Pb-3b, A circular column has to be designed using lateral ties to carry an axial load of 1500kN. Using M25 mix and Fe 415 grade steel, what is the required diameter of column section? (assume longitudinal steel percentage = 1% of gross area) -
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- A column is to support an axial dead load of 900 kN and an axial live load of 1240 kN. Use fy = 414 MPa and f’c = 27.6 MPa. Assume 2% steel ratio and use 28 mm main bars and 10 mm ties/ spiral. Use NSCP 2015- and a 30-mm steel coverAn axially loaded square tied column is to be designed for the following service loads.DL = 2000 kNLL = 900 kNfc’ = 27 MPafy = 345 MPaconcrete cover = 80 mmUse reduction factor φ = 0.65 a. Compute the required size using 2.5% steel ratiob. Compute the maximum moment for axially loaded columnc. Compute the largest size of the column to support the given loadingDesign a building column made of A36 steel (fy=36 ksi) having length of 10ft using AISC table to support a dead load of 50 kips & live load of 80 kips. Assume K=0.9 & E=29,000 ksi. Use LRFD approach for design. Initially assume fcr =0.75 fy. Note: Use “W” section for column member.
- Design a square base plate to support a 10mm thick pipe column with an inside diameter of 300mm. The base plate minimum dimension is 500mm. To design the base plate thickness, convert the circular perimeter of the pipe into an equivalent square with an area equal to the original circle. the loads are Pdl=850 kN and Pll=420 kN . Assume that the Pipe is to rest on a very wide slab. Use A-50 steel and concrete fc'=35 MPa and LRFD.Design a square base plate to support 10 mm thick pipe column with an inside diameter of 300 mm. The base plate minimum dimension is 500 mm. To design plate thickness, convert the circular perimeter of the pipe into an equivalent square with an area equal to the original circle. The loads are PDL = 850 kN and PLL =420 kN. Assume that the pipe is to rest on a very wide slab. Use A-50 steel and concrete fc’ = 35MPa and LRFD.A Wx35 steel column has an unsupported height of 6m. Use A36 steel with Fy = 248 MPa and E = 200,000 MPa. A = 6634 rx=88.9 mm ry = 51.56 mm Determine the safe axial load if one end is fixed and the other is pinned.
- Compute the maximum service axial compression load permitted on the built-up cross-section of the accompanying figure. The load is 30% dead load and 70% live load. The steel used is A992, and the effective lengths are (KL)y = 14 ft and (KL)x = 42 ft.The rods are used as pin-ended column, each with 2.85m effective length. a) Find wall thickness of hollow square rod where both rods have same cross-sectional area. b) Determine Critical load of each rod when E=105GPaFor the floor system shown below, the un-factored live load is 100 psf and un-factored dead load is 150 psf including self-wt. Draw the free body diagram for B2 and G1
- Determine the flexural strength in ft kips of a W12 × 35 of A242 (Fy = 50ksi) steel subject to a. Continuous lateral support. Blank 1 b. An unbraced length of 10 ft with Cb = 1.0 c. An unbraced length of 20 ft with Cb = 1.0topic: Compression Members An A36 (Fy=248MPa) W12x79 fixed-hinged column of 6.13m length is laterally supported at mid-height on its weaker axis. If LL=2DL, calculate the maximum service Liveload in KN that the column can carry using LRFD. Use ASEP Steel Manual for WF section properties. Express your answer in 2 decimal places.For the flat plate shown in the figure below (edge beam are not used), design the hatchedframe (middle and column strips) in the N-S direction.Given:- fc = 28MPa , f y = 420MPa- Super imposed dead load = 4 kN/m^2- Live Load = 3 kN/m^2