jrjones
Mechanical
- Oct 10, 2006
- 38
Folks,
I'm trying to find a source that would describe how to axisymmetrically model a flanged joint. Specifically, I have always made an equivalent plane strain "bolt" by matching both the cross sectional areas and the moments of inertia for a single stud (bolt) and the equivalent axisymmetric "stud". I supposed I could validate by comparing results to a 3D flange section, but I'm looking for a little academic support for what I've done.
So, assume you have a regular ASME flange. Assume the following:
-Stud diameter is "D"
-Equivalent plane strain "stud" has width "d", depth "b"
So, I figure the areas and moments of inertia have to match:
For regular, cylindrical, stud:
-Area, A1 = pi/4*D^2*n (where n is the number of studs in the flange)
-Moment of inertia, I1 = pi/64*D^4*n
For the axisym "bolt":
-A2 = b*d
-I2 = 1/12*b*d^3
If you set A1=A2 (eqn 1), and I1=I2 (eqn 2), you can work through the system of equations:
-I solve for b = eqn 1 and sub into eqn 2.
-Doing all the gymnastics, I get that d/D=sqrt(3/4)
-So basically, the "width" of the axi bolt is ~0.87 x D
This seems directionally correct, and the flanged joint behaves as I would expect, but I'm looking for some validation.
Cheers,
jrjones
I'm trying to find a source that would describe how to axisymmetrically model a flanged joint. Specifically, I have always made an equivalent plane strain "bolt" by matching both the cross sectional areas and the moments of inertia for a single stud (bolt) and the equivalent axisymmetric "stud". I supposed I could validate by comparing results to a 3D flange section, but I'm looking for a little academic support for what I've done.
So, assume you have a regular ASME flange. Assume the following:
-Stud diameter is "D"
-Equivalent plane strain "stud" has width "d", depth "b"
So, I figure the areas and moments of inertia have to match:
For regular, cylindrical, stud:
-Area, A1 = pi/4*D^2*n (where n is the number of studs in the flange)
-Moment of inertia, I1 = pi/64*D^4*n
For the axisym "bolt":
-A2 = b*d
-I2 = 1/12*b*d^3
If you set A1=A2 (eqn 1), and I1=I2 (eqn 2), you can work through the system of equations:
-I solve for b = eqn 1 and sub into eqn 2.
-Doing all the gymnastics, I get that d/D=sqrt(3/4)
-So basically, the "width" of the axi bolt is ~0.87 x D
This seems directionally correct, and the flanged joint behaves as I would expect, but I'm looking for some validation.
Cheers,
jrjones