Mechanics Analysis Composite Materials Episode 4 pps

... express the strains as follows 122 Mechanics and analysis of composite materials \- Fig. 4. I. Laminated structure of a composite pipe. n L Fig. 4. 2. Composite drive shaft with external ... strains, i.e., to materials whose typical stressstrain diagram is shown in Fig. 4. 6. 118 Mechanics and analysis of composite materials that is the basic element...

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... shaft radius, and B 44 is the shear stiffness of the wall. According to Eqs. (5.39), B 44 = A44h, where h = 5 mm is the wall thickness, while A 44 is specified by Eqs. (4. 72) and can ... criterion to the failure analysis of composite materials. J. Composite Mater. 14 (suppl) 2 841 . Tsai, S.W. and Hahn, H.T. (1975). Failure analysis of composite materia...

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... A12&,~ + A22&,,~ - -A74KxyT = Al29 . 4- 0 o h o h A 44? xyT - 4 (A14JGT $. A24K,T) = 0, o h -A 14? .r,,;r +T(AllKxT +A12KvT) = 0, -A 24? :.~ + - (A IZK.XT + A22Ky-r) = 0, AI4Ex~ + A 24& amp;$~ - jA44K.ryT ... solution of Eqs. (7. 74) can be written as (7. 74) (7.75) (7.76) 348 Mechanics and analysis of composite materials 5- 5 10 15 20 25...

Ngày tải lên: 13/08/2014, 17:20

... form { σ } = [ S ] { ε } (2 .46 ) where { σ } = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ σ x σ y σ z τ xy τ xz τ yz ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭ { ε } = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ε x ε y ε z γ xy γ xz γ yz ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭ [ S ] = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ S 11 S 12 S 13 S 14 S 15 S 16 S 21 S 22 S 23 S 24 S 25 S 26 S 31 S 32 S 33 S 34 S 35 S 36 S 41 S 42 S 43 S 44 S 45 S 46 S 5...

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... Advanced mechanics of composite materials Fig. 3 .46 . Two-, four-, and eight-sector test fixtures for composite rings. Fig. 3 .47 . A composite ring on a eight-sector test fixture. Fig. 3 .48 . Failure ... specimens. P P a t 12 x y 1 2 45 ° Fig. 3. 54. Simulation of pure shear in a square frame. Chapter 3. Mechanics of a unidirectional ply 111 0.3 0 .4 0.5 0.6 0.7 0.8 0.9 1 0.3 0...

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... τ 12 = τ 45 12 ,weget σ 1 = 1 2  σ 45 1 +σ 45 2  −τ 45 12 σ 2 = 1 2  σ 45 1 +σ 45 2  +τ 45 12 τ 12 = 1 2  σ 45 1 −σ 45 2  (6. 24) Substitution in Eq. (6.23) yields F  σ 45 1 ,σ 45 2 ,τ 45 12  = 1 4  2 σ 2 0 + 1 τ 2 0    σ 45 1  2 +  σ 45 2  2  + 1 2  2 σ 2 0 − 1 τ 2 0  σ 45 1 σ 45 2 + 2 σ 2 0  τ 45 12  2 = ... B 44 γ 0 xy ,M xy = C 4...

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... to F  σ 45 1 ,σ 45 2 ,τ 45 12  = 1 σ 2 45   σ 45 1  2 +  σ 45 2  2  +  1 τ 2 45 − 2 σ 2 45  σ 45 1 σ 45 2 +  τ 45 12 τ 45  =1 (6.27) where σ 45 and τ 45 are given by 1 σ 2 45 = 1 4  2 σ 2 0 + 1 τ 2 0  , τ 2 45 = 1 2 σ 2 0 Comparing ... form  R 45  = [ T ] T  R 0  [ T ] = ⎡ ⎢ ⎣ R 45 11 R 45 12 0 R 45 12 R 45 11 0 004S 45 12 ⎤ ⎥ ⎦...

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... problem as γ ∗ xy (p) = T ∗ (p) 2πR 2 B ∗ 44 (p) (7.60) Here, B ∗ 44 (p) = A ∗ 44 (p)h, where h is the shell thickness. Let the shell be made of glass–epoxy composite whose mechanical properties ... Advanced mechanics of composite materials 0 0 .4 0.8 1.2 1.6 01020 3 040 50 C(t),% hourt , 3 2 1 Fig. 7.13. Dependence of the moisture content on time for a carbon–epoxy composite...

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... B 11 ε 0 x +B 12 ε 0 y ,N y = B 21 ε 0 x +B 22 ε 0 y ,N xy = B 44 γ 0 xy M x = D 11 κ x +D 12 κ y ,M y = D 21 κ x +D 22 κ y ,M xy = D 44 κ xy (8.86) 44 8 Advanced mechanics of composite materials using the maximum stress ... notations in Eqs. (8.88), we get σ h = P 2πDhδ h c 2 (8.91) 44 4 Advanced mechanics of composite materials 0 0 .4 0.8 1.2 1.6 2 0 153 045 607590 f 0 h...

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... 4. 2.2. 4. 3.1. 4. 3.2. 4. 4.1. 4. 4.2. 4. 5.1. 4. 5.2. 4. 5.3. 4. 2. 4. 3. 4. 4. 4. 5. 4. 6. 4. 7. 4. 8. 4. 9. Isotropic Layer 121 Linear Elastic Model 121 Nonlinear Models I 24 Unidirectional Orthotropic ... 110 40 320 200 180-200 69 I20 240 -310 41 0 360 2 .4- 4.2 2.8-3.8 7-1 1 6.&10 3.2 4. 2 6-8.5 30 3.5 2.8 2.5 2.7 2. 84. 4 3.1-3.8 3.5...

Ngày tải lên: 13/08/2014, 17:20