PERMANENT DISPLACEMENTS OF THE GROUNDS. 25 
point which the line makes with its original unstrained direction. We have represented 
the value of this force by the broken line WG’LH’E in fig. 12. Starting at W where it is 
zero, the shearing force becomes negative; that is, it is directed in a southerly direction, 
reaching a negative maximum at G’, where the displacement curve has a point of inflection ; 
it then diminishes in value, becoming zero at A, where the displacement curve is parallel 
with its original diréction ; it then increases rapidly in value, reaching a positive maximum, 
L, at O, the point of rupture; the shearing force to the east of the rupture has somewhat 
the same value it has at an equal distance to the west, tho symmetry is not required. 
The total shearing force which we have determined is not the force applied at each point 
under consideration, but is equal to the sum of all the forces applied to the east or west of 
the point; the actual force applied at each unit length of the line is proportional to the 
difference in value of the total shearing force at points a unit distance apart; that is, 
to the angle which the line representing the total shearing force makes with the line WOE; 
it is represented by WGDOFHE in fig. 13. Starting with a zero value at W, it first has a 
small negative value but becomes zero again at G; it then becomes positive and increases 
to a maximum at D, where the line of total sheer has a point of inflection —and dies down 
rapidly to zero at O, where the total shear is a maximum; it has somewhat similar but 
opposite values to the east of O. 


Fic. 13. 
Without insisting on accuracy in small details the full line in fig. 13 shows in a general 
way the relative distribution of the forces, applied at the under side of the moved region, 
which brought about the California earthquake. 
The distribution of the total shearing forces shows why in 1906 there was no break at the 
Haywards fault, where the break occurred which caused the earthquake of 1868. This 
fault is about 30 km. (18.5 miles) east of the San Andreas fault; and therefore in the 
neighborhood of C (fig. 12), where the surveys detected no displacement relative to 
Mount Diablo; in this region, as the figure shows, there was practically no shearing force, 
and therefore no break occurred. . For the same reason there was no rupture at the San 
Bruno fault south of San Francisco. This fault is 4 km. (2.5 miles) east of the San 
Andreas fault and at that distance (fig. 5) the shearing force was only about one-third 
as strong as it was where the rupture actually occurred. We have seen that the elastic 
strain was probably accumulating for 100 years; it is quite possible, then, that the 
earthquake of 1868 partially relieved the strain for some distance south of San Francisco 
and that there would have been no fracture in this part of the San Andreas fault if ad- 
ditional strains had not been thrown on it by the rupture of the fault-plane further north. 
It is to be noticed that the distances from O to A and from O to C, beyond which no 
distortion of the rocks occurred, were probably less than 10 km. (6 miles), and the distances 
OG and OH, over which the distorting forces were distributed, were probably ten or more 
times as great, and the total area over which they were applied was many times as great 
as the area of the fault-surface; the applied forces were therefore considerably smaller 
per unit area than the shearing forces at the fault; for the sum of all these forces on each 
side of the fault-plane must have equaled the shearing force at that plane plus the small 
shearing force at G or H, due to the slight reverse curving at this point. 
As the dragging forces are applied at the base of the crust they have a moment about 
its center of gravity which is balanced by the moment due to stronger and greater shears 
near the bottom than near the top at the points G, O, and H (fig. 12); and lines at differ- 
