User Guide for FB-Deep F.C. Townsend

Intro: FB-Deep (formerly SHAFT-SPT) is a combination of SHAFT 98, which uses the FHWA O’Neill & Reese method for drilled shafts and intermediate geomaterials, and SPT97, which is the FDOT version for driven piles. The program is designed such that the SPT boring log only is entered once, and either shafts or piles can be analyzed.

SPT97 Example

SPT 97 Uniform Sand (N=15) Example

Depth N ST Sand ST=3 0 15 3

40 ft N = 15 blows 20 15 3 2 ft x 2 ft 30 15 3 psc 50 15 3 55 0 0

Opening screen Units = English, Type = Pile Unit wt. = 150 pcf, Section = Square (psc) Click – Insert Pile Single, L=40ft, W=24”

Click boring log icon in Toolbar

Boring Log Screen Click “ Insert Layer” Depths = 0 to 50 ft Soil Type = 3 (sand) SPT N = 15 blows Click OK Note: Last entry doesn’t = 0 As for old SPT97 Main Screen Click “Cap. Report” Results

SPT 97 Uniform Sand (N=15) Example

Depth N ST Sand ST=3 0 15 3 40 ft N = 15 blows 20 15 3 2 ft x 2 ft 30 15 3 50 15 3 55 0 0 F. Bearing Capacity

8 B above = 40 – 8(2) = 24ft 3.5 B below = 40 + 3.5(2) = 47ft

Elev N qt (tsf) for sand (ST = 3) qt = 3.2N/3 = 3.2(15)/3=16tsf 24 15 16 40 15 16 Ave[AQPTA] = 2(16)/2 = 16tsf Ave = (16 +16)/2=16tsf 40 15 16 Ave [AQPTB] =2(16)/2 = 16tsf 47 15 16 Check depth of embedment to see if correction required: Dc = 9B = 9(2)=18<Da=40; NO corection required. Maximum BC = 16tsf (2x2) = 64T Corrected CMAXB( no correction) = 64T E. Skin Friction Cpacity

A. Skin friction above bearing layer = 0 (air) B. Untimate skin friction in bearing layer (note: “trick” @ elva = 0 N=15, uniform distribution.)

0ft fs = 0.019N = 0.019(15) = 0.285 tsf 40ft

C. Ult. SFBL = 0.285(2x4)(40ft) = 91.2T Since uniform no corections required. G. Pile Capacity Estimated Davisson (USF + CMAXB) = 91.2T+64T = 155.2T (310k) Allowable = Davisson ÷ 2 = 155.2 /2 = 77.6T Ultimate (for driving) USF + 3(CMAXB) = 91.2T+ 3(64T) = 283.6T Deformation @ Davisson Offset (x) = 0.15 + B/120 = 0.15 + 24”/120 = 0.35”

"058.0)4415(24)1240(4.310

2 ==ksix

AEPL k

∴ δ = 0.35” +0.058” = 0.41” @ 310k

To design a range of pile Lengths and widths, Click “Insert Range” Range 20 to 40 ft long in increments of 5ft. Click “Graph cap.” Graphics screen SHAFT98 Example CLAYS: Example File: Clay1.dat 1. Multi Layer Clay with Casing

g

3 ft

Cl

Cl

6 ft

20 ft

20 ft

Casin

N =8, γ = 100 pcf , qc = 16 tsf

N =12,

γ = 100 pcf , qc = 30 tsf

ay

ay

Opening Screen:Drilled Shaft Units = English Water Table = deep R% = 0.0 Insert Shaft” Case = 6ft, L = 40ft, Diam =36”, Bell L = 0.0ft Bell Diam = 36”

Click Boring Log Icon In upper toolbar to get Boring log. Boring Log Screen Ground Surface = 0.0ft Cu Calcs = CPT (Lower Left) Click “Insert Layer” Depth ST γ CPT

0 1(clay) 100pcf 16tsf 20 1 100 16 20 1 100 30 40 1 100 40 50 1 100 50 Rock Friction – Not included Click “OK” to return to main menu

Click “ Cap. Report” For results Results Settlement

For 0.3”, R % = 0.833 Click “ Cap. Report”

R Results show Cap = 293.15T @ 0.3”

APPENDIX A - Examples CLAYS: Example File: Clay1.dat 2. Multi Layer Clay with Casing 3. Multi Layer Clay with Casing B > 75 ”

150σ−

= cqc

Clay Layer # 1 : c = *16

Clay Layer # 2 : c = *30

Clay Layer # 2 Tip c = 30

1. Multi Layer Clay with

a) Skin Friction Qs = π

= 4.9 = 24

b) End Beari

Qb

g

3 ft

Cl

Cl

6 ft

20 ft

20 ft

Casin

N =8, = 100 pcf , qc = 16 tsf

N =12, = 100 pcf , qc = 30 tsf

)0333.1(67.066,215

100*102000 tsfpsf=−

)90.1(800,315

100*302000 tsfpsf=−

)867.1(3.733,315

100*402000* tsfpsf=−

Casing: Full Capacity (40 ft Shaft)

: [ ])9.1*55.0)(302()033.1*55.0)(602(*0.3* ′−′+′−′

[ ]765.179567.7*248 +

Tons42.2

ng:

= 4

.2bqb

π ,

ay

ay

ucb CNq = , )9(9223

402.01*0.6 useNc >=

+=

Qb = Tonstsf 75.11843).867.1*9(

2

=π

c) Total Capacity = Skin Friction + End Bearing = 242.42 + 118.75 = 361.17 Tons (ultimate) d) Calculation of Skin Friction:

e) Settlement: S = (i) 0.3 ” and (ii) S = 3.0 ”

Qs = 242.42T , Qb = 118.75T , QT = 361.17T

(i)

TstqB

S 70.23442.242*)74.0833.0(115817.0978929.0,74.0833.036

100*3.0100*=−−=>==

T

EEEQq

b

br

60.5875.118*4935.0

(78436.0)833.0(26537.0)833.0(024944.4)833.0(037091.3)833.0(*041832.1 2345

==

+−−+−−−=

QT @0.3" = 234.70 + 58.60 = 293.30 T

1.9*0.55 =1.045 tsf

1.0333*0.55 = 0.568 tsf

. 50 1.0

40

30

35

25

15

20

10

5

0

For range of shaft lengths Click “Insert Range” Min L = 20ft Max L =40ft Increment = 2ft. Click “Cap. Graph” Graphical Results Example: Sand overlying Rock (IGM) Socket

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

40' 45'

Sand N=15 ST=3 γ = 100pcf

Rock qu = 10tsf qt = 1.0tsf qb = 0.5qu Em = 115qu

Diam = 36-in

Opening Screen: Units = English Type = Drilled Shaft γc = 130pcf, slump=6.89” Ec = 57,000(5000psi)1/2=4030 Click “Insert Shaft” L = 45ft, D=36” Click “Boring Log” Icon In top toolbar Enter Boring Log Data Depth ST N γ qu qt 0 3 15 100 40 3 15 100 40 4 10 1 65 4 10 1 Check “Use Default values” Depth qb Em RQD Socket 40 5 default 1.0 0 65 5 default 1.0 0 qb is not default, but ½ qu Em is 115 qu x 2/144 RQD = 1.0, no reduction Socket = 0 = smooth sides McVay’s Friction

= tu qq21

Click “OK”

Set R% = 0.1 Click “ Cap. Report”

Results Case 5) – Shaft in rock under sand with smooth socket (Example file: 5shaftsandsmoothrocksocket.spc)

End Bearing

User defined qb = ½qu = ½ (10tsf) = 5tsf

36” diameter γc = 130pcf Ec = 4030ksi 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

40' 45'

Sand N=15 ST=3 γ = 100pcf

Rock qu = 10tsf qt = 1.0tsf qb = 0.5qu Em = 115qu

Total End Bearing

tonsfttsfAqQ BTT 343.352

3**0.5*2

=

== π

Skin Friction

For soil type 3, sand, skin friction is:

z = 0 to 5 ft

∫=5

0

)100)(2.1(3 zdzQs π

( )( )5

0

2

21203 zQs π=

tonslbsQs 069.7167.14137 ==

z = 5 to 40 ft

∫ −=40

5

)100)(135.05.1(3 zdzzQs π

dzzzQs ∫

−=

40

5

23

5.131503π

40

5

25

2

525.13

21150)3(

−= zzQs π

tonslbslbsQs 570.30012.60113918.1482530.615964 ==−=

Skin friction in sand layer = 7.069tons + 300.570tons = 307.639tons

For soil type 4 skin friction based on McVay et al. (1992):

tsftsftsfqqf tUSU 581.10.11021

21

===

tonsftfttsfLDfQ SUS 509.745*3**581.1*** === ππ

Total skin friction = 307.639tons + 74.509tons =382.148tons

Total Shaft Capacity

Shaft Capacity 35.343tons + 382.148tons = 417.491tons Settlement

FHWA IGM Calculations: (Note: Must enter values for Ec, slump, and Em)

Em = 115 qu = 115 (10.0tsf) = 1150tsf

44.0)log(105.014.12/12/1

−

−

−

=Ω

m

c

EE

DL

DL

( ) ( ) 99679.044.0)1150

288000log(1667.105.0667.114.1 2/12/1 =−−−=Ωtsftsf

13.0)log(115.037.02/12/1

+

−

−

=Γ

m

c

EE

DL

DL

( ) ( ) 50282.013.0)1150

288000log(1667.115.0667.137.0 2/12/1 =+−−=Γtsftsf

tususu

m qqffL

Ew 2

1; =ΓΩ

=π

θ

=wθ 1799.91

)581.1(*50282.0*5*99679.0*1150 −= ft

tsffttsf

π

[ ] [ ] 67.01200

)1()(0134.0

Γ

+Ω−

+=Λ

LE D

LDL

DLDL

m π

[ ] [ ] 67.0

50282.0*5*667.1199769.0667.1200

667.2667.1)1150(0134.0

+−

=Λft

tsfπ

= 71.354tsf ft-0.67

Determine n for deformation criteria (figure4) 576.9044272.1

0.10==

tsftsfq

p

u

σ

4655.0312.640796.1

1150

796.135915.42*130*65.0

65.0)4(,7

5.422540;;

≈∴==∴

===∴

==

=+==

ntsftsfE

tsfpsfftpcf

tableMinslumpaFor

ftZSinceZME

n

m

n

cccnn

m

σ

σ

γσσ

Select values of ‘w’ for calculating

nforqDfkLDQ

wqnforqDfLDQ

bsut

bbsut

>+=

Λ=<+=

θππ

θπθπ

4

;4

2

67.02

Let w = 0.036in = 0.003 ft ; θ / w =91.799 ft-1,

∴θ = 91.799ft-1 * 0.003ft = 0.2754 < n = 0.4655

Sand Layer Above

Load Corresponding to R = 0.1% carried in skin friction:

For R ≤ 0.908333 RRRRff

S

S 15.436.734.616.2 234

max

+−+−=

( ) ( ) ( ) ( ) 3475.01.015.41.036.71.034.61.016.2 234

max

=+−+−=S

S

ff

QS = (0.3475)307.639tons = 106.905tons

( ) ( )

tons

tonstons

ftfttsffttsfftftQt

808.30

29.1051.20

003.0**354.71*43*)581.1(*2754.0*5*3* 67.067.0

2

=

+=

+= −ππ

Total load at R=0.1 30.808 tons + 106.905tons = 137.712 tons

Let w = 0.216 in = 0.018 ft. ; θ / w =91.799 ft-1,

∴θ = 91.799ft-1 * 0.018ft = 1.652 > n = 0.4655

8334.0)1)4655.0(2652.1(

)4655.01)(4655.0652.1(4655.0)12()1)((

=+−

−−+=

+−−−

+=n

nnnkθ

θ

Sand Layer Above

Load Corresponding to R = 0.6% carried in skin friction:

For R ≤ 0.908333 RRRRff

S

S 15.436.734.616.2 234

max

+−+−=

( ) ( ) ( ) ( ) 9781.06.015.46.036.76.034.66.016.2 234

max

=+−+−=S

S

ff

QS = (0.9299)307.639tons = 286.12tons

Total load at R=0.6 96.27 tons + 286.12tons = 382.39 tons

Let w = 0.360 in = 0.030 ft. ; θ / w =91.799 ft-1,

∴θ = 91.799ft-1 * 0.030ft = 2.754 > n = 0.39

8988.0)1)4655.0(2754.2(

)4655.01)(4655.0754.2(4655.0)12()1)((

=+−

−−+=

+−−−

+=n

nnnkθ

θ

.

( ) ( )

tons

tonstons

ftfttsffttsfftftQt

27.96

18.34091.62

018.0**354.71*43*)581.1(*8334.0*5*3* 67.067.0

2

=

+=

+= −ππ

( )

tons

tonstons

ftfttsftftsfftftQt

09.115

13.4896.66

030.0**354.71*43*)581.1(*8988.0*5*3* 67.067.0

2

=

+=

+= −ππ

Sand Layer Above

Load Corresponding to R = 1.0% carried in skin friction:

For R ≤ 0.908333 RRRRff

S

S 15.436.734.616.2 234

max

+−+−=

( ) ( ) ( ) ( ) 9781.00.115.40.136.70.134.60.116.2 234

max

=+−+−=S

S

ff

QS = (0.9781)307.639tons = 300.952tons

Total load at R=1.0 115.09 tons + 300.952tons = 416.05 tons