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