OK, that is a bit more clear to me now. Although Q is also varying, it is a function of position, so you write it in terms of position and integrate in only one variable.
You say that you are calling the position of the differential load "x", and that dQ at "a" is zero, and dQ at "a + c" is w. This is still confusing to me. This would mean that the start point of the triangular distributed load is "a" and the end point is "a+c". However, in Case 14, "c" is defined as the start point.
I would've defined it:
X is the position of a differential load dQ, measured upward from point a to a final point c
dQ varies with X based on the slope of the distribution so, we have dQ = (w/(c-a))*x dx.
Then I follow everything until you substitute for dQ.
If you are defining the linear distributed load going from two points a to c, if you put in a value of (c-a) in for X in the dQ equation you defined, it should return the max value, W, but it doesn't.
Since you are integrating from x=a to x=a+c, I think you should be substituting ((w*x)/(c-a)) dx for dQ. Then integrate.
Maybe I'll make 2D FEM for the generic case this weekend, but TBH I think that might be overkill. There are a plethora of 2D frame analysis spreadhsheets and programs available on the web, many using the matrix force method described by Argyris.
I made a worksheet that performs the ASM calculation for Case 14, as given in the manual. I can confirm it is working correctly with generic inputs for the geometry because when I set b=c=0 it returns the same answers as the equations for the special case.
Then I made a duplicate where instead of the manual's equation for X11, I used your correction (that is the only change).
Then, I used this spreadhseet [
] with the exact same geometry and loading. Again, it matches the answer perfectly for the specialized case. But when I use the same load parameters I did with your corrected version of the ASM calculations, the answers don't come out the same.
These are my parameters for everything:
frame height = 10 units
frame width = 5 units
load goes from 2.5 units to 7.5 units from 0 to 50.
assumed cross section of the frame is 1x1 and the second moment of areas based on bh^3/12.
With the ASM formulas (including your correction of X11) I get horizontal reactions of 36.91 and 88.086 for equilibrium with the 125 distributed.
With the spreadsheet, the horizontal reactions come out to 92.53 and 32.47. Try using a 2D frame too from the web and see if it matches your calcs.
Instead of using Case 4 and adapting it, I'd think it would be better to just get the reactions from scratch using the matrix method
Keep em' Flying
//Fight Corrosion!