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The shaft whose torque varies from 2000 to 6000 in-lbs has 1 ½ inch in diameter and 60,000 psi yield strength. Compute for the shaft mean average stress. A. 6036 psi B. 6810 psi C. 5162 psi D. 5550 psi Please solve the Problem elaborately and Refer Machine Elements and Stresses Equations from the figures or use other shortcut Methods you have or you may Solve both or in more different methods the better ?. Your solution will be use as reference for my Fu

The shaft whose torque varies from 2000 to 6000 in-lbs has 1 ½ inch in diameter and 60,000 psi yield strength. Compute for the shaft mean average stress. A. 6036 psi B. 6810 psi C. 5162 psi D. 5550 psi Please solve the Problem elaborately and Refer Machine Elements and Stresses Equations from the figures or use other shortcut Methods you have or you may Solve both or in more different methods the better ?. Your solution will be use as reference for my Fu

Elements Of Electromagnetics
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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The shaft whose torque varies from 2000 to 6000 in-lbs has 1 ½ inch in diameter and 60,000 psi yield strength. Compute for the shaft mean average stress. A. 6036 psi B. 6810 psi C. 5162 psi D. 5550 psi Please solve the Problem elaborately and Refer Machine Elements and Stresses Equations from the figures or use other shortcut Methods you have or you may Solve both or in more different methods the better ?. Your solution will be use as reference for my Future Board Exam preparation/review/study. Thank you so much dear your work will be appreciated much and rated excellently.
V1 Machine Elements 1. Cylinders Rolling in opposite direction: Á. Tangential speed B. Relation of diameter and speed V, = V2 = T D, N, = r D2 N2 DI N1 = D2 N2 Speed of Driver C. Speed Ratio = Speed of the Driven D. Center Distance = R, + R2 = D, + D2 2. Cylinders Rolling in the same direction Vi = V2 = n D, N; = n D2 N2 D, N1 = D2 N2 A. Tangential speed B. Relation of diameter and speed Speed of Driver Speed of the Driven D2 - D1 C. Speed Ratio = D. Center Distance = R2 -R, = Stresses 1. 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 failure 3. Yield stress(Sy) - maximum stress without causing deformation 4. Allowable stress(Sal) = Ultimate stress/Factor of Safety Sy 5. Design stress(Sa) - stress used in determining the size of a member. Sa = or Sa = FS FS where: FS = factor of safety 1. Tensile Stress (S.) S = For solid circular cross-section: A = A D2 #0,² -D²) For hollow circular cross-section: A = For rectangular cross-section: A = base x height = b x h Fa 2. Compressive Stress(Sc)S. = A 3. Shearing Stress(S.) F A. For single bolt of rivet needed to join to plates together. S, = For single rivet: A = t/4 D² For double riveted joint: A = 2(TT/4 D²) where: D B. 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 square C. Pressure needed to punch a hole, F: F = d x t x 80, tons t = plate thickness t = thickness, in where: A = DL Where: d = hole diameter, in -Shear area 4. Bearing Stress(Sb) Sp = Fb IA a. Based on yield strength FS = Sy / Sall 5. Factor of safety(FS) b. Based on ultimate strength FS = S,/ Sall D 6. Torsional Shear Stress(Ss) Projected Area D
Transcribed Image Text:V1 Machine Elements 1. Cylinders Rolling in opposite direction: Á. Tangential speed B. Relation of diameter and speed V, = V2 = T D, N, = r D2 N2 DI N1 = D2 N2 Speed of Driver C. Speed Ratio = Speed of the Driven D. Center Distance = R, + R2 = D, + D2 2. Cylinders Rolling in the same direction Vi = V2 = n D, N; = n D2 N2 D, N1 = D2 N2 A. Tangential speed B. Relation of diameter and speed Speed of Driver Speed of the Driven D2 - D1 C. Speed Ratio = D. Center Distance = R2 -R, = Stresses 1. 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 failure 3. Yield stress(Sy) - maximum stress without causing deformation 4. Allowable stress(Sal) = Ultimate stress/Factor of Safety Sy 5. Design stress(Sa) - stress used in determining the size of a member. Sa = or Sa = FS FS where: FS = factor of safety 1. Tensile Stress (S.) S = For solid circular cross-section: A = A D2 #0,² -D²) For hollow circular cross-section: A = For rectangular cross-section: A = base x height = b x h Fa 2. Compressive Stress(Sc)S. = A 3. Shearing Stress(S.) F A. For single bolt of rivet needed to join to plates together. S, = For single rivet: A = t/4 D² For double riveted joint: A = 2(TT/4 D²) where: D B. 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 square C. Pressure needed to punch a hole, F: F = d x t x 80, tons t = plate thickness t = thickness, in where: A = DL Where: d = hole diameter, in -Shear area 4. Bearing Stress(Sb) Sp = Fb IA a. Based on yield strength FS = Sy / Sall 5. Factor of safety(FS) b. Based on ultimate strength FS = S,/ Sall D 6. Torsional Shear Stress(Ss) Projected Area D
where: J = polar moment of inertia = T Dª /32 (for solid shaft) c = distance from neutral axis to the farthest fiber T = torque c = r (for circular cross section) 7. Bending Stress(S:) Si = d = diameter For rectangular beam S = 6 M b h? Mc where: M = moment F c = distance of farthest fiber from neutral axis | = moment of inertia about the neutral axis | = bh/12 (for rectangular cross section) I M c Sf L Z = section modulus Y 8. Strain and Elongation Strain = Stress FL Stress = A E Strain FL Y = AE AY where: y = elongation due to applied load L = original length F = force A = area S = stress 9. Thermal Elongation; Stresses y = kL (t2 - t1) S = E! = kE (t2 - t1) where: k = coeficient of thermal expansion, m/m-°C For steel k = 6.5 x 10-6 in/in-F = 11.7 x 106 m/m-C E = 30 x 10° psi Relation between shearing and tensile stress based on theory of failure: Samax = Sty /2 Smax = Sty 10. Variable Stress 1 Sm Sa FS Sy S, Sy = yield point where: FS = factor of safety Sn = endurance limit Smax + Smin S. = variable component stress = Smax - Smin Sm = mean stress = 2 2 Smax = maximum stress 11. Poisson's Ratio(u) = is the ratio of lateral unit deformation to axial unit deformation. Smin = minimum stress E F E = strain =- AE where: G = shear modulus of elasticity 2G W2 – W1 Ey t2 - ty Lateral Strain Ey Ez Ex L2 -L, Ex L1 Ez Longitudin al Strain Ex t,
Transcribed Image Text:where: J = polar moment of inertia = T Dª /32 (for solid shaft) c = distance from neutral axis to the farthest fiber T = torque c = r (for circular cross section) 7. Bending Stress(S:) Si = d = diameter For rectangular beam S = 6 M b h? Mc where: M = moment F c = distance of farthest fiber from neutral axis | = moment of inertia about the neutral axis | = bh/12 (for rectangular cross section) I M c Sf L Z = section modulus Y 8. Strain and Elongation Strain = Stress FL Stress = A E Strain FL Y = AE AY where: y = elongation due to applied load L = original length F = force A = area S = stress 9. Thermal Elongation; Stresses y = kL (t2 - t1) S = E! = kE (t2 - t1) where: k = coeficient of thermal expansion, m/m-°C For steel k = 6.5 x 10-6 in/in-F = 11.7 x 106 m/m-C E = 30 x 10° psi Relation between shearing and tensile stress based on theory of failure: Samax = Sty /2 Smax = Sty 10. Variable Stress 1 Sm Sa FS Sy S, Sy = yield point where: FS = factor of safety Sn = endurance limit Smax + Smin S. = variable component stress = Smax - Smin Sm = mean stress = 2 2 Smax = maximum stress 11. Poisson's Ratio(u) = is the ratio of lateral unit deformation to axial unit deformation. Smin = minimum stress E F E = strain =- AE where: G = shear modulus of elasticity 2G W2 – W1 Ey t2 - ty Lateral Strain Ey Ez Ex L2 -L, Ex L1 Ez Longitudin al Strain Ex t,
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