--- title: "Beam thrust and bending moment diagrams with max moments" subject: "Strength of Materials" chapter: 1 key: "semester_3/strength_of_materials:beam-thrust-bmd-2kn-5kn-1knm" source: https://www.bctengineer.com/q/semester_3/strength_of_materials/beam-thrust-bmd-2kn-5kn-1knm site: BCT Engineering --- # Beam thrust and bending moment diagrams with max moments Strength of Materials · Chapter 1: Axial Forces, Shearing Forces and Bending Moments - Asked in: 2066 Chaitra (old course), 16 marks - Video: [Draw the shear and moment diagrams for the beam - 7-53](https://www.youtube.com/watch?v=Yn279P6dJQU) — Grantstrong - 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-thrust-bmd-2kn-5kn-1knm - Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-1 ## Question Calculate maximum positive and maximum negative bending moment and draw thrust diagram and bending moment diagram for the beam as shown in the figure. [Figure: beam on a hinge support and a roller support; a vertical post of height 1 m carries a 2 kN horizontal force; a 5 kN downward load; a triangular load of 1 kN/m at the right end with a couple applied; dimensions 1.5 m, 1 m, 1.5 m, 1.5 m, 1.5 m.] ## Short answer Reactions: H_A=2 kN leftwards, V_A=0.38 kN upward, R_B=6.12 kN upward. Axial force (thrust) is 2 kN tension from A to the vertical post (0–1.5 m) and zero thereafter. Maximum sagging bending moment = 2.94 kN·m at 2.50 m from A; maximum hogging bending moment = –4.00 kN·m at the roller support (4 m). ## Answer **Assumptions (taken from the figure):** - Beam from x = 0 to 7 m: hinge A at x = 0 and roller B at x = 4 m. A 1 m post at x = 1.5 m carries a 2 kN horizontal force (→) at its top; this is equivalent to a 2 kN axial force on the beam plus a clockwise couple of 2 × 1 = 2 kN·m. - 5 kN downward at x = 2.5 m; triangular load from 0 at x = 4 m to 1 kN/m at x = 7 m (total 1.5 kN, acting at x = 6 m); clockwise couple of 1 kN·m at the free end x = 7 m. Sign convention: N is positive in tension; V is positive when the forces on the part to the left of the section act upward; M is quoted with the tension face named (sagging = bottom, hogging = top). #### Support reactions - $\Sigma F_x = 0$: $H_A$ = 2.00 kN ← - $\Sigma M_A = 0$: $4R_B = 2 + 5(2.5) + 1.5(6) + 1 = 24.5$, so $R_B$ = 6.12 kN ↑ (6.125) - $\Sigma F_y = 0$: $V_A = 6.5 - 6.125$ = 0.38 kN ↑ (0.375) #### Axial force, shear force and bending moment at the key sections | Section | N (kN) | V (kN) | M (kN·m) | |---|---|---|---| | At A (x = 0) | 2.00 | 0.38 | 0.00 | | Just left of post (x = 1.5) | 2.00 | 0.38 | 0.56 (bottom) | | Just right of post | 0.00 | 0.38 | 2.56 (bottom) | | Just left of 5 kN (x = 2.5) | 0.00 | 0.38 | 2.94 (bottom) | | Just right of 5 kN | 0.00 | -4.62 | 2.94 (bottom) | | Just left of roller B (x = 4) | 0.00 | -4.62 | -4.00 (top) | | Just right of roller B | 0.00 | 1.50 | -4.00 (top) | | At x = 5.5 m | 0.00 | 1.12 | -1.94 (top) | | Free end, left of couple (x = 7) | 0.00 | 0.00 | -1.00 (top) | #### Thrust diagram N = +2 kN (tension) from A to the post (x = 0 to 1.5 m) and zero from the post to the free end. #### 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, and M jumps at couples (by 2 kN·m at the post). - 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 and is zero at the hinge support A. #### Salient features - Maximum positive (sagging) M = 2.94 kN·m at x = 2.50 m - Maximum negative (hogging) M = -4.00 kN·m at x = 4.00 m (over the roller B) - Point of contraflexure (M = 0) at x = 3.14 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. The 2 kN horizontal force at the top of the 1 m post is replaced on the beam axis by what? A. A 2 kN vertical force B. A 2 kN axial force and a 2 kN·m couple C. A 1 kN axial force only D. A 2 kN·m couple only Answer: B. Moving the force to the beam axis gives the same 2 kN force plus a couple 2 kN x 1 m = 2 kN·m, which causes a jump in the BMD at the post. (Key point: Post load equivalence) 2. What does the thrust (axial force) diagram show for the beam? A. 2 kN tension from A up to the post, then zero beyond B. 2 kN compression for the whole beam C. Zero throughout D. 2 kN only in the overhang Answer: A. H_A = 2 kN balances the post force; only the segment from A to the post carries 2 kN tension, and the rest of the beam has no axial force. (Key point: Thrust diagram) 3. The maximum negative moment is -4.00 kN·m at the roller. How is it obtained from the right side? A. 1.5 kN triangular load x 2 m plus the 1 kN·m end couple B. 1.5 kN x 3 m C. 5 kN x 1.5 m D. Only the 1 kN·m couple Answer: A. The triangular load of 1.5 kN acts 2 m from the roller (2/3 of 3 m), giving 3 kN·m, plus the 1 kN·m couple, total 4 kN·m hogging. (Key point: Maximum hogging moment) 4. Why is the maximum positive moment 2.94 kN·m at x = 2.5 m? A. The roller reaction acts there B. The shear force changes sign (from +0.38 to -4.62 kN) under the 5 kN load C. The couple acts there D. The axial force vanishes there Answer: B. Shear changes sign at the point load, so the moment peaks there: 0.38 x 2.5 + 2 = 2.94 kN·m. (Key point: Maximum sagging moment) ## Related in Strength of Materials - [Frame BMD with maximum and minimum moments](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aframe-bmd-6knm-5kn-2kn-max) — Strength of Materials, chapter 1 - [T-beam with overhang: max bending stress](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3At-beam-4kn-m-10kn-overhang) — Strength of Materials, chapter 7 - [Frame AFD, SFD, BMD (5 and 10 kN/m, hinge)](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aframe-afd-sfd-bmd-5knm-10knm) — Strength of Materials, chapter 1, asked 2× - [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 - [Flexural rigidity; T-beam bending stresses (60 kN loads)](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aflexural-rigidity-t-beam-60kn) — Strength of Materials, chapter 7 - [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 ## Related in other subjects - [Overhang beam: ILD, max positive/negative moment at D](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aoverhang-ild-moment-d-three-loads) — Theory of Structures I, chapter 5 - [ILD: bending moment at fixed support, 10 m beam](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aild-bm-fixed-support-10m-1) — Theory of Structures II, chapter 4 - [Frame horizontal deflection at roller support](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aframe-virtual-work-roller-2076) — Theory of Structures I, chapter 3 - [Foundation stresses from moments and vertical force](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Agravity-dam-vertical-stress-moments) — Hydropower Engineering, chapter 4