A 36 ft simple span beam is to support two movable service 20 kip loads a distance of 12 ft apart. Assuming a dead load of 1.0 k/ft including the beam self-weight, select a 50 ksi steel section to resist the largest possible moment. Use LRFD method only.
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A 36 ft simple span beam is to support two movable service 20 kip loads a distance of 12 ft apart. Assuming a dead load of 1.0 k/ft including the beam self-weight, select a 50 ksi steel section to resist the largest possible moment. Use LRFD method only.
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- A welded plate girder of 18 m span having flange plates 400x16 mm for each flange and 200x10 mm web. Compute the sectional properties and moment of resistance of the plate girder . Design also the end bearing and intermediate transverse stiffener, if the girder is subjected to U.D.L. of 120 kN/mUse 2015 NSCP A simply supported beam is subjected to a uniform service dead load of 1.0 kip/ft(including the weight of the beam) a uniform service live load of 2.0 kips/ft and a concentrated service dead load of 40 kips. The beam is 40ft long and a concentrated load is located 15 ft from the left end. The beam has continuous lateral support and A36 steel is used. Is W30x108 adequate? a) Use LRFD b) use ASD P₁=40* -15- 25'- WD = 1.0km W₁ = 2.0³m W30 x 108 40'-A W310 x97 steel wide-flange section Supports a uniformly distributed line load on a simple span of 7 m. Assume A36 steel. Calculate the allowable load (kN/m). Neglect the weight of the beam. Use Fv=0.4 Fy Material Steel ASTM A36 or A501 Steel (high-strength low-alloy) ASTM A992 Steel AISI 1020 hot-rolled Steel AISI 1040 hot-rolled Stainless steel (annealed) Cast iron (gray) Aluminum alloy 6061-T6 Magnesium alloy Titanium alloy Brass (rolled) Bronze (cast) Copper (hard drawn) Concrete Density kg/m³ x 10³ 7.85 7.85 7.85 7.85 7.85 7.21 2.64 1.794 4.40 8.56 8.56 8.81 2.40 Mod. of Mod. of Elasticity E MPa Rigidity G MPa x 10³ x 10³ 207 83 207 207 207 200 103. 70 45 114 97 83 103 21.5 83 79.3 79.3 80 41 28 17 45 41 34 41 Tensile Yield Strength F, MPa 250 345 210 290 280 240 210 900 340 170 280 Ultimate Strength MPa Tens. Comp. 400 450 380 520 580 140 260 280 970 410 230 380 550 390 21 Shear Coeff. of Thermal Expansion m/m/c x 10-6 260 11.7 11.7 11.7 Poisson's Ratio μ 0.25 9.9 0.25 11.7 0.25…
- Tutorial 7 Вeams 1. A simply supported beam must carry the factored ultimate loads (two 15kN point loads) shown in the figure. Do not consider the self-weight of the beam. The beam is laterally and torsional supported only at A, B, C and D. For a 203 x 133 x 25 I- section: (Grade 350 W steel) (a) Determine the class of the section. (b) Determine whether the section has sufficient flexural resistance. 15 kN 15 kN 4,0 m 2,0 m 4,0 mQ4. Considering BOTH strength and deflection criteria in approximate truss design, what would be the maximum distributed load, w, that the below truss can carry? d is the distance between the top and bottom boom. Given: alloeable - 200 MPa - Span 350 W 20kN 15m Sallowable E = 200 GPa d= 1.5m Cross-section area of tensile or compressive boom (not together) A= 2500mmIn the built-up girder shown below that has an unbraced length of 25 ft. (L-25 ft) and moment gradient factor of C,-1, determine the maximum flexural design strength. E-29000 ksi Fy-50 ksi PL 25x0.6" PL 35*0.3" PL 25x0.6"
- A built up section of A992 steel, F, = 345 mPa is made from plates fully welded together. The flanges consist of PL16X380 and PL12X500 for the web. Use the NSCP 2015 specifications. PL16x380 WL = 1.5wp %3D A Wu = 3.6wp 10m Which of the following best gives the maximum service dead load, wp? 40.2 kN/m 26.8 kN/m O 10.8 kN/m O 96.6 kN/m PL12x500Determine whether ASD or LRFD provides a larger relative effective ‘factor of safety’ for thefollowing loads on a steel column (show all work):a. The live load is three times greater than the dead load and the snow load is half of the deadload. The safety factor (Ω) is 1.67 and the resistance factor (φ) is 0.90.b. The dead load (does not include self-weight) is 80 kips, the live load is 150 kips and the selfweight of the column is 3 kips. The safety factor (Ω) is 1.75 and the resistance factor (φ) is0.90.A simply supported steel joist of 4 m effective span is laterally supported throughout. It carries a total UDL of 40 kN (inclusive of self weight). Design an appropriate section using steel of grade Fe410. The following two beam sections are available ISMB 150 @ 146 N/m ISMB 175 @ 191 N/m b₁ = tf = 8.6 mm t = 5.5 mm bf = 80 mm = 7.6 mm tf t = 4.8 mm 1₂=726.4 x 104 mmª = 96.9 x 103 mm³ 22 Z₂ R₁ Zp = 9 mm = 110.5 x 103 mm³ = 90 mm = 1272 × 104 mm² Izz Z = 145.4 x 10³ mm³ = 10 mm R₁ Z₁ = 166.1 x 10³ mm³
- PROBLEM 2: The steel beam loaded below has the built-up cross section as shown. Determine the maximum permissible value of the load w so as not to exceed allowable bending stresses of 110 MPa. -125 mm- w (including beam weight) 20 mm 20 mm 100 mm .-- NA A B - 2 m - 6 m 20 mm FINAL ANSWER: Max. allowable w kN/mProblem 2. A simply supported beam is subjected to a uniform service dead load of 1.5 kips/ft (including the weight of the beam), a uniform service live load of 2.2 kips/ft, and a concentrated service dead load of 50 kips. The beam is 45 ft long, and the concentrated load is located 15 ft from the left end. The beam has continuous lateral support, and A572 Grade 50 steel is used. Is W30x124 adequate?The W21 x 201 columns on the ground floor of the 5-story shopping mall project are fabricated by welding a 12.7 mm by 100 mm cover plate to one of its flanges. The effective length is 4.60 meters with respect to both axes. Assume that the components are connected in such a way that the member is fully effective. Use A36 steel. Compute the column strengths in LRFD and ASD based on flexural buckling.