Stiffness matrix: final end moments, 2.5EI frame

Assumptions: A(0,0), B(0,6), C(10,6), D(10,0); the 20 kNm moment at B is clockwise; the 15 kN load acts to the right on AB at 4 m above A; the UDL of 8 kN/m acts downward on BC. A and D are fixed. Axial deformation is neglected.

Unknowns

$\theta_B,\theta_C$ and the sway $\Delta$ of the girder (right +); $\psi_{AB}=\psi_{CD}=\Delta/6$.

Fixed-end moments

• BC: $\mp\dfrac{8\times10^2}{12}=\mp66.67$ kNm
• AB (point load 15 kN at a = 4 m from A, b = 2 m): the load acts perpendicular to the member: $FEM_{AB}=-6.67$, $FEM_{BA}=13.33$ kNm (from $Pab^2/L^2$ and $Pa^2b/L^2$ with the sign for a load to the right)

Stiffness equations of the members

• $M_{AB}=\frac{2(2.5EI)}{6}\left(\theta_{B} - (0.5000)\Delta_1\right) - 6.67$
• $M_{BA}=\frac{2(2.5EI)}{6}\left(2\theta_{B} - (0.5000)\Delta_1\right) + 13.33$
• $M_{BC}=\frac{2(2EI)}{10}\left(2\theta_{B} + \theta_{C}\right) - 66.67$
• $M_{CB}=\frac{2(2EI)}{10}\left(2\theta_{C} + \theta_{B}\right) + 66.67$
• $M_{CD}=\frac{2(2EI)}{6}\left(2\theta_{C} - (0.5000)\Delta_1\right)$
• $M_{DC}=\frac{2(2EI)}{6}\left(\theta_{C} - (0.5000)\Delta_1\right)$

Equilibrium

• Joint B: $M_{BA}+M_{BC}=20$ (applied clockwise moment)
• Joint C: $M_{CB}+M_{CD}=0$
• Storey shear: taking moments about B for column AB (the 15 kN load is 2 m below B) gives the base shear $H_A=\dfrac{M_{AB}+M_{BA}-15\times2}{6}$; for CD, $H_D=\dfrac{M_{CD}+M_{DC}}{6}$. Horizontal equilibrium $H_A+H_D+15=0$ then gives $\dfrac{M_{AB}+M_{BA}}{6}+\dfrac{M_{CD}+M_{DC}}{6}+10=0$. When the member equations (which include the FEMs) are substituted, the constant on the right-hand side becomes $10+\dfrac{-6.67+13.33}{6}=11.11$ kN. This equals the equivalent joint load at B, i.e. the fixed-end reaction of the loaded column at B, $\dfrac{Pa^2(L+2b)}{L^3}=\dfrac{15\times4^2(6+4)}{6^3}=11.11$ kN.

$$EI\begin{bmatrix}2.4667 & 0.4000 & -0.4167 \\ 0.4000 & 2.1333 & -0.3333 \\ -0.4167 & -0.3333 & 0.2500\end{bmatrix}\begin{Bmatrix}\theta_{B} \\ \theta_{C} \\ \Delta_{1}\end{Bmatrix}=\begin{Bmatrix}73.333 \\ -66.667 \\ 11.111\end{Bmatrix}$$

(Rows 1 and 2: joints B and C. Row 3: storey shear, right-hand side = 11.11 kN.)

Solution

$$EI\theta_{B}=49.5343,\quad EI\theta_{C}=-26.1394,\quad EI\Delta_{1}=92.1490$$

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

Checks:

• Joint B: (57.50) + (-37.50) = 20.00 kNm (applied clockwise moment 20.00) ✓
• Joint C: (65.57) + (-65.57) = 0.00 kNm (applied clockwise moment 0.00) ✓

Reactions: A: $H=3.95$, $V=37.19$, $M=3.78$; D: $H=-18.95$, $V=42.81$, $M=48.14$.

Answer: $M_{AB}=-3.78$, $M_{BA}=57.50$, $M_{BC}=-37.50$, $M_{CB}=65.57$, $M_{CD}=-65.57$, $M_{DC}=-48.14$ kNm.