125_208_Spring_2010_Lecture_17_Principal_Stress+_Compatibility+Mode_

125_208_Spring_2010_Lecture_17_Principal_Stress+_Compatibility+Mode_

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1 125:208 Introduction to Biomechanics Lecture 17-18:Principal Stress I & II What happens if we analyze the stress within a material in regards to differing coordinate systems? For this situation, we apply a tensile load, but what is happening to the diamond in the middle? It is experiencing shear. What about the square inside the diamond (back to tension and compression (laterally)). What about an arbitrary plane? Can we quantify normal and shear stresses in any plane? Of course we can!
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2 Multi-axial stress analysis: 3 ways 1. Memorize equations (only reasonably works for 2D) 2. Mohr’s Circle (good for 2D and 3D, but tougher in 3D) 3. Eigenvalues of Stress Tensor (Harder in 3D than 2D, but much preferred over other methods for 3D). 1. Memorize equations
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3 1. Memorize equations What is the normal Stress σ x’ after a rotation of θ ? Assume stress is constant throughout thickness,t. θ τ σ cos sin sin cos sin sin cos cos 0 2 2 ' ' xy xy y x x x dA dA dA dA dA F = sin cos 2 sin cos sin cos 2 sin cos 2 2 ' ' xy y x x xy y x x dA dA dA dA + + = + + = () 2 cos 1 2 1 cos 2 cos 1 2 1 sin 2 2 + = = but 2 sin cos sin 2 = 2 sin 2 cos 2 2 2 sin 2 cos 1 2 1 2 cos 1 2 1 ' ' xy y x y x x xy y x x + + + = + + + =
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4 In order to find σ y’ we can replace θ with θ +90° so that θ τ σ 2 sin 2 cos 2 2 ' xy y x y x y + = Cos(2 θ +180)= -Cos(2 θ ) and Sin(2 θ +180)= -Sin(2 θ ) Now what about the tangential stress, τ x’y’ ? ' ' ' cos cos sin sin sin cos cos sin 0 xy xy y x y x y dA dA dA dA dA F + + = () 2 2 ' ' 2 2 ' ' cos sin cos sin cos sin cos sin xy xy y x y x xy xy y x y x dA dA dA dA + = + = 2 cos 1 2 1 cos 2 cos 1 2 1 sin 2 2 + = = 2 sin cos sin 2 = ( ) ( ) ( ) 2 cos 2 sin 2 2 2 cos 1 2 2 cos 1 2 sin 2 ' ' ' ' xy y x y x xy xy y x y x + = + + =
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5 In summary θ τ σ 2 sin 2 cos 2 2 ' xy y x y x x + + + = () 2 cos 2 sin 2 2 sin 2 cos 2 2 ' ' ' xy y x y x xy y x y x y + = + = So these equations can be used for find stress in any coordinate system if find stress in any coordinate system if it is known in a single coordinate system Also note that y x y x + = + ' ' Why is this useful?? We can use these equation to find the maximum stress anywhere in a material under loading How do you find the maximum stress? Take the derivative of stress with respect to θ and set it to zero so that 2 sin 2 cos 2 2 ' xy y x y x x + + + = The maximum stress occurs when y x xy p xy y x x d d = = + = 2 2 tan 0 2 cos 2 2 sin '
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6 () y x xy p σ τ θ = 2 2 tan τ xy 2 θ ½( σ x - σ y ) The length of the hypotenuse of this triangle is 2 2 2 1 xy y x + 2 2 2 2 2 1 2 1 2 cos 2 1 2 sin xy y x y x xy y x xy + ± = + ± = ( ) ( ) 2 2 2 2 1 2 1 2 cos 1 2 sin xy y x y x p xy y x xy p + ± = + ± = 2 2 2 2 min max, 1 1 2 1 2 2 xy xy y x y x y x + ± + + ± + + = 2 sin 2 cos 2 2 ' xy y x y x x + + + = 2 2 ( ) ( ) 2 2 min max, 2 2 2 2 min max, 2 1 2 2 1 2 2 2 2 xy y x y x xy y x xy y x y x xy y x xy y x + ± + = +
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This note was uploaded on 02/12/2011 for the course 125 208 taught by Professor Shreiber during the Spring '08 term at Rutgers.

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125_208_Spring_2010_Lecture_17_Principal_Stress+_Compatibility+Mode_

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