Question

A sign structure on the NJ Turnpike is to be designed to resist a wind force that produces a moment of 25 k-ft in one direction. The axial load is 30 kips. Soil conditions consist of a normally consolidated clay layer with following properties; su=800 psf, andγsat= 110 pcf. Design for a FOS of 3. Assume frost depth to be 3ft below grade

Answer 1

Finding the cohesion of the soil(c) using the relation:

`$c = (q_u)/(2)$`

Here,
`$q_u$` is the unconfined compression strength of the soil;

`$c = (800)/(2)$`

= 400 psf

∴ The cohesion value is greater than 0

So the use of the angle of internal friction is 0

Referring to the table relation between bearing capacity factors and angle of internal friction.

For the angle of inter friction
`$0^\circ$`

`$N_c = 5.14$`

`$N_q = 1.0$`

`$N_r = 0$`

Therefore,

`$q_(ult) = (400 * 5.14 )+(110 * 3 * 1.0)+(0.5 * 100 * 13 * 0)$`

= 2386 psf

∴ Allowable bearing capacity
`$q_(a) = (Q_(allow))/(A)$`

`$=(30)/(B^2)$`

∴
`$q_a = (q_(ult))/(F.O.S)$`

`$(30)/(B^2) = (2386)/(3)$`

∴ B = 0.2 ft

Therefore, the dimension of the square footing is 0.2 ft x 0.2 ft

`$=0.04 \ ft^2$`