Steps of displacement method for a frame

The displacement (stiffness) method treats the unknown joint displacements (rotations and translations) as the unknowns. The steps are:

1. Find the degree of kinematic indeterminacy (the number of independent joint rotations and joint translations). Neglect axial deformation of members, so joints of a rigid girder have a common sway.
2. Lock all joints (add imaginary restraints on rotation and translation). Every member becomes a fixed-ended member.
3. Compute the fixed-end moments and forces for the loads on each member (tables), and the equivalent joint loads $\{P\}$ = applied joint loads minus the fixed-end actions.
4. Write the member stiffness relations (slope-deflection equations) for every member in terms of the joint displacements:

$$M_{ij}=\frac{2EI}{L}\left(2\theta_i+\theta_j-3\psi\right)+FEM_{ij}$$

5. Write the equilibrium equations, one for each unknown: $\sum M=0$ at each rotating joint and, for sway, the shear (storey) equation. This gives $[K]\{d\}=\{P\}$.
6. Solve for the displacements $\{d\}$.
7. Back-substitute into the member equations to get the end moments, then find the shears, the axial forces and the support reactions from the equilibrium of the members, and draw the BMD, SFD and AFD.
8. Check the joint equilibrium and the overall equilibrium.

Example

Portal frame A-B-C-D, columns 4 m, beam 6 m, EI constant, fixed bases, UDL 12 kN/m on the beam.

Step 1. Unknowns: $\theta_B$, $\theta_C$. The loading and the frame are symmetrical, so no sway occurs.

Step 3. $FEM_{BC}=-\dfrac{12\times6^2}{12}=-36.00$ kNm, $FEM_{CB}=+36$ kNm.

Step 4. With $\theta_A=\theta_D=0$ and $\psi=0$:

• $M_{AB}=\frac{2EI}{4}\left(\theta_{B}\right)$
• $M_{BA}=\frac{2EI}{4}\left(2\theta_{B}\right)$
• $M_{BC}=\frac{2EI}{6}\left(2\theta_{B} + \theta_{C}\right) - 36.00$
• $M_{CB}=\frac{2EI}{6}\left(2\theta_{C} + \theta_{B}\right) + 36.00$
• $M_{CD}=\frac{2EI}{4}\left(2\theta_{C}\right)$
• $M_{DC}=\frac{2EI}{4}\left(\theta_{C}\right)$

Step 5. $M_{BA}+M_{BC}=0$ and $M_{CB}+M_{CD}=0$ give

$$EI\begin{bmatrix}1.6667 & 0.3333 \\ 0.3333 & 1.6667\end{bmatrix}\begin{Bmatrix}\theta_{B} \\ \theta_{C}\end{Bmatrix}=\begin{Bmatrix}36.000 \\ -36.000\end{Bmatrix}$$

Step 6.

$$EI\theta_{B}=27.0000,\quad EI\theta_{C}=-27.0000$$

Step 7. End moments: $M_{AB}=13.50$, $M_{BA}=27.00$, $M_{BC}=-27.00$, $M_{CB}=27.00$, $M_{CD}=-27.00$, $M_{DC}=-13.50$ kNm. The mid-span moment of the beam is $\dfrac{wL^2}{8}-27.00=27.00$ kNm.

Step 8. Joint B: $27.00+(-27.00)=0$ ✓.