Two steel plate tension members have been connected using 0.72” diameter bolts arranged in an equally-spaced four by four square formation.Total plate self-weight is specified as 912 pounds.Design of the elements adhered to the set of values that are twice as much as the minimum requirements. Both plates have a thickness equal to 1/3 in. Take shear fracture stress as 54ksi and the min required edge distance as 0.125ft. Find the maximum allowable service dead load(excluding the self-weight)and live load assuming live load is half as much as dead load including the self-weight of the plates
Two steel plate tension members have been connected using 0.72” diameter bolts arranged in an equally-spaced four by four square formation.Total plate self-weight is specified as 912 pounds.Design of the elements adhered to the set of values that are twice as much as the minimum requirements. Both plates have a thickness equal to 1/3 in. Take shear fracture stress as 54ksi and the min required edge distance as 0.125ft. Find the maximum allowable service dead load(excluding the self-weight)and live load assuming live load is half as much as dead load including the self-weight of the plates
Steel Design (Activate Learning with these NEW titles from Engineering!)
6th Edition
ISBN:9781337094740
Author:Segui, William T.
Publisher:Segui, William T.
Chapter5: Beams
Section: Chapter Questions
Problem 5.6.3P
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Two steel plate tension members have been connected using 0.72” diameter bolts arranged in an equally-spaced four by four square formation.Total plate self-weight is specified as 912 pounds.Design of the elements adhered to the set of values that are twice as much as the minimum requirements. Both plates have a thickness equal to 1/3 in. Take shear fracture stress as 54ksi and the min required edge distance as 0.125ft.
Find the maximum allowable service dead load(excluding the self-weight)and live load assuming live load is half as much as dead load including the self-weight of the plates.
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