06w schuppener worked examples hyd&upl limit states

8/13/2019 06w Schuppener Worked Examples HYD&UPL Limit States

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Background & Applications

GEOTECHNICAL DESIGN wi th worked examples 13-14 June 2013, Dublin

Worked examples an m

Dr. Bernd Schuppener

German Federal Waterways Engineering and

Research Institute, KarlsruhePast Chairman of CEN TC250/SC 7


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w wor e examp es - une, u n

Worked example:

in an excavation

2013 Bernd Schuppener. All rights reserved2


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Example: verification of failure by uplift UPL

=

LSPW =15 m

,

Uk

a = 10 m; b = 5 m

conrete = 24,0 kN/m = 32,5

= 10 kN/m Steel d = 2,34 kN/m


k

w


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EN 1997-1: 2.4.7.4 Verification of uplift

(1)P Verification for uplift (UPL) shall be carried out by checking

that the design value of the combination of destabilisingpermanen an var a e ver ca ac ons dst;d s ess an orequal to the sum of the design value of the stabilisingpermanent vertical actions (Gstb d) and of the design value of any

additional resistance to uplift (Rd):

V G + R (2.8),where

dst,d dst;d dst;d

(2) Additional resistance to uplift may also be treated as a


stb;d .


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UPL partial factors and model factor

Factor type Factor EN 1997-1 German NA

and DIN

1054

Irish NA

Permanent unfavourable

action

G;dst 1.0 1.0 1.0

G;stb . . .

Variable unfavourable action Q;dst 1.5 1.5 1.5

Angle of shearing resistance 1.25 1.25 1.25

Pile tensile resistance s;t 1.4 1.4 2.0

Model factor on wall - 0.8 -


resistance


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UPL- verification with the weight of the structure only

For the verifications the following quantities are needed:

Self-wei ht base-slab:

Base-slab: d = 1,0 m, concrete = 24,0 kN/m:Base-area A = a b = 10,0 5,0 = 50,00 m

concrete = concreteGconcrete = 50 1,0 24,0 = 1200 kN.

e we g s ee -p e-wa : SPW = , m, Steel = , m

Weight per unit area: g = 78,0 0,03 = 2,34 kN/m,

SPW = SPW = , =

Total characteristic weight of the structure:


Gk

= Gconcrete

+ GSPW

= 1200 + 1053 = 2253 kN


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UPL-verification with the weight of the structure only

G

U

V

dst,d

= Ud

Gd

dst,d stb;d d . .

k G,dst k G,stbBut: 5000 1,00 = 5000 kN > 2253 0,90 = 2028 kN


ence requ remen no sa s e


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UPL-verification including wall fr ictional resistance R

GR

Vdst,dGstb;d + Rd +

UdGd + Rd2013 Bernd Schuppener. All rights reserved8

, s ,s


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Verification of failure by upli ft UPL frictional resistance R

The characteristic value of the wall frictional resistance Rk will be

treated as a stabilising action determined as the vertical

a model factor of = 0,80:

Rk = Eah,k tan a

e wa r c on ang e s assume o e k = k. orhorizontal surface area and vertical wall the earth pressurecoefficient is Kah k = 0,25 (from Fig. C 1.1 of EN 1997-1 for k=32.5) and tan a = 0,397. The earth pressure is assumed to actonly on the outer surface of the sheet pile wall:

E = 2 a + b 0 5 L K, ,

Eah,k = 30 0,5 15 10 0,25 = 8437 kN

Rk = Eah k tan k

2013 Bernd Schuppener. All rights reserved9 Rk = 8437 0,397 0,8 = 2680 kN


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UPL - including wall frictional resistance

G

R

U

U G + R Require: UdGd + Rd

, ,But: 5000 1,00 = 5000 > (2253+2680) 0,90 = 4400 kN



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Example: Irish verif ication of uplift failure for UPL

Calculation of Rd

s ng k,inf= , , .e. e n er or c arac er s c va ue, an app y ng

UPL = 1,25 to the tan(k,inf) to obtain to d and d, gives a design

wall frictional resistance Rd value that is slightly greater than the

characteristic value Rk, which is not acceptable.Hence a superior characteristic valuek,sup is needs to be used. This is

o a ne , a er on an arr s , y assum ng a norma

distribution for and a standard deviation of 3o so that k,sup = 32,5o + 2

1.624 3 0 = 42.4o. Then a l in the artial factor of 1.25 as amultiplier to tan(k,sup) gives d,sup = 48.8

o and hence an Rd value of

2580 kN with a margin of safety of 1.33 on the characteristic value.



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Example: Irish verification of failure by uplift UPL

R

U

Uk G,dstGk G,stb + RkG,stbRequire: UdGd + Rd

i.e. 5000 1,00 = 5000 > 2253 0,90 + 2580 = 4608 kN



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GEO partial factors and model factor

DA1.C1

DA1.C2 DA2 + DIN

1054ermanen un avoura e

action, transient situation

G;dst . . .

Permanent favourable G;stb 1.0 1.0 1.0

actionVariable unfavourable

action

Q;dst 1.5 1.3 1.5

Angle of shearing

resistance

1.0 1.25 1.0

Pile tensile resistance s;t 1.25 1.6 1.4

Model factor on wall - - 0.8


resistance


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Irish design for GEO verification

As explained previously,

RG Rd is obtained using thesuperior value so that

for:

UEd Rd = 1,0, Rd = 3440 kN = 1,25, Rd = 2580 kNE

d= U

k

G- G

k

G,inf= 5000

G 2253

G;inf. : d = , , = d =

DA1.C2: Ed = 5000 1,02253 1,0=2747kN < Rd = 2580kN

Hence GEO ULS is not satisfied



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U lift - verification with tension iles

EN 1997-1

7.6.3 Piles - ground tensile resistance

(3)P For tension piles, two failure mechanismsshall be considered:

pull-out of the piles from the ground mass;

uplift of the structure and the block of ground

containing the piles.



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Pile design (GEO) for verif ication of failure by uplift

G

U

F

Ft,d = Uk G - (Gk+ Rk) G,infFt,d: design value of the action on the pilesG = 1.20 for unfavourable actions (according to DIN 1054 for


G,inf= 1.00 for favourable effects of actions


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G

U

Ed = UkG - (Gk+ Rk) G inf E1

Ft,d = 5000 1,20 - (2253 + 2680) 1,00

Ft,d = 6000 4933


Ft,d

= 1067 kN


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E

L

Rt,dDriven iles: L = n L D = 0 5 m = 35 kN/m

Rt,d = Ltot qs,k D / P

, P = 1,40 (DIN 1054)

2013 Bernd Schuppener. All rights reserved19 Rt,d = 39,25 Ltot


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l =3 33 m

1,66 m

x

3,33 m

10,0 ma= , m

3,33 m

la=3,33 m

1,66 m

x

2,50 m2,50 m

5,0 m

Selected:

3 Piles L = 10 m



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Irish pile design for GEO verification of uplift failure

As explained previously,

G Rd is obtained using thesuperior value so that

for:

U

= 1,0, Rd = 3440 kN = 1,25, Rd = 2580 kN

Ed = UkG - Gk G,inf = 5000G 2253 G;inftd d . .

Rt,d = Rt,k / s; t + Rd = Ltot qs,k D / s;t + Rd

= Ltot35 0.5 / s;t + Rd = Ltot 54.98 / s;t + Rd


Hence require: Lto t s;t (5000G 2253 G;inf Rd) / 54.98


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Verification of uplift of structure + ground

oc con a n ng e p es

G RVdst,dGstb;d + Rd (2.8)

UdGd + Gsoil,d + Rd

G

k G,dst k soil,k k G,stb



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Determination of Gsoil,k according to DIN 1054, 8.5.4

+= cotll/LllnG babak,soil2231

: num er o p es, L = engt o p es

la = 5,0 m greater grid distance piles,

=b , ,

= 10,0 kN/m submerged weight soil = 0,80 model factor for the wei ht.

Gsoil,k= 3{5,03,33(10

1/3 cot 32,5)}0,81022 33305 ,, +

soil,k =

Model factor: = 0,80



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Verification of uplift of structure + ground

oc con a n ng e p es

GR

U G + R + G

G

,

Uk G,dst(Gk+ Rk+ Gsoil,k) G,stb5000 6906 kN



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Worked example:



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. . .

(1)P When considering a limit state of failure due to heave

, . ,

be verified, for every relevant soil column, that the design

value of the destabilising total pore water pressure (udst d )

at the bottom of the column, or the design value of theseepage force (Sdst;d) in the column is less than or equal to

e s a s ng o a ver ca s ress stb;d a e o om o

the column, or the submerged weight (Gstb;d) of the same

udst;d stb;d (2.9a)


dst;d stb;d .


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Verification of safety against hydraulic heave

hh =

us ng ormu a y

3 1t

h1 ++

h = 4,4 m

4,1

4,4=rh = 2,13 mh= 1,4 m

60,6

t = 6,60 mi = hr / t = 0,32


u = (t + hr) w


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NN - 1,0 m Seepage force Sk:

Sk = i W V

NN - 4,0 mEffective self weight ofthe soil column Gk:

NN - 5,4 m Gk = V

Sk dst Gk G,stbk

NN - 12,0 mSk



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Table A.17 - Partial factors on actions

FAction Symbol Valuepermanent

un avoura e

favourableb

G;dstG;stb

.

0.90

unfavourablea G;stb 1.50

a

b stabilising



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Safety against hydraulic heave: eq. 2.9bNN - 1,0 m udst;d stb;d

u = t + h

NN - 4,0 m

hr= 2,13m

u = (6.60+2.13)10- , m

uk = 87.3 kN/mstb;k = t(+w)

h= 1,4 m

stb;k = 6.60 20

stb;k

= 132 kN/mt = 6,60 m

NN - 12,0 m ukG,dst stb;kG,stb87.31.35 1320.90

stb;d


. .s ;


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GEOTECHNICAL DESIGN wi th worked examples 13-14 June 2013, Dublin

Geotechnical designw wor ex m l

eurocodes.jrc.ec.europa.eu

06w schuppener worked examples hyd&upl limit states

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