Lecture21 - ME 382 Lecture 21 YIELD OR FRACTURE Yield only occurs along planes of maximum shear Load for yield PY =(b-2a t Y Moment at cracked

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ME 382 Lecture 21 8/iii/06 1 Y IELD OR F RACTURE ? Yield only occurs along planes of maximum shear Load for yield: P Y = ( b -2 a ) t σ Y Moment at cracked section: M = Pa /2 Maximum stress on cracked section (ignoring stress singularity): " = M b # a ( ) 2 12 t b # a ( ) 3 + P b # a ( ) t = 2 a + b ( ) P b # a ( ) 2 t Conservative value to use for load at onset of yield: P Y = b " a ( ) 2 t # Y / 2 a + b ( ) (c) Two possible loads for yield: P Y =(2 b - a ) t σ Y P Y = 3 . (2 b - 2 a ) t σ Y (inverse of hardness test - complex stress between cracks) Ligament yields only if a > 4 b / 5 Example Rectangular plate ( b= 50.0 mm x t = 10.0 mm) with single edge crack (length a ) Made of aluminum alloy (K Ic = 30.0 MPa m, σ Y = 300 MPa) Calculate failure load for different values of a
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ME 382 Lecture 21 8/iii/06 2 (1) Consider yield Load for yield: P Y = b " a ( ) 2 t # Y / 2 a + b ( ) = 300 " 10 6 0.050 # a ( ) 2 " 10 # 2 / 2 a + 0.050 ( ) = 3.00 0.050 " a ( ) 2 / 2 a + 0.050 ( ) MN (if a in m) (2) Consider fracture: Loading parameter for fracture: K I = f ( a / b ) ! " a = f ( a / b ) P bt a where f(a/b) given in Dowling or data book Material property for fracture: K Ic = 30.0 MPa m)
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This note was uploaded on 01/26/2011 for the course ME 382 taught by Professor Garikipati during the Fall '08 term at University of Michigan.

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Lecture21 - ME 382 Lecture 21 YIELD OR FRACTURE Yield only occurs along planes of maximum shear Load for yield PY =(b-2a t Y Moment at cracked

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