Determine the flexural strength of a W14x 68 of A572 Grade 50 steel subject to a. Continuous lateral support. b. An unbraced length of 20 ft with Cb = 1.0. c. An unbraced length of 30 ft with Cb = 1.0.
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Determine the flexural strength of a W14x 68 of A572 Grade 50 steel subject to
a. Continuous lateral support.
b. An unbraced length of 20 ft with Cb = 1.0.
c. An unbraced length of 30 ft with Cb = 1.0.
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- Problem 3b. The built-up shown has Fy=50 ksi MC18X 42.7 W12 X 72 KL = 18.5 ft %3D Find the following: a. The KL/r b. The design strength for LRFD. c. The allowable strength for ASD. d. Is the section? Why?Using the Yield Strength of Grade 60 Steel, what would be the Maximum load that the 25 mm diameter bar can carry? a 204 kN b 203 kN c 205 kN d 206 kNUse 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.
- A beam must be designed to the following specifications: Span length = 35 ft Beam spacing = 10 ft 2-in. deck with 3 in. of lightweight concrete fill (wc=115 pcf) for a total depth of t=5 in. Total weight of deck and slab = 51 psf Construction load = 20 psf Partition load = 20 psf Miscellaneous dead load = 10 psf Live load = 80 psf Fy=50 ksi, fc=4 ksi Assume continuous lateral support and use LRFD. a. Design a noncomposite beam. Compute the total deflection (there is no limit to be checked). b. Design a composite beam and specify the size and number of stud anchors required. Assume one stud at each beam location. Compute the maximum total deflection as follows: 1. Use the transformed section. 2. Use the lower-bound moment of inertia.A W18X 97 steel beam will carry the given loads shown in Figure 1. Using ASD with A-36 steel (Fy = 250 MPa). A. If the beam is laterally supported at the two vertical supports only.a) What is the allowable bending stress in the beam in MPa?b) What is the actual bending stress developed in the beam in MPa?c) What is the actual shearing stress developed in the beam in MPa? B. If the beam is laterally supported at the two vertical supports and at its mid-span.a) What is the allowable bending stress in the beam in MPa?b) What is the allowable shearing stress in the beam in MPa?.Neglect the effect of beam weight.An A36 W14X605 simply supported steel beam with span “L” carries a concentrated service liveload “PLL” at midspan. The beam is laterally supported at supports and its midspan. Consider its beam selfweight to be its service deadload, “w” (use ASEP steel manual for selfweight, w and other section properties). In checking section's compactness, calculate the upper limit for non-compact web category? Express your answer in 2 decimal places.
- Determine the Design Flexural Strength of W460x52 A992 Steel with the following cases: CIVIL ENGINEERING STEEL DESIGN With Continuous lateral bracing Unbraced length = 4m, Cb = 1.0 Unbraced length= 12m, Cb = 1.0 a.) b.) c.) Fy = 345 MPa Properties: d = 450mm tf = 10.8 mm 1x 212x10mm* Cw 306x10 mm" Sx** 944 x10¹mm¹ bf 142mm MY Ag - 6650mm2 Zx 1090x10'mm' Ho439mm ry = 31mm tw* 7.62mm J-211x10¹mm* ly = 6.37x10 mm rls - 38.4mmDetermine the design strength of the beam shown. Assume fy-345MPa and fc'-21 MPa.AW14 x 43 column in a braced frame has a yield stress of 50 ksi and a height of 10 ft. The column is pinned at the top and bottom and has no intermediate bracing. Determine the available strength of the column considering the critical stress for flexural buckling. From AISC Manual Table 1-1, the properties of a W14 x 43 are: A = 12.6 in^2 tw = 0.305 in bf/2tf = 7.54 h/tw = 37.4 rx = 5.82 in ry = 1.89 in Ix = 428 in4 ly = 45.2 in4 G= 11,200 ksi J = 1.05 in^4 Cw=1950 in^6 What is the nominal axial strength considering flexural buckling stress of the column section (in kips). O 310 O 468 O 447 O 375 O 420 O 436 O 517 O 280
- Determine whether the D = 560 kips L = 68 kips compression member shown is adequate to support the given service loads. Take %3D note Pu = 1.4D. 20' W12 x 79 = 3.05 in K = 0.80, r= E = 29000 ksi, Fy = 50 ksi A992 steel Input Yes or No for your final answer. Note: Ag = 23.2 in^2Compute the compressive strength of a WT12 x81 of A992 steel. The effective length with respect to the x-axis is 25 feet 6 inches, the effective length with respect to the y-axis is 20 feet, and the effective length with respect to the z-axis is 20 feet.Q2 1- Check the following weather one way or two way 2- Calculate the design ultimate load Wu on the on the beams B3, B6 Assume slab load: Live load = 5 kN/m2 Dead load = 8 kN/m? Beam size 300x500 mm %3D B5 B6 B7 3.6 m |L 4.8 4 m 5m 5m BI B2 B4