Concrete Footing Design

Transcript

**************************************DESIGN OF FOOTING************************* Carmel B. Sabado Prof. Geronides P. Ancog CE-162 10-Aug-09 I.Column details: 1.(type the letter corresponding to the shape of column) a A. Square B. Rectangular C. Circular 2.Dimensions of column: (put the values in the green colored spaces provided) a=b= 600 mm *length and width of the column are equal *not applicable *not applicable *not applicable II. Information Needed 1.Material Properties: f'c= 27.5 Mpa fy= 414 Mpa 2. Load to be carried: 1580 KN PDL = PLL = 1200 KN 3. Allowable soil Pressure: 285 Kpa 4.Shape of Footing to be Designed: (type the letter corresponding to the shape of footing) A. Square B. Rectangle length= width= B= 5. Diameter of Bar Reinforcement: rebars long φb= rebars short φb= 6. Assumed Footing weight: 3.3 4.4 m 2.4 m 2.4 m 25 30 mm 20 mm b *longer dimension of the footing *shorter dimension of the footing *reinforcement in the longer dimension *reinforcement in the shorter dimension 7 %Qu III. Design Computations: 1.Dependable Ultimate Soil Bearing: 1. 2 D L1.6 L L qa D LL L = 391.21 kPa Q u =1 . 2 DL 1 . 6 LL = 3816 kPa q u= 2.Required Footing Area: 10.44 m² Actual Footing Area 3. Net soil Pressure: = 10.56 m² *design footing area is okay 369.8 kPa 4. Critical Section of the Footing: 1. Footing depth as controlled by BEAM SHEAR: 1 B−a vc =  f c ' −d 6 2 B−a V u =q n L v =q n −d 2 Lv =  φv c = Vu bd ; d=  = 0.87 Mpa a= b= 600 mm 1000 mm * consider a 1 meter strip V u *with these working equations we can get the value of d: φv c b   B−a −d 2 d= φv c  b=1000  qn in the short dimension trial d: computed d: 320 288.71 304.35 296.49 300.42 298.45 299.44 298.94 299.19 299.06 in the long dimensio trial d: 678 643.14 634.38 632.18 2. Footing Depth as controlled by Punching Shear vc = 1  f c' 3 = 1.75 Mpa Vu = qn[Af-(a+d)²] b = 4(a+d) φv c = Vu bd ; d= [  qn A f − d= Vu φv c b  a2d  1000 ] 2 x 1000 *with these working equations we can get the value of d:  φv c  [ 4  ad  ] trial d: computed d: 452.52 471.15 473.19 473.4 473.42 ***beam shear controls! 500 476.26 473.71 473.45 473.42 Reinforcements in the long Direction: Ll'= (L-a)/2= 1.9 MuL = qn(Ll')²/2= 667.48 Trial dl= 680 mm 5.Design as Singly Reinforced Rectangular Beam: [ ρ min = 1.4  f c' , f y 4f y ] Reinforcements in the short Direction: Ls' = (B-a)/2= 0.9 Mus = qn(Ls')²/2= 149.77 Trial ds=dl-((φbl/2)+(φbs/2)) 655 use b=1000mm 0 min ρ max =. 75 ρ= [  [ . 85 f c ' β 1 . 003 E s ] f y . 003 E s  f [  2R u 2ωR u . 85 f c ' 1 = 1− 1− 1− 1− fy .85 f c ' ω fy y ] ] = = Ru= ρ= As= Rebars= = N=int([B or L]As/Ao+1) 0.02 in the long direction: 1.6 0 ok! 2731.61 mm2/mm 30 mm 10 bars Check validity of assumed footing weight of : Total depth : h= d+100 = Wf = (1.40) B² h wc = Actual footing weight : Assumed footing weight : %Qu= in the short direction: 0.39 0 okay! 618.84 mm2/mm 20 mm 9 bars 780 mm 272.14 Kn 267.12 Kn Wc= revision should be done! 23.6 Kn/m3 6. Short Rebars Distribution: Ast= As=Ast(2/(L/B)+1)= central strip= No. of bars at the outer strip= 2827.43 mm2 1995.84 mm2 7 2 -20 mm φbars -20 mm φbars 7 20mm φbars 2 780 mm 2.4 meters 0.8 meters END OF THE PROGRAM!!!=) ******NOTHING FOLLOWS****** in the long dimension computed d: 608.27 625.63 629.99 631.08 mm 20mm φ bars 680 mm 100 mm FOOTING REBARS: #REF! #REF! 100 mm 0 #REF! #REF! #REF! #REF! #REF! FOOTING DIMENSION: #REF! #REF! 0m #REF! #REF! COLUMN DIMENSION: #REF! #REF! 0m #REF! #REF! #REF! 0 mm 100 100 mm mm #REF! #REF! #ΡΕΦ! #ΡΕΦ! #ΡΕΦ! #REF! #REF! #REF! 100 mm #REF! mm 100