Stiffness matrix: frame with 100 kN, end reactions

Assumptions: A(0,0) is a hinge, B(0,6) and C(6,6) are the top corners (AB = 6 m, BC = 6 m, CD = 2 m high, so D is 4 m above A and fixed). EI is the same for all members; the 100 kN load at B acts to the right. Axial deformations are ignored.

Degrees of freedom (coordinates)

1. $\theta_A$ (hinge rotation) 2. $\theta_B$ 3. $\theta_C$ 4. $\Delta$ (horizontal sway of the girder). Chord rotations: $\psi_{AB}=\Delta/6$, $\psi_{CD}=\Delta/2$, $\psi_{BC}=0$.

Member stiffness equations ($M_{ij}=\tfrac{2EI}{L}(2\theta_i+\theta_j-3\psi)$, no member loads)

• $M_{AB}=\frac{2EI}{6}\left(2\theta_{A} + \theta_{B} - (0.5000)\Delta_1\right)$
• $M_{BA}=\frac{2EI}{6}\left(2\theta_{B} + \theta_{A} - (0.5000)\Delta_1\right)$
• $M_{BC}=\frac{2EI}{6}\left(2\theta_{B} + \theta_{C}\right)$
• $M_{CB}=\frac{2EI}{6}\left(2\theta_{C} + \theta_{B}\right)$
• $M_{CD}=\frac{2EI}{2}\left(2\theta_{C} - (1.5000)\Delta_1\right)$
• $M_{DC}=\frac{2EI}{2}\left(\theta_{C} - (1.5000)\Delta_1\right)$

Equilibrium equations

• Hinge A: $M_{AB}=0$
• Joint B: $M_{BA}+M_{BC}=0$
• Joint C: $M_{CB}+M_{CD}=0$
• Storey shear: $\dfrac{M_{AB}+M_{BA}}{6}+\dfrac{M_{CD}+M_{DC}}{2}+100=0$

Stiffness matrix of the frame $[K]\{d\}=\{F\}$

$$EI\begin{bmatrix}0.6667 & 0.3333 & 0.0000 & -0.1667 \\ 0.3333 & 1.3333 & 0.3333 & -0.1667 \\ 0.0000 & 0.3333 & 2.6667 & -1.5000 \\ -0.1667 & -0.1667 & -1.5000 & 1.5556\end{bmatrix}\begin{Bmatrix}\theta_{A} \\ \theta_{B} \\ \theta_{C} \\ \Delta_{1}\end{Bmatrix}=\begin{Bmatrix}0.000 \\ 0.000 \\ 0.000 \\ 100.000\end{Bmatrix}$$

The matrix is symmetric. The only non-zero load is the 100 kN at the sway coordinate.

Solution

$$EI\theta_{A}=44.8598,\quad EI\theta_{B}=-14.0187,\quad EI\theta_{C}=86.9159,\quad EI\Delta_{1}=151.4019$$

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

Checks: $M_{AB}=0.00$ ✓; Joint B: $-19.63+(19.63)=0.00$ ✓; Joint C: $53.27+(-53.27)=0.00$ ✓.

End reactions

Column shears: $H_A=\dfrac{M_{AB}+M_{BA}}{6}$ and $H_D=\dfrac{M_{CD}+M_{DC}}{2}$ (magnitudes), and the vertical reactions from the girder moments.

• Support A: $H_A=3.27$ kN (to the left), $V_A=-12.15$ kN, $M_A=0$
• Support D: $H_D=96.73$ kN (to the left), $V_D=12.15$ kN, $M_D=140.19$ kNm (anticlockwise +)
• Check: $\sum H=100-3.27-96.73=0.00$ ✓, $\sum V=0.00$ ✓

Answer: $EI\Delta=151.402$; $M_{BA}=-19.63$, $M_{BC}=19.63$, $M_{CB}=53.27$, $M_{CD}=-53.27$, $M_{DC}=-140.19$ kNm; reactions $H_A=3.27$ kN, $H_D=96.73$ kN (both to the left).