a.) Determine the maximum normal stress developed in the bar when it is subjected to a tension of P = 16 kN. The stress concentration factors K can be found from the graphs below. b.) If the bar has an ultimate tensile stress Oult = 275 MPa, and using a factor of safety = 2, determine the maximum tension of P that could be applied to the bar.
a.) Determine the maximum normal stress developed in the bar when it is subjected to a tension of P = 16 kN. The stress concentration factors K can be found from the graphs below. b.) If the bar has an ultimate tensile stress Oult = 275 MPa, and using a factor of safety = 2, determine the maximum tension of P that could be applied to the bar.
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
Related questions
Question
![6 mm
80 mm
40 mm
P
r = 20 mm
40 mm
a.) Determine the maximum normal stress developed in the bar when it is subjected to a tension of P
= 16 kN. The stress concentration factors K can be found from the graphs below.
b.) If the bar has an ultimate tensile stress Oult = 275 MPa, and using a factor of safety = 2, determine
the maximum tension of P that could be applied to the bar.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fca9fae2d-ac76-4cbf-b8f5-b3ad2aa9a87d%2Fcb55f2aa-0486-4a31-8ad5-bd4c487bd9cc%2Fzv1a8r8_processed.png&w=3840&q=75)
Transcribed Image Text:6 mm
80 mm
40 mm
P
r = 20 mm
40 mm
a.) Determine the maximum normal stress developed in the bar when it is subjected to a tension of P
= 16 kN. The stress concentration factors K can be found from the graphs below.
b.) If the bar has an ultimate tensile stress Oult = 275 MPa, and using a factor of safety = 2, determine
the maximum tension of P that could be applied to the bar.
![3.0
3.2
2.8
N-
2.6
3.0
N-
2.4
ht
2.8
2.2
K
-4.0.
avg
(w – 2r)t
K
2.0
2.6
-= 2.0–
1.8
-1.5
1.6
1.2-
2.4
1.4
2.2
1.2
1.0
2.0
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
0,1
0.2
0.3
0.4
0.5
1.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fca9fae2d-ac76-4cbf-b8f5-b3ad2aa9a87d%2Fcb55f2aa-0486-4a31-8ad5-bd4c487bd9cc%2Fnyw2evm_processed.png&w=3840&q=75)
Transcribed Image Text:3.0
3.2
2.8
N-
2.6
3.0
N-
2.4
ht
2.8
2.2
K
-4.0.
avg
(w – 2r)t
K
2.0
2.6
-= 2.0–
1.8
-1.5
1.6
1.2-
2.4
1.4
2.2
1.2
1.0
2.0
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
0,1
0.2
0.3
0.4
0.5
1.
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