--- title: "BMD of a simple beam and a frame" subject: "Strength of Materials" chapter: 1 key: "semester_3/strength_of_materials:bmd-simple-beam-and-frame-10knm" source: https://www.bctengineer.com/q/semester_3/strength_of_materials/bmd-simple-beam-and-frame-10knm site: BCT Engineering --- # BMD of a simple beam and a frame Strength of Materials · Chapter 1: Axial Forces, Shearing Forces and Bending Moments - Asked in: 2070 Chaitra, 4 marks - Video: [Strength of Materials~Lec 13~U1~SFD & BMD(Important Question on Frames) by Bharat Kumar Mahawar](https://www.youtube.com/watch?v=KC8ZKLEAeaU) — BHARAT KM Online पाठशाला - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/strength_of_materials/bmd-simple-beam-and-frame-10knm - Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-1 ## Question Draw bending moment diagram in the simple beam and frame shown in figure below: [Figure: (i) simply supported beam of span 4 m + 6 m with a 10 kNm couple applied at the point 4 m from the left support; (ii) frame: a vertical column of height 5 m fixed/hinged at the base with a horizontal member of 4 m at the top carrying a 20 kN downward load at its free end.] ## Short answer (i) Simple beam: reactions 1 kN↓ at left, 1 kN↑ at right. BMD: -4 kN·m left of couple, +6 kN·m right of couple, linear segments. (ii) Frame: horizontal member moment 0 to 80 kN·m hogging; column constant 80 kN·m, axial compression 20 kN. ## Answer #### (i) Simply supported beam (span 4 m + 6 m = 10 m, 10 kN·m couple at 4 m from the left support) Assume the couple is clockwise (if it is anticlockwise, reverse all signs). **Reactions:** $\Sigma F_y = 0 \Rightarrow R_A + R_B = 0$. Moments about A: $R_B(10) - 10 = 0$ (the clockwise couple is balanced by the anticlockwise moment of $R_B$), so $$ R_B = +1\ \text{kN}\ (\uparrow),\qquad R_A = -1\ \text{kN}\ (\text{i.e. }1\ \text{kN}\ \downarrow) $$ **Bending moment** (sagging +ve): - Left of the couple ($0 \le x < 4$): $M = R_Ax = -x$, so $M_A = 0$ and $M = -4$ kN·m just left of the couple (hogging). - The clockwise couple causes a sudden rise of 10 kN·m: just right of the couple $M = -4 + 10 = +6$ kN·m. - Right of the couple: $M = R_B(10 - x)$, falling linearly to 0 at B ($M = 1\times6 = 6$ at $x = 4$ ✓). ```text BMD (kN·m, sagging +ve above the line) +6 |`-._ | `-._ | `-._ A o--------------+------------`-o B `-._ | `-._ | `-._ | -4 | x = 4 m (couple) ``` The shear force is constant, $V = -1$ kN throughout (the couple does not affect it). The BMD is two straight lines with a jump of 10 kN·m at the couple. #### (ii) Frame (column 5 m, horizontal member 4 m, 20 kN at the free end) Assume the column is fixed at the base (a hinged base would make this frame unstable) and the 20 kN load acts downward at the free end of the horizontal member. - **Horizontal member:** $M = -20x$ (hogging, tension on top), from 0 at the free end to $-20\times4 = -80$ kN·m at the corner. The shear force is constant, $V = 20$ kN, and $N = 0$. - **Column:** the load acts at a horizontal distance of 4 m from the column axis, so it causes an axial force $N = 20$ kN (compression) and a constant moment $M = 20\times4 = 80$ kN·m along the full height, with tension on the outer face (the same side as the top of the beam). The shear force in the column is zero. - **Reactions at the base:** $V = 20$ kN (↑), $H = 0$, $M = 80$ kN·m. ```text free end ●───────┐ BMD: beam 0 → -80 (linear) 20 kN │ column -80 (constant) │ │ ▒▒▒ fixed base ``` **Answer: (i) $M = -4$ kN·m (left of the couple) and $+6$ kN·m (right of the couple), with $R_A = 1$ kN ↓ and $R_B = 1$ kN ↑; (ii) the beam moment varies linearly from 0 to 80 kN·m (hogging), and the column carries a constant 80 kN·m.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why are R_A and R_B equal and opposite (1 kN)? A. The beam is symmetric B. The couple creates a vertical load of 1 kN C. The supports are fixed D. There is no vertical load; the couple is balanced by the couple from the reactions, R_B x 10 = 10 Answer: D. Sum Fy = 0 gives R_A = -R_B and moments give R_B = 10/10 = 1 kN. (Key point: Reactions from couple) 2. What happens to BMD at the couple? A. Jumps up by 10 kN·m, from -4 to +6 kN·m B. Stays continuous C. Jumps down by 10 D. Changes slope only Answer: A. A clockwise couple raises the sagging moment diagram by its value. (Key point: Jump at couple) 3. What do the column carry (fixed base, 20 kN at free end 4 m away)? A. Shear 20 kN and zero moment B. Axial force 20 kN and constant moment 80 kN·m, zero shear C. Moment varying 0 to 80 D. Axial force 0 and shear 80 kN Answer: B. The load is eccentric by 4 m, so N = 20 kN and M = 20 x 4 = 80 kN·m along the column. 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