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:
"Macro Instruction"
Explanation:
A macro definition is a rule or pattern that specifies how a certain input sequence should be mapped to a replacement output sequence according to a defined procedure. The mapping process that instantiates a macro use into a specific sequence is known as macro expansion.
It is a series of commands and actions that can be stored and run whenever you need to perform the task. You can record or build a macro and then run it to automatically repeat that series of steps or actions.
Answer:
maximum stress is 2872.28 MPa
Explanation:
given data
radius of curvature = 3 × 
mm
crack length = 5.5 × 
mm
tensile stress = 150 MPa
to find out
maximum stress
solution
we know that maximum stress formula that is express as
......................1
here σo is applied stress and a is half of internal crack and t is radius of curvature of tip of internal crack
so put here all value in equation 1 we get
σm = 2872.28 MPa
so maximum stress is 2872.28 MPa
Answer:
A. smallest wire is No. 12
