Introduction to Contact Mechanics Part 3 pot

... Equilbrium (Stable) U g = 4gc 8c 3 E d 3 h 2 U s = Energy (U) Crack length c d c P h (a) (b) 26 Mechanical Properties of Materials () ()()() [] 2 21 2 13 2 32 2 3 2 2 2 1 2 32 1 6 1 2 1 3 σσσσσσ σσσ σ σσσσσσσ −+−+−= ++= ++=++= ddd o dzdydxdddo ... resolved into a mean, or average compo- nent and the deviatoric components. The mean, or average, stress is found from: 3 3 3...

Ngày tải lên: 11/08/2014, 09:20

... *2125 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 = −= −= −= ∫∫ (6 .3. 2) Equation 6 .3. 2 equates the kinetic energy of the projectile to the ... mean contact pressure p = 31 8 .3 MPa. Results are shown d r 0.05 0.1 0.2 0.4 0.6 Eqn. µ=0 Surface radial displacements Bonded r/a (b) Sphere 0.4 0.0E+0 −1.0E 3 −2.0...

Ngày tải lên: 11/08/2014, 09:20

... mean contact pressure and a is the radius of circle of contact) . Substituting this value of B into Eqs. 8.2.1a to 8.2.1d and normalizing to mean contact pressure p and radius of circle of contact ... complex nature of the stresses and their 0 1 2 3 4 r −4 3 −2 −1 0 z (a) 0 1 2 3 4 r −4 3 −2 −1 0 z (b) 0 1 2 3 4 r −4 3 −2 −1 0 z (c) − 0 . 3 0 0 − 0 . 2 0 0 − 0 .1 0 0...

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... 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 − 0.200 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 .020 − 0.010 − 0.008...

Ngày tải lên: 11/08/2014, 09:20

... 5, 33 -35 , 36 , 41, 42, 49, 61, 129, 130 , 131 , 133 surface tractions, 44, 45, 46 tensile strength, 3, 31 , 35 , 53, 64, 66, tension, 5, 6, 10, 14, 18, 25, 28, 29, 31 , tensor, 8, 12 time to ... repulsion, 1 crack growth, 33 -35 , 43, 46, 49, 50, 54, 56, 58, 61, 124, 131 , 134 crack path, 133 crack resistance, 41, 75 crack tip, 33 -37 , 39 , 40-45, 47, 50-54, 90, 11...

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... Grade CS101 CS101 EE200 EE200 CS101 PH100 1 234 5 67890 1 234 5 22222 33 333 67890 A B C B+ A− C+ Course StudentId Grade CS101 CS101 CS101 1 234 5 67890 33 333 A B A− 18 INTRODUCTION TO COMPUTER SCIENCE: HANDOUT #3. THE RELATIONAL ... follows: The table CDH. StudentId 1 234 5 67890 22222 33 333 Course Day Hour CS101 CS101 CS101 EE200 EE200 EE200 M W F Tu W Th 9AM 9AM 9AM 10AM 1PM 10...

Ngày tải lên: 09/08/2014, 11:21

... available as bond potential energy (neglecting any dissipative losses due to heat, sound, etc.). The resulting two separated atoms have the potential to form bonds with other atoms. The atoms, now ... directions on a particular atom, and the atom takes up an equi- librium position within the material. Now consider an atom “B” on the surface. Such an atom is attracted by the many atoms j...

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... Combining Eqs. 3. 3b and 3. 3c, Brown 24 was able to show that for a specific flaw that leads to failure, the integrated form of Eq. 3. 3c is: () [] SdtRTT f t o n a exp 0 =γ−σ ∫ (3. 3d) where ... () [] RT dt d o n x γ−σΚ= ρ exp A (3. 3c) In Eq. 3. 3c, Κ and n are constants, x represents the displacement of the crack boundary into the material, and the other symbols are a...

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... and σ 3 trajectories, (g) τ max trajectories. Fig. 5 .3. 2 Stress trajectories and contours of equal stress in MPa for Boussinesq “point 0 1 2 3 4 r −4 3 −2 −1 0 z 0 1 2 3 4 r −4 3 −2 −1 0 z 0 ... 4 r −4 3 −2 −1 0 z 0 1 2 3 4 r −4 3 −2 −1 0 z 0 1 2 3 4 r −4 3 −2 −1 0 z 0 1 2 3 4 r −4 3 −2 −1 0 z 0 1 2 3 4 r −4 3 −2 −1 0 z P 0 1 2 3 4 r −4 3 −2 −1 0 z 0 1 2...

Ngày tải lên: 11/08/2014, 09:20