assignment2-solution

assignment2-solution - Problem 2.1 Solution: a) x ' x ' u x...

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Problem 2.1 Solution: a) ' ' (1 ) (1 0.03) 1.03 '( 1 ) ( 1 0 . 0 1 ) 0 . 9 9 ' x xux xx x x yy v y y y y x zz z              b) 0.03 xx du dx   0.01 yy dv dy  0 zz d dz 0 xy dv du dx dy  0 yz dd v dy dz  0 xz dx dz  c) ' ' ' ' 1.03 *0.99 * 1.0197 ' (1.0197 1) 0.0197 0.02 0.03 0.01 0 0.02 xx yy zz Vx y z x y z x y z VVV V xyz xyz xyz      Problem 2.3 Consider the displacement field in a body z y x u 01 . 0 02 . 0 02 . 0 cm z y v 02 . 0 01 . 0 cm z x w 01 . 0 01 . 0 cm Calculate the strain in the body. What is the distance after deformation between the two points that are located at (0,0,0) & (5,0,0) before deformation? Check the solution obtained from the definition or strain xx with that using the geometrical method.
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Solution: Using strain-displacement relations, 02 . 0 x u xx 02 . 0 x v y u xy 01 . 0 y v yy 02 . 0 x w z u xz 01 . 0 z w zz 02 . 0 y w z v yz P 1 P 2 x=0 x=5 y=0 y=0 z=0 z=0 After deformation P 1 & P 2 are P 1 P 2 x = x+u=0 x =x+u= 5+0.1=5.1 y =y+v=0 y = y+v=0 z =z+w=0 z =z+w= -0.05
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Distance between P1 and P2 |P 1 P 2 | = 5 Distance between P1’ and P2’ | P 1 P 2 | = 1 . 5 ) 05 . 0 ( 1 . 5 2 2 0025 We have xx from displacement field as 0.02 Geometrically, '' 12 | P | | | 0.10025 0.02005 || 5 xx PP P 
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This note was uploaded on 11/29/2010 for the course AE 350 taught by Professor Achuthan during the Fall '09 term at Clarkson University .

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assignment2-solution - Problem 2.1 Solution: a) x ' x ' u x...

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