## Frame Analysis with Excel 2 – Single inclined beam

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:

1. Find the fixed-end moments due to the vertical distributed loads using the equations shown above, FEMA(3,1) and FEMA(6,1)
2. If the beam slope, Theta <> 0 then FEMA(x,1) = FEMA(x,1) * Cos(Theta)
3. Find fixed-end moments due to point loads, MomInc(1) and MomInc(2)
4. For vertical point loads MomInc = MomInc * Cos(Theta).  For horizontal loads MomInc = MomInc * Sin(Theta)
5. Find fixed end moments and end reactions due to point moments
6. Find end reactions due to distributed loads
7. Find end reactions due to point loads
8. Adjust end reactions for effect of out of balance moments, allowing for the slope of the beam
9. 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: This entry was posted in Excel, Frame Analysis, Newton, UDFs and tagged , , , . Bookmark the permalink.

### 11 Responses to Frame Analysis with Excel 2 – Single inclined beam

1. susant patra says:

I am bit confused about the result beam1 and beam2.

I applied following loads in program-beam1: 0.707 in X and 0.707 in Y
I applied following loads in program-beam2: 1.0 in X and theta = 45 degree

For easier visualization I tool all properties to be 1 (i.e. L=1, E=1kpa, I=1).
BC in both cases as follows: at end#1: X=Y=0, at end#2 Y=0

I was expecting identical results. But I did not get it. So I am confused. I would appreciate if you can explain why they should not be equal or is there a bug in Beam-2?

Regards,
Susant

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2. dougaj4 says:

Susan – the two load cases are not quite the same. In Beam1 there is an axial load, which is all transferred to the fixed end, and a vertical load which is equally distributed to the two supports (for a point load at mid span). In Beam2 one end is pinned, and the other end is sliding, so there is no horizontal reaction, and hence there can be no axial load. A point load at mid span will be equally distributed to each support as a vertical load. If you calculate the resultant of half the horizontal reaction + the vertical reaction from Beam1 you should find it is equal to 0.5. Also the end rotations (M1 and M2 values in the Deflection column) should be equal in Beam 1 and Beam 2.

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3. R Shameia says:

Hi

Thank you so much for this post.

Is there any reference for this derivation ??
Is there a deriviation for equivalent nodal force for trapezoidal loading in local x-direction.

Thank you again for your help.

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4. dougaj4 says:

R Shameia – thanks for asking, I should have given the references.

The fixed end moments came from The Reinforced Concrete Designer’s Handbook (C.E. Reynolds and J.C. Steedman). In the tenth edition the formulae are in Section 11.1.1. This gives a formula for a partial uniform load and one for a partial triangular load, decreasing from left to right. The triangular load formula can be modified for a load decreasing from right to left (by swapping the alpha and beta coefficients), and these two are then added, that gives the formula for a trapezoidal distribution, as shown in the diagram above.

An alternative approach (which will give the same results) is given in Roark’s Formulas for Stress and Strain (W.C. Young and R.G. Budynas). This gives a formula for a trapezoidal load extending from some point in the span up to the right hand end. By adding a positive load and a shorter negative load you can get the fixed end moments for a partial trapezoidal load with a gap at both ends. The formula is in Table 8.1 – 2c on page 192 in the 7th Edition (International version).

As for the distibution of the local x-direction forces, this is the same as the distribution of the transverse forces, see (for exampe) Table 23 in The Reinforced Concrete Designer’s Handbook. Note that when finding the end forces you have to correct for the effect of the end moments, if these are not equal and opposite.

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5. R Shameia says:

But, I can not understand the distibution of the local x-direction forces.
There will be local forces in x-direction and no end moments.

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6. dougaj4 says:

Maybe an example will make it clearer. In the download spreadsheet, if you set the angle of the beam to zero, and the length to 12, then enter a single point load of 12 units in the X direction, then the load at the right hand end is equal to the distance of the load from the left hand end, and the left hand load is (12 – the right hand load). This distribution is the same as the distribution of a vertical load applied to a simply supported beam, but as you say, there are no fixed end moments.

Doug

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7. sundhoo sagar says:

hello, I am a civil eng student from mauritius. I am working on my final year project at diploma level and I really need help. my problem is that I am renovating a stone house of 9m length, 5m breadth and 3.8m height..rectangularly.. from the height there is another triangular form of 2.7 height and has a inclination of 51 degrees.. in the 5m side there should be an inclined roof of concrete.. it previously has iron sheet and now the new model should be of concrete. I thus seek help from you to help me calculate the reinforcement and calculation appropriate for the building.. there will be a board of engineers and lecturer to assess me.. attached I send photos of the house.. I would be grateful if you could help me advance in my educational career. thanks in advance..

regards sagar

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• dougaj4 says:

Sagar, feel free to ask specific questions about any of the software available on this site, but I am not able to provide general engineering tutorials. You need to make full use of the facilities provided in your school/university.

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