Find the maximum distributed load can be applied to the two fixed ends beam shown below. Wu L=6m
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![Find the maximum distributed load can be applied to the two fixed ends
beam shown below.
Wu
L=6m](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb7a395db-4fcf-4ac0-9256-45fbabb87afa%2F8964d016-c687-454a-96bd-c4449c36be5c%2Fowznh6o_processed.jpeg&w=3840&q=75)
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- A beam is part of the framing system for the floor of an office building. The floor is subjected to both dead loads and live loads. The maximum moment caused by the service dead load is 45 ft-kips, and the maximum moment for the service live load is 63 ft-kips (these moments occur at the same location on the beam and can therefore be combined). a. If load and resistance factor design is used, determine the maximum factored bending moment (required moment strength). What is the controlling AISC load combination? b. What is the required nominal moment strength for a resistance factor of 0.90? c. If allowable strength design is used, determine the required moment strength. What is the controlling AISC lead combination? d. What is the required nominal moment strength for a safety factor of 1.67?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.Compute the nominal shear strength of an M107.5 of A572 Grad 65 steel.
- Determine the smallest value of yield stress Fy, for which a W-, M-, or S-shape from Part 1 of the Manual will become slender. To which shapes does this value apply? What conclusion can you draw from your answer?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.Verify the value of Zx for a W1850 that is tabulated in the dimensions and properties tables in Part 1 of the Manual.
- 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.A column in a building is subjected to the following load effects: 9 kips compression from dead load 5 kips compression from roof live load 6 kips compression from snow 7 kips compression from 3 inches of rain accumulated on the roof 8 kips compression from wind a. If lead and resistance factor design is used, determine the factored load (required strength) to be used in the design of the column. Which AISC load combination controls? b. What is the required design strength of the Column? c. What is the required nominal strength of the column for a resistance factor of 0.90? d. If allowable strength design is used, determine the required load capacity (required strength) to be used in the design of the column. Which AISC load combination controls? e. What is the required nominal strength of the column for a safety factor of 1.67?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.
- 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.Use an elastic analysis and compute the extra load in the weld (in kips per inch of length) caused by the eccentricity.Use an elastic analysis and determine the maximum load in the weld (in kips per inch of length).
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