--- title: "AFD, SFD, BMD of frame with 60 kN/m UDL" subject: "Applied Mechanics" chapter: 6 key: "semester_1/applied_mechanics:frame-afd-sfd-bmd-60knm-20knm-12kn" source: https://www.bctengineer.com/q/semester_1/applied_mechanics/frame-afd-sfd-bmd-60knm-20knm-12kn site: BCT Engineering --- # AFD, SFD, BMD of frame with 60 kN/m UDL Applied Mechanics · Chapter 6: Analysis of Beams and Frames - Asked in: 2069 Chaitra, 14 marks - Video: [SFD and BMD for frame](https://www.youtube.com/watch?v=S9W7_Nv2eVg) — Vise Vision - 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-60knm-20knm-12kn - Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-6 ## Question Draw AFD, SFD and BMD of the given frame loaded as shown in figure below. Indicate the salient feature if any. [Figure: frame with pin support at A and at H; left column A-B-C 3 m + 3 m with 15 kN horizontal at B; top beam C-D-E with 60 kN/m UDL; 20 kN·m moment at E, 10 kN downward load at F; right part G with 12 kN horizontal, heights 2 m + 2 m; horizontal dimensions 4 m, 1 m, 2 m.] ## Short answer Reactions: Ax=3 kN left, Ay=136.6 kN up, Hy=113.4 kN up. Maximum bending moment 128.50 kN·m (tension bottom) at 2.28 m from C in CD. Contraflexure points: 0.75 m above B in BC, 0.21 m from C in CD, 0.38 m from D in DE. Axial forces: AB-BC 136.6 kN compression, EG-GH 113.4 kN compression, CD-DE 12 kN compression. ## Answer #### Reading of the figure - Origin at A: A(0,0) pin, B(0,3), C(0,6); beam C–D(4,6)–E(5,6)–F(7,6); right column from E down: G(5,4), H(5,2). - Loads: 15 kN → at B; UDL 60 kN/m on CD (240 kN, 2 m from C); couple 20 kN·m at E (taken clockwise); 10 kN ↓ at F; 12 kN ← at G. - With pins at both A and H and no internal hinge the frame would have r = 4 and be indeterminate, so H is taken as a roller (vertical reaction only). Then r = 3 with no hinge, and the frame is determinate. #### Reactions ```text 60 kN/m 20kNm 10kN C========D=======E=======F ↓ | | B→15kN G←12kN | | A (pin) H (roller) ``` - ΣFx: Aₓ + 15 − 12 = 0, so **Aₓ = 3.00 kN (←)**. - ΣM_A: 5H_y − 15(3) − 240(2) − 20 − 10(7) + 12(4) = 0, so **H_y = 113.40 kN (↑)**. - ΣFy: A_y + H_y − 240 − 10 = 0, so **A_y = 136.60 kN (↑)**. #### 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, 3.00 m): N = 136.60 kN (compression); V = 3.00 to 3.00 kN; M = 0.00 (–) to 9.00 (right) kN·m, linear. - **BC** (B to C, 3.00 m): N = 136.60 kN (compression); V = −12.00 to −12.00 kN; M = 9.00 (right) to 27.00 (left) kN·m, linear. M changes sign (contraflexure) at 0.75 m above B, from M = 3y − 15(y − 3) = 0 at y = 3.75 m. - **CD** (C to D, 4.00 m): N = 12.00 kN (compression); V = 136.60 to −103.40 kN; M = 27.00 (top) to 39.40 (bottom) kN·m, parabolic. V = 0 at 2.28 m from C, where M = 128.50 kN·m. M changes sign (contraflexure) at 0.21 m from C. - **DE** (D to E, 1.00 m): N = 12.00 kN (compression); V = −103.40 to −103.40 kN; M = 39.40 (bottom) to 64.00 (top) kN·m, linear. M changes sign (contraflexure) at 0.38 m from D. - **EF** (E to F, 2.00 m): N = 0.00 kN (zero); V = 10.00 to 10.00 kN; M = 20.00 (top) to 0.00 (–) kN·m, linear. - **EG** (E to G, 2.00 m): N = 113.40 kN (compression); V = 12.00 to 12.00 kN; M = 24.00 (right) to 0.00 (–) kN·m, linear. - **GH** (G to H, 2.00 m): N = 113.40 kN (compression); V = 0.00 to 0.00 kN; M = 0.00 (–) to 0.00 (–) kN·m, constant. #### Salient features - Maximum bending moment = **128.50 kN·m** (tension at the bottom) at 2.28 m from C in the beam, where the shear is zero. - Hogging moments occur at C (27.00 kN·m, tension on the top/outside) and at E (64.00 kN·m, tension on top). - Points of contraflexure: 0.75 m above B in the column BC, 0.21 m from C in CD and 0.38 m from D in DE. - Joint E check: 64.00 (DE) = 20.00 (EF) + 24.00 (EG) + 20.00 (clockwise couple). - Axial force: AB–BC 136.60 kN compression; EG–GH 113.40 kN compression (= H_y); CD–DE 12.00 kN compression (the net horizontal force of 12 kN passed to the right column). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is this frame determinate? A. It has a hinge and four reactions B. It has pin and roller supports giving r = 3 and no hinge C. It has a fixed support and a roller D. It has no loads on one column Answer: B. A pin (2) and a roller (1) give r = 3, equal to the three equilibrium equations, with no internal hinge. (Key point: Determinacy) 2. How is Ax = 3 kN found? A. Sum of vertical forces B. Moments about H C. ΣFx = 0: Ax + 15 - 12 = 0, so Ax acts to the left D. Moments about A Answer: C. Horizontal forces are 15 kN to the right and 12 kN to the left, so Ax must make up the net 3 kN, acting leftward. (Key point: Horizontal reaction at A) 3. Where does the maximum moment of 128.50 kN·m occur? A. In beam CD at 2.28 m from C, where shear is zero B. At E C. At C D. At the roller H Answer: A. In CD the shear goes from 136.60 to -103.40 kN. It is zero at 2.28 m from C, giving the maximum sagging moment (tension at bottom). (Key point: Maximum bending moment) 4. What does the joint E check 64.00 = 20.00 + 24.00 + 20.00 show? A. Moments balance at the joint, including the applied 20 kN·m couple B. Shear forces balance at E C. The axial force at E is 64 kN D. The couple is ignored at joints Answer: A. The member-end moment in DE equals the sum of the moments in EF and EG plus the applied couple, confirming joint equilibrium. 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