---
title: "Stiffness matrix: BMD of frame, 60 kN"
subject: "Theory of Structures II"
chapter: 3
key: "semester_5/theory_of_structures_ii:stiffness-frame-60kn-a-fixed-d"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/stiffness-frame-60kn-a-fixed-d
site: BCT Engineering
---

# Stiffness matrix: BMD of frame, 60 kN

Theory of Structures II · Chapter 3: Displacement method

- Asked in: 2073 Shrawan, 10 marks
- Video: [Direct Stiffness Method - Frame -1](https://www.youtube.com/watch?v=TYAKMvDHlrI) — The Classroom
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/stiffness-frame-60kn-a-fixed-d
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-3

## Question

Using the stiffness matrix method, draw the bending moment diagram for the frame shown in figure below. Take constant EI. [Figure: frame ABCD; A fixed; beam ABC with 60 kN at B (2 m + 2 m); column CD 5 m high, D hinged.]

## Short answer

Stiffness matrix method: solving for rotations, final end moments are M_AC = -39.38 kNm, M_CA = 11.25 kNm, M_CD = -11.25 kNm, M_DC = 0; maximum sagging moment under the 60 kN load is 34.69 kNm.

## Answer

**Assumptions:** the beam ABC is a horizontal member of 4 m (2 m + 2 m) with the 60 kN load downward at B, its mid-point; A is fixed. Column CD is 5 m high, rigidly connected at C and hinged at D. EI is constant. Because A is fixed, the frame cannot sway. The unknown rotations are $\theta_C$ and $\theta_D$ (clockwise +).

Treat AC as one member (the load at B is a member load).

#### Fixed-end moments (clockwise +)
$FEM_{AC}=-\dfrac{PL}{8}=-\dfrac{60\times4}{8}=-30$ kNm, $FEM_{CA}=+30$ kNm. CD is unloaded.

#### Stiffness equations of the members
- $M_{AC}=\frac{2EI}{4}\left(\theta_{C}\right) - 30.00$
- $M_{CA}=\frac{2EI}{4}\left(2\theta_{C}\right) + 30.00$
- $M_{CD}=\frac{2EI}{5}\left(2\theta_{C} + \theta_{D}\right)$
- $M_{DC}=\frac{2EI}{5}\left(2\theta_{D} + \theta_{C}\right)$

#### Equilibrium
- Joint C: $M_{CA}+M_{CD}=0$
- Hinge D: $M_{DC}=0$

#### Stiffness matrix equation
$$
EI\begin{bmatrix}1.8000 & 0.4000 \\ 0.4000 & 0.8000\end{bmatrix}\begin{Bmatrix}\theta_{C} \\ \theta_{D}\end{Bmatrix}=\begin{Bmatrix}-30.000 \\ 0.000\end{Bmatrix}
$$
Here $K_{11}=\dfrac{4EI}{4}+\dfrac{4EI}{5}=1.8EI$, $K_{12}=K_{21}=\dfrac{2EI}{5}=0.4EI$, $K_{22}=\dfrac{4EI}{5}=0.8EI$.
$$
EI\theta_{C}=-18.7500,\quad EI\theta_{D}=9.3750
$$

#### Final end moments (kNm)
| End | $M$ (kNm) | End | $M$ (kNm) |
|---|---|---|---|
| AC | -39.38 | CA | 11.25 |
| CD | -11.25 | DC | 0.00 |

Checks: Joint C: $11.25+(-11.25)=0.00$ ✓; hinge $M_{DC}=0.00$ ✓.

#### Bending moment diagram
| Member | BM at first end | BM at second end | Max within member |
|---|---|---|---|
| AC | -39.38 | -11.25 | 34.69 at 2.00 m |
| CD | -11.25 | 0.00 | - |

Under the 60 kN load the sagging moment is 34.69 kNm. BM sign: sagging for the beam; for the column, tension on the right when travelling from C to D.

Reactions: A: $H=2.25$, $V=37.03$, $M=39.38$; D: $H=-2.25$, $V=22.97$, $M=0.00$.

**Answer: $M_{AC}=-39.38$ kNm, $M_{CA}=11.25$ kNm, $M_{CD}=-11.25$ kNm, $M_{DC}=0$; BM under the load = 34.69 kNm.**

## Check yourself

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

1. Why are only theta_C and theta_D unknowns, with no sway?
   A. D is fixed
   B. The load is symmetrical
   C. Axial loads cause sway
   D. A is fixed so the frame cannot sway; D is hinged so its rotation is free
   Answer: D. No sway is possible, and the hinge at D lets it rotate, giving two rotations. (Key point: Unknowns)

2. What is the equation for the hinge at D?
   A. M_DC = 0
   B. M_CD = 0
   C. theta_D = 0
   D. M_AC = 0
   Answer: A. A hinge cannot carry a moment, so M_DC = 0, giving a second equilibrium equation. (Key point: Hinge condition)

3. K11 = 1.8EI comes from?
   A. 12EI/L^3
   B. 4EI/4 (AC) + 4EI/5 (CD)
   C. 2EI/4 + 2EI/5
   D. 4EI/9
   Answer: B. At joint C: 4EI/L of AC (4 m) plus 4EI/L of CD (5 m). (Key point: K11)

4. Final moment M_CA and M_CD?
   A. 39.38 and 11.25
   B. 0 and 0
   C. +11.25 and -11.25 kNm
   D. -11.25 and +11.25
   Answer: C. Joint C equilibrium needs M_CA + M_CD = 0; solution gives 11.25 and -11.25. (Key point: Result)

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