Lifting Lug Design Per ASME BTH-1-2005
Input: Nd = t= #= D' = *e = D! = +ured Edge = u =
2,00 0,472441 in!es 2,0% ,0%%&2& in!e !es 1,25 ,25&(4) in!e !es 2,0% ,0%%&2& in!e !es 1,)7 ,)77&5) in!e !es N . /r N )4,0( $si 52,21 $si
Lug P" P"#te T! T!i$ness
M#teri#" .ie"d Stress M#teri#" "ti3#te Stress
Output: *eff1 = 1,(& *eff2 = 1,55 *eff = 1,55 r = 2,75 ,755&0% 8 = 2,75 ,755&0% 9: = 0,00 A = 2,14 Pt = P* = P = P' =
)1,(& )1,(2 )2,%4 12,%(
in!es in!es in!es in!e !es in!e !es in!es s in!es in!es
ASME Eu#ti/n )-4%6 ASME Eu#ti/n )-476
ASME Eu#ti/n +)-26 ASME Eu#ti/n Eu#ti/n )-506 )-506 3/dified 'er 'er +/33ent#r +/33ent#r
$i's $i's $i's $i's
ASME Eu#ti/n )-456 ASME Eu#ti/n )-4(6 ASME Eu#ti/n )-4&6 ASME Eu#ti/n )-516;D';t
3 2 1 0
08/07/2018
Pr Projeto
H.T.C.
MC.
Revisão
Data
Designação
Elaoro!
"eri#i$o!
%&rovo!
Pin Diameter Effect: D!
1,0& . 0,((& 50,2(% 2,2&4 0,000 2,17
Pt = P* = P = P' =
2(,)) 2(,2( )),01 12,%(
M#@ P =
. /r N +!e$ een !en D!= 11 8eduti/n #t/r 8ef 2 Eu#ti/n %6 Degrees 8ef2 Eu#ti/n &6 ?n ?n!es 8ef 2 Eu#ti/n 206 ?n ?n!es 8ef 2 Eu#ti/n 216 s in!es $i's $i's $i's $i's
12,%( $i's
(Input)