Homework Help is Here – Start Your Trial Now!
Engineering
Mechanical Engineering
Two 30 mm thick plate is fastened by two bolts, 25 mm in diameter. If the plate is subjected to 50 KN tension, find the bearing stress in bolts. A. 33333.33 KPa B. 4444.44 KPa C. 5555.55 KPa D. 555555 Kpa

Two 30 mm thick plate is fastened by two bolts, 25 mm in diameter. If the plate is subjected to 50 KN tension, find the bearing stress in bolts. A. 33333.33 KPa B. 4444.44 KPa C. 5555.55 KPa D. 555555 Kpa

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
See similar textbooks
icon
Related questions
Question
100%
Two 30 mm thick plate is fastened by two bolts, 25 mm in diameter. If the plate is subjected to 50 KN tension, find the bearing stress in bolts. A. 33333.33 KPa B. 4444.44 KPa C. 5555.55 KPa D. 555555 Kpa 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.
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,
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
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 2 steps with 1 images

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Similar questions
Recommended textbooks for you
Elements Of Electromagnetics
Elements Of Electromagnetics
Mechanical Engineering
ISBN:
9780190698614
Author:
Sadiku, Matthew N. O.
Publisher:
Oxford University Press
Mechanics of Materials (10th Edition)
Mechanics of Materials (10th Edition)
Mechanical Engineering
ISBN:
9780134319650
Author:
Russell C. Hibbeler
Publisher:
PEARSON
Thermodynamics: An Engineering Approach
Thermodynamics: An Engineering Approach
Mechanical Engineering
ISBN:
9781259822674
Author:
Yunus A. Cengel Dr., Michael A. Boles
Publisher:
McGraw-Hill Education
Control Systems Engineering
Control Systems Engineering
Mechanical Engineering
ISBN:
9781118170519
Author:
Norman S. Nise
Publisher:
WILEY
Mechanics of Materials (MindTap Course List)
Mechanics of Materials (MindTap Course List)
Mechanical Engineering
ISBN:
9781337093347
Author:
Barry J. Goodno, James M. Gere
Publisher:
Cengage Learning
Engineering Mechanics: Statics
Engineering Mechanics: Statics
Mechanical Engineering
ISBN:
9781118807330
Author:
James L. Meriam, L. G. Kraige, J. N. Bolton
Publisher:
WILEY
SEE MORE TEXTBOOKS