | CF | 4.000 | 15 | 44.00 | -0.707 | -8.297 | 0.1333 | | DF | 5.000 | 30 | -55.00 | 0.000 | 0.000 | 0.0000 | | CE | 5.657 | 30 | 8.49 | 1.000 | 1.600 | 0.1886 | | BF | 5.657 | 30 | 0.00 | 1.000 | 0.000 | 0.1886 | | **Sum** | | | | | **-12.438** | **0.8238** | $$ X = -\frac{(-12.438)}{0.8238} = 15.10\ \text{kN} $$ #### Final member forces ($S = S_0 + X S_1$, tension +) AB = 27.00, BC = 16.32, CD = 33.00, EF = -43.68, AE = -45.00, BE = 19.32, CF = 33.32, DF = -55.00, CE = 23.58, BF = 15.10 kN. **Answer: force in BF = 15.10 kN (tension)** for the assumed areas. (Checked by a full direct-stiffness analysis of the 10-member truss.) ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the redundancy of the truss found? A. m + r - 2j = 10 + 3 - 12 = 1 B. m - j = 0 C. r - 3 = 0 D. 2j - m = 2 Answer: A. One extra member, so BF is taken as the redundant. (Key point: Degree of redundancy) 2. The redundant X in BF is given by: A. X = ΣS0S1L/A B. X = -ΣS0S1(L/A)/ΣS1^2(L/A) C. X = ΣS1^2L/A D. X = -ΣS0/ΣS1 Answer: B. Zero relative displacement at the cut gives ΣS0S1L/A + XΣS1^2L/A = 0. (Key point: Compatibility for X) 3. What is S1? A. Forces due to the 30 and 50 kN loads B. Forces from a pair of unit tensile forces along BF C. Reactions D. Forces with BF still carrying load Answer: B. S1 is the member-force pattern for unit loads on the cut BF in the released truss, BF itself being +1. (Key point: Unit load case) 4. Final forces are S = S0 + X S1. For CE with S0 = 8.49, S1 = 1, X = 15.10: A. 23.58 kN B. -6.61 kN C. 8.49 kN D. 15.10 kN Answer: A. 8.49 + 15.10 = 23.58 kN. 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