## Model Sensitivity

Download an interactive exercise to explore the SHALSTAB model.     main muddled stability page Let's see that as a graph. The parabola represents the equation above. Under the parabola, the hillside is stable. However, the hillside is also stable (barring positive pore pressure) if the tangent of the slope is less than half the tangent of the friction angle (phi). Furthermore, the addition of additional water makes no difference after the entire soil column has been saturated. Consider the class of landscape cells at 20°. As area/base rises above 205 meters, overland flow begins. Any greater contributing area, even at values above 250 meters, will have no effect on stability.

Let's look at what happens when we vary single parameters. The black curve shows the base case of typical parameters for a northwest clearcut slope with a heavy steady-state rainfall of 4 inches per day. The region above the black parabola and to the right is unstable. The region above the nearly diagonal line is saturated; increasing the contributing area merely causes increasing overland flow. (Overland flow can eventually cut channels, but it not a cause of landslides.) The region to the left of the point where the diagonal meets the parabola is always stable; the ground saturates before the wetness can rise to the level to cause instability. The region to the left of the vertical line is always stable at any cohesion. The green vertical line goes with the green parabola and the red lines and the blue lines also go together. The vertical black line corresponds to all the other cases. Holding other factors constant, we see the effects of altering the cohesion. A cohesion greater than 7 kPascals will be stable in any topography. At some cohesion values, the model predicts greater stability at higher slopes. Holding other factors constant, we see the effects of altering the friction angle. As in previous graphs, the vertical lines are irrelevant, because they lie to the left of the intersection of the saturation line with the parabola. Everything in the previous graph was proportional to T/Q, so we can pull it out by multiplying by Q/T. Look at another version of the first graph, but with a base case cohesion of 6. The slope in nearly totally stable (cohesion of 7 is always stable), but the contribution of cohesion is inversely proportional to soil depth. Thus doubling the soil depth cancels out the increased stability resulting from doubling the cohesion.

Direct questions to Harvey Greenberg and Dave Montgomery