Problem 3b. The built-up shown has Fy=50 ksi MC18 X 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.
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- If the beam in Problem 5.5-9 i5 braced at A, B, and C, compute for the unbr Cb aced length AC (same as Cb for unbraced length CB). Do not include the beam weight in the loading. a. Use the unfactored service loads. b. Use factored loads.7 7a 7b 7c Alaterally supported beam was designed for flexure. The beam is safe for shear & deflection. The most economical section is structural tubing however the said section is not readily available at the time of the construction. If you are the engineer in charge of the construction what alternative section will be the best replacement? Why? The section is 8" x 8" x 7.94 mm thick: Use Fy=248 MPa: E=200,000 MPa AISC wall thickness Ix 106 S x 103 Jx 103 mm4 mm3 mm4 rx =ry Area Ag (mm2) mm Designation Weight/m 8x8 7.94 47.36 6,039 79.25 37.84 371.99 60.35 expla'n briefly your cho'ce. (transform your comparative analys's 'nto a narative form to support your cho'ce) 8x8 14.29 80.61 10,258 76.2 59.52 585.02 99.06 8x8 72.7 9,290 76.96 54.53 539.13 90.32 8x8 9.53 56.09 7,161 78.48 44.12 432.62 70.76 mm 12.7 Zx 103 mm3 437.53 714.48 650.57 512.92SITUATION 2 Determine the safe service load W permitted for this beam-column that is not part of a frame system. Assume ASTM A992 (Fy = 350 MPa; Fu = 450 MPa) as material. Use LRFD specifications. Servise koada P= 30 kips dead load 80 kips live koad Assume hinged for both principal 10- 0 directions W- 20 dead load 8OG live load 10-o Fined for both principal directions Type your final answer/s in the text box provided below.
- 4.3-4 Determine the available strength of the compression member shown in Figure P4.3-4. in each of the following ways: a. Use AISC Equation E3-2 or E3-3. Compute both the design strength for LRFD and the allowable strength for ASD. 15 HSS 10x6x ASTM A500, Grade B steel (Fy=46 ksi) 2/32) Find the axial stresses of menbers FD, GD, GE State if it is tensile or Compressive. 4M 3M A LE 3m G 20 RN Go KNO 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 condition
- Use AISC Equation E3-2 or E3-3. Compute both the design strength for LRFD and the allowable strength for ASD.PLEASE WITH FBD. THANK YOUASSIGNMENT #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 581
- HOMEWORK-1 3m IF For the truss shown in the figure; Design members 1-2 and 2-3 assuming that out of plane deflections are restrained and Ner>>10, i.e. global buckling is not the critical failure mode and therefore linear static analysis is sufficient. P.S. Use square rectangular hollow sections in Grade 355 steel. You will have to find the related section tables from the internet. Do not forget to refer to the related EN1993-1-1 tables and equations. Later, you will design the connections (the weld lengths, etc.) and the supports for this problem.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 Input Yes or No for your final answer. Blank 1 Blank 1 Add your answer