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'
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- The uniformly distributed live load on the floor plan in the figure given below is 65 lb/ft. Consider the live load reduction if permitted by the ASCE standard. I B2 (B) I G3 6 @ 6.67' = 40 B4 G4 BI GI C2 G2 C3 Establish the loading for girder G4. The loading (P) for girder G4 is 20'- kips. C4 I 2 @ 10' = 20 5 @ 8' 40'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)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 I B2 G3 C-I -6 @ 6.67' = 40 -6 @ 6.67%; B4 G4 B1 40' I G1 C2 G2 C3 Establish the loading for floor beam B4. The loading (W) for floor beam B4 is B3 20'- (3 C4 -I 2 @ 10' = 20' 5 @ 8' = 40' kips/ft.
- 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 B4Problem 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"The steelwork framing plan for a floor system shows decks as D1-D12, beams and girders as M1- M20 and columns as C1-C9. The unit load on the system is 75-psf dead and 45-psf live. What is the total load going into column C2? C3 C1 C2 I. M1 M2 D3 M8 M3 M4 M5 M6 D1 D2 D4 36' M7 D5 C5 C6 C4 M9 M10 D6 M16 M17 M18 M19 M11 M12 D7 D10 D11 D12 40' M13 D8 M14 D9 I. C9 C7 C8 M20 M15 24' 16'
- 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-men²Q2 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)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 Controlled
- Q2) For the slab shown in Fig.(2),find the allowable live load for the section 1-1. Use f'c-25 MPa, fy=350MPa, h= L/24. The beams dimensions are 250mm*600 mm. 8.5 E # # * 8 # 3.5m 4 1 11 # 1 11 1 B 1 1 1 1 4 1 t 1 I I 1 1 I 3.5m 1 1 I 1 1 1 1 I 1 S 1 9 12@500mm 1 10@500mm Fig. (2) 250mmure B 2m2 m C -2 m D 4 m 500 N 700 N 2 of 2 Part B Identify the zero-force members in the truss shown in (Figure 2). Check all that apply. 000 CD DE CG CF FG BC BG DF AG EF AB Submit Provide Feedback Request AnswerWhen the columns at E and H of the floor framing plan shown are deleted, girder BEHK becomes one span fixed ended beam supporting beam DEF at E and beam GHI at H.The following loads on girder BEHK are as follows: Concentrated load at E = 277 kN Concentrated load at H = 266 kN Uniform load throughout its length = 8 kN/m 2.5 m 2.0 m 2.0 m A D G 6m B E H K 6m C F Calculate the maximum positive moment (kN.m) due to the uniformly distributed load only. (TWO DECIMAL PLACES)