Exercício Pórtico Resolvido-2022 2

Dimensionar o ponto espacial abaixo empregar Coulomb: σAT = 5MPa e σNL = 30MPa (despaóstico constante). Seção Circular, S = Z. [Diagram with coordinates: x, y, z] [Diagram with forces: 20kN, 10kN] [Diagram with moments: 40kNm \ Mx, My] [Diagram with moments: 20kNm \ Mz] [Diagram of force and moment interaction] Seção crítica - Engaste [Diagram with circle and moments: My = 40kNm, Mz = 20kNm] [(Estrutura contínua 'atrás' da folha)] σmin = ? [Circle with diagrams of force and moments: Mt, Tmax] MR = [ \sqrt{40^2 + 20^2} = 40,72kNm θ = cos^{-1} \left( \frac{40}{44,72} \right) = 26,56^° [Diagrams with torsion and shear stress] Rebatendo no plano principal: σmin = \frac{44,72 \times 10^6 \times R}{\pi R^4} = \frac{56,8 \times 10^6}{R^3} τmin = \frac{20 \times 10^6 \times R}{\pi R^4} = \frac{12,7 \times 10^6}{R^3} Coulomb: σA + σ3 = 1/5 σAT σNL σPM = [ \left( \frac{56,9}{2} \right)^2 + \left( \frac{56,9}{2} + (12,7) \right)^2 ] \times \frac{10^6}{R^3} σPM = 59,6 \times \frac{10^6}{R^3} = σ1 - 2,72 \times \frac{10^6}{R^3} = σ3 \left( \frac{59,6}{5} + \frac{-2,72}{-30} \right) \times \frac{10^6}{R^3} = 1/2 R = 288\ \text{mm} Para a estrutura dimensionada acima, calcular a se seguir uma carga de 100kN é adicionada desplastar as Coulombs. [Diagram with forces: 20kN, 100kN, 10kN] [Diagram with forces: N, -100kN] SEÇÃO CRÍTICA - ENGASTE Mt = 20 kNm N = 100 kN My = 40 kNm Mz = 20 kNm Mt = 20 kNm N = 100 kN M = 44,72 kNm como |sigma_máx| > sigma_máx ambos pontos devem ser verificados. PONTO 1 tau_máx = 20 x 10^3/((pi/4) x 288) = 0,53 MPa sigma_máx = 100 x 10^3/((pi/4) x 288) = -0,38 + 44 x 10^3/((pi/4) x 288) * 3/4 = 2,34 sigma_máx = 1,96 MPa sigma_PA = (1,96/2) + sqrt((1,96/2)^2 + 0,53^2) = 2,09 MPa = sigma_1 sigma_PA = -0,13 MPa = sigma_3 2,09/5 + -0,13/-30 = 1/S1 -> S1 = 2,37 PONTO 2 tau_máx = 0,53 MPa sigma_máx = -0,38 - 2,39 = -2,72 MPa sigma_PA = (-2,72/2) + sqrt((2,72/2)^2 + 0,53^2) sigma_PA = 0,10 MPa = sigma_1 -2,82 MPa = sigma_3 0,10/5 + -2,82/-30 = 1/S2 -> S2 = 8,77 R: S3 = 2,37