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ExamSecond midtermExam paper only

07 02 13 testi

Study material for Meccanica dei Solidi, shared by the Studwiz community and reviewed by moderators.

Meccanica dei SolidiSecond midterm

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IIcomp1213.001MdS Energetici 2012/13 000000 Accettura Carlotta @ Adolfo Zavelani Rossi, Politecnico di Milano, vers.28.01.13 07.02.13 y,Ty My 124 94 44 24 0 θ,Mt z,N Mx mm x,Tx57473717-17-37-47-57 0 109 114 124 BA C D E F G H I J KL M N O P Calcolo degli sforzi in * con forze baricentriche N = 117000 N Ty = 304000 N Mx = 5270000 Nmm σa = 50 N/mm 2 E = 200000 N/mm 2 G = 80000 N/mm 2 yG = A = Su * = Ju = Jv = σ(N) = σ(Mx)= τ(Ty) = σ = τ = σI = σII = σtresca= σmises= σst.ven= ru = rv = ro = IIcomp1213.002MdS Energetici 2012/13 000000 Agostinelli Matteo @ Adolfo Zavelani Rossi, Politecnico di Milano, vers.28.01.13 07.02.13 y,Ty My 120 94 44 24 0 θ,Mt z,N Mx mm x,Tx57473717-17-37-47-57 0 107 114 120 BA C D E F G H I J KL M N O P Calcolo degli sforzi in * con forze baricentriche N = 122000 N Ty = 256000 N Mx = 5100000 Nmm σa = 50 N/mm 2 E = 200000 N/mm 2 G = 80000 N/mm 2 yG = A = Su * = Ju = Jv = σ(N) = σ(Mx)= τ(Ty) = σ = τ = σI = σII = σtresca= σmises= σst.ven= ru = rv = ro = IIcomp1213.003MdS Energetici 2012/13 000000 Aita Matteo @ Adolfo Zavelani Rossi, Politecnico di Milano, vers.28.01.13 07.02.13 y,Ty My 116 94 44 24 0 θ,Mt z,N Mx mm x,Tx57473717-17-37-47-57 0 105 114 116 BA C D E F G H I J KL M N O P Calcolo degli sforzi in * con forze baricentriche N = 95300 N Ty = 296000 N Mx = 4830000 Nmm σa = 50 N/mm 2 E = 200000 N/mm 2 G = 80000 N/mm 2 yG = A = Su * = Ju = Jv = σ(N) = σ(Mx)= τ(Ty) = σ = τ = σI = σII = σtresca= σmises= σst.ven= ru = rv = ro =

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