Answer:
the elongation of the metal alloy is 21.998 mm
Explanation:
Given the data in the question;
K = σT/ (εT)ⁿ
given that metal alloy true stress σT = 345 Mpa, plastic true strain εT = 0.02,
strain-hardening exponent n = 0.22
we substitute
K = 345 / 
K = 815.8165 Mpa
next, we determine the true strain
(εT) = (σT/ K)^1/n
given that σT = 412 MPa
we substitute
(εT) = (412 / 815.8165 )^(1/0.22)
(εT) = 0.04481 mm
Now, we calculate the instantaneous length
= 
given that
= 480 mm
we substitute
=
× 
= 501.998 mm
Now we find the elongation;
Elongation = 
we substitute
Elongation = 501.998 mm - 480 mm
Elongation = 21.998 mm
Therefore, the elongation of the metal alloy is 21.998 mm
Answer:
PDSA, where you work as a group
Explanation:
Answer:
A) ν = 0.292
B) ν = 0.381
Explanation:
Poisson's ratio = - (Strain in the direction of the load)/(strain in the direction at right angle to the load)
In axial tension, the direction of the load is in the length's direction and the direction at right angle to the load is the side length
Strain = change in length/original length = (Δy)/y or (Δx)/x or (ΔL/L)
A) Strain in the direction of the load = (2.49946 - 2.5)/2.5 = - 0.000216
Strain in the direction at right angle to the load = (7.20532 - 7.2)/7.2 = 0.0007389
Poisson's ratio = - (-0.000216)/(0.0007389) = 0.292
B) Strain in the direction of the load = (2.09929 - 2.1)/2.1 = - 0.0003381
Strain in the direction at right angle to the load = (5.30470 - 5.3)/5.3 = 0.0008868
Poisson's ratio = - (-0.0003381)/(0.0008868) = 0.381