Foundation Engineering Bearing Capacity: § Types of Foundations § Theory of Bearing Capacity § Terzaghi Approach Dr Hussein M. Al.Khuzaie; [email protected]1 Syllabus of Foundation Design Site Investigation, Bearing capacity of Soil, Settlement, Foundation Design, Piles, Lateral Earth Pressure, slope Stability.
29
Embed
3th lecture on bearing Capacity Cont'd€¦ · that the bearing capacity factors of Equations 4.23 and 4.24 are exact bearing capacity solutions for shallow foundations on cohesive-frictional
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Failure in Sand. The approximate limits of types of failure to beexpected at relative depths D/B and relative density of sand DRvary as shown in Figure below. There is a critical relative depth below which only punching shear failure occurs. For circular foundations, this critical relative depth is about D/B = 4 and for long (L » 5B) rectangular foundations around D/B = 8. The limits of the types of failure depend upon the compressibility of the sand. More compressible materials will have lower critical depths (Vesic, 1973).
which gives the following expression for the ultimate bearing capacity of a shallow foun-dation on cohesive-frictional soils:
qu = c cotØ e(π tanØ)tan2π4+Ø
2
! "− 1
# $+ qe(π tanØ)tan2
π4+Ø
2
! "
= cNc + qNq (4.22)
where the bearing capacity factors are given by
Nq = e(π tanØ) tan2π4+Ø
2
! "(4.23)
Nc = cotØ e(π tanØ)tan2π4+Ø
2
! "− 1
# $= cotØ(Nq − 1) (4.24)
Note when Ø= 0, Equation 4.24 has no meaning. In this case,Nc= (2+ π) and its der-ivation are described in Section 4.3. Or, if we plotNc against Ø, you will find that (2+ π)becomes an asymptotic value as Ø approaches zero.
More importantly, Shield (1954) has shown that by extending satisfactorily the plasticstress field associated with the Prandtl mechanism into the remaining rigid regions belowthe shear surface, the upper-bound bearing capacity solution Equation 4.22 is also a lowerbound for soils with an internal friction angle of less than 75◦. It is therefore concludedthat the bearing capacity factors of Equations 4.23 and 4.24 are exact bearing capacitysolutions for shallow foundations on cohesive-frictional soils. This would be true regard-less of the roughness of the soil–foundation interface.
EXAMPLE 4.1
Given
A strip foundation with a width of 3 m is located at a depth of 2 m in a cohesive-fric-tional soil. The cohesion of the soil c is 30 kPa and its internal friction angle Ø is 30◦.The unit weight of the soil is γ= 17 kN/m3. For simplicity, the overburden effect will beaccounted for by using an equivalent surcharge, defined as the unit weight multiplied byembedment depth of the foundation.
Required
The ultimate bearing capacity of the strip foundation.
Note that r0 is linked to the foundation width B through the geometry of the failuremechanism as follows:
r0 = B
4 cosπ4+ Ø
2
! " (4.30)
The total external work is the sum of Equations 4.25 through 4.27 and 4.29. Accordingto the upper-bound theorem, the external work is equal to internal power dissipation δW(which is zero in this case because soil cohesion is zero):
δE = δEOA + δEAOC + δEADE + δEACD = δW = 0 (4.31)
which can be simplified to the following bearing capacity equation:
q u = γB2Nγ (4.32)
and the bearing capacity factor Nγ is
Nγ =14tan
π4+ Ø
2
! "tan
π4+ Ø
2
! "e
32π tanØ( ) − 1
# $
+ 3 sinØ1 + 8 sin2Ø
tanπ4+ Ø
2
! "− cotØ
3
# $e
32π tanØ( ) + tan
π4+ Ø
2
! "cotØ3
+ 1% &
(4.33)
As stated by Chen (1975), the bearing capacity due to soil weight is sensitive to theroughness of soil–foundation interface. Hill’s mechanism is suitable for a perfectly smoothsoil–foundation interface, and Prandtl’s mechanism does not allow soil–foundationslip and is therefore suitable for a perfectly rough soil–foundation interface. In fact, ifwe follow the above upper-bound approach but with Prandtl’s mechanism, the bearingcapacity obtained is exactly twice the solution of Equation 4.33 as derived from Hill’smechanism.
EXAMPLE 4.2
Given
Consider the same foundation problem as in Example 4.1. A strip foundation with awidth of 3 m is located at a depth of 2 m in a cohesive-frictional soil. The cohesionof the soil c is 30 kPa and its internal friction angle Ø is 30◦. The unit weight of thesoil is γ = 17 kN/m3.
Required
The additional bearing capacity of the foundation due to soil weight.
EXAMPLE 1GivenA strip foundation with a width of 3 m is located at a depth of 2 m in acohesive-frictional soil. The cohesion of the soil c is 30 kPa and itsinternal friction angle Ø is 30◦. The unit weight of the soil is γ = 17kN/m3. For simplicity, the overburden effect will be accounted for byusing an equivalent surcharge, defined as the unit weight multiplied byembedment depth of the foundation.RequiredThe ultimate bearing capacity of the strip foundation.