MSE 230
HW5 (due 02/11, 02/12)
Spring 2010
1. Callister 8.7: Suppose that a wing component on an aircraft is fabricated from an aluminum alloy that
has a plane strain fracture toughness of 26 MPa·m
1/2
.
It is determined that fracture results at a stress of
112 MPa when the maximum internal crack length is 8.6 mm.
For this same component and alloy,
compute the stress level at which fracture will occur for a critical internal crack length of 6.0 mm.
2. Callister 8.9: Calculate the maximum internal crack length allowable for a Ti6Al4V titanium alloy
component loaded to a stress onehalf of its yield strength. Assume that the value of
Y
is 1.5.
3.
The yield strength and fracture toughness of 4340 steel, in two different heat treatment conditions
("tempers"), are given below.
(a) For a particular component (
Y
= 1), the largest edge flaw is
a
= 0.5 mm.
If the component were
considered to fail if it yields or fractures, which condition (tempering temperature) would allow the
higher design stress and what would be the value of that stress?
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 Spring '08
 Trice
 Tensile strength, Fracture mechanics, internal crack length

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