Another Roark Question - Unit Step Function
Another Roark Question - Unit Step Function
(OP)
I'm trying to use Table 8.1, case 2, Partial distributed load. The formulas use what is called a "unit step function". I looked on page 131 for the definition.
If x < a, then <x-a> = 0. If x > a, then <x-a> = (x-a).
OK, I get that. But what about when x=a? It says it's undefined. I have just that case to figure out. Can someone explain this to me?
If x < a, then <x-a> = 0. If x > a, then <x-a> = (x-a).
OK, I get that. But what about when x=a? It says it's undefined. I have just that case to figure out. Can someone explain this to me?






RE: Another Roark Question - Unit Step Function
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RE: Another Roark Question - Unit Step Function
another day in paradise, or is paradise one day closer ?
RE: Another Roark Question - Unit Step Function
P/2 - P<x-(L/2)>0
When x = L/2 the formula is undefined, but that is because the shear value is also undefined (within the limitations of the mathematical model being used). The formula is still correct.
As for the dilemma this result might leave you in, what shear do you design the beam to withstand at its very centre? I'd design it to take +P/2 AND -P/2. Just in case the load wanders by an angstrom either way.
RE: Another Roark Question - Unit Step Function
in my day we'd write SF(x) = FA x<a,
= FA-w.... x>=a
another day in paradise, or is paradise one day closer ?
RE: Another Roark Question - Unit Step Function
RE: Another Roark Question - Unit Step Function
another day in paradise, or is paradise one day closer ?