dmorchard
commented
5 years ago Hmmm... Consider this code, which defines a cantilever with it's free end at x=0...
Sub issue3()
'Investigating issue #3: https://github.com/JWock82/xlFrame/issues/3
Dim myModel As New FEModel
'Define the nodes
Call myModel.AddNode("N1", 0, 0)
Call myModel.AddNode("N2", 5, 0)
'Define the members
Call myModel.AddMember("M1", "N1", "N2", 1, 1, 1)
'Define the supports
Call myModel.EditSupport("N2", True, True, True)
'Define member point loads
Call myModel.AddMemberPointLoad("M1", 10, 0, Transverse)
'Results...
With Sheet1
.Cells.Clear
.Range("B1") = "N1"
.Range("C1") = "N2"
'Support reactions
.Range("A2") = "FY"
.Range("B2") = myModel.GetReaction("N1", FY)
.Range("C2") = myModel.GetReaction("N2", FY)
'Support shear
.Range("A3") = "V(x)"
.Range("B3") = myModel.GetShear("M1", 0)
.Range("C3") = myModel.GetShear("M1", 5)
'Get V(x) diagram
.Range("A5") = "x"
.Range("B5") = "V(x)"
Call myModel.GetShearDiagram("M1").PrintEZArray(.Range("A6"))
End With 'Sheet1
End Sub...when graphed it looks right to me.
Now shift the point load elsewhere...
Call myModel.AddMemberPointLoad("M1", 10, 2, Transverse) ...and repeat...
My interpretation of what's going on is that we are expecting a nice vertical step in V(x) at the location of applied P. In order to get that result we'd need two V(x) values at that single location, in this case V(2) = 0 and V(2) = -10, but xlFrame is hardwired to return shear values for 20 consecutive x-values.
In my own stiffness matrix solving spreadsheet I encountered this same issue, and in my diagram generating sub I resorted to spoon-feeding it all the extra V(x) and M(x) points necessary to get the diagrams to step properly. This included forcing it to evaluate at points of interest, i.e. at x=2 in my second attempt above. I found that the concept of singularity functions was particularly useful for this task and was going to suggest them as a feature request actually.
I haven't quite figured out what the segment class does in xlFrame, but my gut feeling is segmenting each member at points of interest like load locations, the roots of V(x) & M(x), etc. might have the desired effect.
JWock82
Describe the bug Member point loads and moments applied at x = 0 along the length of a member are not reported in member shear and moment results. This is because the results are always reported "looking to the left" (or "just to the right") of the cut. This causes the reaction and the point load to cancel each other out at x=0. This is rarely an issue, since it usually makes sense to apply these loads as nodal loads instead of member loads.
To Reproduce Steps to reproduce the behavior:
**Suggested Fix*** The fix is pretty simple: change the code to report member end forces at the starts and ends of members, rather than calculating the internal shear force or moment as you would away from the ends.