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Unformatted text preview: EAS 209-Fall 2010 Instructors: Christine Human 11/1/2010 1 Lecture 24- Transformation of Stress Lecture 24 Transformation of Plane Stress The components of stress depend upon the orientation of the axes. We saw this with both axial loading and torsion. For uni-axial loading, stresses on an incline plane are given by: The max. normal stress occurs when θ =0 o and the max. shear stress occurs when θ =45 and 135 o . It is important to know both the maximum normal and shearing stress to predict where failure may occur. Today’s Objective: develop and use stress transformation equations for state of plane stress. Today’s Homework: cos sin cos sin cos cos cos 2 A P A P A V A P A P A F EAS 209-Fall 2010 Instructors: Christine Human 11/1/2010 2 Lecture 24- Transformation of Stress General State of Stress: The most general state of stress at a point Q may be represented by 6 stress components We are interested in the orientation of the axes that will give us the maximum normal stress or maximum shearing stress. stresses shearing , , stresses normal , , zx yz xy z y x ) , , : (Note xz zx zy yz yx xy We can obtain different stress values, ( x’ , y’ , z’ … ) by rotating the coordinate axes, but they all represent the same state of stress at Q. EAS 209-Fall 2010 Instructors: Christine Human 11/1/2010 3 Lecture 24- Transformation of Stress Plane Stress: Plane Stress is a state of stress in which two faces of the cubic element are free of stress (assume the faces perpendicular to the z axis have zero stress). Example of elements in plane stress For the illustrated example, the state of stress is defined by xy , , y x with zy zx z Thin plate with forces acting in the mid plane of the plate Free surface of structural element or machine part EAS 209-Fall 2010 Instructors: Christine Human 11/1/2010 4 Lecture 24- Transformation of Stress Transformation of Plane Stress Consider an element in plane stress. The conditions for equilibrium of a prismatic element with faces perpendicular to the...
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