Corrigé De L'application
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Corrigé de l'application Déterminons l'angle unitaire de torsion. I0 =
d4 32
AN:
=
Alors
=
=
M t = G I0 4
24.10 80000x I0
x74 32
= 235.7 mm
4
Be tci tech nol ogi es
avec I 0 =
Mt GI0
4
-5
24.10 80000x 235.7
= 127.10 rad/mm
2- Calculons la contrainte tangentielle maximale cas 1:
MAXI
donc
=G. .R
MAXI
Or: R= d/2
=G. .d/2
-5
AN:
MAXI
=(80000 x 127.10 x 3.5)
MAX =355.6 N/mm²
cas 2:
MAXI
AN:
MAXI
=
=
Mt
I0
v
x
24000x3.5 235.7
= 356.38 N/mm².
MAX =356.38 N/mm²
PROF:KOUAME K.M.
LA TORSION R.D.M
DOC : 5/5
d4 32
AN:
=
Alors
=
=
M t = G I0 4
24.10 80000x I0
x74 32
= 235.7 mm
4
Be tci tech nol ogi es
avec I 0 =
Mt GI0
4
-5
24.10 80000x 235.7
= 127.10 rad/mm
2- Calculons la contrainte tangentielle maximale cas 1:
MAXI
donc
=G. .R
MAXI
Or: R= d/2
=G. .d/2
-5
AN:
MAXI
=(80000 x 127.10 x 3.5)
MAX =355.6 N/mm²
cas 2:
MAXI
AN:
MAXI
=
=
Mt
I0
v
x
24000x3.5 235.7
= 356.38 N/mm².
MAX =356.38 N/mm²
PROF:KOUAME K.M.
LA TORSION R.D.M
DOC : 5/5
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