Transactions of the American Society of Civil Engineers (Volume 81) online

Otherwise, if QO is replaced by its components, QN and NO, equilib-
rium is indicated by the closed polygon, OPQNO.

Tor equilibrium, the sum of the components of pa, qb, and the com-
ponents {cl -\- nl tan. 0) and nl, of rl, perpendicular and parallel to AC
must be separately equal to zero.

p a sin. a -\- qh cos. {a -\- i) ^ nl (1)

p a cos. a — qh sin. {a-\- i) =^ nl tan. , but it does occur when t > ^ at a certain depth, which




W = 100 Lb. per cu.ft.
\ C = 100 Lb. per sq.ft.



S a = 640 Lb. per sq.ft.

Fig. 27.



call x^. Thus, substituting p = wx^ cos. i, in the first of Equations
(6), we find the limiting depth at which the surface of rupture is
parallel to the surface,

C COS.

X =: ^ CJ)

^ w COS. i sin. (i — 0) ^ ^

There is no equilibrium for a greater depth, since a cannot increase
farther, no wedge of rupture, ABC, Fig. 26, being formed. Thus slip-
ping is impending at x = x^, and would occur for x > x^, vmless the
earth is confined by walls, natural or otherwise, the resistance of
which thus introduces external forces not contemplated in the theory
of the unlimited mass, subjected to no external force but its own
weight.



DISCUSSION ON EARTH PRESSURES 213

If, in Fig. 26, we drop a perpendicular from B on the "free sur- Mr.

face", and call its length h, we have h = x^ cos. t; whence, substituting

in Equation (7), we derive,

c cos. (b
sm. {i — (P) = -^ (8)

From this equation, for an assumed value of h, the value of i can be
found. This is the maximum inclination of the surface corresponding

to the afiven h. After i has ])eeu thus found, 3?„ = can be com-

cos. i

puted.

2

2c , /,_ , 0, have now been given. When the earth surface slopes down-
ward to the right from D, Fig. 26, the conjugate thrust still acting
parallel to the surface, it can be shown that Equations (3) and (4)
hold, on simply replacing i by ( — i) .

It is not the intention to enter into passive thrust or resistance, but
it may be stated that the solution can be effected along lines similar to
those used in the case of active thrust.*

For active thrust, the most important case is that for which i = 0,

or the free surface horizontal. From Equation (4) we obtain, at

(p
once, cos. (2 a + 0) = ; whence, a = 45° — — , for any x.

It follows that the surface of rupture is plane, and that it bisects the
angle between the vertical and the line making the angle,