1 Introduction Hyperbolic Soil Bearing Capacity This example involves analyzing the bearing capacity of a round footing. The example is useful for illustrating several SIGMA/W features and procedures, and for demonstrating that SIGMA/W has been formulated correctly. The soil is treated as having a nonlinear hyperbolic stress-strain behavior. With this model, the soil stiffness in essence softens as the shear stresses increase; that is, the soil stiffness modulus E diminishes as the shear stress increases. The soil softens until it reaches its shear strength, at which point the deformations become very large and the footing load has reached the maximum available; it has reached the bearing capacity. The hyperbolic model consequently makes it possible to compute a nonlinear load-deformation curve for the footing and to estimate the ultimate bearing capacity. Closed-form bearing capacity formulas are available for a simple configuration like this, making it possible to compare the SIGMA/W results with hand-calculated values. 2 Feature highlights The soil stiffness modulus E is a function of the stress state in the ground, and therefore we must first establish the insitu stress state conditions before we can apply the footing load. This requires a special insitu-type analysis. The undrained shear strength C u and the modulus E can be specified as a function of the overburden (initial vertical stress). These features are illustrated here. 3 Hyperbolic constitutive model It is useful to have the hyperbolic soil model in mind when analyzing and discussing his problem. E t ( σ1 σ3)( 1 sinφ) ( φ) + σ φ R f = 1 2c cos 2 3 sin 2 E i and ( ) 2 Et = 1 Rf 1. E i when at the shear strength 4 Problem configuration The next diagram shows the problem configuration. Note the English system of units. The footing has a radius of 4 ft. This is an axisymmetric problem, so the symmetric axis needs to be at an x-coordinate of zero. The footing load is applied at a constant displacement rate of -.1 ft per load step. The horizontal displacement under the footing area is fixed at zero. Pushing the footing into the ground at an even level rate and fixing the horizontal displacement is like simulating a rigid footing with a very rough base such that there is no slippage at the footing-soil contact. SIGMA/W Example File: Hyperbolic footing.doc (pdf)(gsz) Page 1 of 7
Radius = 4 ft Round footing Axisymmetric axis 35 25 Elevation - feet 15 1 5-5 -5 5 1 15 25 35 5 Analysis 1 Initial insitu stresses Distance - feet This analysis establishes the initial insitu stresses. This best accomplished with the SIGMA/W analysis type called Insitu. The Insitu analysis type uses the specified soil unit weight and the Poisson s ratio. The modulus parameter E is ignored. The unit weight for this problem is specified as 1 pcf and ν is.495. Remember that in a 2D laterally constrained problem, K o is related to ν as follows: K o = ν / (1 ν) =.495 / (1.495) =.98 The insitu analysis results should indicate that the horizontal (x) stresses are.98 times the vertical stress. The vertical stresses will be γ times the depth below the ground surface. The next diagram shows the configuration for the insitu analysis. The cross-hatching indicates that the gravity load is being applied. Note the boundary conditions on the left and right, which allow vertical movement but not horizontal movement. This is equivalent to having rollers on the vertical sides. SIGMA/W Example File: Hyperbolic footing.doc (pdf)(gsz) Page 2 of 7
Need diagram here; painting is not right with gravity turned on. Figure 1 shows the initial insitu vertical and horizontal stress profiles. They are of course very similar, because K o is close to 1.. There are slight bends in the profiles at both ends. This is due to the fact that the stresses are constant within these 4-node rectangular elements. Averaging the stresses from adjacent elements at the nodes results in a nice straight line except for the nodes along the perimeter, which reflect the constant stresses in the elements. Y (ft) 1 1 (psf) Figure 1 6 Analysis 2-Constant C and E In this analysis the undrained strength C u and the E-modulus are specified as constants. Depending on the bearing capacity factor used, the ultimate bearing capacity for an undrained soil with a round footing at the surface is: Q ult = 1.3cN c The value of N c is typically taken as 5.14. With C u at 1 psf and N c = 5.14, the bearing capacity is 667 psf. The load deformation curve in Figure 2 shows that applied load when the settlement has reached one-foot is right around this value. (The load-deformation curve in Figure 2 was created by using the Graph Sum vs. Average Values option. All the Y-Boundary forces at each of the nodes under the footing are tallied and the Y-displacement is averaged for all the nodes. Without this option, a load-deformation curve is produced for each node). SIGMA/W Example File: Hyperbolic footing.doc (pdf)(gsz) Page 3 of 7
Load-Deformation -1 Y-Boundary Force (lbs) - - - -5-6 -7-1. -.8 -.6 -.4 -.2. Y-Displacement (ft) Figure 2 The load deformation curve does not flatten out at some well-defined value. This is in large part due to the fact that the tangent modulus E t can never go to zero in the hyperbolic stress-strain relationship. Radius = 4 ft Round footing etric 35 1 5 25 7 Elevation - feet 15 9 1 5 Figure 3 Tangent E-modulus contours at the end of loading SIGMA/W Example File: Hyperbolic footing.doc (pdf)(gsz) Page 4 of 7
The parameter R f is specified as.7. This means that the E t value will be 9 psf when the applied shear stresses reach the soil shear strength. Figure 3 shows the variation in E t at the end of the loading. Right under the footing where the shear stresses are the highest, E t diminishes to about 1, as it properly should. Worth noting is the fact that C u and E i are constant in this analysis and therefore not dependent on the initial insitu stresses, but E t may be affected by any initial shear in the ground. In this problem there is essentially no initial shear, since the vertical and horizontal insitu stresses are nearly the same. In situations where this is not the case, the initial insitu shear stresses may affect the initial E t value. 7 Analysis 3 Variable C and E In this case we want to make C u and E i a function of the initial overburden stress (function of depth), but then keep these values constant while the footing load is applied. The two specified functions are shown in Figure 4 and Figure 5. C u varies linearly with depth while E i varies with some curved relationship. In the previous analysis E i was a constant 1, psf. Now E i varies about this value. The minimum value is 55, and goes up to about 137, psf at depth. The primary purpose of this analysis is to illustrate the capability of specifying material properties that vary with depth as a function of the initial insitu stress state, but then remain constant while the loading is applied and the soil behaves in an undrained manner. C(u) with depth Total Cohesion (psf) 1 1 Y-Total Stress (psf) Figure 4 Specified C u function SIGMA/W Example File: Hyperbolic footing.doc (pdf)(gsz) Page 5 of 7
E(i) function Total E-Modulus (psf) 1 1 1 11 1 9 8 7 6 5 1 5 Y-Total Stress (psf) Figure 5 Figure 6 shows the SIGMA/W computed variation of C u with depth. The data is plotted for the initial and last Load Step numbers ( and 1 sec where the sec unit is really load-step number). As they should, the curves fall on top of each other; that is, the strength is the same at the end as at the beginning under the simulated undrained condition. Cu with depth sec Y (ft) 1 1 sec 1 15 25 35 Undrained Shear Strength (psf) Note that C u is remains a constant at depth. This is because the insitu vertical stresses fall outside the specified function. Generally, any values outside the function take on the value of the data at the ends of the function. This is graphically illustrated by the straight line segments at the ends of the E i function in Figure 7. SIGMA/W Example File: Hyperbolic footing.doc (pdf)(gsz) Page 6 of 7
E(i) with depth sec Y (ft) 1 1 sec Tangential Modulus (E) (psf) Figure 6 Need to re-do; wrong graph The contours of E t now look vastly different because E i was specified as being a function of depth. Radius = 4 ft Round footing Axisymmetric axis 35 1 6 7 25 9 Elevation - feet 15 1.1e+5 1 5-5 -5 5 1 15 25 35 Distance - feet Figure 7 8 Summary This example illustrates how material properties can be specified as a function of depth, how the properties can remain constant during loading, and confirms that the hyperbolic model is functioning as expected. SIGMA/W Example File: Hyperbolic footing.doc (pdf)(gsz) Page 7 of 7