--- title: "Frame AFD, SFD, BMD (5 and 10 kN/m)" subject: "Strength of Materials" chapter: 1 key: "semester_3/strength_of_materials:frame-afd-sfd-bmd-5knm-2" source: https://www.bctengineer.com/q/semester_3/strength_of_materials/frame-afd-sfd-bmd-5knm-2 site: BCT Engineering --- # Frame AFD, SFD, BMD (5 and 10 kN/m) Strength of Materials · Chapter 1: Axial Forces, Shearing Forces and Bending Moments - Asked in: 2066 Bhadra (old course), 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_3/strength_of_materials/frame-afd-sfd-bmd-5knm-2 - 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 salient points with their values. [Figure: portal frame; left column hinged at the base, 4 m + 2 m high, carrying a triangular horizontal load increasing to 10 kN/m at the base over 4 m; top beam carrying 5 kN/m UDL; a 10 kN horizontal force on the right column at 4 m above the right base (2 m + 4 m heights marked on the right); right base hinged; horizontal dimensions 2 m, 4 m, 2 m.] ## Short answer Determinate by assumng a hinge at the crown. Reactions: H_A=7.78 kN ←, V_A=21.67 kN ↑, H_E=2.22 kN ←, V_E=18.33 kN ↑. Maximum bending moment is 46.67 kN·m at the top of left column (tension on left face). ## Answer **Assumptions (taken from the figure):** - A(0,0) hinged; AB = 6 m (4 m + 2 m); horizontal triangular load 10 kN/m at A to 0 at 4 m above A (→). - Beam B–C–D: 8 m with 5 kN/m downward. Right column DE = 6 m, E hinged; 10 kN horizontal (←) at 4 m above E. - With two hinged supports a hinge is assumed at the crown (mid-span of BD) to make the frame determinate. 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 | | | | | | | | | | | | | | | | | | | | | | | | A E ``` #### 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. - $H_A$ = 7.78 kN ← - $V_A$ = 21.67 kN ↑ - $H_E$ = 2.22 kN ← - $V_E$ = 18.33 kN ↑ Check: the three equilibrium equations of the complete frame (and the hinge condition, where present) are satisfied to rounding (residuals below $10^{-9}$). #### Axial force, shear force and bending moment at the key sections | Section | N (kN) | V (kN) | M (kN·m) | |---|---|---|---| | AB at A | -21.67 | 7.78 | 0.00 | | AB at B | -21.67 | -12.22 | -46.67 (left face) | | BC at B | -12.22 | 21.67 | -46.67 (top) | | BC at C | -12.22 | 1.67 | 0.00 | | CD at C | -12.22 | 1.67 | 0.00 | | CD at D | -12.22 | -18.33 | -33.33 (top) | | DE at D | -18.33 | 12.22 | -33.33 (right face) | | DE at E | -18.33 | 2.22 | 0.00 | #### Shape of the diagrams - AB: V parabolic, M cubic - BC: V linear, M parabolic - CD: V linear, M parabolic - DE: 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 = 46.67 kN·m at B (on AB), tension on the left face - V = 0 at 0.87 m from A along AB: M = 3.26 (right face) kN·m (local maximum) - Point of contraflexure (M = 0) at 1.84 m from A along AB - V = 0 at 0.33 m from C along CD: M = 0.28 (bottom) kN·m (local maximum) - Point of contraflexure (M = 0) at 0.67 m from C along CD - Bending moment is zero at the internal hinge (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. Why is a hinge assumed at the crown of this frame with two hinged supports? A. To make the beam stronger B. Four support reactions exceed the three equilibrium equations, and the hinge supplies M = 0 as a fourth equation C. To remove the axial force D. To make the right column free to rotate Answer: B. Two hinges give four reactions against three equations; the crown hinge (zero moment) provides the missing condition, making the frame determinate. (Key point: Crown hinge assumption) 2. The beam carries 5 kN/m over 8 m. What should V_A + V_E equal? A. 20 kN B. 40 kN C. 10 kN D. 60 kN Answer: B. Vertical equilibrium: total load 5 x 8 = 40 kN, matching V_A = 21.67 kN plus V_E = 18.33 kN. (Key point: Vertical reactions check) 3. Where is the maximum bending moment 46.67 kN·m, and which face is in tension? A. At E, bottom B. At the crown C, top C. At B on column AB, left face D. At the middle of CD Answer: C. Moment at the corner B from the left column: 20 kN triangular load x 4.667 m minus H_A x 6 m = 93.3 - 46.7 = 46.67 kN·m, tension on the left face. (Key point: Maximum bending moment) 4. What are the shapes of V and M along column AB, which carries the triangular horizontal load? A. V parabolic, M cubic B. V linear, M parabolic C. V constant, M linear D. V and M both linear Answer: A. Under a linearly varying load, V is parabolic and M, one degree higher, is cubic. 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