Question

The beam shown in the figure is loaded with a distributed dead load G (including self-weight) and a distributed live load Q. Additionally, a point load Pq is acting on the structure. aPa d. The following values are used for this singly reinforced beam 29 MPa 500 MPa A 280 mm Ag 1350 mm Po 90 kN d 540 mm G 12 kN/m d. 60 mm .Q26 kN/m . b.350 mm I 7m do load factor for Pa .S300 mm a 2m Use the simplified approach: p-0.75, k,-0.15, 8, -36 1. Compute the distributed factored load w and the factored point load doPa at the critical section for shear. Only consider loading pattem depicted in the figure (no load combinations, no shear envelope). w 53.4 kN/m, agPa 135 kN. (1 mark each) 2 Assume that the beam rests on a column (typical beam-column joint). kN. (2 marks) The shear force V at critical section 3 The shear design strength pV 289.087 kN. (5 marks)

Answer 1

mãs B 71B Pe = 90 KN G= 12 kN/m = 26 kN/m &= 7m a=2 m Agv = 280 mm? d= 540 mm dc = 60 mm Agt = 1350 mm? 5= 300 mm by = 350 mm E = 29 MPa Py = Pyz = 500 MPa 0 = 0.75, Kv = 0.15, =36° I w* = 1.86 +1.50 = 1-2(19) + 1.5(26) = 14.4+ 39 = 53.4 kN/m La pe = 1.5x90 = 135 KN 2) EFG =0 VA TVB = W*2 + dapa VA + V8 = 53-4x7+135 VA + 16 = 508.8 kN

ΣΜΑ 20 VBXl = do Paxa + wx 2x de = 135 xq + 53.4x7x3.5 3) Vo = 225.47 KN VA = 283.33 KN Critical Section lies at id from A. V' at oritical section = VA-w*xd = 283.33–53.4*0.54 = 254.494 KN over y = ky budr JFC = 0.15 X0.35 x 0.54*J 29 x106x0.9 = D: 13740 X106 N. = 137.40 KN Vous = Asy Pyt dy cob (@) = 280X 500 X 0.7x 540 Cok (34) - = 0.312163 X 106 N = 312.163 kN QVc = $(Vue + Voz) = 0.75 13.40 + 312.168) = 337.17 kN 280 X 500 X 0.9 x 540 300 S