Find the toughness (or energy to cause fracture) for a metal that experiences both elastic and plastic deformation. Assume Equat
ion 6.5 for elastic deformation, that the modulus of elasticity is 172 GPa (25 ´ 106 psi), and that elastic deformation terminates at a strain of 0.01. For plastic deformation, assume that the relationship between stress and strain is described by Equation 6.19, in which the values for K and n are 6900 MPa (1 ´ 106 psi) and 0.30, respectively. Furthermore, plastic deformation occurs between strain values of 0.01 and 0.75, at which point fracture occurs.
1 answer:
You might be interested in
Answer:
a) -1.46 x 10∧-5, 1.445x 10∧-4, -6.355 x 10∧-4
b) 3.926 x 10∧-4, -2.626 x 10∧-4
c) 6.552 x 10∧-4, 6.5 x 10∧-5
Explanation:
a) -1.46 x 10∧-5, 1.445x 10∧-4, -6.355 x 10∧-4
b) 3.926 x 10∧-4, -2.626 x 10∧-4
c) 6.552 x 10∧-4, 6.5 x 10∧-5
The explanation is shown in the attachment. I hope i have been able to help.
Answer:
Following are the proving to this question:
Explanation:
using the energy equation for entry and exit value :
where
L.H.S = R.H.S