Lug under tranverse load formula question 1

[franc11]

Hi there,

Little question here. Usually, to calculate the allowable load for a lug under transverse load, you have to calculate and average area of the section of the lug.

The formule is : Aavg = 6/(3/A1+1/A2+1/A3+1/A4).

Here's my question. In all the books and references, they show without explaining it, that A1 and A4 are area at a 45 degree from center the lug and I am assuming that A1 is always in direction of the load. Is that a correct assumption? I have a lug which is straight at the bottom and 45 degress on the top and my tranverse load is in the straight direction, so is A1 always the area in the load direction or A1 is always the biggest area because of the factor (3/A1) ?

Thanks!

 

[rb1957]

i would apply the load as components along (and normal to) the lug CL. in your case the lug CL is 22.5deg down.

 

[vonlueke]

franc11: Your statements are correct. The axial direction of a lug is 0 deg, and for a transversely-loaded lug, the transverse load is at +90 deg. Section A1 is always at +135 deg, A2 is always at +90 deg, and A4 is always at -135 deg. A3 is the least of any radial section around the hole, regardless of direction.