ASME VIII Div1 Appendix 24 ( clamp connection )

[hsemus]

Could you please help me to derive the M0, rotational moment equation provided in ASME code.

Mo=0.5*W*(C-G) / tan(theta+mue)

Note:-

I do understand that M=F*Distance (N-mtr)

so the load for single bolt = W/2 =0.5W (N) and (C-G) = the distance between the neutral axis to the inside clamp)

but tan(theta+mue) not making any sense to my lazy brain ...

 

[prex]

Can you be more specific with references to App.24? I don't find that equation, only a similar one, but theta is not defined, phi instead is used as a symbol for clamp shoulder angle.
BTW C-G is two times the distance..., as C and G are diameters.

prex
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[hsemus]

Hi Prex,

thanks a lot for your help.

its in the section 24-5 of this attached document .

the formula (6) is the one which I am struggling to understand.

 

[prex]

W is a tangential load and 1/tan([φ]+[μ]) transforms it into an axial load. The moment arm is (C-G)/2, so the remaining factor is [π]/2 (2 x 0.785). To justify the latter one should look at how the stress is calculated, I can't see an immediate explanation for it.

prex
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[hsemus]

thanks for the informations.

Could you also let me know how the "MH" reaction moment at hub neck is worked out.

the main reason was when I use the ASME calculation the longitudinal stress "S1" is higher than the material yield strength.

So the two factors affecting this is due to the reaction moment & its rotational moment.

But my understanding is that the longitudinal stress is half of hoop stress. but in my calculation it seems that the longitudinal stress is much higher.

any thoughts on this .

once again thanks a lot for your help.