Stiffness matrix: frame with coordinates 1, 2, 3

Assumptions (figure partly illegible): A(0,0), B(0,6), C(3,6), D(3,2). The left column AB (6 m, 3 m + 3 m) carries the 10 kN horizontal load (to the right) at mid-height; the beam BC is 3 m long with 15 kN/m downward; the right column CD is 4 m. A and D are fixed. EI is the same for all members. Coordinates: 1 = rotation of joint B (clockwise), 2 = horizontal translation (sway) of the beam (to the right), 3 = rotation of joint C (clockwise). The additional forces at coordinates 1 and 3 are taken as a 5 kNm and a 4 kNm clockwise moment (a 'kN' at coordinate 3 cannot act on a rotation, so the 4 is read as kNm).

Stiffness matrix $[K]$ (units EI)

Terms: $\psi_{AB}=\Delta/6$, $\psi_{CD}=\Delta/4$.

• $k_{11}=\dfrac{4EI}{6}+\dfrac{4EI}{3}=2.0000EI$; $k_{13}=k_{31}=\dfrac{2EI}{3}=0.6667EI$; $k_{33}=\dfrac{4EI}{3}+\dfrac{4EI}{4}=2.3333EI$
• $k_{12}=-\dfrac{6EI}{6^2}=-0.1667EI$; $k_{32}=-\dfrac{6EI}{4^2}=-0.3750EI$; $k_{22}=\dfrac{12EI}{6^3}+\dfrac{12EI}{4^3}=0.2431EI$

$$[K]=EI\begin{bmatrix}2.0000 & -0.1667 & 0.6667 \\ -0.1667 & 0.2431 & -0.3750 \\ 0.6667 & -0.3750 & 2.3333\end{bmatrix}$$

Equivalent loads from the member loads (fixed-end effects)

Fixed-end moments: BC $\mp\dfrac{15\times3^2}{12}=\mp11.25$ kNm; AB (10 kN at 3 m, horizontal) $\mp\dfrac{PL}{8}=\mp7.5$ kNm (sign for a horizontal load to the right on a column); the equivalent lateral force at the girder level is the fixed-end shear at B, $10/2=5$ kN.

$$\{F_0\}=\begin{Bmatrix}3.750\\5.000\\-11.250\end{Bmatrix}\ (\text{kNm, kN, kNm}),\qquad \{F\}=\{F_0\}+\begin{Bmatrix}5\\0\\4\end{Bmatrix}=\begin{Bmatrix}8.750\\5.000\\-7.250\end{Bmatrix}$$

Analysis

$[K]\{d\}=\{F\}$ gives

$$EI\theta_B=6.7426,\quad EI\Delta=23.1753,\quad EI\theta_C=-1.3090$$

End moments (member equations, kNm, clockwise +)

• $M_{AB}=\frac{2EI}{6}\left(\theta_{B} - (0.5000)\Delta_1\right) - 7.50$
• $M_{BA}=\frac{2EI}{6}\left(2\theta_{B} - (0.5000)\Delta_1\right) + 7.50$
• $M_{BC}=\frac{2EI}{3}\left(2\theta_{B} + \theta_{C}\right) - 11.25$
• $M_{CB}=\frac{2EI}{3}\left(2\theta_{C} + \theta_{B}\right) + 11.25$
• $M_{CD}=\frac{2EI}{4}\left(2\theta_{C} - (0.7500)\Delta_1\right)$
• $M_{DC}=\frac{2EI}{4}\left(\theta_{C} - (0.7500)\Delta_1\right)$

Checks:

• Joint B: (8.13) + (-3.13) = 5.00 kNm (applied clockwise moment 5.00) ✓
• Joint C: (14.00) + (-10.00) = 4.00 kNm (applied clockwise moment 4.00) ✓

Bending moment ordinates (kNm)

Reactions: A: $H=-5.16$, $V=18.88$, $M=9.12$; D: $H=-4.84$, $V=26.12$, $M=9.35$.

Answer: $[K]$ as above; with the loads given, $M_{AB}=-9.12$, $M_{BA}=8.13$, $M_{BC}=-3.13$, $M_{CB}=14.00$, $M_{CD}=-10.00$, $M_{DC}=-9.35$ kNm.