Practice Problem 4 A36 steel beam having a simple span of 7m is laterally unsupported and the section has the following properties: tf = 16 mm tw = 18 mm bf = 210 mm d = 530 mm The beam is restrained against lateral buckling only at the supports. a) Find rt. b) Find the allowable bending stress.
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- a two span beam subject to shear and flexure only is reinforced as follows: Section: TOP BARS @ midspan: 2-ø20mm @ face of supports: 5-ø20mm Section: BOTTOM BARS @midspan: 3-ø20mm @ face of supports: 2-ø20mm GIVEN: Stirrup diameter ds = 10mm concrete f'c= 21 mpa steel rebar fy= 415 mpa stirrup fy = 275 mpa beam size b x h = 270 mm x 450 mm assume all bars laid out in a single layer. Determine the DESIGN moment strength of section at midspan for positive bendingA two span beam subjected to shear and flexure only is reinforced as follows: SECTION @ MIDSPAN @ FACE OF SUPPORTS TOP BARS 2-∅20 mm 5-∅20 mm BOTTOM BARS 3-∅20 mm 2-∅20 mm Given: Stirrup diameter, ds = 10 mm Concrete f'c = 21 MPa Steel rebar fy = 415 MPa Stirrup fy = 275 MPa Beam size b x h = 270 mm x 450 mm Assume all bars laid out in single layer. Calculate the following: Tensile steel ratio in positive bending at midspan = __________ (in 5 decimal places) Design Moment strength of section at midspan for positive bending = ___________ kN·m (nearest whole number) Nominal Moment strength of section at face of support for negative bending = ___________ kN·m (nearest whole number)Use the composite beam tables and select a W-shape and stud anchors for the following conditions: Span length = 18 6 Beam spacing = 9 ft Total slab thickness = 51 2 in. (the slab and deck combination weighs 57 psf). Lightweight concrete with a unit weight of 115 pcf is used Construction load = 20 psf Partition load = 20 psf Live load = 225 psf Fy=50 ksi and fc=4 ksi A cross section of the formed steel deck is shown in Figure P9.8-9. The maximum live-load deflection cannot exceed L/360 (use a lower-bound moment of inertia). a. Use LRFD. b. User ASD.
- Determine the design strength of the beam shown. Assume fy-345MPa and fc'-21 MPa.Design economical spacing of 10 mm diameter stirrups for the beam which carries the loads shown. Design shear reinforcement of prestressed beam with the following properties listed. Properties: f^c=28 MPa fy = 276MPa Aps = 800 mm² Ds = 10mm Dm = 18mm fse = 800 MPa fpu = 1500 MPa fpe = 860 MPa Pe = 250kNSOLVE please Determine the design bending strength of ISLB 350 @ 486 N/m, considering the beam to be laterally supported. Assume factored design shear force will not exceed 0.6 times the design shear strength of the cross-section. If it is used as simply support beam of unsupported length 5 m then determine the factored distributed load the beam can carry. Use Fe410 grade steel For ISLB 350 @ 486 N/m, section properties are as follows: h = 350 mm; b= 165 mm; t₁ = 11.4 mm; t = 7.4 mm; R₁ = 16 mm; Izz = 13158.3 × 104 mm²; I = 631.9 × 10¹ mm²; Zpz = 851.11 × 10³ mm³; Zez = 751.9 x 10³ mm³
- A prestressed T beam shown which is reinforced with a bonded tendon having an area Aps = 1580 mm?, fc' = 34.5 MPa, fpu = 1862 MPa, and the effective stress after losses fse = 1102 MPa, fpy= 1713 MPa. Span of the beam is 6 m. %3! 1220 mm 0.85 fc 804 C NA 600 d-a/2 Aps 60 T=Ape fps 250 Find the live load that can be carry by the beam using USD load combination if the concrete's specific gravity is 2.4. (kN/m)A two span beam subjected to shear and flexure only is reinforced as follows: SECTION @ MIDSPAN @ FACE OF SUPPORTS TOP BARS 2-16 mm diameters 5-16 mm diameters BOTTOM BARS 3-16 mm diameters 2-16 mm diameters Given: Stirrup diameter = 10 mm Concrete f'c = 21 MPa Steel rebar fy = 415 MPa Stirrup fy = 275 MPa Beam size b x h = 250 mm x 450 mm Assume all bars laid out in single layer. Calculate the following: Tensile steel ratio in positive bending at midspan Design Moment Strength of section at midspan for positive bending in kN-m Nominal Moment Strength of section at face of support for negative bending in kN-mProblem 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 W350x90 girder 8m long carries a concentrated dead load, P at every quarter points and a uniform dead load of 5 kN/m (including dead weight) amd a uniform live load of 7.2 kN/m. CIVIL ENGINEERING STEEL Properties: A = 11,500 mm² tw = 10 mm tf = 16 mm mm Ix=1119.7 x 10 mm* Allowable Stresses: Fb = 0.66 Fy d = 350 mm bf = 250 mm tf = 16 mm Fv = 0.40 Fy a.) Determine P base on Flexure. b.) Determine P base on Shear. b.) Determine P base on Deflection. DESIGN Sallow=1/360Is the beam shown below adequate for flexural (moment) stresses? Use ASD.A two span beam subjected to shear and flexure only is reinforced as follows: SECTION TOP BARS ВотTом ВАRS @ MIDSPAN @ FACE OF SUPPORTS 2-020 mm 5-020 mm 3-020 mm 2-020 mm Given: Stirrup diameter, de = 10 mm Concrete fc = 21 MPa Steel rebar fy = 415 MPa Stirrup fy = 275 MPa Beam size b xh = 270 mm x 450 mm Assume all bars laid out in single layer. Calculate the following: Tensile steel ratio in positive bending at midspan = (in 5 decimal places) Design Moment strength of section at midspan for positive bending = kN.m (nearest whole number) Nominal Moment strength of section at face of support for negative bending = kN-m (nearest whole number)