Structural Steel Design (6th Edition)
6th Edition
ISBN: 9780134589657
Author: Jack C. McCormac, Stephen F. Csernak
Publisher: PEARSON
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Question
Chapter 5, Problem 5.6PFS
To determine
The design strength using LRFD and ASD for given column section
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A.) If the F = { - 200i + 400j } N , determine the force in each member and state if its in compressions or tension.
A.) Using the same given, if the members have a square cross-section with a diagonal of 11.31cm and thickness of 1cm, determine the axial stressess and the factors of safety on each of the member, if the maximum allowable tensile strength is 60kPa and the maximum allowable compressive strength is 70kPa
NOTE: Complete solution and units, free body diagrams, table data if possible
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A column is to support an axial dead load of 900 kN and an axial live load of 1240 kN. Use fy = 414 MPa and f’c = 27.6 MPa. Assume 2% steel ratio and use 28 mm main bars and 10 mm ties/ spiral. Use NSCP 2015- and a 30-mm steel cover
Determine the design compressive strength, P in kips, that the column can carry using LRFD. The column is W10X88, A992 Steel with fy=50 ksi. The column has lateral bracing in the weak axis. Use theoretical values for k.
Chapter 5 Solutions
Structural Steel Design (6th Edition)
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- A short rectangular column 300 mm on one side and 400 mm on the other side. It is reinforced with 8-20-mm-diameter (28) longhitudinal bars equally distributed to the shorte sides of the column. Use f'c = 21 MPa and fy = 415 MPa. Calculate the required spacing of 10-mm-diameter ties, s (mm). Calculate the nominal axial strength of the column, Pn (kN). Calculate the maximum ultimate axial load the column can carry, Pu (kN)arrow_forwardDetermine the required plastic moment capacity of the beam loaded as shown below. The service loads are 42.6 kN/m and 16.54 kN/m, respectively for dead and live loads. Use A-50 steel and LRFD. L=11 m. Use Equilibrium Method and FBD’s.arrow_forwardA built-up section consisting of W 350 x 90 with two 12 – mm plates welded to form a box section as shown in figure. The section is used as a column 10 meters long. The column is fixed at both ends, and braced at mid height about the weak axis (Y-axis). Use Fy = 248 MPa. Determine the axial load capacity of the column in kN.arrow_forward
- A rectangular beam has dimensions of 250 mm width and an effective depth of 430 mm. It is subjected to shear dead load of 102 kN and shear live load of 114 kN. Use f'c = 20.7 MPa and fyt = 276 MPa for 12 mm diameter U-stirrup. Design the required spacing of the shear reinforcement.arrow_forward4.4-2 A W21 × 101 is used as a compression member with one end fixed and the other endfree. The length is 10 feet. What is the nominal compressive strength if Fy = 50 ksi?Note that this is a slender-element compression member, and the equations of AISCSection E7 must be used. 4.6-2 A 15-foot long column is pinned at the bottom and fixed against rotation but free totranslate at the top. It must support a service dead load of 100 kips and a service liveload of 100 kips.a. Select a W12 of A992 steel. Use the column load tables.1. Use LRFD.2. Use ASD.b. Select a W16 of A992 steel. Use the trial-and-error approach covered in Section 4.6.1. Use LRFD.2. Use ASD.arrow_forwardA rectangular beam has a width of 300mm and an effective depth of 510mm. The beam is provided with 10mm vertical U-stirrups with yield strength, fy=275MPa. Concrete strength fc’=27MPa. Design the stirrups for the following factored shear a) 160kN b) 250kNarrow_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_forwardA 300 mm x 500 mm simply supported rectangular beam has a span of 4.50 M. Concrete cover is 50 mm. It carries a uniformly distributed Dead Load of 15 Kn/M (incl. its own weight) and a Live Load of 10 Kn/M. fc’ = 24.15 MPa; fy = 275 MPa. a. Find the maximum factored moment of the beam. b. Determine the required tension steel area. c. Find the no. of pcs if 25mmØ is to be used.arrow_forwardA rectangular beam has b = 250 mm, d = 330 mm, fy = 414 MPa, f’c = 20.7 MPa. Determine the required steel area if the beam is required to carry a dead load moment of 50 kN-m and a live load moment of 30 kN-m. Use the 2015 NSCP.arrow_forward
- 4.7-1 AW18× 97 with Fy = 60 ksi is used as a compression member. The length is 13 feet. Compute the nominal strength for Kx = 2.2 and Ky = 1.0.arrow_forwardA rectangular beam has a width of 300mm and an effective depth of 537.50 mm to the centroid of tension steel bars. Tension reinforcement consists of 6 – 28mm dia in two rows, compression reinforcement consists of 2 – 22mm dia, fc’ = 27.6MPa, fy = 414.7MPa, d’ = 60mm, L = 4m, the Unit weight of concrete is 24.56kN/cu.m The ultimate moment capacity of the beam is _____kN.m.arrow_forwardA simply supported rectangular beam of length L=5.4m has a width of b=395mm and an effective depth ofd=311m. Compression steel bars are placed at a distance d′=155mm from the extreme concrete compressivefiber. The beam supports service dead loads of Wd=27.5KN/m and Wl=35.5KN/m, respectively. Use fc′ = 28MPaand fy= 420 MPa. For the reinforcements, use 4 25 mm tension steel bars and 4 12 mmcompression steel bars.arrow_forward
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