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:

Mechanics of Materials (MindTap Course List)
9th Edition
ISBN:9781337093347
Author:Barry J. Goodno, James M. Gere
Publisher:Barry J. Goodno, James M. Gere
Chapter1: Tension, Compression, And Shear
Section: Chapter Questions
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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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