Question

A large 20-story steel structure is planned for an office building the Denver Metropolitan. The exploration program revealed that foundation soil comprises of 25-foot clay with undrained shear strength of 1,000 psf overlying a thick stiff clay shale with the undrained shear strength of 3,000 psf. To avoid construction noise, the architect selected 3-ft diameter drilled shaft foundation. The drilled shafts are socketed 15 feet in the stiff clay shale. Calculate:

a. The ultimate capacity of the drilled shaft.

b. Estimate the total structural dead load.

c. Estimate the office building life load.

d. At a factor of safety of 2,5, how many drilled shafts would be required?

Note: When you suspect some missing information, please make reasonable assumption(s) and process with your solution. Clearly show why assumption(s) is needed.

Answer 1

Ans) Given,

Thickness of clay layer = 25 ft

Undrained shear strength of overlying soil = 1000 psf

Undrained shear strength of below soil = 3000 psf

Diameter of drilled shaft(d) = 3 ft

Length of pile(L) = 15 ft

(a) We know,

Ultimate capacity of pile(Q) = Bearing capacity + Skin friction resistance

Q = q_b A_b + q_s A_s

Q = C N_cA_b + C' A_s

According to Meyerhoff, N_c = 9

C = unit cohesion at base of shaft

C = undrained shear strength/2

C = 1000/2 or 500psf

A_b = (/4) d²

= adhesive factor ( 0.5 for medium stiff clay)

C' =Average Cohesion along pile

C' =  3000/2 or 1500 psf

A_s = dL

Putting values,

Q = 9C (/4)d² + 0.5 C' (dL)

Q = 9(500) (0.785 x 9) + (0.5 x 1500 x 3.14 x 3 x 15)

Q = 31792.5 + 105975

Q = 137767.5 psf

Ans b) Assume 1 story dimension as 20ft x 20 ft x 10ft

Unit weight of steel = 500 pcf

Dead load on single story :

Wall load = 4 x 20 x 10 x 500

= 400000 lb/ft

Roof load = 20 x 20 x 500

= 200000 lb/ft

Floor load = 20 x 20 x 500

= 200000 lb/ft

Total dead load = 800000 lb/ft or 800 kips/ft

Weight of machine, appliance etc contribute 50% of total dead load, therefore factored dead load = 1.5 x 800

DL = 1200 kips/ft

For 20 story , DL = 1200 x 20

DL = 24000 kips/ft

Ans c) Live load Calculation for single story

For office building, live load = 500 psf

Live load per unit length = 500/20 lb/ft

Live load = 25 lb/ft

Factored live load = 1.5 x 25 lb/ft

= 37.5 lb/ft

For 20 story , LL = 37.5 x 20

LL = 750 lb/ft

Ans d) Total load required to be resist = DL+LL

= 24000+0.75

= 24000.75 kips/ft

or 24000750 / 20 psf

or 1200037.5 psf

Q =137767.5 psf

FOS = 2.5

Allowable pressure = Q/FOS

= 55107 psf

Load resisted by one shaft = 55107 psf

Load to be resisted =  1200037.5 psf

Number of drilled shaft required = 1200037.5/55107

= 21.7 or 22