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ExamSecond midtermSolution only

06 02 15 soluzioni

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

Meccanica dei SolidiSecond midterm

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A1.001MdS Energetici 2014/15 Nome: @ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.02.15 y,Ty My 64 24 0 θ,Mt z,N Mx mm x,Tx55300-30-55 t=1.64 t=1.64 t=2.04 t=1.64 t=1.64 t=2.04 t=2.04 110 64.82 A B C F E G D H Calcolo degli sforzi in * con forze baricentriche essendo * il punto medio di BF N = 5000 N Ty = 1290 N Mt = 4600 Nmm Mx = -65500 Nmm σa = 50 N/mm 2 E = 200000 N/mm 2 G = 80000 N/mm 2 yG = 20.63 mm uo = 0 mm vo = -17.02 mm A = 491 mm 2 Su * = 0 mm 3 Cw = 34249904 mm 6 Ju = 194784 mm 4 Jv = 358303 mm 4 Jt = 544.3 mm 4 σ(N) = 10.18 N/mm 2 σ(Mx) = 2.903 N/mm 2 τ(Mt)d = 17.24 N/mm 2 τ(Tyc) = 13.25 N/mm 2 τ(Tyb)d= 0 N/mm 2 τ(Ty)s = 13.25 N/mm 2 τ(Ty)d = 13.25 N/mm 2 σ = 13.09 N/mm 2 τs = -3.995 N/mm 2 τd = 30.49 N/mm 2 σIs = 37.73 N/mm 2 σIIs = -24.64 N/mm 2 σId = 14.21 N/mm 2 σIId = -1.123 N/mm 2 σtresca = 62.37 N/mm 2 σmises = 54.4 N/mm 2 σst.ven = 43.89 N/mm 2 θt = 0.1056 /m ru = 19.92 mm rv = 27.01 mm ro = 37.63 mm Jp = 695240 mm 4 u v 20.63 A1.002MdS Energetici 2014/15 Nome: @ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.02.15 y,Ty My 60 24 0 θ,Mt z,N Mx mm x,Tx55300-30-55 t=1.64 t=1.64 t=2.04 t=1.64 t=1.64 t=2.04 t=2.04 110 60.82 A B C F E G D H Calcolo degli sforzi in * con forze baricentriche essendo * il punto medio di BF N = 5330 N Ty = 1310 N Mt = 3250 Nmm Mx = -66200 Nmm σa = 50 N/mm 2 E = 200000 N/mm 2 G = 80000 N/mm 2 yG = 19.21 mm uo = 0 mm vo = -14.84 mm A = 474.6 mm 2 Su * = 0 mm 3 Cw = 29476892 mm 6 Ju = 165875 mm 4 Jv = 343615 mm 4 Jt = 521.6 mm 4 σ(N) = 11.23 N/mm 2 σ(Mx) = 2.878 N/mm 2 τ(Mt)d = 12.71 N/mm 2 τ(Tyc) = 14.67 N/mm 2 τ(Tyb)d= 0 N/mm 2 τ(Ty)s = 14.67 N/mm 2 τ(Ty)d = 14.67 N/mm 2 σ = 14.11 N/mm 2 τs = 1.958 N/mm 2 τd = 27.38 N/mm 2 σIs = 35.33 N/mm 2 σIIs = -21.22 N/mm 2 σId = 14.37 N/mm 2 σIId =…

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