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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
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 .
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
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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