Stiffness matrix: frame with 100 kN, 15 kNm

Assumptions: A(0,0), B(0,10), C(10,10), D(10,5): AB = 10 m (EI), BC = 10 m (2EI) with 20 kN/m, CD = 5 m (EI); A and D fixed; the 100 kN load at B acts to the right and the 15 kNm moment at C is clockwise. Axial deformation is neglected.

Unknowns

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

Fixed-end moments

$FEM_{BC}=-\dfrac{20\times10^2}{12}=-166.67$ kNm, $FEM_{CB}=+166.67$ kNm.

Stiffness equations

• $M_{AB}=\frac{2EI}{10}\left(\theta_{B} - (0.3000)\Delta_1\right)$
• $M_{BA}=\frac{2EI}{10}\left(2\theta_{B} - (0.3000)\Delta_1\right)$
• $M_{BC}=\frac{2(2EI)}{10}\left(2\theta_{B} + \theta_{C}\right) - 166.67$
• $M_{CB}=\frac{2(2EI)}{10}\left(2\theta_{C} + \theta_{B}\right) + 166.67$
• $M_{CD}=\frac{2EI}{5}\left(2\theta_{C} - (0.6000)\Delta_1\right)$
• $M_{DC}=\frac{2EI}{5}\left(\theta_{C} - (0.6000)\Delta_1\right)$

Equations of equilibrium

Joint B: $M_{BA}+M_{BC}=0$; Joint C: $M_{CB}+M_{CD}=15$; storey shear: $\dfrac{M_{AB}+M_{BA}}{10}+\dfrac{M_{CD}+M_{DC}}{5}+100=0$.

$$EI\begin{bmatrix}1.2000 & 0.4000 & -0.0600 \\ 0.4000 & 1.6000 & -0.2400 \\ -0.0600 & -0.2400 & 0.1080\end{bmatrix}\begin{Bmatrix}\theta_{B} \\ \theta_{C} \\ \Delta_{1}\end{Bmatrix}=\begin{Bmatrix}166.667 \\ -151.667 \\ 100.000\end{Bmatrix}$$ $$EI\theta_{B}=185.9848,\quad EI\theta_{C}=19.6496,\quad EI\Delta_{1}=1072.9167$$

Final end moments (kNm, clockwise +)

Checks:

• Joint B: (10.02) + (-10.02) = 0.00 kNm (applied clockwise moment 0.00) ✓
• Joint C: (256.78) + (-241.78) = 15.00 kNm (applied clockwise moment 15.00) ✓
• Storey shear: $\dfrac{-17.16}{10}+\dfrac{-491.42}{5}+100=0.00$ ✓

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

Reactions: A: $H=-1.72$, $V=75.32$, $M=27.18$; D: $H=-98.28$, $V=124.68$, $M=249.64$.

Answer: $M_{AB}=-27.18$, $M_{BA}=10.02$, $M_{BC}=-10.02$, $M_{CB}=256.78$, $M_{CD}=-241.78$, $M_{DC}=-249.64$ kNm.