Determine the amplification factor for the following braced column (W21x44) (A992 steel). The flexural rigidity, El, is unreduced, Use ASD. ML= 22 ft-k MD= 19 ft-k 20 W21x44 ML= 29 ft-k MD= 33 ft-k PL= 100 k PD= 220 k
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- Problem 3 of 3 A 15 ft long W8x40 wide flange column of A36 steel is fixed at the base and free at the top. The modulus of elasticity for steel E = 29 x 106 PSI Determine: 1. the controlling slenderness ratio, KL/rx or KL/ry 2. the AISC allowable compressive stress Fa (NOT the Euler critical buckling stress) 3. the AISC allowable column load Pa (NOT the Euler critical buckling load) Pa 15'A built-up section shown, 30 feet long, is fixed at both ends and braced in the weak direction at the midheight. A992 steel is used. Determine the following: 11. Ix 12. Ky 13. effective slenderness ratio 14. Fe 15. LRFD design strength 16. ASD allowable strength 18 4 W10 × 49 A 14.4 in² bf 10 in d. 10 in tw 0.340 in tf 0.560 in Ix= 272 in¹ Iy=93 4in¹ W10×54 A 15-3 in² bf 10 in d = 10 lin tw0-370 in tf 0.615 in Ix= 303 in 4 Ty = 103 in 1A column is built up from four (4)- 125 x 125 x 18 angle shapes as shown. The plates are not continuous but are spaced at intervals along the column length and function to maintain the separation of the angles. They do not contribute to the cross-sectional properties. The effective length is 4 m. Compute the allowable design compressive strength based on flexural buckling. E= 250 MPa. Use ASD. k 375 mm 125mm, HPlate 125mm 4 - 4 125 × 125× l8 section 下好业
- Q2) The members of the truss structure shown below is plain concrete. The compressive strength of the concrete is 25 MPa. Compute the maximum load P that can be carried by the structure. (Cross section of each member of the truss is 200 x 200 mm and don't use material factors and do not consider slenderness) Comment on your results briefly. P A& 2m SC 2 m 1380 2m DDetermine the safe load of the column section shown, if it has a yield strength of 25 MPa. E = 200000 MPa. Use NSCP Specifications. Fyz248 mpa Properties of Channel Section d = 305 mm t₂ = 7.2 mm A = 3929 mm² t₁ = 12.7 mm Ix=53.7 x 10mm¹ x = 117 mm Properties of W 460 x 74 A = 9450 mm² b = 190 mm ly= 1.61 x 10 mm x = 17.7 mm tw = 9.0 mm rx = 188 mm ry = 41.9 mm d = 457 mm tr = 14.5 mm Ix = 333 x 10 mm Iy = 16.6 x 10mm* 7.21 When the height of column is 6 m. When the height of column is 10 m. Assume K= 1.0 457 CIVIL ENGINEERING- STEEL DESIGN3. Check the suitability of the W12x65 section made of A992 steel for the axial loads and moments shown in the figure. Mor= 40 k -1.0 1.4 14 14 Mer Pat = 150K Per = 50k Bar %3D
- USE NSCP 2010 A rectangular beam reinforced for both tension and compression barshas an area of 1450 mm² for compression bars and 4350 mm² fortension bars. The tension bars and compression bars are placed at a distance of 600 mm and 62.5 mm respectively from the top of thebeam. The beam width 300 mm, fc’ = 21 MPa, fy = 415 MPa andtension steel covering is 60 mm.If it is 8 m-simply-supported beam that carries three concentratedservice live loads P applied at three quarter points of the beam (exceptat the supports), neglecting the self weight of the beam, determine themaximum value of service load P in kiloNewtons.AW 12 x 65 column section shown in the figure is pinned at each end and has an additional support in the weak direction at a point 4.1 m from the top. Use X- 4.8 m K 1 for both directions. X+Y Properties of W 12 x 65 A 12323 mm2 - 134.11 mm ry 76.71 mm Fy- 248 MPa Determine the allowable compressive strength (kN).AW14 x 43 column in a braced frame has a yield stress of 50 ksi and a height of 10 ft. The column is pinned at the top and bottom and has no intermediate bracing. Determine the available strength of the column considering the critical stress for flexural buckling. From AISC Manual Table 1-1, the properties of a W14 x 43 are: A = 12.6 in^2 tw = 0.305 in bf/2tf = 7.54 h/tw = 37.4 rx = 5.82 in ry = 1.89 in Ix = 428 in4 ly = 45.2 in4 G= 11,200 ksi J = 1.05 in^4 Cw=1950 in^6 What is the nominal axial strength considering flexural buckling stress of the column section (in kips). O 310 O 468 O 447 O 375 O 420 O 436 O 517 O 280
- Determine the design moment capacity of the beam with given properties below. bf = 550mm bw 350mm tf = 100mm %3D d = 370mm %3D d' = 100mm %3D fc' = 34.5 MPa fy = 400 MPa %3D As = 6-32mm bars a USE NSCP 2001, Mu = kN-m b. USE NSCP 2015, Mu = kN-mPROBLEM 01: Considering a W24x104 section for the beam Properties of the W24 x 104 section shown, check whether this member satisfies the appropriate Ag = 19742 mm2 AISC interaction equation. Assume the beam section to be d = 611.12 mm compact and use A36 steel. The horizontal beam-column is subjected to a service live load P=125 kN and Q = 800 kN. bi = 323.85 mm Consider the weight of the beam. The member is laterally t = 19.05 mm braced at its end and the bending is along it major axis. k = 1290 x 10 mm+ Lateral supports are provided at the X marks. Use LRFD and ly = 108 x 10 mm4 AISC specifications tw = 12.70 mm Sx = 4228 x 103 mm3 Sy = 667 x 10° mm nominal weight = 155 kg/m Zx = 4736 x 103 mm3 Zy = 1023 x 103 mm3 IP Q 2m 4m EDIFICAUse A 992 Steel and Select the most economical W shape for the beam shown in the figure below. The beam weight is not included in the service loads shown. Do not check deflection. Assume continuous Lateral support. K 6 IPD = lok PL=20k * 161 1 WD= 3-33 K/F+ WL=6-67 k/Ft -xhul