User:Radinadd/EAS Homework 1

Problem 1
$$ \epsilon_y = \epsilon_a = 0.016 $$

$$ \epsilon_x = (2/3)[\epsilon_c-(1/2)\epsilon_a+\epsilon_b] = (2/3)[0.016+0.008+0.004] = 0.019 $$

$$ \epsilon_{xy} = (1/\sqrt{3})(\epsilon_c-\epsilon_b) = (1/\sqrt{3})(0.016-0/004) = 0.007 $$

$$ \epsilon_x=0.019; \epsilon_y=0.016; \epsilon_{xy}= 0.007 $$

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Problem 2
$$ [\sigma] = \ \begin{bmatrix} -2 & 1 & -3 \\ 1 & 0 & 4 \\ -3 & 4 & 5  \\ \end{bmatrix} x10^7 Pa$$

$$ \epsilon_{xx} = \frac {1}{20x10^{10}Pa}[-2-0.3(5)]x10^7 = -1.75x10^{-4}$$

$$ \epsilon_{yy} = \frac {1}{20x10^{10}Pa}[-0.3(-2+5)]x10^7 = -4.5x10^{-5}$$

$$ \epsilon_{zz} = \frac {1}{20x10^{10}Pa}[5-0.3(-2)]x10^7 = 2.8x10^{-4}$$

$$ \epsilon_{xy} = \frac {2(1+0.3)}{20x10^{10}Pa}1x10^7 = 1.4x10^{-4}$$

$$ \epsilon_{xz} = \frac {2(1+0.3)}{20x10^{10}Pa}-3x10^7 = -3.9x10^{-4}$$

$$ \epsilon_{yz} = \frac {2(1+0.3)}{20x10^{10}Pa}4x10^7 = 5.4x10^{-4}$$

$$ [\epsilon] = \ \begin{bmatrix} -1.75 & 1.4 & -3.9 \\ 1.4 & -.45 & 5.4 \\ -3.9 & 5.4 & 2.8  \\ \end{bmatrix} x10^{-4} $$

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Problem 3
$$ \sigma_l = \frac{pD}{4t} = \frac{(100psi)(20in)}{4(0.1in)} = 5000psi ;    \sigma_h=  \frac{pD}{2t} = \frac{(100psi)(20in)}{2(0.1in)}=10000psi $$

$$ \epsilon_{xx} = \frac {1}{10x10^{6}psi}[5000-(0.3)10000]psi = 2x10^{-4}$$

$$ \epsilon_{yy} = \frac {1}{10x10^{6}psi}[-(0.3)5000+10000]psi = 8.5x10^{-4}$$

$$ \epsilon_{zz} = \frac {-0.3}{10x10^{6}psi}[5000+10000]psi = 4.5x10^{-4}$$

$$ \epsilon_{xx} = 2x10^{-4} ; \epsilon_{yy} = 8.5x10^{-4} ;  \epsilon_{zz} = 4.5x10^{-4} $$

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Problem 4
$$ T_{max} = \frac{\tau_{max}J}{d/2} $$

$$ T_{max} = \frac{(400x10^6Pa)(\pi.02^4m^4/32)}{.02m/2}= 628 kNm $$

$$ \theta = \frac {\tau}{GR} = \frac{(400x10^6Pa)}{(30x10^9Pa)(0.025m)}=.5333rad $$

$$ T_{max}=628 kNm  ;   \theta = 30.6 degrees $$

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Problem 5
$$ T_{max}=\tau_{max}k_2(2b)(2h)^2 = (400x10^6Pa)(2.31)(0.05m)(0.03m)^2 = 41.6 kNm $$

$$ \theta = \frac {T}{GJ} = \frac{(41.6x10^3kNm)}{(30x10^9Pa)(0.196)(.02^4)(m^4)}= 44.2 degrees $$

$$ T_{max}=41.6 kNm  ;   \theta = 44.2 degrees $$

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