validación del shaft98 y sahftspt

8/11/2019 Validacin Del SHAFT98 y SAHFTspt

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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


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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


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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


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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


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Click Cap. ReportFor results

Results

SettlementFor 0.3, R % = 0.833Click Cap. Report

R Results showCap = 293.15T @ 0.3


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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


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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


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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


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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


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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 ===


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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


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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

>+=

=


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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

validación del shaft98 y sahftspt

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