Chapter15

Chapter15 - Mechanics & Materials 1 Chapter 15 Stress and...

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- FAMU FSU College of Engineering Department of Mechanical Engineering Chapter 15 Chapter 15 Stress and Strain Stress and Strain Transformation Transformation
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General State of Stress General State of Stress In general, the three dimensional state of stress at a point in a body can be represented by nine components: σ xx σ yy and σ zz : Normal stresses τ xy τ yx τ xz τ zx τ yz and τ zy : Shear stresses By equilibrium, we can show that there are only six independent components of the stress σ xx , σ yy , σ zz , τ xy , τ xz , and τ yz
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Plane Stress Plane Stress • In much of engineering stress analysis, the condition of plane stress applies. Plane Stress : one of the three normal stresses, usually σ z vanishes and the other two normal stresses σ x and σ y , and the shear stress τ xy are known.
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Plane Stress Transformation: Plane Stress Transformation: Finding Stresses on Various Planes Finding Stresses on Various Planes • General Problem: • * Given two coordinate systems, x- y and x' - y', and a stress state defined relative to the first coordinate system xyz : σ x σ y , τ xy • * Find the stress components relative to the second coordinate system x’y’z’ : σ x σ y , τ xy
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• Consider a triangular block of uniform thickness, t: • For equilibrium: Plane Stress Transformation Plane Stress Transformation
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• Simplifying: • Using : Plane Stress Transformation Plane Stress Transformation
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Transformation Equations for Transformation Equations for Plane Stress Plane Stress
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Special Cases of Plane Stress Special Cases of Plane Stress • 1. Uniaxial Stress State: σ y = τ xy =0 • 2. Biaxial Stress State : τ xy = 0
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• 3. Pure Shear : σ x = σ y = 0 Special Cases of Plane Stress Special Cases of Plane Stress
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Principal Stress Principal Stress σ ' x varies as a function of the angle θ The maximum and minimum values of σ ' x are called the principal stresses. To find the max and min values: Where θ p defines the orientation of the principle planes on which the principle stress act. 0 = θ σ d d x
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Two values of 2 θ p in the range of 0 to 360 satisfy this equation. These two values differ by 180° so that θ p has two values that differ by 90°, one between 0 and 90° and the other between 90° and 180°. For one of the angles
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This note was uploaded on 12/11/2011 for the course EML 3011 taught by Professor Schwarz during the Spring '09 term at FSU.

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Chapter15 - Mechanics & Materials 1 Chapter 15 Stress and...

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