Structural Steel Design (6th Edition)
6th Edition
ISBN: 9780134589657
Author: Jack C. McCormac, Stephen F. Csernak
Publisher: PEARSON
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Chapter 7, Problem 7.7PFS
To determine
Design the right handed and left handed column using both the LRFD and the ASD method
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Use the Effective Length Method, assume elastic behavior, and use both the LRFD and ASD methods. The columns are assumed to have no bending moments
Design W12 columns for the bent shown in the accompanying figure, with 50 ksi steel.The columns are braced top and bottom against sidesway out of the plane of the frame so that Ky = 1.0 in that direction. Sidesway is possible in the plane of the frame, the x-x axis. Design the right-hand column as a leaning column, Kx = Ky = 1.0 and the left-hand column as a moment frame column, determined from the alignment chart. PD = 350 k and PL = 240 k for each column.The beam has a moment connection to the left column, and has a simple or pinned connection to the right column.
Determine the critical buckling load for each of the columns, using the Euler equation. E = 29,000 ksi. Proportional limit = 36,000 psi. Assume simple ends and maximum permissible L/r = 200.
A W12 x 50, L = 22 ft 0 in.
BAL=8 mt = 13 mm-150 mm113 mmFor the steel column with both ends fixed against rotation having a tubularcross-section, determine the allowable load Pall against buckling. E=200 GPa andOy=250 MPa.
Chapter 7 Solutions
Structural Steel Design (6th Edition)
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- Determine the critical buckling load for each of the columns, using the Euler equation. E = 29,000 ksi. Proportional limit = 36,000 psi. Assume simple ends and maximum permissible L/r = 200. The pipe section shown:a. L = 22 ft 0 inb. L = 18 ft 0 inc. L = 12 ft 0 inarrow_forwardYou are to design a column for PD = 750 k and PL = 1000 k, using Fy = 50 ksi and KL = 14 ft. A W12 x 120(for which fcPn = 1290 k and Pn /Æc = 856 k from AISC Manual, Table 4-1) is on hand. Design cover plates to be snug-tight bolted at 6-in spacings to the W section, as shown in Fig. 6.7, to enable the column to support the required load.arrow_forwardUse the Effective Length Method, assume elastic behavior, and use both the LRFD and ASD methods. The columns are assumed to have no bending moments. The frame shown in the accompanying figure is unbraced against sidesway about the x-x axis. Determine Kx for column AB. Support conditions in the direction perpendicular to the frame are such that Ky =1.0. Determine if the W14 x 109 column for member AB is capable of resisting a dead load of 250 k and a live load of 500 k.A992 steel is used.arrow_forward
- Design an axially loaded short spiral column if it is subjected to axial dead load of 430 KN and axial live load of 980 KN. Use f’c = 27.6 MPa, fy = 414 MPa, ρ = 0.025 and 25 mm diameter main bars. Also, use 12 mm dia. ties with fyt = 276 MPa and clear concrete cover of 40 mm. Provide section drawing.arrow_forwardQ. The (10+0.5R)-ft-long steel column is an W8 x 40 section that is fixed at both ends. The midpoint of the column is braced by two cables that prevent displacement in the x-direction. Determine the critical value of the axial load P. Use E = 29 x 106 psi for steel. Where R=37 Kindly answer this question as soon as. Figures are attachedarrow_forwardA T-beam with bf=700mm, hf=100mm, bw=200mm, h=400mm, cc=40mm, stirrups=12mm, fc'=21 MPa, fy=415 MPa is reinforced by 4-32mm diameter bars for tension only. Calculate the depth of the neutral axisarrow_forward
- A 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_forwardDetermine 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.arrow_forwardDesign a square tied column to support an axial dead load of 450kN and an axial live load of 756kN. Assume 2.5% longitudinal steel is desired. Use fc' = 27.6MPa, fy = 276MPa (ties), fy = 414MPa (main bars - 20mm dia.)arrow_forward
- Design a round spiral column to support an axial dead load of 450kN and an axial live load of 756kN. Assume 2.5% longitudinal steel is desired. Use fc' = 27.6MPa, fy = 276MPa (ties), fy = 414MPa (main bars - 20mm dia.)arrow_forwardAn axially loaded square tied column is to be designed for the following service loads.DL = 2000 kNLL = 900 kNfc’ = 27 MPafy = 345 MPaconcrete cover = 80 mmUse reduction factor φ = 0.65 a. Compute the required size using 2.5% steel ratiob. Compute the maximum moment for axially loaded columnc. Compute the largest size of the column to support the given loadingarrow_forwardA.) 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 <up voting!>arrow_forward
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