david roylance mechanics of materials Part 2 pdf

... 1.8 22 –33 aramid 124 3.6 2. 3 1.45 86 2. 5 22 –33 boron 400 3.5 1.0 2. 45 163 1.43 330–440 HS graphite 25 3 4.5 1.1 1.80 140 2. 5 66–110 HM graphite 520 2. 4 0.6 1.85 28 1 1.3 22 0–660 Of course, these materials ... neck. 5 INTRODUCTION TO COMPOSITE MATERIALS David Roylance Department of Materials Science and Engineering Massachusetts Institute of Technology Cambrid...

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... written 7 ATOMISTIC BASIS OF ELASTICITY David Roylance Department of Materials Science and Engineering Massachusetts Institute of Technology Cambridge, MA 021 39 January 27 , 20 00 Introduction The ... ELASTICITY David Roylance Department of Materials Science and Engineering Massachusetts Institute of Technology Cambridge, MA 021 39 January 21 , 20 00 Introduction Th...

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... Strength of Materials, McGraw-Hill, New York, 19 52 12 The Kinematic Equations David Roylance Department of Materials Science and Engineering Massachusetts Institute of Technology Cambridge, MA 021 39 Septemb ... is: U ∗ =  τdγ= 1 2 τγ = τ 2 2G = 1 2G  Tr J  2 This is then integrated over the specimen volume to obtain the total energy: U =  V U ∗ dV =  L  A 1 2G  T...

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... as      σ x  σ y  τ x  y       =    c 2 s 2 2sc s 2 c 2 −2sc −sc sc c 2 − s 2         σ x σ y τ xy      (4) 1 R.M. Jones, Mechanics of Composite Materials, McGraw-Hill, 1975. 2 I 3 =det|σ|= 1 3 σ ij σ jk σ ki (21 ) These ... σ x sin 2 θ + σ y cos 2 θ − 2 xy sin θ cos θ τ x  y  =(σ y −σ x )sinθ cos θ + τ xy (cos 2 θ − sin 2 θ) Or ...

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... sequence of nine equations, since each component of  ij is a linear combination of all the components of σ ij . For instance:  23 = S 23 11 σ 11 + S 23 12 σ 12 + ···+S 23 33  33 Based on each of the ... E 2 =4,G 12 =2. 8(allGPa)and ν 12 =0 .25 . The compliance matrix S in the 1 -2 (material) direction is: S =   1/E 1 −ν 21 /E 2 0 −ν 12 /E 1 1/E 2 0 001/G...

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... diff(U,F); The result of these manipulations yields δ F = LF  12 L 2 G − 12GL 2 cos 2 θ +9Gr 2 2 cos 2 θ +10r 2 2 E−10 r 2 2 E cos 2 θ  9 r 1 r 3 2 Eπ G This displacement is in the direction of the applied ... σ 2 /2E over the specimen volume: U b =  V U ∗ dV =  L  A σ 2 x 2E dA dL =  L  A 1 2E  −My I  2 dA dL =  L M 2 2EI 2  A y 2 dA dL Since ...

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... -0. 122 8D-15 -0.1639E-16 -0 .21 50E-15 0 .20 83E-09 -0.6 022 D-16 -0. 122 8D-15 0 .22 28D-19 0. 721 8E-10 0.1084E-18 -0.6 022 E-16 0.3058D-09 -0. 426 5D-11 -0.1967D-15 0. 621 4E -22 -0. 721 8E-10 -0. 122 8E-15 -0. 426 5D-11 ... − a 2 r 2  + σ 2  1+ 3a 4 r 4 − 4a 2 r 2  cos 2 σ θ = σ 2  1+ a 2 r 2  − σ 2  1+ 3a 4 r 4  cos 2 (19) τ rθ = − σ 2  1 − 3a 4 r 4 + 2a...

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... DOF 1 DOF 2 DOF 3 DOF 4 DOF 5 DOF 6 1 000000 2 0.013333 -0.0 321 9 0000 3 0. 02 -0.084379 0000 4 000000 5 -0.0066667 -0.038856 0000 Element Stresses 1: 4000 2: 20 00 3: -28 28.4 4: 20 00 5: -28 28.4 6: ... k (2) 21 k (1) 44 + k (2) 22 k (2) 23 k (2) 24 00 k (2) 31 k (2) 32 k (2) 33 k (2) 34 00 k (2) 41 k (2) 42 k (2) 43 k (2) 44          ...

Ngày tải lên: 11/08/2014, 02:21

... and divide by the complex conjugate of the denominator) to yield: E ∗ = kω 2 τ 2 1+ω 2 τ 2 + i kωτ 1+ω 2 τ 2 (25 ) In Eq. 25 , the real and imaginary components of the complex modulus are given explicitly; ... (3,1) 13 ENGINEERING VISCOELASTICITY David Roylance Department of Materials Science and Engineering Massachusetts Institute of Technology Cambridge, MA 021 39...

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... differential equation a 2 ¨σ + a 1 ˙σ + a 0 σ = b 2 ¨ + b 1 ˙ + b 0  where a 2 = τ 1 τ 2 ,a 1 = τ 1 + τ 2 ,a 0 =1 b 2 = τ 1 τ 2 (k e + k 1 + k 2 ) ,b 1 = k e (τ 1 + τ 2 )+k 1 τ 1 + k 2 τ 2 ,b 0 = k e 14. ... of C(t)atthattime. 26 Using the given shift factor, we can adjust the time of the second temperature at 50 ◦ Ctoanequivalent time t  2 at 20 ◦ C as foll...

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