Corrigé De L'evalution
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Corrigé de l'évaluation
Durée:5min
1 - Déterminons l'angle unitaire de torsion. I0 =
d4 32
AN:
=
80000 80000x I0
Alors
=
80000 80000x 4970
Mt GI0
=
chn olo gie s
M t = G I0
4
avec I0 =
x15 32
= 4970 mm
4
-4
= 2.10 rad/mm
2- Calculons la contrainte tangentielle maximale cas 1:
MAXI
donc
=G. .R MAXI
Or: R= d/2
=G. .d/2
-4
MAXI
=(80000 x 2.10 x 7.5)
Be tci te
AN:
MAX =120 N/mm²
cas 2:
MAXI
AN:
MAXI
=
=
Mt x I0
v
80000x7.5 4970
= 120.72 N/mm².
MAX =120.72 N/mm²
PROF:KOUAME K.M.
LA TORSION R.D.M
DOC : 2/2
Durée:5min
1 - Déterminons l'angle unitaire de torsion. I0 =
d4 32
AN:
=
80000 80000x I0
Alors
=
80000 80000x 4970
Mt GI0
=
chn olo gie s
M t = G I0
4
avec I0 =
x15 32
= 4970 mm
4
-4
= 2.10 rad/mm
2- Calculons la contrainte tangentielle maximale cas 1:
MAXI
donc
=G. .R MAXI
Or: R= d/2
=G. .d/2
-4
MAXI
=(80000 x 2.10 x 7.5)
Be tci te
AN:
MAX =120 N/mm²
cas 2:
MAXI
AN:
MAXI
=
=
Mt x I0
v
80000x7.5 4970
= 120.72 N/mm².
MAX =120.72 N/mm²
PROF:KOUAME K.M.
LA TORSION R.D.M
DOC : 2/2
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