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" ▼
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- 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"Q # 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)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-men²
- The service load bending moments on a beam are 128 kip-ft for dead load and 107 kip-ft for live load. The beam is 12 in. wide, f'c is 3000 psi and fy is 60 ksi. Determine Mu. Mt. p. R. the depth of the beam and the As of tensile reinforcing required. How many no. 8 bars or No. 10 bars would have the required As?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 B4Calculate the quantity of steel required for stirrups for the following beam. Y -2 - 12 Dia Anchor bars 4 - 16 Dia 8 mm Dia at 200mm C/C 2- 12 & 2 -16 Dia 650 mm 650 mm X Y 8 mm Dia at 350 mm C/C Clear Span 4000mm Clear Cover Bottom= 25mm Top= 25mm Ends=32mm Sp.Wt of Steel =7860kg/Cum 2- 12 mm Dia 2- 12 mm 2-16 mm Dia 2- 16 mm Dia 4- 16 mm Dia 550 mm 230 mm 1944.7 mm 550 mm -
- Problem 1: Determine the allowable service live load, PL assume the dead load is due to the beam weight f'c = 3 ksi, fy = 60 ksi, Dc = 150 pcf. 25" 5" 4-#8 10" 10" PL 10" PL 10"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'Two beams have the same section. Calculate maximum factored loads as shown below for case (a) and (b) aj bj 1 8m 8m 2-20M SO 5 I A Section for (a) and (b): 350mm BB fé= ↓ Pman = ? 0 ooo 4-25M = 30MPa fy=400мра &man = ? 450mm IS A B 650m ↑ 200mm
- load, 2 wat O: A fixed and beam is subjected to a load, 1/302 span from the left support as shown in the figure. The collapse load of the bean is 2W *4/3- X₂ Me 2mp71 Y K4/₂4/₂ Date: ZDesign 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 4Q2 a) This is a concrete beam (the interior lines represent reinforcing bars), If P is live load, draw the BMD and the SFD as a result of ULTIMATE LOAD. P= 50 kips 20 ft. 20 ft. 40 ft. Own weight of the beam = 1.0 kip/ft. (Uniform dead load) b) This is a concrete frame (the interior lines represent reinforcing bars), If P is dead load, draw the BMD and the SFD as a result of ULTIMATE LOAD. 6.0 ft. P= 1.0 Kip Own weight of the beam = 1.0 kip/ft. (Uniform dead load)