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

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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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4-Determine the rotation angle between A and C, if the torque is T=0.9 N.m in point C, if G=80 GPa and diameter of all shafts are D=4 mm. (The radius of small circle are r and the bigger one are 2r) e are Coder 40 T=09N.N
a: What is the magnitude of torque applied to the shaft? b: The magnitude of the Polar Moment of Inertia of the shaft cross-section? c: The magnitude of the shear stress at r=0in (center of shaft) is?
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PLEASE ANSWER PART 4 and 5 Part 1 Calculate the polar moments of inertia in segments (1) and (2). Answer: J₁ = 13.84 in.⁴ J₂= 1.535 in. ⁴ Part 2 Calculate the internal torques T1 and T2 in shaft segments (1) and (2), respectively. On paper, prepare an internal torque diagram for the compound shaft that shows these internal torques. Use the sign convention presented in Section 6-6. Answer: T₁ = 800 Ib-ft, T₂= -800 Ib-ft. Part 3 Calculate the maximum shear stress in each shaft segment. On paper, prepare a diagram that shows the maximum shear stress in segments (1) and (2) of the shaft. Use the sign convention presented in Section 6-6. Answer: τ₁ = 1560.69 psi, τ₂ = -7817.59 psi Part 4 (PLEASE ANSWER) Determine the rotation angle of B with respect to the support at A. Please Answer: φB= rad. Part 5 (PLEASE ANSWER) Determine the rotation angle of C with respect to the support at A. Please Answer: φC= rad.
A solid 0.64-in.-diameter shaft is subjected to the torques shown. The bearings shown allow the shaft to turn freely. Determine the shear stress magnitude in shaft (3).