Solids Mechanics, Including Elastic Wave Propagation.
Professor Julius Kaplunov
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Solids Mechanics, Including Elastic Wave Propagation. Professor - - PowerPoint PPT Presentation
Solids Mechanics, Including Elastic Wave Propagation. Professor Julius Kaplunov Typeset by Foil T EX Part 1. Elementary introduction. 1D rod 1. Equilibrium Equation (Statics) x + x A { ( x + x ) ( x ) } = A
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x
∆x→0
∆x→0 utt(x + θ∆x)
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∆x→0
c2utt = 0 with c =
E ρ
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xuxx = 0
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1∆φ − φtt) + curl(c2 2∆−
1 = λ+2µ β
2 = µ ρ denote the speeds of the dilatation and
1∆φ − φtt = 0
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2∆−
∂ ∂x3 ≡ 0
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c2
1,
c2
2
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2
2
2
2
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2
1
2
2
3
1
3
2
2
3
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3, σ23 = σ∗ 23
3
3 = f∗(x2)eiq(x1−ct)
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2
2
2
2
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1
2
1
2
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h , ζ = x2 h and
h and seek the solution to equations (3.5) in the form
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∂u2 ∂ζ = 0 at ζ = ±1,
∂x1 = 0.
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∂u1 ∂ζ = 0 at ζ = ±1,
∂x1 = 0.
st and Λsh = Λs sh,
st = π(2n − 1)
sh = π
s
st)2](1 + O(K2))
st
st
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s
sh)2] + (1 + O(K2))
sh)
sh
st and Λsh = Λα sh where
st = πκ
sh = π(2n − 1)
α [Ω2 − (Λα st)2] + (1 + O(K2))
st
st
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α [Ω2 − (Λs st)2] + (1 + O(K2))
st)
st
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st ∼ K2)
st)2
st cos(Λs stκζ) + (−1)nκ cos(Λs stζ)
st)3 sin Λs st sin(κΛs stζ),
st)4 sin Λs st cos(κΛs stζ),
st)4sinΛs st cos(κΛs stζ),
st)4 sin Λs st cos(κΛs stζ),
st)3 (
st sin(κΛs stζ) + (−1)n+1 sin(Λs stζ)
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st ∼ K2)
st)2 sin(κΛs st) cos(Λs stζ),
st tan(κΛs st)
stζ) + cos(κΛs st) sin(Λs stζ)
sh)2 tan(κΛs sh)
shζ) − cos(κΛs sh) cos(Λs shζ)
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sh)2 tan(κΛs sh)
shζ) − cos(κΛs sh) cos(Λs shζ)
sh)2 tan(κΛs sh) cos(κΛs shζ),
sh)3 sin(κΛs sh) sin(Λs shζ);
sh ∼ K2)
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sh)2 (1 2 cos Λa sh sin(κΛa shζ) + (−1)n+1κ sin(Λa shζ)
sh)3 cos(Λa sh) cos(κΛa shζ),
sh)4 cos(Λa sh) sin(κΛa shζ),
sh)4 cos(Λa sh) sin(κΛa shζ),
sh)4 cos(Λa sh) sin(κΛa shζ),
sh)3
sh) cos(κΛa shζ) + (−1)n+1 cos(Λa shζ)ight
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sh ∼ K2)
sh)2 cos(κΛa sh) sin(Λa shζ),
sh cot(κΛa sh)
shζ) − sin(κΛa sh) cos(Λa shζ)
sh)2 cot(κΛa sh)
shζ) − sin(κΛa sh) sin(Λa shζ)
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sh)2 cot(κΛa sh)
shζ) + sin(κΛa sh) sin(Λa shζ)) ,
sh)2 cot(κΛa sh) sin(κΛa shζ),
sh)3 cos(κΛa sh) cos(Λa shζ);
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1,
2,
ii
21,
22,
R << 1 - small geometric parameter R - a typical
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h , t = η−2(c2 R)−1τ, i.e. in the
2
22 − η2ν(σ∗ 11 + σ∗ 22),
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1
2
21,
11 =
1
22,
21
11
1
22
21
2
21 = σ∗ 22 = 0
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2
1
2
11 =
1
22 =
1
21
11
22
21
2
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2 = u(0) 2 ,
1 = ζu(1) 1 ,
11 = ζσ(1) 11 ,
22 = ζσ(1) 33 ,
21 = σ(0) 21 + ζ2σ(2) 21 ,
22 = ζσ(1) 22 + ζ3σ(3) 22 .
1
2
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11 =
1
22 =
1
21 = −1
11
21 = −σ(2) 21 ,
22 = −∂σ(0) 21
2
22 = −1
21
22 + σ(3) 22 = 0.
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2
1
2
11 = −
2
1
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22 =
2
1
21 =
2
1
21 = −
2
1
22 =
2
1
2
22 = −
2
1
22 and σ(3) 22
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2
1
2
2 . Then the Kirchhoff equation becomes
1
2Eh3 3(1−ν2)
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