Structural Steel Design (6th Edition)
6th Edition
ISBN: 9780134589657
Author: Jack C. McCormac, Stephen F. Csernak
Publisher: PEARSON
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Chapter 2, Problem 2.10PFS
To determine
To find maximum combined load by recommended AISC for given loading
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1.Explain the LRFD (Load and Resistance Factor Design) method.
What is the LRFD time effect factor, A, for a member designed for the 1.2D+1.6L load
combination where the live load comes from occupancy?
1.0
0.8
1.25
0.7
D E please
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- For the Casper hospital, the weight of each floor is 2500 kips while the roof weighs 2700 kips. The braced frames shown in the Typical Braced Frames Showing Story Height document constitute the dual steel concentrically braced frame for the hospital and allow the Response Modification Coefficient (R) to be 7. The hospital site is classified as Site ClassD for seismic design. T† can be found in ASCE 7-16 Figure 22-14. Calculate the Seismic Base Shear (V) and the lateral equivalent static seismic force (Fr) applied at each story and the roof. Use the ASCE 7-16 document to determine Sps, Ta, Ct, x and I, when doing your calculations. Note that the values in ASCE 7-16 Figure 22-1 need to be divided by 100 to convert from percent to decimal. Alternatively, you can use the Applied Technology Council's Hazards by Location website (https://hazards.atcouncil.org/) but you must still show the calculations by hand. You may use a spread sheet to calculate the force on each floor as the…arrow_forwardINCLUDE FIGURE/DIAGRAM A multistory commercial building is to be designed as a flat plate system with floors of uniform thickness having no beams or drop panels. Columns are laid out on a uniform 20 ft spacing in each direction and have a 16 in. square section and a vertical dimension 10 ft from floor to floor. Specified service live load is 100 psf including partition allowance. Using the direct design method, design a typical interior panel, determining the required floor thickness, size and spacing of reinforcing bars, and bar details including cutoff points. To simplify construction, the reinforcement in each direction will be the same; use an average effective depth in the calculations. Use all straight bars. For moderate spans such as this, it has been determined that supplementary shear reinforcement would not be economical, although column capitals may be used if needed. Thus, slab thickness may be based on Eqs. (13.11 a ), (13.11 b ), and (13.11 c ); or column capital…arrow_forwardDescribe about the local shear failure.arrow_forward
- 10:41 M (55) اس اس a = * 0 Activity no 1 Submit thi... classroom.google.com Pre-stressed Concrete Design CE 515 (MW 10:30 AM - 12:00 PM) Activity no. 1 1. Design a section of a concrete beam reinforced for tension only. The beam is simply supported on a span of 6 m and carries a load of 18 kn/m. What stirrup will be required? Use fc' = 17.2 MPa, fs = 124 MPa, n = 12. Assume b = d/2.arrow_forwardQ4. With the aid of sketches, critically discuss the installation of an independence type of scaffold system in building construction. The discussion should highlight the function of each component members of the system and necessary safety precaution that need to be taken during the installation.arrow_forward(a) The detached servant room of a quarter is having internal size of 3.0mxx4.50m. It is to be covered by a roof of RCC. The dead load on roof 4.5kN//m^(2) and the live load is 1.50kN//m^(2) Design the slab by I.S. Code method. Assume grade of concrete as M25 and grade of Steel as Fe415.arrow_forward
- Activity 2 Considering the given cantilever beam, columns and their cross-sections in figure 2, then justify (p2) the selection of standard rolled steel sections for hot rolled beams (EN 10060:2003 (E)) with the round hollow and solid cross-sections as well as columns by using the data of table 1 & 2 to ensure the stability of the structure has been guaranteed for these beams and columns. Dout 50KN |- 3m Figure 2: Cantilever beam and column using for table 2 Table 1: Steel properties applied for the beam Density (Kg/M3) | Young's Modulus (MPa) Yield Stress (MPa) Ultimate Tensile Strength (UTS) (MPa) 7800 200×10³ 275 420 Table 2: Dimensions of the beams and columns. Round hollow Round Round hollow Round Beam 1 Solid Column 1 Solid Beam 2 Column 2 Dout 130 (mm) 45 (mm) 400 (mm) 500 (mm) Thickness 20 (mm) 0.00 30 0.00 3 kN 1.5 marrow_forwardDetermine the maximum combined loads using the recommendedAISC expressions for LRFD. D = 100 psf, L = 70 psf, R = 12 psf, Lr = 20 psf and S = 30 psfarrow_forwardShow solution and box final answer pls tyarrow_forward
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