---
title: "Frame BMD with maximum and minimum moments"
subject: "Strength of Materials"
chapter: 1
key: "semester_3/strength_of_materials:frame-bmd-6knm-5kn-2kn-max"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/frame-bmd-6knm-5kn-2kn-max
site: BCT Engineering
---

# Frame BMD with maximum and minimum moments

Strength of Materials · Chapter 1: Axial Forces, Shearing Forces and Bending Moments

- Asked in: 2066 Jestha (old course), 12 marks
- 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-bmd-6knm-5kn-2kn-max
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-1

## Question

Draw bending moment diagram and locate the maximum and minimum values of BM for the frame which has hinged support at A and other support at B as shown in figure. [Figure: vertical column from A (hinged/fixed support at the bottom) up 2 m to a horizontal beam; a 2 kN horizontal force acts on the column at 1 m above A; the beam carries a triangular/trapezoidal load of peak 6 kN/m over its first 4 m (2 m + 2 m), rests on a roller support at B and has a 1.5 m overhang with a 5 kN downward point load at the free end.]

## Short answer

Reactions: H_A=2 kN ←, V_A=3.62 kN ↑, R_B=13.38 kN ↑. Maximum sagging moment is 5.76 kN·m at 1.55 m from C on CB (bottom tension), and maximum hogging moment is 7.5 kN·m over support B (top tension).

## Answer

**Assumptions (taken from the figure):**
- A(0,0) hinged at the foot of a 2 m column AC; 2 kN horizontal (→) at 1 m above A. Beam from the column top C: 4 m span to the roller B, then a 1.5 m overhang BD with 5 kN downward at the free end D.
- Load on the 4 m span CB: triangular, symmetrical, peak 6 kN/m at 2 m from the column (total 12 kN).

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
C--------------------------------B-----------D
|
|
|
|
|
|
|
A
```

#### Support reactions
Equilibrium of the whole frame gives $\Sigma F_x = 0$, $\Sigma F_y = 0$ and $\Sigma M = 0$.

- $H_A$ = 2.00 kN ←
- $\Sigma M_A = 0$: $4R_B = 2(1) + 12(2) + 5(5.5) = 53.5$, so $R_B$ = 13.38 kN ↑
- $V_A = 17 - 13.38$ = 3.62 kN ↑

#### Axial force, shear force and bending moment at the key sections
| Section | N (kN) | V (kN) | M (kN·m) |
|---|---|---|---|
| AC at A | -3.62 | 2.00 | 0.00 |
| AC at C | -3.62 | 0.00 | 2.00 (right face) |
| CB at C | 0.00 | 3.62 | 2.00 (bottom) |
| CB at B | 0.00 | -8.38 | -7.50 (top) |
| BD at B | 0.00 | 5.00 | -7.50 (top) |
| BD at D | 0.00 | 5.00 | 0.00 |

#### Shape of the diagrams
- AC: V constant (steps at point loads), M linear
- CB: V parabolic, M cubic
- BD: V constant, 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.

#### Maximum and minimum bending moments
- Maximum positive (sagging) BM = 5.76 kN·m at 1.55 m from C along CB, where V = 0 (tension on the bottom)
- Maximum negative (hogging) BM = -7.50 kN·m over the roller B (tension on the top); this is the largest BM in magnitude
- Point of contraflexure (M = 0) at 3.05 m from C along CB
- In the column, M rises linearly from 0 at A to 2.00 kN·m at the 2 kN load and stays 2.00 kN·m up to C.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. The symmetrical triangular load on the 4 m span has a peak of 6 kN/m. What is its total value?
   A. 24 kN
   B. 12 kN
   C. 6 kN
   D. 18 kN
   Answer: B. Area of the triangle = 1/2 x 4 x 6 = 12 kN, acting at the mid-span (2 m from the column). (Key point: Total load on the span)

2. Which check confirms the vertical reactions V_A = 3.62 kN and R_B = 13.38 kN?
   A. Their sum 17 kN equals 12 kN + 5 kN of vertical load
   B. Their difference equals 5 kN
   C. Their sum equals the 2 kN horizontal force
   D. Their sum equals 12 kN
   Answer: A. Sum of Fy = 0 gives V_A + R_B = 12 + 5 = 17 kN. (Key point: Vertical equilibrium)

3. Why is the maximum bending moment 7.50 kN·m at C, tension on top?
   A. Because the 5 kN load at the end of the 1.5 m overhang gives 5 x 1.5 hogging at C
   B. Because the roller gives a moment
   C. Because of the 2 kN horizontal force
   D. Because the triangular load peaks at C
   Answer: A. The overhang is a cantilever from the support, so M = 5 kN x 1.5 m = 7.5 kN·m, hogging. (Key point: Maximum moment at C)

4. In span BC the local sagging maximum 5.76 kN·m occurs where?
   A. At the roller support
   B. Where V = 0, at 1.55 m from B
   C. At mid-span, 2 m from B
   D. At the free end
   Answer: B. Bending moment is a maximum where the shear force passes through zero; here V = 0 at 1.55 m from B. (Key point: Maximum sagging in span)

## Related in Strength of Materials

- [Beam thrust and bending moment diagrams with max moments](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Abeam-thrust-bmd-2kn-5kn-1knm) — Strength of Materials, chapter 1
- [BMD of a simple beam and a frame](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Abmd-simple-beam-and-frame-10knm) — Strength of Materials, chapter 1
- [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 (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 (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 (20 kNm, 15 kN)](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aframe-afd-sfd-bmd-20knm-15kn) — Strength of Materials, chapter 1

## Related in other subjects

- [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
- [Force method: frame with hinged support](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aforce-method-frame-30knm-50kn-hinged) — Theory of Structures II, chapter 2
- [Fixed beam: end moments from support rotation or settlement](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Afixed-beam-end-moments-rotation-settlement) — Theory of Structures II, chapter 3, asked 4×
- [Moment distribution: frame with SFD, AFD, BMD](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Amoment-dist-frame-sfd-afd-bmd) — Theory of Structures II, chapter 3