Question

2. The displacement vector of a continuous medium is given as follows: \textbf{u} = 0.2\textit{t} \textit{X}\textsubscript{2} \textbf{e}\textsubscript{1} + 0.3\textit{t} \textit{X}\textsubscript{2} \textbf{e}\textsubscript{2} - 0.1\textit{t} \textit{X}\textsubscript{1} \textit{X}\textsubscript{2} \textbf{e}\textsubscript{3} d. For the particle initially located at (1, -1, 2) determine below questions when the particle is at \textit{t} = 2; i. Determine the deformation gradient, Green, Cauchy, Piola, and Finger deformation tensors, as well as the Lagrange and Euler strain tensors. ii. Calculate the direction of the vector \textbf{n} = [1, 1, -1]\before deformation, and the stress and unit strain along this direction.

          2. The displacement vector of a continuous medium is given as follows:
\textbf{u} = 0.2\textit{t} \textit{X}\textsubscript{2} \textbf{e}\textsubscript{1} + 0.3\textit{t} \textit{X}\textsubscript{2} \textbf{e}\textsubscript{2} - 0.1\textit{t} \textit{X}\textsubscript{1} \textit{X}\textsubscript{2} \textbf{e}\textsubscript{3}
d. For the particle initially located at (1, -1, 2) determine below questions when the particle is at \textit{t} = 2;
i. Determine the deformation gradient, Green, Cauchy, Piola, and Finger deformation tensors, as well as
the Lagrange and Euler strain tensors.
ii. Calculate the direction of the vector
\textbf{n} = [1, 1, -1]\before deformation, and the stress and unit strain
along this direction.
        
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2. The displacement vector of a continuous medium is given as follows:
u = 0.2t X2 e1 + 0.3t X2 e2 - 0.1t X1 X2 e3
d. For the particle initially located at (1, -1, 2) determine below questions when the particle is at t = 2;
i. Determine the deformation gradient, Green, Cauchy, Piola, and Finger deformation tensors, as well as
the Lagrange and Euler strain tensors.
ii. Calculate the direction of the vector
n = [1, 1, -1]deformation, and the stress and unit strain
along this direction.

Added by Samuel S.

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University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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Transcript

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00:01 All right, so let's say the x coordinate of an object given has a function of time as a times 1 plus the cosine of 2 omega -t, and then y.
00:12 So, a function of times just a sine of 2 omega -t.
00:19 So we want to find the velocity and accelerations of these.
00:24 So x dot is just going to be equal to negative 2a -o -o -mega times the sine.
00:33 Of omega -t and y dot is just going to equal two a omega cosine of two omega -t so the magnitude of the velocity which is the square root of x -d squared plus y -d squared we're going to have a two a omega term and then that's basically it because we'll have a cosine squared plus a sine squared term so which that adds up to one and then the acceleration so x -a -o -o -term and then the acceleration so x double dot we can see should just be equal to negative 2 a omega squared times the cosine of 2 omega t i think i forgot a 2 here and then y double dot is equal to negative to a omega squared cosine or sorry sign of 2 omega t and so the magnitude of our acceleration we can see is just going to be i'm sorry i this should be a 4.
01:39 It should be 4a.
01:43 Omega squared.
01:45 So that's the magnitude of our acceleration.
01:48 And then we want the tangent and normal components of the acceleration.
01:58 All right...
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