P - Simone Caffè

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Dott. Ing Simone Caffè ____________________________________________________________________________________________________________ 2) Mensola tozza con meccanismo di rottura tirante – puntone (ferri inclinato) α l P z Equilibrio verticale: Nt ⋅ sinα − P = 0 → h P Nt = sinα Equilibrio orizzontale: Nc + Nt ⋅ cosα = 0 → Nc = −P ⋅ cot α Resistenza dell’armatura piegata: = N ⋅ sinα = A ⋅ f ⋅ sinα ≥ P Pw , Rd t sw yd Nc 1 Resistenza biella compressa : Pc , Rd = = 0 . 2 ⋅ d ⋅ b ⋅ fcd ⋅ ≥ P cot α cot α Definendo ~ λ = cot α = l z l = 0. 9 ⋅ d e ricordandosi che sin α = 1 1+ cot 1 Resistenza dell’armatura piegata: Pw , Rd = Asw ⋅ fyd ⋅ ≥ P ~ 2 1+ λ 1 Resistenza biella compressa : Pc , Rd = 0 . 2 ⋅ d ⋅ b ⋅ fcd ⋅ ~ ≥ P λ 2 α
Dott. Ing Simone Caffè ____________________________________________________________________________________________________________ 3) Mensola tozza con meccanismo di rottura tirante – puntone (ferri inclinato + armature tese dritte) • • 1 P + N 2 45° N + N ⋅ cosψ = 0 2 z l P Ψ ⋅ sinψ = 0 N3 N2 ⋅ cosψ + = 0 2 N3 − N2 ⋅ cosψ + N4 + = 0 2 l' z N1 = P ⋅ cotψ → P = − sinψ h → → 45° N l 1 ψ 3 2 4 N = P ⋅ cot ψ N N 1 2 3 4 P = − sinψ = −N 2 = −P ⋅ 1 Pt , Rd = As ⋅ fyd ⋅ cotψ ψ ⋅ sinψ ⋅ P 2 = P ⋅ ( cot ψ + 1) = −P ⋅ N2 → Pc , Rd = 0 . 2 ⋅ b ⋅ d ⋅ fcd ⋅ sinψ N = P ⋅ 2 → 3 l N = −P ⋅ z 4 → fyd Pw , Rd = Asw ⋅ 2 Pc′ , Rd = 0 . 2 ⋅ b ⋅ d ⋅ fcd ⋅ Il meccanismo di rottura in questo caso è la sovrapposizione dei due meccanismi precedenti: Nt 1 Resistenza dell’armatura tesa: Pt , Rd = = As ⋅ fyd ⋅ ≥ P cot ψ cot ψ 1 Pt , Rd = As ⋅ fyd ⋅ ≥ P λ′ l′ λ′ = z z l l z 2
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P - Simone Caffè

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