Stiffness matrix: frame, BMD

Assumptions (figure partly illegible): A(0,0), B(0,6), C(5,6), D(5,0) with both columns 6 m high and EI constant for all members. BC = 5 m carries 2 kN/m downward; 10 kN acts horizontally (to the right) at B; the 20 kNm moment at C is clockwise; a second 10 kN horizontal force acts to the right on CD at E, 2 m above D. A and D are fixed.

Coordinates

1 = $\theta_B$, 2 = $\theta_C$ (clockwise +), 3 = $\Delta$ (sway to the right). $\psi_{AB}=\psi_{CD}=\Delta/6$.

Fixed-end moments (clockwise +)

• BC: $\mp\dfrac{2\times5^2}{12}=\mp4.167$ kNm
• CD (10 kN at $a=4$ m from C, $b=2$ m from D, horizontal): $FEM_{CD}=4.44$, $FEM_{DC}=-8.89$ kNm ($Pab^2/L^2=10\times4\times4/36$, $Pa^2b/L^2=10\times16\times2/36$, sign for a load to the right on a column running downward)

Stiffness matrix of the frame

Member stiffnesses: $4EI/6=0.6667EI$, $2EI/6=0.3333EI$, $6EI/6^2=0.1667EI$, $12EI/6^3=0.0556EI$; beam $4EI/5=0.8EI$, $2EI/5=0.4EI$.

• $K_{11}=0.6667+0.8=1.4667EI$, $K_{12}=0.4EI$, $K_{22}=0.8+0.6667=1.4667EI$
• $K_{13}=K_{23}=-0.1667EI$, $K_{33}=2\times0.0556EI=0.1111EI$

$$[K]=EI\begin{bmatrix}1.4667 & 0.4000 & -0.1667 \\ 0.4000 & 1.4667 & -0.1667 \\ -0.1667 & -0.1667 & 0.1111\end{bmatrix}$$

Load vector

The loads at the coordinates are the applied moment, the applied lateral force and the equivalent joint loads from the member loads:

$$\{F\}=\{4.167,\ 11.389,\ 12.593\}^T$$

(row 1: $-FEM_{BA}-FEM_{BC}$; row 2: $-FEM_{CB}-FEM_{CD}$ plus the applied 20 kNm; row 3: the 10 kN at B plus the equivalent lateral force from the 10 kN on CD).

$$EI\begin{bmatrix}1.4667 & 0.4000 & -0.1667 \\ 0.4000 & 1.4667 & -0.1667 \\ -0.1667 & -0.1667 & 0.1111\end{bmatrix}\begin{Bmatrix}\theta_{B} \\ \theta_{C} \\ \Delta_{1}\end{Bmatrix}=\begin{Bmatrix}4.167 \\ 11.389 \\ 12.593\end{Bmatrix}$$ $$EI\theta_{B}=16.1268,\quad EI\theta_{C}=22.8976,\quad EI\Delta_{1}=171.8699$$

Final end moments (kNm, clockwise +)

Checks:

• Joint B: (-17.89) + (17.89) = 0.00 kNm (applied clockwise moment 0.00) ✓
• Joint C: (28.94) + (-8.94) = 20.00 kNm (applied clockwise moment 20.00) ✓

Bending moment diagram ordinates (sagging/inside +, kNm)

Reactions: A: $H=-6.86$, $V=-4.37$, $M=23.27$; D: $H=-13.14$, $V=14.37$, $M=29.90$.

Answer: $M_{AB}=-23.27$, $M_{BA}=-17.89$, $M_{BC}=17.89$, $M_{CB}=28.94$, $M_{CD}=-8.94$, $M_{DC}=-29.90$ kNm.