--- title: "Use of trusses; cantilever truss member forces" subject: "Applied Mechanics" chapter: 7 key: "semester_1/applied_mechanics:cantilever-truss-bc-be-ef-ab-af-bf" source: https://www.bctengineer.com/q/semester_1/applied_mechanics/cantilever-truss-bc-be-ef-ab-af-bf site: BCT Engineering --- # Use of trusses; cantilever truss member forces Applied Mechanics · Chapter 7: Analysis of Plane Trusses - Asked in: 2071 Chaitra, 2+6 marks - Video: [Truss Method Of Joints | Analysis Of Cantilever Truss | (हिंदी में)](https://www.youtube.com/watch?v=kArxcMX-r9k) — Civil Stuff - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_1/applied_mechanics/cantilever-truss-bc-be-ef-ab-af-bf - Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-7 ## Question Describe the use of trusses in engineering. Determine the force developed in BC, BE, EF, AB, AF and BF members of the cantilever truss loaded as shown in figure below. [Figure: cantilever truss fixed to a wall at A and G (6 m apart vertically); bottom chord G-F-E-D with panels of 4 m + 4 m + 4 m; joints A, B, C, D; loads 8 kN at F, 6 kN at E and 4 kN at D, all downward.] ## Short answer Trusses are light frameworks of triangular units carrying axial forces, used in roofs, bridges, towers, etc. Using method of joints from free end: BC = 8.94 kN (T), BE = 8.49 kN (T), EF = 14.00 kN (C), AB = 15.65 kN (T), AF = 13.22 kN (T), BF = 3.00 kN (C). ## Answer #### Use of trusses in engineering A truss is a light, stiff framework of straight members arranged in triangles, so every member carries only tension or compression and the material is used efficiently over long spans. Trusses are used for roofs of halls and factories, bridges (railway and road), transmission and telecom towers, crane jibs and gantries, floor and wall frames, aircraft and ship structures, and temporary stages and scaffolding. #### Data Wall supports at A(0,6) and G(0,0) (pin-type, 4 reaction components). Bottom chord G(0,0), F(4,0), E(8,0), D(12,0); top chord A(0,6), B(4,4), C(8,2), D(12,0) in one straight line (slope 6 m in 12 m); verticals BF and CE; diagonals AF and BE. Loads: 8 kN ↓ at F, 6 kN ↓ at E, 4 kN ↓ at D. (m = 10, r = 4, j = 7: m + r = 2j, determinate.) Lengths: AB = BC = CD = √(4² + 2²) = 4.472 m (cos 0.8944, sin 0.4472); BE = 5.657 m; AF = 7.211 m. #### Method of joints, starting at the free end (no need for reactions) **Joint D**: ΣFy: −4 + 0.4472F_CD = 0, so **F_CD = +8.94 kN (T)**. ΣFx: −0.8944(8.94) − F_ED = 0, so F_ED = −8.00 kN (C). **Joint C**: CB and CD are collinear and CE is the only other member (no load at C), so **F_CE = 0** and **F_BC = F_CD = +8.94 kN (T)**. **Joint E**: ΣFy: −6 + 0.7071F_BE = 0, so **F_BE = +8.49 kN (T)**. ΣFx: −8.00 − 0.7071(8.49) − F_EF = 0, so **F_EF = −14.00 kN (C)**. **Joint B** (BC = 8.94 T, BE = 8.49 T): ΣFx: 0.8944(8.94) + 0.7071(8.49) − 0.8944F_AB = 0, so 8.00 + 6.00 = 0.8944F_AB and **F_AB = +15.65 kN (T)**. ΣFy: −0.4472(8.94) + 0.4472(15.65) − F_BF − 0.7071(8.49) = 0, so −4.00 + 7.00 − 6.00 = F_BF and **F_BF = −3.00 kN (C)**. **Joint F**: ΣFy: −8 − 3.00 + 0.8321F_AF = 0, so F_AF = +13.22 kN (T). #### Result | Member | Force (kN) | Nature | |---|---|---| | BC | 8.94 | tension | | BE | 8.49 | tension | | EF | 14.00 | compression | | AB | 15.65 | tension | | AF | 13.22 | tension | | BF | 3.00 | compression | Also CD = 8.94 T, DE = 8.00 C, CE = 0 and GF = 21.33 C (from ΣFx at F). Reactions: GF is the only member at G and it is horizontal, so the wall at G gives only a horizontal reaction of 21.33 kN →; A gives 21.33 kN ← and 18 kN ↑ (the total load). Check ΣM about G: 4(8) + 8(6) + 12(4) = 128 = 6 × 21.33 ✓. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why can the method of joints start at joint D without finding reactions? A. D has only two members and a known load, so two unknowns can be solved B. D is a support C. D has no load D. Reactions are zero Answer: A. Starting at the free end, each joint has at most two unknown members, so reactions are not needed. (Key point: Starting at the free end) 2. Why is F_CE = 0? A. CE is horizontal B. CB and CD are collinear and no load acts at C, so the third member CE is zero C. C is a support D. CE is the longest member Answer: B. At an unloaded joint with two collinear members, the third member carries no force. (Key point: Zero-force member CE) 3. What is the force in EF? A. 14.00 kN tension B. 8.00 kN compression C. 14.00 kN compression D. 8.49 kN tension Answer: C. At joint E, ΣFx gives F_EF = -14.00 kN, compression. (Key point: Force in EF) 4. Why does the support at G take only a horizontal pull? A. G has no vertical support B. The only member at G, GF, is horizontal, so it can transmit only horizontal force C. G is free D. The loads are horizontal Answer: B. Joint G has only the horizontal member GF, so the reaction there must be horizontal (21.33 kN), while A takes the vertical load. 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