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, 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 , in which the values for K and n are 6900 MPa and 0.30, respectively. Furthermore, plastic deformation occurs between strain values of 0.01 and 0.75, at which point fracture occurs.
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Answer:
attached below
Explanation:
Answer:
7.7 kN
Explanation:
The capacity of a material having a crack to withstand fracture is referred to as fracture toughness.
It can be expressed by using the formula:
where;
fracture toughness K = 137 MPa
geometry factor Y = 1
applied stress = ???
crack length a = 2mm = 0.002
∴
Now, the tensile impact obtained is:
P = A × σ
P = 1728.289 × 4.5
P = 7777.30 N
P = 7.7 kN