The lap joint of a tension member is shown in the figure. The plate is 252 mm wide and 12 mm thick. The A325 bolts are 20 mm in diameter and the holes are 3 mm larger than the bolt diameter. Steel is A36. It is required to determine the ultimate capacity of the joint based on gross area, net area, and block shear. Use LRFD specifications and assume that live load to dead load ratio is equal to 3.0. A325 Bolts Properties: Fnt = 620 MPa Fnv = 469 MPa A36 Properties: Fy = 248 MPa Fu = 400 MPa 1.Determine the total service load (DL + LL).
The lap joint of a tension member is shown in the figure. The plate is 252 mm wide and 12 mm thick. The A325 bolts are 20 mm in diameter and the holes are 3 mm larger than the bolt diameter. Steel is A36. It is required to determine the ultimate capacity of the joint based on gross area, net area, and block shear. Use LRFD specifications and assume that live load to dead load ratio is equal to 3.0. A325 Bolts Properties: Fnt = 620 MPa Fnv = 469 MPa A36 Properties: Fy = 248 MPa Fu = 400 MPa 1.Determine the total service load (DL + LL).
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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The lap joint of a tension member is shown in the figure. The plate is 252 mm wide and 12 mm thick. The A325 bolts are 20 mm in diameter and the holes are 3 mm larger than the bolt diameter. Steel is A36. It is required to determine the ultimate capacity of the joint based on gross area, net area, and block shear. Use LRFD specifications and assume that live load to dead load ratio is equal to 3.0.
A325 Bolts Properties:
Fnt = 620 MPa
Fnv = 469 MPa
A36 Properties:
Fy = 248 MPa
Fu = 400 MPa
1.Determine the total service load (DL + LL).
Transcribed Image Text:P
38 mm
63 mm
*
63 mm
63 mm
↓
63 mm
63 mm 63 mm
O
O
O O O
O O O
t=12mm
38 mm
252 mm
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