A plate is fastened to a fixed member by four 20-mm-diameter rivets arranged as shown in Fig. P-333. Compute the maximum and minimum shearing stress developed

International Edition---engineering Mechanics: Statics, 4th Edition
4th Edition
ISBN:9781305501607
Author:Andrew Pytel And Jaan Kiusalaas
Publisher:Andrew Pytel And Jaan Kiusalaas
Chapter7: Dry Friction
Section: Chapter Questions
Problem 7.30P: Solve Prob. 7.29 if =0.
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A plate is fastened to a fixed member by four 20-mm-diameter rivets arranged as shown in Fig. P-333. Compute the maximum and minimum shearing stress developed.
radial distance p from the centroid of the rivet group.
2_53032313900897..
Solution 332
The shearing stress on each rivet is P/A
T = Tp/J
Where:
T = PRn
p= R
J = EAp? = AR'n
%3D
PRn(R)
T =
AR?n
T=
- ok!
A
This shows that 7 = Tp/J can be used to find the shearing stress at the center of any rivet.
Solution to Problem 333 | Flanged bolt couplings
A plate is fastened to a fixed member by four 20-mm-diameter rivets arranged as shown in Fig. P-333.
Compute the maximum and minimum shearing stress developed.
14 kN
mm
mm
14 kN
Figure P-333
Solution 333
Tp
T =
120
40 40.
120
14 kN
14 kN
Where:
T = 14(1000)(120) = 1 680 000N . mm
J = EAp? = {#(20)² [ 2(40²) + 2(120²) ]
J = 3 200 000 mm
m shearing stress (p = 120 mm):
1 680 000(120)
3 200 000
Tmax = 20.05 MPa answer
Minimum shes ri n stress (p = 40 mm):
1 680 000(40)
Tuin =
Transcribed Image Text:radial distance p from the centroid of the rivet group. 2_53032313900897.. Solution 332 The shearing stress on each rivet is P/A T = Tp/J Where: T = PRn p= R J = EAp? = AR'n %3D PRn(R) T = AR?n T= - ok! A This shows that 7 = Tp/J can be used to find the shearing stress at the center of any rivet. Solution to Problem 333 | Flanged bolt couplings A plate is fastened to a fixed member by four 20-mm-diameter rivets arranged as shown in Fig. P-333. Compute the maximum and minimum shearing stress developed. 14 kN mm mm 14 kN Figure P-333 Solution 333 Tp T = 120 40 40. 120 14 kN 14 kN Where: T = 14(1000)(120) = 1 680 000N . mm J = EAp? = {#(20)² [ 2(40²) + 2(120²) ] J = 3 200 000 mm m shearing stress (p = 120 mm): 1 680 000(120) 3 200 000 Tmax = 20.05 MPa answer Minimum shes ri n stress (p = 40 mm): 1 680 000(40) Tuin =
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