Problem 2:Given: A 25"x25" column is subject to afactored axial load of Pu=1,200 kips, andfactored design moment of Mu-354 kips-ft.f'c 4,000psi, fy = 60,000psi.Determine the required steel ratio (p) andties. Sketch the design.2.0P=0.08INTERACTION DIAGRAM R4-60.9fc-4 ksi1.81,- 60 ksi0.077=0.91.6160.06Kmax0.051.4f/f, = 00.0490K₁ = P₁/f'Ag1.20.030.250.021.00.010.80.60.4€= 0.00350.2€,= 0.00500.0hyh0.500.751.0.0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50R₁ = P₁e/f'Ah

Step 1: Step 2: Step 3: Step 4:

Answered: Problem 2: Given: A 25"x25" column is…

Steel Design (Activate Learning with these NEW titles from Engineering!)

6th Edition

ISBN:9781337094740

Author:Segui, William T.

Publisher:Segui, William T.

Chapter4: Compression Members

Section: Chapter Questions

Problem 4.3.3P

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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.
If the beam in Problem 5.5-9 i5 braced at A, B, and C, compute for the unbr Cb aced length AC (same as Cb for unbraced length CB). Do not include the beam weight in the loading. a. Use the unfactored service loads. b. Use factored loads.
The given beam is laterally supported at the ends and at the 1 3 points (points 1, 2, 3, and 4). The concentrated load is a service live load. Use Fy=50 ksi and select a W-shape. Do not check deflections. a. Use LRFD. b. Use ASD.
Use the composite beam tables and select a W-shape and stud anchors for the following conditions: Span length = 18 6 Beam spacing = 9 ft Total slab thickness = 51 2 in. (the slab and deck combination weighs 57 psf). Lightweight concrete with a unit weight of 115 pcf is used Construction load = 20 psf Partition load = 20 psf Live load = 225 psf Fy=50 ksi and fc=4 ksi A cross section of the formed steel deck is shown in Figure P9.8-9. The maximum live-load deflection cannot exceed L/360 (use a lower-bound moment of inertia). a. Use LRFD. b. User ASD.
A beam must be designed to the following specifications: Span length = 35 ft Beam spacing = 10 ft 2-in. deck with 3 in. of lightweight concrete fill (wc=115 pcf) for a total depth of t=5 in. Total weight of deck and slab = 51 psf Construction load = 20 psf Partition load = 20 psf Miscellaneous dead load = 10 psf Live load = 80 psf Fy=50 ksi, fc=4 ksi Assume continuous lateral support and use LRFD. a. Design a noncomposite beam. Compute the total deflection (there is no limit to be checked). b. Design a composite beam and specify the size and number of stud anchors required. Assume one stud at each beam location. Compute the maximum total deflection as follows: 1. Use the transformed section. 2. Use the lower-bound moment of inertia.

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