Stiffness matrix: frame, flexural deformations only

Assumptions (figure partly illegible): A(0,0), B(0,3), C(5,3), D(5,0): both columns are 3 m high (left column AB = 2EI, right column CD = EI), the beam BC = 5 m (EI). The 15 kN/m load acts horizontally to the right on AB, the 10 kN load acts downward at 3 m from B and the 15 kNm moment at C is clockwise. A and D are fixed. Only flexural deformation is considered (axial and shear deformations are neglected).

Unknowns

$\theta_B,\theta_C$ and the sway $\Delta$ of the beam: $\psi_{AB}=\psi_{CD}=\Delta/3$.

Fixed-end moments (clockwise +)

• AB: $\mp\dfrac{wL^2}{12}=\mp\dfrac{15\times3^2}{12}=\mp11.25$ kNm, with the sign for a horizontal load on a vertical member: $FEM_{AB}=-11.25$, $FEM_{BA}=11.25$ kNm
• BC ($a=3$, $b=2$): $FEM_{BC}=-\dfrac{10\times3\times2^2}{5^2}=-4.80$, $FEM_{CB}=+\dfrac{10\times3^2\times2}{5^2}=7.20$ kNm

Stiffness equations of the members

• $M_{AB}=\frac{2(2EI)}{3}\left(\theta_{B} - (1.0000)\Delta_1\right) - 11.25$
• $M_{BA}=\frac{2(2EI)}{3}\left(2\theta_{B} - (1.0000)\Delta_1\right) + 11.25$
• $M_{BC}=\frac{2EI}{5}\left(2\theta_{B} + \theta_{C}\right) - 4.80$
• $M_{CB}=\frac{2EI}{5}\left(2\theta_{C} + \theta_{B}\right) + 7.20$
• $M_{CD}=\frac{2EI}{3}\left(2\theta_{C} - (1.0000)\Delta_1\right)$
• $M_{DC}=\frac{2EI}{3}\left(\theta_{C} - (1.0000)\Delta_1\right)$

Equilibrium

• Joint B: $M_{BA}+M_{BC}=0$
• Joint C: $M_{CB}+M_{CD}=15$
• Storey shear: the 15 kN/m load on AB is replaced by its fixed-end reaction at the girder level (the fixed-end shear $wL/2=22.5$ kN at B), giving the sway equation with right-hand side 22.50 kN.

$$EI\begin{bmatrix}3.4667 & 0.4000 & -1.3333 \\ 0.4000 & 2.1333 & -0.6667 \\ -1.3333 & -0.6667 & 1.3333\end{bmatrix}\begin{Bmatrix}\theta_{B} \\ \theta_{C} \\ \Delta_{1}\end{Bmatrix}=\begin{Bmatrix}-6.450 \\ 7.800 \\ 22.500\end{Bmatrix}$$ $$EI\theta_{B}=9.0133,\quad EI\theta_{C}=11.9186,\quad EI\Delta_{1}=31.8476$$

Final end moments (kNm, clockwise +)

Checks:

• Joint B: (-7.18) + (7.18) = 0.00 kNm (applied clockwise moment 0.00) ✓
• Joint C: (20.34) + (-5.34) = 15.00 kNm (applied clockwise moment 15.00) ✓

Reactions: A: $H=-38.79$, $V=-1.50$, $M=41.70$; D: $H=-6.21$, $V=11.50$, $M=13.29$.

Answer: $M_{AB}=-41.70$, $M_{BA}=-7.18$, $M_{BC}=7.18$, $M_{CB}=20.34$, $M_{CD}=-5.34$, $M_{DC}=-13.29$ kNm.