... A36 (Fy=36ksi; Fu=58ksi) steel is used for the tension member shown. a. Determine the design strength based on the gross area. b. Determine the design strength based on the net area. - PL % ×12 2" 5/8" 2" 4" 12" 4" 2" 4-in.-diameter bolts Cross Sectional area of PL3/8x6 a) Blank 1 b) Blank 2 Blank 1 Add your answer Blank 2 Add your answer

... A36 (Fy=36ksi; Fu=58ksi) steel is used for the tension member shown. a. Determine the design strength based on the gross area. b. Determine the design strength based on the net area. - PL % ×12 2" 5/8" 2" 4" 12" 4" 2" 4-in.-diameter bolts Cross Sectional area of PL3/8x6 a) Blank 1 b) Blank 2 Blank 1 Add your answer Blank 2 Add your answer

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

6th Edition

ISBN:9781337094740

Author:Segui, William T.

Publisher:Segui, William T.

Chapter3: Tension Members

Section: Chapter Questions

Problem 3.4.3P

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A Channel C6x13 is used as a tension member. The holes are for 5/8 inch in diameter of bolts. Using A36 steel with Fy=36 ksi and Fu=58 ksi. The gross area of C6x13 is 3.81 in? and Assume U =0.85 Using LRFD what is the design strength based on Tensile Rupture/Fracture 4 @ 2" a 1 1/2 Ob 3" Oc 1 1/2 " I d e C6 x 13 Round your answer to 3 decimal places.
A double-angle shape is shown in the figure. The steel is A36, and the holes are for 2-inch-diameter bolts. Assume that A₂ = 0.75An. Ae 1 2 Determine the design tensile strength for LRFD. Determine the allowable strength for ASD. CIVIL ENGINEERING - STEEL DESIGN AL Section 2L5 x 3 x 516 LLBB
... A36 (Fy=36ksi; Fu=58ksi) steel is used for the tension member shown. a. Determine the design strength based on the gross area. b. Determine the design strength based on the net area. - PL % x12 2" 5/8" 2" 4" 12" 4" 2" 4-in.-diameter bolts Cross Sectional area of PL3/8x6 a) Blank 1 b) Blank 2 Blank 1 Add your answer Blank 2 Add your answer
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