According to ASME B&PV Sec VIII Div 1 App.2,

Fot integral Flanges:hd=R+0.5g1 and ht=0.5( R+g1+hg).

My questions are:

1) Why hd should not be R+g1 in this case?

2) If I consider that it is R+0.5g1 as the thrust load is acting on a diameter greater than B, then Hd should be (pi/4)*(B+x)^2 where x is a positive quantity.

3)For integral flanges if we consider ht as arithmetic mean of hg and hd and hd=R+g1 then only we arrive at the prescribed value of ht ( Table 2-6) =0.5(R+g1+hg). However if hd is =R+0.5g1 and consider ht as average of hg and hd, we cannot get the value of ht as prescribed in the same table.I want to know how the value of this moment arm is arrived at.

Anindya Bhattacharya

Hello Anindya

There are three forces acting on the flange, which are:

Hg, Hd, and Ht. These all add up to Wm1 (the operating bolt force), therefore Wm1 = Hg + Hd + Ht. Now, the confusion arises because it is not clearly understood why there is a force Hp. To clarify this, consider a typical flange, with the forces acting:

Now, Hg is the force acting on the gasket over a width 2b. That is why this force is given by Hg = 2b*3.14*m*P which is actually Hp for the operating case. At the bore of the flange (Diameter B). there is a hydraulic force Hd = 0.785*B2*P. Between the gasket diameter G, and the inside bore B, there is an annular strip over which the internal pressure also acts. This is called Ht. Now, Ht can be calculated as 0.785(G2 – B2)*P which is the annular area times the internal pressure. Thus, Ht = 0.785*G2 – 0.785*B2 = H – Hd. ( 0.785 = Pi / 4)

Once we have the three forces, all we need is the three moment arms, and then the total moment can be computed.

The problem arises often, because it is not clear what the term Ht means.

I hope this clarifies the situation.