STABILITY AGAINST CRUSHING. The preceding discussion of the stability against overturning is on the assumption that the masonry does not crush. This method of failure will now be con sidered. When the reservoir is full, the thrust of the water concen trates the pressure upon the down-stream edge of a horizontal joint or section; and it is proposed to find the law of the distribution of the pressure on any horizontal section when the reservoir is full. Whcn the reservoir is empty, there is no external force to affeot the distribution of the pressure upon a horizontal section; but the vertical cross section of a dam, particularly a high one, is unsymmetrical, being roughlysomewhat similar to a right angled triangle with the water against the vertical side, and there fore the vertical through the center of gravity of the dam does not pass through the center of the base, and hence the pressure upon the sec tion is not uniform, being a maximum at the heel and a minimum at the toe. Therefore, there are two cases that must be considered, viz.: I, reservoir full; and II, reservoir empty.
M = the moment about A of the water pressure = — (see equations 7 and 8, pages 464 and 465).
tiV = the weight of a section of the dam 1 foot Tong.
V = the vertical component of the water pressure.
P = the maximum pressure, per unit of area, at A.
p = the change in unit pressure, per unit of distance, from A towards B.
x = any distance from A towards B.
P — px = the pressure per unit at a distance x from A towards B.
Y = a general expression for a vertical force.
1 = the length of the base of the section considered = AB.
d = the distance the center of pressure deviates from the cen ter of the base = rm.
= the distance from the down-stream edge of any horizontal joint or section to the point in which the vertical through the center of gravity pierces the section = Ag.
Taking moments about A gives For equilibrium, the sum of the forces normal to AB must also be equal to zero; or If the overturning moment is determined algebraically, i.e., by equations 7 and 8, pages 464 and 465, then M is known; and therefore P can be computed by equation 18.
If the stability against overturning is determined graphic ally by resolution of forces, equation 18 can not be employed to de termine the stability against crushing, since M is not then known.' To meet this case, equation 18 may be transformed as follows: M, the moment of the water pressure, is equal to the moment of the weight of the dam minus the moment of the reaction of the soil; or taking moments about A, M = W. Ag — (W + V) Ar. From Fig. 100, page 470, Ag = and = 1 — d.
Substituting the above values of x and M in equation 18, gives Equation 19 is suitable for computing P when the stability against overturning is determined by resolution of forces.
Discussion of Equations 18 and 19. Equation 18 is the equivalent of equation 1, page 354, except that (1) in the latter case there no vertical component of the external force, and (2) equation 1 is applicable to any form of cross section while equation 18 is limited to a rectangular cross section.
If the wall is symmetrical, z = 1, and if V = 0, equation 18 becomes which is the same as equation 3, page 354, except that equation 20 is for a unit length of the wall, and hence the dimension cor responding to b in equation 3 does not appear.
If there is no external overturning force, V = 0 and M = 0, and then equation 18 becomes In equation 21 if z = 1, that is, if the resultant vertical force passes through the center of the base, P = W _ 1, as it should, since the pressure on the base should then be uniform. If i = 1, P = 2 W - 1, which shows that the maximum unit pressure on the base of a right-angled triangle cross section is twice the average pressure. If x = 1, P = 0, as it should, since this is the case of a dam having a right-angle at B, and consequently the pressure per unit of area at A is zero.