--- title: "Member forces in truss with 20, 10, 12, 50 kN" subject: "Applied Mechanics" chapter: 7 key: "semester_1/applied_mechanics:truss-forces-20kn-10kn-12kn-50kn" source: https://www.bctengineer.com/q/semester_1/applied_mechanics/truss-forces-20kn-10kn-12kn-50kn site: BCT Engineering --- # Member forces in truss with 20, 10, 12, 50 kN Applied Mechanics · Chapter 7: Analysis of Plane Trusses - Asked in: 2069 Asar, 8 marks - Video: [Trusses Method of Joints | Mechanics Statics | Learn to Solve Questions](https://www.youtube.com/watch?v=_rK02neOF18) — Question Solutions - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_1/applied_mechanics/truss-forces-20kn-10kn-12kn-50kn - Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-7 ## Question Determine the forces in all members of the truss shown in figure below. [Figure: truss with pin support at A and roller at D; joints A, B, C, D, E; height 4 m; spans 5 m and 5 m; loads 20 kN downward at B, 10 kN downward at C, 12 kN horizontal at C, 50 kN downward at E.] ## Short answer Using method of joints, all member forces are: AB 44.46 kN (C), BC 34.62 kN (C), CD 49.88 kN (C), AE 35.56 kN (T), ED 26.44 kN (T), BE 20.87 kN (T), CE 38.09 kN (T). ## Answer #### Data and assumptions A(0,0) is a pin, D(10,0) a roller, E(5,0) lies on the bottom chord; the top joints are B(2.5,4) and C(7.5,4) (taken at the middle of AE and ED, so BC = 5 m). Members: AB, BC, CD, AE, ED, BE, CE (m = 7, j = 5, r = 3, so m + r = 2j: determinate and stable). Loads: 20 kN ↓ at B; 10 kN ↓ and 12 kN → at C; 50 kN ↓ at E. Length AB = BE = CE = CD = √(2.5² + 4²) = 4.717 m, so sin = 4/4.717 = 0.848 and cos = 2.5/4.717 = 0.530. #### Reactions - ΣM_A = 0: 10D_y − 20(2.5) − 10(7.5) − 50(5) − 12(4) = 0, so **D_y = 42.3 kN ↑**. - ΣFy: A_y = 80 − 42.3 = **37.7 kN ↑**; ΣFx: **A_x = 12 kN ←**. #### Method of joints (tension +, compression −) **Joint A**: ΣFy: 37.7 + 0.848F_AB = 0, so **F_AB = −44.46 kN (C)**. ΣFx: −12 + 0.530(−44.46) + F_AE = 0, so **F_AE = +35.56 kN (T)**. **Joint D**: ΣFy: 42.3 + 0.848F_CD = 0, so **F_CD = −49.88 kN (C)**. ΣFx: −0.530(−49.88) − F_ED = 0, so **F_ED = +26.44 kN (T)**. **Joint B** (AB pushes B with 23.56 kN → and 37.70 kN ↑): - ΣFy: 37.70 − 20 − 0.848F_BE = 0, so **F_BE = +20.87 kN (T)**. - ΣFx: 23.56 + F_BC + 0.530(20.87) = 0, so **F_BC = −34.62 kN (C)**. **Joint C**: BC is in compression and pushes C with 34.62 kN to the right; CD is in compression and pushes C with 26.44 kN to the left and 42.30 kN upward; the loads are 12 kN → and 10 kN ↓. CE goes down-left from C. - ΣFx: 34.62 − 26.44 + 12 − 0.530F_CE = 0 (tension in CE pulls C towards E, i.e. down-left), so **F_CE = +38.09 kN (T)**. - ΣFy (check): 42.30 − 10 − 0.848(38.09) = 0 ✓. #### Result | Member | Force (kN) | Nature | |---|---|---| | AB | 44.46 | compression | | BC | 34.62 | compression | | CD | 49.88 | compression | | AE | 35.56 | tension | | ED | 26.44 | tension | | BE | 20.87 | tension | | CE | 38.09 | tension | ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With m = 7, j = 5 and r = 3, what does m + r = 2j tell us? A. The truss is a mechanism B. The truss is statically indeterminate C. The truss is statically determinate and stable, so joint equations can solve all forces D. Reactions are not needed Answer: C. m + r = 2j gives exactly as many equations (2j) as unknowns, so the truss is determinate. (Key point: Determinacy check) 2. Which moment equation about A gives D_y? A. 10D_y - 20(2.5) - 10(7.5) - 50(5) - 12(4) = 0, so D_y = 42.3 kN B. 10D_y = 80 x 5, so D_y = 40 kN C. 10D_y - 20(2.5) - 10(7.5) - 50(5) = 0, so D_y = 37.5 kN D. D_y = 80 - 12 = 68 kN Answer: A. The 12 kN horizontal load at height 4 m contributes a moment 12 x 4 about A and must be included. (Key point: Roller reaction D_y) 3. At A, A_y = 37.7 kN up and AB has vertical component 0.848F_AB. What is F_AB? A. 44.46 kN tension B. 44.46 kN compression C. 37.7 kN compression D. 32.0 kN tension Answer: B. AB must push down on A to balance A_y: 37.7 + 0.848F_AB = 0 gives F_AB = -44.46 kN (compression). (Key point: Joint A force in AB) 4. After finding BC and CD, how is F_CE found at joint C? A. From ΣFx at C using BC, CD, the 12 kN load and the horizontal part of CE, giving 38.09 kN tension B. By assuming CE equals BE C. By taking CE as a zero-force member D. From ΣFy alone, ignoring the 12 kN load Answer: A. All other forces at C are known, so ΣFx gives F_CE = 38.09 kN (T); the ΣFy check then closes the joint. 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