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Home Work: The lap joint shown in the figure is fastened by four rivets of 19 mm diameter. Find the maximum loạd P that can be applied if the working stresses are 96 MPa for shear in the rivet and 124 MPa for bearing in the plate. Solution:

Home Work: The lap joint shown in the figure is fastened by four rivets of 19 mm diameter. Find the maximum loạd P that can be applied if the working stresses are 96 MPa for shear in the rivet and 124 MPa for bearing in the plate. Solution:

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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Home Work: The lap joint shown in the figure is fastened by four rivets of 19 mm diameter. Find the maximum load P that can be applied if the working stresses are 96 MPa for shear in the rivet and 124 MPa for bearing in the plate. Solution: For safe load P= 108.8 KN
Transcribed Image Text:Home Work: The lap joint shown in the figure is fastened by four rivets of 19 mm diameter. Find the maximum load P that can be applied if the working stresses are 96 MPa for shear in the rivet and 124 MPa for bearing in the plate. Solution: For safe load P= 108.8 KN
Calculate the parameters for force P=16000 N. Step2 b) P 16000 N 10 MPa A. 1600 mm² ΔL 0.4 = 0.010 40 Lo 10 MPa E= 1000 MPa 0.010 ( 40) ( 1600) L,A, Ac= Lo+AL 1584. 458 mm? 40+0.4 16000 N 10.1 MPа OT= Ac 1584.458 mm2 &T=ln(1+e)= ln(1+0.010)= 0.00995 Calculate the parameters for force P-20000 N. 20000 N 12.5 MPа A. 1600 mm2 ΔL E= 0.45 : 0.01125 40 12.5 MPa 0.01125 E- 1111. 11 MPa L,A, ( 40) ( 1600) Ac= Lo+AL 1582. 2 mm² %3D 40+0.45 20000 N OT= = 12. 64 MPa 1r000
Transcribed Image Text:Calculate the parameters for force P=16000 N. Step2 b) P 16000 N 10 MPa A. 1600 mm² ΔL 0.4 = 0.010 40 Lo 10 MPa E= 1000 MPa 0.010 ( 40) ( 1600) L,A, Ac= Lo+AL 1584. 458 mm? 40+0.4 16000 N 10.1 MPа OT= Ac 1584.458 mm2 &T=ln(1+e)= ln(1+0.010)= 0.00995 Calculate the parameters for force P-20000 N. 20000 N 12.5 MPа A. 1600 mm2 ΔL E= 0.45 : 0.01125 40 12.5 MPa 0.01125 E- 1111. 11 MPa L,A, ( 40) ( 1600) Ac= Lo+AL 1582. 2 mm² %3D 40+0.45 20000 N OT= = 12. 64 MPa 1r000
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