The engineering stress-engineering tensile curve for a stainless steel typ alloy is shown. Calculate the consta n for the true stress and true strain c before necking (o: = K (&)").
The engineering stress-engineering tensile curve for a stainless steel typ alloy is shown. Calculate the consta n for the true stress and true strain c before necking (o: = K (&)").
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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Transcribed Image Text:19:48
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قبل 6 دقائق
حسن فلاح
جنة بوسط من
200
The engineering stress-engineering strain
tensile curve for a stainless steel type 304
alloy is shown. Calculate the constants K and
n for the true stress and true strain curve
before necking (G = K (&)").
Engineering Stress (MPa)
1000
900
800 -
700-
600-
500-
400-
300-
200-
100-
0+
0.0
0.2
ده
04
Enginering Strain (mm/mm)
0.8
10
OO REDMI NOTE 9S
AI QUAD CAMERA
|||
![From the stress-strain diagram
Engineering
strain (e)
0.2
0.6
568
Ot
From the relation of the stress and strain
ot = K (et)"
.. (1)
From the table values
558 K (0.1823)"
908.8 = K (0.47)"
From the eq(2)/eq(3)
(0.1823)"
558
908.8
(0.47)"
=
Engineering stress(True stress(ot = o(1 + ε))
o) (MPa)
(MPa)
465
=
Ot=465(1+0.2) = 558
True strain (ln(1 + ε))
Et = ln(1 +0.2) = 0.1823
568 (1+0.6) 908.8&t= ln(1 +0.6) = 0.47
=
⇒n [ln (0.38787)] = ln 0.613996
⇒n=0.515
⇒ (0.38787)" = 0.613996
. (2)
(3)
From the eq(2)
558 = K (0.1823)0.515 ⇒ K = 1340.69
From the eq(1)
ot = 1340.69 (t) 0.515
=](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa935b5b3-550a-4564-9870-ed0c06de3fa7%2F3040ed90-30d8-4662-8837-cb2578da785e%2Fj0mf8tb_processed.jpeg&w=3840&q=75)
Transcribed Image Text:From the stress-strain diagram
Engineering
strain (e)
0.2
0.6
568
Ot
From the relation of the stress and strain
ot = K (et)"
.. (1)
From the table values
558 K (0.1823)"
908.8 = K (0.47)"
From the eq(2)/eq(3)
(0.1823)"
558
908.8
(0.47)"
=
Engineering stress(True stress(ot = o(1 + ε))
o) (MPa)
(MPa)
465
=
Ot=465(1+0.2) = 558
True strain (ln(1 + ε))
Et = ln(1 +0.2) = 0.1823
568 (1+0.6) 908.8&t= ln(1 +0.6) = 0.47
=
⇒n [ln (0.38787)] = ln 0.613996
⇒n=0.515
⇒ (0.38787)" = 0.613996
. (2)
(3)
From the eq(2)
558 = K (0.1823)0.515 ⇒ K = 1340.69
From the eq(1)
ot = 1340.69 (t) 0.515
=
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