6 Answers to the exercises of chapter 6 Exercises 61 \u03b5 x 2ax b 62 u x 1 3a x3 1

6 answers to the exercises of chapter 6 exercises 61

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6 Answers to the exercises of chapter 6 Exercises 15
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16 Answers to the exercises of chapter 6 6.4
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Exercises 17 6.6 The distributed load q is neglected and A = 1 4 π ( D 2 - d 2 ). (a) Integrating once leads to: EA du dx = c 1 = - P Note the negative sign in front of P (compressive force)! σ = - P A = - P 1 4 π ( D 2 - d 2 ) (b) u ( x ) = - P EA x + c 2 u ( ) = 0 c 2 = P‘ EA u ( x ) = P ( - x ) 1 4 πE ( D 2 - d 2 ) Supplementary, the strain can be determined: ε = du dx = - P 1 4 πE ( D 2 - d 2 ) Note that the strain is negative! 6.7 (a) Everywhere in the muscle tendon complex the internal force is F . (b) Between A and B: σ 1 = F/A 1 . Between B and C: σ 2 = F/A 2 . (c) Forces and stresses remain unchanged. (d) u B = F‘ 1 E 1 A 1 u C = F‘ 1 E 1 A 1 + F‘ 2 E 2 A 2
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7 Answers to the exercises of chapter 7 Exercises 18
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Exercises 19 7.5 L = - α 2 2 ( ~e x ~e y + ~e y ~e x )
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8 Answers to the exercises of chapter 8 Exercises 8.1 σ xy = - 2 axy + 2 a 8.2 (a) σ = 10 ~e x ~e x + 20 ~e y ~e y + 5( ~e x ~e y + ~e y ~e x ) (b) ~s = 1 2 2(15 ~e x + 25 ~e y ) (c) ~s n = 10 2 ( ~e x + ~e y ) ~s t = 5 2 2( - ~e x + ~e y ) 8.3 (a) σ a = 10 0 0 - 20 ; σ b = 0 5 5 0 (b) ~s a = 10 cos( α ) ~e x - 20 sin( α ) ~e y ~s b = 5 sin( α ) ~e x + 5 cos( α ) ~e y (c) | ~s a | = 10 q cos 2 ( α ) + 4 sin 2 ( α ) | ~s b | = 5 20
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Exercises 21
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9 Answers to the exercises of chapter 9 Exercises 9.1 v ( x ) = a + b ( y - at ) a 0 9.2 a ( x , t ) = 1 (1 + αt ) 2 bcx - αay - αbz 0 - αcx + acy + bcz 9.3 t = 3 / 2 [s] 9.4 ~a = - ω 2 x~e x - ω 2 y~e y 9.5 ~a = c 2 x~e x + c 2 y~e y 9.6 x P = 1 2 + 1 6 sin ( 2 π t T ) 1 3 0 22
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10 Answers to the exercises of chapter 10 Exercises 10.1 x 0 Q = - 1 9 6 10.2 ~a 0 · ~a = | ~a 0 || ~a | cos( φ ) ~a 0 · F · ~a 0 = p ~a 0 · ~a 0 q ( ~a 0 · F T ) · ( F · ~a 0 ) cos( φ ) cos( φ ) = ~a 0 · F · ~a 0 q ( ~a 0 · ~a 0 )( ~a 0 · F T · F · ~a 0 ) 10.3 ~x Q = 11 ~e x + 6 ~e y + 3 ~e z 10.4 J = 2 10.5 = 3 e 0 . 02 t [cm] 23
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24 Answers to the exercises of chapter 10 10.6 ˙ F = 1 2 ˙ F · F T + F · ˙ F T = 1 2 ˙ F · F - 1 · F · F T + F · F T · F - T · ˙ F T = 1 2 L · B + B · L T
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11 Answers to the exercises of chapter 11 Exercises 11.1 ρ A V A = ρ B V B 11.2 ρ = 1500 1 + 0 . 02 t expressed in [kg/m 3 ] 25
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12 Answers to the exercises of chapter 12 Exercises 12.1 ε v = 1 / 8 12.2 F v = - 65 24 G‘ 2 12.3 E = 5 h K v 6 π δ R 2 12.4 ν = 0 . 2 12.5 (a) ε yy = 0 . 1 (b) ε xx = - 0 . 067 (c) σ yy = 0 . 8 [MPa] 12.6 ε = 1 2 αR L ( ~e x ~e z + ~e z ~e x ) σ = - E 2(1 + ν ) αR L ( ~e x ~e z + ~e z ~e x ) σ M = E 2(1 + ν ) αR L 3 26
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13 Answers to the exercises of chapter 13 Exercises 13.1 The surfaces with a normal in y -direction are stress free. So at these surfaces σ yz = σ zy = 0. This contradicts the assumptions
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