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Lady_Fox [76]
3 years ago
6

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.

Engineering
1 answer:
PtichkaEL [24]3 years ago
7 0

Answer:

Detailed solution is attached below in three simple steps the problem is solved.

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See attachment below

Explanation:

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Rubber block is not shown. I have attached an image of it.

Answer:

A) ε_x = 0.0075

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Answer:

Hope it helps...

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PLEASE THANK MY ANSWER

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