An A36 tension member shown in thefigure is connected with three 19 mmdiameter bolts. Dia. Of bolt hole = 22 mm.50, 10010099.45T-50.55100Properties of 150 mm x 100 mm x 12.5mm angleA = 3065 mm2y = 50.55 mmFy = 248 MPaFu = 400 MPaa. Determine the design strength due toyielding in the gross section. (kN)b. Determine the allowable strength due toyielding in the gross section. (kN)c. Determine the design strength due tofracture in the net section. Use U = 0.85.(kN)

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An A36 tension member shown in the figure is connected with three 19 mm diameter bolts. Dia. Of bolt hole = 22 mm. 50, 100 100 99.45 T- 50.55 100 Properties of 150 mm x 100 mm x 12.5 mm angle A = 3065 mm2 %D y = 50.55 mm Fy = 248 MPa Fu = 400 MPa %3D a. Determine the design strength due to yielding in the gross section. (kN) b. Determine the allowable strength due to yielding in the gross section. (kN) c. Determine the design strength due to fracture in the net section. Use U = 0.85. (kN)

An A36 tension member shown in the figure is connected with three 19 mm diameter bolts. Dia. Of bolt hole = 22 mm. 50, 100 100 99.45 T- 50.55 100 Properties of 150 mm x 100 mm x 12.5 mm angle A = 3065 mm2 %D y = 50.55 mm Fy = 248 MPa Fu = 400 MPa %3D a. Determine the design strength due to yielding in the gross section. (kN) b. Determine the allowable strength due to yielding in the gross section. (kN) c. Determine the design strength due to fracture in the net section. Use U = 0.85. (kN)

Steel Design (Activate Learning with these NEW titles from Engineering!)

6th Edition

ISBN:9781337094740

Author:Segui, William T.

Publisher:Segui, William T.

Chapter4: Compression Members

Section: Chapter Questions

Problem 4.9.9P

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