Designation Serial size 457x191 67 Calculate the design tension resistance of the 457 × 191 × 67 UB. Given information from steel section table (assuming S275): Elastic modulus Wel Axis y-y cm³ 1296 Axis Z-Z cm³ 153 per metre Mass Depth of Width of section section h b mm 189.9 kg/m mm 67.1 453.4 Plastic modulus Wpl Axis y-y cm³ 1471 Axis Z-Z cm³ 237 Thickness Thickness Root of web of flange radius tw tr mm r mm mm 8.5 12.7 10.2 Buckling parameter U 0.872 Depth between fillets d mm 407.6 Torsional Warping index constant Ilw X 37.9 dm 0.705 Flange C₁/t₁ Ratios for local buckling 6.34 Torsional constant IT cm¹ 37.1 Web Cw/tw 48.0 Area of section cm² 85.5 Second moment of area I Axis y-y cm4 29380 Indicative values for $355 steel Mc.y.Rd kNm Axis Z-Z cmª 1452 522 Nb.z.Rd* for Lcr=3.5m kN 1600 Radius of gyration i Axis y-y cm 18.5 Axis Z-Z cm 4.12: Designation Serial size 67 457x191

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I need a clear answer by hand, not by keyboard and fast answer within 20 minutes. Thank you | dybala For the beam loaded as shown, find the maximum tensile and compressive bending stresses and maximum shearing stress.. IN.A=4*10mm* 80mm 100mm 60mm 60mm 1.5KN 2kN/m 1mal A |-----|- | Im ++ Im->> 3m 3kN/m

The aluminum W 310 x 202 (wide flange) column is subjected to an eccentric axial load P as shown. The maximum compressive stress in the column is given by the so-called secant formula: where --- [¹ + ( √5)] ec o 1+ sec max E ō = P/A = average stress A = 25800 mm² = cross-sectional area of the column e = 85 mm = eccentricity of the load c = 170 mm = half depth of the column r = 142 mm = radius of gyration of the cross section L = 7100 mm = length of the column E = 71 x 10⁹ Pa = modulus of elasticity Determine the maximum load P that the column can carry if the maximum stress is not to exceed 120 × 10^6 Pa. Terminate the computation when Σa < 0.0005%.