EMA4223 HW 8 Solutions

EMA4223 HW 8 Solutions - at the voids equals one-half of...

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8.2 Explain why FCC metals show a ductile fracture even at low temperatures, while BCC metals do not. FCC metals show ductile fracture at low temperatures while BCC metals do not because of the fact that there are many more slip systems in FCC metals that can be active than in BCC metals. The numerous slip systems lead to greater dislocation mobility and allow for cross-slipping between planes which leads to ductile fracture even at low temperatures where dislocations are not as abundant.
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8.7 A ceramic with K Ic =4 MPa m 1/2 contains pores with radii a=5 μm due to incomplete sintering. These pores lead to a decrease in the failure stress of the material in both tension and compression. One in every ten grain- boundary junctions contains a void; the grain size of the ceramic is 50 μm. The ceramic fails in compression when the length l of each crack generated
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Unformatted text preview: at the voids equals one-half of the spacing between the voids. (a) Determine the compression strength of the ceramic, using the equation from Sammis and Ashby MPa L L a K m m L L L K a c c Ic c Ic c 9 . 55 ) 51 ( ) 1 ( 1 . 1 50 10 5 10 4 ) 1 ( ) 1 . 2 1 ( 1 . 1 50 5 ) 500 2 1 ( ) 1 ( ) 1 . 2 1 ( 1 . 1 1 3 . 3 2 / 1 6 6 3 . 3 2 / 1 3 . 3 2 / 1 (b) Determine the tensile strength of the ceramic, assuming flaws with size a. GPa a K Ic 009 . 1 10 5 10 4 6 6...
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EMA4223 HW 8 Solutions - at the voids equals one-half of...

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