Problem #2 A channel shown is attached to a 12 mm gusset plate with 9 - 22 mm diameter A 325 botls as shown Fax = 300 MP
Posted: Sun May 08, 2022 8:57 pm
- Problem 2 A Channel Shown Is Attached To A 12 Mm Gusset Plate With 9 22 Mm Diameter A 325 Botls As Shown Fax 300 Mp 1 (30.43 KiB) Viewed 42 times
- Problem 2 A Channel Shown Is Attached To A 12 Mm Gusset Plate With 9 22 Mm Diameter A 325 Botls As Shown Fax 300 Mp 2 (30.74 KiB) Viewed 42 times
- Problem 2 A Channel Shown Is Attached To A 12 Mm Gusset Plate With 9 22 Mm Diameter A 325 Botls As Shown Fax 300 Mp 3 (18.24 KiB) Viewed 42 times
Problem #2 A channel shown is attached to a 12 mm gusset plate with 9 - 22 mm diameter A 325 botls as shown Fax = 300 MPa. (nominal shear stress) F, = 400 MPa (uitmate tensile stress of channel) F, = 248 MPa Ag = 3354 mm? gusset plate 12 mm thick 64 64 30 Channel 12 mat trick 75 to o OO Gusset plak 22 from bolts Determine the capacity of the channel based on the gross area and effective net area. U = 0.60. Determine the capacity of the channel based on the shearing capacity of the bolts. Determine the capacity of the channel based on the bearing strength of the connection.
Capacity of the channel based on the shear strength of the bolts: P.FA A (22)?(9) (there are 9 bolts) A = 34212 mm Solution: Capacity of channel based on gross area and effective net area: Gross area: P. F, A P. = 248(3554) P. = 881392 N 12 mm P. = 881,39 KN 251 Effective net area: Dia of hole = 22 +3 257 Dia of hole = 25 mm A A, Ane A = 3554 - 2(25)(12) A = 2954 P. = 330(34212) P = 1128996 N P. = 1129 KN P=0.75(1129) P = 846.75 KN AUA A. = 0.60(2954) A. = 1772.4 mm? P. EF, A. P = 400(1772.4) P. = 708960 N P = 708.96 kN Use P. = 708.96 KN Capacity of channel: P = 0.75(708.96) P. = 531.72 KN
(4) Determine the allowable strength due to fracture. Use ASD Method. (5) Determine the allowable strength due to yielding. Use ASD Method. (6) Determine the capacity of the channel based on the shearing capacity of the bolts