HW_Tutorial_2_Sp_2011

HW_Tutorial_2_Sp_201 - Homework Tutorial Problems 2 MME 412 Advanced Mechanics of Materials 1 Problem 2.2 Given y y y 30 MPa y 4 0 MPa x x x 30o xy

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Homework Tutorial Problems # 2 MME 412: Advanced Mechanics of Materials - 1 - 1. Problem 2.2 Given: Determine the complete state of stress for original and transformed co-ordinates. The transformed state of stress can be determined by the following transformation, cos sin sin cos cos sin sin cos y yx xy x y x y y x x T σ leading to the following equations, 2 2 2 2 2 2 sin cos cos sin cos sin 2 cos sin cos sin 2 sin cos xy y x y x xy y x y xy y x x By substituting the given values, MPa 40 x , MPa 30 y , MPa 20 y x and o 30 in the above equations we get, o o xy o o x 30 cos 30 sin 2 30 sin 30 30 cos 40 2 2 (1) o o xy o o y 30 cos 30 sin 2 30 cos 30 30 sin 40 2 2 (2) x y y x o 30 MPa 40 x MPa 30 y x y MPa 20 y x
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Homework Tutorial Problems # 2 MME 412: Advanced Mechanics of Materials - 2 - o o xy o o 30 sin 30 cos 30 cos 30 sin 30 40 20 2 2 (3) From (3) we have, 625 . 100 5 . 0 311 . 30 20 xy xy By substituting this value in (1) and (2) we get, 6 . 109 x and 6 . 99 y Hence the complete state of stresses is: Solution check by MATLAB State of Stress in the original co-ordinate S=[40 -100.625; -100.625 -30] S = 40.0000 -100.6250 -100.6250 -30.0000 Given rotation for the new co-ordinate th=-pi/6 th = -0.5236 x y y x o 30 MPa 40 x MPa 30 y x y MPa 20 y x MPa 625 . 100 xy MPa 6 . 109 x x MPa 6 . 99 y
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Homework Tutorial Problems # 2 MME 412: Advanced Mechanics of Materials - 3 - Transformation matrix T=[cos(th) sin(th); -sin(th) cos(th)] T = 0.8660 -0.5000 0.5000 Transformed state of stress Spr=T*S*T' Spr = 109.6438 -20.0016 -99.6438
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MME 412: Advanced Mechanics of Materials - 4 - 2. Problem 2.5 and check the stress invariants relative to both the co-ordinates Given: State of stress: kpsi 3 0 1 0 2 1 1 1 2 z yz xz yz y yx xz xy x σ Direction Cosines for transformation: 6 6 6 2 3 2 3 2 3 2 2 3 2 3 0 6 1 z z y z x z z y y y x y z x y x x x n n n n n n n n n (a) By using the following equations and substituting the given values we can obtain the transformed state of stress. Transformed state of stress: kpsi 5 . 1 707 . 0 598 . 2 707 . 0 0 . 1 225 . 1 598 . 2 225 . 1 5 . 0 z
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This document was uploaded on 11/04/2011 for the course MME 512 at Miami University.

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HW_Tutorial_2_Sp_201 - Homework Tutorial Problems 2 MME 412 Advanced Mechanics of Materials 1 Problem 2.2 Given y y y 30 MPa y 4 0 MPa x x x 30o xy

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