Structural Steel Design (6th Edition)
6th Edition
ISBN: 9780134589657
Author: Jack C. McCormac, Stephen F. Csernak
Publisher: PEARSON
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Chapter 7, Problem 7.2PFS
To determine
The effective length factors for all columns of the given frame.
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Q1- the reinforced flat plate roof slab system that shownin figure. Supports a service live load of 1.5 kN / m2 and a service superimposed dead load of 3.4 kN / m? in addition to its own weight. The slab is supported along its outer edges by a brick wall. Slab thickness is 200mm and wall thickness is 240mm. all interior columns are 300x600 mm. Using the ACI direct design method, design and detail the column strip top bars required at the continuous edges of strip A shown in the plan. Use fe-22 MPa andf-400MP.
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Structural Steel Design (6th Edition)
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- A flitch beam is composed of two 50x300mm planks reinforced with a 12.5mm steel plate as shown. Assume that continuous lateral support is furnished by a floor system supported by the beam. Use A36 steel with an allowable bending stress of 150 MPa, Fy = 250 MPa. Es = 200,00MPa, Ew = 7692MPa, fw = 4.3 MPa. The beam is on a simple span of 4.8m and supports a uniformly distributed load of 7.3 kN/m which includes the weight of the beam. Allowable stresses for shear: Steel: fv = 0.60Fy MPa Wood: fv = 0.58 MPa Determine the following: a.) The resisting moment of the flitch beam b.) The applied moment to the flitch beam c.) The shear capacity of the flitch beamarrow_forwardSelect a cover-plated W section limited to a maximum depth of 21.00 in to support the service loads shown in the accompanying figure. Use A36 steel and assume the beam has full lateral bracing for its compression flange.arrow_forwardAn interior column is to be designed as a part of an unbraced frame system, located at ground floor. It is to be pin-supported at the foundation end, and it is to be connected by two W14x68 girders and W18x106 column at the other end. All of the members, including the column described, are made of A36. Architectural specifications require that the column is to be designed as a rolled W-section with depth of 21 inches. The column is expected to carry axial compressive load of 300 kN dead load and 350 kN live load. Length of girders is 4 meters and length of columns is 3.0 meters. Select the best section that should be used for the said column. For determining effective length factors, neglect inelastic action; and use major axis moment of inertia for girders and minor axis moment of inertia for columns. Use LRFD specifications from 2015 NSCP Vol. 1.arrow_forward
- Refer to the elevation plan and schedule of beam shown. Use girder 2B1 to layout the rebar. Compute and estimate: a. Create a rebar cutting list for stirrups needed for the three spans shown using 2B1. Consider the restrictions in tension zones.arrow_forwardA (6 INCH × 36 INCH) SIMPLY SUPPORTED BEAM IS RESTING ON TWO(6 INCH ×18 INCH). CLEAR SPAN BETWEEN COLUMN IS 12 FT WITH CLEAR HEIGHT 8 FT. DRAW PICTORIAL VIEW OF THIS ARRANGEMENT.arrow_forwardA beam with b=200mm, h=400mm, cc=40mm, stirrups=10mm, fc'=32 MPa, fy=415 MPa is reinforced by 3-32mm diameter bars. Calculate the depth of the neutral axisarrow_forward
- Q Given table displays the bending moments and shear force values for beam B2 at the supports and midspan.Beam B2: Clear span, 28 inches wide by 33 inches deep, l_n = 25.17, f_c' = 4000 psi, f_x = 60000 g.In B2, there is hardly any axial force. Required. Frame Beam B2's schematic design and detail using the provisions of ACI 318-99.arrow_forwardA rectangular beam has a width of 300mm and depth of 330 mm with a steel cover of 70 mm. Using f'c=28 Mpa, fy=415 Mpa and width applied loadings of the following: • Wo = 50 kN/m • L = 15 m 1. Determine the required area of the bottom bars in mm2. 2. Determine the required area of the top bars (if needed) in mm2.arrow_forwardDetermine the effective length factor for the specified columns in the following unbraced frame:arrow_forward
- A reinforced concrete beam 550mmx300mm withtensile reinforcements of 5-28mm is simply supported over a span of 6.495m with additional 2m overhang from the right support. Using steel covering of 65mm, fc’=28MPa and fs=138Mpa.Calculate the following requirements: a)Check whether the beam is safe from the following loadings: Wdl=30KN/m (beam weight included)– placed throughout the length of the beam Wll=20KN/m (along the simple span only) Wll=10KN/m (along the cantilever only) Pdl=50KN (located 2m from the left support) Pll= 75KN (located at the free end)arrow_forwardAnswer problem 5.11-9: Same as problem 5.11-8, but let all the loads on the girder act as a uniform load (be sure to include the weight of the beam).arrow_forwardA 6.96.9 mm long simply supported wood beam carries a uniformly distributed load of 10.810.8 kN/mkN/m, as shown in Figure A. The cross-sectional dimensions of the beam as shown in Figure B are b=195b=195 mmmm, d=490d=490 mmmm, yH=80yH=80 mmmm, and yK=158yK=158 mmmm. Section a–a is located at x=1.4x=1.4 mm from B. (a) At section a–a, determine the magnitude of the shear stress in the beam at point H. (b) At section a–a, determine the magnitude of the shear stress in the beam at point K. (c) If the allowable shear stress for the wood is 950950 kPakPa, what is the largest distributed load ww that can be supported by the beam?arrow_forward
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