Introduction to Contact Mechanics Part 2 docx
Introduction to Contact Mechanics Part 2 docx
... 2 1 2 1 cossin2sincos 22 xyyxyx xyyx +−++= ++= (1 .2. 5i) and the shear stress across the plane is found from: ( ) ( ) () θτθσσ θθτθθσστ θ 2cos2sin 2 1 cossin cossin 22 xyyx xyyx −−= −+−= (1 .2. 5j) ... (1 .2. 9.3a) and also, 0 ;0 ;0 === yzxzz τ τ ε (1 .2. 9.3b) 1 .2 Elasticity 21 radial stress σ r and a tangential stress σ θ , and the principal stresses are found fr...
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Introduction to Contact Mechanics Part 3 pot
... xz zx zy yz yx xy zxx z yz z y xyy x ∂∂ γ∂ ∂ ε∂ ∂ ε∂ ∂∂ γ∂ ∂ ε∂ ∂ ε∂ ∂∂ γ∂ ∂ ε∂ ∂ ε∂ 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 =+ =+ =+ (1 .2. 11.1b) The compatibility relations imply that the displacements within the material vary smoothly throughout the specimen. Solutions to problems ... Equilbrium (Stable) U g = 4gc 8c 3 E d 3 h 2 U s = Energy (U) Crack length c d c P h (a)...
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Introduction to Contact Mechanics Part 4 ppt
... of obtaining 0, 1, or 2 heads in two tosses of a coin. f(X=x) x 0 1 2 0.5 0 .25 F(X£ x) x 0 1 2 1.0 0.5 0 .25 Delayed Fracture in Brittle Solids 58 () 2 1 22 1 2 1 2 − − ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ −πσ=σ n n C n afp YnDKt ... normal to the crack. The total stress intensity factor is given by integrating Eq. 2. 5.2c with F replaced by dF = σ(b)db. 12 1 12 22 0 2( )σ = π...
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Introduction to Contact Mechanics Part 5 pdf
... cylindrical polar coordinates by Timoshenko and Goodier 12 : () ()() () ()() () () 25 22 2 25 22 3 23 22 21 22 2 2 25 22 2 21 22 2 2 2 3 2 3 1 21 2 31 21 2 zr rzP zr zP zr z zrr z r P zr zr zrr z r P rz z r + −= + −= ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ + + + +−−= ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ + − ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ + −−= π τ π σ ν π σ ν π σ θ ... 4 r −4 −3 2 −1 0 z P 0 1 2 3 4 r −4 −3 2 −1 0...
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Introduction to Contact Mechanics Part 7 potx
... 109 P dt d m −= 2 2 δ (6.3.1) Equating Eqs. 6 .2. 1f and 6.3.1, 6 multiplying both sides by velocity and inte- grating, we obtain: *21 25 2 *21 23 *21 23 *21 23 3 4 5 2 2 1 3 4 3 4 3 4 ERmv dERdvmv dt d ER dt dv mv ER dt dv m o v v o δ δδ δ δ δ = −= −= −= ∫∫ ... distribution r < a Sphere 1 2 3 21 2 2 ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ −−= a r p m z σ Cylinder 2- D 1 2 21 2...
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Introduction to Contact Mechanics Part 8 pot
... 8 .2. 2, the stresses are given by: () () () () () 2 2 3 12 22 cos 45 cos 2 2 r PB rr =−−− +− −− σ ννθ νθν π (8 .2. 1a) ( ) () () 2 2 2 3 12 cos 21 2 cos 1cos 2 PB rr − =−− + θ νθ σ νθ θ π (8 .2. 1b) ... () () () θν r a . θ θν r a p σ m θ 2 3 2 2 cos2 120 60 cos1 cos21 2 1 − ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − + − ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ = (8 .2. 1f) ( ) () () θν θ θ ν σ 2 3 2 cos 322 120 60...
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Introduction to Contact Mechanics Part 9 pot
... Reference 22 ). 0 1 2 3 4 −4 −3 2 −1 0 z/a o 0 1 2 3 4 −4 −3 2 −1 0 z/a o 0 1 2 3 4 r/a o −4 −3 2 −1 0 z/a o −4 −3 2 −1 0 r/a o −4 −3 2 −1 0 −4 −3 2 −1 0 −4 −3 2 −1 0 −4 −3 2 −1 0 −4 −3 2 −1 0 (a) (b) (c) 0.005 − 0 . 3 00 − 0 .20 0 0.005 0 . 0 1 0 0.010 0 .0 05 0 . 020 0. 02 0 0.00 5 − 0.0 25 − 0.050 − 0. 1 00 0.010 0.0 05 − 0 .20 0 − 0.100 − 0 .060 − 0 .040 − 0 ....
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Introduction to Contact Mechanics Part 10 ppsx
... full plasticity, then: 2. 2 23 .0 1 2 = ∴= C p Y m (9.3.4.2a) The theory appears to be inconsistent with their requirement that the pressure distribution across the contact area be unchanged ... finite value of the contact radius at the initiation of yield, we obtain: () () () 2 2 2 2 21 * 1 1ln 4 12 32 61 paaaa EY YaR aa ⎡ ⎤ ⎛⎞ ⎛⎞ ⎡⎤ ⎛⎞ ⎢ ⎥ ⎜⎟ =+ − +− ⎜⎟ ⎢⎥ ⎜⎟ ⎜⎟ ⎜⎟ ′−...
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Introduction to Contact Mechanics Part 12 pot
... diameter 116 0.10 2. 85 0.79 Contact 127 22 7 0. 12 3.43 Suggested # Actual diameter 21 9 3 92 0.14 4.00 154 348 622 0.16 4.57 22 9 520 929 0.18 5.14 326 741 1 322 0 .20 5.71 448 1016 1813 0 .22 6 .28 596 13 52 2414 0 .24 ... 173, 186, 20 3 Nomarski, 20 1, 20 2 indentation hardness, 21 7 169, 1 72, 177, 20 3, 20 4, 20 5, 20 6 20 2 -20 6 187, 20 8 inert strength, 21...
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