Question

o Design the shown cantilevered beam , fc'=28MPa and fy = 420 MPa. Use p=0.5 pmax DL = 14 KN / m LL = 16 KN/m BIN 3.5 m

Answer 1

solution 프 X 3.5m - 16 KNIM DiL 14 KNIM L-L fy 420 MPa fé Clear span - 8 28 M Per Assuming Calum wirdth 스 0.5m Assume, ba 0.25 m = 250m :. left Clear span & Column width. 3.575 =4m. = 12 for Cantileverdest depth left. = 12 x modification factor mezdification fallor = 0.75 4 12* 0.75= 0.444 m 448mm Effective cover, d'= Nominal covent Shrrups edia of main bay) - (45 +10 +(2) 67.5 . deff Depth = d+d' = 444+67.50 = 54.5- assomm D-d'- (550- 675) em het in 482,50m 0.4825 m load Calculation ; selft wt = bxdefft 24 = 0.95 x 0.4725724 - 2.9 KN 14 KN L-L D'L √6 KN 32.9 kn total load =

= 49.35 KN. Fachored lead & 1.5x 32.9 moment in beam 2قيا 99.358 42 2 2 My 394.8 KN-M Shear in beam 99.35x4 به رقبا - 19 7.70 KN Mulim 0.36 fcx š Kulima bx (d-0.416 Hulim). stalim =0.5 desse - o'yx 0.48455 = 0.241 = کمر پر ہر Mulim = 0.36 * 288 841 x 050 ( 482.5~0.416 * 241) 232144426 de mim 232.1 KNM = hence we here, Mulim < Mu (over reinforced) we wse Mulim for design Purpose: 0.5 feat bo 4.6 Merlin Ast = fek bola 4.6X232.1428 8o5o X20X250X 492-5, PELA 28 X250X Ass = 420 (482.572 Ź 1660 mm (Asphrey (Asting 1660 No. of bars Area of Ikar EX X 252 ~ 3.38 ~ 4 bars Ast provided = 4x 1 x 252 = 1963.5 Mesut 100x Art % of steel bxd

% of steel Loo X 1963 280x 982.5 a 1.62% Check for shear : Vu bel 197.4 x los Tv =1:6 - 250X 482.5 Temane = The Temen , 3.5 (for fck= 28 ) , 7c = 0.76 But, minimum reing to provided A sv .. Minimum teny. Spacing = 0:4 bers 0.87fy O-4 100.18 250Xsy 0.87X420 ~ 360 mm Minimum Sv num Spacing Spacing Reinforcement : is 300mm is 0-75xd 50-75% 482.5 362 mm minimum reinforcement spacing = 300 mm Deflection calculation! 8:3 Span - Actual > deft 4000 482-5 Required Span Depth = 124 modification factor =12 X 0.75 =9 Actual /< Rey (6), hence * eur section is safe and not deflect