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STEEL DESIGN

Using Fy=50 ksi (345 MPa) and Fu = 65 ksi (448 MPa), unless otherwise noted.
Select the lightest S section that will safely support the service tensile loads D = 92 kips and L = 56 kips. The member is to be 20 ft long and is assumed to have one line of holes for ¾-in diameter bolts in each flange. Assume that there are at least three holes in each line 4 inches on center. Use A36 steel.
Use LRFD and ASD design expressions and show your complete solution. Select an alternative section if none of the specified section is adequate.

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Main answer:The given problem requires the lightest S section that can support the tensile loads with safety. Therefore, for this problem, we need to use the LRFD (Load and Resistance Factor Design) and ASD (Allowable Strength Design) design expressions.As given in the problem, Fy = 50 ksi (345 MPa) and Fu = 65 ksi (448 MPa). The member is 20 ft long and has one line of holes in each flange. Also, it is assumed that there are at least three holes in each line 4 inches on center. The steel to be used is A36 steel.Let's first calculate the service load stress due to the given tensile loads. The member is subjected to a service load stress in the longitudinal direction due to the tensile loads D and L. Therefore, the service load stress on the section is given by;fs = D/A + L/AWhere,fs = service load stress on the sectionD = 92 kipsL = 56 kipsA = cross-sectional area of the sectionWe need to select the lightest S section that can safely support the given tensile loads. Therefore, we must compare the required area of the section with the actual area of different sections of S shape. We will consider different sections with varying dimensions and thicknesses and then calculate the required section properties to determine which section satisfies the required conditions with minimum weight. LRFD Design Method:Using LRFD design expressions, the required area of the section is given by;Ar = φ.(D + L)/FyWhere,φ = Load factorD = 92 kipsL = 56 kipsFy = 50 ksiAr = φ.(D + L)/FyAr = (1.2*92+1.6*56)/(50) = 3.28 sq. in.ASD Design Method:Using ASD design expressions, the required area of the section is given by;Ar = (D + L)/0.6FyWhere,D = 92 kipsL = 56 kipsFy = 50 ksiAr = (92+56)/(0.6*50) = 3.47 sq. in.The lightest S section that can safely support the given tensile loads is the one that has the minimum weight and has an area of at least 3.28 sq. in. The standard S section 3" × 5.7# has an area of 3.36 sq. in., which is more than the required area. Therefore, this section will safely support the given tensile loads.Conclusion:Using the LRFD and ASD design expressions, the required area of the section is calculated. It is then compared to the actual area of different S sections to determine the lightest section that satisfies the required conditions with minimum weight. The standard S section 3" × 5.7# with an area of 3.36 sq. in. satisfies the required conditions and can safely support the given tensile loads.

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