---
title: "Frame AFD, SFD, BMD (t/m loads, hinge)"
subject: "Strength of Materials"
chapter: 1
key: "semester_3/strength_of_materials:frame-afd-sfd-bmd-tonne-hinge"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/frame-afd-sfd-bmd-tonne-hinge
site: BCT Engineering
---

# Frame AFD, SFD, BMD (t/m loads, hinge)

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

- Asked in: 2069 Asar, 16 marks
- Video: [Frame Analysis || Shear Force & Bending Moment Diagram](https://www.youtube.com/watch?v=0pG1to_L44M) — Civil Engineering
- 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-tonne-hinge
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-1

## Question

Draw axial force, shearing force and bending moment diagrams for the following frame loaded as shown. Indicate salient points if any. [Figure: left column hinged at the base, 4 m high (2 m + 2 m) with a horizontal distributed load of 5 t/m; top beam with an internal hinge at the middle and a triangular load rising to 1 t/m towards the right end; right column 3 m high with a 3 t/m horizontal distributed load over 1 m; horizontal dimensions 3 m and 3 m; both bases hinged.]

## Short answer

Support reactions: H_A=12.57 t ←, V_A=3.18 t ↓, H_E=4.43 t ←, V_E=6.18 t ↑. Maximum bending moment 15.80 t·m at 2.51 m from A on AB (tension right face). Bending moment zero at internal hinge C. Diagrams: AB V linear, M parabolic; BC and CD V parabolic, M cubic; DE V linear, M parabolic.

## Answer

**Assumptions (taken from the figure):**
- Units: tonne (t), t/m, t·m. A(0,0) hinged; AB = 4 m with 5 t/m horizontal (→) over the full height (20 t).
- Beam B–C–D = 3 m + 3 m with internal hinge at C; triangular load increasing uniformly from 0 at B to 1 t/m at D (total 3 t).
- Right column DE = 3 m, E hinged at (6, 1); 3 t/m horizontal (←) over the top 1 m of DE (3 t).

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
|                                      |
|                                      |
|                                      |
|                                      |
|                                      |
|                                      |
|                                      |
|                                      |
|                                      |
|                                      E
|
|
A
```

#### 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.

- $H_A$ = 12.57 t ←
- $V_A$ = 3.18 t ↓
- $H_E$ = 4.43 t ←
- $V_E$ = 6.18 t ↑

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 (t) | V (t) | M (t·m) |
|---|---|---|---|
| AB at A | 3.18 | 12.57 | 0.00 |
| AB at B | 3.18 | -7.43 | 10.29 (right face) |
| BC at B | -7.43 | -3.18 | 10.29 (bottom) |
| BC at C | -7.43 | -3.93 | 0.00 |
| CD at C | -7.43 | -3.93 | 0.00 |
| CD at D | -7.43 | -6.18 | -14.79 (top) |
| DE at D | -6.18 | 7.43 | -14.79 (right face) |
| DE at E | -6.18 | 4.43 | 0.00 |

#### Shape of the diagrams
- AB: V linear, M parabolic
- BC: V parabolic, M cubic
- CD: V parabolic, M cubic
- DE: V linear, M parabolic
- 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 = 15.80 t·m at 2.51 m from A on AB, tension on the right face
- V = 0 at 2.51 m from A along AB: M = 15.80 (right face) t·m (local maximum)
- 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. Triangular load 0 to 1 t/m over 6 m is:
   A. 6 t, parabolic V
   B. 3 t, linear V
   C. 3 t, giving parabolic V and cubic M
   D. 1 t, cubic V
   Answer: C. 1/2 x 6 x 1 = 3 t. (Key point: Triangular load)

2. Maximum bending moment 15.80 t·m is:
   A. At the hinge C
   B. At D
   C. At E
   D. On AB at 2.51 m from A, where V = 0
   Answer: D. Under the UDL on the column, V = 0 at 2.51 m, so M is maximum there. (Key point: Maximum moment)

3. V_A = 3.18 t down and V_E = 6.18 t up. Why?
   A. Net upward is 3 t, equal to the triangular load
   B. Both are up
   C. They cancel exactly
   D. They balance 20 t
   Answer: A. 6.18 - 3.18 = 3 t = beam load; the couple from the horizontal loads makes A pull down. (Key point: Vertical reactions)

## Related in Strength of Materials

- [Frame AFD, SFD, BMD (100 kN, 25 kN/m, hinge)](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aframe-afd-sfd-bmd-100kn-25knm) — Strength of Materials, chapter 1
- [Frame AFD, SFD, BMD (5 kN/m, 10 kN, hinge)](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aframe-afd-sfd-bmd-5knm-10kn) — Strength of Materials, chapter 1, asked 2×
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## Related in other subjects

- [AFD, SFD, BMD of frame with hinge at D](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Aframe-afd-sfd-bmd-8knm-hinge-50kn) — Applied Mechanics, chapter 6
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