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francois |
941 |
Solution exercice 2 du chapitre introduction
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Resolution du probleme statique
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Ray=375 N
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Rcy=1375 N
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Calcul des diagrammes V et M
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entre A et B
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Vab =
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-375
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Mab =
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375*x
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entre B et C
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Vbc =
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875
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Mbc =
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2500 - 875*x
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entre C et D
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Vcd =
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-500
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Mcd =
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500*x - 3000
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Calcul de la contrainte normale maximale
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sur la fibre superieure en x=4 sigma=4.166667e+007 Pa
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sur la fibre inferieure en x=4 sigma=-4.166667e+007 Pa
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Calcul de la contrainte tangeantielle maximale
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sur la fibre neutre entre B et C superieure tau=546875 Pa |
