Mathcad - Drilled Shaft.pdf

NORTON UNIVERSITY

Dept of Civil Engineering

Bored Bore d Pile Capacity Capa city Calculation

( B a s e d o n D r. r. L i m S o k T a y 's 's b o o k )

I . Base Beari ng Capaci Capaci ty I . 1 Cohes Cohesi ve Soil s (Cl ay)

= Ab × qb

Qb

Base capacity of pile:

if L ≥ 3 D qb = cub × N cb  q = 2 1 + (1/ 6) L  if L < 3 D  b 3  D 

Unit base capacity of pile:

 24  cux := 48    96  INPUT:

; qb ≤ 4000 kPa

 6.5  Ncx := 8    9 

cub := 300 k P a L := 27 m D := 1

m

2

A b :=

πD 4

= 0.785

m 2

2

W p :=

πD 25L⋅ 4

= 530.144

k N

Nc := linterp( cux , Ncx , cub) Ncb :=

Nc if 6.5

≤ Nc ≤ 9

"Err "Error or-R -Ret etry ry"" if Nc

Ncb

=9

< 6.5

9 otherw otherwise ise

Unit base capacity

q b

:= ( cub⋅ Ncb)

if L

≥ 3D

 1 + 1 ⋅ L  c ⋅ N 3  6 D  ub cb 2

qB :=

if L

< 3D

q b if q b ≤ 4000 4000 otherwise otherwise

OUTPUT:

Base capacity of pile:

q b = 2700

k P a

qB = 2700

k P a

Q b := A b ⋅ q B = 2120.575

Drilled Shaft Capacity Calculation

k N

09-Nov-13

page: 1

Calculated by HOR Heak

NORTON UNIVERSITY


I I . Ski n Bear in g Capacity I I . 1 Ski n F r icti on in Cohesi ve Soil s (Cl ay) Unit skin friction: f s

•

= α × cu

Please see the following table for the value of  and c u : Table 1: Drilled shaft: Adhesion factor,  (AASHTO, 1998 reported by Gunaratne, 2006)

α

Undrained strength, su or c u (kPa)

Undrained strength, su or c u (kPa)

α

< 192

0.55 479 - 575

0.35

192 - 287

0.49 575 - 670

0.33

287 - 383

0.42 670 - 766

0.32

383 - 479

0.38 766 - 862

0.31

For the undrained strength su > 862 kPa, it is treated as rock

INPUT:

1.

Layer of Clay 1

cu1 := 66

k P a

Lclay1 := 1 Asclay1 :=

OUTPUT:

Skin friction of Clay1:

INPUT:

Table 1 gives

α1 := 0.55

m

π D⋅ Lclay1

= 3.142

m 2

f sclay1 := α1⋅ c u1

= 36.3

f sclay1 = 36.3

k P a

Qsclay1 := Asclay1 ⋅ f sclay1

k P a

= 114.04

k N

2. Layer of Clay 2 cu2 := 114 k P a

Table 1 gives

α2 := 0.55

Lclay2 := 1.3 m Asclay2 := π D⋅ Lclay2

OUTPUT:

= 4.084

m 2


= 62.7

f sclay2 = 62.7

k P a


k P a

09-Nov-13

page: 2


NORTON UNIVERSITY


Skin friction of Clay 2: Qsclay2 := Asclay2 ⋅ f sclay2 INPUT:

= 256.071

k N

3. Layer of Clay 3 cu3 := 90.8 k P a

Table 1 gives

α3 := 0.55

Lclay3 := 1.5 m Asclay3 := π D⋅ Lclay3 OUTPUT:

= 4.712

m 2


= 49.94

f sclay3 = 49.94

k P a


k P a

= 235.337

k N

4. Layer of Clay 4 cu4 := 112.6 k P a

Table 1 gives

α4 := 0.55


= 13.823


OUTPUT:

f sclay4 = 61.93

m 2

= 61.93 k P a

Skin friction of Clay 4: Qsclay4 := Asclay4 ⋅ f sclay4

INPUT:

k P a

= 856.059

k N

5. Layer of Clay 5 cu5 := 110 k P a

Table 1 gives

α5 := 0.55


OUTPUT:

= 7.226

m 2


= 60.5

f sclay5 = 60.5

k P a


k P a

= 437.153

k N

6. Layer of Clay 6 cu6 := 300

k P a


Table 1 gives

α6 := 0.42 09-Nov-13

page: 3


NORTON UNIVERSITY



= 16.965


OUTPUT:

f sclay6

= 126

m 2

= 126

k P a

k P a

Skin friction of Clay 4: Q sclay6 := Asclay6 ⋅ f sclay6

= 2137.54

k N

I I . 2 Ski n F r icti on i n Cohession less Soil s (Sand) S u m m a r y o f Fo r m u l a e:

Unit skin friction:

• •

0 ≤ z < L ′ L ′

≤ z < L

⇒ ⇒

f s f s

= K × σ v′ × tan δ = f s ( z = L′)

where, L` =15 x Diameter ! = (3/4)"` K = 1-sin"`

INPUT: 1. Layer of sand 1

Lsand1 := 11.1

m

L'sand1 := 15D = 15

φ'sand1 := 42

m

3 π δsand1 := ⋅ φ'sand1 ⋅ 4 180

= 0.55

m

d eg

tan( δsand1 ) = 0.613

•

psand1 := π D = 3.142

At the TOP of s and layer 1: z =-13.5

K sand1 := 1

⋅ − sin φ'sand1



 = 0.331 180  π

m

γw1 :=

γw1

= 10

kN/m 3

γ1_1 := 18 kN/m 3

γ1_2 := 18.5 kN/m 3

γ1_3 := 20.9 kN/m 3

γ1_4 := 20.6 kN/m 3

γ1_5 := 20.2 kN/m 3

γ1_6 := 20 kN/m 3

γ1_7 := 21 kN/m 3

γ1_8 := 0

γ1_9 := 0

kN/m 3

kN/m 3

γ1_10 := 0 kN/m 3


09-Nov-13

page: 4


NORTON UNIVERSITY


h1_1 := 1.8 m

h1_2 := 2.2 m

h1_3 := 1.3 m

h1_4 := 1.5 m

h1_5 := 4.4 m

h1_6 := 2.3 m

h1_7 := 0

h1_8 := 0

h1_9 := 0

h1_10 := 0

γ1 := γ1_1 γ5 := γ1_5

− γw1

γ9 := γ1_9

− γw1

σ'vsand1_1 := γ1 ⋅ h 1_1

+

m

m

γ2 := γ1_2

− γw1

γ3 := γ1_3

− γw1

γ4 := γ1_4

− γw1

γ6 := γ1_6

− γw1

γ7 := γ1_7

− γw1

γ8 := γ1_8

− γw1

m

γ10 := γ1_10 − γw1

γ2 ⋅ h 1_2

σ'vsand1_1 = 149.05

+

γ3 ⋅ h 1_3

+

γ4 ⋅ h 1_4

+

γ5 ⋅ h 1_5

+

γ6 ⋅ h 1_6

+

γ7 ⋅ h 1_7

+

γ8 ⋅ h 1_8

+

γ9 ⋅ h 1

k Pa

f ssand1_1 := K sand1 ⋅ σ'vsand1_1⋅ tan( δsand1 )

•

m

At the BOTTOM of sand layer 1: z = -24.6

f ssand1_1 = 30.221

k P a

m

γw1 :=

γw1

= 10

kN/m 3

γ1_1 := 18 kN/m 3

γ1_2 := 18.5 kN/m 3

γ1_3 := 20.9 kN/m 3

γ1_4 := 20.6 kN/m 3

γ1_5 := 20.2 kN/m 3

γ1_6 := 20 kN/m 3

γ1_7 := 21 kN/m 3

γ1_8 := 0

kN/m 3

γ1_9 := 0

kN/m 3

γ1_10 := 0 kN/m 3

h1_1 := 1.8 m

h1_2 := 2.2 m

h1_3 := 1.3 m

h1_4 := 1.5 m

h1_5 := 4.4 m

h1_6 := 2.3 m

h1_7 := 11.1 m

h1_8 := 0

h1_9 := 0

h1_10 := 0

γ1 := γ1_1 γ5 := γ1_5

− γw1

γ9 := γ1_9

− γw1

σ'vsand1_2 := γ1 ⋅ h 1_1

+

γ2 := γ1_2

− γw1

γ3 := γ1_3

− γw1

γ4 := γ1_4

− γw1

γ6 := γ1_6

− γw1

γ7 := γ1_7

− γw1

γ8 := γ1_8

− γw1

+

γ3 ⋅ h 1_3

+

γ4 ⋅ h 1_4

+

γ5 ⋅ h 1_5

+

γ6 ⋅ h 1_6

m

+

γ7 ⋅ h 1_7

+

γ8 ⋅ h 1_8

+

γ9 ⋅ h 1

k Pa

f ssand1_2 := K sand1 ⋅ σ'vsand1_2⋅ tan( δsand1 ) f ssand1 := 0.5⋅ Lsand1⋅ ( f ssand1_1


m

γ10 := γ1_10 − γw1

γ2 ⋅ h 1_2

σ'vsand1_2 = 271.15

m

f ssand1_2 = 54.978

k P a

+ f ssand1_2) 09-Nov-13

page: 5


NORTON UNIVERSITY


OUTPUT:

f ssand1 = 472.852

Skin friction of Sand 1:

Qssand1 := p sand1⋅ f ssand1

k P a

= 1485.507

k N

INPUT: 2. Layer of sand 2

Lsand2 := 0

psand2 := π D = 3.142

m

L'sand2 := 15D = 15

φ'sand2 := 42

m

3 π δsand2 := ⋅ φ'sand2 ⋅ 4 180

d eg

tan( δsand2 ) = 0.613

•

m

K sand2 := 1

At the TOP of s and layer 2: z =-13.5

= 0.55

− sin φ'sand2 ⋅



 = 0.331 180  π

m

γw2 :=

γw2

= 10

kN/m 3

γ2_1 := 18 kN/m 3

γ2_2 := 18.5 kN/m 3

γ2_3 := 20.9 kN/m 3

γ2_4 := 20.6 kN/m 3

γ2_5 := 20.2 kN/m 3

γ2_6 := 20 kN/m 3

γ2_7 := 21 kN/m 3

γ2_8 := 0

kN/m 3

γ2_9 := 0

kN/m 3

γ2_10 := 0 kN/m 3 h2_1 := 1.8 m

h2_2 := 2.2 m

h2_3 := 1.3 m

h2_4 := 1.5 m

h2_5 := 4.4 m

h2_6 := 2.3 m

h2_7 := 11.1 m

h2_8 := 0

h2_9 := 0

h2_10 := 0

γL1 := γ2_1 γL5 := γ2_5

− γw2

γL9 := γ2_9

− γw2

σ'vsand2_1 := γL1⋅ h 2_1

+

m

m

m

γL2 := γ2_2

− γw2

γL3 := γ2_3

− γw2

γL4 := γ2_4

− γw2

γL6 := γ2_6

− γw2

γL7 := γ2_7

− γw2

γL8 := γ2_8

− γw2

+

γL5⋅ h 2_5

γL6⋅ h 2_6

09-Nov-13

page: 6

γL10 := γ2_10 − γw2

γL2⋅ h 2_2

+

γL3⋅ h 2_3

+

γL4⋅ h 2_4

+

+

γL7⋅ h 2_7

+

γL8⋅ h 2_

σ'vsand2_1 = 271.15 k P a f ssand2_1 := K sand2 ⋅ σ'vsand2_1⋅ tan( δsand2 )



NORTON UNIVERSITY


f ssand2_1 = 54.978

•

k P a

At the BOTTOM of sand layer 2: z = -24.6

m

γw2 :=

γw2

= 10

kN/m 3

γ2_1 := 18 kN/m 3

γ2_2 := 18.5 kN/m 3

γ2_3 := 20.9 kN/m 3

γ2_4 := 20.6 kN/m 3

γ2_5 := 20.2 kN/m 3

γ2_6 := 20 kN/m 3

γ2_7 := 21 kN/m 3

γ2_8 := 0

kN/m 3

γ2_9 := 0

kN/m 3

γ2_10 := 0 kN/m 3 h2_1 := 1.8 m

h2_2 := 2.2 m

h2_3 := 1.3 m

h2_4 := 1.5 m

h2_5 := 4.4 m

h2_6 := 2.3 m

h2_7 := 11.1 m

h2_8 := 0

h2_9 := 0

h2_10 := 0

γL1 := γ2_1 γL5 := γ2_5

− γw2

γL9 := γ2_9

− γw2

σ'vsand2_2 := γL1⋅ h 2_1

+

m

m

m

γL2 := γ2_2

− γw2

γL3 := γ2_3

− γw2

γL4 := γ2_4

− γw2

γL6 := γ2_6

− γw2

γL7 := γ2_7

− γw2

γL8 := γ2_8

− γw2

γL5⋅ h 2_5

γL6⋅ h 2_6

γL10 := γ2_10 − γw2

γL2⋅ h 2_2

+

γL3⋅ h 2_3

+

γL4⋅ h 2_4

+

+

+

γL7⋅ h 2_7

+

γL8⋅ h 2_

σ'vsand2_2 = 271.15 k P a f ssand2_2 := K sand2 ⋅ σ'vsand2_2⋅ tan( δsand2 ) f ssand2 := 0.5⋅ Lsand2⋅ ( f ssand2_1

f ssand2_2 = 54.978

k P a

+ f ssand2_2)

=0

OUTPUT:

f ssand2

k P a

Skin friction of Sand 2:

Qssand2 := p sand2⋅ f ssand2

=0

k N

OUTPUT:

Total Skin friction: Qs := Qsclay1

+ Qsclay2 + Qsclay3 +

Base Capacity: Q = 2120.575 b


Qsclay4

+ Qsclay5 + Qsclay6 + Qssand1 + Qssand2 = 5521.706

k N

k N

09-Nov-13

page: 7


NORTON UNIVERSITY

Weight of Pile: W = 530.144 p

Dept of Civil Engineering k N

Preferred Safety Factor: F := 2.5 s Ultimate Bearing Capacity of P ile:

Qult := Qs

Allowable Bearing Capacity of Pile: Q := all


Qs

+ Q b − W p = 7112.138 + Q b Fs

− W p = 2526.769

09-Nov-13

page: 8

k N

k N


NORTON UNIVERSITY


IMPORTANT NOTES :

cu cu cu

= 24 kPa → N c = 6.5 = 48 kPa → N c = 8 ≥ 96 kPa → N c = 9

Ab = Area of pile base (m2 ) qb = Unit base resistance of pile (kPa) c ub = Mean undrained shear strength of clay withi n 2 Dim eters below tip of pil e (kPa) Ncb = Bearing capacity factor at base L = Length of pile (m) D = Diameter of pile (m) W p = Weight of pile THIS COLOUR

= Data To Be Entered

THIS COLOUR

= Important Result

Axm :=

 13.64   18.04 

Aym :=

 1228.55   1629.64 

Louver material := linterp(Axm , Aym , 27.3) = 2473.752

Axl :=

 13.64   18.04 

Ayl :=

 307.14   407.41 

Louver labor := linterp(Axl , Ayl , 21.35) = 482.84

Axe :=

 13.64   18.04 

Aye :=

 221.14   293.34 

Louver expenses := linterp(Axe , Aye , 21.35) = 347.654


09-Nov-13

page: 9


NORTON UNIVERSITY


IMPORTANT NOTES:

f s = Unit skin friction (kPa)  = Adhesion factor

c u= Undrained shear strength (kPa) Lclay(i) = Thickness of the i th clay layer


09-Nov-13

page: 10


NORTON UNIVERSITY


IMPORTANT NOTES :

z = The c oordinate axis along the depth direction (m) L` = Critical depth = 15D K = Coefficient of lateral earth pressure #`v = Vertical effective stress at the depth under consideration (kPa) ! = Frictional angle between soil and pile material = (3/4)"` "` = Angle of internal friction of soil before installing pile (deg)

Lsand(i) = Thickness of t he ith sand layer (m) p = The perimeter of pile (m)


09-Nov-13

page: 11


NORTON UNIVERSITY


IMPORTANT NOTES:

hi = Thickness of ith layer of soils

9

+

γ10⋅ h 1_10

9

+

γ10⋅ h 1_10


09-Nov-13

page: 12


NORTON UNIVERSITY

+

γL9⋅ h 2_9

+


γL10⋅ h 2_10


09-Nov-13

page: 13


NORTON UNIVERSITY

+

γL9⋅ h 2_9

+


γL10⋅ h 2_10


09-Nov-13

page: 14


Mathcad - Drilled Shaft.pdf

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