This post continues from – Frame Analysis with Excel 1 – Single beam
The stiffness matrix shown in the previous post was for a single beam with loads applied either perpendicular to the beam, or along the longitudinal beam axis, which directions were aligned with the Y and X axes respectively. In this post the stiffness matrix will be modified for loading in other directions. It would be possible to simply resolve the applied loads into perpendicular and axial components, but since our intention is to derive an analysis method suitable for analysing frames with members intersecting at any angle it makes more sense to change the stiffness matrix to work in the global coordinate system so that both the applied loads and the resulting translations are specified with respect to the global X and Y axes.
It is now necessary to specify a direction for the beam, which is the angle of the beam to the Global X axis (horizontal), and is positive anti-clockwise. The input is otherwise identical to the horizontal beam case.
The modified stiffness matrix coefficients are shown below, where k_1 to k_4 are defined as before, and c and s are the Cosine and Sine of the beam angle respectively.
It is also necessary to adjust the calculation of the fixed end moments and reactions to account for the slope of the beam.
The derivation of the fixed end moments for a horizontal beam subject to partial trapezoidal loading is shown below:
The spreadsheet uses the UDF FEMact to solve these equations, and also to calculate the corresponding end reactions. The output from the UDF is a column array of the six fixed end actions. The algorithm for the UDF is:
- Find the fixed-end moments due to the vertical distributed loads using the equations shown above, FEMA(3,1) and FEMA(6,1)
- If the beam slope, Theta <> 0 then FEMA(x,1) = FEMA(x,1) * Cos(Theta)
- Find fixed-end moments due to point loads, MomInc(1) and MomInc(2)
- For vertical point loads MomInc = MomInc * Cos(Theta). For horizontal loads MomInc = MomInc * Sin(Theta)
- Find fixed end moments and end reactions due to point moments
- Find end reactions due to distributed loads
- Find end reactions due to point loads
- Adjust end reactions for effect of out of balance moments, allowing for the slope of the beam
- Return the array of beam end actions, FEMA
The rest of the process is as for a horizontal beam, with the output being the translations and rotations of the active end freedoms, and the reactions at the fixed freedoms. Full source code for the function is available in the download file: Beam2.zip
To check the output of the program an example of an inclined beam loaded with distributed loads, vertical and horizontal point loads, and point moments was analysed in the finite element package Strand7. The results are compared with the spreadsheet output in the screen shots below, showing near exact agreement: