--- title: "AFD, SFD, BMD of frame with slanted column" subject: "Applied Mechanics" chapter: 6 key: "semester_1/applied_mechanics:frame-afd-sfd-bmd-30knm-50kn-bc-slanted" source: https://www.bctengineer.com/q/semester_1/applied_mechanics/frame-afd-sfd-bmd-30knm-50kn-bc-slanted site: BCT Engineering --- # AFD, SFD, BMD of frame with slanted column Applied Mechanics · Chapter 6: Analysis of Beams and Frames - Asked in: 2067 Asar, 12 marks - Video: [ANALYSIS OF FRAME SHEAR BENDING & AXIAL THRUST DIAGRAM | EX 03](https://www.youtube.com/watch?v=VhZ1_8AFJrI) — Structural Funda by:- KheNiSiR - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_1/applied_mechanics/frame-afd-sfd-bmd-30knm-50kn-bc-slanted - Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-6 ## Question Draw bending moment, shear force and axial force diagrams for the given figure. And also give ordinates of the salient points, if any. [Figure: frame with pin support at A and roller at E; slanted member A-B (3 m horizontal run, 4 m rise); top member B-C of 4 m with 30 kN/m UDL; column C-D-E with D 4 m below C and E 2 m below D; 50 kN horizontal load at D.] ## Short answer Reactions: Ax=50 kN right, Ay=34.29 kN up, Ey=85.71 kN up. Maximum bending moment 200.00 kN·m at corner C. Beam BC: 97.14 kN·m at B, 200 kN·m at C, shear zero at 1.14 m from B with 77.55 kN·m. Slanted AB: axial 57.43 kN compression, shear 19.43 kN, moment 0 to 97.14 kN·m. CD: shear 50 kN, moment 200 to 0. DE: no bending, axial 85.71 kN compression. ## Answer #### Reading of the figure - Origin at A: A(0,0) pin; B(3,4), so AB is slanted (3 m run, 4 m rise, length 5 m); beam BC is 4 m long at y = 4, ending at C(7,4); the column runs down from C with D(7,0) 4 m below C and the roller E(7,−2) 2 m below D. - Loads: UDL 30 kN/m on BC (120 kN, 2 m from B); 50 kN ← at D. - r = 3, determinate. #### Reactions ```text 30 kN/m B=========C / | / | / | A (pin) D←50kN | E (roller) ``` - ΣFx: Aₓ − 50 = 0, so **Aₓ = 50.00 kN (→)**. - ΣM_A: 7E_y − 120(5) + 50(0) = 0 (the 50 kN load acts along the level of A), so **E_y = 85.71 kN (↑)**. - ΣFy: A_y + E_y − 120 = 0, so **A_y = 34.29 kN (↑)**. - For the slanted member AB (cos θ = 0.6, sin θ = 0.8), resolve the reaction at A: axial force = Aₓcos θ + A_y sin θ = 50(0.6) + 34.29(0.8) = 57.43 kN (compression); shear V = Aₓ sin θ − A_y cos θ = 50(0.8) − 34.29(0.6) = 19.43 kN; moment at B = 19.43 × 5 = 97.14 kN·m. #### Member forces and diagrams Sign notes: N is + for tension and − for compression; V follows the member axis in the order of the letters (e.g. A→B); for M the side in brackets is the tension side, on which the BMD is drawn. - **AB** (A to B, 5.00 m): N = 57.43 kN (compression); V = −19.43 to −19.43 kN; M = 0.00 (–) to 97.14 (top-left) kN·m, linear. - **BC** (B to C, 4.00 m): N = 50.00 kN (compression); V = 34.29 to −85.71 kN; M = 97.14 (top) to 200.00 (top) kN·m, parabolic. V = 0 at 1.14 m from B, where M = −77.55 kN·m (hogging). - **CD** (C to D, 4.00 m): N = 85.71 kN (compression); V = 50.00 to 50.00 kN; M = 200.00 (right) to 0.00 (–) kN·m, linear. - **DE** (D to E, 2.00 m): N = 85.71 kN (compression); V = 0.00 to 0.00 kN; M = 0.00 (–) to 0.00 (–) kN·m, constant. #### Salient features - Maximum bending moment = **200.00 kN·m** at the corner C (tension on the outside, i.e. the top of the beam and the right face of the column). The moment at B is 97.14 kN·m. - In BC the shear is zero at 1.14 m from B, where the hogging moment has its smallest value, 77.55 kN·m. The beam is in hogging throughout (97.14 kN·m at B, 200.00 kN·m at C), so there is no sagging and no contraflexure. - CD carries a constant shear of 50 kN and DE carries none; DE has no bending, only 85.71 kN compression (= E_y). - AB is an inclined member with axial compression 57.43 kN and constant shear 19.43 kN; its bending moment is linear from 0 at A to 97.14 kN·m at B. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the axial force in the slanted member AB found? A. It equals the vertical reaction Ay B. Resolve Ax and Ay along the member: Ax cosθ + Ay sinθ C. It equals Ax only D. It is zero because the member is inclined Answer: B. For an inclined member the reactions are resolved along and normal to it: the axial force is Ax cosθ + Ay sinθ = 68.86 kN (compression). (Key point: Slanted member resolution) 2. Where does the maximum bending moment of 100 kN·m occur? A. At the corner C B. At B C. At D D. At A Answer: A. The moment at C is 100 kN·m (tension outside), larger than 54.29 kN·m at B. (Key point: Maximum bending moment) 3. Why is there no contraflexure in beam BC? A. The shear is never zero B. The beam is in hogging throughout, from 54.29 to 100 kN·m C. The load is zero D. The beam is a roller Answer: B. Even at the zero-shear point the moment is still hogging (14.97 kN·m), so the moment never changes sign. (Key point: Beam BC moment behaviour) 4. What does DE carry? A. Constant shear of 50 kN B. A bending moment of 100 kN·m C. No bending, only 71.43 kN compression (= Ey) D. Tension from the roller Answer: C. The horizontal load acts at D, so below D there is no shear or moment. DE only carries the roller's vertical reaction as compression. 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