Question

15.3. psi. Design a single-column footing (including dowels) to support an 11 in. square column reinforced with eight No. 9 (No. 29) bars centered 2.5 in. from the column faces (equal number of bars on each face). The unfactored axial dead load = 135 kips, and the unfactored axial live load=125 kips. For the column, f=4000 psi and f=60,000 psi. The base of the footing will be 3 ft below grade. The allowable soil bearing pressure is 3000 lb/ft2. Material strengths for the footing are f' = 3000 psi and f, = 60,000 psi.

Answer 1

2.7 Assume overall depth of footing = 24 in da 24-4.5=19.5 in V 3000 - R4X 150 = 2400 psf = 2.7 KSE Area required 155 +125 = 96.8 ft 9.8 ft x 9.8 ft footing = 96ff Bearing pressure for strength design - 1.2135 +1.6x125 = 3.77 ksf 96 Depth required for two way shears in use vo

19419.5= 30.5n, =20542 와 없 + 더

bo = 4x30.5 = 122 in (96 - 2.542") * 3.777 = 337.56K Vuz q=0.75 0.47 337560 0.75 X 4X1X √3000 x 122 nal (Normal weight conerete) fe - 2000 psi = 16.84 in < 19.5 in . Ok. vuz d = f P (asa bo 2) arti so bo X = 40 for interior column 337560 d. 0.45 ( 40319.5. +2)x1 x V 3000 x 122 122 = 8.02 in. < 19.5 in ok. Punching shear is ok.

Development length Y t = Ye = Ys Borrs spaced 14 in on center leaving & in on each side. G = bottom cover = 3.5 in de = 1.128 A G = 3.5 in (#9). Assume kao G = x 14 = zin . C & ker 3.5t o = 3.172.5 1.128 use 2.5 3 fo 40 do afker di (XIX) 3 40 60000 1x3000 2.5 32.86 diameter As, read As, prov 32.86 x do 7.56 8 = 3 diameter da = 31% 1.128 = 35 in. Available l& 4.9x12 - 1 3 2 & 50.3 in >l = 35 in ok

Design of dowels Bearing force at the base of columns = 1.2 x 135 tl.68 125 & 362K Allowable bearing force in concrete at base of column = 8 x 0.85 tét = 0.65 X 0.85X (4x 267.41 k < 362 k, Not good. x (11x11) Design bearing strength on footing concrete = 8* 0.85&A (SALA) = 10.6992. A2/A 9.8 x 9.8 MIX use 2. = 0.65 X0.85*3 X 11 X 1 X 2. 11 401K 7362k Froting bearing is ok ,

the dowels must be designed for excess load Excess load 362-267.41 94.sk 94.6 -1.75 in 0.9 x 60 (1.77 in ina). Ą of dowels inn ube 4# 6 bar

ld Development lengths for dowels for the columm use a=l, f = 4000 psi 0.02d, fx 0.02x61)*60000 1x27000 = 14.23 in for the footing R=1, t = 3000 psi 0.02 X 0.75 X 60000 = 18.43 in 1 x 3000 In addition, the development must not be less than either lg = 0.0003 de ty = 0.0003 X 0.75 X60000 = 13.5 in la = 8 in. 4 #6 dowels extending 15 in up into the columm and it in down into the footing. ld use