【Example 5-8 】There is a shaped base,width b = 3m,embedment depth d = 1.5m.Foundation soil γ= 19kN/m3，?Find:
(1) ground-based plastic loadP1andP1；
43sat
=20KN/m3,c=10kPa,??10?。
(2) If the groundwater rises to the underside of the foundation, and its value changed ? [Solution]
（1）??10?，Look-up table 5.2 we have:
3 4（5-31）have
P1??bN1??odNd?cNc?19?3?0.18?19?1.5?1.73?10?4.17?101.3kPa4 4
P1??dN1??odNd?cNc?19?3?0.25?19?1.5?1.73?10?4.17?105.3kPa3 3
N1?0.18 N1?0.25
Nd=1.73 Nc=4.17 Into Eq. （5-32）and
（2）when the groundwater level rose to base，If it is assumed that
the shear strength index c 、φvalue unchanged，, And thus the capacity factors follows above,under the underground water level using heavy
effectively.
????sat??w?20?10?10kN/m3
P1???bN1??odNd?cNc?10?3?0.18?19?1.5?1.73?10?4.17?94.4kPa
44
P1???bN1??odNd?cNc?10?3?0.25?19?1.5?1.73?10?4.17?98.5kPa3 3 From the calculation results shows that The local water level rises,The foundation of plastic load (here the foundation bearing capacity)will be reduced.
【Example 5-9 】A square base of a column,due to the limitation of conditions,Base width shall not be greater than 2m.The foundation soil
γ=18.8kN/m3，c=110kPa,，??4?，Seeking at least how much depth to safety under vertical center load of 1800 kn? [Solution]
P1??bN1/4??odNd?cNc
4According to ??4?Look-up table 5.2 we have:N1?0.06、Nd=1.25、Nc=3.51.
4Basal pressure should meet the following conditions:
p?p1??bN1/4??0dNd?cNc
4p?F Ab=2m、F=1800kN、γ=18.8kN/m3、c=110kpa、N?0.06、Nd=1.25、Nc=3.51
14Into the above equation was:
1800?0.06?18.8?2?1.25?18.8d?3.51?11022 d?2.6m
5.9 The ultimate bearing capacity of the foundation
When the load on the foundation is too large,, soil in the plastic zone development for the continuous penetration of slip surface,Loss of foundation stability and destruction.At this point, the load acting on the foundation known as the ultimate load.Which corresponds to the P-S curve local shear failure load stage corresponding to the overall shear failure stage cut-off point b.Known as the ultimate bearing capacity of foundation on the basis of the reaction force,with Ρu to represent.Determine the ultimate bearing capacity of the foundation,You must first determine the type of foundation shear failure
.
Figure of the P-S curve
Load test and model test results analysis shows,The foundation has three different failure modes,
The overall shear failure.the Local shear failure and the Punching shear failure,as the picture 5.29 shows:
Figure 5.29 (1)The overall shear failure.
The damage types of P - S curve has obvious turning point,
Continuous complete that destroyed the foundation of the sliding surface above the ground,Based on both sides of the ground has a hump phenomenon. when the foundation occurs the destruction of this type,Buildings will suddenly dumped.
Before the Damage of the foundation , the foundation settlement is smaller.For smaller soil compression,Such as hard close-grained sand or clay,When the pressure P is large enough, Usually this type of damage occurs.For the low bearing capacity,the ground under the foundation with small relative depth (d/b) ,This destruction may occur.Design to the foundation strength as controlling factors,This is a typical soil strength damage,Damage has certain suddenness. (2) the Local shear failure:
The damage type of P - S inflexion point of curve are difficult to determine,Destruction of the sliding surface of the foundation is not complete,Also does not extend to the ground,There may be slight bulge on the ground,But the foundation is not significantly tilt,The foundation settlement is large compared with the foundation overall shear failure,because of the large settlement,the foundation Loses to continue carrying capacity.For loose sand and soft clay,The base with the relatively larger depth has such a destructive type.Designed mainly to ground deformation as a control factor,It is a kind of deformation as the main characteristics of the failure mode. (3) the Punching shear failure:
Such the destruction types of P-S curve inflection point can not be determined.ground foundation damage is due to the soft soil deformation below the foundation and vertical shear along the base perimeter,so that Making the foundation sink,
Destroyed the foundation settlement is very big,Also called Pierce shear failure.For loose sand and soft soil which have large compression,or the foundation with the large relative depth situation will appear this kind of the damage types.Design on foundation deformation as the control factors,It is a kind of typical failure mode that is characterized by deformation .
At present the theory calculation formula of foundation bearing capacity is mainly according to the overall shear failure type.Then two kinds of damage type, there is no theoretical formula to follow.
The following describes the Ultimate Bearing Capacity of the limit equilibrium theory to establish the Prang del Reisner’s ultimate bearing capacity formula and through the basic model test,Research foundation overall shape of the sliding surface shear failure mode,And simplify for given slip surface,According to the sliding soil static equilibrium
conditions established of Terzaghi, Wei táng grams and Hansen limit bearing cross-sectional force formula.
5.9.1 Prang Bender (Prandtl) theoretical solution
In 1920, l. Prandtl (Prandtl) according to limit equilibrium theory, studies The rigid impact to moulde into massless semi-infinite rigid-plastic media,Derived the shape of the sliding surface and the ultimate compressive stress formula,when the media reaches the destruction .His solution is applied to the subject of the ultimate bearing capacity.Considering the rigid plate bar that the bottom is smooth with the uniformly distributed vertical load to put on the surface of the weightless ( γ= 0) homogeneous (d = 0) foundation ,The foundation material has a rigid-plastic nature.. When the soil under the load board is in a state of limit equilibrium,The shape of the sliding surface shown in Figure 5.30 (a) above.
Figure 5.30
Plastic zone is divided into five areas, a Area I, two Area II and two Area III. Area I
Below the basement (rigid bar) at the center of the wedge is called active Rankine's (Rankine) area. Because the base is smooth,There is no friction between the bottom of foundation and soil,The horizontal (side) as the major principal stress plane,, vertical surface is a principal stress
?plane . Rupture surface and the horizontal angle into 45??
2Area II
Slip lines in two groups, a set of logarithmic spiral (or),Its curve equation for
r?roexp(?tan?)(r0for the radius vector,ro?ab?a?b),Another group is the
ray starting point ofa? and a . Area III
Small primary stress surface is horizontal.Rupture surface and the