CHAUDY’S THEOREM.
169
it exercises its greatest effect. He proceeds to do this by resolving the
pressure, Q (fig. 92), into its component parts, Q sin 7, and Q cos 7, along,
and perpendicular to, the direction of the oblique thrust, assumed to make
an angle, 7, with the horizontal, and, in this way, he determines the amount
of the oblique pressure as
P = Q sin 7 - Q cos y tan p = Q sin y (1 -
tan p\
tan y)’
(27)
the last term being the deduction due to friction.
Considering, now, an element, x, of the surface, A 0, as undergoing an
intensity of pressure, q, and noting that y, the corresponding element of
the surface exposed to the oblique intensity, p, is as sin y, we can derive
from the above equation—
py = px sin y = yæ sin y 1 1 -
whence,
tan p\
tan y/’
tan p'
tan y,
(28)
which gives the relative intensities of the two pressures.
Applying this to the case of a retaining wall, A B C F (fig. 93), we see
that the vertical force for each element of surface is the weight of a strip
of earth, wxa, and, therefore, that
P = wxSasa x
sin y
tan p'
tan y.
= area F CE x w sin y( 1 -
tan p
tan y.
Now, the area F OE = JFG.CE,
in which FG = F C cos (y- ß) — à sec ß cos (y - ß),
and CE = À cosec y;
^2
. • . the area F 0 Ê = — cosec y sec ß cos (y — ß),
and P = ^ . sec ß cos(y-ß)^1 - ^^y • (29)
When the back of the wall is vertical, ß = 0, and the equation simplifies
into
wh2
P = — ^— cos y
tan p\
tan y/’
(30)
To determine the value of y, which will give the maximum value to the
•equation, differentiate the variable factors, as before, and equate to zéro : —
d cos
1 -
dy
tan p'
tan y.