---
title: "AFD, SFD, BMD for frame with UDL and couple"
subject: "Applied Mechanics"
chapter: 6
key: "semester_1/applied_mechanics:frame-afd-sfd-bmd-30kn-4knm-8kn"
source: https://www.bctengineer.com/q/semester_1/applied_mechanics/frame-afd-sfd-bmd-30kn-4knm-8kn
site: BCT Engineering
---

# AFD, SFD, BMD for frame with UDL and couple

Applied Mechanics · Chapter 6: Analysis of Beams and Frames

- Asked in: 2072 Kartik, 14 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_1/applied_mechanics/frame-afd-sfd-bmd-30kn-4knm-8kn
- Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-6

## Question

Draw AFD, SFD and BMD of the given frame loaded as shown in figure below. Indicate also the salient features if any. [Figure: frame with pin support at A (bottom of left column) and roller/hinge at H; left column A-B-C 3 m + 3 m with 10 kN at 45° at B and 30 kN at 30° at C; beam C-D-E-F with 4 kN/m UDL, 20 kN·m moment at E, 5 kN downward at F; right column from E/F down to G (8 kN horizontal) and H with 2 m + 2 m; horizontal dimensions 3 m, 1 m, 1 m.]

## Short answer

Reactions: A_x = 25.05 kN ←, A_y = 2.95 kN ↓, H_y = 42.02 kN ↑. Maximum bending moment is 129.10 kN·m at knee C (tension inside). Contraflexure in DE at 0.97 m from D.

## Answer

#### Reading of the figure
- Origin at A, x to the right, y upward: A(0,0) pin, B(0,3), C(0,6), D(3,6), E(4,6), F(5,6), G(4,4), H(4,2) roller (vertical reaction only).
- Loads: 30 kN at C (30° below horizontal, towards right) = 25.98 kN → and 15.00 kN ↓; 10 kN at B (45°) = 7.07 kN → and 7.07 kN ↓; UDL 4 kN/m on CD (12 kN at 1.5 m from C); couple 20 kN·m at E (taken anticlockwise); 5 kN ↓ at F; 8 kN ← at G.
- r = 3, so the frame is statically determinate.

#### Reactions
```text
 30kN\_  4 kN/m   20kN·m  5kN
      C=======D==E====F ↓
      |          |
 10kN\_B         G←8kN
      |          |
      |          H (roller)
      A (pin)
```
- ΣFx = 0: Aₓ + 7.07 + 25.98 − 8 = 0, so **Aₓ = 25.05 kN (←)**.
- ΣM_A = 0 (anticlockwise +): 4H_y − 7.07(3) − 25.98(6) − 12(1.5) + 20 − 5(5) + 8(4) = 0, so 4H_y = 168.10 and **H_y = 42.02 kN (↑)**.
- ΣFy = 0: A_y + 42.02 − 7.07 − 15 − 12 − 5 = 0, so **A_y = 2.95 kN (↓)**.

#### Member forces and diagrams
Sign notes: N is + for tension and − for compression; V follows the member axis in the order of the letters (e.g. A→B); for M the side in brackets is the tension side, on which the BMD is drawn.

- **AB** (A to B, 3.00 m): N = 2.95 kN (tension); V = 25.05 to 25.05 kN; M = 0.00 (–) to 75.16 (right) kN·m, linear.
- **BC** (B to C, 3.00 m): N = 10.02 kN (tension); V = 17.98 to 17.98 kN; M = 75.16 (right) to 129.10 (right) kN·m, linear.
- **CD** (C to D, 3.00 m): N = 8.00 kN (compression); V = −25.02 to −37.02 kN; M = 129.10 (bottom) to 36.02 (bottom) kN·m, parabolic.
- **DE** (D to E, 1.00 m): N = 8.00 kN (compression); V = −37.02 to −37.02 kN; M = 36.02 (bottom) to 1.00 (top) kN·m, linear. M changes sign (contraflexure) at 0.97 m from D.
- **EF** (E to F, 1.00 m): N = 0.00 kN (zero); V = 5.00 to 5.00 kN; M = 5.00 (top) to 0.00 (–) kN·m, linear.
- **EG** (E to G, 2.00 m): N = 42.02 kN (compression); V = 8.00 to 8.00 kN; M = 16.00 (right) to 0.00 (–) kN·m, linear.
- **GH** (G to H, 2.00 m): N = 42.02 kN (compression); V = 0.00 to 0.00 kN; M = 0.00 (–) to 0.00 (–) kN·m, constant.

#### Salient features
- Maximum bending moment = **129.10 kN·m** at the knee C (tension on the inside of the frame). Maximum axial force = 42.02 kN compression in the right column.
- Shear in CD never becomes zero, so the moment falls steadily from C to D; in DE the moment changes sign 0.97 m from D (point of contraflexure).
- The roller at H gives no horizontal reaction, so the whole horizontal load is resisted at A. The 8 kN load is carried by compression in beam CDE.
- Moment is carried round the rigid corner C unchanged (129.10 kN·m on both members, tension on the inner fibres). Joint E check: 1.00 + 20 (couple) = 5.00 + 16.00.

## Check yourself

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

1. How is the 30 kN load at 30° below horizontal treated?
   A. Split into 25.98 kN horizontal and 15 kN vertical
   B. Used as a 30 kN horizontal load
   C. Used as a 30 kN vertical load
   D. Split into 15 kN horizontal and 25.98 kN vertical
   Answer: A. 30 cos30° = 25.98 and 30 sin30° = 15. (Key point: Resolving inclined loads)

2. Why does A carry all the horizontal reaction?
   A. A is a fixed support
   B. The roller at H gives no horizontal reaction
   C. The 8 kN load acts at A
   D. H is a hinge
   Answer: B. The roller gives only vertical reaction, so ΣFx must be balanced at A. (Key point: Horizontal reaction)

3. Where is the maximum bending moment?
   A. At A
   B. At D
   C. At the knee C, 129.10 kN·m
   D. At E
   Answer: C. M rises from A to C (129.10 kN·m) and then falls. (Key point: Maximum moment)

4. What does the sign change of M in DE show?
   A. Zero shear
   B. Zero axial force
   C. A point load
   D. A point of contraflexure about 0.97 m from D
   Answer: D. M changes from bottom tension to top tension, so M passes through zero there. (Key point: Contraflexure in DE)

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