---
title: "AFD, SFD, BMD of frame with hinge"
subject: "Applied Mechanics"
chapter: 6
key: "semester_1/applied_mechanics:frame-afd-sfd-bmd-hinge-4knm-20knm"
source: https://www.bctengineer.com/q/semester_1/applied_mechanics/frame-afd-sfd-bmd-hinge-4knm-20knm
site: BCT Engineering
---

# AFD, SFD, BMD of frame with hinge

Applied Mechanics · Chapter 6: Analysis of Beams and Frames

- Asked in: 2072 Chaitra, 14 marks
- Video: [Three hinged frames, problem 001 [Structural Mechanics]](https://www.youtube.com/watch?v=73B-iN7mcWg) — Engineering Handbooks
- Page: https://www.bctengineer.com/q/semester_1/applied_mechanics/frame-afd-sfd-bmd-hinge-4knm-20knm
- Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-6

## Question

Draw the Axial Force, Shear Force and Bending Moment diagram for the given frame shown in figure below. Also show the salient features. [Figure: frame with pin support at A and roller/pin support at G; left column A-B-C with 10 kN horizontal load at B, heights 2 m and 2 m; top beam with 4 kN/m UDL and a hinge, 15 kN downward load at E, 20 kN·m moment at F, 20 kN horizontal load at H; horizontal dimensions 2 m, 2 m, 1 m, 1 m, 2 m; right column height 3 m.]

## Short answer

Reactions: A_x = 6.71 kN ←, A_y = 0.57 kN ↑, G_x = 3.29 kN ←, G_y = 42.43 kN ↑. Maximum negative moment is 40.00 kN·m at F. Moment is zero at hinge D and free end H. Maximum positive moment in CD is 6.90 kN·m at 0.14 m from C.

## Answer

#### Reading of the figure
- Origin at A, x right, y up: A(0,0) pin, B(0,2), C(0,4), D(2,4) internal hinge, E(3,4), F(4,4), H(6,4); right column FG goes down 3 m to the pin G(4,1).
- Loads: 10 kN → at B; UDL 4 kN/m on CD (8 kN at 1 m from C); 15 kN ↓ at E; couple 20 kN·m at F (taken anticlockwise); 20 kN ↓ at the overhang end H.
- r = 4 reactions, 3 equilibrium equations + 1 hinge condition (M_D = 0): statically determinate.

#### Reactions
```text
 4kN/m        15kN  20kNm    20kN
 C=====D(hinge)==E==F=====H ↓
 |                  |
 B→10kN             |
 |                  G (pin)
 A (pin)
```
- ΣFx: Aₓ + Gₓ + 10 = 0.  ΣFy: A_y + G_y − 8 − 15 − 20 = 0.
- ΣM_A: 4G_y − 1·Gₓ − 10(2) − 8(1) − 15(3) − 20(6) + 20 = 0, i.e. 4G_y − Gₓ = 173.
- Hinge D (left part, M_D = 0): 4Aₓ − 2A_y + 10(2) + 8(1) = 0.
- Solving: **Aₓ = 6.71 kN (←), A_y = 0.57 kN (↑), Gₓ = 3.29 kN (←), G_y = 42.43 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, 2.00 m): N = 0.57 kN (compression); V = 6.71 to 6.71 kN; M = 0.00 (–) to 13.43 (right) kN·m, linear.
- **BC** (B to C, 2.00 m): N = 0.57 kN (compression); V = −3.29 to −3.29 kN; M = 13.43 (right) to 6.86 (right) kN·m, linear.
- **CD** (C to D, 2.00 m): N = 3.29 kN (compression); V = 0.57 to −7.43 kN; M = 6.86 (bottom) to 0.00 (–) kN·m, parabolic. V = 0 at 0.14 m from C, where M = 6.90 kN·m.
- **DE** (D to E, 1.00 m): N = 3.29 kN (compression); V = −7.43 to −7.43 kN; M = 0.00 (–) to 7.43 (top) kN·m, linear.
- **EF** (E to F, 1.00 m): N = 3.29 kN (compression); V = −22.43 to −22.43 kN; M = 7.43 (top) to 29.86 (top) kN·m, linear.
- **FH** (F to H, 2.00 m): N = 0.00 kN (zero); V = 20.00 to 20.00 kN; M = 40.00 (top) to 0.00 (–) kN·m, linear.
- **FG** (F to G, 3.00 m): N = 42.43 kN (compression); V = 3.29 to 3.29 kN; M = 9.86 (right) to 0.00 (–) kN·m, linear.

#### Salient features
- Bending moment is zero at the hinge D and at the free end H. Maximum negative moment in the beam is **40.00 kN·m** (tension on top) at F, due to the 20 kN overhang load.
- Maximum positive moment in CD is 6.90 kN·m at 0.14 m from C (V = 0 there); moment in the column is at most 13.43 kN·m at B.
- The beam from D to F has hogging moment (tension on top). Joint F check: 29.86 (EF) + 20 (couple) = 40.00 (FH) + 9.86 (FG).
- Axial force is largest in the right column FG (42.43 kN compression); the left column carries only 0.57 kN compression.

## Check yourself

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

1. What extra equation is used with the hinge at D?
   A. ΣFx = 0 on the right part
   B. M_D = 0 for the left part
   C. ΣFy = 0 at C
   D. M_A = 0 again
   Answer: B. r = 4 needs a fourth equation, given by M = 0 at the hinge. (Key point: Extra equation)

2. Why is the moment at F maximum negative at 40 kN·m?
   A. The 20 kN overhang load acts 2 m from F
   B. The UDL acts at F
   C. The hinge is at F
   D. The couple alone gives 40
   Answer: A. 20 kN x 2 m = 40 kN·m hogging at F. (Key point: Overhang moment)

3. What is the significance of V = 0 at 0.14 m from C in CD?
   A. The moment is zero there
   B. Axial force is zero there
   C. The sagging moment is maximum there
   D. The hinge is located there
   Answer: C. Zero shear marks a maximum of M (6.90 kN·m). (Key point: Zero shear in CD)

4. Which statement of equilibrium at F is correct?
   A. 29.86 = 40.00 + 9.86
   B. 29.86 + 40 = 20 + 9.86
   C. 29.86 = 9.86
   D. 29.86 + 20 = 40.00 + 9.86
   Answer: D. The couple at F is included in the joint moment balance. (Key point: Joint F check)

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