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
Input: what is put in, taken in, or operated on by any process or system.
Output: the amount of something produced by a person, machine, or industry.
Answer:
shear plane angle Ф = 26.28°
shear strain 2.20
Explanation:
given data
angle = 16°
chip thickness t1 = 0.32 mm
cut yields chip thickness t2 = 0.72 mm
solution
we get here first chip thickness ratio that is
chip thickness ratio = 
................. 1
put here value
chip thickness ratio = 
chip thickness ratio r = 0.45
so here shear angle will be Ф
tan Ф = 
............2
tan Ф = 
tan Ф = 0.4938
Ф = 26.28°
and
now we get shear strain that is
shear strain r = cot Ф + tan (Ф - α ) ................3
shear strain r = cot(26.28) + tan (26.28 - 16 )
shear strain r = 2.20
Answer:
Steady state temperature is approximately 801K
Explanation:
Detailed explanation and calculation is shown in the images below.
