pile caps


16 يناير 2012
مجموع الإعجابات
الرجاء المساعدة بتزويدي بملفات لتصميم البايل كاب وليس بالضرورة ملفات اكسل لانها لم تساعدني للضرورة القصوى
وجزاكم الله عني كل خير

مواضيع مماثلة

مهندس سمير

عضو معروف
11 أغسطس 2007
مجموع الإعجابات

[h=1]Design of pile cap[/h][h=2]General[/h][h=2]Pile layout pattern:[/h]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:


[h=3]Pile spacing, edge distance, and pile cap thickness:[/h]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 diameter
Pile spacing
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.
[h=2]Theory:[/h][h=3]Punching shear[/h]The punching shear strength according to ACI is
fv[SUB]c[/SUB] = 4Öf[SUB]c[/SUB]’
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.
[h=3]Direct shear or beam shear[/h]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[SUB]c[/SUB] =0.85[1.9Öf[SUB]c[/SUB]’+2500r[SUB]w[/SUB](V[SUB]u[/SUB]d/M[SUB]u[/SUB])] ³ 0.85(2Öf[SUB]c[/SUB]’)
where r[SUB]w[/SUB] (» 0.002) is reinforcement ratio, V[SUB]u[/SUB] is factored shear stress, M[SUB]u[/SUB] is factored moment at the critical section. For r[SUB]w[/SUB] » 0.002 and f[SUB]c[/SUB]’ between 3000 psi and 4000 psi,
fv[SUB]c[/SUB] =0.85[1.9Öf[SUB]c[/SUB]’+0.1Öf[SUB]c[/SUB]’(V[SUB]u[/SUB]d/M[SUB]u[/SUB])] ³ 0.85(2Öf[SUB]c[/SUB]’)
[h=3]Deep beam shear[/h]Deep beam shear is evaluated at face of column when w < d and V[SUB]u[/SUB]*d/M[SUB]u[/SUB] ³ 1
The shear strength is calculated as follows:
fv[SUB]c[/SUB] =0.85{(d/w)[3.5-2.5(M[SUB]u[/SUB]/V[SUB]u[/SUB]d)][1.9Öf[SUB]c[/SUB]’+2500r[SUB]w[/SUB](V[SUB]u[/SUB]*d/M[SUB]u[/SUB])]} ³ 0.85(10Öf[SUB]c[/SUB]’)
where w is the distance from face of column to the nearest pile. For r[SUB]w[/SUB] » 0.002 and f[SUB]c[/SUB]’ between 3000 psi and 4000 psi,
fv[SUB]c[/SUB] =0.85{(d/w)[3.5-2.5(M[SUB]u[/SUB]/V[SUB]u[/SUB]d)][1.9Öf[SUB]c[/SUB]’+0.1Öf[SUB]c[/SUB]’ (V[SUB]u[/SUB]*d/M[SUB]u[/SUB])]} ³ 0.85(10Öf[SUB]c[/SUB]’)
[h=3]Flexural reinforcement[/h]Design of flexural reinforcement is the same as spread footing design. The critical section is at face of column.
[h=2]Pile load calculation[/h]Pile load can be calculated as
p[SUB]i[/SUB] = P/n+M[SUB]x[/SUB]*d[SUB]x[/SUB]/I[SUB]y[/SUB]+ M[SUB]y[/SUB]*d[SUB]y[/SUB]/I[SUB]x[/SUB]
where p[SUB]i[/SUB] 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[SUB]x[/SUB] and I[SUB]y[/SUB] are moment of inertia of pile group in x and y directions. I[SUB]x[/SUB] and I[SUB]y[/SUB] are calculated as
I[SUB]x[/SUB] = S d[SUB]y[/SUB][SUP]2[/SUP], I[SUB]y[/SUB] = S d[SUB]x[/SUB][SUP]2[/SUP].
[h=2]Design procedure[/h]
  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.
[h=2]Design Examples[/h]Pile cap design example:
Design Data:
Column dead load: P[SUB]D[/SUB] = 300 kip
Column live load: P[SUB]L[/SUB] = 350 kip
Column dead load moment: M[SUB]DX[/SUB] = 40 ft-kip, M[SUB]DY[/SUB] = 80 ft-kip
Column Live load moment: M[SUB]LX[/SUB] = 35 ft-kip, M[SUB]LY[/SUB] = 65 ft-kip
Column size: 18"x18" concrete column
Type of pile: 16 in diameter concrete pile
Allowable pile compression capacity: P[SUB]c[/SUB] = 125 kip
Allowable pile tension capacity: P[SUB]t[/SUB] = 50 kip
Compressive strength of concrete: f[SUB]c[/SUB]’ = 3000 psi
Tensile strength of reinforcing steel: f[SUB]y[/SUB] = 60 ksi
[h=4]Requirement: Design pile group and pile cap[/h]Solution:
1. Estimate number of pile and select pile layout pattern
Total service pile vertical load: P = P[SUB]D[/SUB]+P[SUB]L[/SUB] = 650 kip
Estimate number of pile: n = P/P[SUB]c[/SUB] = 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[SUB]x1[/SUB] = -4 ft, d[SUB]x2[/SUB] = -4 ft, d[SUB]x3[/SUB] = 0 ft, d[SUB]x4[/SUB] = 0 ft, d[SUB]x5[/SUB] = 4 ft, d[SUB]x6[/SUB] = 4 ft
I[SUB]y[/SUB] = d[SUB]x1[/SUB][SUP]2[/SUP]+ d[SUB]x2[/SUB][SUP]2[/SUP]+ d[SUB]x3[/SUB][SUP]2[/SUP]+ d[SUB]x4[/SUB][SUP]2[/SUP]+ d[SUB]x5[/SUB][SUP]2[/SUP]+ d[SUB]x6[/SUB][SUP]2[/SUP] = 64 ft[SUP]2[/SUP].
d[SUB]y1[/SUB] = -2 ft, d[SUB]y2[/SUB] = 2 ft, d[SUB]y3[/SUB] = -2 ft, d[SUB]y4[/SUB] = 2 ft, d[SUB]y5[/SUB] = -2 ft, d[SUB]y6[/SUB] = 2 ft
I[SUB]x[/SUB] = d[SUB]x1[/SUB][SUP]2[/SUP]+ d[SUB]x2[/SUB][SUP]2[/SUP]+ d[SUB]x3[/SUB][SUP]2[/SUP]+ d[SUB]x4[/SUB][SUP]2[/SUP]+ d[SUB]x5[/SUB][SUP]2[/SUP]+ d[SUB]x6[/SUB][SUP]2[/SUP] = 24 ft[SUP]2[/SUP].
Column service load moment:
M[SUB]x[/SUB] = M[SUB]DX[/SUB]+M[SUB]LX[/SUB] = 75 ft-kip
M[SUB]y[/SUB] = M[SUB]Dy[/SUB]+M[SUB]Ly[/SUB] = 145 ft-kip
Maximum pile compression load:
P[SUB]1[/SUB] = P/n+(M[SUB]x[/SUB]*d[SUB]y1[/SUB]/I[SUB]x[/SUB])+(M[SUB]y[/SUB]*d[SUB]x1[/SUB]/I[SUB]y[/SUB]) = 93 kip
P[SUB]2[/SUB] = P/n+(M[SUB]x[/SUB]*d[SUB]y2[/SUB]/I[SUB]x[/SUB])+(M[SUB]y[/SUB]*d[SUB]x2[/SUB]/I[SUB]y[/SUB]) = 105 kip
P[SUB]3[/SUB] = P/n+(M[SUB]x[/SUB]*d[SUB]y3[/SUB]/I[SUB]x[/SUB])+(M[SUB]y[/SUB]*d[SUB]x3[/SUB]/I[SUB]y[/SUB]) = 125 kip
P[SUB]4[/SUB] = P/n+(M[SUB]x[/SUB]*d[SUB]y4[/SUB]/I[SUB]x[/SUB])+(M[SUB]y[/SUB]*d[SUB]x4[/SUB]/I[SUB]y[/SUB]) = 114.5 kip
P[SUB]5[/SUB] = P/n+(M[SUB]x[/SUB]*d[SUB]y5[/SUB]/I[SUB]x[/SUB])+(M[SUB]y[/SUB]*d[SUB]x5[/SUB]/I[SUB]y[/SUB]) = 111.1 kip
P[SUB]6[/SUB] = P/n+(M[SUB]x[/SUB]*d[SUB]y6[/SUB]/I[SUB]x[/SUB])+(M[SUB]y[/SUB]*d[SUB]x6[/SUB]/I[SUB]y[/SUB]) = 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[SUB]u[/SUB] = 1.4*P[SUB]D[/SUB]+1.7*P[SUB]L[/SUB] = 1015 kip
Factored column moment:
M[SUB]ux[/SUB] = 1.4M[SUB]DX[/SUB]+1.7M[SUB]LX [/SUB]= 115.5 ft-kip
M[SUB]uy[/SUB] = 1.4M[SUB]Dy[/SUB]+1.7M[SUB]Ly [/SUB]= 222.5 ft-kip
Factored pile load:
P[SUB]u1[/SUB] = P[SUB]u[/SUB]/n+(M[SUB]ux[/SUB]*d[SUB]y1[/SUB]/I[SUB]x[/SUB])+(M[SUB]uy[/SUB]*d[SUB]x1[/SUB]/I[SUB]y[/SUB]) = 145.6 kip
P[SUB]u2[/SUB] = P[SUB]u[/SUB]/n+(M[SUB]ux[/SUB]*d[SUB]y2[/SUB]/I[SUB]x[/SUB])+(M[SUB]uy[/SUB]*d[SUB]x2[/SUB]/I[SUB]y[/SUB]) = 164.8 kip
P[SUB]u3[/SUB] = P[SUB]u[/SUB]/n+(M[SUB]ux[/SUB]*d[SUB]y3[/SUB]/I[SUB]x[/SUB])+(M[SUB]uy[/SUB]*d[SUB]x3[/SUB]/I[SUB]y[/SUB]) = 159.5 kip
P[SUB]u4[/SUB] = P[SUB]u[/SUB]/n+(M[SUB]ux[/SUB]*d[SUB]y4[/SUB]/I[SUB]x[/SUB])+(M[SUB]uy[/SUB]*d[SUB]x4[/SUB]/I[SUB]y[/SUB]) = 178.7 kip
P[SUB]u5[/SUB] = P[SUB]u[/SUB]/n+(M[SUB]ux[/SUB]*d[SUB]y5[/SUB]/I[SUB]x[/SUB])+(M[SUB]uy[/SUB]*d[SUB]x5[/SUB]/I[SUB]y[/SUB]) = 168.6 kip
P[SUB]u6[/SUB] = P[SUB]u[/SUB]/n+(M[SUB]ux[/SUB]*d[SUB]y6[/SUB]/I[SUB]x[/SUB])+(M[SUB]uy[/SUB]*d[SUB]x6[/SUB]/I[SUB]y[/SUB]) = 192.6 kip
The factored shear force at the critical section is V[SUB]u[/SUB] = P[SUB]u5[/SUB]+P[SUB]u6 [/SUB]= 361.3 kip
The factored moment at one d from face of column is
M[SUB]u[/SUB] = (P[SUB]u5[/SUB]+P[SUB]u6[/SUB])(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[SUB]c[/SUB] =0.85[1.9Öf[SUB]c[/SUB]’+0.1Öf[SUB]c[/SUB]’(V[SUB]u[/SUB]d/M[SUB]u[/SUB])]bd = 367.5 kip > 361.3 kip O.K.
3. Check deep beam shear in the short direction.
Factored shear force: V[SUB]u[/SUB] = P[SUB]u2[/SUB]+P[SUB]u4[/SUB]+P[SUB]u6[/SUB] = 536.3 kip
The factored moment face of column is
M[SUB]u[/SUB] = (P[SUB]u2[/SUB]+ P[SUB]u4[/SUB]+P[SUB]u6[/SUB])(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[SUB]u[/SUB]*d/M[SUB]u[/SUB] = 2.26 > 1
fV[SUB]c[/SUB] =0.85{(d/w)[3.5-2.5(M[SUB]u[/SUB]/V[SUB]u[/SUB]d)][1.9Öf[SUB]c[/SUB]’+0.1Öf[SUB]c[/SUB]’ (V[SUB]u[/SUB]*d/M[SUB]u[/SUB])]}bd=2524 kip
fV[SUB]c[/SUB] =0.85(10Öf[SUB]c[/SUB]’)bd = 2184 kip > 536.3 kip O.K.
4. Design reinforcement in short direction: M[SUB]u[/SUB] = 670.5 ft-kip
Factor: R[SUB]n[/SUB] = M[SUB]u[/SUB]/(0.9*b*d[SUP]2[/SUP]) = 56 ksi, m = f[SUB]y[/SUB]/0.85f[SUB]c[/SUB]’ = 23.5
Reinforcement ratio: r[SUB]w [/SUB]= (1/m)[1-Ö(1-2mR[SUB]n[/SUB]/f[SUB]y[/SUB])] = 0.00094
Check minimum reinforcement: r[SUB]min [/SUB]=r[SUB]w [/SUB]*4/3 = 0.0012 or r[SUB]min [/SUB]= 0.002
Area of reinforcement: As = 0.002*b*d = 9.4 in[SUP]2[/SUP].
Use 10#9 bar, A[SUB]s[/SUB] = 10 in[SUP]2[/SUP].
Design reinforcement in longitudinal direction: b = 7.5 ft
M[SUB]u[/SUB] = (P[SUB]u5[/SUB]+P[SUB]u6[/SUB])(4 ft – 9 in) = 1174 ft-kip
Factor: R[SUB]n[/SUB] = M[SUB]u[/SUB]/(0.9*b*d[SUP]2[/SUP]) = 150.4 ksi
Reinforcement ratio: r[SUB]w [/SUB]= (1/m)[1-Ö(1-2mR[SUB]n[/SUB]/f[SUB]y[/SUB])] = 0.0026
Check minimum reinforcement: r[SUB]min [/SUB]=r[SUB]w [/SUB]*4/3 = 0.0034
Area of reinforcement: A[SUB]s[/SUB] = 0.0034*b*d = 10.5 in[SUP]2[/SUP].
Use 11#9 bar, A[SUB]s[/SUB] = 11 in[SUP]2[/SUP].

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