9.7 Determine the allowable load capacity (Pallowable) for an Fy = 36 ksi steel column, W12 × 65, when L18'and a. the base and top are both fixed. b. the base is fixed and the top is pinned. c. both the top and bottom are pinned.
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- 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.1. Check the beam shown for compliance with the AISCS. Lateral support is provided only at the ends, and A992 steel (E345 MPa and Fu= 450 MPa). The only uniform service dead load is the weight of the beam. The 70 KN service loads are 30%DL and 70%LL. Use LRFD. IR₁ 0.90m 70,KN * W14 x 68 1.20m 70 KN * 0.90m R₂ W14x68 70 KNQ2) 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 D
- The given girder has beams framing into it at the ends and at every L/3 point. The beam carries a service live load of 20 kips as shown and superimposed uniformly distributed service dead load of 10 kip/ft. Select the lightest A992 W-section that can carry the load. Do not check for deflection. P, = 20 kips LL PLL W. = 10 kip/ft DL = 20 ft WDL L/3 BEAMS FRAMING INTO GIRDER LA W14x54 steel column rests on a concrete pedestal. Column base plate is to be provided and designed using the full capacity of the column for axial load. The properties of the steel column are shown in the table. Yield strength of the column and base plate is 248 MPa. For concrete; f’c = 21 MPa. a. Determine the axial capacity of the column.(LRFD)b. Determine the smallest dimension of the squarebase plate so that the allowable bearing for the concrete support is not exceeded? (Increment of 5mm)c. Determine the thickness of the base plate1. A steel column 10 m long is fabricated from a cover plate and C section arranged as shown. Determine the safe compressive load. Fy = 248 MPa, E= 200 GPa. Use AISC/NSCP Specs. 450 mm -cover plate 'I 12 mm y2 IP d2 10 m C 310 x 37 A = 4720 mm? d = 305 mm bf = 77 mm tf = 12.7 mm tw = 9.8 mm C 310 X 37 a) Both ends of column are fixed b) Both ends of column are hinged c) One end fixed, the other end hinged Use design values of k. tw d=305- Ix = 59.9x10° mm ly = 1.85x10° mm x = 17.1 mm x=17.1
- Q2 a) This is a concrete beam (the interior lines represent reinforcing bars), If P is live load, draw the BMD and the SFD as a result of ULTIMATE LOAD. P= 50 kips 20 ft. 20 ft. 40 ft. Own weight of the beam = 1.0 kip/ft. (Uniform dead load) b) This is a concrete frame (the interior lines represent reinforcing bars), If P is dead load, draw the BMD and the SFD as a result of ULTIMATE LOAD. 6.0 ft. P= 1.0 Kip Own weight of the beam = 1.0 kip/ft. (Uniform dead load)For the Column in Shown in Figure, Design the Base Plate with A36 steel, if the service dead loads PD=240 and service live loads PL=345 kips. The concrete has compressive strength f’c=3.5 ksi .The Base Plate cover entire concrete pedestal underneath. (The Steel I- section is W14x120)5. The beam-column has a height of 7.5m and is a member of a braced frame. It is subjected to an axial load of DL-65KN and LL= 1OOKN. It is pinned at both ends. A W12X35 section was used and was made up of A-36 steel with Fy-250MPA. Compute the interaction value for beam- column based on NSCP 2015 code provisions. Ngelect the weight of the section. A = 6645 mm2 Zx- 839x10^3 mm3 d- 317.50mm Sx- 747x10^3 mm3 tw= 7.62 Rx= 133.35 bf= 166.62 ly= 10x10^6 mm4 tf- 13.21 Sy= 122x10^3 mm3 Ix= 119x10^6 mm4 Ry3 39.12mm 3.5 Note: Use C=D1.00%--0.2
- An HSS 10 x 6 x 5/16 with Fy = 46 ksi is used as a column. The length is 16 feet. Both ends are pinned, and there is support against weak axis buckling at a point 6 feet from the top. Determine a. the design strength for LRFD. b. the allowable stress for ASD. |·+· 10' 1 y-axis 16' J-axisA plate girder must be designed for the conditions shown in Figure P10.7-4. The given loads are factored, and the uniformly distributed load includes a conservative estimate of the girder weight. Lateral support is provided at the ands and at the load points. Use LRFD for that following: a. Select the, flange and web dimensions so that intermediate stiffeners will he required. Use Fy=50 ksi and a total depth of 50 inches. Bearing stiffeners will be used at the ends and at the load points, but do not proportion them. b. Determine the locations of the intermediate stiffeners, but do not proportion them.If the beam in Problem 5.5-9 i5 braced at A, B, and C, compute for the unbr Cb aced length AC (same as Cb for unbraced length CB). Do not include the beam weight in the loading. a. Use the unfactored service loads. b. Use factored loads.