---
title: "AFD, SFD, BMD of frame with 60 kN/m UDL"
subject: "Applied Mechanics"
chapter: 6
key: "semester_1/applied_mechanics:frame-afd-sfd-bmd-60knm-hinge"
source: https://www.bctengineer.com/q/semester_1/applied_mechanics/frame-afd-sfd-bmd-60knm-hinge
site: BCT Engineering
---

# AFD, SFD, BMD of frame with 60 kN/m UDL

Applied Mechanics · Chapter 6: Analysis of Beams and Frames

- Asked in: 2070 Asar, 13 marks
- Video: [SFD and BMD for frame](https://www.youtube.com/watch?v=S9W7_Nv2eVg) — Vise Vision
- 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-60knm-hinge
- Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-6

## Question

Draw the Axial Force, Shear Force and Bending Moment diagram of the given frame. Also show the salient features if any. [Figure: frame with pin supports at A and E; column A-B-C with 5 m + 5 m, 10 kN at 30° at the upper part and 40 kN·m moment at B; top beam C-D of 8 m with 60 kN/m UDL and an internal hinge; column D-E of 5 m.]

## Short answer

Using equilibrium and hinge condition at C, reactions are Ax = 1.50 kN→, Ay = 244.60 kN↑, Ex = 6.50 kN←, Ey = 244.06 kN↑. The maximum bending moment is 463.89 kN·m at 3.93 m from C on beam CD (bottom tension). The 40 kN·m couple at B causes a 40 kN·m jump in column moment; moment is zero at hinge C.

## Answer

#### Reading of the figure
- Origin at A: A(0,0) pin, B(0,5), C(0,10) internal hinge, D(8,10), E(8,5) pin. Column AC 10 m, beam CD 8 m, column DE 5 m.
- Loads: 10 kN at B at 30° to the vertical, pointing down-right: 5.00 kN → and 8.66 kN ↓; clockwise couple 40 kN·m at B; UDL 60 kN/m on CD (480 kN, 4 m from C).
- r = 4 and a hinge at C: determinate.

#### Reactions
```text
       60 kN/m
  C(hinge)==============D
  |                      |
  |\10kN 30°             |
  B (40kNm, clockwise)   E (pin)
  |
  A (pin)
```
- Hinge C (left part, M_C = 0): 10Aₓ + 5(5) − 40 = 0, so **Aₓ = 1.50 kN (→)**.
- ΣFx: Aₓ + Eₓ + 5 = 0, so **Eₓ = 6.50 kN (←)**.
- ΣM_A: 8E_y − 5Eₓ − 5(5) − 40 − 480(4) = 0, so **E_y = 244.06 kN (↑)**.
- ΣFy: A_y + E_y − 480 − 8.66 = 0, so **A_y = 244.60 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, 5.00 m): N = 244.60 kN (compression); V = −1.50 to −1.50 kN; M = 0.00 (–) to 7.50 (left) kN·m, linear.
- **BC** (B to C, 5.00 m): N = 235.94 kN (compression); V = −6.50 to −6.50 kN; M = 32.50 (right) to 0.00 (–) kN·m, linear.
- **CD** (C to D, 8.00 m): N = 6.50 kN (compression); V = 235.94 to −244.06 kN; M = 0.00 (–) to 32.50 (top) kN·m, parabolic. V = 0 at 3.93 m from C, where M = 463.89 kN·m. M changes sign (contraflexure) at 7.86 m from C.
- **DE** (D to E, 5.00 m): N = 244.06 kN (compression); V = 6.50 to 6.50 kN; M = 32.50 (right) to 0.00 (–) kN·m, linear.

#### Salient features
- The beam behaves almost as a simply supported beam: maximum moment = **463.89 kN·m** (tension at the bottom) at 3.93 m from C, where V = 0. At D the moment is −32.50 kN·m (small hogging, tension on the outside of the corner).
- The columns carry very large axial compression (244.60 kN in AB and 244.06 kN in DE) but small bending moments: at most 32.50 kN·m at B.
- The 40 kN·m couple makes the column BMD jump by 40 kN·m at B (7.50 kN·m below B and 32.50 kN·m above, with opposite sides in tension).
- Both horizontal reactions are small (1.50 kN and 6.50 kN).

## Check yourself

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

1. In this frame, which equation is used to find the horizontal reaction Ax at A?
   A. Sum of moments about the hinge C for the left part (M_C = 0)
   B. Sum of vertical forces for the whole frame
   C. Sum of moments about E for the whole frame
   D. Sum of moments about the hinge C for the whole frame, ignoring the hinge
   Answer: A. An internal hinge cannot carry moment, so M_C = 0 for the left part gives an extra equation: 10Ax + 5(5) - 40 = 0, so Ax = 1.50 kN. (Key point: Hinge condition)

2. What does the 40 kN·m couple at B do to the bending moment diagram of the column?
   A. It changes the axial force in the column
   B. It produces a sudden jump of 40 kN·m in the BMD at B
   C. It changes the shear force in the beam CD
   D. It has no effect on the BMD
   Answer: B. An applied couple causes a sudden jump in the BMD equal to its value: 7.50 kN·m below B and 32.50 kN·m above B, on opposite tension sides. (Key point: Effect of the 40 kN·m couple)

3. Where does the maximum bending moment of 463.89 kN·m occur?
   A. At the corner D
   B. At the hinge C
   C. At the fixed-type joint B
   D. In beam CD at 3.93 m from C, where shear is zero (tension at bottom)
   Answer: D. Bending moment is maximum where the shear force is zero. In CD this is 3.93 m from C, and the beam acts almost like a simply supported beam. (Key point: Maximum bending moment)

4. Why do columns AB and DE carry very large axial forces (about 244 kN)?
   A. The vertical reactions at the supports carry almost the whole 480 kN UDL, so the columns are in compression
   B. The 40 kN·m couple produces axial force
   C. The horizontal reactions are very large
   D. The hinge at C transfers the load as tension
   Answer: A. The vertical reactions (Ay = 244.60 kN, Ey = 244.06 kN) carry the UDL and travel down the columns as compression, while the horizontal reactions are small. (Key point: Axial force in columns)

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