4. A W 14 x 82 column is 10 feet long and has pinned end conditions (K 1.0). If it is braced at its mid-height in the weak direction, determine its design capacity using the ASD and LRFD 1 methods. Use Fy 50 ksi and E = 29,000 ksi. 5. Using the Column Tables found in the Appendix, select a W 12 section to carry a factored load, Pu, of 1000 kips given that the column unbraced and has a length of 14 feet. Use K 1.0 and Fy 50 ksi.(LRFD) 12
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- The given girder has beams framing into it at the ends and at every L/3 point. The beam carries a service live load of 20 kips as shown and superimposed uniformly distributed service dead load of 10 kip/ft. Select the lightest A992 W-section that can carry the load. Do not check for deflection. P, = 20 kips LL PLL W. = 10 kip/ft DL = 20 ft WDL L/3 BEAMS FRAMING INTO GIRDER LVerify the adequacy of column AB, part of a sway frame structure, to carry the loads shown in the figure. The column is made of ASTM A36 steel (Fy = 250 MPa). Use ASD specifications. For determining its effective length factor, neglect inelastic effect. %3D W21X62 C W21X62 L=5 m L=5 m Flanges of Columns and web of girders are in the plane of the frame NOTE: Take note of the moment of inertia to be used (ly) for columns. W18X50 W18X50 L=5 m L=5 m P(DL) = 40 kN NM 001 = (10)d NM 00S = (1)d NY SZI = (1)d W21X62 L-4.2m W21X62 L=4.2 mP = 800 kips The beam column shown is braced against sidesway with bending about the strong axis. The beam-column is made of a W 30 x 108 section of A992 steel. If lateral support is provided only at both ends and both ends are fixed, what is the maximum permissible applied distributed load (@,) that this member can carry? 20 ft @y = ? k/ft W 30 x 108 A 992 Steel P= 800 kips
- TASK 3 250 180 14 I 30 24 Weld 450 A compound girder consists of a steel joist with steel plates welded onto each flange as shown. If the ends are simply supported and the effective span is 10 m, what is the maximum UDL which can be supported by the girder? [kN/m] Allowable longitudinal stress in plates = 110 MN/m² Allowable load in shear for each weld = 60 kN/m Allowable shearing stress in web of girder = 75 MN/m² Note: 1.0 MN/m² = 1.0 N/mm² 50Determine the safe load P that the member below can carry. Use A-36 steel and NSCP 2001 PL = 1/8x 10 Fy = 50ksi kL = 20ft MC 8 X 8.5A plate girder must be designed for the conditions shown in Figure P10.7-4. The given loads are factored, and the uniformly distributed load includes a conservative estimate of the girder weight. Lateral support is provided at the ands and at the load points. Use LRFD for that following: a. Select the, flange and web dimensions so that intermediate stiffeners will he required. Use Fy=50 ksi and a total depth of 50 inches. Bearing stiffeners will be used at the ends and at the load points, but do not proportion them. b. Determine the locations of the intermediate stiffeners, but do not proportion them.
- 1. A steel column 10 m long is fabricated from a cover plate and C section arranged as shown. Determine the safe compressive load. Fy = 248 MPa, E= 200 GPa. Use AISC/NSCP Specs. 450 mm -cover plate 'I 12 mm y2 IP d2 10 m C 310 x 37 A = 4720 mm? d = 305 mm bf = 77 mm tf = 12.7 mm tw = 9.8 mm C 310 X 37 a) Both ends of column are fixed b) Both ends of column are hinged c) One end fixed, the other end hinged Use design values of k. tw d=305- Ix = 59.9x10° mm ly = 1.85x10° mm x = 17.1 mm x=17.1Design the flexural reinforcement of the cantilever beam shown below for the given loading and data. The loads shown include the self-weight. f. = 28 MPa ; fy 420 MPa Main Reinforecemnt : 020mm ; Stirrups : Øø10mm P= 70kN -1000mm W= 100 kN/m 150mm 450mm -2.0m 400mm Page 4Determine whether a W24 x 117 of A992 steel is adequate for the beam shown. The uniform load does not include the weight of the beam. Lateral support is provided at A, B, and C. Use LRFD. P= 12 k P = 36 k WD = 1 k/ft WL = 3 k/ft 10 20- 30
- Determine the nominal axial compressive strength, Pn for the following cases: a. L = 15 ft b. L = 20 ft Note: Pinned both end support W10× 33 r = 1.94 in E = 29000 ksi Fy = 50ksi A992 steel %3D Note: Ag = 9.71 in^2 a) Blank 1 b) Blank 2 Blank 1 Add your answer Blank 2 Add your answerA 16-foot-long single overhang beam is loaded as shown. Assuming a W8 x 35 determine the maximum bending stress developed. The informally distributed load of 3 k/ft include the weight of the beam. Given: A992 steel Fb = 30 ksi.For W8 x 35 section properties see Table A3 (Textbook Appendix) (Provided)Compute for the force in members FH, GH, and GI (USE METHOD OF SECTIONS) *For load P, take the last digit of your ID number example: ID # : 2187525, then P= 5 kN IF ZERO IS THE LAST DIGIT, TAKE THE 2ND TO THE LAST DIGIT example: ID #: 2187520, then P= 2 kN