---
title: "AFD, SFD, BMD of frame with UDL and triangular load"
subject: "Applied Mechanics"
chapter: 6
key: "semester_1/applied_mechanics:frame-afd-sfd-bmd-10knm-5kn-triangular"
source: https://www.bctengineer.com/q/semester_1/applied_mechanics/frame-afd-sfd-bmd-10knm-5kn-triangular
site: BCT Engineering
---

# AFD, SFD, BMD of frame with UDL and triangular load

Applied Mechanics · Chapter 6: Analysis of Beams and Frames

- Asked in: 2068 Shrawan, 12 marks
- Video: [Frame 2 Analysis// Draw AFD SFD and BMD// Assign Inclined Loads](https://www.youtube.com/watch?v=tqUsUQSGFM4) — Engineer's Home
- 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-10knm-5kn-triangular
- Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-6

## Question

Draw bending moment diagram, shear force diagram and axial force diagram for the given figure below. And also indicate the salient points if any. [Figure: frame with fixed/pin support at A; column A-B-C of 4 m total height with 5 kN horizontal load at B 2 m above A; top beam C-D of 4 m carrying 10 kN/m UDL; right column D-E of 2 m with a triangular load.]

## Short answer

Reactions: Ax=5 kN right, Ay=40 kN up, MA=56.67 kN·m anticlockwise. Maximum bending moment 86.67 kN·m at knee C (tension outside). Beam CD moment varies parabolically from 86.67 to 6.67 kN·m; DE moment cubic from 6.67 to 0. Axial forces: AC 40 kN compression, CD 10 kN compression, DE zero.

## Answer

#### Reading of the figure
- Origin at A: A(0,0) fixed support (the column ends in a ground hatch with no free end), B(0,2), C(0,4), D(4,4), E(4,2). Column AC is 4 m, beam CD 4 m, and the right column DE 2 m with a free end at E.
- Loads: 5 kN → at B; UDL 10 kN/m on CD (40 kN, 2 m from C); triangular load on DE acting to the left, 10 kN/m at D and zero at E (total 10 kN, acting 0.67 m below D).
- E has no support, so the frame is a cantilever frame fixed at A: r = 3, determinate.

#### Reactions
```text
 10 kN/m
 C================D
 |                 |<- 10 kN/m
 |                 | triangular
 B→5kN             E (free)
 |
 A (fixed)
```
- ΣFx: Aₓ + 5 − 10 = 0, so **Aₓ = 5.00 kN (→)**.
- ΣFy: A_y − 40 = 0, so **A_y = 40.00 kN (↑)**.
- ΣM_A: M_A − 5(2) − 40(2) + 10(3.33) = 0, so **M_A = 56.67 kN·m (anticlockwise)**.

#### 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, 2.00 m): N = 40.00 kN (compression); V = −5.00 to −5.00 kN; M = 56.67 (left) to 66.67 (left) kN·m, linear.
- **BC** (B to C, 2.00 m): N = 40.00 kN (compression); V = −10.00 to −10.00 kN; M = 66.67 (left) to 86.67 (left) kN·m, linear.
- **CD** (C to D, 4.00 m): N = 10.00 kN (compression); V = 40.00 to 0.00 kN; M = 86.67 (top) to 6.67 (top) kN·m, parabolic.
- **DE** (D to E, 2.00 m): N = 0.00 kN (zero); V = 10.00 to 0.00 kN; M = 6.67 (right) to 0.00 (–) kN·m, cubic.

#### Salient features
- Maximum bending moment = **86.67 kN·m** at the knee C (tension on the outside, i.e. the left face of the column and the top of the beam). The moment at the fixed base is 56.67 kN·m.
- The shear in the beam falls linearly from 40.00 kN at C to zero at the corner D (no vertical load acts on DE), so the beam moment varies as a parabola from 86.67 kN·m at C to 6.67 kN·m at D, with zero slope at D.
- DE acts as a cantilever under the triangular load: the moment falls from 6.67 kN·m at D to zero at E along a cubic curve, and the shear falls from 10 kN at D to zero at E.
- Axial force: AC 40.00 kN compression; CD 10.00 kN compression (equal to the 10 kN resultant of the triangular load); DE zero.

## Check yourself

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

1. Why is this frame determinate?
   A. End E is free, so the frame is a cantilever fixed at A with r = 3
   B. It has an internal hinge
   C. It has a roller at E
   D. It has more members than reactions
   Answer: A. E carries no support, so only the fixed support at A gives r = 3 reaction components, which equilibrium can solve. (Key point: Why determinate)

2. What is Ay and why?
   A. 10 kN, from the triangular load
   B. 40 kN up, balancing the 40 kN UDL on CD
   C. 5 kN, from the horizontal load
   D. 56.67 kN, the moment reaction
   Answer: B. ΣFy: the only vertical load is the UDL 10 x 4 = 40 kN, so Ay = 40 kN upward. (Key point: Vertical reaction)

3. Where is the maximum bending moment of 86.67 kN·m?
   A. At the fixed base A
   B. At the free end E
   C. At the knee C
   D. At the middle of the beam
   Answer: C. The moment grows up the column from 56.67 kN·m at A to 86.67 kN·m at C, then drops along the beam, so the maximum is at the knee C. (Key point: Maximum bending moment)

4. What are the axial forces in CD and DE?
   A. CD zero, DE 10 kN compression
   B. CD 40 kN, DE 10 kN
   C. Both 10 kN compression
   D. CD 10 kN compression, DE zero
   Answer: D. The triangular load (10 kN) acts horizontally, so it compresses the beam CD. DE is loaded transversely, so its axial force is zero. (Key point: Axial forces)

## Related in Applied Mechanics

- [AFD, SFD, BMD of frame with triangular load](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Aframe-afd-sfd-bmd-30knm-10knm-triangular-hinge) — Applied Mechanics, chapter 6, asked 2×
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- [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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## Related in other subjects

- [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 (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
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