Determine the ultimate moment capacity of the beam in kN-m.
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- HiW@ Draw S.f.n and BiMin for beam shown by Naphial methds Anst hinge %3D M= 9 W.m 3m 2m 3m Mmny = -6 IW.mA For the truss shown, find veyticat deflection bylunit lead Methed. EA 15 _atfd) Ioku for all members increase in temp is (50€) in (ad) short in (ad)is (3 EmfrThe prestressed I beam shown in cross section is pre tensioned using seven ordinary strands grade 250 (fpu = 1728 MPa) carrying an effective prestress fpe = 988 MPa, fpy =1480 MPa. %3D 300 0.85fc' 150 a 300 440 600 (d-a2) 300 T=Aps fps Aps =650 mm2 Total depth of beam = 600 mm Distance from center to tendons to the top of the beam = 440 mm Aps = 650 mm² %3D Average flange thickness = 150 mm Width of flange = 300 mm %3D Thickness of web = 100 mm %3D fc' = 27.6 MPa Determine the value of the stress in the prestressed reinforcement at nominal strength. (MPa)
- I the Design the reinforcements of the given T beam below. bf=900mm bw=420mm tf=120mm d=550mm d'=80mm fc'=34MPA fy=415MPA USE NSCP 2015 mm2 A.Mu = 1400KN-m, As = mm2 B.Mu = 1700kN-m, As = mm2, As' = CS Scanned with CamScannerTABLE 8.5 Allowable Spans in Feet and inches for Low-Slope Raftern, Slope 3 in 12 or Less (No Celling Siat Spacing IN) Allowable Extreme Fiber Stress in Bending F (pai). 1200 500 600 700 800 900 1000 1100 1300 1400 1500 1600 1700 1800 1900 11-7 13-0 0.94 13-7 109 11-9 0.94 9.7 14-2 124 12-4 14-9 154 15-11 16-5 191 14-2 16-11 17-5 2.28 15-1 9-2 10-0 17-10 10-10 0.55 9-5 12-4 0.67 O80 10-8 1.56 13-3 12.0 2.09 033 7-11 0.29 6-6 0.44 140 12-10 1.73 13-9 247 156 10-0 058 8-2 113 0.82 9-2 8-8 14-8 1.35 10-10 0.70 107 10-0 121 10-5 0.99 19-6 1.50 1.65 11-7 181 11-11 1.97 12-4 214 12-8 2x6 160 0.38 7-1 0,48 7-8 0.39 8-8 11-3 1.22 20-11 L73 18-2 0.24 12-1 0.33 10-6 0.29 8-7 024 15-5 0.33 13-4 0.31 13-3 0.44 11-6 0.48 15-3 0.57 16-3 0.67 17-1 0.77 17-11 0.88 18-9 L10 20-3 1.35 21-7 148 22-3 2.09 1.75 22-11 23-7 2.28 24.0 1.61 144 067 13-3 0.94 14-10 1.09 15-6 124 16-3 140 16-10 121 13-9 0.99 24-10 1.56 17-6 12.0 0.55 12-5 0.48 10-1 0.39 0.80 14-0 191 18-9 2.47 19-10 20-5 19-4 0.70 I1-6 2x8 16.0…For the column shown what is the nominal compressive strength (Pn) ? Pn=? W12x50 10 12X50 THT X combilevered WI
- "messenger_media?thread_id%=D100002917920802&attachment_id%3891210971774069&message_id%3Dmid.%24CAAAA Question 3 For the truss shown, determine the forces in members (a) BD; and (b) BF. 4 m4 m4 m-4 m- A B C 6 m E F G H O00 20 kN 20 kN 20 kN A BD-40kN (C); BF-33.33kN (T)A76 x 76 x6 mm angular section shown is welded to an 8 mm thick gusset plate. The length L1 is 65 mm, L2 is 125 mm, and the cross sectional area of the angle is 929 sq.mm. Fy-248MPA and Fu -400MPA. Gusset Plate Alowable Sresse Alowable tensle stress igross ara)0.00 Alowable tereile stres net area)0SOF. Allowable shear stress (net ares)0.30F. Determine the value of P based on net area using a strength reduction coefficient of 85% O 138.24 KN O 140.14 kN O 157.93 kN O 304.00 N NextEQuiz No.3 584nkkz1 Cc7KLzOsdr9YUFgvcfFSqpsdlyljtewQ/formResponse Choose the correct answer What will be the value of load applied parallel to the diagonal of (100x100x100))mm cube since the splitting tensile strength=3 Mpa 57.8 KN O 46.7 KN O 47.1 KN O (28)=27.5 Mpa, then the fc(180)=-- for concrete produced using moist
- Locate the centroid (x, y) of the composite area. y -3 ft -3 ft- - 1.5 ft 1 ft O a. 2.1ft, 3.2ft O b. 2.1 ft, 3.1ft O c. 1.2ft, 1.2ft O d. 2.11ft, 1.34ft O e. 1.2ft, 3 ftPabacboadcom Question Campletion Stutn MOD LE 4 A Movtng to another question will save this response Question 1 A hwo span beam subjected to shear and flexure only is reinforced as follows: @ MIDSPAN 2-016 mm 3-016 mm SECTION @ FACE OF SUPPORTS TOP BARS BOTTOM BARS 5-016 mm 2-016 mm Given: Stirrup diameter, dg - 10 mm Concrete fe=21 MPa Steel rebar fy 415 MPa Stirrup fy= 275 MPa Beam sizebxh=250 mm x 450 mm Assume all bars laid out in single layer. Calculate the following: Tensile steel ratio in positive bending at midspan = (in 5 decimal places) Design Moment strength of section at midspan for positive bending = kN-m (nearest whole number) kN m (nearest whole number) Nominal Moment strength of section at face of support for negative bending =AW18 x 40 standard steel shape is used to support the loads shown on the beam. Assume P = 19 kips, w = 3.8 kips/ft, LAB = 5.2 ft, LBc = 5.2 ft, and LCD= 14.4 ft. Determine the magnitude of the maximum bending stress in the beam. %3D D B LAB LBC LCD Answer: ksi Omax