Find the maximum axial load that the W200 x 46 column can safely carry if its effective length is 6m. Fy = 248 MPa and E = 200 GPa. Use ASD. Section Properties A = 5860 mm2 1 = 15.3 x 106 mm4 r = 51.1 mm Final Answers: 703.87 1-67 Pn 2 = 421.481
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- Situation 4: A W12x106 simply supported beam carries a uniformly distributed load. The properties of the section relevant to the problem are as follow: Use A50 steel and Cb - 1.14 d - 328 mm bf-310 mm tf = 25.1 mm tw = 15.5 mm kdes - 40.4 mm ry - 79.0 mm lx = 388 x 10^6 mm^4 ly - 125 x 10^6 mm^4 Sx - 2380 x 10^3 mm^3 Zx=2690 x 10^3 mm^3 J-3800 x 10^3 mm^3 Cw=2870 x 10^9 mm^6 1. Which of the following nearly gives the Design Strength Moment (LRFD) if the span of the beam is 2.5m? 878 KN-m 793 KN-m 836 KN-m 811 KN mSituation 3: A W12x22 simply supported beam carries a uniformly distributed load. The properties of the section relevant to the problem are as follow: Use A50 steel and Cb - 1.14 d-312 mm bf - 102 mm tf = 10.8 mm tw = 6.60 mm kdes - 18.4 mm ry-21.5 mm Ix - 64.5 x 10^6 mm^4 ly-1.94 x 10^6 mm^4 Sx - 416 x 10^3 mm^3 Zx 480 x 10^3 mm^3 J-122 x 10^3 mm^3 Cw-44.0 x 10^9 mm^6 1. Which of the following nearly gives the maximum safe uniform load (Wu) that the beam could if the span of the beam is 0.800m? Neglect weight of the beam. 960 KN/m 933 KN/m 922 KN/m 975 KN/m1) Column AB is made of solid 410, cold worked Steel with the following specification Sy=85 KSI, E-29000 KSI Length: 9.5 in. Diameter: 425 in. Ends: Free & Fixed For the above column calculate: a) Type of column failure (show proof) b) The critical load that would cause failure. c) Allowable force with Factor of Safety of 2.5. Answer Unit Value a 13 14 14
- O 88 130% v - + I Annotate T| Edit Trial expired Unlock Full Version ENGR 263 + A A A T O 4.10 Member AB is the beam under consideration. As shown in the illustration of the loading condition, member AB is an overhanging beam that supports a uniformly distributed roof load of 500 lb/ft. It also carries concentrated loads from a rooftop HVAC unit (4000 lb), an interior hanging display support (2000 lb), and a marquee sign (3000 lb). Marquee overnang Displau SLIPPort II Raof Kiots Ol I W = 500 Ib/Ft A B 4000 b 2000 000 I Steel beam !! I1 3 FE 5 FE - Ft RoOFtop HVAC unit 10 FE 4 Ft Marquee sign< R R2 Steel beam (negligible weight) Free-body diagram Dispiay Support Loading conditionSituation 3: A W12x22 simply supported beam carries a uniformly distributed load. The properties of the section relevant to the problem are as follow: Use A50 steel and Cb - 1.14 d=312 mm bf - 102 mm tf 10.8 mm tw = 6.60 mm kdes 18.4 mm ry-21.5 mm Ix- 64.5 x 10^6 mm^4 ly- 1.94 x 10^6 mm^4 Sx416 x 10^3 mm^3 Zx 480 x 10^3 mm^3 J-122 x 10^3 mm^3 Cw-44.0 x 10^9 mm^6 1. Which of the following nearly gives the Design Strength Moment (LRFD) if the span of the beam is 2.5m? 89 KN-m O 114 KN-m O 123 KN-m O 101 KN-mA 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 simply supported beam is subjected to a uniform service dead load of 15kN/m(including the weight of the beam), a uniform service live load of 30kN/m. the beam is 12 meters long and is laterally supported at the midspan, and A572 Gr.50 steel is used. Is W30x108 adequate? Assume Cb=1 by 4 d fore k h h Z S Ly 267 TH mm 19.3 757 13.8 min m 35.8 685.4 54.6 737.7 5670 x10^3 4900 x103 mm 60.8 x10 6 mm 2080 x10 3 8300 x10^9 mm wwwsteel Design Civil Engineering Please show solutions: AW 310 x 118 section with a length of 8.0 m is used as a column. Determine the safe axial load the column can carry using AISC specifications with Fy = 345 MPa when: Properties of W 310 x 118 A = 15000 mm² rx = 136 mm ry = 77.6 mm A. column ends are fixed B. one end of the column is fixed; the other free Use LRFD Answers: A.] For Ag = 248.06X15000 100 3720,9kw $en=0.90x3720.9 = 3348.21 kW B.JPn= For Ag =40.72 x 15 000 LOW Pr= 0.90x610,81 = 549.729 KNDesignation Serial size 457x191 ■ Calculate the design tension resistance of the 457 × 191 × 67 UB. Given information from steel section table (assuming S275): Axis y-y cm³ 1296 67 Elastic modulus Wel Axis Z-Z cm³ 153 Mass per metre kg/m 67.1 Depth of section h mm 453.4 Width of section b mm 189.9 Plastic modulus Wpl Axis y-y cm³ 1471 Axis Z-Z cm³ 237 Thickness Thickness Root of web of flange radius tw tf mm r mm mm 8.5 12.7 Buckling parameter U 0.872 Torsional index X 37.9 10.2 Depth between fillets d mm 407.6 Ratios for local buckling Flange Web Cf/tf Cw/tw dm6 0.705 6.34 Warping Torsional constant lw constant IT cm4 37.1 48.0 Area of section cm² 85.5 Second moment of area Axis y-y cm4 29380 Axis Z-Z cm4 kNm 522 1452 Indicative values for S355 steel Mc.y.Rd 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
- ... Determine whether the D = 560 kips L = 68 kips compression member shown is adequate to support the given service loads. Take note Pu = 1.4D. 20' W12 × 79 K= 0.80, r= 3.05 in E = 29000 ksi, Fy = 50 ksi %3D A992 steel %3D Input Yes or No for your final answer. Note: Ag = 23.2 in^2 Blank 1 Blank 1 Add your answerASSIGNMENT #5 BIAXIAL BENDING Problem: Check the beam shown for compliance with the NSCP Specification. Lateral support is provided only at the ends, and A992 steel is used, F, = 345 mPa. The 90 kN service loads are 30% dead load and 70% live load. a. Use LRFD b. Use ASD d tw by ty 1x(106) S(10³) mm mm mm mm³ mm mmª W410x100 415 10 260 16.9 398 1920 SECTION Tx mm 177 1m 90 kN 90 KN 1.5m W410x100 I Loaded Section Ly(106) Sy(10³) Ty mmª mm³ mm 49.5 381 62.4 1m Zx(10³) Zy(10³) mm³ mm³ 2130 5818. H / The beam shown below has across seclion of channel Shape with width 360mm.and haghs h= loomm, the web thickness ist= 14mm.Deler- Mine ihe Max imum tensile aud Compressive stressin the beam due to unifom Load? 360Mm 3.6KN/M 14m tiom Please Sir 5T CASIC f-991ES PL DATURAL u CALC (-)