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Properties and Dimension Ag=35.30 in^2 X = 6.24 in ry = 3.74 in d=14.5 in tf=0.94 in bf=14.7 in tw=0.59 in Y k1=1.5 I tf=0.94 1.54 -tw=0.59 bf=14.7

Properties and Dimension Ag=35.30 in^2 X = 6.24 in ry = 3.74 in d=14.5 in tf=0.94 in bf=14.7 in tw=0.59 in Y k1=1.5 I tf=0.94 1.54 -tw=0.59 bf=14.7

Mechanics of Materials (MindTap Course List)
Mechanics of Materials (MindTap Course List)
9th Edition
ISBN:9781337093347
Author:Barry J. Goodno, James M. Gere
Publisher:Barry J. Goodno, James M. Gere
Chapter8: Applications Of Plane Stress (pressure Vessels, Beams, And Combined Loadings)
Section: Chapter Questions
Problem 8.3.5P: An inflatable structure used by a traveling circus has the shape of a half-circular cylinder with...
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Situation 1: A-D A W14X120 is used as a tension member in atruss. The flanges of the member are connected to a gusset plate by ¾ inch boltas shown below. Use A36 steel with Fy=36 ksi and Fu=58 ksi Properties and Dimension Ag=35.30 in^2 rx = 6.24 in ry = 3.74 in d=14.5 in tf=0.94 in bf=14.7 in tw=0.59 in Y k1=1.5 k=1.54 I tf=0.94 d=14.5 tw=0.59 bf=14.7 A. Determine the Yielding Capacity of the section based on LRFD (kips) Round your answer to 2 decimal places. From Situation 1: b. Determine the effective net area of the section. For shear lag Factor U used the recommended values depending on the dimension of the section for W shapes with 3 or more fasteners per line. Note: use hole diameter if bf < d;U=0.85 else U =0.9 3 Round your answer to 3 decimal places. From Situation 1: c. Determine the Tensile Rupture capacity of the section based on LRFD From Situation 1: d. Determine the Demand to Governing Capacity Ratio (based on yielding and rupture only) if the Demand load carried by the section are DL=200 kips LL=400 kips use LRFD Round your answer to 3 decimal places.
Transcribed Image Text:Situation 1: A-D A W14X120 is used as a tension member in atruss. The flanges of the member are connected to a gusset plate by ¾ inch boltas shown below. Use A36 steel with Fy=36 ksi and Fu=58 ksi Properties and Dimension Ag=35.30 in^2 rx = 6.24 in ry = 3.74 in d=14.5 in tf=0.94 in bf=14.7 in tw=0.59 in Y k1=1.5 k=1.54 I tf=0.94 d=14.5 tw=0.59 bf=14.7 A. Determine the Yielding Capacity of the section based on LRFD (kips) Round your answer to 2 decimal places. From Situation 1: b. Determine the effective net area of the section. For shear lag Factor U used the recommended values depending on the dimension of the section for W shapes with 3 or more fasteners per line. Note: use hole diameter if bf < d;U=0.85 else U =0.9 3 Round your answer to 3 decimal places. From Situation 1: c. Determine the Tensile Rupture capacity of the section based on LRFD From Situation 1: d. Determine the Demand to Governing Capacity Ratio (based on yielding and rupture only) if the Demand load carried by the section are DL=200 kips LL=400 kips use LRFD Round your answer to 3 decimal places.
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