The floor structural system shown in Fig 1 comprise W-section steel beams and girders supporting a 5-inch thick concrete slab which provides full restraint to the flanges of the beams. The density of concrete is 150 lbs./ft³. and the uniformly distributed live load L- 50 lbs/ft². L-22¹ 12' A. Beam B2 Gl GI [BI] FIG. 1-Steel Floor Beam 12' + (a) Strength Design i. Flexure (Bending) ii. Shear iii. Serviceability Design - Deflection --I 1) Produce a sketch of the floor plan showing the tributary load distribution areas for the beams and girders [B2] 2) Using AISC LRFD methodology design the following structural members using W-section A992 steel (Fy = 50 ksi). Assume the beams are cambered to offset dead load deflection. Use Design Tables in your solution.

The floor structural system shown in Fig 1 comprise W-section steel beams and girders supporting a 5-inch thick concrete slab which provides full restraint to the flanges of the beams. The density of concrete is 150 lbs./ft³. and the uniformly distributed live load L- 50 lbs/ft². L-22¹ 12' A. Beam B2 Gl GI [BI] FIG. 1-Steel Floor Beam 12' + (a) Strength Design i. Flexure (Bending) ii. Shear iii. Serviceability Design - Deflection --I 1) Produce a sketch of the floor plan showing the tributary load distribution areas for the beams and girders [B2] 2) Using AISC LRFD methodology design the following structural members using W-section A992 steel (Fy = 50 ksi). Assume the beams are cambered to offset dead load deflection. Use Design Tables in your solution.

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

See similar textbooks

Similar questions

Answer the following for the section at Point D Only Calculate the distributed load "w" that: Will cause the section crack Will cause the reinforcement to yield. Material Properties: F'c = 5000 psi Fy = 60000 psi Es = 29000000 psi Ln = 27 ft L wl₂² 16 wl,² 14 CD L wl,2 vl₁² 10 11 win² 16 h: 28 in A=4 in² b=14 in n d: 25 in
Answer untel step 4 Please I need clear Answer with full details. DO NOT COPPY THE ANSWER IF YOU DO NOT WANT TO ANSWER MY QUESTION, DO NOT REJECTED IT. SOMEONE ELSE WILL SOLVE IT If you are copy from other websites solution I will give you down vote
Design the reinforcement of a rectangular beam to resist a dead load moment of 200kN.m (including its own weight) and a live load moment of 300kN.m. The beam is limited in size to 350mm by 600mm overall depth. Steel covering from centroid of bars to outermost fiber is 100mm for both tension and compression reinforcement. Use fc'=27.5Mpa, fy = 414MPa,20mm barsA. As'= 7 of 20mm and As=14 of 20mmB. As' = 9 of 20mm and As=18 of 20mmC. As'= 5 of 20mm and As=10 of 20mmD. As' = 3 of 20mm and As=6 of 20mm Please provide neat and clear solutions
Please answer with a complete solution. Thank you! The roof truss is supported by a hinge at A and a horizontal roller support at B. The wind loads are perpendicular to BC. Determine the reactions at A and B.
For a composite section attached, derive the equation for Y-bar assuming that the PNA is located in the flange of the steel beam. Hint: Balance the resultant forces through the section. Recall that Y-bar is the distance from the top of the steel flange to the PNA for any location with the PNA below the top of the steel flange.
A laminated beam is composed of four (4) planks, each of 150 mm x 30 mm, glued together to form a section 150 mm x 150 mm. The allowable shear stress in the glue is 850 kPa, the allowable shear stress in the wood is 1050 KPA, and the allowable flexure stress in the wood is 10 MPa. Determine the maximum uniformly distributed load which can be carried by the beam on a 45 m simple span.
A rectangular tied braced column is reinforced with 6-28mm bars as shown in the figure below. Check the adequacy of the column for the following gravity loads: - - - - - - Service dead load = 1050kN Service live load = 850 kN Service dead load moment = 39 kN-m Service live load moment = 27 kN-m top=bott. = 1.0 40 lu = 5.0m f = 27.6 N/mm², fy = 414 N/mm² axis of bending 500mm Ties 10mm@300c/c } 350mm At eccentricity of 40mm P M +40 + 500mm MU +40- P Top end Bott. end
A reinforced concrete cantilever beam is shown below. The beam cross section is a rectangle, 12 inches deep by 18 inches wide. Material properties are: - Concrete weight: wt = 150 pcf - Concrete compressive strength: f'c = 4000 psi - Concrete tensile strength: f't = 400 psi - Steel tensile strength: fy = 60 ksi - Stirrup consists of #4 bar - Beam is not exposed to weather or in contact with soil. Determine if the beam will crack under its own weight, if its self-weight is 250lb/ft. 10 ft -18" BEAM SECTION beam will crack beam will not crack unknown, not enough information to calculate
The answer is between 18.25 and 20.25
t 4#9 1) fé= 5000 psi r Grade 60 steel; 4 #9 Rebars, 5-9" t 4#9 NWC, 5-9" L = 25 ft & Mn = ? 4#9 14" $ 21" #3" Simple span 39
For the following plan it is required to: 1- Design all beams for flexure and shear, where they have additional 30 kN/m distributed factored load 2- Design the columns. The concrete has a strength of 30 MPa, and the yield strength of the steel is 420 MPa. Show full calculations and neat sketches. All dimensions are in m. I 03 B2 0,3 7 B1 Wu=20 kN/m2 B1
The floor structural system shown in Fig 1 comprise W-section steel beams and girders supporting a 5-inch thick concrete slab which provides full restraint to the flanges of the beams. The density of concrete is 150 lbs. /ft³. and the uniformly distributed live load L=50 lbs/ft². L-22¹ B B. Girder G1: - [82] 12' Gl GI [BI] FIG. 1-Steel Floor Beam a) Strength - Bending and Shear 12' + I 1) Produce a sketch of the floor plan showing the tributary load distribution areas for the beams and girders [zul