Stiffness matrix: frame, 25 kN, 100 kN

Assumptions: A(0,0), B(0,10), C(8,10), D(8,5). Left column AB (2I) is 10 m high (3 m + 7 m) with the 100 kN horizontal force (to the right) acting 7 m above A; beam BC (2I) is 8 m and carries 30 kN/m (the '30 kNm' of the figure is read as a UDL of 30 kN/m); right column CD (I) is 5 m; A and D are fixed. The 25 kN vertical load at B only produces axial force in AB (it acts through the column axis) and no bending moment. Axial deformation is neglected.

Unknowns

$\theta_B,\theta_C$ and the sway $\Delta$: $\psi_{AB}=\Delta/10$, $\psi_{CD}=\Delta/5$.

Fixed-end moments (clockwise +)

• AB (100 kN at $a=7$ m from A, $b=3$ m): $FEM_{AB}=-63.00$, $FEM_{BA}=147.00$ kNm (from $Pab^2/L^2=100\times7\times9/100$ and $Pa^2b/L^2=100\times49\times3/100$)
• BC: $\mp\dfrac{30\times8^2}{12}=\mp160$ kNm

Stiffness equations

• $M_{AB}=\frac{2(2EI)}{10}\left(\theta_{B} - (0.3000)\Delta_1\right) - 63.00$
• $M_{BA}=\frac{2(2EI)}{10}\left(2\theta_{B} - (0.3000)\Delta_1\right) + 147.00$
• $M_{BC}=\frac{2(2EI)}{8}\left(2\theta_{B} + \theta_{C}\right) - 160.00$
• $M_{CB}=\frac{2(2EI)}{8}\left(2\theta_{C} + \theta_{B}\right) + 160.00$
• $M_{CD}=\frac{2EI}{5}\left(2\theta_{C} - (0.6000)\Delta_1\right)$
• $M_{DC}=\frac{2EI}{5}\left(\theta_{C} - (0.6000)\Delta_1\right)$

Equations of equilibrium

Joint B: $M_{BA}+M_{BC}=0$; joint C: $M_{CB}+M_{CD}=0$; sway ($\sum H=0$, with $H_A=\dfrac{M_{AB}+M_{BA}-100\times3}{10}$ and $H_D=\dfrac{M_{CD}+M_{DC}}{5}$): $\dfrac{M_{AB}+M_{BA}}{10}+\dfrac{M_{CD}+M_{DC}}{5}+\dfrac{100\times7}{10}=0$. When the fixed-end moments are moved to the right-hand side, the load term of this row becomes the fixed-end reaction of the 100 kN load at B, $\dfrac{Pa^2(L+2b)}{L^3}=\dfrac{100\times49\times16}{1000}=78.40$ kN $\left(=70+\dfrac{147-63}{10}\right)$.

$$EI\begin{bmatrix}1.8000 & 0.5000 & -0.1200 \\ 0.5000 & 1.8000 & -0.2400 \\ -0.1200 & -0.2400 & 0.1200\end{bmatrix}\begin{Bmatrix}\theta_{B} \\ \theta_{C} \\ \Delta_{1}\end{Bmatrix}=\begin{Bmatrix}13.000 \\ -160.000 \\ 78.400\end{Bmatrix}$$ $$EI\theta_{B}=56.5023,\quad EI\theta_{C}=-13.5535,\quad EI\Delta_{1}=682.7287$$

Final end moments (kNm, clockwise +)

Checks:

• Joint B: (110.27) + (-110.27) = 0.00 kNm (applied clockwise moment 0.00) ✓
• Joint C: (174.70) + (-174.70) = 0.00 kNm (applied clockwise moment 0.00) ✓

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

Reactions: A: $H=-31.21$, $V=136.95$, $M=122.33$; D: $H=-68.79$, $V=128.05$, $M=169.28$.

Answer: $M_{AB}=-122.33$, $M_{BA}=110.27$, $M_{BC}=-110.27$, $M_{CB}=174.70$, $M_{CD}=-174.70$, $M_{DC}=-169.28$ kNm.