validación del shaft98 y sahftspt
TRANSCRIPT
-
8/11/2019 Validacin Del SHAFT98 y SAHFTspt
1/16
-
8/11/2019 Validacin Del SHAFT98 y SAHFTspt
2/16
Boring Log ScreenClick Insert Layer
Depths = 0 to 50 ftSoil Type = 3 (sand)SPT N = 15 blows
Click OK Note: Last entry doesnt = 0As for old SPT97
Main ScreenClick Cap. Report
Results
-
8/11/2019 Validacin Del SHAFT98 y SAHFTspt
3/16
SPT 97 Uniform Sand (N=15) Example
Depth N STSand ST=3 0 15 3
40 ft N = 15 blows 20 15 32 ft x 2 ft 30 15 3
50 15 355 0 0
F. Bearing Capacity8 B above = 40 8(2) = 24ft3.5 B below = 40 + 3.5(2) = 47ft
Elev N q t (tsf) for sand (ST = 3) q t = 3.2N/3 = 3.2(15)/3=16tsf24 15 1640 15 16 Ave[AQPTA] = 2(16)/2 = 16tsf
Ave = (16 +16)/2=16tsf40 15 16 Ave [AQPTB] =2(16)/2 = 16tsf47 15 16Check depth of embedment to see if correction required: D c = 9B = 9(2)=18
-
8/11/2019 Validacin Del SHAFT98 y SAHFTspt
4/16
To design a range of pileLengths and widths,Click Insert RangeRange 20 to 40 ft longin increments of 5ft.
Click Graph cap.
Graphics screen
SHAFT98 ExampleCLAYS:
Example File: Clay1.dat
1. Multi Layer Clay with Casing
N =8,= 100 pcf ,
qc = 16 tsf
N =12,= 100 pcf , q c = 30 tsf
Casin
3 ft
Cla
Cla
6 ft
20 ft
20 ft
-
8/11/2019 Validacin Del SHAFT98 y SAHFTspt
5/16
Opening Screen:Drilled ShaftUnits = EnglishWater Table = deepR% = 0.0Insert Shaft
Case = 6ft, L = 40ft,Diam =36, Bell L = 0.0ftBell Diam = 36
Click Boring Log IconIn upper toolbar to get
Boring log.
Boring Log ScreenGround Surface = 0.0ftCu Calcs = CPT (Lower Left)Click Insert LayerDepth ST CPT0 1(clay) 100pcf 16tsf20 1 100 1620 1 100 3040 1 100 4050 1 100 50
Rock Friction Not included
Click OK to return to mainmenu
-
8/11/2019 Validacin Del SHAFT98 y SAHFTspt
6/16
Click Cap. ReportFor results
Results
SettlementFor 0.3, R % = 0.833Click Cap. Report
R Results showCap = 293.15T @ 0.3
-
8/11/2019 Validacin Del SHAFT98 y SAHFTspt
7/16
APPENDIX A - Examples
CLAYS:Example File: Clay1.dat
2. Multi Layer Clay with Casing
3. Multi Layer Clay with Casing B > 75
N =8, = 100 pcf ,
qc = 16 tsf
N =12, = 100 pcf ,
qc = 30 tsf
150 = cqc
Clay Layer # 1 : c = )0333.1(67.066,215
100*102000*16
tsf psf =
Clay Layer # 2 : c = )90.1(800,315
100*302000*30tsf psf =
Clay Layer # 2 Tip c = )867.1(3.733,315
100*402000*30tsf psf =
1. Multi Layer Clay with Casing : Full Capacity (40 ft Shaft)
a) Skin Friction:Qs = [ ])9.1*55.0)(302()033.1*55.0)(602(*0.3* +
= [ ]765.179567.7*4248.9 + = Tons42.242
b) End Bearing:
Q b =4
.2b
qb
,
Casin
3 ft
Cla
Cla
6 ft
20 ft
20 ft
-
8/11/2019 Validacin Del SHAFT98 y SAHFTspt
8/16
-
8/11/2019 Validacin Del SHAFT98 y SAHFTspt
9/16
For range of shaft lengthsClick Insert RangeMin L = 20ft Max L =40ftIncrement = 2ft.Click Cap. Graph
Graphical Results
Example: Sand overlying Rock (IGM) SocketDepth N ST q u q t
0 15 35' 15 3
10' 15 315' 15 320' 15 325' 15 330' 15 335' 15 3
40'- 15 340'+ 4 10tsf 1.0tsf45 4 10tsf 1.0tsf50 4 10tsf 1.0tsf55' 4 10tsf 1.0tsf60' 4 10tsf 1.0tsf65 4 10tsf 1.0tsf
40'45'
Sand N=15ST=3 = 100pcf
Rockqu = 10tsfqt = 1.0tsfq b = 0.5q uEm = 115q u
Diam = 36-in
-
8/11/2019 Validacin Del SHAFT98 y SAHFTspt
10/16
Opening Screen:Units = English
Type = Drilled Shaft
c = 130pcf, slump=6.89Ec = 57,000(5000psi) 1/2=4030Click Insert ShaftL = 45ft, D=36
Click Boring Log Icon
In top toolbar
Enter Boring Log DataDepth ST N qu q t
0 3 15 10040 3 15 100
40 4 10 1
65 4 10 1
Check Use Default valuesDepth q b Em RQD Socket40 5 default 1.0 065 5 default 1.0 0
q b is not default, but q u Em is 115 q u x 2/144RQD = 1.0, no reductionSocket = 0 = smooth sides
McVays Friction
= t u qq21
Click OK
-
8/11/2019 Validacin Del SHAFT98 y SAHFTspt
11/16
Set R% = 0.1Click Cap. Report
Results
Case 5) Shaft in rock under sand with smooth socket(Example file: 5shaftsandsmoothrocksocket.spc)
End Bearing
User defined q b = q u = (10tsf) = 5tsf
36 diameterc = 130pcf E c = 4030ksi Depth N ST q u q t 0 15 35' 15 3
10' 15 315' 15 320' 15 325' 15 330' 15 335' 15 340'- 15 3
40'+ 4 10tsf 1.0tsf45 4 10tsf 1.0tsf50 4 10tsf 1.0tsf55' 4 10tsf 1.0tsf60' 4 10tsf 1.0tsf65 4 10tsf 1.0tsf
40'45'
Sand N=15ST=3 = 100pcf
Rock
qu = 10tsfqt = 1.0tsfq b = 0.5q uEm = 115q u
-
8/11/2019 Validacin Del SHAFT98 y SAHFTspt
12/16
Total End Bearing
tons ft
tsf AqQ BT T
343.352
3**0.5*
2
=
==
Skin Friction
For soil type 3, sand, skin friction is:
z = 0 to 5 ft
=5
0
)100)(2.1(3 zdz Qs
( )( )5
0
2
21203
z Qs =
tonslbsQs 069.7167.14137 ==
z = 5 to 40 ft
=40
5
)100)(135.05.1(3 zdz z Qs
dz z z Qs
=
40
5
23
5.131503
40
5
25
2
52
5.1321
150)3( = z z Qs
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):
tsf tsf tsf qq f t U SU 581.10.11021
21 ===
-
8/11/2019 Validacin Del SHAFT98 y SAHFTspt
13/16
tons ft ft tsf L D f Q SU S 509.745*3**581.1*** ===
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 E c, slump, and E m)
Em = 115 q u = 115 (10.0tsf) = 1150tsf
44.0)log(105.014.12/12/1
=
m
c
E E
D L
D L
( ) ( ) 99679.044.0)1150
288000log(1667.105.0667.114.1 2/12/1 ==
tsf tsf
13.0)log(115.037.02/12/1
+
=
m
c
E E
D L
D L
( ) ( ) 50282.013.0)1150
288000log(1667.115.0667.137.0 2/12/1 =+=tsf
tsf
t u su su
m qq f f L
E w 2
1; =
=
=w 1799.91
)581.1(*50282.0*5*99679.0*1150 = ft
tsf ft tsf
[ ] [ ]67.0
1200)1(
)(0134.0
++
= L
E D
L D L
D L
D L
m
[ ] [ ] 67.050282.0*5*
667.1199769.0667.1200
667.2667.1
)1150(0134.0 +
= ft
tsf
= 71.354tsf ft -0.67
-
8/11/2019 Validacin Del SHAFT98 y SAHFTspt
14/16
Determine n for deformation criteria (figure4) 576.9044272.1
0.10 ==tsf
tsf q
p
u
4655.0312.640796.1
1150
796.135915.42*130*65.065.0)4(,7
5.4225
40;;
==
=====
=+==
ntsf tsf E
tsf psf ft pcf table M in slumpa For
ft Z Since Z M E
n
m
n
cccnn
m
Select values of w for calculating
n for q D f k L DQ
wqn for q D
f L DQ
b sut
bb sut
>+=
=
-
8/11/2019 Validacin Del SHAFT98 y SAHFTspt
15/16
-
8/11/2019 Validacin Del SHAFT98 y SAHFTspt
16/16
Sand Layer Above
Load Corresponding to R = 1.0% carried in skin friction:
For R 0.908333 R R R R f
f
S
S 15.436.734.616.2 234max
++=
( ) ( ) ( ) ( ) 9781.00.115.40.136.70.134.60.116.2 234max
=++=S
S
f f
QS = (0.9781)307.639tons = 300.952tons
Total load at R=1.0 115.09 tons + 300.952tons = 416.05 tons