NEW 375 Lecture07

# NEW 375 Lecture07 - BIAXIAL STRESS SITUATION x σ x σ Y σ...

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Unformatted text preview: BIAXIAL STRESS SITUATION x σ x σ Y σ Y σ = ≠ ≠ z y x σ σ σ = ≠ ≠ z y x σ σ σ HOOKE’S LAW FOR BIAXIAL STRESS SITUATION E E E E E E y x z y x y y x x σ ν σ ν ε σ σ ν ε σ ν σ ε-- = +- =- = x σ x σ Y σ Y σ 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 ν = 0.33 OBSERVE THE PLATE IS SUBJECTED TO BIAXIAL LOADING ksi 18 ksi 12 z y x = σ- = σ + = σ ( 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-- + =-- = ε- + +- = σ + σ ν- = νσ- νσ- = ε 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-- + = + = ε = δ GENERAL STRESS SITUATION ≠ ≠ ≠ z y x σ σ σ ≠ ≠ ≠ xz yz xy τ τ τ x y z GENERAL STRESS SITUATION y σ x σ z σ yx τ zx τ zy τ xz τ xy τ yz τ SHEAR STRESS NORMAL STRESS τ in plane z in direction y RELATION BETWEEN...
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NEW 375 Lecture07 - BIAXIAL STRESS SITUATION x σ x σ Y σ...

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