NEW 375 Lecture07 - BIAXIAL STRESS SITUATION x x Y Y = z y...

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Unformatted text preview: BIAXIAL STRESS SITUATION x x Y Y = z y x = z y x HOOKES LAW FOR BIAXIAL STRESS SITUATION E E E E E E y x z y x y y x x -- = +- =- = x x Y Y EXAMPLE PROBLEM 2 Determine the change in the thickness of the aluminum plate due to the normal stresses applied. 1 8 k s i 1 2 k s i x y z Aluminum: E = 10 x 10 3 ksi = 0.33 OBSERVE THE PLATE IS SUBJECTED TO BIAXIAL LOADING ksi 18 ksi 12 z y x = - = + = ( 29 ( 29 ( 29 [ ] ( 29 [ ] in in 10 x 98 . 1 . 6 10 x 33 . ksi 18 ksi 12 ksi 10 x 10 33 . E E E 4 4 z 3 y x y x z-- + =-- = - + +- = + - = - - = Normal strain in the z direction Change in thickness of plate ( 29 ( 29 . in 10 x 99 . . in 50 . 10 x 98 . 1 t 4 4 z-- + = + = = GENERAL STRESS SITUATION z y x xz yz xy x y z GENERAL STRESS SITUATION y x z yx zx zy xz xy yz SHEAR STRESS NORMAL STRESS in plane z in direction y RELATION BETWEEN...
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NEW 375 Lecture07 - BIAXIAL STRESS SITUATION x x Y Y = z y...

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