For the given riveted connection shown, having the allowable stresses: Fy= 248 MPa, F, = 150 MPa Tensile Stress = 0.60 Fy Bearing Stress = 1.35 Fy Rivet hole has a diameter 2 mm greater than the 1/2 300mm P rivet diameter. The main plate has width of 300mm and thickness of 20 mm. PI P/2 P12 t=20mm PM a. Compute the diameter of rivets such the shear capacity of the rivets is equal to its bearing capacity. b. Compute the max. value of P so as not to exceed the allowable shearing stress c. Compute the max. value of P so as not to exceed the allowable tensile stress of plate.
For the given riveted connection shown, having the allowable stresses: Fy= 248 MPa, F, = 150 MPa Tensile Stress = 0.60 Fy Bearing Stress = 1.35 Fy Rivet hole has a diameter 2 mm greater than the 1/2 300mm P rivet diameter. The main plate has width of 300mm and thickness of 20 mm. PI P/2 P12 t=20mm PM a. Compute the diameter of rivets such the shear capacity of the rivets is equal to its bearing capacity. b. Compute the max. value of P so as not to exceed the allowable shearing stress c. Compute the max. value of P so as not to exceed the allowable tensile stress of plate.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Subject: Steel design- Bolted Steel Connection
*Use NSCP 2015 formula/guide to solve this problem
*Use Handwritten
Transcribed Image Text:For the given riveted connection shown, having the allowable stresses:
Fy = 248 MPa, F, = 150 MPa
Tensile Stress = 0.60 Fy
Bearing Stress = 1.35 Fy
Rivet hole has a diameter 2 mm greater than the
1/2
300mm
P
P
rivet diameter.
The main plate has width of 300mm and thickness of 20 mm.
t/21
t/2
O
P/2
P12
P12
t=20mm
Pr
a. Compute the diameter of rivets such the shear capacity of the rivets is equal to its
bearing capacity.
b. Compute the max. value of P so as not to exceed the allowable shearing stress
c. Compute the max. value of P so as not to exceed the allowable tensile stress of plate.
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