---
title: "Frame AFD, SFD, BMD (15 kN/m, 75 kN)"
subject: "Strength of Materials"
chapter: 1
key: "semester_3/strength_of_materials:frame-afd-sfd-bmd-15knm-75kn"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/frame-afd-sfd-bmd-15knm-75kn
site: BCT Engineering
---

# Frame AFD, SFD, BMD (15 kN/m, 75 kN)

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

- Asked in: 2080 Bhadra, 14 marks
- Video: [ANALYSIS OF FRAME SHEAR BENDING & AXIAL THRUST DIAGRAM | EX 03](https://www.youtube.com/watch?v=VhZ1_8AFJrI) — Structural Funda by:- KheNiSiR
- 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-15knm-75kn
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-1

## Question

Draw axial force, shear force and bending moment diagram for the given frame. [Figure: left column A-B of height 8 m with A hinged at the base; inclined member B-C rising 2 m over a horizontal span of 7 m and carrying a 15 kN/m UDL; right column C-D of height 8 m (D hinged at the base); a 75 kN horizontal force acts on the right column at 3 m from the base; overall width 7 m.]

## Short answer

The frame has a hinge at mid-BC. Support reactions: H_A=25.55 kN→, V_A=91.94 kN↑, H_D=49.45 kN→, V_D=13.06 kN↑. Axial force constant on each member (e.g., AB -91.94 kN). Shear constant on AB (-25.55 kN), linear on BC, steps at 75 kN load on CD. Maximum bending moment 204.38 kN·m at B; local maximum on BC 34.44 kN·m. BMD is drawn on tension side.

## Answer

**Assumptions (taken from the figure):**
- A(0,0) hinged; AB = 8 m, so B = (0, 8). B to C (7, 10) is inclined (rises 2 m in 7 m, length 7.28 m). The right column CD is 8 m high, so the hinged support D = (7, 2) is 2 m above the level of A.
- With two hinged supports and no internal hinge the frame is indeterminate, so a hinge is assumed at the middle of BC, (3.5, 9), giving a three-hinged frame.
- 15 kN/m acts downward per metre of horizontal projection of BC (total 105 kN); the 75 kN horizontal force acts to the left on CD, 3 m above D, i.e. at (7, 5).

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  (2 m above A)
|
|
A
```

#### Support reactions
Moments about A: $7V_D - 2H_D - 105(3.5) + 75(5) = 0$. Hinge at the middle of BC (left part): $-3.5V_A + 9H_A + 52.5(1.75) = 0$. With $\Sigma F_x = 0$ ($H_A + H_D = 75$) and $\Sigma F_y = 0$ ($V_A + V_D = 105$):

- $H_A$ = 25.55 kN →
- $V_A$ = 91.94 kN ↑
- $H_D$ = 49.45 kN →
- $V_D$ = 13.06 kN ↑

Check: moments of the right part about the hinge: $13.06(3.5) + 49.45(7) - 75(4) - 52.5(1.75) \approx 0$ ✓

#### Axial force, shear force and bending moment at the key sections
| Section | N (kN) | V (kN) | M (kN·m) |
|---|---|---|---|
| AB at A | -91.94 | -25.55 | 0.00 |
| AB at B | -91.94 | -25.55 | -204.38 (left face) |
| BC at B | -49.82 | 81.39 | -204.38 (top) |
| BC at hinge (3.64 m from B) | -35.40 | 30.91 | 0.00 |
| BC at C | -20.98 | -19.57 | 20.63 (bottom) |
| CD at C | -13.06 | 25.55 | 20.63 (left face) |
| CD just above the 75 kN load | -13.06 | 25.55 | 148.36 (left face) |
| CD just below the 75 kN load | -13.06 | -49.45 | 148.36 (left face) |
| CD at D | -13.06 | -49.45 | 0.00 |

#### Shape of the diagrams
- AB: V constant, M linear
- BC: V linear, M parabolic
- CD: V constant on each side of the 75 kN load (it steps by 75 kN there), M linear on each side
- 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 = 204.38 kN·m at joint B (left face of AB, top of BC)
- V = 0 at 5.87 m from B along BC (5.64 m horizontally): M = 34.44 kN·m (bottom), the local maximum
- The moment on BC changes sign only at the internal hinge (3.64 m from B), where M = 0
- On CD the moment is 20.63 kN·m at C and 148.36 kN·m at the 75 kN load, falling to 0 at D (tension on the left, inner face)

## Check yourself

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

1. Why is a hinge assumed at the middle of BC?
   A. Because two hinged supports with no internal hinge give four reactions, which is one more than the three equilibrium equations, so the frame is indeterminate
   B. Because BC is inclined
   C. Because the 75 kN force is horizontal
   D. Because the UDL is on BC
   Answer: A. With a hinge the condition M = 0 at that point supplies the extra equation, making the three-hinged frame determinate. (Key point: Why assume a hinge)

2. The 15 kN/m acts per metre of horizontal projection of BC (7 m). What is the total load?
   A. 109.2 kN
   B. 15 kN
   C. 105 kN
   D. 75 kN
   Answer: C. 15 x 7 = 105 kN, using the horizontal projection. (Key point: Load from UDL)

3. Where is the maximum bending moment (181.67 kN·m) and which face is in tension?
   A. At C, bottom
   B. At B on column AB, tension on the left face
   C. At D, right face
   D. At the hinge, top
   Answer: B. The moment grows linearly along AB from zero at the hinge A and reaches its largest value at the knee B. (Key point: Maximum moment)

4. Under the UDL on BC, what shapes do V and M have?
   A. V constant, M linear
   B. V linear, M parabolic
   C. V parabolic, M cubic
   D. Both constant
   Answer: B. A UDL gives linear shear and parabolic moment. V = 0 at 5.42 m from B gives the local maximum of 21.94 kN·m. (Key point: Shape on BC)

## Related in Strength of Materials

- [Frame AFD, SFD, BMD (100 kN, inclined member)](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aframe-afd-sfd-bmd-100kn-20knm) — Strength of Materials, chapter 1
- [Frame AFD, SFD, BMD (inclined column, triangular)](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aframe-afd-sfd-bmd-30knm-20kn) — 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
- [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 (50 kN, 15 kN/m)](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aframe-afd-sfd-bmd-50kn-15knm) — Strength of Materials, chapter 1
- [Frame AFD, SFD, BMD (30 kN, 20 kN/m)](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aframe-afd-sfd-bmd-30kn-20knm) — Strength of Materials, chapter 1

## Related in other subjects

- [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
- [AFD, SFD, BMD of frame with 60 kN/m UDL](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Aframe-afd-sfd-bmd-60knm-20knm-12kn) — Applied Mechanics, chapter 6
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- [Moment distribution: multi-bay frame, AFD, SFD, BMD](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Amoment-dist-multi-column-frame-afd) — Theory of Structures II, chapter 3