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The solid cylinders AB and BC are bonded together at B and are attached to fixed supports at A and C. The modulus of rigidity is 3.3 × 106 psi for aluminum and 5.9 × 106 psi for brass. If T = 14 kip·in, a=1.1 in. and b=2.2 in., determine (a) The reaction torque at A (b) Maximum shear stress in BC
The solid cylinders AB and BC are bonded together at B and are attached to fixed supports at A and C. The modulus of rigidity is 3.3 × 106 psi for aluminum and 5.9 × 106 psi for brass. If T = 14 kip·in, a=1.1 in. and b=2.2 in., determine (a) The reaction torque at A (b) Maximum shear stress in BC
ThereactionTorqueat A and themaximum shear stressin BC are respectively; T_ab = 697 lb.in; T_bc = 13302 lb.in and τ_bc = 6.352 KsiWhat is Torque and Maximum Shear Stress?Thetorquesin cylinders AB and BC arestatically indeterminate.Matching therotationφ_b for each cylinder.For Cylinder AB; c = ¹/₂d = ¹/₂ * 1.1 = 0.55 inPolar Moment of Inertia; J = ¹/₂ *πc⁴ = ¹/₂ * π * 0.55⁴ = 0.1437 in⁴Rotation; φ_b = (T_ab * L)/GJWe are given;G = 3.3 * 10⁶L = 12 inThus;φ_b = (12/(3.3 * 10⁶ * 12))T_abφ_b = 2.53 × 10⁻⁵ T_abFor Cylinder BC; c = ¹/₂d = ¹/₂ * 2.2 = 1.1 inPolar Moment of Inertia; J = ¹/₂ *πc⁴ = ¹/₂ * π * 1.1⁴ = 2.2998 in⁴Rotation; φ_b = (T_bc * L)/GJWe are given;G = 5.9 * 10⁶L = 18 inThus;φ_b = (18/(5.9 * 10⁶ * 2.2998))T_bcφ_b = 1.3266 × 10⁻⁶ T_bcMatching expressions for φ_b, we have;2.53 × 10⁻⁵ T_ab = 1.3266 × 10⁻⁶ T_bcThus; T_bc = 19.0717T_ab -----(eq 1)Equilibriumof connection at B;T_ab + T_bc - T = 0We are given T = 14 kip·in = 14 × 10³ lb.inThus;T_ab + T_bc - (14 × 10³) = 0T_ab + T_bc = 14000 ----(eq 2)Combining equation 1 and 2, we have;T_ab = 697 lb.inT_bc = 13302 lb.inB) Formula formaximum shear stressis;τ = Tc/JMaximum shear stressin BC is;τ_bc = (13302 * 1.1)/2.2998τ_bc = 6.352 KsiRead more aboutTorque and Maximum Shear stressat;brainly.com/question/13507543...