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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- 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 DA 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 plate
- 1. 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=17.1SHOW CLEAR AND COMPLETE SOLUTION. SOLVE FOR 1 HRSolve by LRFD (AISC ) masnual
- ASAP PLSS I GIVE UPVOTEE ? ♥A 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.
- Can you help me answer this problem pls? I need it asap. Thank you.Using A-36 steel, determine the safe load Pu based on the capacity of the bottom strut. Section 2 L100 x 100 x 8 3m | Pu 10 mm 4m CS Scanned with CamScannerAn 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-axis