---
title: "Frame AFD, SFD, BMD (40 kN at 45 deg)"
subject: "Strength of Materials"
chapter: 1
key: "semester_3/strength_of_materials:frame-afd-sfd-bmd-30kn-10knm"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/frame-afd-sfd-bmd-30kn-10knm
site: BCT Engineering
---

# Frame AFD, SFD, BMD (40 kN at 45 deg)

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

- Asked in: 2069 Chaitra, 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-afd-sfd-bmd-30kn-10knm
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-1

## Question

Draw axial force, shear force and bending moment diagram indicating salient points for the frame loaded as shown. [Figure: left column 4 m high with a hinge at its top and a triangular horizontal load increasing to 20 kN/m at the hinged base; top beam with a 30 kN point load, a 10 kN/m UDL over 2 m and a 40 kN force at $45^\circ$ at the right end; inclined right member descending to a hinged support; horizontal dimensions 1 m, 1 m, 2 m, 1 m.]

## Short answer

H_A=26.67 kN ←, V_A=4.36 kN ↑, H_F=41.62 kN ←, V_F=73.92 kN ↑. Maximum bending moment 92.55 kN·m at joint E. Local maximum 20.53 kN·m at 1.69 m from A on AB. Contraflexure at 1.17 m from B. Zero moment at hinge B. Diagrams: AB V parabolic, M cubic; BC, CD V constant, M linear; DE V linear, M parabolic; EF V constant, M linear.

## Answer

**Assumptions (taken from the figure):**
- A(0,0) hinged at the base; AB = 4 m, with an internal hinge at its top B; horizontal triangular load from 20 kN/m at A to 0 at B (→), total 40 kN acting 1.33 m above A.
- Beam B–C–D–E: BC = 1 m, CD = 1 m, DE = 2 m; 30 kN downward at C; 10 kN/m downward over DE (20 kN); 40 kN at 45° (down and to the right) at E, with components 28.28 kN → and 28.28 kN ↓.
- The inclined member EF descends 4 m over 1 m horizontally to the hinged support F = (5, 0).

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
Bo----C------D------------E
|                         |
|                          |
|                          |
|                           |
|                           |
|                            |
|                            |
|                            ||
|                             |
|                             ||
|                              |
|                              ||
A                               F
```

#### Support reactions
- Hinge at B (moments of column AB about B): $4H_A + 40(4 - 1.33) = 0 \Rightarrow H_A = -26.67$ kN.
- Moments about A: $5V_F = 40(1.33) + 30(1) + 20(3) + 28.28(4) + 28.28(4) = 369.61 \Rightarrow V_F = 73.92$ kN.
- $\Sigma F_y = 0$: $V_A = 30 + 20 + 28.28 - 73.92 = 4.36$ kN.
- $\Sigma F_x = 0$: $H_F = -(40 + 28.28) + 26.67 = -41.62$ kN.

Results:
- $H_A$ = 26.67 kN ←
- $V_A$ = 4.36 kN ↑
- $H_F$ = 41.62 kN ←
- $V_F$ = 73.92 kN ↑

Check: moments of EF about E from F: $1(73.92) - 4(41.62) = -92.55$ kN·m, which matches the beam moment at E below ✓

#### Axial force, shear force and bending moment at the key sections
| Section | N (kN) | V (kN) | M (kN·m) |
|---|---|---|---|
| AB at A | -4.36 | 26.67 | 0.00 |
| AB at B | -4.36 | -13.33 | 0.00 |
| BC at B | -13.33 | 4.36 | 0.00 |
| BC at C | -13.33 | 4.36 | 4.36 (bottom) |
| CD at C | -13.33 | -25.64 | 4.36 (bottom) |
| CD at D | -13.33 | -25.64 | -21.27 (top) |
| DE at D | -13.33 | -25.64 | -21.27 (top) |
| DE at E | -13.33 | -45.64 | -92.55 (top) |
| EF at E | -81.81 | 22.45 | -92.55 (right face) |
| EF at F | -81.81 | 22.45 | 0.00 |

#### Shape of the diagrams
- AB: V parabolic, M cubic
- BC: V constant, M linear
- CD: V constant, M linear
- DE: V linear, M parabolic
- EF: V constant, M linear
- AFD: constant between loads on a member, changing only where a load component acts along the member.
- The 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 = 92.55 kN·m at joint E (top of DE, right face of EF)
- V = 0 at 1.69 m from A along AB: M = 20.53 kN·m (right face), the local maximum
- Maximum sagging moment on the beam = 4.36 kN·m at C
- Point of contraflexure (M = 0) at 1.17 m from B along the beam (0.17 m to the right of C)
- The bending moment is zero at the internal hinge B (from both sides).

## Check yourself

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

1. AB has a triangular horizontal load. V and M:
   A. V linear, M parabolic
   B. V constant, M linear
   C. V zero, M constant
   D. V parabolic, M cubic
   Answer: D. Triangular load gives parabolic V and cubic M. (Key point: Shapes on AB)

2. 10 kN/m over DE (2 m) is:
   A. 20 kN, V linear, M parabolic
   B. 10 kN, V constant
   C. 20 kN, V constant, M linear
   D. 40 kN, V cubic
   Answer: A. 10 x 2 = 20 kN; UDL gives linear V. (Key point: UDL on DE)

3. The moment at hinge B is:
   A. 20.53 kN·m
   B. 1.64 kN·m
   C. 0
   D. 13.33 kN·m
   Answer: C. Hinges carry no moment; read as 0 from both sides. (Key point: Hinge B)

## Related in Strength of Materials

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