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?
This is the end of the preview. Sign up
to
access the rest of the document.
This note was uploaded on 11/11/2010 for the course MSE 230 taught by Professor Trice during the Spring '08 term at Purdue.
 Spring '08
 Trice

Click to edit the document details