Stiffness method: final end moments, frame

Assumptions (figure partly unclear): A(0,0), B(4,0), C(9,0) lie on one horizontal girder: AB = 4 m (3EI), BC = 5 m (2EI) with 8 kN/m downward. Column BE (2EI) is 4 m long and carries the 20 kN horizontal load (to the right) at its mid-height; E is a hinge. Column CD (EI) is 7 m long and fixed at D. A is fixed. The 20 kNm moment at B is clockwise. Because A is fixed, the girder cannot sway, so the unknowns are only the joint rotations $\theta_B,\theta_C$ and the hinge rotation $\theta_E$.

Fixed-end moments (clockwise +)

• BC: $\mp\dfrac{wL^2}{12}=\mp\dfrac{8\times5^2}{12}=\mp16.67$ kNm
• BE (point load 20 kN at mid-height): $\mp\dfrac{PL}{8}=\mp\dfrac{20\times4}{8}=\mp10$ kNm (for the load acting to the right on this downward member: $FEM_{BE}=10.00$, $FEM_{EB}=-10.00$ kNm)
• AB and CD: no loads, so zero.

Slope-deflection (stiffness) equations

• $M_{AB}=\frac{2(3EI)}{4}\left(\theta_{B}\right)$
• $M_{BA}=\frac{2(3EI)}{4}\left(2\theta_{B}\right)$
• $M_{BC}=\frac{2(2EI)}{5}\left(2\theta_{B} + \theta_{C}\right) - 16.67$
• $M_{CB}=\frac{2(2EI)}{5}\left(2\theta_{C} + \theta_{B}\right) + 16.67$
• $M_{BE}=\frac{2(2EI)}{4}\left(2\theta_{B} + \theta_{E}\right) + 10.00$
• $M_{EB}=\frac{2(2EI)}{4}\left(2\theta_{E} + \theta_{B}\right) - 10.00$
• $M_{CD}=\frac{2EI}{7}\left(2\theta_{C}\right)$
• $M_{DC}=\frac{2EI}{7}\left(\theta_{C}\right)$

Equilibrium

• Joint B: $M_{BA}+M_{BC}+M_{BE}=20$ (applied clockwise moment)
• Joint C: $M_{CB}+M_{CD}=0$
• Hinge E: $M_{EB}=0$

$$EI\begin{bmatrix}6.6000 & 0.8000 & 1.0000 \\ 0.8000 & 2.1714 & 0.0000 \\ 1.0000 & 0.0000 & 2.0000\end{bmatrix}\begin{Bmatrix}\theta_{B} \\ \theta_{C} \\ \theta_{E}\end{Bmatrix}=\begin{Bmatrix}26.667 \\ -16.667 \\ 10.000\end{Bmatrix}$$ $$EI\theta_{B}=4.7900,\quad EI\theta_{C}=-9.4402,\quad EI\theta_{E}=2.6050$$

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

Checks:

• Joint B: (14.37) + (-16.55) + (22.18) = 20.00 kNm (applied clockwise moment 20.00) ✓
• Joint C: (5.39) + (-5.39) = 0.00 kNm (applied clockwise moment 0.00) ✓
• Joint E: (0.00) = 0.00 kNm (applied clockwise moment 0.00) ✓
• Hinge: $M_{EB}=0.00$ ✓

Bending moment ordinates (sagging/inside + according to member direction):

Reactions: A: $H=-14.39$, $V=-5.39$, $M=-7.18$; E: $H=-4.45$, $V=27.62$, $M=0.00$; D: $H=-1.16$, $V=17.77$, $M=2.70$.

Answer: $M_{AB}=7.18$, $M_{BA}=14.37$, $M_{BC}=-16.55$, $M_{CB}=5.39$, $M_{BE}=22.18$, $M_{EB}=0$, $M_{CD}=-5.39$, $M_{DC}=-2.70$ kNm.