A water tank is resting on 1.5 ft x 1.5 ft concrete column. Neglecting the later

Question

A water tank is resting on 1.5 ft x 1.5 ft concrete column. Neglecting the lateral soil pressure and using the attached drawing, Meyerhof's Bearing Capacity Equation ( General Bearing Capacity Equation ), and the following data, determine the size of square footing that can safely transmit the loads to the soil.
- Tank is made of 1/2 in thick steel 10 ft by 12 ft walls and 12 ft by 12 ft base. Neglect the weigth of the cover and the support structure for the tank.
- Unit weigth of concrete = 150 pcf ( to be used to calculate the weigth of the column)
- Wind pressure on the tank ( East- West Direction Only) = 13.89 psf

- FS = 6

Explanation / Answer

Given :

A water tank is resting on 1.5 ft x 1.5 ft concrete column.Tank is made of 1/2 in thick steel 10 ft by 12 ft walls and 12 ft by 12 ft base.

Unit weigth of concrete = 150 pcf.

Wind pressure on the tank ( East- West Direction Only) = 13.89 psf. and factor of safety, FS = 6

for layer (1), upto the base of footing ;

unit weight = 100 pcf, c=0, = 300 , Df = 10 ft.

and for layer (2), 15 ft below from the base of footing ;

unit weight = 165.36 pcf, c=0, = 300

water table at base of the footing.

take, unit weight of steel = 490 pcf and unit weight of water = 62.4 pcf.

partial safety factor for combination of DL + WL+ load due to weight of water = 1.5

Step :(1) :calculation of ultimate bearing capacity, qu of soil :

from the Meyerhof's Bearing Capacity Equation (General Bearing Capacity Equation) ;

qu = cNcscdcic + q'0Nqsqdqiq + (1/2)LNsdi

where,

c =unit adhesion

Nc ,Nq ,N = bearing capacity factors

sc, sq, s = shape factors

dc, dq, d = depth factors

ic, iq, i = load inclination factors

L = width of footing

q'0 = effective overburden pressure at the base lavel of the foundation.

asuume L < 15 ft. (because for shallow foundation Meyerhof's Bearing Capacity Equation is valid upto L ft below from the base of footing).

here, c = 0

therefore, qu = q'0Nqsqdqiq + (1/2)LNsdi

at = 300

Nq= 18.4 ; N = 15.1

depth of foundation = 10 ft and unit weight = 100 pcf

therefore, q'0 = 100*10 = 1000 pcf

ic=iq=i= 1 because there is no load inclination.

sc = 1+0.2tan2(45+/2)B/L = 1+0.2*3*1 = 1.6

sq = 1+0.1tan2(45+/2)B/L = 1+0.1*3*1 = 1.3 for > 100

s = sq for > 100

dq = 1+ 0.1tan(45+/2)Df/L = 1+ 0.1*3*10/L = 1+ 3/L   for > 100

d = dq for > 100

= 165.36 pcf - 62.4 pcf = 102.96 pcf

put all the values in the formula, we get ;

qu = 23920*(1+3/L) + 66.924*L(1+3/L)

= 24035.92 + 41430.66/L + 66.924*L

Step: (2) : calculation of allowable bearing capacity, qa

FS = 6

therefore, allowable bearing capacity, qa = qu/FS

qa =4006 + 6905.11/L + 11.154*L

Step : (3) : calculation of loads on foundation due to column weight, water tank weight, wind load and due to weight of water stored :

load = volume * unit weight

load due to column weight = 1.5*1.5*20*150 = 6750 lb.

load due to tank which is made of 1/2 inch (=1/24 ft) thick steel = (1/24)*2(12+12)*10*490 = 9800 lb.

load due to weight of water stored in a tank = 12*12*10*62.4 = 89856 lb.

load due to wind load (East- West Direction Only) = facing area*WL

= ((1.5*20) + (12*10))*13.89 = 2083.5 lb

total load = 6750+9800+89856+2083.5 = 108489.5 lb

factored load = 1.5*(DL+WL+load due to weight of water stored) = 1.5*108489.5 = 162734.25 lb.

Step : (4) : calculation of size of square footing :

area of footing = L*L

therefore, total load/area = 162734.25/(L*L)

this much of capacity must be bear by soil, therefore,

162734.25/L2 = allowable soil bearing capacity.

this implies,

162734.25/L2 = 4006 + 6905.11/L + 11.154*L

arranging the L, we get ;

11.154*L3 + 4006*L2 + 6905.11*L - 162734.25 = 0

by solving this cubic equation, we get ;

L = 5.53, -357.3, -7.38

so, value of L should be = 5.53 ft

take, L= 5.6 ft. or 6 ft.

check :

total load/area = 162734.25/(5.6*5.6) = 5189.23 pcf

put the value of L = 5.6 ft in equation of allowable bearing capacity

we get ;

qa =4006 + 6905.11/5.6 + 11.154*5.6 = 5301.52 pcf > 5189.23 pcf

hence it is safe.

hence,

the size of footing = 5.6 ft * 5.6 ft

or another dimention we can choose = 6 ft * 6 ft

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