A shaft when subjected to pure torsion developed a stress of 50 Mpa. If polar moment ofinertia is 6.1359 x 10-7 m4, determine the maximum torque the shaft could handle. A. 1.23 KN.m B. 1.68 KN.m C. 1.84 KN.m D. 2.48 KN.m where: J = polar moment of inertia = T Dª /32 (for solid shaft)c = distance from neutral axis to the farthest fiberT = torquec = r (for circular cross section)7. Bending Stress(S:) Si =d = diameterFor rectangular beam S = 6 Mb h?Mcwhere: M = momentFc = distance of farthest fiber from neutral axis| = moment of inertia about the neutral axis| = bh/12 (for rectangular cross section)I Mc SfLZ = section modulusY8. Strain and Elongation Strain =StressFLStress =AEStrainFLY =AEAYwhere: y = elongation due to applied loadL = original lengthF = forceA = areaS = stress9. Thermal Elongation; Stressesy = kL (t2 - t1) S = E! = kE (t2 - t1)where: k = coeficient of thermal expansion, m/m-°CFor steel k = 6.5 x 10-6 in/in-F = 11.7 x 106 m/m-CE = 30 x 10° psiRelation between shearing and tensile stress based on theory of failure:Samax = Sty /2Smax = Sty10. Variable Stress1 Sm SaFS Sy S,Sy = yield pointwhere: FS = factor of safetySn = endurance limitSmax + SminS. = variable component stress =Smax - SminSm = mean stress =22Smax = maximum stress11. Poisson's Ratio(u) = is the ratio of lateral unit deformation to axial unit deformation.Smin = minimum stressEFE = strain =-AEwhere: G = shear modulus of elasticity2GW2 – W1Eyt2 - tyLateral StrainEy EzExL2 -L,ExL1EzLongitudin al StrainExt, V1Machine Elements1. Cylinders Rolling in opposite direction:Á. Tangential speedB. Relation of diameter and speedV, = V2 = T D, N, = r D2 N2DI N1 = D2 N2Speed of DriverC. Speed Ratio =Speed of the DrivenD. Center Distance = R, + R2 =D, + D22. Cylinders Rolling in the same directionVi = V2 = n D, N; = n D2 N2D, N1 = D2 N2A. Tangential speedB. Relation of diameter and speedSpeed of DriverSpeed of the DrivenD2 - D1C. Speed Ratio =D. Center Distance = R2 -R, =Stresses1. Stress (S) = a total resistance that a material offers to an applied load, Ib/in? , kg/cm² , KN/m?2. Ultimate stress (Su ) - is the stress that would cause failure3. Yield stress(Sy) - maximum stress without causing deformation4. Allowable stress(Sal) = Ultimate stress/Factor of SafetySy5. Design stress(Sa) - stress used in determining the size of a member. Sa =or Sa =FSFSwhere: FS = factor of safety1. Tensile Stress (S.)S =For solid circular cross-section: A =AD2#0,² -D²)For hollow circular cross-section: A =For rectangular cross-section: A = base x height = b x hFa2. Compressive Stress(Sc)S. =A3. Shearing Stress(S.)FA. For single bolt of rivet needed to join to plates together. S, =For single rivet: A = t/4 D²For double riveted joint: A = 2(TT/4 D²)where:DB. Shearing due to punching of hole.S =where A = 1 Dt (for punching a hole) A = 4 St (for square hole)Where: S = length of side of squareC. Pressure needed to punch a hole, F: F = d x t x 80, tonst = plate thicknesst = thickness, inwhere: A = DLWhere: d = hole diameter, in-Shear area4. Bearing Stress(Sb)Sp = Fb IAa. Based on yield strength FS = Sy / Sall5. Factor of safety(FS)b. Based on ultimate strength FS = S,/ SallD6. Torsional Shear Stress(Ss)Projected AreaD

Answered: Image /qna-images/answer/5edf461b-de74-4ac9-89c9-f045067776a8.jpg

A shaft when subjected to pure torsion developed a stress of 50 Mpa. If polar moment of inertia is 6.1359 x 10 -7 m4 , determine the maximum torque the shaft could handle. A. 1.23 KN.m B. 1.68 KN.m C. 1.84 KN.m D. 2.48 KN.m

A shaft when subjected to pure torsion developed a stress of 50 Mpa. If polar moment of inertia is 6.1359 x 10 -7 m4 , determine the maximum torque the shaft could handle. A. 1.23 KN.m B. 1.68 KN.m C. 1.84 KN.m D. 2.48 KN.m

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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