Q2 Figure Q2 Shows A Part Of Floor System Consist Of Beams And Suspended Slabs The Beam Sizes 250 Mm X 500 Mm Are Conti 1 (19.28 KiB) Viewed 33 times
Q2 Figure Q2 Shows A Part Of Floor System Consist Of Beams And Suspended Slabs The Beam Sizes 250 Mm X 500 Mm Are Conti 2 (14.23 KiB) Viewed 33 times
answer for 2a
Q2 Figure Q2 Shows A Part Of Floor System Consist Of Beams And Suspended Slabs The Beam Sizes 250 Mm X 500 Mm Are Conti 3 (30.87 KiB) Viewed 33 times
Q2 Figure Q2 shows a part of floor system consist of beams and suspended slabs. The beam sizes 250 mm x 500 mm are continuously spanning 6 m between columns. The characteristic material strengths are f= 30 N/mm² and f = 500 N/mm². Determine the effective flange width of beam A/1-2 and B/1-2. (b) (c) (14 marks) Based on the analysis in Q2(a), design the main reinforcement for the heam A/1-2 if the bending moment, M is 800 kNm, effective depth is 430 mm and the depth to compression reinforcement is 50 mm. (14 marks) Verify deflection of the beam A/1-2 according to EC2. (7 marks)
6000 mm. X # (A) h = 100 mm 3000 mm (B (B) X-X FIGURE Q2 3500 mm C X
Consider beam edge L-bean. h = 100 mm L = 6000 mm bw 4/2 sparing = 250 mm effective flange width in smallest of = 250 + 6000 4 250 + 6×100 bw + 3) bw + Ghe - (Clean, distance) = effective flange Consider A/1-2 bw 250mm cle spacing (left) Internal T-beam =100mm he L = 6000 mm 14/ 1. c/c spacing (Right) Effective flange beam B/1-2 % e d. bw + 16 hp 3000 mm = 250 +1 A width be 3000 mm 3500mm 6000 = 1500 mm + (2000-250) = - 1750 mm = 850mm width is smallest of 3. c/c spacing (minimum) (тілінии) Effective flange = 850mm 250+ 16x 100 = =1625mm 3000 mm width (be) = 1500 m 1850 mm
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