∑  (L/B;0)=0.  can be interpolated from the table taking the right value of z/B in place of H/B.   W.Brząkała: GEOENGINEERING-FOUNDATIONS (M.Sc.,CEB) – Design Project #1 (week No.5)

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W.Brząkała: GEOENGINEERING-FOUNDATIONS (M.Sc.,CEB) – Design Project #1 (week No.5)

1. Selection of the linear elastic model of the subsoil

H – total thickness of all subsoil layers situated between the founding level and the undeformable bedrock (or a

“very rigid” soil), B – foundation width a) H  ~1,5B  the Winkler model,

b) ~1,5B < H < ~6B  the elastic layer (or layers),

c) H  ~6B  the elastic half-space (homogeneous or horizontally layered).

2.

Determination of values of elastic parameters. Assumptions

 For simply design purposes, parameters of the subsoil can be expressed by making use of standard con- stants of elasticity theory: Eo – the Young modulus (primary),  - the Poisson coefficient,   0.3 generally;

in case of the homogeneous half-space both parameters happen as (1-²)/Eo therefore for the elastic half- space only one stiffness modulus Es = Eo/(1-²) is in use,

 it is assumed that foundation settlement is equal to a settlement wo of a corresponding elastic subsoil with the unknown parameters Eo, (half-space, layer) or C (Winkler); from such an equation, the values of the model parameters can be found using the inverse analysis;

 the following formula can be used ¹⁾

wo1 = average settlement of a rectangu- lar area BxL, uniformly loaded

(q=const), due to elastic half-space compressibility – but only in the considered interval of depth 0H  ,

 if there are for example two different subsoil layers of thicknesses H1,H2 within 0H under the founding level zo = 0, i.e. their bottom levels are z1 = H1 and z2 = H1+H2 = H respectively, so:

Analogously won for n>2 layers; coefficients śr can be interpolated from the table taking the right value of z/B in place of H/B.

Comment:

The above expressions for wo1, wo2, … assume the same vanishing rate of vertical stresses z = q(z) with depth under the loaded rectangle BxL (the Steinbrenner integral) and such a feature is only approximate;

both the presence of a (shallow) undeformable bedrock and (significantly) different elastic parameters of sub- sequent layers can change - more or less - the shape of the function z(z) under a loaded place.

3. Determination of values of elastic parameters. Calculations

a) The Winkler model: here w = q/C and by assumption w = wo1, therefore q/C = qBśr(L/B,H/B)/Es

and the unknown value of C can be found. For multi-layers, won is used instead of wo1, n>1.

It is clear that such a value of C depends on both H and B i L.

b) Elastic half-space (generally for n >1): for elastic stiffness moduli Esi , i=1,2…,n, the equivalent

homogeneous elastic half-space has a certain modulus

E

_s^¿ to be found from the equation wo1(*) = won, so:

where zo=0, zn=H =, śr(L/B;0)=0.

If not only Es* but also the Young modulus is of interest, so Eo* = Es*(1-²)  Es*(1-0.3²) taking   0.3 . c) Finite elastic layer – practically the same approach as in b) but for H <  ¹⁾.

1⁾ for „small” H this can be unsatisfactory because boundary conditions on z = H appear to be significant;

Z.Wiłun (Zarys Geotechniki, WKŁ ed.) recommends another coefficient called h

śr(L/B,H/B) H/B L/B=

1 L/B=1

0 L/B=2 0

L/B=



0 0 0 0 0

0,2 5

0,22 0,25 0,25 0,25

0,5 0

0,39 0,46 0,46 0,46

0,7 5

0,53 0,63 0,63 0,64

1,0 0

0,62 0,77 0,77 0,79

1,5 0

0,72 1,00 1,01 1,03

2,0 0

0,77 1,15 1,16 1,20

w

_o1=q⋅B⋅

ω

_śr(

L/ B, H / B) E

_s

w

_o2

=q⋅B⋅ [ ^ω

^śr

⁽ ^{L/ B ,z} ^E

^s1 ¹

^/ ^{B )} ⁺ ^ω

^śr

⁽ ^{L/B , z}

²

^/ ^B)−ω ^E

^{s 2} ^śr

⁽ ^{L /B , z}

¹

^/ ^{B )} ]

w

_o1(∗)=q⋅B⋅

ω

_śr(

L/B , ∞/B )

E

_s^¿ =q⋅B⋅

∑

i=1

n

ω

_śr(L /B , z_i/

B )−ω

_śr(

L/ B ,z

_i−1/

B )

E

_si =w_on

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Final conclusion:

only in a first approximation, the Winkler model and the homogeneous elastic half-space (with correspondingly increased stiffness) can be used as “an equivalent” o the real finite elastic layer; average settlements are indeed the same – as assumed – but deformations outside the loaded area and contact stresses under foundations are a little bit different.

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4. Examples

1) A long foundation beam 2x20m rests on the elastic layer Es=25 MPa which is H = 8m thick.

The value of Es* for a corresponding elastic half-space has to be found assuming the same values of settlements.

For the foundation settlement, the expression w01 for H = 8m and Es=25 MPa is used; for the corresponding half-space the same w01 but for H = + and an unknown Es*. Such the equality leads to

E

_s^¿=

E

_s⋅

ω

_śr(

L/B,∞/B )

ω

_śr(

L/B, H /B )

=25⋅

ω

_śr(10, ∞)

ω

_śr(10,4)=25⋅2,25

1,50=37,5>25=E_s .

Note that for H/B = 4 the subsoil is not enough thick to be replaced by infinite half-space (H/B  ).

2) The subsoil consists of several thick layers and a shallow footing 3mx3m is considered, loaded by a central force 1800 kN (q=const=0.200 MPa); according to a code requirements, settlement has been calculated as wo = 0.012m.

If replace the real case by an elastic virtual half-space of a stiffness Es*

0,012=w_o=w_o1=q⋅B⋅ω_śr(L /B, H /B )

E_s^¿ =0,200⋅3⋅ω_śr(1, ∞)

E_s^¿ =0,200⋅3⋅0,95 E^¿_s its value should be Es* = 47.5 MPa.

If replace the real case by an elastic virtual layer of the thickness, let’s say, H=6m 0,012=w_o=w_o1=q⋅B⋅ω_śr(L /B, H /B )

E_s^¿ =0,200⋅3⋅ω_śr(1,2 )

E_s^¿ =0,200⋅3⋅0,77 E_s^¿ it should be Es* = 38.5 MPa.

3) There are 2 layers of deformable soils (FSa and CL) H = 3m thick, situated on a stiff gravel (Gr):

 directly under foundation in 0.01.5m: FSa, Es1 = 40 MPa,

 still deeper in 1.53.0m: Cl, Es2 = 20 MPa.

For a shallow footing 2mx4m the Winkler coefficient C has to be found – ignoring the gravel (undeformable).

z1 = H1 = 1.5m and z1/B = 1.5/2.0 = 0.75

z2 = H1 + H2 = H = 3.0m and z2/B = 3.0/2.0 = 1.50 .

Linear interpolation of śr from the table for L/B = 2 is as follows:

- between 0.53 and 0.63 the result is: 0.53+(0.63-0.53)(2-1)/(10-1)  0.54 - between 0.72 and 1.00 the result is: 0.72+(1.00-0.72)(2-1)/(10-1)  0.75.

q

C=w=w_o2=q⋅2,0⋅

[

^0,54−0⁴⁰ ⁺^0,75−0,54²⁰

]

Thus C = 20.8 MPa/m.

If Gr is not ignored (7.0m thick and Es3 = 200 MPa), there is:

q

C =w=w

_{o 3}

= q ∙2.0 ∙ [ ^0.54−0 ⁴⁰ ⁺ ^0.75−0.54 ²⁰ ⁺ ^0.95−0.75 ²⁰⁰ ]

^, so C = 20.0 MPa/m  20.8 MPa/m.

If the soils happen in reversed order (Cl resting on FSa) - ignoring the gravel (undeformable):

q

C=w=w_o2=q⋅2,0⋅

[

^0,54−0²⁰ ⁺^0,75−0,54⁴⁰

]

Thus C = 15.5 MPa/m.

If Gr is not ignored (7.0m thick and Es3 = 200 MPa), there is:

q

C =w=w

_{o 3}

= q ∙2.0 ∙ [ ^0.54−0 ²⁰ ⁺ ^0.75−0.54 ⁴⁰ ⁺ ^0.95−0.75 ²⁰⁰ ]

^, so C = 15.0 MPa/m  15.5 MPa/m.

Note that:

- although both soils are of the same thickness, the method favors the upper layer since the weighting coefficients are 0.54 > 0.21=0.75-0.54 (that is realistic),

- the method is sensitive to the order of layers (that is realistic),

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- ignoring Gr looks acceptable, C is almost the same.

4) Under a founding level there are 3 layers of deformable soils (FSa, CL, MSa) H = 10m thick, situated on a rigid bedrock:

 in 0.01.5m: FSa, Es1 = 40 MPa,

 in 1.53.0m: Cl, Es2 = 20 MPa,

 in 3.010.0m: MSa, Es3 = 60 MPa.

For a shallow footing 2mx4m the elastic half-space modulus Es* has to be found.

z1 = H1 = 1.5m and z1/B = 1.5/2.0 = 0.75, z2 = H1 + H2 = 3.0m and z2/B = 3.0/2.0 = 1.50,

z3 = H1 + H2 + H3 = H = 10.0m and z3/B = 10.0/2.0 = 5.00.

Linear interpolation of śr from the table for L/B = 2 is as follows:

- between 0.53 and 0.63 the result is: 0.53+(0.63-0.53)(2-1)/(10-1)  0.54, - between 0.72 and 1.00 the result is: 0.72+(1.00-0.72)(2-1)/(10-1)  0.75, - between 0.87 and 1.63 the result is: 0.87+(1.63-0.87)(2-1)/(10-1)  0.95.

w

_{o 3}

=q ∙ 2.0 ∙ [ ^0.54−0 ⁴⁰ ⁺ ^0.75−0.54 ²⁰ ⁺ ^0.95−0.75 ⁶⁰ ]

- between 0.95 and 2.25 for the half-space the result is: 0.95+(2.25-0.95)(2-1)/(10-1)  1.09.

w

_{o 1}

=q ∙ 2.0 ∙ [ ^1.09−0 ^E

s

¿

]

Both expressions are equal if Es* = 39.9 MPa.

W.Brząkała, WUST, Wroclaw/Poland

∑  (L/B;0)=0.  can be interpolated from the table taking the right value of z/B in place of H/B.   W.Brząkała: GEOENGINEERING-FOUNDATIONS (M.Sc.,CEB) – Design Project #1 (week No.5)

E

w

ω

L/ B, H / B) E

w

=q⋅B⋅ [ ω

( L/ B ,z E

/ B ) + ω

( L/B , z

/ B)−ω E

( L /B , z

/ B ) ]

w

ω

L/B , ∞/B )

E

∑

ω

B )−ω

L/ B ,z

B )

E

E

E

ω

L/B,∞/B )

ω

L/B, H /B )

ω

ω

[

]

q

C =w=w

= q ∙2.0 ∙ [ 0.54−0 40 + 0.75−0.54 20 + 0.95−0.75 200 ]

[

]

q

C =w=w

= q ∙2.0 ∙ [ 0.54−0 20 + 0.75−0.54 40 + 0.95−0.75 200 ]

w

=q ∙ 2.0 ∙ [ 0.54−0 40 + 0.75−0.54 20 + 0.95−0.75 60 ]

w

=q ∙ 2.0 ∙ [ 1.09−0 E

]

=q⋅B⋅ [ ^ω

⁽ ^{L/ B ,z} ^E

^/ ^{B )} ⁺ ^ω

⁽ ^{L/B , z}

^/ ^B)−ω ^E

⁽ ^{L /B , z}

^/ ^{B )} ]

= q ∙2.0 ∙ [ ^0.54−0 ⁴⁰ ⁺ ^0.75−0.54 ²⁰ ⁺ ^0.95−0.75 ²⁰⁰ ]

= q ∙2.0 ∙ [ ^0.54−0 ²⁰ ⁺ ^0.75−0.54 ⁴⁰ ⁺ ^0.95−0.75 ²⁰⁰ ]

=q ∙ 2.0 ∙ [ ^0.54−0 ⁴⁰ ⁺ ^0.75−0.54 ²⁰ ⁺ ^0.95−0.75 ⁶⁰ ]

=q ∙ 2.0 ∙ [ ^1.09−0 ^E