--- //[[honeypot@handbasket.org|David Wagner]] 2009/10/25 22:49//
== √ME 5463-001 Fracture Mechanics, Homework #15 ==
<box 100%>
**6.7**
A 1/4 inch-thick wide aluminum (sigma_ys = 40 ksi) tension panel has an edge crack
0.80 inch long.  Based on fracture tests with a similar material for the same thickness,
the fracture toughness has been measured as 45 ksi sqrt(in) for slowly applied loads.

**(a)**
Compute the diameter of the plastic zone for both plane-stress and plane-strain conditions.

**(b)**
Based on your engineering judgment and the information available to you, 
which of the two plane conditions prevails?  Why?

**(c)**
What is the magnitude of the largest remotely applied steady-state stress that the panel can support without failure?

**(d)**
What would be the consequence of an impact load on your answer to part (c)?
</box>

====== Part (a) ======

Plane Stress: <m>r_b = {1/{2 pi}} (K_I/{sigma_{ys}})^2</m>
=<m>1/{2pi} (45000/40000)^2</m> = 0.20 in.

Plane Strain: <m>r_y</m> = 0.067 in.

====== Part (b) ======

For plane strain: <m>0.25 = B > 2.5(K_c/sigma_{ys})^2 = 3.16</m>

This is not so, so plane stress conditions prevail.

====== Part (c) ======

<m>sigma_c = K_c/{1.12 sqrt{pi a}}</m>
=
<m>sigma_c = 45000/{1.12 sqrt{0.8*pi}}</m>
= 2534 psi

====== Part (d) ======
The specimen could withstand a higher impact load.