--- title: "AFD, SFD, BMD of frame with 40 kN/m UDL" subject: "Applied Mechanics" chapter: 6 key: "semester_1/applied_mechanics:frame-afd-sfd-bmd-40kn-frame" source: https://www.bctengineer.com/q/semester_1/applied_mechanics/frame-afd-sfd-bmd-40kn-frame site: BCT Engineering --- # AFD, SFD, BMD of frame with 40 kN/m UDL Applied Mechanics · Chapter 6: Analysis of Beams and Frames - Asked in: 2074 Asoj, 12 marks - Video: [Frame Analysis || Shear Force & Bending Moment Diagram](https://www.youtube.com/watch?v=0pG1to_L44M) — Civil Engineering - Page: https://www.bctengineer.com/q/semester_1/applied_mechanics/frame-afd-sfd-bmd-40kn-frame - Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-6 ## Question Calculate and draw the axial force, shear force and bending moment diagram, with its salient features, for the given frame. [Figure: frame with hinge/pin support at A and roller support at E; members A-B inclined, B-C horizontal carrying 40 kN/m UDL, D-E vertical; 10 kN inclined load at B, 60 kN horizontal load at D; dimensions 2.5 m, 1.5 m, 3 m, 4 m, 2 m as printed.] ## Short answer Reactions: A_x=52 kN, A_y=66.18 kN, R_E=99.82 kN. All axial forces compressive (AB 84.14, BC 60, CE 99.82 kN). Max hogging moment 120 kN·m at C; BC has a small sagging peak 4.55 kN·m at 1.5 m from B where shear zero. ## Answer **Reading of the figure:** A is a hinge at (0, 0), B is at (3, 4) (the member AB is 5 m long, inclined; vertical rise 2.5 + 1.5 = 4 m), BC is horizontal from x = 3 to x = 7 m (4 m) and carries 40 kN/m. C–D–E is the vertical column at x = 7 m, with E a roller at the base (0 m) and D 2 m above E. Loads: 10 kN acting perpendicular to AB at 1.875 m from A (the point at height 1.5 m), and 60 kN horizontal, towards the left, at D. The 10 kN has components 10(0.8, −0.6) = (8 kN →, 6 kN ↓), since AB makes 53.13° with the horizontal. #### Support reactions ```text ΣM_A = 0: R_E × 7 + 60 × 2 − 160 × 5 − 6 × 1.125 − 8 × 1.5 = 0 (160 = 40 × 4 acts at x = 5) 7 R_E = 800 + 6.75 + 12 − 120 = 698.75 -> R_E = 99.82 kN ↑ ΣFx = 0: A_x + 8 − 60 = 0 -> A_x = 52.00 kN → ΣFy = 0: A_y + 99.82 − 6 − 160 = 0 -> A_y = 66.18 kN ↑ ``` #### Internal forces (tension positive for N; sagging positive for M) **Member AB** (inclined, direction cosines 0.6, 0.8) - Axial: N = −(52 × 0.6 + 66.18 × 0.8) = **−84.14 kN** (compression), constant, since the 10 kN is perpendicular to AB. - Shear (perpendicular to AB): V = 1.89 kN from A to the 10 kN load, then 11.89 kN beyond it (both in the same sense). - Moment: M_A = 0; at the load point (1.875 m from A): 1.89 × 1.875 = **3.55 kN·m**; at B: 3.55 + 11.89 × 3.125 = **40.71 kN·m** (tension on the outer, upper-left face). **Member BC** (horizontal, 4 m) - Axial: N = **−60.00 kN** (compression), constant. - Shear: V_B = +60.18 kN, V(x) = 60.18 − 40x, V_C = **−99.82 kN**. V = 0 at x = 1.50 m from B. - Moment: M(x) = −40.71 + 60.18x − 20x²: M_B = **−40.71**, zero at x = 1.03 m and 1.98 m, maximum sagging **+4.55 kN·m** at x = 1.50 m, M_C = **−120.00 kN·m** (hogging, tension on top). **Member CE** (vertical) - Axial: N = **−99.82 kN** (compression) over the whole length. - Shear: V = 0 from E to D; V = **60 kN** from D to C. - Moment: M = 0 from E to D; then increases linearly to **120 kN·m at C** (tension on the right, outer face). #### Diagram values | Point | N (kN) | V (kN) | M (kN·m) | |---|---|---|---| | A | −84.14 | 1.89 | 0 | | Load on AB | −84.14 | 1.89 → 11.89 | 3.55 | | B (end of AB) | −84.14 | 11.89 | 40.71 | | B (start of BC) | −60.00 | 60.18 | −40.71 | | 1.5 m from B | −60.00 | 0 | +4.55 | | C (end of BC) | −60.00 | −99.82 | −120.00 | | C (top of CE) | −99.82 | 60 | 120 | | D | −99.82 | 0 / 60 | 0 | | E | −99.82 | 0 | 0 | ```text BMD (tension side) AFD SFD of BC B 40.71 -- 120 C all compressive +60.18 \ (+4.55) | -84.14 AB \ 3.55 | -60.00 BC \_ 0 at 1.5 m A D (0) -99.82 CE \ -99.82 ``` **Salient features:** M is maximum at C (120 kN·m, hogging); M is zero at A, at E, and along ED; the member EC carries no shear below the 60 kN load; the BMD is parabolic under the UDL, with a small sagging peak of 4.55 kN·m; the shear passes through zero where the sagging moment peaks; the joint moments are in equilibrium (at C, 120 kN·m in the column = 120 kN·m in the beam, at B 40.71 kN·m in both members). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is R_E obtained? A. From sum of vertical forces only B. From sum of horizontal forces only C. Taking moments about A, since A_x and A_y pass through A D. By assuming R_E equals the UDL Answer: C. Moments about the hinge A eliminate A_x and A_y, leaving R_E as the only unknown. (Key point: Moment equation about A) 2. Why is the axial force in AB constant at -84.14 kN? A. The 10 kN load is perpendicular to AB, so it adds no axial component B. There is no load on AB C. The member is horizontal D. The shear is zero Answer: A. A perpendicular load changes shear and moment but not axial force. (Key point: Axial force in AB) 3. Where is the sagging moment in BC maximum? A. At B B. Where shear is zero, 1.5 m from B C. At C D. At mid-span regardless of shear Answer: B. Moment is maximum where V = 0; V = 60.18 - 40x = 0 gives x = 1.5 m. (Key point: Location of maximum sagging moment in BC) 4. Where is the overall maximum moment? A. At B, 40.71 kN·m B. At 1.5 m from B, 4.55 kN·m C. At D D. At C, 120 kN·m (hogging) Answer: D. M_C = -120 kN·m, the largest magnitude, with tension on top. 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