Design of pile cap

General

Pile layout pattern:

Pile under pile cap should be layout symmetrically in both directions.  The column or wall on pile cap should be centered at the geometric center of the pile cap in order to transferred load evenly to each pile.  Example of pile layout pattern are shown below:

Pile spacing, edge distance, and pile cap thickness: 

In general, piles should be spacing at 3 times of pile diameter in order to transfer load effectively to soil.  If the spacing is less than 3 times of diameter, pile group settlement and bearing capacity should be checked.

Pile cap thickness is normal determined by shear strength.  For smaller pile cap, the thickness is normally governed by deep beam shear.  For large pile cap, the thickness is governed by direct shear.  When necessary, shear reinforcement may be used to reduced thickness pile cap. 

The edge distance is normally governed by punching shear capacity of corner piles.

Theory:

Punching shear

The punching shear strength according to ACI is

fv_c = 4Öf_c’

where f = 0.85 is strength reduction factor, fc’ is compressive strength of concrete.

The critical section of punching shear stress is at a distance, d/2, from edge of pile, d is the effective depth of pile cap.  For corner pile, the critical section normally extends to the corner edge of pile cap since it gives less shear area.

Direct shear or beam shear

The critical section of direct shear is at a distance, d, from edge of column or pile.

The direct shear shrength according to ACI is

fv_c =0.85[1.9Öf_c’+2500r_w(V_ud/M_u)] ³ 0.85(2Öf_c’)               

where r_w (» 0.002) is reinforcement ratio, V_u is factored shear stress, M_u is factored moment at the critical section.  For r_w » 0.002 and f_c’ between 3000 psi and 4000 psi,

fv_c =0.85[1.9Öf_c’+0.1Öf_c’(V_ud/M_u)] ³ 0.85(2Öf_c’)    

Deep beam shear

Deep beam shear is evaluated at face of column when w < d and V_u*d/M_u ³ 1

The shear strength is calculated as follows:

fv_c =0.85{(d/w)[3.5-2.5(M_u/V_ud)][1.9Öf_c’+2500r_w(V_u*d/M_u)]} ³ 0.85(10Öf_c’)                 

where w is the distance from face of column to the nearest pile. For r_w » 0.002 and f_c’ between 3000 psi and 4000 psi,

fv_c =0.85{(d/w)[3.5-2.5(M_u/V_ud)][1.9Öf_c’+0.1Öf_c’ (V_u*d/M_u)]} ³ 0.85(10Öf_c’)                 

Flexural reinforcement

Design of flexural reinforcement is the same as spread footing design.  The critical section is at face of column.

Pile load calculation

Pile load can be calculated as

p_i = P/n+M_x*d_x/I_y+ M_y*d_y/I_x

where p_i is axial load for individual pile, P is column load, M is moment from column moment and/or from eccentricity between center of column and center of pile group, n is total number of piles, dx and dy are x and y distance from center of pile group, I_x and I_y are moment of inertia of pile group in x and y directions. I_x and I_y are calculated as

I_x = S d_y², I_y = S d_x².

Design procedure

1. Estimate number of pile needed. Selection pile layout pattern. Calculate individual pile load.  The maximum pile load shall not exceed allowable pile capacity.
2. Calculate factored pile load. Assume a depth of pile cap, calculate factored moment and shear at critical section, check direct shear
3. Calculate moment and shear at face of column, check deep beam shear.
4. Check punching shear and edge distance.
5. Design flexural reinforcement.

Design Examples

Pile cap design example:

Design Data:

Column dead load: P_D = 300 kip

Column live load: P_L = 350 kip

Column dead load moment: M_DX = 40 ft-kip, M_DY = 80 ft-kip

Column Live load moment: M_LX = 35 ft-kip, M_LY = 65 ft-kip

Column size: 18"x18" concrete column

Type of pile: 16 in diameter concrete pile

Allowable pile compression capacity: P_c = 125 kip

Allowable pile tension capacity: P_t = 50 kip

Compressive strength of concrete: f_c’ = 3000 psi

Tensile strength of reinforcing steel: f_y = 60 ksi

Requirement:  Design pile group and pile cap

Solution:

1. Estimate number of pile and select pile layout pattern

Total service pile vertical load: P = P_D+P_L = 650 kip

Estimate number of pile: n = P/P_c = 5.2  

Try a six-pile layout pattern, n = 6

Minimum spacing of pile: s = 16 in x 3 = 4 ft

2. Check pile capacity:

d_x1 = -4 ft, d_x2 = -4 ft, d_x3 = 0 ft, d_x4 = 0 ft, d_x5 = 4 ft, d_x6 = 4 ft

I_y = d_x1²+ d_x2²+ d_x3²+ d_x4²+ d_x5²+ d_x6² = 64 ft².

d_y1 = -2 ft, d_y2 = 2 ft, d_y3 = -2 ft, d_y4 = 2 ft, d_y5 = -2 ft, d_y6 = 2 ft

I_x = d_x1²+ d_x2²+ d_x3²+ d_x4²+ d_x5²+ d_x6² = 24 ft².

Column service load moment:

M_x = M_DX+M_LX = 75 ft-kip

M_y = M_Dy+M_Ly = 145 ft-kip

Maximum pile compression load:

P₁ = P/n+(M_x*d_y1/I_x)+(M_y*d_x1/I_y) = 93 kip

P₂ = P/n+(M_x*d_y2/I_x)+(M_y*d_x2/I_y) = 105 kip

P₃ = P/n+(M_x*d_y3/I_x)+(M_y*d_x3/I_y) = 125 kip

P₄ = P/n+(M_x*d_y4/I_x)+(M_y*d_x4/I_y) = 114.5 kip

P₅ = P/n+(M_x*d_y5/I_x)+(M_y*d_x5/I_y) = 111.1 kip

P₆ = P/n+(M_x*d_y6/I_x)+(M_y*d_x6/I_y) = 123.6 kip

3. Assume a pile cap of 3'6" depth, the top of pile is at 6" above bottom of pile cap and the reinforcement is at 2" above top of pile, the effective depth is d = 34 in

Since the effective depth d is less than 4 ft, check direct shear in the longitudinal direction.

Factored column load, P_u = 1.4*P_D+1.7*P_L = 1015 kip

Factored  column moment:

M_ux = 1.4M_DX+1.7M_LX = 115.5 ft-kip

M_uy = 1.4M_Dy+1.7M_Ly = 222.5 ft-kip

Factored pile load:

P_u1 = P_u/n+(M_ux*d_y1/I_x)+(M_uy*d_x1/I_y) = 145.6 kip

P_u2 = P_u/n+(M_ux*d_y2/I_x)+(M_uy*d_x2/I_y) = 164.8 kip

P_u3 = P_u/n+(M_ux*d_y3/I_x)+(M_uy*d_x3/I_y) = 159.5 kip

P_u4 = P_u/n+(M_ux*d_y4/I_x)+(M_uy*d_x4/I_y) = 178.7 kip

P_u5 = P_u/n+(M_ux*d_y5/I_x)+(M_uy*d_x5/I_y) = 168.6 kip

P_u6 = P_u/n+(M_ux*d_y6/I_x)+(M_uy*d_x6/I_y) = 192.6 kip

The factored shear force at the critical section is V_u = P_u5+P_u6 = 361.3 kip

The factored moment at one d from face of column is

M_u = (P_u5+P_u6)(4 ft – d – 9 in) = 150.5 ft-kip

Assume an edge distance of 1'9", the width of pile cap is b = 7.5 ft

The shear strength of pile cap is

fV_c =0.85[1.9Öf_c’+0.1Öf_c’(V_ud/M_u)]bd = 367.5 kip > 361.3 kip O.K.

3. Check deep beam shear in the short direction.

Factored shear force: V_u = P_u2+P_u4+P_u6 = 536.3 kip

The factored moment face of column is

M_u = (P_u2+ P_u4+P_u6)(2 ft – 9 in) = 670.4 ft-kip

The deep beam shear strength of concrete is as follows:

The distance from pile to face of column, w = 24 in – 9 in = 15 in

The length of pile cap is b = 11 ft

The ratio, V_u*d/M_u = 2.26   > 1

fV_c =0.85{(d/w)[3.5-2.5(M_u/V_ud)][1.9Öf_c’+0.1Öf_c’ (V_u*d/M_u)]}bd=2524 kip

fV_c =0.85(10Öf_c’)bd = 2184 kip > 536.3 kip O.K.                

4. Design reinforcement in short direction: M_u = 670.5 ft-kip

Factor: R_n = M_u/(0.9*b*d²) = 56 ksi, m = f_y/0.85f_c’ = 23.5

Reinforcement ratio: r_w = (1/m)[1-Ö(1-2mR_n/f_y)] = 0.00094

Check minimum reinforcement: r_min =r_w *4/3 = 0.0012 or r_min = 0.002

Area of reinforcement: As = 0.002*b*d = 9.4 in².

Use 10#9 bar, A_s = 10 in².

Design reinforcement in longitudinal direction: b = 7.5 ft

M_u = (P_u5+P_u6)(4 ft – 9 in) = 1174 ft-kip

Factor: R_n = M_u/(0.9*b*d²) = 150.4 ksi

Reinforcement ratio: r_w = (1/m)[1-Ö(1-2mR_n/f_y)] = 0.0026

Check minimum reinforcement: r_min =r_w *4/3 = 0.0034

Area of reinforcement: A_s = 0.0034*b*d = 10.5 in².

Use 11#9 bar, A_s = 11 in².