1. Determine the available strength of the compression member HSS 6 x 4 x 1/2 shown in the figure below. The member is Use AISC equations to compute design strengths for both LRFD and ASD. 15
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- Problem2. The compression member is shown in figure. Find the following: a. The Euler stress Fe. b. The buckling stress Fcr c. The design strength d. The allowable strength e. Does the member satisfactorily meet the design requirements? Why? HSS 8x 8x4 ASTM AS00, Grade B steel (Fy = 46 ksi) 15'Problem 1 Considering flexural buckling, determine the allowable and design compressive strength using L = 5m (A = 6,265mm2, rx = 106.43mm, ry = 49.28mm %3D %3D W10× 33 L A992 steelwrite legibly Indicate all necessary FBDs, concepts and formulas needed in the solution. The members of the truss are 25mmx50mm in cross section. Determine the maximum load P that can be applied so that the stresses will not exceed 100 MPa in tension or 70 MPa in compression. See figure 101.
- Compression Member Analysis and Design 2Stress Analysis of Trusses using method of sections. pls answer within 30 minsThe beam has a cross section shown in the figure. fc’ = 26 MPa, fy = 390MPa, determine the following:a. The minimum steel area permitted by the NSCP Specs.b. The flexural design strength Mu if it is reinforced with minimum steelarea.c. The maximum area of flexural steel that can be used in reinforced thesection.
- AW18 x 40 standard steel shape is used to support the loads shown on the beam. Assume P = 18 kips, w = 4.6 kips/ft, LAB = 3.4 ft, LBc = 3.4 ft, and LCD= 16.6 ft. Determine the magnitude of the maximum bending stress in the beam. A P Answer: Omax LAB = B Save for Later LBC eTextbook and Media C ksi W LCD D X Attempts: 0 of 3 used Submit AnswerAn under-reinforced beam section given below. What could be the maximum design load (Pa) on the beam. Materials C30/37 and B420C. Ignore beams' own weight! Pa Pd H A 2 m 2 m, 2 m B As 1500 mm² 300 mm > Cross-section 470 mm 30 mm2. A column HP 14 x 102 of A572 Gr. 55 steel has a length of 15 ft is fixed at both ends. Compute the design compressive strength for LRFD and the allowable compressive strength for ASD. (Steel section properties are provided in the next page) ASTM Designation A572 Gr. 42 Gr. 50 Gr. 55 Gr. 60⁰ Gr. 65⁰ Yield Stress (ksi) 42 50 55 60 65 Fu Tensile Stressa (ksi) 60 65 70 75 80
- A simply supported beam is reinforced with 4-ø28 mm at the bottom and 2-ø20 mm at the top of the beam, Concrete covering to centroid of reinforcement is 70 mm at the top and 64 mm at the bottom of the beam. The beam has a gross depth of 450 mm and gross width of 300 mm. fc'=28 MPa, fy=415 MPa. Assume bars laid out in single layer. Calculate the following if the limitin tensile steel strains is 0.004 for a ductile failure: Depth of the neutral axis from the extreme concrete compression fiber to the nearest whole number = _____________mm Design strength of the beam section to the nearest whole number =____________ kN-m Maximum service uniform live load over the entire span in addition to a DL = 20 kN/m (including the weight of the beam) if it has a span of 6 m = _____________ kN/m (to the nearest whole number)AW18 x 40 standard steel shape is used to support the loads shown on the beam. Assume P = 19 kips, w = 3.8 kips/ft, LAB = 5.2 ft, LBc = 5.2 ft, and LCD= 14.4 ft. Determine the magnitude of the maximum bending stress in the beam. %3D D B LAB LBC LCD Answer: ksi OmaxDetermine 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 DESIGN