--- title: "Beam ABC with wire support: BMD and SFD" subject: "Strength of Materials" chapter: 1 key: "semester_3/strength_of_materials:beam-wire-support-sfd-bmd" source: https://www.bctengineer.com/q/semester_3/strength_of_materials/beam-wire-support-sfd-bmd site: BCT Engineering --- # Beam ABC with wire support: BMD and SFD Strength of Materials · Chapter 1: Axial Forces, Shearing Forces and Bending Moments - Asked in: 2068 Chaitra, 10 marks - Video: [Shear Force and Bending Moment Made EASY!](https://www.youtube.com/watch?v=dnIhP4ndtoE) — Math Physics Engage - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/strength_of_materials/beam-wire-support-sfd-bmd - Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-1 ## Question Draw bending moment and shear force diagrams for the beam ABC, which has hinged support at 'A' and other support at B, supported by wire (Tension member) as shown in figure. [Figure: beam A-B-C hinged at A on a vertical wall 6 m high; a wire from the top of the wall (6 m above A) to B; 2 kN/m UDL over A to B (4 m); 4 kN/m triangular load over B to C (2 m + 6 m shown as 4 m, 2 m, 6 m); C is the free end.] ## Short answer Reactions: H_A=5.78 kN →, V_A=3.33 kN ↑, wire tension T=10.42 kN. Maximum sagging moment 2.78 kN·m at 1.67 m from A, maximum hogging moment -2.67 kN·m at B. Contraflexure at 3.33 m from A. Axial force: 5.78 kN compression from A to B, zero beyond. SFD: linear under UDL, parabolic under triangular load; BMD: parabolic and cubic. ## Answer **Assumptions (taken from the figure):** - A(0,0) hinged on the wall; B at x = 4 m, C at x = 6 m (BC = 2 m). - The wire runs from the top of the wall (0, 6) to B, so its direction from B is $(-4, 6)/\sqrt{52}$ (tension only). - 2 kN/m downward on AB; triangular load on BC from 4 kN/m at B to 0 at C (total 4 kN, acting 2/3 m from B). 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 Moments about A (the horizontal wire component passes through the beam axis): $T_y\times4 = 8\times2 + 4\times4.667 = 34.667$, so $T_y = 8.667$ kN and $T = T_y\sqrt{52}/6$. - $H_A$ = 5.78 kN → - $V_A$ = 3.33 kN ↑ - $T$ = 10.42 kN (tension in the wire; components 5.78 kN ← and 8.67 kN ↑ on the beam) Check: $\Sigma F_y$: $3.33 + 8.67 = 8 + 4$ ✓; $\Sigma F_x$: $5.78 - 5.78 = 0$ ✓ #### Axial force, shear force and bending moment at the key sections | Section | N (kN) | V (kN) | M (kN·m) | |---|---|---|---| | AB at A | -5.78 | 3.33 | 0.00 | | AB at B | -5.78 | -4.67 | -2.67 (top) | | BC at B | 0.00 | 4.00 | -2.67 (top) | | BC at C | 0.00 | 0.00 | 0.00 | #### Shape of the diagrams - Between loads V is constant; under a UDL it varies linearly and under a triangular load parabolically. M is one degree higher than V (linear, parabolic, cubic respectively). V jumps at point loads and reactions; M jumps at couples. - AFD: a constant compression of 5.78 kN from A to B and zero from B to C. - The BMD is zero at the hinge A and at the free end C. #### Salient features - Maximum positive (sagging) M = 2.78 kN·m at x = 1.67 m - Maximum negative (hogging) M = -2.67 kN·m at x = 4.00 m (at B, where the wire is attached) - V = 0 at 1.67 m from A: M = 2.78 kN·m (bottom), the local maximum - Point of contraflexure (M = 0) at 3.33 m from A #### Sketches (sagging +ve above the line) SFD: ```text ** ** ***** | *** ***** | ***** ------------*****----------------------------******* ***** | ***** | ***** | ***| ``` BMD: ```text ********** **** *** *** *** ** ** *---------------------------**-------------********* * **** ** ** * ** * ``` ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. When taking moments about A to find the wire force, why does the horizontal component of the wire force not appear in the equation? A. Its line of action passes through the beam axis, so its moment arm about A is zero B. Horizontal forces never produce moments C. It is cancelled by the vertical component D. It acts at the hinge A itself Answer: A. The horizontal wire component acts along the beam axis through A, so its moment about A is zero and only the vertical component T_y enters. (Key point: Moment equation about A) 2. Taking moments about A, T_y x 4 = 8 x 2 + 4 x 4.667. What does this give for the vertical component of the wire pull? A. T_y = 3.33 kN B. T_y = 5.78 kN C. T_y = 8.67 kN D. T_y = 10.42 kN Answer: C. 34.667 / 4 = 8.667 kN. The 5.78 kN is the horizontal component and 10.42 kN is the full wire tension. (Key point: Wire vertical component) 3. In span AB the maximum sagging bending moment occurs at 1.67 m from A. Why there? A. Because the bending moment is zero there B. Because the shear force is zero there, so the moment is at a local maximum C. Because that is the midpoint of AB D. Because the axial force changes sign there Answer: B. dM/dx = V, so M is maximum where V = 0. With V_A = 3.33 kN and the 2 kN/m UDL, V = 0 at x = 3.33/2 = 1.67 m. (Key point: Maximum sagging moment) 4. How is the hogging moment of 2.67 kN·m at B obtained from the overhang BC? A. 2 kN/m x 2 m x 1 m B. The 4 kN triangular load acting at 2/3 m from B (4 x 0.667) C. The 4 kN triangular load acting at 1 m from B D. The wire force times 6 m Answer: B. The triangular load (4 kN/m to 0 over 2 m) totals 4 kN and acts at 2/3 m from the heavy end B, giving 4 x 0.667 = 2.67 kN·m with tension on top. 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