Structural Steel Design (6th Edition)
6th Edition
ISBN: 9780134589657
Author: Jack C. McCormac, Stephen F. Csernak
Publisher: PEARSON
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Chapter 4, Problem 4.7PFS
To determine
The lightest section of
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Select sections for the conditions described, using Fy = 50 ksi and Fu = 65 ksi, unless otherwise noted, and neglecting block shear.Select the lightest MC12 section that will resist a total factored load of 372 kips and a total service load of 248 kips.The member is to be 20 ft long and is assumed to be welded on the end as well as to each flange for a distance of 6 in along the length of the channel. A36 steel is used.
Select sections for the conditions described, using Fy = 50 ksi and Fu = 65 ksi, unless otherwise noted, and neglecting block shear.
Select the lightest W12 section available to support working tensile loads of PD =120 k and PW = 288 k. The member is to be 20 ft long and is assumed to have two lines of holes for 3/4-in Ø bolts in each flange. There will be at least three holes in each line 3 in on center.
A W12 x 30 tension member with no holes is subjected to an axial load, P, which is 40 percent dead load and 60 percent live load and a uniform service wind load of 2.40 k/ft.The member is 14 ft long, laterally braced at its ends only and bending is about the x axis.Assume Cb = 1.0, Fy = 50 ksi and Fu = 65 ksi.What is the maximum value of P for this member to be satisfactory?
Chapter 4 Solutions
Structural Steel Design (6th Edition)
Ch. 4 - Prob. 4.1PFSCh. 4 - Prob. 4.2PFSCh. 4 - Prob. 4.3PFSCh. 4 - Prob. 4.4PFSCh. 4 - Prob. 4.5PFSCh. 4 - Prob. 4.6PFSCh. 4 - Prob. 4.7PFSCh. 4 - Prob. 4.8PFSCh. 4 - Prob. 4.9PFSCh. 4 - Prob. 4.10PFS
Ch. 4 - Prob. 4.11PFSCh. 4 - Prob. 4.12PFSCh. 4 - Prob. 4.13PFSCh. 4 - Prob. 4.14PFSCh. 4 - Prob. 4.15PFSCh. 4 - Prob. 4.16PFSCh. 4 - Prob. 4.17PFSCh. 4 - Prob. 4.18PFSCh. 4 - Prob. 4.19PFSCh. 4 - Prob. 4.20PFSCh. 4 - Prob. 4.21PFSCh. 4 - Prob. 4.22PFSCh. 4 - Prob. 4.23PFSCh. 4 - Prob. 4.24PFSCh. 4 - Prob. 4.25PFSCh. 4 - Prob. 4.26PFS
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- Select sections for the conditions described, using Fy = 50 ksi and Fu = 65 ksi, unless otherwise noted, and neglecting block shear. Select the lightest W10 section that will resist a service tensile load, PD = 175 k and PL = 210 k.The member is to be 25 ft long and is assumed to have two lines of holes in each flange and two lines of holes in the web.Assume there are four bolts in each line 3 in on center. All holes are for 7/8 -in Ø bolts. Use A992 – Grade 50 steel.arrow_forwardSelect sections for the conditions described, using Fy = 50 ksi and Fu = 65 ksi, unless otherwise noted, and neglecting block shear. Select the lightest WT7 available to support a factored tensile load Pu = 250 k, Pa = 160 k. Assume there are two lines of 7/8 -in Ø bolts in the flange (at least three bolts in each line 4 in on center). The member is to be 30 ft long.arrow_forwardWhat are the design moment capacities of the beam at sections A-A' and B-B', without the necessity of verifying the minimum steel area requirements and associated details (e.g., minimum bar spacing, cover, etc.).arrow_forward
- Determine the LRFD design strength and the ASD allowable strength for each of the compression members shown. Use the AISC Specification and a steel.arrow_forwardFor the beam loaded as shown, calculate the values of maximum shearing force and maximum bending moment. Also, determine the bending moment at E. (Please provide detailed solution with FBD please, thank you) Vmax = ? kip Mmax = ? kip-ft ME = ? kip-ftarrow_forwardInstructions: For the beam below, design the beam usingtrial sectionW21x147. If the section is insufficient,select other properties from the W21 table to a.) satisfy adequacy, local buckling, design capacity, b.) shear, and deflection. Use the LRFD method. Latera I bracings are found at supports. Also, use A36. Also, utilize the W21 section property table that is found below.arrow_forward
- Determine the compressive load carrying capacity (both LRFD and AD) for a W14x21 1 section using the following given table of AlSC manual. The column has overall length of 30 ft, fixed at one end and hinged at other end. Assume Fy= 50 ksiarrow_forwardQ/ Using both LRFD and ASD, select the most economical sections, with F, : unless otherwise specified, and assuming full lateral bracing for the com flanges. Working or service loads are given for each case, and beam weight included. P, =12k LD =10Kit - L=15k/ftarrow_forwardA built-up beam made from fastening two channels to a wide flange section carriesloads` as shown. The fasteners are 12mm∅ bolts with allowable shear stress τ= 110 MPa. Forbearing, σb= 220MPa for single shear, and σb= 275 MPa for double shear. E=200 GPA. Fy=420MPa. 1. Determine the horizontal shearing stress at 95 mm above the Neutral Axis.2. Determine the spacing of the bolts.arrow_forward
- For the floor system shown below, the un-factored live load is 100 psf and un-factored dead load is 150 psf including self-wt. Draw the free body diagram for B2 Draw the free body diagram for G1 Assuming Fy = 50 ksi, wu = 2.5 k/ft and the floor beam B2, is only braced at the supports, select the lightest W section that will safely carry the load in bending, shear and deflection. Use l/360. Assuming Fy = 50 ksi, l = 20 ft, pin-pin connection and PDES = 104 k, select the lightest, rectangular HSS section for column C1 that will safely carry the load in compression. Using the design load of wu = 2.5 k/ft for the floor beam B2, select the lightest W section for girder G1 that will safely carry the load in bending and shear. The girder is supported throughout its length.arrow_forwardPLEASE WRITE THE COMPLETE SOLUTION CLEARLY Design a square column for the given concentric axial loads. The column is appliedwith total dead load of 1000 kN and live load of 1400 kN. Use gross steel ratio ofapproximately 3%. fc’ = 28 Mpa , fy = 415 Mpa. Use 28mm bars for main bars and 10mm bars for ties.a. Compute for the required area of steel reinforcement,b. Recommend the number of 28 mm bars.c. Calculate the required spacing of ties.d. Draw/sketch the cross-sectional detail of the column.arrow_forwardCalculate the nominal flexural and shear strength of an HSS 8×4×1/8 section. ? y =50 ksi . Please use hand calculations to solvearrow_forward
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