14. The plate shown in the figure is fastened to the fixed member by seven 10-mm-diameter rivets. Compute the value of the loads P so that the average shearing stress in any rivet does not exceed 70 MPа. 60 mm 60 mm. P 120 mm 80 mm 40 mm 40 mm 80 mm 100 mm 120 mm P.
14. The plate shown in the figure is fastened to the fixed member by seven 10-mm-diameter rivets. Compute the value of the loads P so that the average shearing stress in any rivet does not exceed 70 MPа. 60 mm 60 mm. P 120 mm 80 mm 40 mm 40 mm 80 mm 100 mm 120 mm P.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Question
Please refer to the image, and a similar one but with different number of rivets for your reference in solving.
My question is with seven rivets, anf the reference has only five. Thank you!
Transcribed Image Text:14. The plate shown in the figure is fastened to the
fixed member by seven 10-mm-diameter rivets.
Compute the value of the loads P so that the
average shearing stress in any rivet does not
exceed 70 MPa.
60 mm 60 mm,
120 mm
80 mm
40 mm
40 mm
80 mm
100 mm
P.
120 mm
P.
![Problem 335
The plate shown in Fig. P-335 is fastened to the fixed member by five 10-mm-diameter
rivets. Compute the value of the loads P so that the average shearing stress in any rivet
does not exceed 70 MPa. (Hint: Use the results of Prob. 332.)
120 mm
80 mm
40 mm
40 mm
80 mm
100 mm
120 mm
Figure P-335
Solution 335
Solving for location of centroid of rivets:
A XG = Eax
A = (80 + 160)(80)
= 9600 mm?
a, = az = az = (80)(80) = 3200 mm?
Where
80
80
X-X,= 글 (80) 3 80/3 mm
80
X2 = (80) = 160/3 mm
80 XG 'e
9600XG = 3200(80/3) + 3200(160/3) + 3200(80/3)
XG = 320/9 mm
r1 = (320/9)2 + 80? = 87.54 mm
80
12 = J(80 – 320/9)2 + 402 = 59.79 mm
80
J= ZAP2 = 1(10²)(2r1² + 2r2² + XGA)
J= r(102)[2(87.54)² + 2(59.79)² + (320/9)²]
12
40
c.g
the
40
80
J=1 864 565.79 mm²
T= (120 + 100)P = 220P
The critical rivets are at distance r, from centroid:
Tp
220P(87.54)
70 =
1 864 565.79
P = 6777.14 N](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe6757be2-9f4c-4ee9-a5fe-42990c5a203e%2F9cc3fe1b-0556-4d5d-bbc0-835c85c39451%2Foqjf36i_processed.png&w=3840&q=75)
Transcribed Image Text:Problem 335
The plate shown in Fig. P-335 is fastened to the fixed member by five 10-mm-diameter
rivets. Compute the value of the loads P so that the average shearing stress in any rivet
does not exceed 70 MPa. (Hint: Use the results of Prob. 332.)
120 mm
80 mm
40 mm
40 mm
80 mm
100 mm
120 mm
Figure P-335
Solution 335
Solving for location of centroid of rivets:
A XG = Eax
A = (80 + 160)(80)
= 9600 mm?
a, = az = az = (80)(80) = 3200 mm?
Where
80
80
X-X,= 글 (80) 3 80/3 mm
80
X2 = (80) = 160/3 mm
80 XG 'e
9600XG = 3200(80/3) + 3200(160/3) + 3200(80/3)
XG = 320/9 mm
r1 = (320/9)2 + 80? = 87.54 mm
80
12 = J(80 – 320/9)2 + 402 = 59.79 mm
80
J= ZAP2 = 1(10²)(2r1² + 2r2² + XGA)
J= r(102)[2(87.54)² + 2(59.79)² + (320/9)²]
12
40
c.g
the
40
80
J=1 864 565.79 mm²
T= (120 + 100)P = 220P
The critical rivets are at distance r, from centroid:
Tp
220P(87.54)
70 =
1 864 565.79
P = 6777.14 N
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