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A 2.5 in diameter by 1.8 in long journal bearing is to carry 5000 lb load at 320 rpm using SAE 40 lube oil at 200 oF through a single hole at 30 psi. Compute the bearing pressure. A. 1111.11 psi B. 142.23 psi C. 123.34 psi D. 197.34 psi Please solve the Problem elaborately. Your solution will be use as reference for my studies. Thank you so much your work will be appreciated much!

A 2.5 in diameter by 1.8 in long journal bearing is to carry 5000 lb load at 320 rpm using SAE 40 lube oil at 200 oF through a single hole at 30 psi. Compute the bearing pressure. A. 1111.11 psi B. 142.23 psi C. 123.34 psi D. 197.34 psi Please solve the Problem elaborately. Your solution will be use as reference for my studies. Thank you so much your work will be appreciated much!

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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A 2.5 in diameter by 1.8 in long journal bearing is to carry 5000 lb load at 320 rpm using SAE 40 lube oil at 200 oF through a single hole at 30 psi. Compute the bearing pressure. A. 1111.11 psi B. 142.23 psi C. 123.34 psi D. 197.34 psi Please solve the Problem elaborately. Your solution will be use as reference for my studies. Thank you so much your work will be appreciated much!
C. Pressure needed to punch a hole E are 80. tons e thickness Machine Elements 1. Cylinders Rolling in opposite direction: A. Tangential speed V = Va = n D, N == D, Na D. N. = D. Na B. Relation of diameter and speed Speed of Driver Speed of the Driven Speed Ratio D. Center Distance- R, + R, D, + D2 2 2. Cylinders Rolling in the same direction A. Tangential speed VVn D, N = D, N, B. Relation of diameter and speed D: N = D, Na Speed of Driver Speed of the Driven . C. Speed Ratio = D. Center Distance = R, -R,- D-D, 2 Stresses 1. Stress (S) = a total resistance that a material offers to an applied load, Ibin, kg/cm, KN/m? 2. Ultimate stress (S.) - is the stress that would cause failure 3. Yield stress(S,) - maximum stress without causing deformation 4. Allowable stress(Sa) = Ultimate stress/Factor of Safety 5. Design stress(S.) - stress used in determining the size of a member. S= or S= where: FS - factor of safety 1. Tensile Stress (S) S = 4 For solid circular cross-section: A = D For hollow circular cross-section: A = (D, -D) For rectangular cross-section: A = base x height = b xh 2. Compressive Stress(S:)S. = 4 3. Shearing Stress(S.) A. For single bolt of rivet needed to join to plates together. S, = For single rivet: A= z4 D For double riveted joint: A= 2(m/4 D) where: B. Shearing due to punching of hole. S= where A DI (for punching a hole) A = 4St (for square hole) Where: S= length of side of square C. Pressure needed to punch a hole, F: F=dxtx 80, tons t= thickness, in where: A = DL t= plate thickness Where: d= hole diameter, in 4. Bearing Stress(S) S, F IA She we 5. Factor of safety(FS) a. Based on yield strength FS = S, /S. b. Based on ultimate strength FS = s./ Sa 6. Torsional Shear Stress(S.) Projected Area where: J = polar moment of inertia = T D' /32 (for solid shaft) T= torque c - distance from neutral axis to the farthest fiber Cr (for circular cross section) d- diameter 7. Bending Stress(S) S 6 M For rectangular beam S= bh where: M = moment C= distance of farthest fiber from neutral axis - moment of inertia about the neutral axis = bh/12 (for rectangular cross section) Z = section modulus = c S E. Stress . FL Strain 8. Strain and Elongation Strain = ! Stress = Y= FL - AX * AE where: y- elongation due to applied load force L- original length S- stress A area 9. Thermal Elongation; Stressos y = kL (t - t:) S = EL = kE (t - t) where: k = coefficient of thermal expansion, m/m-C For steel k- 6.5 x 104 inin-F- 11.7 x 104 m/m-C E = 30 x 10 psi Relation betwoen shearing and tensile stress based on theory of failure: Swa- S 10. Variable Stress 1 S. 5. FS S, S S, - yield point S - mean stress Suma- Sy 12 where: FS = factor of safety S. endurance limit Smax + Sn S. = variable component stress= S S.. maximum stress Sen minimum stress 11. Poisson's Ratio(u) = is the ratio of lateral unit deformation to axial unit deformation. where: G = shear modulus of elasticity F Estrain AE 2G AE EyE E, Ex Lateral Strain Wa-W Longitudin al Strain E, 2-
Transcribed Image Text:C. Pressure needed to punch a hole E are 80. tons e thickness Machine Elements 1. Cylinders Rolling in opposite direction: A. Tangential speed V = Va = n D, N == D, Na D. N. = D. Na B. Relation of diameter and speed Speed of Driver Speed of the Driven Speed Ratio D. Center Distance- R, + R, D, + D2 2 2. Cylinders Rolling in the same direction A. Tangential speed VVn D, N = D, N, B. Relation of diameter and speed D: N = D, Na Speed of Driver Speed of the Driven . C. Speed Ratio = D. Center Distance = R, -R,- D-D, 2 Stresses 1. Stress (S) = a total resistance that a material offers to an applied load, Ibin, kg/cm, KN/m? 2. Ultimate stress (S.) - is the stress that would cause failure 3. Yield stress(S,) - maximum stress without causing deformation 4. Allowable stress(Sa) = Ultimate stress/Factor of Safety 5. Design stress(S.) - stress used in determining the size of a member. S= or S= where: FS - factor of safety 1. Tensile Stress (S) S = 4 For solid circular cross-section: A = D For hollow circular cross-section: A = (D, -D) For rectangular cross-section: A = base x height = b xh 2. Compressive Stress(S:)S. = 4 3. Shearing Stress(S.) A. For single bolt of rivet needed to join to plates together. S, = For single rivet: A= z4 D For double riveted joint: A= 2(m/4 D) where: B. Shearing due to punching of hole. S= where A DI (for punching a hole) A = 4St (for square hole) Where: S= length of side of square C. Pressure needed to punch a hole, F: F=dxtx 80, tons t= thickness, in where: A = DL t= plate thickness Where: d= hole diameter, in 4. Bearing Stress(S) S, F IA She we 5. Factor of safety(FS) a. Based on yield strength FS = S, /S. b. Based on ultimate strength FS = s./ Sa 6. Torsional Shear Stress(S.) Projected Area where: J = polar moment of inertia = T D' /32 (for solid shaft) T= torque c - distance from neutral axis to the farthest fiber Cr (for circular cross section) d- diameter 7. Bending Stress(S) S 6 M For rectangular beam S= bh where: M = moment C= distance of farthest fiber from neutral axis - moment of inertia about the neutral axis = bh/12 (for rectangular cross section) Z = section modulus = c S E. Stress . FL Strain 8. Strain and Elongation Strain = ! Stress = Y= FL - AX * AE where: y- elongation due to applied load force L- original length S- stress A area 9. Thermal Elongation; Stressos y = kL (t - t:) S = EL = kE (t - t) where: k = coefficient of thermal expansion, m/m-C For steel k- 6.5 x 104 inin-F- 11.7 x 104 m/m-C E = 30 x 10 psi Relation betwoen shearing and tensile stress based on theory of failure: Swa- S 10. Variable Stress 1 S. 5. FS S, S S, - yield point S - mean stress Suma- Sy 12 where: FS = factor of safety S. endurance limit Smax + Sn S. = variable component stress= S S.. maximum stress Sen minimum stress 11. Poisson's Ratio(u) = is the ratio of lateral unit deformation to axial unit deformation. where: G = shear modulus of elasticity F Estrain AE 2G AE EyE E, Ex Lateral Strain Wa-W Longitudin al Strain E, 2-
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