The ultimate shearing stress of the plate is , = 300 MPa The compressive stress in the punch is o̟ = 400MPA . The diameter of punched hole is d=100 mm. The thickness of plate is t=10 mm . %3D
The ultimate shearing stress of the plate is , = 300 MPa The compressive stress in the punch is o̟ = 400MPA . The diameter of punched hole is d=100 mm. The thickness of plate is t=10 mm . %3D
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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![Given data:
The ultimate shearing stress of the plate is t, = 300 MPa .
The compressive stress in the punch is o̟ = 400 MPa .
The diameter of punched hole is d=100 mm.
The thickness of plate is t=10 mm .](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F94458cf2-014b-466a-bada-fa68de68f3b9%2F13ca9e99-cee6-400b-ae87-33f57008762a%2Fquecb5_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Given data:
The ultimate shearing stress of the plate is t, = 300 MPa .
The compressive stress in the punch is o̟ = 400 MPa .
The diameter of punched hole is d=100 mm.
The thickness of plate is t=10 mm .
![Grop: B1
a)
In this problem, we will complete the table for four
forces.
Calculate the parameters for force P=16000 N.
b)
16000 N
P
A.
10 MPa
1600 mm2
ΔL
E=
Lo
0.4
0.010
40
10 MPa
E=% =
1000 MPa
0.010
L,A,
( 40) (1600)
Ac=
Lo+AL
1584. 458 mm?
40+0.4
16000 N
10.1 MPa
OT=
Ac
1584.458 mm²
€r=ln(1+e)= ln(1+0.010)= 0. 00995
Calculate the parameters for force P=20000 N.
20000 N
12.5 MPa
1600 mm²
ΔL
E=
Lo
0.45
= 0.01125
40
12,5 MPa
E=
1111. 11 MPa
0.01125
LoA.
( 40) (1600)
Ac=
1582. 2 mm²
Lo+AL
40+0.45
20000 N
12. 64 MPa
OT=
Ac
1582.2 mm2
ET=In(1+e)= In(1+0.01125)= 0.011187](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F94458cf2-014b-466a-bada-fa68de68f3b9%2F13ca9e99-cee6-400b-ae87-33f57008762a%2F7zlza0u_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Grop: B1
a)
In this problem, we will complete the table for four
forces.
Calculate the parameters for force P=16000 N.
b)
16000 N
P
A.
10 MPa
1600 mm2
ΔL
E=
Lo
0.4
0.010
40
10 MPa
E=% =
1000 MPa
0.010
L,A,
( 40) (1600)
Ac=
Lo+AL
1584. 458 mm?
40+0.4
16000 N
10.1 MPa
OT=
Ac
1584.458 mm²
€r=ln(1+e)= ln(1+0.010)= 0. 00995
Calculate the parameters for force P=20000 N.
20000 N
12.5 MPa
1600 mm²
ΔL
E=
Lo
0.45
= 0.01125
40
12,5 MPa
E=
1111. 11 MPa
0.01125
LoA.
( 40) (1600)
Ac=
1582. 2 mm²
Lo+AL
40+0.45
20000 N
12. 64 MPa
OT=
Ac
1582.2 mm2
ET=In(1+e)= In(1+0.01125)= 0.011187
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