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terms of a friction coefficient alone, supposed to allow for the com-
bined cohesion and true friction, is untenable, as the writer has con-
clusively shown.* The error can likewise be realized by a numerical
example. Thus, on dividing Equation (1) by A, and putting

Q P

q = — — = unit total resistance to sliding, and j9 = — ? = unit
A A

normal pressure, we have,

(l = f Pn-\- C (2)

From Table 8, the lowest value of c, for the "sandy yellow clay" is



"Cohesion in Earth", Transactions, Am. Soc. C. E., Vol. LXXX, pp. 1332-1.334.



1048 discussion: reconstruction of stony river dam

Mr. 389, say 400, lb. per sq. ft. On arbitrarily assuming = 0.213, Equation (2) reduces to,

q = 0.213 Pn + 400.
Now, if it is assumed tliat a friction coefficient, / = -— , with c = 0,

can sufficiently represent tlie value of q, which denote by q', we have,
from Equation (2),

It will be found, for p„ = 3 333 lb. per sq. ft., that q '^ q' ; but, for
Pn < 3 333, q' < q; whereas for Pn > B 333, q" > q.

Thus, for Pn = l 000 lb. per sq. ft., q' = 333, q = 613 lb. per sq. ft. ;
but when p„ = 6 000 lb. per sq. ft., then q^ = 2 000 and q = 1 678 lb.
per sq. ft.

The differences would be much more marked had large values of c
been taken, corresponding to the 'Vhite clay" or "gumbo.'' As to the
value = 12°, arbitrarily assumed, it seems probable, from the experi-
mental values of Bell, given in Table 7, that lies somewhere between
7° and 16°, and thus an average was taken as a probable value. Like-
wise, from the author's Table 2, the yellow clay shows a high, com-
parative value of the "coefficient of frictional resistance", including
"adhesion", so that, possibly, ^ should lie nearer to 16° than 7°, and
12° is possibly a little below the true value.

In connection with the foregoing values of p„, it is of interest to

know that, ignoring the cut-off wall and taking the area of the base

of the dam as 52 X 15 sq. ft. for one 15-ft. bay, the average unit

, . , . 2 030 400
pressure on the foundation of the old dam is - - ■ - — = 2 600 lb.

per sq. ft.; also, if the earth below it weighs 100 lb. per cu. ft., the
unit pressure at 15 ft. below the foundation (at the bottom of the
"anchoring wall", or at the level, A B, of Plate XII), amounts to 4 100
lb. per sq. ft. The values of q and (f, as found from the foregoing
equations, do not differ greatly for these values, 2 600 and 4 100 for
Pn, but this is only due to the particular values of / and assumed.

It is not pretended that these are near the true values, as the com-
putation was simply made to show what differences might occur by the
two methods of estimating the resistance to sliding, and that the value
of Pn, at the foundation or below it, is an important factor in the
computation.

Although the data as to c and (/> are "not sufficient in number and
reliability" to lead to an exact determination of the passive resistance
to sliding of this dam, by the theory of coherent earth, nevertheless,
some interesting conclusions can be drawn, by finding, by this theory,



DISCUSSION : RECONSTRUCTION OF STONY RIVER DAM



1049



probable values of c and , with the normal ae, since
/ ^ tan. cf>.

Hence, if we lay off, to scale, ah = G, then draw hg ^ cl parallel
to I A; from g, draw a horizontal, gd, to the intersection, d, with the
line from a, making the angle, , above the normal ae, then gd = E,
which is equal and opposed to the horizontal passive resistance to
impending sliding up the plane, A I. This follows because the polygon
of forces, ah g da or ah g de a, is a closed polygon, where, ah = G,
hg =^ cl, gd ^= Ej de = N f, ea = N, represent all the forces acting
on the mass of earth, A I E D.

E can be found from the diagram or from an equation easily
deduced.

Thus, E = gh + hd

= c I cos. a + (G -\- cl sin. a) tan. (a -{- cf)).
E = c.AB-\-{G-\-c.BI) tan. (a + «^) (3)

It is evident from this equation,* regarding G as the load on A 7
for any position of A 7 between the one shown in the figure and A B,
that E diminishes with a and is least for a = 0, or along A B.

Now, to be conservative, take tan. 2 ^45° + ^^ + 2 c tan. (^45° + ~\ ,

where 6 = the weight, in pounds per square foot, acting on a horizontal
plane at the point considered. As the distribution of stress on P ^ is
not given, assume it to be uniform, so that at D, h = 149 300 -^ 52 =
2 870 lb. per sq. ft. ; and at A, I = 2 870 + 15 X 100 = 4 370 lb. per
sq. ft. Hence, taking, as before, = 12°, c = 300 lb. per sq. ft., the
equation gives the unit passive resistance, acting horizontally to the left,

at D : 6 120 lb. per sq. ft. ;
at A: 7 400 lb. per sq. ft.

As the flow of the earth is supposed to occur at 4 000 lb. per sq. ft.,

it is seen that sliding up the dotted plane will not occur when the

maximum permissible stress on AD is taken at 4 000 lb. per sq. ft., as

given by the author. If the permissible value to prevent flow had

been assumed to be greater than 7 400 lb. per sq. ft., then the maximum

unit compression at A must be taken at 7 400 lb. per sq. ft. or less ;

otherwise the earth at A would tend to slide up the dotted plane, which


makes an angle, 45° H — — = 51°, with the vertical.

* Cain's "Earth Pressure", p. 192.



1052 discussion: keconsteuction of stony rivee dam

Mr. It now remains to compute the passive resistance, exerted on F J

^'°' or F C, of the earth to sliding up the proper planes of rupture, J N or
C L, corresponding. The graphical construction for effecting this will
be given presently; but, to complete the present investigation, the
result may be anticipated that the passive resistance on either F J or
F C exceeds the safe compressive stress of the earth, so that the latter
will limit the available resistance. The unit passive resistance at the
free surface is not zero, but a very appreciable quantity, as will be
shown eventually, and reasons will be given why an average pressure
on F J or F G oi S 000 lb. per sq. ft. can be allowed without fear of the
compressive strength of the earth being exceeded, or flow being im-
minent. Thus, the total compression, acting horizontally, that will be
allowed on F J, 9 ft. in depth, is, ^

3 000 X 9 = 27 000 lb., ' 'i

and on F 0, 16 ft. in depth,

3 000 X 16 = 48 000 lb.

It has been shown previously, omitting any resistances along I J
and B C, that the horizontal component of the resistance to sliding
along A C is 64 000 lb., and along A J, 90 000 lb. ; hence the total
horizontal component of the resistance,

■'''• ' along A C and C i^ = 64 000 + 48 000 = 112 000 lb.
and along A J and J F = 90 000 + 27 000 = 117 000 lb.

Both of these totals exceed the horizontal component of the water
pressure, 81 000 lb. ; so that, if the assumed values of, c = 300 lb.
per sq. ft. and ^ = 12°, are safe values, the dam is secure against slid-
ing, for the conditions assumed, along the planes taken. The con-
clusion will not be altered if any plane intermediate between A J and
A C is taken in place of either of these planes. From what precedes,
it seems that c =^ 300 lb. per sq. ft. is a safe value, but there is uncer-
tainty as to the value of



Online LibraryAmerican Society of Civil EngineersTransactions of the American Society of Civil Engineers (Volume 81) → online text (page 87 of 167)