530*150 is used as a compression member with one end fixed and the other end free. The length of the column is 3.2 m. What is the nominal compressive strength if fy=350 MPa?
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A 530*150 is used as a compression member with one end fixed and the other end free. The length of the column is 3.2 m. What is the nominal compressive strength if fy=350 MPa? (please help my assigment)
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- A W 360 x 44 of A992 steel column (Fy =345 MPa) having a length of 4m is fixed at both ends. Compute the effective width, be, of the slender web.Compute the value of reduction factor, Q, due to slender elementsin compression members.Compute the nominal compressive strength of the W-section.Properties of W 360 x 44Ag = 5710 mm2d = 351 mmbf = 171 mmtf = 9.78 mm tw = 6.86 mmkdes = 19.9 mmk1 = 19.1 mmrx = 146 mmry = 37.8 mmCheck the shear strength of an interior beam-column joint. The columns have 20 in. square cross-section, and 12ft clear height. The maximum probable moment strength of columns is (Mpr)ed. 520 ft-kips. The framing beams have the following geometry and reinforcement: Mpr = 345 ft-kips and Mpr = 213 ft-Kips; Top Bar = 5- #8 & Bot. Bar = 3 #8 Additional given data: fe= 4,000 psi; fy=60,000 psiDesign the spacing of the main bars and temperature bars of a one-way slabhaving a total span of 5 m. The slab is supported with hinge in the left support anda vertical cable located 1.5 m from the right support. The distance between thecables of the slab is 2.3 m. The slab is to carry a live load pressure of 2.4 kPa anda dead load pressure (including self-weight) of 4 kPa. Assume f’c = 27.6 MPa andfy = 276 MPa for main and temperature bars. Use 12 mm ∅ for main bars and 10mm ∅ for temperature bars. The slab is not exposed to earth or weather. Usethickness of slab of 120 mm. Use NSCP 2001.
- A cantilever of rectangular section is 70 mm wide and 250 mm deep at the fixed end andtapers uniformly to 75 mm wide and 100 deep at the free end. The projecting length is1800 mm, and there is a load of 1800 N at the free end. Calculate the deflection of thefree end and the maximum bending stress. Take E = 14000 N/mm2. The cantilever is madeof timber.A cantilever beam AB of length L = 6.5 ft supportsa trapezoidal distributed load of peak intensity q,and minimum intensity q/2, that includes the weight ofthe beam (see figure). The beam is a steel W12 X14wide-flange shape (see Table F-1(a), Appendix F).Calculate the maximum permissible load q basedupon (a) an allowable bending stress σallow =18 ksiand (b) an allowable shear stress τallow = 7.5 ksi.Note: Obtain the moment of inertia and section modulusof the beam from Table F-1(a).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.
- Question (1): A rectangular beam has a width b = 400 mm, and effective depth d = 850 mm and a total height h = 900 mm. The beam is subjected to an ultimate moment Mu = 1500 kN.m and an ultimate shear force Vu = 700 kN. 1- ) Design the beam for flexure to calculate the required area of steel. 2-) Using stirrups Φ 10 mm ( No.10) diameter, calculate the required spacing (s) between the stirrups at the ultimate shear force section. For all questions, Use f’c= 28 MPa and Fy= 420 MPaA W610 x 82 beam of A 36 steel has end reaction of 302 kN and is supported on a plate with a length of bearing of 139 mm. Fy = 248 MPa. Properties: d = 599 mm tw = 11 mm tf = 12.8 mm K = 33.76 Determine the compressive stress at the toe of the fillet (web yielding) for end reaction.A rectangular beam has the following properties: fc'=28MPa fy=270MPa b=300mm d=450mm As=4 of 36mm diameter bars Compute For: 1.) The magnitude of the ultimate moment capacity is Blank 1 kN.m 2.) If the fc' is reduced by 50%, the depth of the concrete stress block is Blank 1 mm.
- A short rectangular column 300 mm on one side and 400 mm on the other side. It is reinforced with 8-20-mm-diameter (28) longhitudinal bars equally distributed to the shorte sides of the column. Use f'c = 21 MPa and fy = 415 MPa. Calculate the required spacing of 10-mm-diameter ties, s (mm). Calculate the nominal axial strength of the column, Pn (kN). Calculate the maximum ultimate axial load the column can carry, Pu (kN)A simply supported rectangular beam of length L=5.4m has a width of b=395mm and an effective depth ofd=311m. Compression steel bars are placed at a distance d′=155mm from the extreme concrete compressivefiber. The beam supports service dead loads of Wd=27.5KN/m and Wl=35.5KN/m, respectively. Use fc′ = 28MPaand fy= 420 MPa. For the reinforcements, use 4 25 mm tension steel bars and 4 12 mmcompression steel bars.Determine the safe span that a 100 * 150 mm purlin would have if it has a spacing 800 mm on centers, to carry a roof load of 60 Pa and a wind load of 1.4 kPa acting on the vertical surface. Allowable bending stress is 10 MPa and the allowable horizontal shearing stress is 0.4 MPa. The allowable normal deflection is 1/360 of span. Weight of purlin is 5 kN/m3 and E= 17.3GPa