User Guide for FB-Deep - Bridge Software Institute User Guide.pdfUser Guide for FB-Deep . F.C....
Transcript of User Guide for FB-Deep - Bridge Software Institute User Guide.pdfUser Guide for FB-Deep . F.C....
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 tsfN =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 tsfN =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