ii - bsi.ce.ufl.edu · 12/27/2002 · 2 Case 1A) – Pile in Clay (Example file: 1APileclay.spc)...
Transcript of ii - bsi.ce.ufl.edu · 12/27/2002 · 2 Case 1A) – Pile in Clay (Example file: 1APileclay.spc)...
i
Overview
Within the following pages, you will find 14 complete validation problems. Each problem
exactly models the computational method and formulas used by FB-Deep to determine the axial
capacity of drilled shafts and driven piles. Thus, it is a hand-worked version of the analysis FB-
Deep does electronically.
These problems encompass a full range of shaft and pile geometries, and soil layer types and
characteristics. Each validation problem is accompanied by labeled diagrams, a complete list of
all data input, and a thorough step-by-step solution. There is also a separate .in file that can be
run in FB-Deep for each of these problems (see download area).
By looking at this validation set, it is our hope you will gain a precise understanding of how FB-
Deep arrives at its solutions.
ii
Table of Contents
1. Simplified Problems
a) All clay (N = 15 blows)
I. Pile ..............................................................................................................1
II. Shaft ............................................................................................................4
b) All sand (N = 15 blows)
I. No water table (shaft & pile)
1. Pile ..................................................................................................7
2. Shaft ..............................................................................................10
II. Water table at surface – Shaft ...................................................................14
2. Tip in sand and underlying depth < 3.5D (N = 15 blows)
a) Layer below the sand is clay (N = 10 blows)
I. Pile .............................................................................................................18
II. Shaft ...........................................................................................................21
b) Layer below the sand is clay (N = 30 blows) – Shaft ...........................................25
3. Tip in sand and overlying height < 8.0D (N = 15 blows)
a) Layer above the sand is clay (N = 10 blows)
I. Pile .............................................................................................................29
II. Shaft ...........................................................................................................32
b) Layer above the sand is clay (N = 30 blows) – Shaft ............................................37
4. Tip in clay and underlying depth < D (N = 20 blows)
a) Layer below the clay is clay (N = 10 blows) – Shaft .............................................42
b) Layer below the clay is sand (N = 15 blows) – Shaft ............................................45
5. Rock socket (smooth) (shaft only) .....................................................................................48
(Overlaying sand (N = 15) Rock: qu = 10 tsf, qt = 1.0 tsf and qb = 0.5 qu)
1
Case 1A) – Pile in Clay (Example file: 1APileclay.in)
End Bearing
For soil type 1, clay, end bearing is:
0.7 0.7(15)( ) 3.5
3 3T
Nq tsf tsf= = =
8B above tip: 8*2ft=16ft or depth of 29ft.
3.5B below tip: 3.5*2ft=7ft or depth of 52ft.
,8
,3.5
16 *3.53.5
16
7 *3.53.5
7
3.5 3.53.5
2
T B
T B
T
ft tsfq tsf
ft
ft tsfq tsf
ft
tsf tsfq tsf
= =
= =
+= =
Check embedment:
DC for soil type 1 = 2B = 2 * 2ft = 4ft
DA = 45ft > 4ft No correction needed
Total End Bearing
* 3.5 *2 *2 14T T BQ q A tsf ft ft tons= = =
Depth N ST 0 15 1
5' 15 1 10' 15 1
15' 15 1 20' 15 1 25' 15 1
30' 15 1 35' 15 1
40' 15 1 45' 15 1 50' 15 1
55' 15 1
60’ 15 1
Depth qT
29 3.5tsf 8B Above 30 3.5tsf 35 3.5tsf 40 3.5tsf 45 3.5tsf Pile Tip
50 3.5tsf 52 3.5tsf 3.5B Below
24” x 24”
45’
2
Skin Friction
For soil type 1, clay, skin friction is:
2.0 (110 ) 2.0(15)(110 15)( ) 0.7113
4006.6 4006.6s
N Nf tsf tsf
− −= = =
Skin friction must be corrected for overlying layer (air) bearing weaker. Since DA > DC,
only the skin friction between the top of the bearing layer and critical depth must be corrected:
Corrected Skin Friction at Critical Depth:
[ 0.5( )]LC CD LC
CD
USFACDCSFACD q q q
q= = + −
USFACD = Ultimate skin friction at critical depth:
* 0.7113 *4 *2 *4 22.76USFACD fs SA tsf ft ft tons= = =
At surface, z = 0, determine qLC:
,8
,3.5
0
7 *3.53.5
7
3.5
T B
T B
T
q tsf
ft tsfq tsf
ft
q tsf
=
= =
=
At critical depth, z = 4, find qCD:
,8
,3.5
4 *3.53.5
4
7 *3.53.5
7
3.5 3.53.5
2
T B
T B
CD
ft tsfq tsf
ft
ft tsfq tsf
ft
tsf tsfq tsf
tsf
= =
= =
+= =
Calculate Corrected Skin Friction at Critical Depth
22.76 3.5 0.5 3.5 3.5 22.76
3.[ )]
5(
tonsCSFACD tsf tsf tsf tons
tsf= + − =
Depth qT
-16 0tsf 8B Above -15 0tsf -10 0tsf -5 0tsf 0 3.5tsf “Pile Tip”
5 3.5tsf 7 3.5tsf 3.5B Below
Depth qT
-12 0tsf 8B Above -10 0tsf 0 3.5tsf 4 3.5tsf “Pile Tip” 5 3.5tsf
10 3.5tsf 11 3.5tsf 3.5B Below
3
Total Skin Friction
22.76 0.7113 *2 *41 *4 256.08SQ tsf ft ft tons= + =
Pile Capacity
Davisson Capacity = QT + QS = 14tons + 256.08tons = 270.08tons
Allowable Capacity = ½ Davisson = 135.04tons
Ultimate Capacity = 3QT + QS = 3(14tons) + 256.08tons = 298.08tons
FB-Deep v3.0.0 Prediction
Pile Capacity
Estimated Davisson capacity = 270.08(tons)
Allowable pile capacity = 135.04(tons)
Ultimate pile capacity = 298.08(tons)
4
Case 1A) – Shaft in Clay (Example file: 1AShaftclay.in)
End Bearing
For soil type 1, clay:
1.125
, where 6.0 1 0.2 9,
456.0 1 0.2 24 9
3
9(1.125 ) 10.125
u
T c u c
c c
T
C tsf
Lq N C N
B
ftN N
ft
q tsf tsf
=
= = +
= + = =
= =
Total End Bearing
23
* 10.125 * * 71.572
T T B
ftQ q A tsf tons
= = =
Depth Cu ST 0 1.125 1
5' 1.125 1 10' 1.125 1
15' 1.125 1 20' 1.125 1 25' 1.125 1
30' 1.125 1 35' 1.125 1
40' 1.125 1 45' 1.125 1 50' 1.125 1
55' 1.125 1
60’ 1.125 1
36” Diameter
45’
5
Skin Friction
For soil type 1, clay, skin friction is:
2
1
2
1
2
1
2
1
2.75 SU u SU
z
SUz
z
SUz
z
SU z
z
u z
f c f tsf
Q f dAs
Q Df dzs
Q Df zs
Q D c zs
=
=
=
=
=
z = 0 to 5 ft 3(0.0)(1.125)5 3(0.0)(1.125)0 0Qs tons = − =
z = 5 to 42 ft 3(0.55)(1.125)42 3(0.55)(1.125)5 215.769Qs tons = − =
z = 42 to 45 ft 3(0.0)(1.125)45 3(0.0)(1.125)42 0Qs tons = − =
Total skin friction = 0tons + 215.769tons + 0tons = 215.769tons
Shaft Capacity
Total Shaft Capacity = QT + QS = 71.57tons + 215.769tons = 287.34tons
Settlement
Determine load corresponding to shaft top deflection of 0.216in and 2.16in.
*100
settlementR
shaft diameter=
R(0.216in) = 0.6
R(0.216in) = 0.6
Load Corresponding to R = 0.6:
Load carried in skin friction:
For R = 0.6
( )
( )
max
max
0.095155 0.892937*
0.60.950996
0.095155 0.892937*0.6
(0.950996)215.769 205.196
s
s
s
s
s
f R
f R
f
f
Q tons tons
=+
= =+
= =
45’
=0 for top
5ft and
bottom 3ft
(B).
6
Load carried in end bearing:
For R = 0.6
4 5 3 4
T max
2 3 2
4 5 3 4
T max
2 3 2
1.1823 10 3.7091 10
4.4944 10 0.26537 0.78436
1.1823 10 (0.6) 3.7091 10 (0.6)
4.4944 10 (0.6) 0.26537(0.6) 0.78436(0.6) 0.384
T
T
qx R x R
q
x R R R
qx x
q
x
− −
−
− −
−
= − +
− +
= − +
− + =
QT = (0.384)71.57tons = 27.50tons
Total load at R = 0.6 is 27.50tons + 205.19tons = 232.697tons
Load Corresponding to R = 6.0:
Load carried in skin friction:
For R > 2.0
max
0.833s
s
f
f=
QS = (0.833)215.769tons = 179.74tons
Load carried in end bearing:
4 5 3 4
T max
2 3 2
1.1823 10 (6.0) 3.7091 10 (6.0)
4.4944 10 (6.0) 0.26537(6.0) 0.78436(6.0) 0.973
Tqx x
q
x
− −
−
= − +
− + =
QT = (0.973)71.56tons = 69.635tons
Total load at R = 6.0 is 69.635tons + 179.74tons = 249.38tons
FB-Deep v3.0.0 Prediction
Shaft Capacity
Ultimate Shaft capacity = 287.34(tons)
R(%) Settl.(in) Qs(tons) Qb(tons) Qt(tons)
0.6 0.216 205.20 27.51 232.70
6.0 2.160 179.74 69.64 249.38
7
Case 1B) – Pile in Sand (Example file: 1BPilesand.in)
End Bearing
For soil type 3, sand, end bearing is:
3.2 3.2(15)( ) 16
3 3T
Nq tsf tsf= = =
8B above tip: 8*2ft=16ft or depth of 29ft.
3.5B below tip: 3.5*2ft=7ft or depth of 52ft.
,8
,3.5
16 *1616
16
7 *1616
7
16 1616
2
T B
T B
T
ft tsfq tsf
ft
ft tsfq tsf
ft
tsf tsfq tsf
= =
= =
+= =
Check embedment:
DC for soil type 3 = 9B = 9 * 2ft = 18ft
DA = 45ft > 18ft No correction needed
Total End Bearing
* 16 *2 *2 64T T BQ q A tsf ft ft tons= = =
Depth N ST 0 15 3
5' 15 3 10' 15 3
15' 15 3 20' 15 3 25' 15 3
30' 15 3 35' 15 3
40' 15 3 45' 15 3 50' 15 3
55' 15 3
60’ 15 3
Depth qT
29 16tsf 8B Above 30 16tsf 35 16tsf 40 16tsf 45 16tsf Pile Tip
50 16tsf 52 16tsf 3.5B Below
24” x 24”
45’
8
Skin Friction
For soil type 3, sand, skin friction is:
( ) 0.019 0.019(15) 0.285sf tsf N tsf= = =
Skin friction must be corrected for overlying layer (air) being weaker. Since DA > DC,
only the skin friction between the top of the bearing layer and critical depth must be corrected:
Corrected Skin Friction at Critical Depth:
[ 0.5( )]LC CD LC
CD
USFACDCSFACD q q q
q= = + −
USFACD = Ultimate skin friction at critical depth:
* 0.285 *18 *2 *4 41.04USFACD fs SA tsf ft ft tons= = =
At surface, z = 0, determine qLC:
,8
,3.5
0
7 *1616
7
16
T B
T B
T
q tsf
ft tsfq tsf
ft
q tsf
=
= =
=
At critical depth, z = 18, find qCD:
,8
,3.5
16 *1616
16
7 *3.516
7
16 1616
2
T B
T B
CD
ft tsfq tsf
ft
ft tsfq tsf
ft
tsf tsfq tsf
tsf
= =
= =
+= =
Calculate Corrected Skin Friction at Critical Depth
1[16 (16 )41.04
0.5 16 4 .04]16
tonsCSFACD tsf tsf tsf tons
tsf= + − =
Depth qT
-16 0tsf 8B Above -15 0tsf -10 0tsf -5 0tsf 0 16tsf “Pile Tip”
5 16tsf 7 16tsf 3.5B Below
Depth qT
2 16tsf 8B Above 5 16tsf 10 16tsf 15 16tsf 18 16tsf “Pile Tip”
20 16tsf 25 16tsf 3.5B Below
9
Total Skin Friction
41.04 0.285 *2 *(45 18) *4 102.6SQ tsf ft ft tons= + − =
Pile Capacity
Davisson Capacity = QT + QS = 64tons + 102.6tons = 166.6tons
Allowable Capacity = ½ Davisson = 83.3tons
Ultimate Capacity = 3QT + QS = 3(64tons) + 102.6tons = 294.6tons
FB-Deep v3.0.0 Prediction
Pile Capacity
Estimated Davisson capacity = 166.60(tons)
Allowable pile capacity = 83.30(tons)
Ultimate pile capacity = 294.60(tons)
10
Case 1B) – Shaft in Sand – water table below of shaft (Example file: 1BShaftsand.in)
End Bearing
For soil type 3, sand, end bearing is:
( ) 0.60 , where 45
0.60(15) 9
T SPT T
T
q tsf N q tsf
q tsf
=
= =
1.5B above tip: 1.5*3ft=4.5ft or depth 40.5ft.
2.0B below tip: 2*3ft=6ft or depth of 51ft.
,1.5
,2.0
4.5 *99
4.5
6 *99
6
9 99
2
T B
T B
T
ft tsfq tsf
ft
ft tsfq tsf
ft
tsf tsfq tsf
= =
= =
+= =
Total End Bearing
23
* 9 * * 63.6172
T T B
ftQ q A tsf tons
= = =
Depth N
40.5 15 1.5B Above 45 15 Shaft Tip 50 15 51 15 2.0B Below
Depth N ST 0 15 3
5' 15 3 10' 15 3
15' 15 3 20' 15 3 25' 15 3
30' 15 3 35' 15 3
40' 15 3 45' 15 3 50' 15 3
55' 15 3
60’ 15 3
36” Diameter
45’ Sand
= 100pcf
11
Skin Friction
For soil type 3, sand, skin friction is:
'S vf =
1.5 0.135 ( )z ft = −
1.2 > > 0.25 thus, if z < 5ft, then =1.2 and if z > 86ft, then = 0.25
2
1'
z
vz
Qs dA=
2
1'
z
vz
Qs D dz =
( )2
1
z
Z w wz
Qs D z dz = −
z = 0 to 5ft
( )( )
( )( )
5
0
52
0
3 1.2 100
3 1202
14137.17 7.068
Qs zdz
zQs
Qs lbs tons
=
=
= =
z = 5 to 45ft
( )( )
( )
( )
45
5
453/2
5
45
2 5/2
5
3 1.5 0.135 100
3 150 13.5
1 23 150 13.5
2 5
740033.57 14826.51 725208.39 362.61
Qs z zdz
Qs z z dz
Qs z z
Qs lbs lbs lbs tons
= −
= −
= −
= − = =
Total skin friction = 7.06tons + 362.61tons = 369.67tons
Shaft Capacity
Total Shaft Capacity = QT + QS = 63.617tons + 369.67tons = 433.39tons
Settlement
Determine load corresponding to shaft top deflection of 0.216in and 2.16in.
*100
settlementR
shaft diameter=
R(0.216in) = 0.6
12
R(0.216in) = 0.6
Load Corresponding to R = 0.6:
Load carried in skin friction:
For R < 0.908333
( ) ( ) ( ) ( )
4 3 2
max
4 3 2
max
2.16 6.34 7.36 4.15
2.16 0.6 6.34 0.6 7.36 0.6 4.15 0.6 0.9299
(0.9299)369.67 343.75
s
s
s
s
s
fR R R R
f
f
f
Q tons tons
= − + − +
= − + − + =
= =
Load carried in end bearing:
For R = 0.6
4 3
T max
2
4 3
T max
2
T max
0.0001079 0.0035584
0.045115 0.34861
0.0001079(0.6) 0.0035584(0.6)
0.045115(0.6) 0.34861(0.6)
0.1937
T
T
T
qR R
q
R R
q
q
q
q
= − +
− +
= − +
− +
=
QT = (0.1937)63.617tons = 12.321tons
Total load at R = 0.6 is 343.75tons + 12.321tons = 356.071tons
13
Load Corresponding to R = 6.0:
Load carried in skin friction:
For R > 0.908333
max
0.978112s
s
f
f=
QS = (0.978112)369.67tons = 361.58tons
Load carried in end bearing:
4 3
T max
2
4 3
T max
2
T max
0.0001079 0.0035584
0.045115 0.34861
0.0001079(6.0) 0.0035584(6.0)
0.045115(6.0) 0.34861(6.0)
1.0963
T
T
T
qR R
q
R R
q
q
q
q
= − +
− +
= − +
− +
=
QT = (1.0963)63.617tons = 69.743tons
Total load at R = 6.0 is 69.743tons + 361.58tons = 431.323tons
FB-Deep v3.0.0 Prediction
Shaft Capacity
Ultimate Shaft capacity = 433.74(tons)
R(%) Settl.(in) Qs(tons) Qb(tons) Qt(tons)
0.6 0.216 344.18 12.32 356.50
6.0 2.160 362.02 69.74 431.76
14
Case 1B) – Shaft in Sand – water table at ground surface (Example file: 1BShaftsand_wt.in)
End Bearing (same as for water table below tip)
For soil type 3, sand, end bearing is:
( ) 0.60 , where 45
0.60(15) 9
T SPT T
T
q tsf N q tsf
q tsf
=
= =
1.5B above tip: 1.5*3ft=4.5ft or depth 40.5ft.
2.0B below tip: 2*3ft=6ft or depth of 51ft.
,1.5
,2.0
4.5 *99
4.5
6 *99
6
9 99
2
T B
T B
T
ft tsfq tsf
ft
ft tsfq tsf
ft
tsf tsfq tsf
= =
= =
+= =
Total End Bearing
23
* 9 * * 63.6172
T T B
ftQ q A tsf tons
= = =
Depth N
40.5 15 1.5B Above 45 15 Shaft Tip 50 15 51 15 2.0B Below
Depth N ST 0 15 3
5' 15 3 10' 15 3
15' 15 3 20' 15 3 25' 15 3
30' 15 3 35' 15 3
40' 15 3 45' 15 3 50' 15 3
55' 15 3
60’ 15 3
36” Diameter
45’ Sand
= 100pcf
15
Skin Friction
For soil type 3, sand, skin friction is:
'S vf =
1.5 0.135 ( )z ft = −
1.2 > > 0.25 thus, if z < 5ft, then =1.2 and if z > 86ft, then = 0.25
2
1'
z
vz
Qs dA=
2
1'
z
vz
Qs D dz =
( )2
1
z
Z w wz
Qs D z dz = −
z = 0 to 5ft
( )( )
( )( )
5
0
52
0
3 1.2 100 62.4
3 45.122
5315.57 2.658
Qs zdz
zQs
Qs lbs tons
= −
=
= =
z = 5 to 45ft
( )( )
( )
( )
45
5
453/2
5
45
2 5/2
5
3 1.5 0.135 100 62.4
3 56.4 5.076
1 23 56.4 5.076
2 5
278255.42 5574.73 272680.69 136.34
Qs z zdz
Qs z z dz
Qs z z
Qs lbs lbs lbs tons
= − −
= −
= −
= − = =
Total skin friction = 2.658tons + 136.34tons = 138.99tons
Shaft Capacity
Total Shaft Capacity = QT + QS = 63.617tons + 138.99tons = 202.607tons
16
Settlement
Determine load corresponding to shaft top deflection of 0.216in and 2.16in.
*100
settlementR
shaft diameter=
R(0.216in) = 0.6
R(0.216in) = 0.6
Load Corresponding to R = 0.6:
Load carried in skin friction:
For R < 0.908333
( ) ( ) ( ) ( )
4 3 2
max
4 3 2
max
2.16 6.34 7.36 4.15
2.16 0.6 6.34 0.6 7.36 0.6 4.15 0.6 0.9299
(0.9299)138.99 129.246
s
s
s
s
s
fR R R R
f
f
f
Q tons tons
= − + − +
= − + − + =
= =
Load carried in end bearing:
For R = 0.6
4 3
T max
2
4 3
T max
2
T max
0.0001079 0.0035584
0.045115 0.34861
0.0001079(0.6) 0.0035584(0.6)
0.045115(0.6) 0.34861(0.6)
0.1937
T
T
T
qR R
q
R R
q
q
q
q
= − +
− +
= − +
− +
=
QT = (0.1937)63.617tons = 12.321tons
Total load at R = 0.6 is 12.321tons + 129.246tons = 141.567tons
17
Load Corresponding to R = 6.0:
Load carried in skin friction:
For R > 0.908333
max
0.978112s
s
f
f=
QS = (0.978112)138.99tons = 135.948tons
Load carried in end bearing:
4 3
T max
2
4 3
T max
2
T max
0.0001079 0.0035584
0.045115 0.34861
0.0001079(6.0) 0.0035584(6.0)
0.045115(6.0) 0.34861(6.0)
1.0963
T
T
T
qR R
q
R R
q
q
q
q
= − +
− +
= − +
− +
=
QT = (1.0963)63.617tons = 69.743tons
Total load at R = 6.0 is 69.743tons + 135.948tons = 205.691tons
FB-Deep v3.0.0 Prediction
Shaft Capacity
Ultimate Shaft capacity = 202.78(tons)
R(%) Settl.(in) Qs(tons) Qb(tons) Qt(tons)
0.6 0.216 129.41 12.32 141.73
6.0 2.160 136.12 69.74 205.86
18
Clay N=10
ST=1
Case 2A) – Pile in Sand (N=15) over Clay (N=10) (Example file: 2ApileSandClayN10.in)
End Bearing
For soil type 3, sand, end bearing is:
3.2 3.2(15)( ) 16
3 3T
Nq tsf tsf= = =
For soil type 1, clay, end bearing is:
0.7 0.7(10)( ) 2.33
3 3T
Nq tsf tsf= = =
8B above tip: 8*2ft=16ft or depth of 24ft.
3.5B below tip: 3.5*2ft=7ft or depth of 47ft.
,8
,3.5
16 *1616
16
4 *16 3 *2.33310.143
7
16 10.14313.071
2
T B
T B
T
ft tsfq tsf
ft
ft tsf ft tsfq tsf
ft
tsf tsfq tsf
= =
+= =
+= =
Check embedment:
DC for soil type 3 = 9B = 9 * 2ft = 18ft
DA = 40ft > 18ft No correction needed
Depth qT
24 16tsf 8B Above 25 16tsf 30 16tsf 35 16tsf 40 16tsf Pile Tip
44- 16tsf 44+ 2.333tsf 45 2.333tsf 47 2.333tsf 3.5B Below
Depth N ST 0 15 3
5' 15 3 10' 15 3
15' 15 3 20' 15 3 25' 15 3
30' 15 3 35' 15 3
40' 15 3 44'- 15 3 44’+ 10 1 45’ 10 1 50' 10 1
55' 10 1
60’ 10 1
36” Diameter
Sand N=15
ST=3
44’ 40’
19
Total End Bearing
* 13.071 *2 *2 52.286T T BQ q A tsf ft ft tons= = =
Skin Friction
For soil type 3, sand, skin friction is:
( ) 0.019 0.019(15) 0.285sf tsf N tsf= = =
Skin friction must be corrected for overlying layer (air) being weaker. Since DA > DC,
only the skin friction between the top of the bearing layer and critical depth must be corrected:
Corrected Skin Friction at Critical Depth:
[ 0.5( )]LC CD LC
CD
USFACDCSFACD q q q
q= = + −
USFACD = Ultimate skin friction at critical depth:
* 0.285 *18 *2 *4 41.04USFACD fs SA tsf ft ft tons= = =
At surface, z = 0, determine qLC:
,8
,3.5
0
7 *1616
7
16
T B
T B
T
q tsf
ft tsfq tsf
ft
q tsf
=
= =
=
At critical depth, z = 18, find qCD:
,8
,3.5
16 *1616
16
7 *3.516
7
16 1616
2
T B
T B
CD
ft tsfq tsf
ft
ft tsfq tsf
ft
tsf tsfq tsf
tsf
= =
= =
+= =
Calculate Corrected Skin Friction at Critical Depth
1[16 (16 )41.04
0.5 16 4 .04]16
tonsCSFACD tsf tsf tsf tons
tsf= + − =
Depth qT
-16 0tsf 8B Above -15 0tsf -10 0tsf -5 0tsf 0 16tsf “Pile Tip”
5 16tsf 7 16tsf 3.5B Below
Depth qT
2 16tsf 8B Above 5 16tsf 10 16tsf 15 16tsf 18 16tsf “Pile Tip”
20 16tsf 25 16tsf 3.5B Below
20
Total Skin Friction
41.04 0.285 *2 *(40 18) *4 91.2SQ tsf ft ft tons= + − =
Pile Capacity
Davisson Capacity = QT + QS = 52.286tons + 91.2tons = 143.49tons
Allowable Capacity = ½ Davisson = 71.74tons
Ultimate Capacity = 3QT + QS = 3(58.286tons) + 91.2tons = 248.06tons
FB-Deep v3.0.0 Prediction
Pile Capacity
Estimated Davisson capacity = 143.49(tons)
Allowable pile capacity = 71.74(tons)
Ultimate pile capacity = 248.06(tons)
21
Clay cu=0.75tsf
ST=1
Case 2A) – Shaft in Sand (N=15) over Clay (N=10) (Example file: 2AShaftSandClayN10.in)
End Bearing
For soil type 3, sand, end bearing is:
( ) 0.60 , where 45
0.60(15) 9
T SPT T
T
q tsf N q tsf
q tsf
=
= =
For soil type 1, clay, end bearing is:
( ) , where 40T c u Tq tsf N c q tsf=
( ) 0.750uc tsf tsf=
9(0.750 ) 6.75Tq tsf tsf= =
1.5B above tip: 1.5*3ft=4.5ft or depth 35.5ft.
2.0B below tip: 2*3ft=6ft or depth of 44ft.
,1.5
,2
4.5 *99
4.5
4 *9 2 *6.758.25
6
9 8.258.62
2
T B
T B
T
ft tsfq tsf
ft
ft tsf ft tsfq tsf
ft
tsf tsfq tsf
= =
+= =
+= =
Total End Bearing 2
3* 8.62 * * 60.967
2T T B
ftQ q A tsf tons
= = =
Depth N cu ST 0 15 3
5' 15 3 10' 15 3
15' 15 3 20' 15 3 25' 15 3
30' 15 3 35' 15 3
40' 15 3 44'- 15 3 44’+ 0.75 1 45’ 0.75 1 50' 0.75 1
55' 0.75 1
60’ 0.75 1
Depth qT
35.5 9tsf 1.5B Above 40 9tsf Shaft Tip
44- 9tsf
44+ 6.75tsf
45 6.75tsf
46 6.75tsf 2B Below
36” Diameter
Sand N=15
ST=3
= 100pcf
44’ 40’
22
Skin Friction
For soil type 3, sand, skin friction is:
'S vf =
1.5 0.135 ( )z ft = −
1.2 > > 0.25 thus, if z < 5ft, then =1.2 and if z > 86ft, then = 0.25
2
1'
z
vz
Qs dA=
2
1'
z
vz
Qs D dz =
( )2
1
z
Z w wz
Qs D z dz = −
z = 0 to 5ft
( )( )
( )( )
5
0
52
0
3 1.2 100
3 1202
14137.17 7.068
Qs zdz
zQs
Qs lbs tons
=
=
= =
z = 5 to 40ft
( )( )
( )
( )
40
5
403/2
5
40
2 5/2
5
3 1.5 0.135 100
3 150 13.5
1 23 150 13.5
2 5
615964.30 14826.41 601137.89 300.57
Qs z zdz
Qs z z dz
Qs z z
Qs lbs lbs lbs tons
= −
= −
= −
= − = =
Total skin friction = 7.068tons + 300.57tons = 307.63tons
Shaft Capacity
Total Shaft Capacity = QT + QS = 60.967tons + 307.63tons = 368.597tons
23
Settlement
Determine load corresponding to shaft top deflection of 0.216in and 2.16in.
*100
settlementR
shaft diameter=
R(0.216in) = 0.6
R(0.216in) = 0.6
Load Corresponding to R = 0.6:
Load carried in skin friction:
For R < 0.908333
( ) ( ) ( ) ( )
4 3 2
max
4 3 2
max
2.16 6.34 7.36 4.15
2.16 0.6 6.34 0.6 7.36 0.6 4.15 0.6 0.9299
(0.9299)307.63 286.07
s
s
s
s
s
fR R R R
f
f
f
Q tons tons
= − + − +
= − + − + =
= =
Load carried in end bearing:
For R = 0.6
4 3
T max
2
4 3
T max
2
T max
0.0001079 0.0035584
0.045115 0.34861
0.0001079(0.6) 0.0035584(0.6)
0.045115(0.6) 0.34861(0.6)
0.1937
T
T
T
qR R
q
R R
q
q
q
q
= − +
− +
= − +
− +
=
QT = (0.1937)60.967tons = 11.81tons
Total load at R = 0.6 is 11.81tons + 286.07tons = 297.88tons
24
Load Corresponding to R = 6.0:
Load carried in skin friction:
For R > 0.908333
max
0.978112s
s
f
f=
QS = (0.978112)307.63tons = 300.90tons
Load carried in end bearing:
4 3
T max
2
4 3
T max
2
T max
0.0001079 0.0035584
0.045115 0.34861
0.0001079(6.0) 0.0035584(6.0)
0.045115(6.0) 0.34861(6.0)
1.0963
T
T
T
qR R
q
R R
q
q
q
q
= − +
− +
= − +
− +
=
QT = (1.0963)60.967tons = 66.838tons
Total load at R = 6.0 is 69.838tons + 300.90tons = 367.738tons
FB-Deep v3.0.0 Prediction
Shaft Capacity
Ultimate Shaft capacity = 369.01(tons)
R(%) Settl.(in) Qs(tons) Qb(tons) Qt(tons)
0.6 0.216 286.45 11.81 298.26
6.0 2.160 301.30 66.84 368.14
25
Clay cu=2.255tsf
ST=1
Case 2B) – Shaft in Sand (N=15) over Clay (N=30) (Example file: 2BShaftSandClayN30.in)
End Bearing
For soil type 3, sand, end bearing is:
( ) 0.60 , where 45
0.60(15) 9
T SPT T
T
q tsf N q tsf
q tsf
=
= =
For soil type 1, clay, end bearing is:
( ) , where 40T c u Tq tsf N c q tsf=
( ) 2.250uc tsf tsf=
9(2.250 ) 20.25Tq tsf tsf= =
1.5B above tip: 1.5*3ft=4.5ft or depth 35.5ft.
2.0B below tip: 2*3ft=6ft or depth of 50.5ft.
(The layer below the bearing layer has a qT that exceeds the unit end bearing of the
bearing layer and therefore the bottom layer is excluded from the calculation.)
,1.5
,2
4.5 *99
4.5
4 *99
4
9 99
2
T B
T B
T
ft tsfq tsf
ft
ft tsfq tsf
ft
tsf tsfq tsf
= =
= =
+= =
Depth qT
35.5 9tsf 1.5B Above 40 9tsf Shaft Tip
44- 9tsf
44+ 20.25tsf Bottom Elevation
Depth N cu ST 0 15 3
5' 15 3 10' 15 3
15' 15 3 20' 15 3 25' 15 3
30' 15 3 35' 15 3
40' 15 3 44'- 15 3 44’+ 2.25 1 45’ 2.25 1 50' 2.25 1
55' 2.25 1
60’ 2.25 1
36” Diameter
Sand N=15
ST=3
= 100pcf
44’ 40’
26
Total End Bearing 2
3* 9 * * 63.617
2T T B
ftQ q A tsf tons
= = =
Skin Friction
For soil type 3, sand, skin friction is:
'S vf =
1.5 0.135 ( )z ft = −
1.2 > > 0.25 thus, if z < 5ft, then =1.2 and if z > 86ft, then = 0.25
2
1'
z
vz
Qs dA=
2
1'
z
vz
Qs D dz =
( )2
1
z
Z w wz
Qs D z dz = −
z = 0 to 5ft
( )( )
( )( )
5
0
52
0
3 1.2 100
3 1202
14137.17 7.068
Qs zdz
zQs
Qs lbs tons
=
=
= =
z = 5 to 40ft
( )( )
( )
( )
40
5
403/2
5
40
2 5/2
5
3 1.5 0.135 100
3 150 13.5
1 23 150 13.5
2 5
615964.30 14826.41 601137.89 300.57
Qs z zdz
Qs z z dz
Qs z z
Qs lbs lbs lbs tons
= −
= −
= −
= − = =
Total skin friction = 7.068tons + 300.57tons = 307.63tons
Shaft Capacity
Total Shaft Capacity = QT + QS = 63.617tons + 307.63tons = 371.25tons
27
Settlement
Determine load corresponding to shaft top deflection of 0.216in and 2.16in.
*100
settlementR
shaft diameter=
R(0.216in) = 0.6
R(0.216in) = 0.6
Load Corresponding to R = 0.6:
Load carried in skin friction:
For R < 0.908333
( ) ( ) ( ) ( )
4 3 2
max
4 3 2
max
2.16 6.34 7.36 4.15
2.16 0.6 6.34 0.6 7.36 0.6 4.15 0.6 0.9299
(0.9299)307.63 286.07
s
s
s
s
s
fR R R R
f
f
f
Q tons tons
= − + − +
= − + − + =
= =
Load carried in end bearing:
For R = 0.6
4 3
T max
2
4 3
T max
2
T max
0.0001079 0.0035584
0.045115 0.34861
0.0001079(0.6) 0.0035584(0.6)
0.045115(0.6) 0.34861(0.6)
0.1937
T
T
T
qR R
q
R R
q
q
q
q
= − +
− +
= − +
− +
=
QT = (0.1937)63.617tons = 12.32tons
Total load at R = 0.6 is 12.32tons + 286.07tons = 298.39tons
28
Load Corresponding to R = 6.0:
Load carried in skin friction:
For R > 0.908333
max
0.978112s
s
f
f=
QS = (0.978112)307.63tons = 300.90tons
Load carried in end bearing:
4 3
T max
2
4 3
T max
2
T max
0.0001079 0.0035584
0.045115 0.34861
0.0001079(6.0) 0.0035584(6.0)
0.045115(6.0) 0.34861(6.0)
1.0963
T
T
T
qR R
q
R R
q
q
q
q
= − +
− +
= − +
− +
=
QT = (1.0963)63.617tons = 69.74tons
Total load at R = 6.0 is 69.74tons + 300.90tons = 370.264tons
FB-Deep v3.0.0 Prediction
Shaft Capacity
Ultimate Shaft capacity = 371.66(tons)
R(%) Settl.(in) Qs(tons) Qb(tons) Qt(tons)
0.6 0.216 286.45 12.32 298.77
6.0 2.160 301.30 69.74 371.04
29
Case 3A) – Pile in Sand (N=15) under Clay (N=10) (Example file: 3ApileSandClayN10.in)
End Bearing
For soil type 3, sand, end bearing is:
3.2 3.2(15)( ) 16
3 3T
Nq tsf tsf= = =
For soil type 1, clay, end bearing is:
0.7 0.7(10)( ) 2.33
3 3T
Nq tsf tsf= = =
8B above tip: 8*2ft=16ft or depth of 34ft.
3.5B below tip: 3.5*2ft=7ft or depth of 57ft.
,8
,3.5
6 *2.333 10 *1610.875
16
7 *1616
7
10.875 1613.438
2
T B
T B
T
ft tsf ft tsfq tsf
ft
ft tsfq tsf
ft
tsf tsfq tsf
+= =
= =
+= =
Check embedment:
DC for soil type 3 = 9B = 9 * 2ft = 18ft
DA = 10ft < 18ft Correction needed
( )ATC LC T LC
C
Dq q q q
D= + −
Depth qT
34 2.333tsf 8B Above 35 2.333tsf 40- 2.333tsf 40+ 16tsf 45 16tsf Pile Tip 50 16tsf 55 16tsf 57 16tsf 3.5B Below
Depth N ST 0 10 1
5' 10 1 10' 10 1
15' 10 1 20' 10 1 25' 10 1
30' 10 1 35' 10 1
40'- 10 1 40'+ 15 3 45’ 15 3 50' 15 3
55' 15 3
60’ 15 3
24” x 24”
Clay N=10
ST=1
Sand N=15
ST=3
40’ 50’
30
Find qLC
,8
,3.5
16 *2.3332.333
16
7 *1616
7
2.333 169.167
2
LC B
LC B
LC
ft tsfq tsf
ft
ft tsfq tsf
ft
tsf tsfq tsf
= =
= =
+= =
( )10
9.167 13.437 9.167 11.53918
TC
ftq tsf tsf tsf tsf
ft= + − =
Total End Bearing
* 11.539 *2 *2 46.157T TC BQ q A tsf ft ft tons= = =
Skin Friction
For soil type 1, clay, skin friction is:
2.0 (110 ) 2.0(10)(110 10)( ) 0.4992
4006.6 4006.6s
N Nf tsf tsf
− −= = =
1 0.4992 *2 *40 *4 159.736SQ tsf ft ft tons= =
For soil type 3, sand, skin friction is:
( ) 0.019 0.019(15) 0.285sf tsf N tsf= = =
1 0.285 *2 *10 *4 159.736SQ tsf ft ft tons= =
Skin friction must be corrected for overlying layer being weaker. Since DA < DC, only the
skin friction between the top of the bearing layer and critical depth must be corrected:
Corrected Skin Friction in the Bearing Layer:
( )2*
ALC T LC
T C
DSFBLCSFBL q q q
q D
= + −
22.80 109.167 (13.437 9.167 ) 17.566
13.438 2*18
tons ftCSFBL tons tons tons tons
tons ft
= + − =
Depth qT
24 2.333tsf 8B Above 25 2.333tsf 30 2.333tsf 35 2.333tsf 40- 2.333tsf Layer Change 40+ 16tsf 45 16tsf 47 16tsf 3.5B Below
31
Total Skin Friction
159.736 17.566 177.303SQ tons tons tons= + =
Pile Capacity
Davisson Capacity = QT + QS = 46.157tons + 177.303tons = 233.460tons
Allowable Capacity = ½ Davisson = 111.730tons
Ultimate Capacity = 3QT + QS = 3(46.157tons) + 177.303tons = 315.774tons
FB-Deep v3.0.0 Prediction
Pile Capacity
Estimated Davisson capacity = 223.46(tons)
Allowable pile capacity = 111.73(tons)
Ultimate pile capacity = 315.78(tons)
32
Clay cu=0.75tsf
ST=1
Case 3A) – Shaft in Sand (N=15) under Clay (N=10) (Example file: 3AShaftSandClayN10.in)
End Bearing
For soil type 3, sand, end bearing is:
( ) 0.60 , where 45
0.60(15) 9
T SPT T
T
q tsf N q tsf
q tsf
=
= =
For soil type 1, clay, end bearing is:
( ) , where 40T c u Tq tsf N c q tsf=
( ) 0.750uc tsf tsf=
9(0.750 ) 6.75Tq tsf tsf= =
1.5B above tip: 1.5*3ft=4.5ft or depth 45.5ft.
2.0B below tip: 2*3ft=6ft or depth of 56ft.
,1.5
,2
4.5 *99
4.5
6 *99
6
9 99
2
T B
T B
T
ft tsfq tsf
ft
ft tsfq tsf
ft
tsf tsfq tsf
= =
= =
+= =
Total End Bearing 2
3* 9 * * 63.62
2T T B
ftQ q A tsf tons
= = =
Depth qT
45.5 9tsf 1.5B Above 50 9tsf Shaft Tip
55 9tsf
56 9tsf 2.0B Below
Depth N cu ST 0 0.75 1
5' 0.75 1 10' 0.75 1
15' 0.75 1 20' 0.75 1 25' 0.75 1
30' 0.75 1 35' 0.75 1
40'- 0.75 1 40'+ 15 3 45’ 15 3 50' 15 3
55' 15 3
60’ 15 3
36” Diameter
Sand N=15
ST=3
= 100pcf
50’ 40’
33
Skin Friction
For soil type 1, clay, skin friction is:
2
1
2
1
2
1
2
1
2.75 SU u SU
z
SUz
z
SUz
z
SU z
z
u z
f c f tsf
Q f dAs
Q Df dzs
Q Df zs
Q D c zs
=
=
=
=
=
z = 0 to 5 ft
3(0.0)(0.750 )5 3(0.0)(0.750 )0 0Qs tsf ft tsf ft tons = − =
z = 5 to 40 ft
3(0.55)(0.750 )40 3(0.55)(0.750 )5 136.070Qs tsf ft tsf ft tons = − =
For soil type 3, sand, skin friction is:
'S vf =
1.5 0.135 ( )z ft = −
1.2 > > 0.25 thus, if z < 5ft, then =1.2 and if z > 86ft, then = 0.25
2
1'
z
vz
Qs dA=
2
1'
z
vz
Qs D dz =
( )2
1
z
Z w wz
Qs D z dz = −
z = 40 to 50ft
( )( )
( )
( )
50
40
503/2
40
50
2 5/2
40
3 1.5 0.135 100
3 150 13.5
1 23 150 13.5
2 5
867462.079 615964.303 251497.773 125.749
Qs z zdz
Qs z z dz
Qs z z
Qs lbs lbs lbs tons
= −
= −
= −
= − = =
Total skin friction = 136.070tons + 125.749tons = 261.819tons
=0 for top
5ft and
bottom 3ft
(B).
34
Shaft Capacity
Total Shaft Capacity = QT + QS = 63.62tons + 261.819tons = 325.439tons
Settlement
Determine load corresponding to shaft top deflection of 0.216in and 2.16in.
*100
settlementR
shaft diameter=
R(0.216in) = 0.6
R(0.216in) = 0.6
Load Corresponding to R = 0.6:
Load carried in skin friction for clay layer z = 0 to 40ft:
For R = 0.6
( )
( )
max
max
0.095155 0.892937*
0.60.950996
0.095155 0.892937*0.6
(0.950996)136.070 129.402
s
s
s
s
s
f R
f R
f
f
Q tons tons
=+
= =+
= =
Load carried in skin friction for sand layer z = 40 to 50ft:
For R < 0.908333
( ) ( ) ( ) ( )
4 3 2
max
4 3 2
max
2.16 6.34 7.36 4.15
2.16 0.6 6.34 0.6 7.36 0.6 4.15 0.6 0.9299
(0.9299)125.749 116.934
s
s
s
s
s
fR R R R
f
f
f
Q tons tons
= − + − +
= − + − + =
= =
35
Load carried in end bearing for sand layer z = 40 to 50ft:
For R = 0.6
4 3
T max
2
4 3
T max
2
T max
0.0001079 0.0035584
0.045115 0.34861
0.0001079(0.6) 0.0035584(0.6)
0.045115(0.6) 0.34861(0.6)
0.1937
T
T
T
qR R
q
R R
q
q
q
q
= − +
− +
= − +
− +
=
QT = (0.1937)63.62tons = 12.323tons
Total load at R = 0.6 is 12.323tons + 116.934tons + 129.402tons = 258.659tons
Load Corresponding to R = 6.0:
Load carried in skin friction for clay layer z = 0 to 40ft:
For R > 2.0
max
0.833s
s
f
f=
QS = (0.833)136.070tons = 113.355tons
Load carried in skin friction for sand layer z = 40 to 50ft:
For R > 0.908333
max
0.978112s
s
f
f=
QS = (0.978112)125.749tons = 122.997tons
36
Load carried in end bearing:
4 3
T max
2
4 3
T max
2
T max
0.0001079 0.0035584
0.045115 0.34861
0.0001079(6.0) 0.0035584(6.0)
0.045115(6.0) 0.34861(6.0)
1.0963
T
T
T
qR R
q
R R
q
q
q
q
= − +
− +
= − +
− +
=
QT = (1.0963)63.62tons = 69.747tons
Total load at R = 6.0 is 69.747tons + 122.997tons + 113.355tons = 306.098tons
FB-Deep v3.0.0 Prediction
Shaft Capacity
Ultimate Shaft capacity = 325.51(tons)
R(%) Settl.(in) Qs(tons) Qb(tons) Qt(tons)
0.6 0.216 246.41 12.32 258.73
6.0 2.160 236.42 69.74 306.16
37
Clay cu=2.25tsf
ST=1
Case 3B) – Shaft in Sand (N=15) under Clay (N=30) (Example file: 3BShaftSandClayN30.in)
End Bearing
For soil type 3, sand, end bearing is:
( ) 0.60 , where 45
0.60(15) 9
T SPT T
T
q tsf N q tsf
q tsf
=
= =
For soil type 1, clay, end bearing is:
( ) , where 40T c u Tq tsf N c q tsf=
( ) 0.075(30) 2.250uc tsf tsf= =
9(2.250 ) 20.25Tq tsf tsf= =
1.5B above tip: 1.5*3ft=4.5ft or depth 45.5ft.
2.0B below tip: 2*3ft=6ft or depth of 56ft.
,1.5
,2
4.5 *99
4.5
6 *99
6
9 99
2
T B
T B
T
ft tsfq tsf
ft
ft tsfq tsf
ft
tsf tsfq tsf
= =
= =
+= =
Total End Bearing 2
3* 9 * * 63.62
2T T B
ftQ q A tsf tons
= = =
Depth qT
45.5 9tsf 1.5B Above 50 9tsf Shaft Tip
55 9tsf
56 9tsf 2.0B Below
Depth N cu ST 0 2.25 1
5' 2.252.25
1 10' 2.25 1
15' 2.25 1 20' 2.25 1 25' 2.25 1
30' 2.25 1 35' 2.25 1
40'- 2.25 1 40'+ 15 3 45’ 15 3 50' 15 3
55' 15 3
60’ 15 3
Sand N=15
ST=3
= 100pcf
50’ 40’
36” Diameter
38
Skin Friction
For soil type 1, clay, skin friction is:
2
1
2
1
2
1
2
1
2.75 SU u SU
z
SUz
z
SUz
z
SU z
z
u z
f c f tsf
Q f dAs
Q Df dzs
Q Df zs
Q D c zs
=
=
=
=
=
z = 0 to 5 ft
3(0.0)(2.250 )5 3(0.0)(2.250 )0 0Qs tsf ft tsf ft tons = − =
z = 5 to 40 ft
3(0.55)(2.250 )40 3(0.55)(2.250 )5 408.211Qs tsf ft tsf ft tons = − =
For soil type 3, sand, skin friction is:
'S vf =
1.5 0.135 ( )z ft = −
1.2 > > 0.25 thus, if z < 5ft, then =1.2 and if z > 86ft, then = 0.25
2
1'
z
vz
Qs dA=
2
1'
z
vz
Qs D dz =
( )2
1
z
Z w wz
Qs D z dz = −
z = 40 to 50ft
( )( )
( )
( )
50
40
503/2
40
50
2 5/2
40
3 1.5 0.135 100
3 150 13.5
1 23 150 13.5
2 5
867462.079 615964.303 251497.773 125.749
Qs z zdz
Qs z z dz
Qs z z
Qs lbs lbs lbs tons
= −
= −
= −
= − = =
Total skin friction = 408.21tons + 125.749tons = 533.960tons
=0 for top
5ft and
bottom 3ft
(B).
39
Shaft Capacity
Total Shaft Capacity = QT + QS = 63.62tons + 533.960tons = 597.500tons
Settlement
Determine load corresponding to shaft top deflection of 0.216in and 2.16in.
*100
settlementR
shaft diameter=
R(0.216in) = 0.6
R(0.216in) = 0.6
Load Corresponding to R = 0.6:
Load carried in skin friction for clay layer z = 0 to 40ft:
For R = 0.6
( )
( )
max
max
0.095155 0.892937*
0.60.950996
0.095155 0.892937*0.6
(0.950996)408.21 388.206
s
s
s
s
s
f R
f R
f
f
Q tons tons
=+
= =+
= =
Load carried in skin friction for sand layer z = 40 to 50ft:
For R < 0.908333
( ) ( ) ( ) ( )
4 3 2
max
4 3 2
max
2.16 6.34 7.36 4.15
2.16 0.6 6.34 0.6 7.36 0.6 4.15 0.6 0.9299
(0.9299)125.749 116.934
s
s
s
s
s
fR R R R
f
f
f
Q tons tons
= − + − +
= − + − + =
= =
40
Load carried in end bearing for sand layer z = 40 to 50ft:
For R = 0.6
4 3
T max
2
4 3
T max
2
T max
0.0001079 0.0035584
0.045115 0.34861
0.0001079(0.6) 0.0035584(0.6)
0.045115(0.6) 0.34861(0.6)
0.1937
T
T
T
qR R
q
R R
q
q
q
q
= − +
− +
= − +
− +
=
QT = (0.1937)63.62tons = 12.323tons
Total load at R = 0.6 is 12.323tons + 116.934tons + 388.206tons = 517.463tons
Load Corresponding to R = 6.0:
Load carried in skin friction for clay layer z = 0 to 40ft:
For R > 2.0
max
0.833s
s
f
f=
QS = (0.833)408.211tons = 340.04tons
Load carried in skin friction for sand layer z = 40 to 50ft:
For R > 0.908333
max
0.978112s
s
f
f=
QS = (0.978112)125.749tons = 122.997tons
41
Load carried in end bearing:
4 3
T max
2
4 3
T max
2
T max
0.0001079 0.0035584
0.045115 0.34861
0.0001079(6.0) 0.0035584(6.0)
0.045115(6.0) 0.34861(6.0)
1.0963
T
T
T
qR R
q
R R
q
q
q
q
= − +
− +
= − +
− +
=
QT = (1.0963)63.62tons = 69.747tons
Total load at R = 6.0 is 69.747tons + 340.04tons + 113.355tons = 532.782tons
FB-Deep v3.0.0 Prediction
Shaft Capacity
Ultimate Shaft capacity = 597.65(tons)
R(%) Settl.(in) Qs(tons) Qb(tons) Qt(tons)
0.6 0.216 505.21 12.32 517.53
6.0 2.160 463.11 69.74 532.85
42
Case 4A) – Shaft in Clay (N=20) over Clay (N=10) (Example file: 4AshaftclayclayN10.in)
End Bearing
For soil type 1, clay, end bearing is:
( ) , where 40T c u Tq tsf N c q tsf=
45
6.0 1 0.2 24 93
c c
ftN N
ft
= + = =
For 0ft to 46ft: ( ) 0.075(20) 1.500uc tsf tsf= =
For 46ft to 60ft: ( ) 0.075(10) 0.750uc tsf tsf= =
9(1.500 ) 13.50
9(0.750 ) 6.75
T
T
q tsf tsf
q tsf tsf
= =
= =
1.5B above tip: 1.5 * 3ft = 4.5ft or depth 40.5ft.
3.0B below tip: 3.0 * 3ft = 9ft or depth of 54ft.
,1.5
,3
4.5 *13.513.5
4.5
1 *13.5 8 *6.757.5
9
13 7.87510.5
2
T B
T B
T
ft tsfq tsf
ft
ft tsf ft tsfq tsf
ft
tsf tsfq tsf
= =
+= =
+= =
Total End Bearing 2
3* 10.5 * * 74.22
2T T B
ftQ q A tsf tons
= = =
Depth qT
40.5 13.5tsf 1.5B Above 45 13.5tsf Shaft Tip
46- 13.5tsf
46+ 6.75tsf
50 6.75tsf
54 6.75tsf 3.0B Below
Depth cu ST 0 1.50 1
5' 1.50 1 10' 1.50 1
15' 1.50 1 20' 1.50 1 25' 1.50 1
30' 1.50 1 35' 1.50 1
40' 1.50 1 45' 1.50 1 46’- 1.50 1 46'+ 0.75 1
50' 0.75 1
55’ 0.75 1
60’ 0.75 1
36” Diameter
Clay cu=1.50
ST=1
= 100pcf
Clay cu=0.75
ST=1
= 100pcf
46’ 45’
43
Skin Friction
For soil type 1, clay, skin friction is:
2
1
2
1
2
1
2
1
2.75 SU u SU
z
SUz
z
SUz
z
SU z
z
u z
f c f tsf
Q f dAs
Q Df dzs
Q Df zs
Q D c zs
=
=
=
=
=
z = 0 to 5 ft
3(0.0)(1.50 )5 3(0.0)(1.50 )0 0Qs tsf ft tsf ft tons = − =
z = 5 to 40 ft
3(0.55)(1.50 )42 3(0.55)(1.50 )5 287.691Qs tsf ft tsf ft tons = − =
z = 42 to 45 ft
3(0.0)(1.50 )45 3(0.0)(1.50 )42 0Qs tsf ft tsf ft tons = − =
Total skin friction = 0tons + 287.691tons + 0tons = 287.691tons
Shaft Capacity
Total Shaft Capacity = QT + QS = 74.22tons + 287.691tons = 361.911tons
Settlement
Determine load corresponding to shaft top deflection of 0.216in and 2.16in.
*100
settlementR
shaft diameter=
R(0.216in) = 0.6
R(0.216in) = 0.6
Load Corresponding to R = 0.6:
Load carried in skin friction:
For R = 0.6
=0 for top
5ft and
bottom 3ft
(B). 45’
44
( )
( )
max
max
0.095155 0.892937*
0.60.950996
0.095155 0.892937*0.6
(0.950996)287.691 273.593
s
s
s
s
s
f R
f R
f
f
Q tons tons
=+
= =+
= =
Load carried in end bearing:
For R = 0.6
4 5 3 4
T max
2 3 2
4 5 3 4
T max
2 3 2
1.1823 10 3.7091 10
4.4944 10 0.26537 0.78436
1.1823 10 (0.6) 3.7091 10 (0.6)
4.4944 10 (0.6) 0.26537(0.6) 0.78436(0.6) 0.384
T
T
qx R x R
q
x R R R
qx x
q
x
− −
−
− −
−
= − +
− +
= − +
− + =
QT = (0.384)74.22tons = 28.50tons
Total load at R = 0.6 is 28.50tons + 273.593tons = 302.093tons
Load Corresponding to R = 6.0:
Load carried in skin friction:
For R > 2.0
max
0.833s
s
f
f=
QS = (0.833)287.691tons = 239.647tons
Load carried in end bearing:
4 5 3 4
T max
2 3 2
1.1823 10 (6.0) 3.7091 10 (6.0)
4.4944 10 (6.0) 0.26537(6.0) 0.78436(6.0) 0.973
Tqx x
q
x
− −
−
= − +
− + =
QT = (0.973)74.22tons = 72.216tons
Total load at R = 6.0 is 239.647tons + 72.216tons = 311.863tons
45
FB-Deep v3.0.0 Prediction
Shaft Capacity
Ultimate Shaft capacity = 361.91(tons)
R(%) Settl.(in) Qs(tons) Qb(tons) Qt(tons)
0.6 0.216 273.59 28.52 302.12
6.0 2.160 239.65 72.22 311.87
46
Case 4B) – Shaft in Clay (N=20) over Sand (N=15) (Example file: 4bshaftclayclayN15.in)
End Bearing
For soil type 1, clay, end bearing is:
( ) , where 40T c u Tq tsf N c q tsf=
For 0ft to 46ft: ( ) 0.075(20) 1.500uc tsf tsf= =
9(1.500 ) 13.50Tq tsf tsf= =
For soil type 3, sand, end bearing is:
( ) 0.60 , where 45
0.60(15) 9
T SPT T
T
q tsf N q tsf
q tsf
=
= =
1.5B above tip: 1.5 * 3ft = 4.5ft or depth 40.5ft.
3.0B below tip: 3.0 * 3ft = 9ft or depth of 54ft.
,1.5
,3
4.5 *13.513.5
4.5
1 *13.5 8 *9.09.50
9
13 9.011.5
2
T B
T B
T
ft tsfq tsf
ft
ft tsf ft tsfq tsf
ft
tsf tsfq tsf
= =
+= =
+= =
Total End Bearing 2
3* 11.5 * * 81.289
2T T B
ftQ q A tsf tons
= = =
Depth qT
40.5 13.5tsf 1.5B Above 45 13.5tsf Shaft Tip
46- 13.5tsf
46+ 6.75tsf
50 6.75tsf
54 6.75tsf 3.0B Below
Depth N cu ST 0 1.50 1
5' 1.50 1 10' 1.50 1
15' 1.50 1 20' 1.50 1 25' 1.50 1
30' 1.50 1 35' 1.50 1
40' 1.50 1 45' 1.50 1 46’- 1.50 1 46'+ 15 3
50' 15 3
55’ 15 3
60’ 15 3
Clay cu=1.50
ST=1
= 100pcf
Sand N=15
ST=3
= 100pcf
46’ 45’
36” Diameter
47
Skin Friction
For soil type 1, clay, skin friction is:
2
1
2
1
2
1
2
1
2.75 SU u SU
z
SUz
z
SUz
z
SU z
z
u z
f c f tsf
Q f dAs
Q Df dzs
Q Df zs
Q D c zs
=
=
=
=
=
z = 0 to 5 ft
3(0.0)(1.50 )5 3(0.0)(1.50 )0 0Qs tsf ft tsf ft tons = − =
z = 5 to 40 ft
3(0.55)(1.50 )42 3(0.55)(1.50 )5 287.691Qs tsf ft tsf ft tons = − =
z = 42 to 45 ft
3(0.0)(1.50 )45 3(0.0)(1.50 )42 0Qs tsf ft tsf ft tons = − =
Total skin friction = 0tons + 287.691tons + 0tons = 287.691tons
Shaft Capacity
Total Shaft Capacity = QT + QS = 81.289tons + 287.691tons = 368.98tons
Settlement
Determine load corresponding to shaft top deflection of 0.216in and 2.16in.
*100
settlementR
shaft diameter=
R(0.216in) = 0.6
R(0.216in) = 0.6
Load Corresponding to R = 0.6:
Load carried in skin friction:
For R = 0.6
=0 for top
5ft and
bottom 3ft
(B). 45’
48
( )
( )
max
max
0.095155 0.892937*
0.60.950996
0.095155 0.892937*0.6
(0.950996)287.691 273.593
s
s
s
s
s
f R
f R
f
f
Q tons tons
=+
= =+
= =
Load carried in end bearing:
For R = 0.6
4 5 3 4
T max
2 3 2
4 5 3 4
T max
2 3 2
1.1823 10 3.7091 10
4.4944 10 0.26537 0.78436
1.1823 10 (0.6) 3.7091 10 (0.6)
4.4944 10 (0.6) 0.26537(0.6) 0.78436(0.6) 0.384
T
T
qx R x R
q
x R R R
qx x
q
x
− −
−
− −
−
= − +
− +
= − +
− + =
QT = (0.384)81.289tons = 31.215tons
Total load at R = 0.6 is 31.218tons + 273.593tons = 304.808tons
Load Corresponding to R = 6.0:
Load carried in skin friction:
For R > 2.0
max
0.833s
s
f
f=
QS = (0.833)287.691tons = 239.647tons
Load carried in end bearing:
4 5 3 4
T max
2 3 2
1.1823 10 (6.0) 3.7091 10 (6.0)
4.4944 10 (6.0) 0.26537(6.0) 0.78436(6.0) 0.973
Tqx x
q
x
− −
−
= − +
− + =
QT = (0.973)81.289tons = 79.094tons
Total load at R = 6.0 is 239.647tons + 79.094tons = 318.741tons
49
FB-Deep v3.0.0 Prediction
Shaft Capacity
Ultimate Shaft capacity = 368.98(tons)
R(%) Settl.(in) Qs(tons) Qb(tons) Qt(tons)
0.6 0.216 273.59 31.24 304.83
6.0 2.160 239.65 79.10 318.75
50
Sand N=15
ST=1
= 100pcf
36” Diameter
c =130pcf Ec=4030ksi
Case 5) – Shaft in rock under sand with smooth socket
(Example file: 5shaftsandsmoothrocksocket.in)
End Bearing
User defined: qb = ½qu = ½(10tsf) = 5tsf
Total End Bearing
23
* 5.0 * * 35.3432
t t b
ftQ q A tsf tons
= = =
Skin Friction
For soil type 3, sand, skin friction is:
z = 0 to 5ft
( )( )
( )( )
5
0
52
0
3 1.2 100
3 1202
14137.17 7.068
Qs zdz
zQs
Qs lbs tons
=
=
= =
z = 5 to 40ft
( )( )
( )
( )
40
5
403/2
5
40
2 5/2
5
3 1.5 0.135 100
3 150 13.5
1 23 150 13.5
2 5
615964.30 14826.41 601137.89 300.57
Qs z zdz
Qs z z dz
Qs z z
Qs lbs lbs lbs tons
= −
= −
= −
= − = =
Skin friction in sand layer = 7.069tons + 300.570tons = 307.639tons
Depth N ST qu qt
0' 15 3
5' 15 3
10' 15 3
15' 15 3
20' 15 3
25' 15 3
30' 15 3
35' 15 3
40'- 15 3
40'+ 4 10tsf 1.0tsf
45' 4 10tsf 1.0tsf
50' 4 10tsf 1.0tsf
55' 4 10tsf 1.0tsf
60' 4 10tsf 1.0tsf
65' 4 10tsf 1.0tsf
Rock
qu=10tsf
qt=1.0tsf
qb=0.5qu
Em=115qu
45’ 40’
51
For soil type 4 skin friction based on McVay et al. (1992):
1 110 1.0 1.581
2 2
* * * 1.581 * *3 *5 74.509
SU U t
S SU
f q q tsf tsf tsf
Q f D L tsf ft ft tons
= = =
= = =
Total skin friction = 7.069tons + 300.570tons + 74.509tons = 382.148tons
Shaft Capacity
Total Shaft Capacity = QT + QS = 382.148tons + 35.343tons = 417.491tons
Settlement
FHWA IGM Calculations: (Note: Must enter values for Ec, slump, and Em)
Em = 115*qu = 115(10.0tsf) = 1150tsf
( ) ( )( )
1/2 1/2
1/2 1/2
1.14 0.05 1 log 0.44, 2
40301.14 2 0.05 2 1 log 0.44 1.12
1150
c
m
EL L Lwhere
D D E D
ksi
tsf
= − − −
= − − − =
( ) ( )( )
1/2 1/2
1/2 1/2
0.37 0.15 1 log 0.13
40300.37 2 0.15 2 1 log 0.13 0.504
1150
c
m
EL L
D D E
ksi
tsf
= − − +
= − − + =
1
; 1.581
1150 *1.122103.117
*5 *0.504*1.581
f mSU
t SU
f
t
Ef tsf
w L f
tsfft
w ft tsf
−
= =
= =
( )
( )( )
( )( ) ( )( )
( )( )
( )( ) ( )
0.670.5
0.670.5
2/3
200 / 1 //0.0134
/ 1
200* 3 1.122 * 320.0134*1150 80.931
3 *5 *0.504
m
L D L DL DE
LL D
tsftsf
ft ft
− + = +
− = =
52
Determine n for deformation criteria (Figure 4) 10.0
9.5761.044272
u
p
q tsf
tsf= =
5
; ; since 40 42.52
mn c c c
n
EM Z Z ft
= = + =
For a slump = 7in, M(table4)=0.65
0.65*130 *42.5 1.796
1150640.446 0.4843
1.796
n
m
n
pcf ft tsf
E tsfn
tsf
= =
= =
Select values of ‘w’ for calculating
20.67
2
for ; 4
4
t SU b b
t SU b
DQ DL f q n q w
DQ DLkf q
= + =
= +
Let R=0.1%
w = 0.1%*B = 0.001*3ft = 0.003ft; /w = 103.117 ft-1,
1103.117 *0.003 0.309 0.4843ft ft n − = = =
22/3 2/3
,
(3 )*3 *5 *0.309*1.581 *(80.931 * )*(0.003 )
4
23.049 11.899
34.949
t rock
ftQ ft ft tsf tsf ft ft
tons tons
tons
−= +
= +
=
Sand Layer Above
Load Corresponding to R = 0.1% carried in skin friction:
For R < 0.908333
( ) ( ) ( ) ( )
4 3 2
max
4 3 2
max
2.16 6.34 7.36 4.15
2.16 0.1 6.34 0.1 7.36 0.1 4.15 0.1 0.3475
(0.3475)307.639 106.912
s
s
s
s
s
fR R R R
f
f
f
Q tons tons
= − + − +
= − + − + =
= =
Total load at R = 0.1 is 106.912tons + 34.949tons = 141.861tons
53
Let R=0.6%
w = 0.6%*B = 0.0006*3ft = 0.018ft; /w = 103.117 ft-1,
1103.117 *0.018 1.856 0.4843ft ft n − = = =
( )( )( )
( )( )( )
1 1.856 0.4843 1 0.48430.4843 0.859
2 1 1.856 2*0.4843 1
n nk n
n
− − − −= + = + =
− + − +
22/3 2/3
,
(3 )*3 *5 *0.859*1.581 *(80.931 * )*(0.018 )
4
64.011 39.291
103.302
t rock
ftQ ft ft tsf tsf ft ft
tons tons
tons
−= +
= +
=
Sand Layer Above
Load Corresponding to R = 0.6% carried in skin friction:
For R < 0.908333
( ) ( ) ( ) ( )
4 3 2
max
4 3 2
max
2.16 6.34 7.36 4.15
2.16 0.6 6.34 0.6 7.36 0.6 4.15 0.6 0.9299
(0.9299)307.639 286.075
s
s
s
s
s
fR R R R
f
f
f
Q tons tons
= − + − +
= − + − + =
= =
Total load at R = 0.6 is 286.075tons + 103.302tons = 389.377tons
Let R=1.0%
w = 1.0%*B = 0.01*3ft = 0.03ft; /w = 103.117 ft-1,
1103.117 *0.03 3.093 0.4843ft ft n − = = =
( )( )( )
( )( )( )
1 3.093 0.4843 1 0.48430.4843 0.915
2 1 3.093 2*0.4843 1
n nk n
n
− − − −= + = + =
− + − +
22/3 2/3
,
(3 )*3 *5 *0.915*1.581 *(80.931 * )*(0.03 )
4
68.168 55.232
123.4
t rock
ftQ ft ft tsf tsf ft ft
tons tons
tons
−= +
= +
=
Sand Layer Above
Load Corresponding to R = 1% carried in skin friction:
For R > 0.908333
54
max
0.978112s
s
f
f=
QS = (0.978112)307.639tons = 300.905tons
Total load at R = 1.0 is 300.905tons + 123.4tons = 424.306tons
FB-Deep v3.0.0 Prediction
Shaft Capacity
Ultimate Shaft capacity = 418.80(tons)
R(%) Settl.(in) Qs(tons) Qb(tons) Qt(tons)
0.1 0.036 130.42 11.79 142.21
1.0 0.360 370.36 55.16 425.51