Determine the maximum axial force in member FG.
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- Compute the design moment strength of the beam section describedbelow if fy = 420 MPa, fc’ = 21 MPa.d= 650 mmd’= 70 mmb = 450 mmAs’: 3-28mm dia. As: 4-36mm diaCalculate the ultimate moment capacity (kN - m) of the T - section with the given parameters: PARAMETERS: fc = 25 MPa fy = 345 MPa stirrups dia. = 10 mm concrete cover, cc = 40 mm BEAM DIMENSIONS: bf = 900mm hf = 100mm h = 500mm (total height) bw = 300 mm As = 10 - 28mm diameter rebars at two layers with vertical spacing of 28mm (LOCATED at bw portion)Draw a shear diagram Cx = 0 Cy=30 Mc = -210 V {0<=x<8} = 0 V { 8< x <=14} = -5 (x-8)
- A W 16 x 50 is used as a beans to carry a uniform we and dead load including own weight of 39.7 kN/m and an axial tension load of T acting throughthe centroid, of the member. It has a simple span of 5.8 m. The compression flange of the member is laterally supported against local buckling. Use A 36 steel, Fy = 248 MPa.Properties of W 16 x50A= 9483.85 mm^2d= 412.75 mmbf= 179.65 mmtf= 15.95 mmtw= 9.65 mmSx= 1324 × 10^3 mm^3Sy = 172 × 10^3 mm^3 Determine the allowable axial stress.Higher Order L.D.E (D3+6D2+5D-12)y=0A steel column carries an axial load of 785 kN and a moment of 72 kN-m at the top and moment at the bottom which is only70% of the moment at the top. The two moments are in opposite direction and applied about the x-axis and the steel section has the following properties: A = 12818 mm2rx =109 mmry= 94 mmSx = 1200 x 103 mm3K= 1.0L= 3.6 mE = 200,000 MpaFy = 248 MpaUse NSCP specification for compressive stress and Fb =148 Mpa. Questions:From the Interaction Formula, compute the interaction value. Assume that the column is braced against joint translation (sidesway).
- Given W 16x36 section as shown below (Fy = 50 ksi, Fu = 65ksi). The holes are 3/4 inch in diameter. The connection is bottled at the web only. Assume x= 0.76 inch. 2. The shear lag coefficient U and the effective area AeRepeat Prob. 3-5 using singularity functions exclusively (including reactions).i want handwritten and correct I have attached final sol pls do it fullBAL=8 mt = 13 mm-150 mm113 mmFor the steel column with both ends fixed against rotation having a tubularcross-section, determine the allowable load Pall against buckling. E=200 GPa andOy=250 MPa.
- Determine the strength of a A992 steel column (Fy = 50 ksi), W 12 X 79 with a total length of 50 feet (both fix-end connection) for the following cases: Design axial compressive strength (LRFD) using AISC Requirement. Allowable axial compressive strength (ASD) using AISC Requirement. A = 23.2 in.^2 d = 12.4 in. tw = 0.47 in. bf = 12.1 in. tf = 0.735 in. T = 9-1/8 in. k = 1.33 in. k1 = 1.0625 in. gage = (2) 5-1/2 (2) in. rt = 3.31 in. d/Af = 1.39 Ix = 662 in.^4 Sx = 107 in.^3 rx = 5.34 in. Iy = 216 in.^4 Sy = 35.8 in.^3 ry = 3.05 in.Steel Design Two channels having the given properties shown is placed at a distance of 300 mm to back and is properly connected by a pair of lacings as shown. Properties of one channel A = 5595 mm2 d = 305 mm x = 17mm Ix = 67.3 x 106 mm4 Iy = 2.12 x 106 mm4 rx = 19.3 mm Assume K = 1.0 Determine the safe axial load in kN, that the column section could carry. Unsupported height of column is 6m.A W 16 x50 is used as a beans to carry a uniform we and dead load including own weight of 35.7 kN/m and an axial tension load of T acting through the centroid, of the member. It has a simple span of 5.9 m. The compression flange of the member is laterally supported against local buckling.Use A 36 steel, Fy = 248 MPa. Properties of W 16 x 50A= 9483.85 mm^2d= 412.75 mmbf= 179.65 mmtf= 15.95 mmtw= 9.65 mmSx= 1324 x 10^3 mm^3Sy = 172 × 10^3 mm^3 Questions:Determine the Safe axial load T that the beam could carry.Hint Answer: 399.65KN