At first we will calculate the influence areas for the beams. The we have to calculate the imposed udl. Then we will design the steel beam according to the plastic modulus.
1. The floor framing plan is shown below. Design the beams B1, B2, and Ba Check...
A partial plan of an office building is shown. All structural steel is A36 steel el AM Partial floor plan (office building) Loads: Concrete floor: 150 pef thidh 1" finish wood floor: 2.5 psf Suspended fire-resistant ceiling: 3.0 psf Live load: 70 psf A36 steel: modulus of elasticity E -29x10° ksi The deflection caused by the live load only should be less than L/360; The total deflection caused by the dead load and live load should be less than L/240...
6. The figure below shows the floor framing plan for an office building. The roof has the same framing plan as the floor. Select a W12 section for Column C1 in the first floor and assume the same column will be used for Column C2 in the second floor. Floor loads (applied on the 2 floor; dead load of 200 psf and live load of 300 psf Roof loads (applied on the roof: dead load of 100 psf and roof...
Could you please help me on question 7. Thank you very much. The floor system shown below consists of normalweight concrete (150 pcf). Beams C2 ! LT 1 7.0 inch ! h=20 in G2! 12 in Girders BI 1 7.0 inch Slab 8 ft span h 24 in ----- - -- - --- 15 in 15 ft 15 ft Use 4000 psi concrete and Gr. 60 reinforcing steel. Assume 0.75 inch clear cover. Include concrete self-weight and a superimposed dead...
is the second floor of a four-story building. a flat Consider the floor framing plan shown below. Thi The design floor live load is 90 psf. The design roof live load is 20 psf. The structure has roof G1 G4 /2 G2 GS 2o G3 FLOOR FRAMING PLAN Assume that the floor between Beams B7 and B8 is supported directly by these two beams only (not by Girders G2 and G3). Beam B7 has a self-weight of 30 plf. The...
1. For the given framing plan, first identify all unique beams and columns. After that, find out the final load transferred on each beam and girder. Finally draw load diagrams for all beams and girders, and determine the support reactions. 50 ft 50 ft 50 ft 25 ft I(2 B1 25 ft 3 1 25 ft 4 25 ft A B D Use: Live load 110 PSF ead load excluding weight of the slab Thickness of concrete slab = 5...
Steel Design 3. Given a. W27X84 A992 Steel beam is simply supported over a 25' FT beam length. b. The compression flange of the beam is fully braced along the beam length. c. A uniformly distributed service dead load of 2.5 KLF. d. A uniformly distributed service live load of 3.5 KLF. e. A live load deflection limit of L/360. A dead live load deflection limit of L/240. f. Neglect the self-weight of the beam in all calculations. Determine: The...
Design a W-shape floor beam for the intermediate beam marked A on the floor plan shown below. The beam is loaded uniformly by the floor with a live load of 70psf and a dead load in addition to beam self-weight of 60psf. The beam has a full lateral support provided by the floor deck. Use A992 steel (50ksi) and provisions of LRFD 壬 26 ft L-trib
. Design a beam for a 30-ft simple span to support the working uniform loads of wo 1.25 k/ft (includes beam self-weight) and wi-1.75 k/ft. The maximum permissible total load deflection under working loads is 1/360 of the span. Use 50 ksi steel and consider moment, shear and deflection. The beam is to be braced laterally at its ends and the mid-span only. (25-POINTS)
show work please!! 1. For the given framing plan, first identify all unique beams and columns. After that, find out the final load transferred on each beam and girder. Finally draw load diagrams for all beams and girders, and determine the support reactions (70 Points) 50 ft 50 ft 50 ft 25 ft 25 ft 0 25 ft 25 ft Use Live load 110 PSF . Dead load excluding weight of the slab 15 PSF .Thickness of concrete slab 6...
steel design 2-160 points] A992 steel (F, 50ksi) is used and LRFD method is to be followed. a) Through calculations determine if w14x 38 is compact or not. b) The beam shown below is laterally braced only at the ends. The applied load is a service live load. Through calculations determine if W14x38 is adequate. Note: Shear forces are of no concern. You are not allowed to use the beam design chart included at the end of the exam. 302...