6ft o = 32° Y = 110 pcf G.W.T 2ft + = 30° Y = 125 pcf 0 = 10° Y = 126 pcf 9ft c = 600 pcf Oft $ = 0° 8ft c = 800 pcf Y = 120 pcf $ = 20° Y = 120 pcf 5ft c = 400 pcf A 十十
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For the shown figure below. Plot the pressure diagram and find the resultant force F and its location (from A) under active conditions.
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- A line is recorded as 472.90 m long. It is measured with a0.65 kg tape which is 30.005 m long at 20 °C under a 50N pull supported at both ends. During measurement, thetemperature is 5 °C and the tape is suspended under a 75N pull. The line is measured on 3% grade. E = 200 GPa,cross-sectional area of tape is 3 mm2 and the coefficientof linear expansion is 0.0000116 /°C. Compute:a. The actual length of tape during measurement.b.The total error to be corrected for the inclined distancec.The true horizontal distancec. Calculate "P" in KN considering failure on bearing on projected area. Two A36 16mm Thick Steel Plates Are Connected By Four Rivets With Fv=152 MPa As Shown. Given: Rivet Dia 20 mm x=100 mmThe cantilever beam consists of a rectangular structural steel tube shape [E = 200 GPa; I = 425 × 106 mm4]. Assume P = 35 kN, w = 10 kN/m, LAB = 1.50 m, and LBC = 2.00 m. Using Appendix C and method of superposition, determine:(a) the beam deflection at point A. [vA = -4.54 mm](b) the beam deflection at point B. [vB = -2.07 mm]
- Steel Design Given: S = 2m L = 10m Superimposed dead load = 5 kPa Live load = 3.6 kPa Properties of Beam BFA section = W 460 x 97 kg/mArea (A) = 12320 mm^2Depth (d) = 465 mmFlange Width (bf) = 193 mmFlange Thickness (tf) = 19mmWeb Thickness (tw) = 11 mm Moment of InertiaIx = 445 x 10^6 mm^4Iy = 23 x 10^6 mm^4Yield Strength, Fy = 248 MPar (radius of gyration of a section compromising the compression flange plus 1/3 of the compression web area of compression flange = 50 mm. Consider bending about the x-axis. 1. The compression flanges are laterally supported only at midspan. With this condition, what is the permissible flexural stress if Cb = 1.0?Calculate the member force on member HF in Kips? Specify if T or C. 87.6 C 105.9 C 79.7 C 97.5 C 112.1 Cnote: SN is equal to 18, 100 + SN for topload will now be 118kN
- note: SN = 18, to the topload will be 118kNPlease Answer it fast on Sheet of paper. F1=1.5KN ,F2=3.8KN, a=3.6m ,b=2.0m ,c=1.3m. You need to find Forces in member (CH,AH and CD) And whether it's +ve or -ve.At-beam has a width of flange equal to 700mm and a thickness of 100mm. It has an effective depth of 450mm and the width of the web section is 350mm. It is reinforced at the bottom with a steel area of 2925sq.mm. fc'=17.24MPa, fy=413.7MPa. The depth of the compression block is _mm. A. 136.00mmB. 146.00mmC. 126.00mmD. 156.00mm
- -6,00 Design a short square tied column to carry a factored axial design load Pu of 750 kips an a factored design moment Mu of 250 ft.-kips. Place the longitudinal reinforcing uniformly in the four faces. Use f, = 5000 psi and fy = 60,000 psi. o. Then P 350-1.9 h2 1.86 30 aoA rectangular beam has b=300mm, d=500mm, As=4D22mm, fc’=34.2MPa and fy=420MPa. Determine the design moment, kN-m. Determine the balanced distance of the neutral axis to the outermost compression fiber, mm.A rectangular reinforced concrete beam 350 x 550 mm (b x d) is provided with 3H25 bar (As = 1473 mm2) as compression steel. Determine the area of tension steel needed for the beam to attain its full moment capacity. Calculate the corresponding ultimate moment of resistance. Assume fck = 35 N/mm2and fyk = 500 N/mm2and d’ = 60mm. Take until 2 decimal point for each calculation. Notes : Internal forces Fcc = 0.567fck(b x 0.8x) = 0.454 fck bx Fst = 0.87 fyk As Moment of resistance with respect to steel, M = Fcc.z Moment of resistance with respect to concrete, M = Fst.z Mbal = Kbal fck bd2 where Kbal = 0.167 As ‘ = (M-Mbal) / 0.87 fyk (d-d’) As = Mbal / 0.87 fyk zbal + As’