Chapter 1 Introduction
5.2 Point Force Validation
space to the value of a Poisson solid. The following are variables to be determined by the Neighbourhood Algorithm: Layer 1 thickness, VP, VS; Layer 2 thickness, VP, VS. For the second layer, the obtained ratio for VP and VS is very close to that of a Poisson solid, and for simplicity it is assumed that the second layer is a Poisson solid. Tables 5.3 and 5.4 summarize the two solutions obtained for the optimal 2 layer velocity models.
Depth(m) Density(Kg/m3) Vp(m/s) Vs(m/s)
45 1907 710 376
375 1957 1410 814
1500 2500 5505 3178
Table 5.3 Material properties obtained from the use of the Neighbourhood algorithm for the first of the two converging solutions. There is no linear interpolation between the given data points, as this model is a layered model. This model will be referred to as model Layer2A.
Depth(m) Density(Kg/m3) Vp(m/s) Vs(m/s)
45 1907 830 440
375 1957 1590 843
1500 2500 5505 3178
Table 5.4 Material properties obtained from the use of the Neighbourhood algorithm for the second of the two converging solutions. There is no linear interpolation between the given data points, as this model is a layered model. This model will be referred to as model Layer2B.
couple (point force scenario, even though point couple requires two forces at two dif- ferent nodes). When the 9 node distributed force scenario is used, there is a complete symmetry for the model. This is not the case when the point force scenario is used, as 2 nodes are required for the applied couple. Due to the need for two nodes, the first is chosen to be at the model’s geometric center and the second node is either to the East or North of the first node for an EW or a NS shake, respectively. For the distributed force scenario, extra nodes are added at the center of the model (as shown in Figure 5.5) such that the nodes where the forces are applied are 25 m apart, while the rest of the surface elements remain in a very similar configuration to the mesh used in the point force scenario. This node positioning in turn creates an equivalent building area 4 times larger than the actual floor area of Millikan Library. The shear force (P) and overturning moment (M) were distributed amongst the 9 nodes in a manner proportional to the tributary area of the node (total element area connected to the node), as shown in Figure 5.5. Moreover, as this is only a test, it will only be performed for an EW shake, as the results should be similar for the NS shake.
a) b) c)
d) e) f )
h)
g) i)
25m
Figure 5.5 Distributed node locations for source. Dark lines delineate the larger building floor plan, whereas the actual floor plan is 14 of the size. Node spacing for source are 25 m.
Assuming uniform shear stress, τxx, from the horizontal displacements for the area covered by the new building “floor plan”, and utilizing the node distance, l, as a
length scale, the total shear force is
P = 4τxxl2
The shear force at each node is then equal to the sum over its elements of the product of the shear stress and the tributary area for the node, resulting in the following horizontal node forces
Pxa,c,g,i = 1 24P
Pxb,d,e,f,h = 1 6P
For the calculation of the vertical forces to apply to the nodes, it is assumed that the vertical stress, σzz, varies linearly with distance from the centerline and the moment arm is 2l. The total moment associated with the vertical stress is
M = 4
3σzz,ml3
where σzz,m is the maximum stress at the EW edges for an EW shake. For an EW shake, due to symmetry, it is known that the total vertical force at nodes b,e, and h is zero. The vertical force at each of the remaining nodes is then equal to the sum over its elements of the product of the vertical stress and the tributary area for the node, which results in the following forces for the nodes
Fzb,e,h = 0
Fza,c,g,i = 3M 32l
Fzd,f = 5M 16l
The total forces for an EW shake were applied to the model as either a horizontal point force and a vertical force couple or the distributed force case described above.
This was done for the various force scenarios described in the list below, utilizing the velocity model given in Table 5.3. In the following list, as mentioned before, “point moment” refers to two vertical point forces applied at two adjacent nodes, and the force is a horizontal shear force. The amplitudes of the applied force and moment are computed using the method shown in Section 3.6, and to compute the vertical forces applied at the nodes, divide the moment by the node spacing.
• Point Force and Point Moment together
• Point Force only
• Distributed Force only
• Point Moment only
• Distributed Moment only
The resulting displacements along EW and NS lines in the model are presented in Figures 5.6 and 5.7. As can be clearly seen, the point forces generate very similar displacements to those of the distributed forces. Furthermore, the displacements from the shear force are significantly larger than those of the overturning moment, and generate the overall pattern in the observed displacement field, as is shown in Figure 5.7. This is true for both the EW and NS shaking directions, as the ratio of moment to shear force is approximately 30 for both cases. Moreover, these figures also show the extent for which the point forces directly influence the nodes at which the forces are applied and those directly adjacent to them (the very near-field). It is found that nodes closer than 150 meters to the model’s geometric center are directly influenced by the applied point forces. As a result, any data points within 150 meters of the source are discarded. It is unfortunate that this distance is so large, as it forces us to discard a significant portion of our data points, and limits 4 out of 6 lines to 2 seismometers per line. Since the East and North instrument lines used the most
instruments and are the longest, these are the only two lines that are used to compute the errors described in the previous and following section.
Distance (m)) Point Force
Distributed Force
Distance (m)) EW Profile, Z Comp
NS Profile, T Comp EW Profile, R Comp
Point Couple Distributed Force
Log Displ. (microns)10Log Displ. (microns)10Log Displ. (microns)10
-1500 -1000 -500 0 500 1000 1500 10- 2
10- 1 100 101
Point Force Vs. Distributed Force
-1500 - 1000 - 500 0 500 1000 1500 10- 2
10- 1 100
-1500 -1000 - 500 0 500 1000 1500 10-1
100 101
-1500 -1000 -500 0 500 1000 1500 10- 2
10- 1 100
Point Couple Vs. Distributed Couple
-1500 -1000 -500 0 500 1000 1500 10- 2
10- 1 100
- 1500 - 1000 - 500 0 500 1000 1500 10- 3
10- 2 10- 1 100
EW Profile, Z Comp
NS Profile, T Comp EW Profile, R Comp
Figure 5.6 Displacement comparisons for point forces vs. dis- tributed forces as described in the text for an equivalent EW shake. The left 3 plots present the computed displacements from the FEC for a point force scenario and a distributed force scenario, whereas the right 3 plots show them for a point cou- ple and a distributed couple. The top 2 plots on each side show the non-nodal component displacements for a complete EW profile passing through the model’s geometric center, while the bottom plot shows the displacements for an equivalent NS profile.
-1500 - 1000 -500 0 500 1000 1500 10- 2
10- 1 100
EW Profile, Z Comp
-1500 -1000 - 500 0 500 1000 1500 10- 1
100 101
NS Profile, T Comp
Distance (m))
-1500 -1000 -500 0 500 1000 1500 10- 2
10- 1 100 101
Point Force and Couple Vs. Point Force
EW Profile, R Comp
-1500 -1000 -500 0 500 1000 1500 10- 2
10- 1 100 101
Point Force Vs. Point Couple
-1500 -1000 -500 0 500 1000 1500 10- 2
10- 1 100
- 1500 -1000 -500 0 500 1000 1500 10- 3
10- 2 10- 1 100 101
Point Force Point Couple
Distance (m)) EW Profile, Z Comp
NS Profile, T Comp EW Profile, R Comp
Point Force & Couple Point Force
Log Displ. (microns)10Log Displ. (microns)10Log Displ. (microns)10
Figure 5.7 Same as Figure 5.6, except that the left 3 plots cor- respond to a comparison of 1) an applied point force and 2) a point couple and the right 3 plots compare those of 1) a jointly applied point force and couple, and 2) a point force by itself.