Design a simply supported beam of span 4.2 m carrying reinforce concrete floor in which top compression flange is embedded. Beam is carrying 20 kN/m dead load and 20 kN/m imposed load, resume Fe 410 grade steel.
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- 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 B4ASSIGNMENT 4 (Prelim) Ultimate Strength Design 1. A simply supported beam having a span of 7.5m carries a uniform dead load of 16KN/m and a live load of 19KN/m. i has a width of 400mm, fc-21.7 MPa and Fy 275 MPa. Steel percentage is 0.8% and clear concrete cover is S0mm. Diameter of stirrups 10mm, a Compute the effective depth of the beam. b. Compute the number of 20mm0 bars. c. Compute the maximum spacing of the 20mm0 bars center to center. d=570 h=630.0 20mari 00000 60.0 400Q # 2. Design an interior span slab of a concrete floor system with the following description: Span = 10 m Imposed dead load= 766 N/m² Live load= 4788 N/m? fc'= 27.58 MPa fy = 413.64 MPa (CLO2-C5/PLO3) (15)
- Total load 87 psf = B1W16 x 57 G1W24 x 68 Compute Vmax and Mmax for Beam 1 and Girder 1 in the roof framing plan. 28' H I H- B1 6 @ 6' = 36' B1 B1 B1 B1 | G1 I- I-The uniformly distributed live load on the floor plan in the figure given below is 65 lb/ft². Consider the live load reduction if permitted by the ASCE standard. A (B) B2 G3 -6 @ 6.67' = 40- I B4 G4 B1 40' G1 C2 G2 C3 Establish the loading for girder G3. The loading (P) for girder G3 is [ B3 20' kips. (3) 2 @ 10' = 20' I+ 5 @ 8'=40'Consider the same beam as in Problem 1. It carries a live load of 3 kips/ft. 8" 8" V (kips) wl M (kips) 9 w/² 128 6" 6 #6 7.5 ft. 7.5 ft. A WL =3 kips/ft. 1 =20 ft. 6 #9 500 8 5 ft. B wl 8 -W/2 B
- A singly reinforced rectangular beam section 250 mm wide, 450 mm deep is reinforced with 2-28-mm-dia. rebars, grade G420 enclosed with 10-mm-dia. stirrups. What is the design condition of the beam by USD method? Use f'c = 21 MPa A. Transition Failure B. Balance Strain Failure C. Compression Controlled D. Tension ControlledDesign a short tied column service dead load moment = 400 MPa and steel ratio Pg =0.03. Use 2015 NSCP specs. also carry a for a service DL = 1200 KN and a service LL = 890 KN. It will of 35 KN-m and a service live load moment of 28 KN-m. Use f'c = 30 MPa, Fy1. Check the beam shown for compliance with the AISCS. Lateral support is provided only at the ends, and A992 steel (E 345 MPa and Fy= 450 MPa). The only uniform service dead load is the weight of the beam. The 70 KN service loads are 30%DL and 70%LL. Use LRFD. 70 KN 70, KN 70 KN W 14 x 68 0.90m 1.20m 0.90m WI4 x68 IR, R2
- Note: please use Set B data thank you 2° M₂ A B * 4 m 2m The cantilever beam above is made up of concrete with E- 70 MPa, refer to the table below for the properties and loads for the beam: PROPERTIES LOADS SET b (mm) h (mm) w (kN/m) Mb (kN-m) A (1,3,5,7,9) 300 500 20 40 B (2,4,6,8,0) 250 400 15 45 Using area moment method, determine the following: Flexural rigidity of the beam in N-men²Design the lightest A992 W section possible for the temporary construction loading only. Tkae the beam span to be 24-ft and simply supported. These typical floor beams are 5 feet on the center. The concrete slab is 4 inches, normal weight, f'c = 4ksi and will be placed directly on the beam. Assume that the formwork and the beam self-weight is 10 psf. Apply the concrete ponding factor that steel deck institute (SDI) recommends during the concrete placement and treat the concrete as live load. Take the additional live load during the construction as 20 psf. Assume that the beam has continuous support support along its compression flange. Check the beam for flexure, shear and a maximum LL deflection of L/360.Problem 2: Determine the allowable service live load, WL on a cantilever beam assume the dead load is due to the beam weight f'c = 4 ksi, fy= 60 ksi, Dc= 150 pcf. 5" 25" 4-#8 10" 10" W₁ ▼ ▼ 10"