--- title: "Frame AFD, SFD, BMD (5 and 10 kN/m, hinge)" subject: "Strength of Materials" chapter: 1 key: "semester_3/strength_of_materials:frame-afd-sfd-bmd-5knm-10knm" source: https://www.bctengineer.com/q/semester_3/strength_of_materials/frame-afd-sfd-bmd-5knm-10knm site: BCT Engineering --- # Frame AFD, SFD, BMD (5 and 10 kN/m, hinge) Strength of Materials · Chapter 1: Axial Forces, Shearing Forces and Bending Moments - Asked in: 2068 Baisakh (old course), 16 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_3/strength_of_materials/frame-afd-sfd-bmd-5knm-10knm - Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-1 ## Question Draw axial force, shear force and bending moment diagrams for the frame shown, indicating the principal numerical values at salient points. [Figure: left column hinged at the base carrying a triangular horizontal load increasing to 10 kN/m at the base; top beam carrying 5 kN/m UDL with an internal hinge near the left end and a 10 kN downward point load at the right end; right column with a 10 kN horizontal force and a hinged support at its base; horizontal dimensions 3 m, 3 m, 2 m; vertical dimensions 2 m, 3 m, 1 m.] ## Short answer H_A=15.33 kN ←, V_A=9.72 kN ↑, H_F=0.33 kN →, V_F=30.28 kN ↑. Max bending moment 38.33 kN·m at D (top); local max 13.37 kN·m at 1.89 m from A. Contraflexure at 4.30 m from A, 0.89 m from B, 1.90 m from D. Zero moment at hinge C. Diagrams: AB V parabolic, M cubic; BC, CD V linear, M parabolic; DE V constant, M linear; DF V constant, M linear. ## Answer **Assumptions (taken from the figure):** - A(0,0) hinged; AB = 5 m; horizontal triangular load 10 kN/m at A to 0 at B (→). Beam B–C–D–E: BC = CD = 3 m, DE = 2 m; hinge at C. - 5 kN/m downward over B to D; 10 kN downward at E (overhang, equivalent to 10 kN plus a 20 kN·m hogging moment at D). Right column DF = 5 m, F hinged; 10 kN horizontal (←) at 2 m below D, i.e. 3 m above F (the 2 m and 3 m dimensions on the right column). Sign convention: N is positive in tension; V is positive when the forces on the part before the section (travelling A → end) point along the left-hand normal of the member axis (upward on a left-to-right beam); M is quoted as a magnitude with the tension face named. Support reactions are drawn on the free-body of the whole frame. ```text B---------------C--------------D---------E | | | | | | | | | | | | | | | | | | | | | | | | A F ``` #### Support reactions Equilibrium of the whole frame gives $\Sigma F_x = 0$, $\Sigma F_y = 0$ and $\Sigma M = 0$; the internal hinge adds the condition $M_{hinge} = 0$ (moments of all forces on one side of the hinge), which makes the reactions determinate. - $\Sigma M_A = 0$: $6V_F = 25\left(\tfrac{5}{3}\right) + 30(3) + 10(8) - 10(3) = 181.67$, so $V_F = 30.28$ kN ↑ - $\Sigma F_y = 0$: $V_A = 40 - 30.28 = 9.72$ kN ↑ - $M_C = 0$ (left part, $H_A$ taken positive to the right): $5H_A = 3V_A - 25\left(\tfrac{10}{3}\right) - 15(1.5) = -76.67$, so $H_A = -15.33$, i.e. $H_A$ = 15.33 kN ← - $\Sigma F_x = 0$: $H_F$ = 0.33 kN → #### Axial force, shear force and bending moment at the key sections | Section | N (kN) | V (kN) | M (kN·m) | |---|---|---|---| | AB at A | -9.72 | 15.33 | 0.00 | | AB at B | -9.72 | -9.67 | -6.67 (left face) | | BC at B | -9.67 | 9.72 | -6.67 (top) | | BC at C | -9.67 | -5.28 | 0.00 | | CD at C | -9.67 | -5.28 | 0.00 | | CD at D | -9.67 | -20.28 | -38.33 (top) | | DE at D | 0.00 | 10.00 | -20.00 (top) | | DE at E | 0.00 | 10.00 | 0.00 | | DF at D | -30.28 | 9.67 | -18.33 (right face) | | DF at F | -30.28 | -0.33 | 0.00 | Joint D check: the beam moment 38.33 kN·m on CD is balanced by 20.00 kN·m in the overhang DE plus 18.33 kN·m in the column DF. #### Shape of the diagrams - AB: V parabolic, M cubic - BC: V linear, M parabolic - CD: V linear, M parabolic - DE: V constant (10 kN), M linear from 20 kN·m (top) at D to 0 at E - DF: V constant (steps at point loads), M linear - AFD: constant between loads on a member, changing only where a load component acts along the member. - BMD is drawn on the tension side of each member; it is continuous at rigid joints (moment is transferred) and zero at hinges and at free ends. #### Salient features - Maximum bending moment = 38.33 kN·m at D (on CD), tension on the top - V = 0 at 1.89 m from A along AB: M = 13.37 kN·m (right face), a local maximum - Point of contraflexure (M = 0) at 4.30 m from A along AB - V = 0 at 1.94 m from B along BC: M = 2.79 kN·m (bottom), a local maximum - Point of contraflexure (M = 0) at 0.89 m from B along BC - Point of contraflexure (M = 0) at 1.90 m from D along DF - Bending moment is zero at the internal hinge C (value read from both sides: 0.00). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What extra condition does the internal hinge in the beam supply, making the reactions of the frame determinate? A. Shear force is zero at the hinge B. Bending moment is zero at the hinge C. Axial force is zero at the hinge D. Deflection is zero at the hinge Answer: B. A hinge cannot transmit moment, so taking moments of all forces on one side of it about the hinge gives zero, a fourth equation for the four reactions. (Key point: Internal hinge condition) 2. The 10 kN downward load at the end E, 2 m beyond D, is moved to D for analysis. What must be added at D? A. A 20 kN·m hogging couple B. A 5 kN·m sagging couple C. An extra 10 kN upward force D. Nothing, a load can be moved without a couple Answer: A. Shifting a force parallel to itself needs a couple equal to force x distance = 10 x 2 = 20 kN·m, hogging (tension on top) at D. (Key point: Overhang load at E) 3. For column AB carrying the triangular horizontal load, what are the shapes of the shear force and bending moment diagrams? A. V linear, M parabolic B. V constant, M linear C. V parabolic, M cubic D. V cubic, M parabolic Answer: C. A triangular (linearly varying) load makes V parabolic, and M is one degree higher, i.e. cubic. (Key point: Shape of diagrams under triangular load) 4. Where is the maximum bending moment of 38.33 kN·m found and which face is in tension? A. At A, on the left face B. At B, on the left face C. At D on member CD, tension on the top D. At the hinge C, tension at the bottom Answer: C. The overhang couple and the beam load give the largest moment at the joint D, hogging with tension on top of CD. (Key point: Maximum bending moment) ## Close variants set in other exams - [Frame AFD, SFD, BMD (5 kN/m, 10 kN, hinge)](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aframe-afd-sfd-bmd-5knm-10kn) — Strength of Materials, chapter 1, asked 2× ## Related in Strength of Materials - [Frame AFD, SFD, BMD (100 kN, 25 kN/m, hinge)](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aframe-afd-sfd-bmd-100kn-25knm) — Strength of Materials, chapter 1 - [Frame AFD, SFD, BMD (10 kN/m, 60 kN, hinge)](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aframe-afd-sfd-bmd-10knm-60kn) — Strength of Materials, chapter 1 - [Frame AFD, SFD, BMD (30 kN/m, 50 kNm, hinge)](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aframe-afd-sfd-bmd-30knm-50knm) — Strength of Materials, chapter 1 - [Frame AFD, SFD, BMD (5 and 10 kN/m)](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aframe-afd-sfd-bmd-5knm-2) — Strength of Materials, chapter 1 - [Frame AFD, SFD, BMD (12 kN, 25 kNm, hinge)](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aframe-afd-sfd-bmd-12kn-25knm) — Strength of Materials, chapter 1 - [Frame AFD, SFD, BMD (80 kN, 30 kNm, hinge)](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aframe-afd-sfd-bmd-80kn-30knm) — Strength of Materials, chapter 1 ## Related in other subjects - [AFD, SFD, BMD of frame with hinge at D](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Aframe-afd-sfd-bmd-8knm-hinge-50kn) — Applied Mechanics, chapter 6 - [AFD, SFD, BMD of frame with hinge](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Aframe-afd-sfd-bmd-hinge-4knm-20knm) — Applied Mechanics, chapter 6 - [AFD, SFD, BMD of frame with hinge at C](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Aframe-afd-sfd-bmd-hinge-5knm-10knm-triangular) — Applied Mechanics, chapter 6, asked 2× - [AFD, SFD, BMD for frame with internal hinge](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Aframe-afd-sfd-internal-hinge-5kn-per-m) — Applied Mechanics, chapter 6