AMME2301/AMME5301 Mohr’s Circle – Plane Stress • Lecturer: Dr Li Chang • Room s503, Building J07 • Tel: 9351-5572 • e-mail: [email protected] Mohr’s Circle – Plane Stress τsn s Ѳ y σnn n Ѳ [ ]2 [ ]2 [ ]2 [ ]2 x [c, 0] τsn σnn Mohr’s Circle – Plane Stress Transformation [σss,-τsn] σss τsn Ѳ s y σnn [σyy,-τxy] [c, 0] n σ Ѳ x [σxx, τxy] 2Ѳ [σnn,τsn] τ Mohr’s Circle – Principal Stresses & Maximum In-Plane Shear Stress [σavg, -τmax] [σyy, -τxy] [σ11, 0] [σ22, 0] [c, 0] 2Ѳp a) Principal stresses 2Ѳs [σxx,τxy] [σavg, τmax] b) Maximum in-plane shear stress τ Ѳp - Ѳs= 45o σ Mohr’s Circle – Plane Stress Transformation 9.18 A point on a thin plate is subjected to two successive states of stress as shown. Determine the resulting state of stress with reference to an oriented as shown on the bottom. (p460) I 45 kPa II 50 kPa 50o 30o 18 kPa τxy σxx σyy Mohr’s Circle – Plane Stress Transformation 9.18 A point on a thin plate is subjected to two successive states of stress as shown. Determine the resulting state of stress with reference to an oriented as shown on the bottom. (p460) Step 1: the resulting stresses from part I I 1) Mohr’s circle τxy 50 kPa σxx 30o σyy [σyy1, -τxy1] 2) Plane Stress Transformation [50, 0] [kPa] [c, 0] [0, 0] [kPa] [kPa] τsn 60o [σxx1, τxy1] σnn Mohr’s Circle – Plane Stress Transformation 9.18 A point on a thin plate is subjected to two successive states of stress as shown. Determine the resulting state of stress with reference to an oriented as shown on the bottom. (p460) Step 2: the resulting stresses from part II 1) Mohr’s circle 45 kPa II τxy σxx 50o 18 kPa σyy 2) Plane Stress Transformation [0,-45] [σyy2, -τxy2] [c, 0] β α 100o [-18, 45] τsn σnn [σxx2, τxy2] Mohr’s Circle – Plane Stress Transformation 9.18 A point on a thin plate is subjected to two successive states of stress as shown. Determine the resulting state of stress with reference to an oriented as shown on the bottom. (p460) Step 3: the total resulting stresses the resulting stresses from part I [kPa] I 50 kPa 45 kPa II [kPa] [kPa] 30o 50o the resulting stresses from part II 18 kPa 22.69 kPa the total resulting stresses 74.38 kPa 42.38 kPa Mohr’s Circle – Principal Stresses & Maximum In-Plane Shear Stress 9-15 Determine (a) the principal stresses and (b) the maximum in-plane shear stress and average normal stress at the point. Specify the orientation of the element in each case. (p. 459) 30MPa a) Principal stresses 45MPa 60MPa b) Maximum in-plane stress 14.9o Mohr’s Circle – Principal Stresses & Maximum In-Plane Shear Stress 9-15 Determine (a) the principal stresses and (b) the maximum in-plane shear stress and average normal stress at the point. Specify the orientation of the element in each case. (p. 459) 60MPa 30MPa 1) Mohr’s circle 45MPa 2) Principal stresses [σavg, -τmax] [-60, -30] [σ11, 0] 2Ѳp2 [σ22, 0] [c, 0] 2Ѳp1 3) Maximum in-plane stress 2Ѳs [σavg, τmax] τsn σnn [45, 30] Mohr’s Circle – Principal Stresses & Maximum In-Plane Shear Stress 9-15 Determine (a) the principal stresses and (b) the maximum in-plane shear stress and average normal stress at the point. Specify the orientation of the element in each case. (p. 459) 60MPa 30MPa 1) Mohr’s circle 2) Principal stresses 45MPa 3) Maximum in-plane stress [σavg, -τmax] [-60, -30] 4) Draw infinitesimal elements [σ11, 0] 7.5 MPa [σ22, 0] 14.9o 30.1o [c, 0] 60.5 MPa 2Ѳp1 2Ѳs [σavg, τmax] τsn σnn [45, 30] AMME2301/AMME5301 Theories of Failure • Lecturer: Dr Li Chang • Room s503, Building J07 • Tel: 9351-5572 • e-mail: [email protected] Stress & Strain: Mechanical Properties of Materials σult σY y x Maximum-Normal-Stress Theory: Brittle Materials [σavg, -τmax] [σyy, -τxy] [σ22, 0] [σ11, 0] σ [c, 0] 2Ѳ [σxx,τxy] τ 2Ѳs lσ22l, lσ11l < σult p [σavg, τmax] τ Brittle materials σult σult σY y 0 x Maximum-Normal-Stress Theory: Ductile Materials Ductile materials Maximum-Normal-Stress Theory: Ductile Materials σY = 2 τmax Ductile materials Tresca’s Yield Criterion (Maximum-Shear-Stress Theory) [σavg, -τmax] σY y [0, 0] [c, 0] [σ11, 0] σ x Lüders lines σY = 2τmax τ [σavg, τmax] Ductile materials Tresca’s Yield Criterion (Maximum-Shear-Stress Theory) [σavg, -τmax] [σyy, -τxy] [σ22, 0] [σ11, 0] σ [c, 0] 2Ѳ [σxx,τxy] τ z 2Ѳs σY = 2τmax p [σavg, τmax] σ11 σ22 y x σ11 z x von Mises Yield Criterion (Maximum-Distortion-Energy Theory) z σ33 σ11 x σ22 y z σavg = σavg σavg x y z + -σ +σ avg 33 -σavg+ σ11 x -σavg+ σ22 y (textbook p.526-7) Plane Stresses: Thin-Walled Pressure Vessels 10-89 The gas tank is made from A-36 steel and has an inner diameter of 1.5 m. If the tank is designed to withstand a pressure of 5 MPa, determine the required minimum wall thickness to the nearest milimeter using (a) the maximum-shear-stress theory, and (b) maximumdistortion-energy theory. Apply a factor of safety of 1.5 against yielding. (p.535) A-36 steel: σY = 250MPa F.S. = 1.5 σY‘= 250MPa/1.5 = 166.7MPa (a) the maximum-shear-stress theory t > Pr/σY = ( 5 MPa x 0.75 m)/166.7 MPa = 22 mm σxx y x σθθ Plane Stresses: Thin-Walled Pressure Vessels 10-89 The gas tank is made from A-36 steel and has an inner diameter of 1.5 m. If the tank is designed to withstand a pressure of 5 MPa, determine the required minimum wall thickness to the nearest milimeter using (a) the maximum-shear-stress theory, and (b) maximumdistortion-energy theory. Apply a factor of safety of 1.5 against yielding. (p.535) A-36 steel: σY = 250MPa F.S. = 1.5 σY‘= 250MPa/1.5 = 166.7MPa (b) maximum-distortion-energy theory σxx y x t > Pr/(2σY) = ( 5 MPa x 0.75 m)/(2 x 166.7 Mpa) = 19 mm σθθ
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