Stiffness matrix: frame with 50 kNm moment

Assumptions: the 10 kN/m UDL acts downward on BC, the 40 kN load at B acts to the right and the 50 kNm moment at C is clockwise. EI is constant, axial deformation is neglected, A and D are fixed.

Degrees of freedom

$\theta_B,\theta_C$ (joint rotations, clockwise +) and $\Delta$ (horizontal sway of beam BC, right +). Since BC is axially rigid, B and C move equally, so $\psi_{AB}=\psi_{CD}=\Delta/4$ and $\psi_{BC}=0$.

Member relation: $M_{ij}=\dfrac{2EI}{L}(2\theta_i+\theta_j-3\psi)+FEM_{ij}$ (clockwise end moments +).

Fixed-end moments

$FEM_{BC}=-\dfrac{wL^2}{12}=-\dfrac{10\times8^2}{12}=-53.33$ kNm, $FEM_{CB}=+53.33$ kNm. The columns carry no load between joints, so their FEMs are zero.

Stiffness equations for the members

• $M_{AB}=\frac{2EI}{4}\left(\theta_{B} - (0.7500)\Delta_1\right)$
• $M_{BA}=\frac{2EI}{4}\left(2\theta_{B} - (0.7500)\Delta_1\right)$
• $M_{BC}=\frac{2EI}{8}\left(2\theta_{B} + \theta_{C}\right) - 53.33$
• $M_{CB}=\frac{2EI}{8}\left(2\theta_{C} + \theta_{B}\right) + 53.33$
• $M_{CD}=\frac{2EI}{4}\left(2\theta_{C} - (0.7500)\Delta_1\right)$
• $M_{DC}=\frac{2EI}{4}\left(\theta_{C} - (0.7500)\Delta_1\right)$

Equilibrium conditions

• Joint B: $M_{BA}+M_{BC}=0$
• Joint C: $M_{CB}+M_{CD}=50$ (the applied clockwise moment)
• Storey shear: the column shears resist the 40 kN load, $\dfrac{M_{AB}+M_{BA}}{4}+\dfrac{M_{CD}+M_{DC}}{4}+40=0$

Substituting the member equations gives $[K]\{d\}=\{F\}$:

$$EI\begin{bmatrix}1.5000 & 0.2500 & -0.3750 \\ 0.2500 & 1.5000 & -0.3750 \\ -0.3750 & -0.3750 & 0.3750\end{bmatrix}\begin{Bmatrix}\theta_{B} \\ \theta_{C} \\ \Delta_{1}\end{Bmatrix}=\begin{Bmatrix}53.333 \\ -3.333 \\ 40.000\end{Bmatrix}$$

(First two rows: joint B and C. Third row: sway equation. The load vector holds the applied loads minus the fixed-end effects.)

Solution

$$EI\theta_{B}=87.6667,\quad EI\theta_{C}=42.3333,\quad EI\Delta_{1}=236.6667$$

Final end moments (kNm, clockwise on the member end +)

Checks:

• Joint B: (-1.08) + (1.08) = 0.00 kNm (applied clockwise moment 0.00) ✓
• Joint C: (96.42) + (-46.42) = 50.00 kNm (applied clockwise moment 50.00) ✓
• Storey shear: $\dfrac{(-46.00)+(-114.00)}{4}+40=0.00$ ✓

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

Reactions: A: $H=-11.50$, $V=27.81$, $M=44.92$; D: $H=-28.50$, $V=52.19$, $M=67.58$ (M anticlockwise +).

Answer: $M_{AB}=-44.92$, $M_{BA}=-1.08$, $M_{BC}=1.08$, $M_{CB}=96.42$, $M_{CD}=-46.42$, $M_{DC}=-67.58$ kNm; sway $EI\Delta=236.67$.