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- University
- Politecnico di Milano
- Degree programme
- Energy Engineering
- Subject
- Meccanica dei Solidi
- Academic year
- 2011-2012
- Classification
- Exam · Full exam
- Content
- Solution only
- Original format
- Text
- Searchable text
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25 6 12 soluzioni
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Meccanica dei SolidiFull exam
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250612ENE.001EQUILIBRIO Nome: @ Adolfo Zavelani Rossi, Politecnico di Milano, vers.02.05.12 21.06.12 A B D E F G VCB W CB F W HD VG q q EQUAZIONI DI EQUILIBRIO Rotazione intorno a A: aste AB BC BE EF FC FG CD -HDb +2VGb = -2Fb +W +9/2qb 2 Rotazione intorno a B: aste BC 2VCBb +WCB = 0 Traslazione verticale: aste EF FC FG CD VG -VCB = 3qb Rotazione intorno a F: aste FC FG CD HDb -WCB = W +1/2qb 2 Matrice di equilibrio HDb V Gb V CBb W CB Fb W qb 2 ϕAB -1 2 0 0 -2 1 9/2 ϕBC 0 0 2 1 0 0 0 vEB 0 1 -1 0 0 0 3 = ϕFE 1 0 0 -1 0 1 1/2 Soluzione del sistema Fb W qb 2 HDb 1 0 1 VCBb -1/2 1/2 -1/4 VGb -1/2 1/2 11/4= WCB 1 -1 1/2 250612ENE.001REAZIONI Nome: @ Adolfo Zavelani Rossi, Politecnico di Milano, vers.02.05.12 21.06.12 F 1/4F F 1/4F Fb A B 1/4F 1/4F 1/2Fb B C 2F F 1/2Fb 2F C D F Fb F 2Fb B E F 2Fb F 2F E F 2F 3/4F Fb 2F 3/4F Fb C F 11/4F Fb 11/4F Fb F G 250612ENE.001DEFORMATA E AZIONI INTERNE Nome: @ Adolfo Zavelani Rossi, Politecnico di Milano, vers.02.05.12 21.06.12 12 Fb3/EJ A B C D E F G -1/4 0 2 2 0 1 1 3/411/4 F 1 1/4 1 0 1 0 -2 -2 0 F 0 1 0 1/2 -1/2 0 1 2 2 0 1 -1-1 -1 Fb 250612ENE.001PROCEDIMENTO E RISULTATI Nome: @ Adolfo Zavelani Rossi, Politecnico di Milano, vers.02.05.12 21.06.12 A = 612. mm 2 Ju = 225968. mm 4 Jv = 40716. mm 4 yg = 32.65 mm N = 2325. N Ty = -6200. N Mx = 1426000. Nmm xm = 12. mm um = -9. mm vm = -32.65 mm σm = N/A-Mv/Ju = 209.8 N/mm 2 xc = 21. mm yc = 9. mm vc = -23.65 mm σc = N/A-Mv/Ju = 153. N/mm 2 τc = 16.71 N/mm 2 σo = √σ 2 +3τ 2 = 155.7 N/mm 2 S * = 3655. mm 3 mm 0 12 18 24 30 42x 0 6 48 54 y 9 σc,τc σm u v 250612ENE.002EQUILIBRIO Nome: @ Adolfo Zavelani Rossi, Politecnico di Milano, vers.02.05.12 21.06.12 A B D E F G HCB VCB F W HD VG q q EQUAZIONI DI EQUILIBRIO Rotazione intorno a A: aste AB BC BE EF FC FG CD -HDb +2VGb = -2Fb +W +9/2qb…
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