For the beam show below, draw A.F.D, S.F.D, B.M.D A 2 N M 10 kN.m B 2 M C
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
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Step by step
Solved in 2 steps with 2 images
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- 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.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.Note For Problems 9.6-1 through 9.6-5, use the lower-bound moment of inertia for deflection of the composite section. Compute this as illustrated in Example 9.7. 9.6-3 For the beam of Problem 9.3-1, a. Compute the deflections that occur before and after the concrete has cured. b. If the live-load deflection exceeds L360 , select another steel shape using either LRFD or ASD.
- A W1422 acts compositely with a 4-inch-thick floor slab whose effective width b is 90 inches. The beams are spaced at 7 feet 6 inches, and the span length is 30 feet. The superimposed loads are as follows: construction load = 20 psf, partition load = 10 psf, weight of ceiling and light fixtures = 5 psf, and live load = 60 psf, A992 steel is used, and fc=4 ksi. Determine whether the flexural strength is adequate. a. Use LRFD. b. Use ASD.Same as Problem 5.10-2, except that lateral support is provided only at the ends and at the concentrated load.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.
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