Stiffness matrix: frame, end moments

Assumptions: columns AB = 5 m (I) and CD = 5 m (2I), girder BC = 6 m (2I); A and D fixed. The 60 kN horizontal load at B acts to the right, the 50 kN load acts downward at mid-span of BC and the 10 kNm moment at C is clockwise. Axial deformation is neglected.

Coordinates

1 = $\theta_B$, 2 = $\theta_C$ (clockwise +), 3 = $\Delta$ (sway to the right). $\psi_{AB}=\psi_{CD}=\Delta/5$.

Fixed-end moments

$FEM_{BC}=-\dfrac{PL}{8}=-\dfrac{50\times6}{8}=-37.5$ kNm, $FEM_{CB}=+37.5$ kNm. Other members unloaded.

Stiffness equations of each member

• $M_{AB}=\frac{2EI}{5}\left(\theta_{B} - (0.6000)\Delta_1\right)$
• $M_{BA}=\frac{2EI}{5}\left(2\theta_{B} - (0.6000)\Delta_1\right)$
• $M_{BC}=\frac{2(2EI)}{6}\left(2\theta_{B} + \theta_{C}\right) - 37.50$
• $M_{CB}=\frac{2(2EI)}{6}\left(2\theta_{C} + \theta_{B}\right) + 37.50$
• $M_{CD}=\frac{2(2EI)}{5}\left(2\theta_{C} - (0.6000)\Delta_1\right)$
• $M_{DC}=\frac{2(2EI)}{5}\left(\theta_{C} - (0.6000)\Delta_1\right)$

Stiffness matrix of the frame

Joint B: $K_{11}=\dfrac{4EI}{5}+\dfrac{4(2EI)}{6}$, joint C: $K_{22}=\dfrac{4(2EI)}{6}+\dfrac{4(2EI)}{5}$, coupling $K_{12}=\dfrac{2(2EI)}{6}$, sway terms $K_{13}=-\dfrac{6EI}{5^2}$, $K_{23}=-\dfrac{6(2EI)}{5^2}$, $K_{33}=\dfrac{12EI}{5^3}+\dfrac{12(2EI)}{5^3}$.

$$[K]=EI\begin{bmatrix}2.1333 & 0.6667 & -0.2400 \\ 0.6667 & 2.9333 & -0.4800 \\ -0.2400 & -0.4800 & 0.2880\end{bmatrix}$$

Load vector (applied loads minus fixed-end effects): $\{F\}=\{37.500,\ -27.500,\ 60.000\}^T$ (rows 1 and 2: $-\sum FEM$ plus the applied moment at B and C; row 3: the 60 kN lateral force).

$$EI\begin{bmatrix}2.1333 & 0.6667 & -0.2400 \\ 0.6667 & 2.9333 & -0.4800 \\ -0.2400 & -0.4800 & 0.2880\end{bmatrix}\begin{Bmatrix}\theta_{B} \\ \theta_{C} \\ \Delta_{1}\end{Bmatrix}=\begin{Bmatrix}37.500 \\ -27.500 \\ 60.000\end{Bmatrix}$$ $$EI\theta_{B}=41.2829,\quad EI\theta_{C}=28.8240,\quad EI\Delta_{1}=290.7758$$

End moments using the member stiffness equations (kNm, clockwise +)

Checks:

• Joint B: (-36.76) + (36.76) = 0.00 kNm (applied clockwise moment 0.00) ✓
• Joint C: (103.45) + (-93.45) = 10.00 kNm (applied clockwise moment 10.00) ✓
• Storey shear: $\dfrac{-90.03+(-209.97)}{5}+60=0.00$ ✓

Reactions: A: $H=-18.01$, $V=1.63$, $M=53.27$; D: $H=-41.99$, $V=48.37$, $M=116.51$.

Answer: $M_{AB}=-53.27$, $M_{BA}=-36.76$, $M_{BC}=36.76$, $M_{CB}=103.45$, $M_{CD}=-93.45$, $M_{DC}=-116.51$ kNm.