---
title: "Frame AFD, SFD, BMD (100 kN, inclined member)"
subject: "Strength of Materials"
chapter: 1
key: "semester_3/strength_of_materials:frame-afd-sfd-bmd-100kn-20knm"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/frame-afd-sfd-bmd-100kn-20knm
site: BCT Engineering
---

# Frame AFD, SFD, BMD (100 kN, inclined member)

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

- Asked in: 2076 Asoj, 12 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-100kn-20knm
- 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 a given loaded frame. Also write the salient features. [Figure: hinged support at A at the lower left; inclined member A-B rising to B (slope 3 vertical to 4 horizontal; horizontal projection 2 m, rise 2 m shown) and carrying a 20 kN/m distributed load along it; horizontal member B-C-D with a 100 kN point load at C; B to C 2 m, C to D 2 m; D is the right end, supported as drawn.]

## Short answer

Support reactions: HA=0, VA=80.47 kN upward, VD=76.09 kN upward. Axial, shear and bending moment diagrams: AB parabolic, BC and CD linear. Maximum bending moment 152.19 kN·m at C on BC, tension on bottom side.

## Answer

**Assumptions (taken from the figure):**
- A(0,0) hinged; AB inclined, horizontal projection 2 m and rise 2 m, so B = (2, 2); BC = CD = 2 m horizontal.
- 20 kN/m acts vertically downward on the full inclined length AB (2.83 m, total 56.57 kN); 100 kN downward at C.
- D is taken as a roller (vertical reaction only), which makes the frame determinate.

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
            ///
          ///
        ///
       //
     //
   //
 //
A
```

#### Support reactions
Equilibrium of the whole frame gives $\Sigma F_x = 0$, $\Sigma F_y = 0$ and $\Sigma M = 0$.

- $H_A$ = 0.00 kN ←
- $V_A$ = 80.47 kN ↑
- $V_D$ = 76.09 kN ↑

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 (kN) | V (kN) | M (kN·m) |
|---|---|---|---|
| AB at A | -56.90 | 56.90 | 0.00 |
| AB at B | -16.90 | 16.90 | 104.38 (right face) |
| BC at B | 0.00 | 23.91 | 104.38 (bottom) |
| BC at C | 0.00 | 23.91 | 152.19 (bottom) |
| CD at C | 0.00 | -76.09 | 152.19 (bottom) |
| CD at D | 0.00 | -76.09 | 0.00 |

#### Shape of the diagrams
- AB: V linear, M parabolic
- BC: V constant (steps at point loads), M linear
- CD: V constant (steps at point loads), M linear
- 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 = 152.19 kN·m at C (on BC), tension on the bottom

## Check yourself

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

1. Why is D taken as a roller?
   A. So the frame has three reactions and is statically determinate
   B. Because rollers carry the 100 kN load
   C. Because D must have a fixed moment
   D. Because AB is inclined
   Answer: A. Hinge A (2 reactions) plus a roller at D (1 reaction) gives three reactions, solvable by the three equilibrium equations. (Key point: Support assumption)

2. Why is H_A = 0?
   A. Because A is fixed
   B. Because the roller at D resists all horizontal force
   C. Because the loads are all vertical and D has only a vertical reaction, so sum Fx = 0 gives H_A = 0
   D. Because the inclined member cancels it
   Answer: C. No load or reaction other than H_A acts horizontally, so H_A must be zero. (Key point: Horizontal reaction)

3. The 20 kN/m acts on the full inclined length AB. What is its total?
   A. 40 kN
   B. 56.57 kN
   C. 80 kN
   D. 20 kN
   Answer: B. Length AB = sqrt(2^2 + 2^2) = 2.83 m, so the load is 20 x 2.83 = 56.57 kN. (Key point: Load on inclined member)

4. Where is the maximum bending moment (152.19 kN·m) in this frame?
   A. At B
   B. At A
   C. At C under the 100 kN load, tension at the bottom
   D. At D
   Answer: C. BC and CD have constant shear with a step at C, so the linear BMD peaks under the point load. (Key point: Maximum moment)

## 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 (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
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- [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
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- [Frame AFD, SFD, BMD (15 kN/m, 75 kN)](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aframe-afd-sfd-bmd-15knm-75kn) — Strength of Materials, chapter 1

## Related in other subjects

- [AFD, SFD, BMD of frame with 40 kN/m UDL](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Aframe-afd-sfd-bmd-40kn-frame) — Applied Mechanics, chapter 6
- [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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