---
title: "AFD, SFD, BMD of frame with hinge at D"
subject: "Applied Mechanics"
chapter: 6
key: "semester_1/applied_mechanics:frame-afd-sfd-bmd-8knm-hinge-50kn"
source: https://www.bctengineer.com/q/semester_1/applied_mechanics/frame-afd-sfd-bmd-8knm-hinge-50kn
site: BCT Engineering
---

# AFD, SFD, BMD of frame with hinge at D

Applied Mechanics · Chapter 6: Analysis of Beams and Frames

- Asked in: 2071 Chaitra, 14 marks
- Video: [Frame Bending Moment Diagram Prob. No. 1](https://www.youtube.com/watch?v=Lhbuvw9xeXY) — Dr. Rajeeth T J
- 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-8knm-hinge-50kn
- Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-6

## Question

A frame is loaded as shown in figure below. Draw the AFD, SFD and BMD and also show the salient features of each diagram. [Figure: frame with pin supports at A and G; left column A-B-C with 2 m + 2 m, 50 kN horizontal load and 20 kN·m moment at B; top beam C-D-E-F with 8 kN/m UDL on C-D, internal hinge at D, 9 kN downward load at E; horizontal dimensions 2 m, 1 m, 1 m.]

## Short answer

Using equilibrium and hinge condition at D, reactions are Ay = 5.75 kN↓, Ax = 36.88 kN←, Gy = 30.75 kN↑, Gx = 13.13 kN←. Maximum bending moment is 73.75 kN·m at B in the left column (inside tension). The 20 kN·m couple causes a 20 kN·m jump in column moment, and moment is zero at hinge D.

## Answer

#### Reading of the figure
- Origin at A: A(0,0) pin, B(0,2), C(0,4); beam C(0,4), D(2,4) internal hinge, E(3,4), F(4,4); right column FG down to the pin G(4,0).
- Loads: 50 kN → and a couple of 20 kN·m (taken anticlockwise) at B; UDL 8 kN/m on CD (16 kN, 1 m from C); 9 kN ↓ at E.
- r = 4 and a hinge at D (M_D = 0): determinate.

#### Reactions
```text
 8 kN/m (hinge)   9kN
 C=====D===E=====F
 |                |
 B 50kN→, 20kNm   |
 |                |
 A (pin)          G (pin)
```
- ΣFx: Aₓ + Gₓ + 50 = 0.  ΣFy: A_y + G_y − 16 − 9 = 0.
- ΣM_A: 4G_y + 20 − 50(2) − 16(1) − 9(3) = 0, so **G_y = 30.75 kN (↑)** and **A_y = 5.75 kN (↓)**.
- Hinge D (left part): 4Aₓ − 2A_y + 50(2) + 20 + 16(1) = 0, so **Aₓ = 36.88 kN (←)** and **Gₓ = 13.13 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 = 5.75 kN (tension); V = 36.88 to 36.88 kN; M = 0.00 (–) to 73.75 (right) kN·m, linear.
- **BC** (B to C, 2.00 m): N = 5.75 kN (tension); V = −13.13 to −13.13 kN; M = 53.75 (right) to 27.50 (right) kN·m, linear.
- **CD** (C to D, 2.00 m): N = 13.13 kN (compression); V = −5.75 to −21.75 kN; M = 27.50 (bottom) to 0.00 (–) kN·m, parabolic.
- **DE** (D to E, 1.00 m): N = 13.13 kN (compression); V = −21.75 to −21.75 kN; M = 0.00 (–) to 21.75 (top) kN·m, linear.
- **EF** (E to F, 1.00 m): N = 13.13 kN (compression); V = −30.75 to −30.75 kN; M = 21.75 (top) to 52.50 (top) kN·m, linear.
- **FG** (F to G, 4.00 m): N = 30.75 kN (compression); V = 13.13 to 13.13 kN; M = 52.50 (right) to 0.00 (–) kN·m, linear.

#### Salient features
- The maximum moment is **73.75 kN·m** at B in the left column (tension on the inside); at C it is 27.50 kN·m, and it is zero at the hinge D.
- The moment in the right column is largest at F (52.50 kN·m, tension on the outside) and zero at G. The beam portion DF has a hogging moment (52.50 kN·m at F).
- The 20 kN·m couple at B causes a step of 20 kN·m in the column BMD at B (73.75 below, 53.75 above).
- The shear in AB (36.88 kN) is the largest in the frame; the vertical reaction at A is downward, so the left column is in tension.

## Check yourself

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

1. How is G_y found?
   A. From ΣFx = 0
   B. Moments about A, since A_x and A_y pass through A
   C. From the hinge D
   D. From symmetry
   Answer: B. ΣM_A involves only G_y (horizontal G_x passes through the line of A at the same level). (Key point: Reaction G_y)

2. How does the 20 kN·m couple at B show up in the BMD?
   A. A step of 20 kN·m (73.75 below, 53.75 above)
   B. A step in the shear diagram
   C. A change in axial force
   D. No change
   Answer: A. A couple gives a jump in moment only. (Key point: Couple at B)

3. What is the moment at hinge D?
   A. 27.50 kN·m
   B. 21.75 kN·m
   C. Zero
   D. 52.50 kN·m
   Answer: C. A hinge cannot carry moment. (Key point: Hinge at D)

4. Where is the maximum moment?
   A. At B in the left column, 73.75 kN·m
   B. At F, 52.50 kN·m
   C. At C, 27.50 kN·m
   D. At D
   Answer: A. Shear in AB is 36.88 kN, giving M = 36.88 x 2 = 73.75 kN·m at B. (Key point: Maximum moment)

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