Whether a
Answer to Problem 11.1PFS
Satisfied
Explanation of Solution
Given:
Calculation:
The sectional properties are as follows:
Refer Table 3-2, “W shapes properties” in the AISC steel construction manual.
The ultimate load by LRFD is
The design moment is
Refer Table 1-1, “W shapes properties” in the AISC steel construction manual.
The available tensile strength is
The design strength is
Since,
The nominal moment strength is
Check:
The section is subjected to both the bending and axial tension:
Hence, it’s OK.
The ultimate load by ASD is
The design moment is
The design strength is
Since,
The nominal moment strength is
Check:
The section is subjected to both the bending and axial tension:
Hence, it’s OK.
Conclusion:
Therefore, the tension member
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Chapter 11 Solutions
Structural Steel Design (6th Edition)
- A plate girder must be designed for the conditions shown in Figure P10.7-4. The given loads are factored, and the uniformly distributed load includes a conservative estimate of the girder weight. Lateral support is provided at the ands and at the load points. Use LRFD for that following: a. Select the, flange and web dimensions so that intermediate stiffeners will he required. Use Fy=50 ksi and a total depth of 50 inches. Bearing stiffeners will be used at the ends and at the load points, but do not proportion them. b. Determine the locations of the intermediate stiffeners, but do not proportion them.arrow_forwardThe steel frame (E=200 GPa) shown has a diagonal brace BD with an area of 1920 mm2. Determine thelargest allowable load P if the change in length of member BD is not to exceed 2.5 mmarrow_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_forward
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- The given frame is unbraced, and bending is about the x axis of each member. The axial dead load supported by column AB is 204 kips, and the axial live load is 408 kips. Fy = 50 ksi. Determine Kx for member AB. Use the stiffness reduction factor if possible. a. Use LRFD. b. Use ASDarrow_forwardDetermine if the cantilevered W10x77 beam (A992 steel) has adequate shear strength for the following loading condition. The 90 kip LL is a service load and the uniform DL includes the self-weight. Use LRFD only. Solve manually.arrow_forwardUsing LRFD select the lightest W 14 section to carry a service load P of 100 kips dead load and 400 kips live load. The compression load acts with an eccentricity of 12-inches with respect to the strong axis. Use A992 steel (Fy = 50 ksi). Take the unbraced length to be L. This beam-column is part of a braced frame (BF) system. Please determine Cb and use it. Take Cm = 1.0. Since this can involve many iterations, start the problem by selecting a column meant to resist an ‘equivalent’ axial load of Pu plus 140% of the Mu. (use kips as the unit for Mu). I.E.: Pu,eq = Pu + 1.4*Mu (gives result in kips). Also, note that regardless of the results of this first attempt, the beam-column must meet the AISC criteria of H1-1a or H1-1b as usual. Note: This ‘equivalent’ method was first published in the AISC Engineering Journal by Uang, Watter, and Leet.arrow_forward
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- Steel Design (Activate Learning with these NEW ti...Civil EngineeringISBN:9781337094740Author:Segui, William T.Publisher:Cengage Learning