Strength of Materials

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Strength of Material complete stress strain torsion etc etc

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  • Examples of structures. While introducing these point out the salient features of these: purpose of structure, names of these class of structures, loads they carry, etc.
  • Examples of structures
  • Examples of structures
  • Examples of structures
  • Examples of structures
  • More the force, more is the deformation. Develop the argument that the resisting force increases with deformation.
  • Introduce elasticity
  • Dimensions and units of stress. Show how small in a unit Pascal
  • Dimensions and units of strain and elasticity.
  • Dimensions and units of elasticity.
  • Spend some time on this troerxploain the significance of the relative values.
  • Introducingtensionile, compressive and shear stresses.
  • Add formula for spring constant here
  • Discuss which members are in tension
  • Leading to St Venant’s principle
  • Example calculations. Could be omitted.
  • Photoelastic determination of stresses. Stress concentration near the grips. Uniform distribution only in the middle.
  • Discuss stress concentration due to notches and holes. Effect of these on design of components. Introduce factor of safety here.
  • Basis foundation
  • Discuss where compression
  • Introduce the concept of factor of safety here again in the context of variations in the strength of soil. Discuss how soil tests are conducted to determine the bearing capacity of soil.
  • Discuss occurrence of shear, both in rivets as well as in the plates: the tearing action.
  • Explain the stress mechanisms. In each talk of what area does the stress act on.
  • Explain the stress mechanism. In each talk of what area does the stress act on.
  • Explain how these calculations are used to determine the rating of the blanking press. Also discuss how tapering the punch would reduce the blanking force. Discuss a simple pap[er punch in the context.
  • Discuss compression and on what area does the stress act.
  • Discuss belt drive. Discuss if driver pulley or driven pulley. For one type, discuss role of key. Stresses in key
  • Why smallest area is taken?
  • Each virtual disc slips on the others, causing shear. All shears point in the same direction and contribute to moment that resists twisting moment. More the shear strain, more the shear stress, and more the resisting moment. Thus, increasing twisting moment leads to more twist till the resisting moment builds up to balance the increased twisting moment.
  • How compressive stresses are set up near the top and tensile stresses are set up near the bottom.
  • Three trypes of FBDs to determine all tensions (or compressions).
  • Take a small portion around the pin as FBD.
  • Each pole is a two-force member and hence the reaction must be in line.
  • Some more examples of simple supports
  • Explain degrees of freedom.
  • The ladder at right cannot be in equilibrium if there is no friction
  • Draw FBD with friction and calculate
  • Clamped support has no degree of freedom
  • Can find all reactions
  • Unstable frame. Stable frame now. Triangulation. Diagonal member in tension prevents articulation.
  • Bending at A and B introduces moments that resist the moment due to load. A diagonal member introduces tension to resist P for very small deflections, so that the moment build-up at A and B is small, hence negligible. So can assume the structure to be a pinned structure. No moments anywhere! Very useful simplifying assumption.
  • Take a free body by cutting the structure and taking the left part. The tensions (or compressions) in two force members must be along the length of straight members are now external forces, so should be shown.
  • Each truss member is a 2-force member
  • Draw FBD assuming pinned supports
  • FBD of inclined bar
  • FBD of pin
  • Ask students for the sign of stress components before showing them
  • δx and δy must be small
  • δx and δy must be small
  • The point to make here is that P vs. δ depends upon differing length, and area, but stress vs, strain depends only on material and becomes independent of the grometry.
  • Basic strategy of mechanics of material clarified
  • Explain this using the strategy of the previous slide.
  • Explain again using the fundamental strategy. The new thing here is to introduce differential strategy when parameters are not constant.
  • Area of tape does not play a role. Explain why. Draw attention to differing material properties.
  • Picture on the right is an approximation, where the arcs are replaced by perpendiculars.
  • Basic concept of indeterminacy.
  • Strategy to solve statically indeterminate problems
  • Here we have to find the value of x for which the beam remains horizontal
  • These problems do not need to be solved completely. Discuss the strategy and set up the required number of equations.
  • When external tension is removed, the concrete which is in tension contracts (along with the embedded steel) so that compression is built up in steel to balance the tension in concrete.
  • Explain the shape of stress-strain curve for ductile materials through necking.
  • Strength of Materials

    1. 1. 1. Structures, loads and stresses Vijay Gupta: An Introduction to Mechanics of Materials, Narosa, 2013
    2. 2. This course is concerned with structures: A structure is a solid object or assembly. A structure connects components, carries loads, provides form and integrity.
    3. 3. F22, 2002 Wright Flyer, 1903
    4. 4. Burj al Arab Qutab
    5. 5. • Distance between any two points does not change when a force is applied on it.
    6. 6. An upward reaction force at support Thus, an unbalanced moment results. a moment at But equilibrium requires a force and the support. Where does it come from?
    7. 7. Bar Resisting Force As force increases, elongation increases till the equilibrium is restored again. This implies that there is a force resisting the deformation, and that force increases with deformation.
    8. 8. Stress, ζ = P/A Strain, ε = δ/L Area A P Robert Hooke in 1678 showed l E is the elastic modulus, or simply the Elasticity δ P P
    9. 9.  Dimensions of stress: F/Area = F/L2 Units of stress = N/m2 = (Pa)scal, same as that of pressure. A very small unit. Standard atmospheric pressure = 1.03×105 Pa MPa and GPa (106 Pa and 109 Pa, respectively) are commonly used
    10. 10.  Strain δ/L is dimensionless, hence NO UNITS.
    11. 11.   E = Stress / strain, and therefore, has dimensions of stress, i.e., F/L2. Units of E are, accordingly, Pa(scal).
    12. 12. Material Aluminium 2024-T3 Aluminium 6061-T6 Aluminium 7075-T6 Concrete Copper Glass fibre Cast iron Steel, High strength Steel, Structural Titanium Wood Value of E in GPa 70 70 70 20 – 35 100 65 100 200 200 100 10-15
    13. 13. The external forces acting on a structure result in strains. The strains so produced result in stresses within the material of the members. The stresses, for the most part, are proportional to the strains. The constant of proportionality is termed as the modulus of elasticity.
    14. 14. Stresses due to various loads Vijay Gupta: An Introduction to Mechanics of Materials, Narosa, 2013
    15. 15. Bar F L δ
    16. 16. Tension in the belt
    17. 17. P P P P
    18. 18. Stress: Force Intensity 20 mm 5 mm ζ = F/A = 300N 20×5×10− 6 m2 = 3 MPa 300 N
    19. 19. Uniform stress is an approximation. Valid only in simple loadings. Away from ends.
    20. 20. 105 Steel Concrete Soil N Bearing strengths Steel >> concrete >> soil Required areas Steel << concrete << soil
    21. 21. 105 N Permissible compressive stress in steel is about 400 MPa. Steel Concrete Soil So the area of steel required is 105 N/400 MPa, or 2.5 10−4, or about 16 mm 16 mm
    22. 22. 105 N Permissible compressive stress concrete is about 60 MPa. Steel Concrete Soil So the area of concrete required is 105 N/ 60 MPa, or 1.67 10−3, or about 41 mm 41 mm
    23. 23. Permissible bearing strength 105 N of soils varies widely. For good cohesive soil, it could be between 100 to 400 kPa, if it is above the water table. Steel Concrete Soil So the area of the footing required is 105 N/ 200 kPa (say), or 0.5 m2, or about 710 mm 710 mm
    24. 24. Shear Stresses
    25. 25. Bearing (Compressive) Shear
    26. 26. Bearing load Shear load
    27. 27. Blanking force = shear strength Shear area = perimeter shear area sheet thickness
    28. 28. Compression Shear Compression Shear
    29. 29. P P P Shear Shear area
    30. 30. Shear stresses on the back face of the shaft The stresses result in a moment that balances the twisting moment
    31. 31. Compression near top Extension near bottom
    32. 32. Bending of Beams Net tensile force is zero!
    33. 33. • Forces that tend to reduce the size of a structural member produce compressive strains which, in turn, produce compressive stresses.  Forces that tend to distort the shape of a member produce shear strains which in turn produce shear stresses.
    34. 34. • A twisting moment applied to a shaft produces shear strains. These shear strains give rise to shear stresses which result in a moment that balances the external twisting moment.  A bending moment produces both tensile and compressive strains and stresses. These give rise to a resisting moment which balances the bending moment.
    35. 35. Tensile members and trusses Vijay Gupta: An Introduction to Mechanics of Materials, Narosa, 2013
    36. 36. 10 kN 10 kN 10 kN 10 kN 10 kN 10 kN 10 kN 10 kN 10 kN
    37. 37. A member on which external forces act only at two distinct points (and there is no external torque acting on it) is termed as a two-force member. The forces acting on a two-force member are equal and opposed. But is it enough?
    38. 38. The forces acting on a two-force member are equal, opposed and collinear.
    39. 39. Two-force members 60o 30o F1cos30o – F2cos60o = 0 500 N F1sin30o + F2sin60o – 500 = 0 F1 = 125 N; F2 = 433 N
    40. 40. Simply supported: No moment. Only reaction force. • Cannot translate, • Can rotate.
    41. 41. Load Pinned Support: Reaction could be inclined. (1 DoF) Roller Support: Normal reaction only. (2 DoF)
    42. 42. FRICTION must act
    43. 43. RH2X4 – 600X0.5 = 0 or RH2 = 75 N RH2 4m Or, N RV1 600 N RH1 1m y x z
    44. 44. While in the pinned support, the member is restrained from translating, in the clamped support, the member cannot even rotate.
    45. 45. • Cannot translate, • Cannot rotate. No DoF at all. Also called built-in support
    46. 46. P Fy Fx Mz L y Statically determinate x z
    47. 47. Resisting moment Tension will build up faster than the moments due to bending, and therefore, can treat the joint as pinned
    48. 48. Two-force members
    49. 49. 3m θ cos = 4/5, sin = 3/5
    50. 50. Method of Joints Symmetry: FGB FGC FGF FGA 15 kN 2FGC sinθ – 15 kN = 0 FGC = (15 X 5)/(2 X 3) = 12.5 kN
    51. 51. Method of Joints -FCGsinθ - FCFsin θ = 0 FCF = - FCG = -12.5 kN FCB FCG FCF - FCB – FCGcosθ + FCFcosθ = 0 FCB = –2FCGcosθ = 10 kN
    52. 52. FFC ∑FV = RB,V +(3/5)FFC = 0 or, RB,V = - (3/5)FFC = + 7.5 kN ∑FH = -(4/5)FFC - FFG = 0 or, FFG = - (4 /5)FFC = 10 kN FFG RB,V
    53. 53. B D C 20 kN
    54. 54. B B A two-force member D D C 20 kN FBD,D
    55. 55. F2 B F1 D 20 kN C D 20 kN ∑Fy = 0
    56. 56. • Co-planar • Concurrent
    57. 57. Notation for stresses Vijay Gupta: An Introduction to Mechanics of Materials, Narosa, 2013
    58. 58. First index: normal to the plane on which acting. z x Second index: direction of y the stress component itself
    59. 59. Stress Intensity of Force
    60. 60. The stress vector t depends upon the location as well as the direction of the surface.
    61. 61. z x y The sign of a stress component depends on the direction of normal and the direction of force: If both have same sign then the stress component is positive, if the two have different signs, then the stress component is negative.
    62. 62. ζyy is negative ηyx is positive ηyz is negative z x y
    63. 63. ζxx is positive ηxy is negative ηxz is negative z x y
    64. 64. Face Area Force xy(direction of (assume outward unit depth) compone compone nt nt normal) x δy•1 σxx• δy τxy• δy −x δy•1 −σxx• δy −τxy•δy y δx•1 σyy• δx τyx• δx −y δx•1 −σyy• δx −τyx•δx Consider the moment balance about the mid-point:
    65. 65. Thin-walled cylinders are used extensively in industry and homes because they are very efficient structures. • Oil storage tanks are cylindrical • So are oxygen bottles, cooking gas cylinders • Deodorant bottles are pressurized cylinders. • So are beer cans.
    66. 66.   Cylindrical and spherical pressure vessels are commonly used for storing gas and liquids under pressure. A thin cylinder is normally defined as one in which the thickness of the metal is less than 1/20 of the diameter of the cylinder.
    67. 67.   In thin cylinders, it can be assumed that the variation of stress within the metal is negligible, and that the mean diameter, Dm is approximately equal to the internal diameter, D. At mid-length, the walls are subjected to hoop or circumferential stress, and a longitudinal stress, .
    68. 68.   The internal pressure, p tends to increase the diameter of the cylinder and this produces a hoop or circumferential stress (tensile). If the stress becomes excessive, failure in the form of a longitudinal burst would occur.
    69. 69. C o n sid e r th e h a lf cylin d e r sh o w n . F o rce d u e to in te rn a l p re ssu re , p is b a la n ce d b y th e fo rce d u e to h o o p stre ss,  h . i.e . h o o p stre ss x a re a = p re ssu re x p ro je cte d a re a h x 2 L t = P x d L h = (P d ) / 2 t W h e re : d is th e in te rn a l d ia m e te r o f cylin d e r; t is th e th ickn e ss o f w a ll o f cylin d e r.
    70. 70. T h e in te rn a l p re s s u re , P a ls o p ro d u c e s a te n s ile s tre s s in lo n g itu d in a l d ire c tio n a s s h o w n a b o v e .  d F o rc e b y P a c tin g o n a n a re a lo n g itu d in a l s tre s s ,   d t  L  d 4 L is b a la n c e d b y a c tin g o v e r a n a p p ro x im a te a re a , (m e a n d ia m e te r s h o u ld s tric tly b e u s e d ). T h a t is : x d t  P x  L 4 2  P d 4t 2
    71. 71. 1. Since hoop stress is twice longitudinal stress, the cylinder would fail by tearing along a line parallel to the axis, rather than on a section perpendicular to the axis.  The equation for hoop stress is therefore used to determine the cylinder thickness.  Allowance is made for this by dividing the thickness obtained in hoop stress equation by efficiency (i.e. tearing and shearing efficiency) of the joint. 
    72. 72. Take section of a pressurized cylinder And the upper half FBD of the lower half σθθ p σθθ
    73. 73. FT FR p We can show by symmetry arguments that: (a) Both shear should be inwards or outwards (b) Shear should be ZERO
    74. 74. FT FBD of the ‘contents’ p Net forced on the curved surface = p×2r×δl Equilibrium: FT = ζ 2δl t = p×2r×δl This gives: Hoop stress
    75. 75. Forces on the rim Pressure on ζ the back cap Axial stresses are lone-half of hoop stresses
    76. 76. Forces on the rim Pressure on the ‘content’ p Maximum stress in a spherical vessel is one half that of a cylindrical vessel of same radius and thickness
    77. 77. Shaped structures
    78. 78. Arch Keystone All stones are subjected to compressive forces only.
    79. 79. Towers Load bearing cables Deck of bridge Cables support the bridge through tension. Towers carry compression,
    80. 80. The main span of the Golden Gate suspension bridge is 1.287 km long. The sag in the cables is 140 m. The design loading is 400 kN/m.
    81. 81. Tension in the cable at the lowest point is; To = 2.96×108 N Max tension = 3.23×108 N Each cable consists of 27,572 strands of 4.88-mm diameter wires bundled parallel. Cross-sectional area of the cable = 27,572×[π×0.004882/4] = 0.516 m2 So stress = 625.5 MPa
    82. 82. z x y
    83. 83. Take section of a pressurized cylinder And the upper half FBD of the lower half σθθ p σθθ
    84. 84. FT FR p We can show by symmetry arguments that: (a) Both shear should be inwards or outwards (b) Shear should be ZERO
    85. 85. p
    86. 86. 2. Deformations, strains and material properties Vijay Gupta: An Introduction to Mechanics of Materials, Narosa, 2013
    87. 87. P/A L δ/L Stress = E×Strain P
    88. 88. Deformation depends on loading, material and geometry Strain depends on stress AND material. NOT on geometry Stress depends on loading and geometry
    89. 89. Load Equilibrium Stress Macro Geometry Micro Strain Micro Geometry Material Property Deformation Macro
    90. 90. Cross-section: 6 cm2 Section Tension, N Stress, kPa Strain Length, m Elongation, m AB BC CD 250 500 800 416.7 4.16×10- 3 833.3 8.35×10- 3 1333.3 13.33×10- 3 1.5 2.0 1.5 6.24×10- 3 16.70×10- 3 20.02×10- 3 DE EF 550 300 916.7 9.20×10- 3 500.0 5.00×10- 3 1.5 2.0 13.78×10- 3 10.00×10- 3
    91. 91. W = ρAxg = T ζ = T/A = ρxg 72 m ε = ζ/E = ρxg/E dδ = εdx = ρgxdx/E T dx x W(x)
    92. 92. or For a Nylon wire: density, ρ ~ 0.8X103 kg/m3, and E ~ 400 MPa. We get δ ~ 52 mm For steel: density is 7.6X103 kg/m3, and E is 200 GPa. We get δ ~ 1 mm
    93. 93. RA,y RA,x A B RB,y Member Force kN Lengt h m Area m2 AC 28.3 1.41 1.77×10−4 BC − 20 1 1.77×10−4 RA,y = 20 kN RA,x = − 20 kN RB,y = 20 kN C 20kN Stress MPa TAC = 28.8 kN TBC = − 20 kN Strain Elongation m 7.6×10−4 1.07×10−3 −113.2 −5.4×10−4 −0.54×10−3 160.1
    94. 94. A A y x 45o B E C D C1 45o B E C D 45o G F C1 X-displacement of C ~ shortening of BC =−0.54 mm y-displacement of C ~ EF + FC1 = CD/cos45o + FG(=EC) ~ 1.25 mm
    95. 95. Statically-indeterminate structures Vijay Gupta: An Introduction to Mechanics of Materials, Narosa, 2013
    96. 96. PP P R1 R2 R3 Reaction at middle support (and hence, at all supports depends on the bending of plank.
    97. 97. 1. Consideration of static equilibrium and determination of loads 2. Consideration of relations between loads and deformations, (first converting loads to stresses, then transforming stresses to strain using the properties of the material, and then converting strains to deformations), 3. Considerations of the conditions of geometric compatibility
    98. 98. Indeterminate because x is not known! 1.3 m 2.6 m F2 F1 150 kN 1m
    99. 99. R1 Taking moments about the pivot point, 2R1 + 2R2 – P = 0 P Indeterminate because 4 unknown forces and only three equations to determine them. R2 Geom. Comp. δ1 = δ2 R1L1/E1A1 = R2L2/E2A2
    100. 100. R2 - F - R1 = 0; R1L – Fx = 0 Geom. Comp. h + δ = 2(h - δ ) 1 2 F h R1 = kδ1 R2 = kδ2 L x
    101. 101. P P = R1+ R2 R2 R1 R1 = (E1A1/L1)δ1 R2 = (E2A2/L2)δ2 Geom. Comp. δ1 = δ2
    102. 102. (a) Tendon being stressed during casting. Tension in tendon, no stress in concrete. (b) After casting, the force is released and the structure shrinks. (c) FBD of tendon. The concrete does not let the tendon shrink as much as it would on its own. This results in residual tension in the tendon. (d) FBD of concrete. The residual force in the tendon is trying to compress the concrete..
    103. 103. A concrete beam of cross-sectional area 5 cm 5 cm and length 2 m be cast with a 10 mm dia mild steel rod under a tension of 20 kN. The external tension in steel released after the concrete is set. What is the residual compressive stress in the concrete? Calculate the extension of steel under the tension of 20 kN T = 20 kN →σ = 255 Mpa →ε = 1.21 10- 3 →δ = 2.42 mm
    104. 104. 2.42 mm δs δc δs + δc = 2.42 mm
    105. 105. Lateral strains: Poisson ratio, ν Vijay Gupta: An Introduction to Mechanics of Materials, Narosa, 2013
    106. 106. σxx εxx = σxx/E εyy = - ν εxx ν is Poisson ratio σxx Another material property
    107. 107. Let us consider εxx. σxx produces an εxx = σxx /E σyy produces an εyy = σyy /E, which through Poisson ratio gives εxx = -νεyy = - νσyy/E. Similarly for σzz .
    108. 108. Shear stresses do not cause any normal strain Therefore, εxx = ζxx/E – νζyy/E - ν ζzz/E = [ζxx – ν(ζyy + σzz)]/E Similarly for εyy and εzz
    109. 109. F ζyy = −F/A, ζzz = 0 What is ζxx and εyy y x Geometric compatibility: εxx = 0 εxx = [ζxx – ν(ζyy + ζzz)]/E 0 = [ζxx – ν(ζyy + 0)]/E, → ζxx =νζyy = − ν F/A εyy = [ζyy – ν(ζxx + ζzz)]/E = [−F/A + ν F/A]/E = −(1− ν)F/AE
    110. 110. σyy Steel: εx = 0.6×10−4 εy = 0.3×10−4 Find σxx and σyy : Plug in: εxx = [σxx – ν(σyy + σzz)]/E εyy = [σyy – ν(σzz + σxx)]/E E = 200 GPa, ν = 0.3 σxx
    111. 111. Shear strains and stresses Vijay Gupta: An Introduction to Mechanics of Materials, Narosa, 2013
    112. 112. Apply shear stresses to a block: Shear strain γ is π/2 − θ Shear strain is also seen as: θ1 + θ2 θ2 θ θ1 Since angles are measured positive counter-clockwise, the angle θ2 above is a negative angle. In general terms, then, γ = θ1 − θ2 with θs measured positive when counter-clockwise
    113. 113. A square blocks 0.2 mm × 0.2 mm deforms under shear ystress Coordinates after C D deformation (in mm) are: θ2 θ1 A A(0,0), B(0.194, 0.013), and D(−0.012, 0.196). B x θ1 = 0.013/0.2 = 0. 065 θ2 = 0.012/0.2 = 0. 06 γxy = 0.65 − 0.60 = 0.05 radians
    114. 114. Shear strain γ is related to shear stress τ by γxy = τxy/G, where G is shear modulus It can be shown that γxy does not depend on other components of stress.
    115. 115. Material Aluminium G, GPa 25 Steel 80 Glass 26-32 Soft Rubber 0.003- 0.001
    116. 116. 8,000 N 4,000 N Shear stress τ = 4,000 N/ (0.1 m)(0.12 m) = 3.33×105 Pa Shear strain γ = τ/G 3.33×105 Pa/1 MPa = 0.33 Wall Wall Rubber blocks 10 cm × 10 cm with 12 cm height
    117. 117. Consider the rubber block on the left: 8,000 N γ =0.33 Wall And therefore, The vertical deflection of load = 0.33×0.10 m = 33 mm Wall Rubber blocks 10 cm × 10 cm with 12 cm height
    118. 118. We have so far introduced three elastic properties of materials. Material ν E, GPa G, GPa 70 25 0.33 Steel 200 80 0.27 Glass 50-80 26-32 0.21-.27 0.00080.004 0.0030.001 0.50 Aluminium Soft Rubber
    119. 119. Thermal strains and stresses Vijay Gupta: An Introduction to Mechanics of Materials, Narosa, 2013
    120. 120. On heating, there is linear expansion: There is no thermal shear strain Material Steel Aluminium α (×10-6/oC) ~ 10 ~ 20
    121. 121. Putting Hooke law, Poisson effect and thermal strains all together, εxx = [σxx – ν(σyy + σzz)]/E + αΔT εyy = [σyy – ν(σzz + σxx)]/E+ αΔT εzz = [σzz – ν(σxx + σyy)]/E+ αΔT γxy = τxy/G, γyz = τyz/G, and γzx = τzx/G
    122. 122. Aluminium rod, rigid supports. Temperature raised by ΔT. What are the stresses? εxx = 0 = [σxx/E + αΔT] σxx = −αEΔT x
    123. 123. Tank is flush when empty. Find end forces when pressure is p Due to p: σzz = pr/ 2t, σθθ = pr/ t z If end forces F, axial stress due to it is F/2πrt εzz = [(pr/ 2t − F/2πrt) −νpr/t ]/E Equate it to 0 and determine F p
    124. 124. Determining stress-strain relations Vijay Gupta: An Introduction to Mechanics of Materials, Narosa, 2013
    125. 125.  A material property. Tensile Test Machine, UTM
    126. 126. σ (= F/Ao) Ductile Brittle ε (=∆L/Lo) Ductile Failure cup-and-cone Necking Brittle Failure
    127. 127. Y σ (= F/A0) Yield stress, σY 0.02% Permanent set ε (= ΔL/L0)
    128. 128. Y1 Ultimate stress σ (= F/A0) Y B ε (= ΔL/L0)
    129. 129. σ σ σ (a) Rigid ε ε (b) Perfectly elastic σ ε (c) Elastic-Plastic σ Increase in yield strength ε (d) Perfectly plastic ε (e) Elastic- Plastic (strain hardening)
    130. 130. = 9.82×10-6 m4
    131. 131. Let us check on the stresses: Quite safe
    132. 132. Φ 10 cm Φ 5 cm F Φ 2 cm Φ 6 cm 1m F 0.6 m 150 N.m −250 N.m 150 N.m
    133. 133. Φ 10 cm Φ 5 cm F Φ 2 cm Φ 6 cm 1m F 0.6 m 150 N.m
    134. 134. Φ 10 cm Φ 5 cm F Φ 2 cm Φ 6 cm 1m F 0.6 m 150 N.m Angle θ2 which represents the counter-clockwise movement of the smaller gear due to gearing alone is 10/6 of θ1 or 0.0085 rad counter-clockwise Rotation of the right end of second shaft wrt stationary wall is, therefore, 0.0085 rad + 0.12 rad = 0.1285 rad or 7.36 degree.
    135. 135. By geometry: γθz= rdΦ/dz Therefore, τθz = GrdΦ/dz r varies from R1 to R2, and θ varies from 0 to 2π
    136. 136. 2 1.8 1.6 1.4 1.2 1 0.8 0.6 0.4 0.2 0 Weight 0 0.2 0.4 0.6 0.8 1
    137. 137. 2 1.8 1.6 1.4 1.2 1 0.8 0.6 0.4 0.2 0 0 0.2 0.4 0.6 0.8 Weight Stiffness 1
    138. 138. 2 1.8 1.6 1.4 1.2 1 0.8 0.6 0.4 0.2 0 Strength to weight Stiffness Weight 0 0.2 0.4 0.6 0.8 1
    139. 139. M1 Mo M2 TMD Equilibrium Condition: - M1 + Mo – M2 = 0 Geometric Condition: Φ1 +Φ2 = 0
    140. 140. Shear flow on horizontal surfaces is same as on the vertical surfaces q1 = q2
    141. 141. Relating q to twisting moment T h o qds dT at O = qds×h = q×2×Grey area q = T/2A
    142. 142. T 100 Nm R 20 mm R 16 mm This gives τ = (49 N/m)/0.004 m = 12.25 MPa

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