---
title: "AFD, SFD, BMD of frame with triangular load"
subject: "Applied Mechanics"
chapter: 6
key: "semester_1/applied_mechanics:frame-afd-sfd-bmd-30knm-10knm-triangular-hinge"
source: https://www.bctengineer.com/q/semester_1/applied_mechanics/frame-afd-sfd-bmd-30knm-10knm-triangular-hinge
site: BCT Engineering
---

# AFD, SFD, BMD of frame with triangular load

Applied Mechanics · Chapter 6: Analysis of Beams and Frames

- Asked in: 2070 Chaitra, 13 marks
- Video: [2 1, Internal Actions + AFD SFD BMD 8;19](https://www.youtube.com/watch?v=DD8Dn3xwPUg) — Spoon Feed Me
- 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-30knm-10knm-triangular-hinge
- Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-6

## Question

Draw axial force, shear force and bending moment diagram for the loaded frame as shown in figure below. Also indicate the salient features if any. [Figure: frame with fixed/pin support at A and roller at E; left column A-B with 5 kN/m UDL; 30 kN·m moment at B; top beam B-C of 6 m with triangular load up to 10 kN/m; hinge at C; 20 kN horizontal load at D; right column C-D-E with 4 m + 4 m.]

## Short answer

With pin supports at A and E, reactions are Ax = 50 kN←, Ay = 35 kN↓, Ex = 10 kN←, Ey = 65 kN↑. Maximum moment 270 kN·m just right of B (bottom tension) equals column moment 240 kN·m plus the 30 kN·m couple. Zero moment at hinge C, and right column moment peaks at 40 kN·m at D.

## Answer

#### Reading of the figure and assumptions
- Origin at A: A(0,0), left column AB 8 m up to B(0,8); beam BC 6 m to C(6,8) with an internal hinge at C; right column CDE down 8 m, D(6,4), E(6,0).
- Loads: UDL 5 kN/m acting to the right on AB (40 kN, at 4 m above A); clockwise couple 30 kN·m at B; triangular load on BC, 0 at B rising to 10 kN/m at C (30 kN, 4 m from B); 20 kN → at D.
- Supports are drawn as ground hatching at A and a small circle at E. A roller at E gives only a vertical reaction, and its line of action passes through the hinge C directly above it, so the column CDE could not resist the 20 kN horizontal load at D (it would rotate about C), even with a fixed base at A. For the frame to be stable and determinate, A and E are therefore taken as **pin supports** (r = 4, with one hinge condition).

#### Reactions
```text
 triangular 10kN/m  (hinge)
 B================C
 |^30kNm          |
 | 5kN/m→         |
 |                D→20kN
 |                |
 A (pin)          E (pin)
```
- Hinge C (right part CDE, M_C = 0): 8Eₓ + 20(4) = 0, so **Eₓ = 10.00 kN (←)**.
- ΣFx: Aₓ + Eₓ + 40 + 20 = 0, so **Aₓ = 50.00 kN (←)**.
- ΣM_A: 6E_y − 40(4) − 30(4) − 30 − 20(4) = 0, so **E_y = 65.00 kN (↑)**.
- ΣFy: A_y + E_y − 30 = 0, so **A_y = 35.00 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, 8.00 m): N = 35.00 kN (tension); V = 50.00 to 10.00 kN; M = 0.00 (–) to 240.00 (right) kN·m, parabolic.
- **BC** (B to C, 6.00 m): N = 10.00 kN (tension); V = −35.00 to −65.00 kN; M = 270.00 (bottom) to 0.00 (–) kN·m, cubic.
- **CD** (C to D, 4.00 m): N = 65.00 kN (compression); V = −10.00 to −10.00 kN; M = 0.00 (–) to 40.00 (right) kN·m, linear.
- **DE** (D to E, 4.00 m): N = 65.00 kN (compression); V = 10.00 to 10.00 kN; M = 40.00 (right) to 0.00 (–) kN·m, linear.

#### Salient features
- The maximum bending moment is **270.00 kN·m**, just to the right of B (tension at the bottom of the beam, i.e. inside). It equals the column moment at the top of AB (240.00 kN·m) plus the 30 kN·m couple. The beam moment falls to zero at the hinge C.
- Left column: the moment is 240.00 kN·m at B (tension on the right, i.e. inside); the 30 kN·m couple at B makes it jump to 270.00 kN·m in the beam at the joint.
- Right column CDE: the moment is zero at C and E, with a maximum of 40.00 kN·m at D (tension on the right, i.e. the outside face); the shear is a constant 10 kN, because the 20 kN load is shared equally by the hinge C and the pin E.
- The shear in AB is 50.00 kN at A and drops to 10.00 kN at B; the shear in the beam does not change sign, so there is no extra peak in BC.
- The vertical reaction at A is downward (35.00 kN), so AB is in tension (35.00 kN).

## Check yourself

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

1. Why are A and E taken as pin supports in this frame?
   A. A fixed base with a roller at E would make the column CDE unstable under the horizontal load
   B. Pins always give smaller bending moments
   C. The figure shows pins explicitly
   D. A fixed support cannot be used with an internal hinge
   Answer: A. For the frame with one hinge to be determinate and stable under the horizontal 20 kN load, A and E must give r = 4 (one hinge condition), so both are taken as pins. (Key point: Support assumption)

2. How is Ex = 10 kN found?
   A. Sum of vertical forces for the right part
   B. Sum of moments about A for the whole frame
   C. M_C = 0 for the right part CDE: 8Ex + 20(4) = 0
   D. Sum of horizontal forces for the whole frame
   Answer: C. The moment at the hinge C is zero, so taking moments about C for the right part CDE directly gives Ex. (Key point: Hinge equation for Ex)

3. Why is the maximum moment 270 kN·m just right of B?
   A. The triangular load peaks there
   B. The column moment of 240 kN·m at B plus the 30 kN·m couple
   C. The shear is zero there
   D. The hinge moment is added to the column moment
   Answer: B. The couple at B adds a jump of 30 kN·m to the 240 kN·m column moment, giving 270 kN·m in the beam at the joint. (Key point: Maximum bending moment)

4. Ay is 35 kN downward. What is the axial force in column AB?
   A. 35 kN compression
   B. 50 kN tension
   C. 35 kN tension
   D. Zero
   Answer: C. The downward vertical reaction at A means the column is pulled, so AB carries 35 kN tension. (Key point: Axial force in AB)

## Close variants set in other exams

- [AFD, SFD, BMD of frame with hinge at C](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Aframe-afd-sfd-bmd-hinge-5knm-10knm-triangular) — Applied Mechanics, chapter 6, asked 2×

## 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-10kn-per-m-triangular) — Applied Mechanics, chapter 6
- [AFD, SFD, BMD of frame with UDL and triangular load](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Aframe-afd-sfd-bmd-10knm-5kn-triangular) — Applied Mechanics, chapter 6
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