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Agenda

Agenda. 1. Areas 2. Stress and strain 3. Axial loading 4. Torsion loading 5. Beam loading 6. Engineering materials 7. Vibration 8. Fatigue 9. Thermal loading. 1. Areas. Area First moment of an area Centroid of an area Moment of inertia of an area Parallel axis theorem

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Agenda

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  1. Agenda 1. Areas 2. Stress and strain 3. Axial loading 4. Torsion loading 5. Beam loading 6. Engineering materials 7. Vibration 8. Fatigue 9. Thermal loading

  2. 1. Areas • Area • First moment of an area • Centroid of an area • Moment of inertia of an area • Parallel axis theorem • Polar moment of inertia 1. Areas

  3. Area (1 of 2) dA A = area = y 3 dA = (y-1) dx y = -0.5x + 4 1 x 2 6 6 (-0.5x + 3)dx = 4 A = area = 2 Area by integrating in x direction 1. Areas

  4. Area (2 of 2) y 3 dA = (x-2) dx x = -2y + 8 1 x 2 6 3 (-2y + 6)dy = 4 A = area = 1 Area by integrating in y direction 1. Areas

  5. First moment of an area (1 of 3) Qy = first moment of area with respect to the y-axis = x dA Qx = first moment of area with respect to the x-axis = y dA Definitions of first moment of area 1. Areas

  6. First moment of an area (2 of 3) y 3 dA = (y-1) dx y = -0.5x + 4 x 1 x 2 6 6 Qy = (-0.5x + 3) x dx = 13.33 2 Qy 1. Areas

  7. First moment of an area (3 of 3) y 3 dA = (y-1) dx y = -0.5x + 4 1 y x 2 6 3 Qx = (-2y + 6) y dy = 6.67 1 Qx 1. Areas

  8. Centroid of an area (1 of 2) • xc = Qy/A • yc = Qx/A Centroids of area in terms of moments 1. Areas

  9. Centroid of an area (2 of 2) • xc = Qy/A = 13.33/4 = 3.33 • yc = Qx/A = 6.67/4 = 1.67 Centroid of an area for previous examples 1. Areas

  10. Moment of inertia of an area (1 of 3) Iy = first moment of area with respect to the y-axis = x2 dA Ix = first moment of area with respect to the x-axis = y2 dA Definitions of moment of inertia 1. Areas

  11. Moment of inertia of an area (2 of 3) 6 Iy = (-0.5x + 3) x2 dx = 48 2 Iy for previous example 1. Areas

  12. Moment of inertia of an area (3 of 3) y 3 y = -0.5x + 7/3 1 x -4/3 8/3 10/3 8/3 Iy = (-0.5x + 4/3) x2 dx = 3.55 -4/3 Iy for previous example with axis shifted 10/3 to centroid 1. Areas

  13. Parallel axis theorem • Parallel axis theorem -- Iparallel axis = Ic + A d2 • From previous example -- 3.55 + 4 x (10/3)2 = 48 Using parallel axis theorem to compute Iy 1. Areas

  14. Polar moment of inertia (1 of 2) J = polar moment of inertia = (x2 +y2 ) dA Definition of polar moment of inertia 1. Areas

  15. Polar moment of inertia (2 of 2) dA=2 rdr dr r R R J = r2 2 rdr 0 = R4/2 Example of computing J for a circle 1. Areas

  16. 2. Stress and strain • Physical requirements • Free-body diagrams • Stress and strain • Hooke’s law • Poisson’s ratio • Stress concentrations • Combined stresses 2. Stress and strain

  17. Physical requirements • Works • Doesn’t break 2. Stress and strain

  18. Free-body diagrams • A diagram that illustrates all the forces acting on a body • If the forces are balanced, the body does not accelerate; otherwise it does • Free-body diagram should include the cross section of interest 2. Stress and strain

  19. Stress and strain • Stress is force per unit area • Normal stress • Area is normal to force •  = F/A • Shear stress • Area is parallel to force •  = F/A • Strain is elongation expressed as a fraction or percentage of basis 2. Stress and strain

  20. Hooke’s law (1 of 2) • Relationship between stress and strain •  = E  • E = modulus of elasticity •  = normal strain •  = G  • G = shear modulus •  = shear strain Definition of Hooke’s law 2. Stress and strain

  21. Hooke’s law (2 of 2)  normal shear    G  Lo = 1 E   Hooke’s law for normal and shear stress 2. Stress and strain

  22. Poisson’s ratio (1 of 2) •  = Poisson’s ratio = ratio of lateral strain to axial strain • x = 1/E (x -  y) • y = 1/E (y -  x) • 0 <  < 0.5 • liquids = 0.50 • aluminum = 0.32 - 0.34 • steel = 0.26 - 0.29 • brass = 0.33 - 0.36 • rubber = 0.49 Definition of Poisson’s ratio 2. Stress and strain

  23. Poisson’s ratio (2 of 2) • Problem • x = 22,000 psi • y = -14,000 psi • E = 30,000,000 psi •  = 0.3 • Solution • x = 1/ 30,000,000 (22,000 - 0.3 (-14,000 )) = 0.00087 • y = 1/ 30,000,000 (-14,000 - 0.3 22,000 ) = 0.00069 Example using Poisson’s ratio 2. Stress and strain

  24. Stress concentrations (1 of 2) • Concentrations occur wherever there is a discontinuity or non-uniformity in an object • Stepped shafts • Plates with holes and notches • Shafts with key ways Stress occurs at discontinuities 2. Stress and strain

  25. Stress concentrations (2 of 2) • Concentrations can be thought of as streamlines • Where there are concentrations, the streamlines are closer together • Near concentrations, stress increases over what would normally be calculated • c = k , where 1 k  3 Stress can be visualized as streamlines 2. Stress and strain

  26. Combined stresses (1 of 7) y xy x xy x xy xy  y 2. Stress and strain

  27. Combined stresses (2 of 7) ds nt n x dy xy xy  y dx 2. Stress and strain

  28. Combined stresses (3 of 7) n ds (1) - x dy (1) sin  - y dx (1) cos  + xy dy (1) cos  + xy dx (1) sin  = 0 n = x dy/ds sin  + y dx/ds cos  - xy dy/ds cos  - xy dx/ds sin  dx/ds = cos  dy/ds = sin  n = x sin2  + y cos2  - 2 xy sin  cos  n = (x + y )/2 + (y - x )/2 cos 2 - xy sin 2 2. Stress and strain

  29. Combined stresses (4 of 7)  nt ds (1) + x dy (1) cos  - y dx (1) sin  + xy dy (1) sin  - xy dx (1) cos  = 0  nt = - x dy/ds cos  + y dx/ds sin  - xy dy/ds sin  + xy dx/ds cos   nt = (y - x )/2 sin 2 + xy cos 2 2. Stress and strain

  30. Combined stresses (5 of 7) Set d n/d  = 0 tan 2 = - xy /[(y - x )/2] Minimum and maximum axial stress n = (x + y )/2  sqrt { [(y - x )/2]2 + xy2} 2. Stress and strain

  31. Combined stresses (6 of 7) Set d  nt/d  = 0 tan 2 = [(y - x )/2]/ xy Minimum and maximum shear stress on a plane at 45 degrees to normal stress  nt =  sqrt { [(y - x )/2]2 + xy2} 2. Stress and strain

  32. Combined stresses (7 of 7)  max (y, xy ) 2 min max  0 (x, -xy ) min Mohr’s circle 2. Stress and strain

  33. Failures (1 of 2) U  ductile Y P U brittle U P ductile without yield P  Stress vs strain for ductile, ductile-without-yield, and brittle materials 2. Stress and strain

  34. Failures (2 of 2) • Definitions • Elastic limit -- Maximum stress at which all strain disappears when stress goes away • Proportional limit (P) -- Maximum stress for which stress is proportional to strain • Yield point (Y) -- Stress at which strain increases without increase of stress. Most materials don’t have a yield point • Ultimate strength (U) -- Maximum stress that can be applied Definitions for previous stress-vs-strain curves 2. Stress and strain

  35. 3. Axial loading • Axial force • Transverse force • Deformation • Strain energy • Spring constant 3. Axial loading

  36. Axial force (1 of 2) • An axial force is force in the direction of the axis of the body 3. Axial loading

  37. Axial force (2 of 2) 414 lb 32o10’ 6’ 550 lb 1040 lb 300 lb 8’ A A 30o 5’ 10’ F = 1040 sin(30o)-300 F H 1040 lb M 300 lb A A 30o 3. Axial loading

  38. Transverse force (1 of 2) • A transverse force is force perpendicular to the axis of the body 3. Axial loading

  39. Transverse force (2 of 2) 414 lb 32o10’ 6’ 550 lb 1040 lb 300 lb 8’ A A 30o 5’ 10’ H = 1040 cos(30o) F H 1040 lb M 300 lb A A 30o 3. Axial loading

  40. Deformation • Elastic deformation =  •  = Lo  = Lo  / E = Lo F / E A, where • F = axial force • A = area of cross section 3. Axial loading

  41. Strain energy • Total strain energy = U • U = 1/2 F  = F2Lo / 2E A 3. Axial loading

  42. Spring constant • Stiffness, or spring constant, is the ratio of force to the displacement caused by the force • Stiffness = k = F/  • Stiffness is an important consideration in accommodating vibration 3. Axial loading

  43. 4. Torsion loading • Torsion • Shear 4. Torsion loading

  44. Torsion • Torsion is twist • Usually, the cross section of bar warps in torsion • The shear stress at a point on a boundary is parallel to the boundary • The shear stress at a corner is zero 4. Torsion loading

  45. Shear (1 of 2)  = r  /L r    = G  = G r  /L L T =  dA r = G  /L r2 dA = G J /L  = TL/(JG)  = Tr/J 4. Torsion loading

  46. Shear (2 of 2) • Problem • Solid bar • Diameter = 2 inches • G = 12,000,000 psi • T = 50,000 in-lbf • L = 4 inches • Find • Maximum shear • Maximum twist J =  r4 /2 = 1.57 in4  = Tr/J = 50,000 x 1/1.57 = 31,800 psi  = TL/(JG) = 50,000 x 4/(1.57 x 12,000,000) = 0.6o 4. Torsion loading

  47. 5. Beam loading • Shear and moment • Stress • Deflection • Failures 5. Beam loading

  48. Shear and moment (1 of 3) • Shear at any point on a beam is the sum of all forces from the point to the left end • V=dM/dx • Upward loads are positive Shear 5. Beam loading

  49. Shear and moment (2 of 3) • Moment at any point on a beam is the sum of all moments and couples from the point to the left end • Clockwise moments are positive • M= V dx • Maximum moments occur where V=0 Moment 5. Beam loading

  50. Shear and moment (3 of 3) 2F/L L V F FL/2 M Diagrams

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