---
title: "Beam AFD, SFD, BMD with hinge (80 kN)"
subject: "Strength of Materials"
chapter: 1
key: "semester_3/strength_of_materials:beam-afd-sfd-bmd-20knm-80kn"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/beam-afd-sfd-bmd-20knm-80kn
site: BCT Engineering
---

# Beam AFD, SFD, BMD with hinge (80 kN)

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

- Asked in: 2075 Chaitra, 12 marks
- Video: [Beam with Hinge | Concepts and a solved example|GATE 2020| Calculating reaction and bending moment](https://www.youtube.com/watch?v=7FjMsdWHhVQ) — Concepts in 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/beam-afd-sfd-bmd-20knm-80kn
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-1

## Question

Draw AFD, SFD and BMD for following beam. Also indicate the salient features. [Figure: beam A to B, fixed at B on the right end, with a support at 2 m from A; triangular load of peak 20 kN/m (rising from zero at A to 20 kN/m and falling) on the first 5 m; an 80 kN point load at 1 m further; a 50 kN force at $30^\circ$ to the beam; a 15 kN/m UDL over the last 3 m; an internal hinge; dimensions 2 m, 3 m, 1 m, 2 m, 3 m.]

## Short answer

Reactions: roller 41.67 kN up; fixed support HB=43.30 kN left, VB=158.33 kN up, MB=592.50 kN·m clockwise. Maximum sagging moment 8.02 kN·m at 3.56 m; maximum hogging -592.50 kN·m at B. Contraflexure at 2.5 m and 5 m (hinge).

## Answer

**Assumptions (taken from the figure):**
- Beam of 11 m: A (x = 0) free end; roller at x = 2 m; internal hinge at x = 5 m; fixed support at B (x = 11 m).
- Triangular load on 0–5 m, symmetrical, peak 20 kN/m at x = 2.5 m (total 50 kN); 80 kN downward at x = 6 m; 50 kN at 30° to the beam at x = 8 m, acting down and to the right (components 43.30 kN along the beam towards B and 25 kN down); 15 kN/m downward on 8–11 m (45 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
Hinge condition: $M = 0$ at x = 5 m. Moments of the left part (0–5 m) about the hinge: $3R_{roller} = 50\times2.5$, which gives the roller reaction. The fixed support then follows from $\Sigma F_x = 0$, $\Sigma F_y = 0$ and $\Sigma M = 0$ for the whole beam.

- $R_{roller}$ = 41.67 kN ↑
- $H_B$ = 43.30 kN ←
- $V_B$ = 158.33 kN ↑
- $M_B$ = 592.50 kN·m (clockwise)

Check: the hinge passes $50 - 41.67 = 8.33$ kN to the right part, and $V_B = 8.33 + 80 + 25 + 45 = 158.33$ kN, $M_B = 8.33(6) + 80(5) + 25(3) + 45(1.5) = 592.5$ kN·m ✓

#### Axial force, shear force and bending moment at the key sections
| Section | N (kN) | V (kN) | M (kN·m) |
|---|---|---|---|
| At A | 0.00 | 0.00 | 0.00 |
| Just left of roller | 0.00 | -16.00 | -10.67 (top) |
| Just right of roller | 0.00 | 25.67 | -10.67 (top) |
| At hinge (x = 5 m) | 0.00 | -8.33 | 0.00 |
| Just left of 80 kN | 0.00 | -8.33 | -8.33 (top) |
| Just right of 80 kN | 0.00 | -88.33 | -8.33 (top) |
| Just left of 50 kN | 0.00 | -88.33 | -185.00 (top) |
| Just right of 50 kN | -43.30 | -113.33 | -185.00 (top) |
| At B | -43.30 | -158.33 | -592.50 (top) |

The segment from x = 8 m to B is in compression (43.30 kN), because the horizontal component of the 50 kN load pushes it towards the fixed 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; M jumps at couples.
- AFD: zero from A to x = 8 m, then a constant compression of 43.30 kN up to B.
- The BMD is zero at the free end A and at the hinge.

#### Salient features
- Maximum positive (sagging) M = 8.02 kN·m at x = 3.56 m
- Maximum negative (hogging) M = -592.50 kN·m at x = 11.00 m (fixed end B)
- V = 0 at 3.56 m from A: M = 8.02 kN·m (bottom), the local maximum
- Point of contraflexure (M = 0) at 2.50 m from A
- M = 0 again at 5.00 m from A (the internal hinge), where it changes from sagging to hogging

#### 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. How is the roller reaction found in this beam with an internal hinge?
   A. Take moments of the part left of the hinge about the hinge and set M = 0
   B. Use only sum Fy = 0 for the whole beam
   C. Take moments about the fixed support B and ignore the hinge
   D. Assume the roller carries half the total load
   Answer: A. The hinge cannot carry moment, so M = 0 at x = 5 m. Moments of the left part about it give the roller reaction directly. (Key point: Hinge condition)

2. Under the triangular load, what are the shapes of the SFD and BMD?
   A. SFD linear, BMD parabolic
   B. SFD parabolic, BMD cubic
   C. SFD constant, BMD linear
   D. SFD cubic, BMD parabolic
   Answer: B. M is one degree higher than V; under a triangular load V is parabolic so M is cubic. (Key point: Triangular load shapes)

3. Why is H_B = 43.30 kN?
   A. It balances the vertical component 50 sin30°
   B. It balances the 80 kN point load
   C. It balances the axial (horizontal) component of the 50 kN force at 30°, i.e. 50 cos30°
   D. It comes from the 15 kN/m UDL
   Answer: C. Only the 50 kN inclined force has a component along the beam, 50 cos30° = 43.30 kN, so the fixed support supplies H_B equal to it. (Key point: Axial reaction)

4. Where does the maximum (hogging) bending moment of 592.50 kN·m occur?
   A. At the roller, x = 2 m
   B. At the internal hinge, x = 5 m
   C. Where V = 0 at x = 3.56 m
   D. At the fixed support B, x = 11 m
   Answer: D.  The fixed end carries the full moment of all loads, M_B = 592.50 kN·m; the hinge and the free end have zero moment. (Key point: Maximum moment)

## Related in Strength of Materials

- [Frame AFD, SFD, BMD (80 kN, 30 kNm, hinge)](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aframe-afd-sfd-bmd-80kn-30knm) — 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
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## Related in other subjects

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