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:
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
Problem 1.9.3P: A tie-down on the deck of a sailboat consists of a bent bar boiled at both ends, as shown in the...
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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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F94458cf2-014b-466a-bada-fa68de68f3b9%2F8ada16ef-e7b6-4b84-8c2b-a70432bac42d%2F1899058_processed.jpeg&w=3840&q=75)
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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F94458cf2-014b-466a-bada-fa68de68f3b9%2F8ada16ef-e7b6-4b84-8c2b-a70432bac42d%2Fwn50vp7_processed.jpeg&w=3840&q=75)
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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