^ LIBRARY

UNIVERSITY Ot CALIFORNIA.
' Deceived WAR 15JR93 . i8g .
Accessions No.
forks of Professor Mansfield Merriman.
Published by JOHN WILEY & SONS, 53 E. Tenth
Street, New York.
A TREATISE ON HYDRAULICS.
Designed as a Text Book for Technical Schools and for the
use of Engineers. By Professor Mansfield Merriman, Lehigli
University. Fourth edition, revised 8vo, cloth, $3 50
"As a whole this book is the most valuable addition to the literature of
hydraulic science which has yet appeared in America, and we do not know
of any of equal value anywhere else." Railroad Gazette.
"With a tolerably complete knowledge of what has been written on
Hydraulics in England, France, Germany, United States, and to some extent
Italy, I have no hesitation in saying that I hold this book to be th^ best
treatise for students, young or old, yet written. It better presents the
primary essentials of the art." From CLEMENS HERSCHEL, Hydraulic
Engineer of the Holyoke Water Power Company.
A TEXTBOOK ON THE METHOD OP LEAST SQUARES.
By Mansfield Merriman, C.E., Ph.D., Professor of Civil
Engineering in Lehigh University. Fifth revised edition.
8vo, cloth, 2 00
This work treats of the law of probability of error, the ad
justment and discussion of observations arising in surveying,
geodesy, astronomy and physics, and the methods of compar
ing their degrees of precision. Its rules and tables will assist
all who wish to make accurate measurements.
"This is a very useful and much needed textbook." Science.
"Even the casual reader cannot fail to be struck with the value which
such a book must possess to the working engineer. It abounds in illustra
tions and problems drawn directly from surveying, geodesy and eugineer
ing." Engineering Ntws.
THE MECHANICS OF MATERIALS AND OF BEAMS,
COLUMNS, AND SHAFTS.
By Professor Mansfield Merriman, Lehigh University, South
Bethlehem, Pa.
Fourth edition revised and enlarged. 8vo, cloth, interleaved, 3 50
"We cannot commend the book too highly to the consideration of all
Professors of Applied Mechanics and Engineering arid Technical Schools
and Colleges, and we think a general introduction of the work will mark an
advance in the rational of technical instruction." American Engineer.
"The mathematical deductions of the laws of strength and stiffness of
beams, supported, fixed, and continuous, under compression, tension and
torsion, and of columns, are elegant and complete. As in previous books
by the same author, plenty of practical original and modern examples are
introduced as problems." Proceedings Engineers' Club (/Philadelphia.
A TEXTBOOK ON ROOPS AND BRIDGES.
Being the course of instruction given by the author to the
students of civil engineering in Lehigh University.
To be completed in four parts.
PART!. STRESSES IN SIMPLE TRUSSES. By Professor
Mansfield Merriman. Third edition. 8vo, cloth $250
" The author gives the most modern practice in determining the stresses
due to moving loads, taking actual typical locomotive wheel loads, and
reproduces the Phoenix Bridge Co's diagram lor tabulating wheel move
ments. The whole treatment is concise and very clear and elegant." Rail
road Gazette.
PART II. GRAPHIC STATICS. By Professors Mansfield Mer
riman and Henry S. Jacoby. Second edition. 8vo, cloth, 2 50
" The plan of this book is* simple and easily understood ; and a? the treat
ment of all problems is graphical, mathematics can scarcely be said to enter
into its composition. Judging from our own correspondence, it is a work
for which there is a decided demand outside of technical schools."
Engineering News.
PART III. BRIDGE DESIGN. In Preparation.
This volume is intended to include the design of plate girders,
lattice trusses, and pinconnected bridges, together with the
proportioning of details, the whole being in accordance with
the best modern practice and especially adapted to the needs
of students.
THE FIGURE OF THE EARTH. An Introduction to
Geodesy.
By Mansfield Merriman, Ph.D., formerly Acting Assistant
United States Coast and Geodetic Survey. 12mo, cloth 1 50
" It is so far popularized, that there are few persons of ordinary intelli
gence who may not read it with profit and certainly great interest. "Engi
neering News.
"A clear and concise introduction to the science of geodesy. The book
is interesting and deals with the subject in a useful, and to some extent,
popular manner." London Engineering.
A TEXTBOOK ON RETAINING WALLS AND MASONRY
DAMS.
By Professor Mansfield Merriman, Lehigh University.
8vo, cloth, 2 00
This work is designed not only as a text book for students
but also for the use of civil engineers. It clearly sets forth
the methods of computing the thrust of earth against walls,
and the investigation and design of walls and dams in the
most economic manner. The principles and formulas are
illustrated by numerous numerical examples.
A TEXTBOOK
ON
RETAINING WALLS
AND
MASONRY DAMS.
BY
MANSFIELD MERRIMAN,
j
PROFESSOR OF CIVIL ENGINEERING IN LEHIGH UNIVERSITY.
In scientiis ediscendis prosunt exempla magis quam praecepta. NEWTON.
NEW YORK:
JOHN WILEY & SONS,
53, EAST TENTH STREET.
1892.
COPYRIGHT, 1892,
BY
MANSFIELD MERRIMAN.
ROBERT DRUMMOND, FERRIS BROS.,
Electrotyper, Printers,
444 and 446 Pearl St., 326 Pearl Street,
New York. New York.
CONTENTS.
CHAPTER I.
EARTHWORK SLOPES.
PAGE
Article I. Equilibrium of Loose Earth I
2. The Cohesion of Earth 4
3. Equilibrium of Cohesive Earth q
4. Stability of Slopes in Cohesive Earth 12
5. Curved Slopes and Terraces 16
6. Practical Considerations 21
CHAPTER II.
THE LATERAL PRESSURE OF EARTH.
Article 7. Fundamental Principles 24
8. Normal Pressure against Walls 27
9. Inclined Pressure against Walls 31
** 10. General Formula for Lateral Pressure 35
**n. Computation of Pressures. ..... 36
* 12. The Centre of Pressure 39
CHAPTER III.
INVESTIGATION OF RETAINING WALLS.
Artie! e 13. Weight and Friction of Stone 42
14. General Conditions regarding Sliding. ... 44
15. Graphical Discussion of Sliding 47
16. Analytical Discussion of Sliding 50
17. General Conditions regarding Rotation 54
18. Graphical Discussion of Rotation 57
19. Analytical Discussion of Rotation 59
20. Compressi ve Stresses in the Masonry 63
iii
IV CONTENTS.
CHAPTER IV.
DESIGN OF RETAINING WALLS.
* PAGE
Article 21. Data and General Considerations 68
22. Computation of Thickness. ... 71
23. Security against Sliding , 75
24. Economic Proportions 77
25. The Line of Resistance. 83
26. Design of a Polygonal Section 86
27. Design and Construction 90
CHAPTER V.
MASONRY DAMS.
Article 28. The Pressure of Water 93
29. Principles and Methods 95
30. Investigation of a Trapezoidal Dam 98
31. Design of a Trapezoidal Section 103
32. Design of a High Trapezoidal Section 107
33. Economic Sections for High Dams ... 109
34. Investigation of a Polygonal Section 112
35. Design of a High Economic Section 115
36. Additional Data and Methods 120
SLOPES, WALLS AND DAMS
CHAPTER I.
EARTHWORK SLOPES.
ARTICLE i. EQUILIBRIUM OF LOOSE EARTH.
Earthwork slopes are the surfaces formed when excava
tions, embankments, terraces, mounds, and other construc
tions are made in or with the natural earth. The earth is to
be regarded in discussion as homogeneous and inelastic, and
as consisting of particles more or less united by cohesion be
tween which friction is generated whenever exterior forces
tend to effect a separation. As some kinds of earth when
dry are destitute of cohesion, these will first be considered
under the term " loose earth."
The friction of earth upon earth will be taken to be gov
erned by the same approximate laws as for other materials,
namely: first, the force of friction between two surfaces is
directly proportional to the normal pressure ; second, it varies
2 EARTHWORK SLOPES. [CHAP. I.
with the nature of the material ; and third, it is independent of
the area of contact. These laws may be expressed by the
equation
F=fN, (i)
in which N is the normal pressure, F the force of friction
perpendicular to N, and f is a quantity called the coefficient
of friction which varies with the kind of material. As F and
N are both in pounds, f is an abstract number ; its value for
earth ranges from about 0.5 to i.o.
If a mass of earth be thoroughly loosened so as to destroy
all cohesion between its particles, and then be poured verti
cally upon the point D in the horizontal plane BC, it will form
a cone BAG, all of whose elements AB, AC, etc., make equal
o
FIG. i.
angles with the horizontal. This angle ABC is called the
" angle of repose," or sometimes " the angle of natural slope,"
and it is found by experiment that eacli l.ind of earth has its
own constant angle. The particlepof earth on such a slope
are held in equilibrium by the forces of gravity and friction.
Let <p be the angle of repose A13D, and f the coefficient of
friction. In the figure draw W vertically to represent the
weight of a particle, and let N and F be its components nor
ART. i.] EQUILIBRIUM OF LOOSE EARTH. 3
mal and parallel to the slope. Now since motion is about to
begin,
F = fN.
Also since the angle between A 7 " and J^is equal to the angle
of repose 0, the rightangled triangle NO Ogives
Therefore results the important relation
/= tan 0, ........ (2)
that is, the coefficient of friction of earth is equal to the
tangent of the angle of repose. It is hence easy to determine
/when has been found by experiment.
In building an embankment of joose earth it is necessary
that its slope, or angle of inclination to the horizontal, should
not be greater than the angle of repose. When making an
excavation it is often possible, on account of the cohesion of
the earth, to have its slope at first greater than the angle of
repose, but as the cohesion disappears under atmospheric in
fluences the particles roll down and its slope finally becomes
equal to 0.
The following table gives rough average values of the
angles of repose and coefficients of friction of different kinds
of earth. In the fourth column the inclination or slope is ex
pressed in the manner customary among engineers by the
EARTHWORK SLOPES.
[CHAP. I.
ratio of its horizontal to its vertical projection. In the last
column average values of the weight of the material .are given.
Kind of Earth.
Angle
of
Repose.
*
Coefficient
of
Friction.
f
Inclination,
cot
Weight.
Kilos per
cu. met.
Pounds
per cu. ft.
Gravel round
30
40
35
40
30
40
45
32
0.58
0.84
0.70
0.84
0.58
0.84
1. 00
0.62
.7 to
.2 to
.4 to
.2 to
7tO
.2 tO
to
.6 to
1600
1700
1600
1700
2000
1440
1520
1840
100
IIO
100
IIO
125
90
95
H5
Gravel sharp. ... ..... .
Sand, dry
Earth dry
Earth moist . . .
It will be noticed that the natural slope and specific
gravity of earth undergo quite wide variations as its degree
of moisture varies. In collecting data for the discussion of
particular cases it is hence necessary to determine limits as
well as average values.
Problem i. A bank of loose earth is 16 feet high, and its
width, measured on the slope, is 28 feet. Compute the co
efficient of friction and the angle of repose.
ARTICLE 2. THE COHESION OF EARTH.
Cohesion is a force uniting particles of matter together.
If, for instance, two surfaces have been for some time in con
tact, they become to a certain extent glued or fastened to
ART. 2.] THE COHESION OF EARTH. $
gether so that any attempt to separate them is met by a
resistance. Friction only resists the separation of surfaces
when motion is attempted which produces sliding, but cohe
sion resists their separation whether the motion be attempted
parallel or perpendicular to the plane of contact. Particles of
rock are held together by strong cohesive forces, while parti
cles of dry sand have little, if any, cohesion.
By experiment the following are found to be the laws of
cohesion : first, the force of cohesion between two surfaces is
directly proportional to the area of contact ; second, it depends
upon the nature of the surfaces ; and third, it is independent
of the normal pressure. These laws may be expressed by the
equation
C=cA (3)
in which C is the resisting force of cohesion between two
surfaces, A the area of contact, and c a quantity called the
coefficient of cohesion depending upon the nature of the
material.
The value of c for homogeneous earth may be found as
follows : Dig in the ground several trenches of considerable
length compared with their width, and of different depths.
After a few days it will be observed that all those over a cer
tain depth have caved along some plane such as BM in Figure
2. Let H be the value of this certain depth. Let w be the
weight of a cubic unit of earth, and its angle of repose when
EARTHWORK SLOPES. [CHAP. I.
devoid of cohesion. Then the coefficient of cohesion of the
earth may be computed from the expression
_ Hw( i sin 0)
4 cos0
This formula will now be demonstrated.
Let the plane BM in the figure make an angle x with the
horizontal.' The prism BAM, whose length perpendicular to
the drawing will be taken as unity, tends to slide down the
M
#
FIG. 2.
plane. Let W be the weight of this prism, and P and N its
components parallel and normal to BM. P tends to cause
motion down the plane, and this is resisted by the combined
forces of friction and cohesion, acting in the plane. The force
of friction is fN, and that of cohesion is c/ } if / be the length,
or area, of BM. At the moment of rupture
from which the value of c is
c =
PfN
(4)
v
,
Rput* , ^ ' JW***^ ^r^
A : ,] T/inSSSm^ WE$m+
y Sd^U^Sxt^ Jl^X^^ r ZCOU^
In order to determine the angle ^r, consider that the equations
just written are for the plane of rupture, and that for any
P
other plane./ 5 is less than fN\ cl, or    is less than c.
' l
The value of x for the plane of rupture is hence that which
renders
~~T~~ ~ a maximum ...... (5)
Equation (5) will determine the value of x, and then c will be
found from (4).
To do this insert first for P its value W sin x, for N its
value Fcos x, and for fits value tan0. Then (5) becomes
^& ffi**j[X N ' r*'A*"')L
'*' w
(smx tan cos^) = a maximum.

This may be written
W sin (x 0)
 ,   = a maximum.
/cos
Next express W in terms of H and ^r, and the weight of a
unit of volume of earth w. Thus
and then (5) becomes
sin (90 x) sin (x 0)
2 COS0
= a maximum.
7
Z' *2^.
.' X /^ r2 V f^? /
8 EARTHWORK SLOPES. [CHAP. I.
This expression is a maximum when the two variable factors
' *: are equal; or when
** = * ^
Thus the value of x for the plane of rupture is
 jY . (6)
Now to find <:, insert in formula (4) the values of P, N, and
f in terms of x, and it becomes
Hw sin (90 x) sin (# 0)
' = ~ 2COS0 ~~>
which by virtue of (6) reduces to
=
Since sin 8 (45 J0) equals J(i sin0), this value becomes
Hw(i sin 0)
C =^ ' , ........ (7)
4 cos
which is the formula that was to be demonstrated.
From this formula the numerical value of c can be com
puted when H, w, and are known. For earth weighing 100
pounds per cubic foot and whose angle of natural slope is 30
degrees, the value of c becomes i^.^H. If the vertical ruptur
ing depth H is one foot, c is 14.4 pounds per square foot ; but
if H is ten feet, then c is 144 pounds per square foot.
ART. 3.]
EQUILIBRIUM OF COHESIVE EARTH.
Problem 2. A certain bank of earth, which has a natural
slope when loose of 1.25 to i, stands by virtue of its cohesion
with a vertical face when H = 3 feet. If this bank fails, find
the slope immediately after rupture.
ARTICLE 3. EQUILIBRIUM OF COHESIVE EARTH.
If the particles of earth be united by cohesion, a slope may
exist steeper than the angle of repose. Let Figure 3 rep
resent the practical case of an excavation ABC whose slope
AB makes an angle 6 with the horizontal greater than the
angle of repose 0. AM is the natural surface of the ground
making with the horizontal an angle tf less than 0. It is re
quired to determine the relation between the slope 6 and the
vertical depth h in order that rupture may just occur.
Let BM be the plane along which rupture occurs, and x its
inclination to the horizontal. The weight of the prism BAM
tends to urge it down the plane, and this is resisted by the
forces of friction and cohesion actinga thfi^lane. Let Wbz
UNIVERSITY
10 EARTHWORK SLOPES. [CHAP. I.
the weight of the prism for a length unity, and P and N its.
components parallel and normal to the plane. P is the force
causing the downward sliding, fN is the resisting force of
friction, and cl that of cohesion, if / be the area, or length, of
BM. At the moment of rupture P fN\cl, which may be
written *P<tf+&, / fk
Now as x varies the forces, P and N vary; and for any other
P fN
plane except that of rupture  ~  is less than c. Hence
the condition which will determine the value of x is
/fi tw *** y / , ^ inw*
PfN
When x has been found from (9) its value is to be inserted in
(8) and thus the relation between and h be established.
To do this insert in (9) for P and N their values W sin x
and W CQSX, and for /its value tan ,0. Then jit takes the form
"U~<, W sin (x 0)
7 = a maximu
/COS0
The value of W is the volume of the prism BAM, multiplied:
by the weight of a unit of volume w, or
W = \BA . BM. w sin (0  x) = \hlw ^L_i
v
P . 
ART. 3.] EQUILIBRIUM OF COHESIVE EARTH.
and hence the expression becomes
hw sin (0 x) sin (x
II
= a maxmum.
2 cos sin
This is a maximum with respect to x when
/i j>
that is, the plane of rupture bisects the angle between the
lines of natural slope and excavated slope, or
Now if (8) be expressed in terms of x y it becomes
hw sin (6 x) sin (x 0) = 2c cos sin 0,
and by virtue of (10) this reduces to
hw sin 8 J(0 0) = 2 cos sin 0;
and substituting for sin 2 (0 0) its value f(i cos(0 0)),
and fm its value from (7), there is found
( I cos (0  0)) = H(\  sin 0) sin 0, . . (11)
, 4Jfc ^^0>J^\<s^^
and this is the equation of condition between h and 0.
This discussion shows that both the angle of rupture x and
the relation between h and are independent of the slope d
made by the natural surface of the ground with the horizontal.
12 EARTHWORK SLOPES. [CHAP. I.
By the help of formula (n) the limiting height h may be
found when 0, and H are given. For instance, let it be
required to build a slope of I to i, or # = 45, and let the
earth be such that = 30 and ff=6 feet. Then the depth
at which rupture will occur is
For stability the depth must of course be less than 62 feet,
and precautions be taken that the cohesion of the earth be not
destroyed by the action of the weather.
Problem 3. Let a bank whose height is 30 feet and slope
45 degrees be of earth for which = 34 and H = 3 feet.
How much higher can it be raised, keeping the same slope,
before failure will occur ? ^ 5'
ARTICLE 4. STABILITY OF SLOPES IN COHESIVE EARTH.
In practice it is desired to determine the slope of a bank
so that it may be stable and permanent. To deduce an equa
tion for this case consider again Figure 3, and let BM be any
plane through the foot of the slope making an angle x with
the horizontal. As x varies the forces P and N vary, and it is
easy to see that the weakest plane is that for which the expres
P fN 3
sion r is a maximum. As in Art/J, the value of x ren
ART. 4.] STABILITY OF SLOPES IN COHESIVE EARTH. 13
dering this a maximum is x = j(#  0). Now it is required
that rupture shall not occur along this plane, hence
and PfN<cl\
or if n be a number greater than unity, called the factor of
security,
n(PfN) = cl. ...... (12)
Rupture can now occur only when the weight W becomes n
times that of the prism of earth above the weakest plane. To
adapt this equation to practical use it is only necessary to sub
stitute for P and N their values in terms of x, and then to make
x equal to \(9 f 0). The substitution is performed exactly
as before, and leads to the following result :
n h(\ cos (8 0)) = H(i sin 0) sin 0, . . (13)
which is the required equation of stability.
>e^^^^ +&(4
If n is unity, this of course reduces to the case of rupture
as given by (i i). The value to be assigned to the factor n can
only be determined by observation and experiment on existing
slopes. Probably about 2 or 3 will prove to be sufficient.
When 6 is given, the value of h is derived at once directly
from (13), thus:
, _ H(i sin 0) sin 6
~?z(icos(00)) ......
14 EARTHWORK SLOPES. [CHAP. I
This shows that the height for security should be  of that
n
for rupture. Thus it was found in Article 3, if H 6 feet, and
= 30, jthat the limiting height for a slope of 45 would be
62 feet. Hence with a factor of security of 2 the height would
be 31 feet, and with a factor of 3 the height would be 20 feet.
When h is given and is required, the formula for stability
may be written in the form
I cos(<?  0) _ H(i sin 0)
sin 6 nh
The second member is here a known quantity and may be
called a. By developing the numerator in the first member
and then substituting for sin 6 and cos 6 their values in terms
of tan#, a quadratic expression results whose solution gives
This determines the slope 6 for a factor of security n.
For example, let it be required to find the slope for a
bank 25 feet high with a factor of security of 1.5, the value of
being 30 and that of H being t feet. Here
= =  667 '
ART. 4] STABILITY OF SLOPES IN COHESIVE EARTH.
and then from the formula,
tan \Q 0.304 f V 0.0718 + 0.0922 = 0.447.
Hence %6 is about 24 and is about 48, or a slope of 0.9 to I.
The slope when built must of course be protected from the
action of the weather in order to preserve the cohesion of the
earth.
The security of a bank may be investigated by measuring
its height h and slope 6, and finding by experiment the angle
of repose <p and vertical rupturing depth H. Then from (13)
there is found
H(i sin0)sin0
=
For example, let it be required to find the factor n when
h = 30 feet, = 45, = 34, and H= 3 feet. Substituting,
3 X 0.441 X 0.707
30 X 0.0184
If such a slope had existed many years, and if the values of
and //"were the most unfavorable that could occur, it might
be concluded that the factor of security deduced is sufficiently
high ; but if such a slope should be observed to fail, it would
be necessary to conclude that the factor is too low.
Problem 4. A certain slope has h = 25 feet, = 30, and
H = 5 feet. At what angle 6 will rupture occur ? What is its
factor of security if be 48 degrees ?
i6
EARTHWORK SLOPES.
[CHAP. L
ARTICLE 5. CURVED SLOPES AND, TERRACES.
The preceding articles clearly show that the angle of slope
6 of a bank of cohesive earth increases as its vertical height k
decreases, and, conversely, that as h becomes greater be
comes smaller. It would hence appear that the upper part of
a bank may be steeper than the lower part, and its liability to
rupture be the same throughout. To determine the form of
M
a d
vv'v' .vu<
FIG. 4.
such a curve, let D in the figure be any point upon it whose
ordinate Dd is y. Let DM be the weakest plane making an
angle x with the horizontal. The prism of which aDM is a
section, by virtue of its weight W, tends to slide down the
plane. Let P and N be the components of ^parallel and
normal to the plane. If n be the factor of security, the condi
tion, as in Art. 4, is
n(PfN)cl=o.
By inserting for P, N and / their values in terms of x, this
becomes
nW(\ /cot x) cy(\ + cot* x) = o.
ART. 5.] CURVED SLOPES AND TERRACE s". IJ
The value of W for a prism one unit in length is found from
the difference of the areas dDM and dDa, or if A represent
the surface dDa,
W = w^y* cot * A).
By inserting this the equation of stability becomes
cot* A) (i /cot*) cy(i { cot* *fy<=  ( 1 7)
This expression equals zero for the weakes>^lane, but for any
other plane its value is less than zero. Hence it must be a
maximum with respect to x or cot *, and its first derivative
must vanish. Thus, also,
cot x A) ( /) + nw(i f cot ^)i(/ 2 ) zcy co^ x = o. ( 1 8)
By eliminating cot x from (17) and (18), the following value
of A is found :
A = ~ nfWy + * c ~ 2 V2^(/a7 + 2^(i +/"), (19)
and this is the practical equation of the required curve, A
being the area between the curve and any ordinate whose
value is y.
For example, let it 'be required to construct a curve of
equal stability in a bank of 40 feet height with a factor of
security of 1.5, the earth having a natural slope of 31, a verti
cal rupturing depth of 5 feet, and weighing 100 pounds per
Cubic foot. Here, from (2) and (7), there is first found
f = 0.6, c = 71 pounds per square foot,
1 8 EARTHWORK SLOPES. [CHAP. I.
and formula (19) becomes
A =
2842 ^193(907 +142)].
From this are computed the following special values :
For y = 10 feet A = 27 square feet ;
For y = 20 feet, A = 159 square feet ;
For y = 30 feet, A = 421 square feet ;
For y = 40 feet, A = 809 square feet.
These are the areas between the slope and the given ordinate,
and may be practically regarded as consisting of trapezoids,
as shown in Figure 5. The first area is that of the triangle
abB, hence
. 10 . ab = 27, or ab 5.4 feet.
The second area comprises the triangle AbB and the trapezoid
bBCc, hence
In a similar manner cd = 10.5 feet and de = n.i feet, and the
four points B, C, D, and E are thus located. In Figure 5 the
portions of the slope are drawn as straight lines ; it may be so
built, or intermediate points of the curve be established by the
eye.
ART. 5.]
CURVED SLOPES AND TERRACES.
It is not difficult to deduce from (19) the coordinate equa
tion giving the relation between the abscissa and ordinate for
every point of the curve, but it is of such a nature as to be of
little practical use. In the manner just explained, as many
points upon the curve may be located as required. It is seen
from (19) that A is negative for small values of y, or theoreti
A al> c d e
FIG. 5.
cally the curve overhangs the slope. Practically, of course,
the equation should not be used for values of y less than //,
and it will usually be found advisable and necessary that the
upper part of the curve should be reversed in direction so as
to form an ogee, as shown by the broken line AB in Figure 5.
When terraces are to be constructed, it is evident that the
upper one may have the greatest slope and the lower one the
least slope. Formula (19) may be used for this purpose, since
the area A is not necessarily bounded by a curved line, but
may be disposed in any form desired.
For example, take a bank 30 feet high in which it is
desired to build three terraces, as in Figure 6, with a factor of
20
EARTHWORK SLOPES.
[CHAP. I.
safety of 1.5. The height of each terrace is 10 feet, and there
are two steps BC and DE, each 4 feet wide. Let w = 100
FIG. 6.
pounds, 0= 31, and //= 5 feet, as found by experiments.
Then/"= 0.6 and c = 71, and formula (19) becomes
A =
From this, when y = 10, ^ = 27; when jj/ = 20, ^4 = 159; and
when y = 30, 4 = 421. The abscissas are now found to be
ab = 5.4, cd 6.1, and */"= 8.9 feet. The three slopes are
hence as follows :
For aB,
For CD,
For EF,
SB 23 and
10
=  and
10
= ?2 and
ART. 6.] PRACTICAL CONSIDERATIONS. 21
To insure the permanency of these slopes they should be
protected from the weather by sodding.
Problem 5. Design a terrace of four planes, the upper one
being 6 feet in vertical height, the lowest 10 feet, and the
others 8 feet ; the steps to be 5 feet in width. The earth is
such that cot = 1.5 to i, and H = 3 feet.
ARTICLE 6. PRACTICAL CONSIDERATIONS.
The preceding theory and formulas can be usually applied
to the construction of embankments as well as to excavations,
provided that care be taken to compact the earth to a proper
degree of cohesion and the slopes be protected from the action
of the elements. The height h is always given, and it is re
quired to find the slope 0. Unless h be very large the weak
est plane will intersect the roadway ; but if not, the application
of the formulas can only err on the side of safety. The load
upon the roadway can be regarded as a mass of earth uniformly
distributed over it and thus increasing the height h.
For instance, if w = 100 pounds per cubic foot, = 34 and
H=4 feet, let it be required to find the slope for an em
bankment 30 feet high. For security the weight of the loco
motive should be taken high, say 6000 pounds per linear foot
of track, or about 5CXD pounds per square foot of surface for a
12foot roadbed, which would be equivalent in weight to a mass
of earth about 4 feet "high. Then the value of h to be used
22 EARTHWORK SLOPES. [CHAP. I.
in formula (15) is 34 feet. If the factor of security be 2, the
value of a is 0.0259, and
tan # = 0.320 + V 0.093 5+0. 1024 = 0.414.
Hence %0 is about 22^ degrees and is about 45, or the slope
is I to i. The proposed embankment with this slope contains
47 cubic yards per linear foot, while with the natural slope of
34 it would contain 62 cubic yards per linear foot. A saving
in cost of construction will hence result if the expense of pro
tecting the slopes to preserve the cohesion be not too great.
The degree of moisture of earth exercises so great an influ
ence upon its specific gravity and angle of repose that special
pains should be taken to ascertain the values of those quanti
ties which are the most unfavorable to stability. In general
a high degree of moisture increases w and decreases 0. These
causes alone would tend to increase the cohesion, but at such
times H usually becomes so small that c is greatly diminished.
The determination of H is awkward and there seem to be few
recorded experiments concerning it. Care should be taken
that the trench is long, or that transverse cuts be made at its
ends so that lateral cohesion may not prevent rupture, and a
considerable time should be allowed to elapse so that the co
hesion may be subject to unfavorable weather.
The general conclusions of the above theory are valuable,
but it should be applied with caution to particular cases, not
only on account of the variability in the data but on account
of our ignorance of the proper factor of security. Numerical
ART. 6.] PRACTICAL CONSIDERATIONS. 2$
computations, however, may often prove useful as guides in
assisting the judgment. As shown above, a great saving in the
cost of moving material will result if slopes be built in accord
ance with the theory, but evidently the cost of properly pro
tecting the slopes will be increased. Should the latter cost
prove to be the smaller, the theory will ultimately become of
real practical value.
The preceding theory is not new, having long since been
set forth in many French and German books, but the author
is unaware to what extent it is practically used in those* coun
tries. The introduction of a factor of security is, however,
believed to be novel in this connection, and by proper experi
ments for determining its value the practical application of the
formulas here given may perhaps be rendered possible.
Problem 6. A railroad cut is to be made in material for
which w = 100 pounds per cubic foot, @)= 32, ff 5 feet.
If h is 40 feet and the roadbed 16 feet wide, find the quantity
of material necessary to excavate when the slopes have a fac
tor of security of 3. ^4.
24 THE LATERAL PRESSURE OF EARTH. [CHAP. II.
CHAPTER II.
THE: LATERAL PRESSURE OF EARTH.
ARTICLE 7. FUNDAMENTAL PRINCIPLES.
A retaining wall is a structure, usually nearly vertical,
which sustains the lateral pressure of earth. In investigating
the amount of this pressure it is generally regarded best to neg
lect the cohesion of the earth, and to consider it as loose
(Article i). This is done, partly because the effect of cohesion
is difficult to estimate and partly because the results thus ob
tained are on the side of safety for the wall, the entire inves
tigation in fact being undertaken for the purpose of using the
results in designing walls. The values given in Article I for
the weight of earth and for the angles of repose will be used
in this chapter, but it is again mentioned that they are subject
to much variation, and that in practical problems the values
most dangerous to stability should be selected.
The pressures against a retaining wall are least near the
top and greatest near the base. The resultant of all these
pressures is called the " resultant pressure," or simply the pres
sure, and is designated by the letter P. The determination of
ART. 7.]
FUNDAMENTAL PRINCIPLES.
formulas for the values of P for different cases is the object of
this chapter.
Let the resultant pressure P against the back of a wall be
resolved into a component N acting normal, and a component
.F acting parallel, to the back of a wall. Let z be the angle
between N and the direction of P', then
F = N tan z.
Let /be the coefficient of friction between the earth and the
wall, then for the case of incipient motion,
F=Nf.
Therefore, since /is the tangent of the angle of friction, the
angle z cannot be larger than the angle of friction between the
earth and the wall unless the earth is moving along the wall.
Various views are held by authors regarding the direction of
the pressure P, or the value of the angle z. Some take z as
zero, or regard the thrust as normal to the wall ; others take z
26 THE LATERAL PRESSURE OF EARTH. [CHAP. II.
as equal to the angle of repose of the earth, ; while a few
take z as intermediate between these values.
In Article 8 the friction of the earth against the wall will be
neglected, or the angle z will be taken as o. The value of the
pressure determined under this supposition will be called the
"normal pressure/' and will be designated by P r It is not to
be forgotten that the actual pressure against the back of a re
taining wall cannot, like the pressure of water, be determined
with certainty. The formulas to be deduced are such that, in
general, they give limiting maximum values under the different
conditions, and the hypothesis here adopted has the practical
advantage of erring on the. side of safety for the wall. In an
unlimited mass of earth with horizontal surface, the pressure
against any imaginary vertical plane must evidently be normal
to that plane ; now if a wall is to be designed to replace the
earth on one side, the pressure against its back will also be
normal. It would seem then, that the most satisfactory de
gree of stability of the earth will be secured by designing the
wall under the assumption of normal pressure.
The views just expressed are, howevet, not accepted by
some engineers who claim that the actual normal pressure is
usually less than the values theoretically deduced for P l , par
ticularly for walls that have been observed to fail. In Article
9 there will hence be investigated formulas for the pressure
supposing that it is inclined to the normal to the back of the
wall at an angle 0; the value of the pressure thus derived will
be called the " inclined pressure," and be designated by P v
ART. 8.]
NORMAL PRESSURE AGAINST WALLS.
Hence either P l or P^ can be used in investigating the wall as
the engineer thinks best.
Problem 7. Let the wall in Figure 7 be vertical, 12 feet in
height, its thickness uniformly 2 feet, and its weight 3600
pounds. Let the point of application of P be 4 feet above
the base. Compute the value of P to cause rotation, (a) when
the angle ft is o ; (b) when the angle ft is 30.
ARTICLE 8. NORMAJ. PRESSURE AGAINST WALLS.
In Figure 8 is shown a wall which sustains the lateral press
ure of a bank of earth. The back of the wall BA is inclined to
the horizontal at the angle 6, an^i the surface of the earth. AM
FIG. 8.
is inclined at the angle 8. The line BCf represents the natural
slope of the earth with the inclination <p. It is required to
find the lateral pressure of the earth, supposing that its direc
28 THE LATERAL PRESSURE OF EARTH. [CHAP. II.
tion is normal to the back of the wall, or that, in Figure 7,
the angle 2 is zero.
Draw BM making any angle x with the horizontal, and
consider that the prism ABM in attempting to slide down the
plane B M exerts a pressure upon AB. Let IV be the weight
of this prism, represented by the line OW, and let it be re
solved into a component P^ acting normal to the back of the
wall, and a component R acting opposite to the direction of
the reaction of the earth below BM. Let ON be normal to
the plane BM, then the angle NOR will be equal to the
angle of friction of earth upon earth, if the prism ABM is just
on the point of sliding down BM\ (for, as ON and NR are
components of OR the triangle gives NR = ON. tan NOR,
but from the law of friction NR=f. ON\ hence /= tan =
tan NOR, and accordingly NOR = 0.)
Now in the triangle IVOR the side OW represents the
weight W, and the side WR the resultant normal pressure P, .
Hence
p  TT/ sm WOR
'"" sin WRO'
w But the angle WOR is x and the angle WRO is + x.
Let h be the vertical height of the wall, and w the weight of a
cubic unit of earth ; then the value of W for one unit in length
of the wall is
,,.. . , sin (0 #) sin (9 x)
W = iv . BA . BM. sin ABM = %wfr . v / A__ I
X^x  
;
ART. 8.] NORMAL PRESSURE AGAINST WALLS. '"' '' 29
TU K i re
The above value of P l then can be written
P  JL./ sin ( g ~ <*) sin (g  x) sin (^  0)
which expresses the normal pressure due to any prism whose
plane BM makes an angle x with the horizontal. This expres
sion becomes o, both when x = 6 and when x = 0, and
between those limits it has a maximum value which is to be
taken as the pressure against the wall, since such can occur if
the earth is about to slide down the corresponding plane.
In order to find the value of x which renders (20) a
maximum it is best to write it in the form
in which y = cot (8 x\ a = cot (8 0), b = cot (8 <?),
c = cot (p, A ^w/f sin (0 0) and B =. sin 8 sin 0. Dif
ferentiating this with respect to y and putting the first deriva
tive equal to zero, there is found
y = a + )/(a _ fi) (J+Tj; .... (22)
and inserting this in (21) there results for the maximum
p ^ _A
Thus /\ is expressed in terms of the data w, k,8,d and 0, but
to obtain a more convenient form it is well to replace the
3O THE LATERAL PRESSURE OF EARTH. [CHAP. 1L
cotangents by their equivalents in terms of sines. Then after
reduction it becomes
P,= r=2fc , (*!}
sm
which is the formula for the lateral normal pressure of a bank
of earth against the back of a retaining wall.
This formula is valid for any value of greater than 0, and
for any value of 3 less than 0. By its discussion simpler
formulas for special cases can be deduced.
The greatest value of d will be that of the natural slope 0.
For this case formula (23) becomes
(0
which is the greatest normal thrust that can be caused by a
sloping bank ; if the wall be vertical 6 = 90 and this reduces
to the simple form P l = fywh* cos 2 0.
The most common case is that where the surface AM Is
horizontal ; for this 6 =o and (23) becomes
<2 rt
ART. g.] INCLINED PRESSURE AGAINST WALLS. 31
Avhich is the normal pressure of a level bank of earth against
an inclined wall. If in this 6 = 90, there results the formula
for the pressure of a level bank against a vertical wall,
tan' (45  J0),
which is the wellknown expression first deduced by COULOMB
in 1773.
Problem 8. Prove from (22) that, when # = o, the plane
BM bisects the angle between BA and BC. Prove it also by
making d = o in (21), and then equating the first derivative to
.zero, thus deducing x = (#( 0).
ARTICLE 9. INCLINED PRESSURE AGAINST WALLS.
In Figure 9 is shown a wall which sustains the lateral pres
sure of a bank of earth, the back of the wall being inclined to
the horizontal at an angle 0, and its vertical height being h.
The upper surface of the earth has an inclination 8 to the
horizontal, which is not greater than the natural slope 0. It
is required to find the lateral pressure of the earth supposing
that its direction makes an angle with the normal to the
back of the wall.
Draw BM making any angle x with the horizontal, and
consider that the prism BAM in attempting to slide down
this plane is sustained by the reactions of the wall and of the
earth below BM. Let OW represent the weight of this prism,
THE LATERAL PRESSURE OF EARTH. [CHAP. II.
and let it be resolved into components OP^ and OR opposite
in direction to these reactions. Let OL be normal to the
back of the wall, and ON be normal to the plane BM. Then
if motion is just about to occur, the angle NOR is equal to the
angle of friction of earth upon earth, and LOP 9 is equal to
the angle of friction 0' of earth upon masonry. Although ft
is perhaps in general greater than 0, it is customary to take it
as the same, thus erring on the side of safety ; accordingly
LOP, = <f>.
Let Wbe the total weight of the earth in the prism BAM,
and w its weight per cubic foot. Let P^ represent the inclined
resultant pressure against the wall. In the triangle ROW, the
angle ROW'isx 0, and ORWis0 + 20  x\ hence
= W
sin (x 0)
sin(0{ + 20 x) '
fu/v
ART. 9.] INCLINED PRESSURE AGAINST WALLS. 33
The weight H^ for one unit in length of the wall is w X area
BAM X I. The area of BAM equals i^yi . BM . sin
the side BA is /* r sin 0, the angle y^J/ is #, and
sm y sin . sin *
The value of W is thus expressed in terms of # and the
given data, and P 9 becomes
P  Iwtf sin (0  (?) sin (6  x) sin (x  0) , .
1 sin' sin (*(?) sin (0 + 20  *)'
which gives the pressure due to any prism BAM.
The greatest possible value of P 2 is to be regarded as the
actual value of the inclined pressure. By proceeding as in
Article 8 it can be shown that this obtains when
cot (6  x) = cot (6  0)
+ V [cot (0 0) cotT(0 )] [cot (0  0) + cot 20] > (28)
and that the maximum value is
^
a / ru i * \ / sin 2 . sin
sin 2 6^ sin (0 + 0)1 I + A / .
; y sm
lfi , ,, . , .
(0 \ 0) sin (6 tf
which is the general formula for the socalled inclined pressure
and from which the results for all special cases can be deduced.
34 THE LATERAL PRESSURE OF EARTH. [CHAP. II.
The greatest slope 8 will be the natural slope 0. For this
case the formula reduces to

3 sin 2 B sin (0+0)'
which is the pressure due to a ba$k of maximum slope against
an inclined wall. If the back of the wall be vertical, = 90
and the expression takes the simple form P a %wh* cos 0.
The most common case is that where the surface AM is
horizontal ; for this d = o and (29) becomes
y~rf sin 2 (<9 0)
/^sTn20sin0 y (31)
Vsin(
sin'
which is the inclined pressure of a level bank of earth against
an inclined wall. If in this 6 90, there results the formula
for a level bank of earth retained by a vertical wall.
(32)
which is the wellknown expression deduced by PONCELET.
Problem 9. In formula (27) make = 90 and d o.
Then find the value of x which renders it a maximum, and
deduce the corresponding value of P v
ART. 10.] FORMULA FOR LATERAL PRESSURE. 35
ARTICLE 10. GENERAL FORMULA FOR LATERAL
PRESSURE.
Let a wall whose back is AB sustain a bank of earth BAM
as in Figure 9. If the earth be loose, the weakest plane BM
wil be that along which rupture is about to occur, so that the
angle NOR = 0, as in the two preceding articles. Let the
resultant lateral pressure be designated by g, and let its direc
tion make an angle z with the normal to the wall so that
LOP^ = s. By the same reasoning and methods as before
used, it is found that the expression for the pressure due to
any prism AB M and the value of cot (d x) which renders it
a maximum are the same as given by (27) and (28) if the
single term 20 be replaced by f z, and then results
p = _ _ f (33)
sin 2 6 sin (8 + z](i + . / sin(0 + *) sin(0  tf)V '
\ \/ sin(0 + *)sin(0tf)/
which is a general formula for the lateral pressure in terms of
the unknown angle z. If z = o, the direction of P is normal
to the back of the wall and (33) reduces to (23). If z 0, the
pressure is inclined to the normal at the angle of friction and
(33) reduces to (28).
From the above formula a number of theories of earth
pressure can be deduced by making different assumptions with
regard to the angle z. For instance, it seems to some authors
36 THE LATERAL PRESSURE OF EARTH. [CHAP. IL
a reasonable theory which makes the pressure upon a vertical
wall parallel to the earth surface A M; for this case 9 = 90,
and z = 90 \ 0, and inserting these in (33) it reduces to
p __ _ cos 3 _
= "
cos
which is RANKINE'S formula for the lateral pressure against a
vertical wall. In like manner several other formulas, more or
less reasonable, can be established. But probably everything
necessary for the practical engineer is given in Articles 8 and 9.
Problem 10. Deduce formulas for the earth pressure
under the supposition that its direction is horizontal.
ARTICLE 11. COMPUTATION OF PRESSURES.
In computing the lateral pressure of earth from the above
formulas it is customary to take h in feet and w in pounds per
cubic foot ; then the value of P will be in pounds per running
foot of the wall. On account of the uncertainty in the data
the trigonometric functions need be taken only to three or
four decimal places, or, if logarithms be used, as will be found
most convenient, a fourplace table will be amply sufficient.
The values of the pressures for several cases will now be com
pared, the walls all being 18 feet in vertical height, the earth
weighing 100 pounds per cubic foot and having a natural slope
= 34 degrees. Here the value of ^.wh* is 16200 pounds.
ART. ii.] COMPUTATION OF PRESSURES. 37
For a level bank of earth and a wall whose back slopes
backward with the inclination = 80, formulas (25) and (31)
give the pressures
PI = 3 57 ?* = 2 590 pounds.
For a level bank of earth and the back of the wall vertical, for
mulas (26) and (32) give
P l = 4 580, P 3 4 210 pounds.
For a level bank of earth and the back of the wall sloping
forward so that = 100, formulas (25) and (31) give
P l = 5 760, P z = 5 670 pounds.
Here it must be remembered that the direction of P is
normal to the wall, while the direction of P t makes an angle of
34 with the normal to the wall.
For the same walls sustaining earth whose upper surface
has the slope # = 10 degrees, the following values are found
from formulas (23) and (29):
For 6 = 80, P, = 3 920, P 2 = 3 400 pounds.
For = 90, P l = 5 080, P 2 4 960 pounds.
For 8 = 100, P, = 6 469, P 2 = 6 480 pounds.
For the same walls sustaining earth whose upper surface
has the angle of repose 3 = 34, formulas (24) and (30) give :
For 6 80, P, = 8 780, P 2 = 9 460 pounds.
For 6 90, P, = ii 130, P z = 13 430 pounds.
For 100, P, = 14 1 60, P 2 = 19 380 pounds.
38 THE LATERAL PRESSURE OF EARTH. [CHAP. II.
A comparison of the above values shows that the pressure
increases both with and d. For a level bank of earth the
values of P^ are less than those of /J, but for a large value of
d the values of P 2 become greater than those of P r Whether
the true thrust against the wall is P l or P 2 , or some inter
mediate value, cannot be determined theoretically, and hence
the best procedure for the engineer will be to use those
values which are the most unfavorable to stability.
For the case of water <p = o and d = o, and all the for
mulas for pressures reduce to
p fyj? + sin 0, (35)
in which w is the weight of a cubic foot of water, or 62%
pounds. The pressure of water against a wall 18 feet in verti
cal height will hence be 10 280 pounds when = 80 degrees,
10 125 pounds when = 90 degrees, and 10 280 pounds when
6 = 100 degrees, its direction being always normal to the back
of the wall.
Problem n. Compute the pressures against a wall 9 feet
in vertical height for earth weighing 100 pounds per cubic foot
and having an angle of repose = 34 degrees, (a) for the case
when # = 10 degrees and 6 = 80 degrees ; (b) for the case when
d = 20 degrees and 6 = 80 degrees.
ART. 12.] THE CENTRE OF PRESSURE. 39
ARTICLE 12. THE CENTRE OF PRESSURE.
For all the above cases the formulas for the resultant
lateral pressure of the earth may be written
P = \wtf . k,
in which k is a function of the angles #, $, and 0. If y repre
sent any vertical depth measured downward from the top, the
resultant pressure on the part of the wall corresponding to
this depth is
which shows that the resultant pressure varies as the square of
the height of the wall. The pressure per square unit at any
point on the wall varies, however, directly as the height, for
if y be increased by dy the increase in P is dP or wkydy, and
the pressure per square unit over the area I X dy is
dp
The laws governing the distribution of earth pressures are
hence the same as for water, the unitpressure at any point
varying as the depth, and the total pressure as the square of
the depth.
4O THE LATERAL PRESSURE OF EARTH. [CHAP. II.
The point where the resultant pressure P is applied to the
wall is called the centre of pressure, and this is at a vertical
distance from the top of the wall equal to twothirds its height.
This may be proved by Figure 10, which gives a graphical
representation of the pressure against the back of the wall, the
unitpressures falling into a triangular shape, since each is
proportional to its distance below the top. The point of
FIG. 10.
application of the resultant pressure P hence passes through
the centre of gravity of this triangle and cuts the back at
twothirds its length from the top.
For another proof the principle of moments may be used.
Let y beany vertical distance from the top, and y' the vertical
distance from the top to the centre of pressure. Then taking
the top of the back of the wall as the origin of moments,
Inserting in this the values of P and dP given above and
Integrating betwen the limits y o and y = //, there results
x
/ = **; ........ (36)
ART. 12.] THE CENTRE OF PRESSURE. 41
that is, the centre of pressure is at a vertical distance \h below
the top of the wall, or at \h above the base.
Problem 12. Let a level bank of earth have a load of q
pounds per square foot upon the surface AM. Show that the
resultant normal pressure due to both bank and load is
and that the depth of the centre of pressure below the top of
the wall is
_ 2wh + 3? , ( }

Find the position of the centre of pressure when h = 18 feet,
<w = 100 pounds per cubic foot, and q = 300 pounds per
square foot.
INVESTIGATION OF RETAINING WALLS. [CHAP. III.
CHAPTER III.
INVESTIGATION OF RETAINING WALLS.
ARTICLE 13. WEIGHT AND FRICTION OF STONE.
The lateral pressure of the earth against a retaining wall
tends to cause failure in two ways, namely, by sliding and by
rotation. This tendency is resisted by the friction between
the stones and by the weight of the wall. The following table
gives average values of the unitweights and the coefficients of
friction for different kinds of masonry:
Kind of Masonry.
Coefficient
of
Friction.
Angle
of
Friction.
Weight.
Pounds per
cubic foot.
Kilos per
cubic meter.
Limestone and Granite :
Ashlar Masonry
0.6
0.6
0.65
31
31
33
165
150
125
150
130
no
100
2640
2400
2000
2400
2100
1760
1600
Large Mortar Rubble.. . .
Small Dry Rubble . .
Sandstone :
Ashlar Masonry .
Large Mortar Rubble
Small Dry Rubble
Coarse Brickwork . . . .
ART. 13.] WEIGHT AND FRICTION OF STONE. 43
In the investigation of masonry walls the influence of the
mortar is generally neglected, on account of its uncertain
character and because the error is then on the side of safety.
The above coefficients of friction are hence stated for dry
masonry, and will probably be somewhat increased when mor
tar is used. For rubble masonry the coefficient of friction is
often somewhat higher than for ashlar ; but its value is so un
certain that no figure is given in the table.
The coefficient of friction of stone upon stone is deter
mined by placing two plane surfaces together and then
gradually inclining the surface of contact until the upper
stone begins to slide upon the lower. The angle made by the
plane with the horizontal is the angle of friction, and its tan
gent is the coefficient of friction, as shown by equation (2).
The word " batter " means the inclination of the face or
back of a wall, measured by the ratio of its horizontal to its
vertical projection, or in inches of horizontal projection per
foot of vertical height. Let be the angle of inclination of
the back of the wall to the horizontal, as in Figure 8. Then
cot 6 is the batter of the back, and the values of 0, sin 8,
and cos for different batters are given in the following table.
If the back of the wall leans backward, is less than 90 de
grees, and cos and cot are positive ; if it leans forward,
6 is greater than 90 degrees and cos and cot are negative ;
sin is positive in both cases. These values will be useful
in many computations.
INVESTIGATION OF RETAINING WALLS. [CHAP. Ill,
Batter in
inches
per foot.
Angle & for
Backward
Batter.
Angle
for
Forward Batter.
sin 0.
cos 6.
Batter
cot 6.
O
90 oo'
90 oo'
I.OOOO
O.OOOO
0.0000
i
87 37
92 23
0.9991
0.0461
0.0417
I
85 14
94 46
0.9965
O.C83I
0.0833
Ii
82 52
97 08
0.9923
0.1239
o. 1250
2
80 32
99 28
0.9864
0.1645
0.1667
2*
78 14
101 46
0.9790
0.2039
0.2083
3
75 58
104 02
0.9702
0.2425
O.25OO
3*
73 45
106 15
o . 9600
0.2797
0.2916
4
71 34
108 26
0.9487
0.3162
03333
5
67 29
112 31
0.9239
0.3828
0.4144
6
63 26
116 34
0.8944
0.4472
O.5OOO
Problem 13. Compute the values of sin 6, cos 8, tan 0, and
cot 6 for a batter of 4^ inches per foot ; also for a batter of 5
inches per foot.
ARTICLE 14. GENERAL CONDITIONS REGARDING SLIDING.
A retaining wall may fail by sliding on its base or on some
joint above the base. When a wall is just on the point of
failure it is in the state of equilibrium, that is, the weight of the
wall just balances the pressure and reaction of the earth. The
proper state of a wall is, of course, stability ; and failure brings
disgrace upon its designer.
The degree of stability of a wall against sliding may be in
dicated by a number called the factor of security which ranges
in value from unity to infinity. This factor will be designated
iby n ; when n = i equilibrium exists and the wall will fail ,
ART. 14.] GENERAL CONDITIONS REGARDING SLIDING. 45
when n > I the wall is stable and its degree of stability varies
with n ; when n = oo the highest possible state of stability
exists.
The analytical conditions of equilibrium and stability for
the case of sliding are the following. Let Figure II represent
two bodies having a plane surface of contact, N the total force
normal to that plane, F the total force parallel to it, and /the
coefficient of friction between the surfaces. Then the condi
tion of equilibrium is, as in (i),
F = fN, ....... (39)
and the condition of stability is
F<fN or nF = ftf, .... (40)
in which n is a number greater than unity called the factor of
security. The equation (40) may be used for the discussion of
all cases of sliding, .Fand N being the sum of the components
46 INVESTIGATION OF RETAINING WALLS. [CHAP. IIL
in the directions parallel and normal to the plane of all the
forces exerted by one body on another. If R be the resultant
of all these forces and C be the angle which it makes with the
normal ON, the value of F is R sin C, and that of N is R cos C
Inserting this in (40) it becomes
n tan =/,.". ..... (41)
which is another form of the condition of stability against
sliding.
The graphical conditions of equilibrium and stability for
sliding are simple. In Figure 1 1 let ON be normal to the
plane of contact and NOF be the angle of friction 0, that
is, the angle whose tangent is f. Let R make an angle C with
the normal ON, then equilibrium obtains when C equals 0,
and stability occurs if C is less than 0. Draw NF parallel to
the plane of contact, and let T be the point where it inter
sects the line of direction of R. The position of T indicates
the degree of stability against sliding ; if the distances NT
and NF be determined, the factor of security is the ratio of
the latter to the former, or
NF
(42 >
for, it is seen that this formula is the same as (41), NF being/,
and NT being tan C if the distance ON be unity, and their
ratio being the same as these tangents whatever be the length
of ON.
ART. 15.] ' GRAPHICAL DISCUSSION OF SLIDING. 47
Problem 14. A plane surface is inclined at an angle of 40
to the horizontal, and on it is a block weighing 125 pounds,
against which, to prevent it from sliding, a horizontal force
of 300 pounds acts. If the angle of friction of the block upon
the plane is 18, compute the factor of security against sliding.
ARTICLE 15. GRAPHICAL DISCUSSION OF SLIDING.
Let Figure 12 represent the section of a wall whose dimen
sions and weight are given. Let BC be any joint extending
through the wall, and let P be the lateral pressure of the earth
above B. It is required to investigate the security of the wall
against sliding.
The pressure Pis applied on the back of the wall at one
third of its height above B, and its direction depends on the
hypothesis adopted in its computation ; if Article 8 is used, it is
normal to the back of the wall ; if Article 9, it is inclined at an
angle equal to the angle of natural slope of the earth.
A drawing of the given crosssection is made to scale, and
its centre of gravity found : this is G in the figure. The area
of this crosssection is then determined and called A ; if this
be multiplied by ^, the weight of a cubic unit of masonry,
the product is V, the weight of a wall one unit in length, or
V= vA.
Through G a vertical line is drawn, and the direction of P
is produced to intersect this in O. Lay off OP to scale equal
4 8
INVESTIGATION OF RETAINING WALLS: [CHAP. III.
to the earth pressure P, and OV equal to the weight of the
wall, V. Complete the parallelogram of forces OPRV, thus
finding OR as the resultant of P and V.
Produce OR to meet the joint BC in T. Through O draw
ON normal to BC, and then draw OF, making the angle NOF
equal to the angle of friction of stone upon stone. This com
pletes the graphical work.
If the point T falls between N and F, the wall will not fail
by sliding, and its stability is the greater the nearer T is to N*
If ^coincides with F 9 the wall is just on the point of sliding
along the joint BC, and much more so is this the case if T
falls beyond F. As explained in the last Article, a numerical
expression of the degree of stability can be obtained by divid
ing the distance NF by NT, or if n be the factor of security
against sliding,
_NF
n ~ NT
ART. 15.] GRAPHICAL DISCUSSION OF SLIDING. 49
This becomes unity when NT equals NF, and infinity when
NT is zero, the first value indicating the failure of the wall
and the second the greatest possible degree of stability against
sliding., It is recommended that for firstclass work n should
not be less than 3.0, and fortunately it is always easy in build
ing a wall to make its value greater than this by properly in
clining the joints (Article 23).
The above method applies either to the base of the wall or
to any joint that extends through it, whether the joint be hori
zontal or inclined. Owing to the uncertainty regarding the
weight and angle of repose of the earth, the direction of P, and
the angle of friction of the stone, it will not always be possible
to obtain values of the factor of security which are perfectly
satisfactory. Still the investigation will generally determine
if danger exists, and of course unfavorable values of the data
should be used in the analysis. If the wall have no joints ex
tending through it, an anafysis for sliding need not be made.
Problem 14. Prove that the centre of gravity of a quadri
lateral abed can be found as follows : Draw the diagonal ac and
bisect it in e ; join be, and take ef equal to \be ; through /
draw fk parallel to bd. Draw the other diagonal bd and bisect
it in g; join g<?rand take gh equal to%; through g draw gk
parallel to ac. The centre of gravity is at k, the intersection
of fk
INVESTIGATION OF RETAINING WALLS. [CHAP, III.
ARTICLE 16. ANALYTICAL DISCUSSION OF SLIDING. *
Let A BCD represent the crosssection of a wall whose
dimensions and weight are given, being the inclination of
its back to the horizontal. Let BC be any joint extending
through the wall, and a its inclination to the horizontal. Let
P be the lateral pressure of the earth above this joint, and 2
the angle between its direction and the normal to the wall ;
=Z2%gF^
FIG. 13.
if Pbe the pressure computed by the formulas in Article 8, the
value of z is simply zero ; if by those in Article 9, its value
is the angle of repose of the earth ; if z be assumed at any
intermediate value, P is computed from (33). Let V be the
weight of the wall. It is required to investigate the degree of
stability against sliding.
Let .Fand N be the sum of the components of P and V
respectively parallel and normal to the joint, and f the coeffi
ART. 1 6.] ANALYTICAL DISCUSSION OF SLIDING. 51
cient of friction. Then if n be the factor of security, nF = fN,
and
(43)
Now by resolving P and V parallel and normal to the joint
there is found
F= Psm(6+a + z) Fsin a,
N= F cos or />cos(# + <* + )
.and if these be inserted in (43), the value of n is expressed in
terms of the given data. The entire analytical discussion of '.
the sliding of a wall along a joint consists in computing n from
these formulas. If n is greater than 3, the security against ;
sliding is ample ; if n is less than 3, the wall does not have
proper security for firstclass work; if n = I, failure will
occur.
For example, consider a sandstone wall 18 feet high, 3 feet
wide at the top and 6 feet wide at the base, the back being
vertical. The weight of the masonry is taken at 140 pounds
per cubic foot, and the coefficient of friction on the horizontal
joint at the base is 0.5. This wall supports a level bank of
earth weighing 100 pounds per cubic foot and having an angle
of natural slope of 34 degrees. It is required to find its factor
of security against sliding.
First, let the pressure P and its direction be taken from
Article 8. Here h = 18 feet, w = 100, # = 90, 0=34,
52 INVESTIGATION OF RETAINING WALLS. [CHAP, III.
\ 1^
d = o, and z o. Then by formula (26) there is computed
j
P = 4580 pounds. The weigut of the wall is
V= 140 X 1 8 X 44 = 11340 pounds.
Now since a = o, F= 4580 and ^V = 11340, hence the factor
of security is
11340 X 0.5
which indicates a ver low degree of stability.
Secondly, let the pressure P and its direction be taken from
Article 9. Here z = 34, and using formula (32) there is found
P= 4210 pounds. Fis 11340 pounds as before. From (44),
F = 4210 sin (90 + 34) = 3490,
N= 1 1340 4210 cos (90 + 34) = 13690,
and then from (43) there results the factor
_ 0.5 X 1 3690 _
3490
which indicates a degree of stability too low for firstclass
work.
Unfortunate indeed it is that the theory of earth pressure
is not sufficiently explicit to determine the exact value and
ART. 16.] ANALYTICAL DISCUSSION OF SLIDING. 53
direction of P. He who believes the theory of Article 8 must
conclude that this wall is in a very dangerous condition and
almost about to slide ; he who defends the theory of Article 9
might conclude that it is not in great danger, and that its
degree of security is fair. It is well, however, not to forget
that the given data are liable to variations fully as serious as
the defects in the theory. Imagine a heavy rainfall to increase
w and decrease ; this causes P to become larger, and as F
usually would be smaller in wet weather, it is seen that the
degree of stability of the wall would be greatly diminished.
If the factor of security be computed for both theories as is
done above, and the variation in the data be regarded, a fair
conclusion can generally be made regarding the security of the
wall. The effect of the variable data, however, is often so
great that a ripe judgment, based upon experience, may be
more reliable than computations.
Problem 16. Owing to a heavy rainfall the earth behind
the above wall is increased in weight to 120 pounds per cubic
foot and the angle of natural slope is decreased to 32 degrees,
while the coefficient of friction at the base of the wall becomes
0.45. Compute the factor of security of the wall against
sliding, (a) using the theory of Article 8, and (b) using that of
Article o.
54
INVESTIGATION OF RETAINING WALLS. [CHAP. III.
ARTICLE 17. GENERAL CONDITIONS REGARDING ROTATION.
Let Figure 14 represent two bodies having the plane of
contact BC. Let M be the middle point of BC. Let R be
the resultant of all the forces which each body exerts on the
other, and let T be its point of application on BC. It is
clear that rotation or overturning will instantly occur if T falls
without BC, that equilibrium obtains if T coincides with C,
D
FIG.
and that stability, more or less secure, will result if T falls
within BC. The nearer the point of application T is to the
middle of the base M the greater is the degree of stability
against rotation.
To investigate the degree of security of a given wall
against rotation it is only necessary to find the distance MT
either graphically or analytically. Let n be the factor of
security of the wall, then
n =
MC_
MT
(45)
ART. 17.] CONDITIONS REGARDING ROTATION. 5$
If MT equals MC, the value of n is unity and failure by rota,
tion is about to occur ; if MT is less than MC, the value of n
is greater than unity and the wall is more or less stable ; if
MT is zero, n is infinity and the wall has the greatest possible
degree of stability.
The factor of security n should not have a value less than
three for proper stability. To demonstrate this, consider the
distribution of pressures in a joint as represented in Figure 15.
In the first diagram the resultant pressure R is applied at the
middle of BC\ here the pressure will be uniformly distributed
over the joint, and the unitstress 5, at B will be equal to the
unitstress 5 at C. In the second diagram the resultant R is
applied so that MT has a small value ; then the pressure is
56 INVESTIGATION OF RETAINING WALLS. [CHAP. III.
not uniformly distributed over the joint, but the unitstress
S t at B becomes smaller than in the first diagram, while the
unitstress S at C becomes greater, and the unitstresses be
tween B and C are taken as varying proportionally. In the
third diagram the distance MT is such that the unitstress at
B becomes zero ; this occurs when CT is onethird of CB
(since the line of direction of R passes through the centre
of gravity of the stress triangle) or when MT is onethird of
MC* In the last diagram MT has become greater than one
third of MC, so that the pressure is only distributed over
CB' and the portion BB' is either brought into tension or the
joint opens. As masonry joints cannot take tension this last
is a dangerous condition. Therefore the ratio of MC to MT,
or the factor of security, should not be less than 3.0.
If the joint BC be divided into three equal parts, so that
BD = DE = EC, the portion DE is called the " middle third,"
and the above requirement is otherwise expressed by saying
that for proper security against rotation the resultant of all the
forces above any joint must be within the middle third of that
joint.
Problem 17. In Figure 15 let BC be horizontal, and let
ABCD be a cubical block weighing 625 pounds. Compute the
factor of security against rotation when a horizontal force of
250 pounds is applied at A.
ART. 18.] GRAPHICAL DISCUSSION OF ROTATION. $?
ARTICLE 18. GRAPHICAL DISCUSSION OF ROTATION.
Let Figure 12 represent the crosssection of a wall whose
dimensions and weight are given. Let BC be any joint ex
tending through the wall, and let P be the lateral pressure of
the earth above B. It is required to investigate the security
of the wall against rotation.
The pressure P is applied on the back of the wall at one
third of its height above B (Article 12), and its direction is
either normal to the wall (Article 8), inclined to the normal at
the angle of natural slope of the earth (Article 9), or it has a
direction between these two limits (Article 10).
A drawing of the given crosssection is made to scale, and
its centre of gravity found ; this is at G. The area of this cross
section is next determined and called A ; then the weight of
the wall for one unit in length is V = vA y where v is the weight
of the masonry per cubic unit.
Through G draw a vertical line and produce P to intersect
it in O. Lay off OP to scale equal to the earth pressure P,
and OV equal to the weight V. Complete the parallelogram
of forces OPRV, thus finding OR as the resultant of P and V.
Produce OR to meet the joint BC in 71 Mark M as the
middle point of BC, and measure MT and MC. This com
pletes the graphical work.
58 INVESTIGATION OF RETAINING WALLS. [CHAP. III.
If T falls at C, the wall is on the point of rotation ; and if
at M y it has the highest possible degree of stability. If BC
be divided into three equal parts and T is found within the
middle one, the wall has proper security against rotation. If
it falls without the middle third, it is deficient in security
(Article 17). Dividing MC by MT the factor of security is
found, or
If this is unity or less, the wall fails ; if it be smaller than 3, the
wall is stable but not secure ; if it be greater than 3, the degree
of security is sufficient as far as rotation alone is concerned ; if
it be infinity, nothing more can be desired.
By this method but one construction is needed for the
investigation of a wall against both sliding and rotation. It
will usually be found for ordinary cases that the factor of
security against rotation is least for the base of the wall or for
the lowest joint. For the general discussion the base of the
wall is drawn inclined in Figure 12, but in the actual drawing
it will be best to take it as horizontal.
Problem 1 8. Let a wall with vertical .back support a level
bank of sand weighing 100 pounds per cubic foot and having
an angle of repose of 34 degrees. Let the top of the wall be
2 feet thick, its base 7.57 feet, its vertical height 20 feet, and
its weight per cubic foot 165 pounds. Determine the factors
of stability against sliding and rotation for the horizontal
base, taking the earth pressure from Article 8.
ART. 19.] ANALYTICAL DISCUSSION OF ROTATION.
59
ARTICLE 19. ANALYTICAL DISCUSSION OF ROTATION.
Let ABCD be a crosssection of a wall with vertical back,
AB being 24 feet, the top AD being 3 feet and the base BC
being 8 feet. Let the weight per cubic foot of the masonry
be 150 pounds, and let it be required to determine the factor
of security against rotation for a horizontal earth pressure P
of 4000 pounds.
Let T be the point where the resultant pierces the base,
and let CT be represented by t ; then the factor of security is
MC_ 4
MT t*
60 INVESTIGATION OF RETAINING WALLS. [CHAP. III.
in which t is to be determined. To do this, drop Dd perpen
dicular to BC, dividing the crosssection into a rectangle of
weight V l and a triangle of weight F 3 . The value of V l is
15 X 3 X 24 or 10,800 pounds, and its horizontal distance
from T is (6 t) feet. The value of V is 150 X 5 X 12 or
9000 pounds, and its horizontal distance from T is (f X 5 /)
feet. The lever arm of P with reference to T is 8 feet, and as
R passes through T its lever arm is o. Then the equation of
moments is
8000 X 8 =z io8oo(6J t) + 9000(3^ /),
from which / is found to be 1.83 feet, and then the factor of
security against rotation is
* = IT; = '
which is not sufficient for proper stability.
A general discussion applicable to any trapezoidal cross
section will now be given. Let h be the vertical height of the
wall, a the thickness of the top AD, b the thickness of its base
BC, and v its weight per cubic foot. Let be the angle
which the back of the wall makes with the horizontal, and z the
angle which the earth pressure P makes with the normal to
the back of the wall. The point of application of P is at a
vertical height of \ h above B.
Let V be the weight of the wall acting through the centre
of gravity of the crosssection, and let 5 be the point where its
direction cuts the base. Let R be the resultant of P and V
ART. 19.] ANALYTICAL DISCUSSION OF ROTATION.
61
acting at some point T on the base. Let s represent the dis
tance and t the distance CT. Let the point T be taken
as a centre of moments, and let the leverarm of P with
N
FIG. 17.
reference to it be /. The leverarm of V is b s t, and
that of R is zero. Then the equation of moments is
Pp =V (bs t\
(46)
which is the fundamental formula for the investigation of re
taining walls. This may be written
which is sometimes a more convenient form for use, since Vb
and Vs can be treated like single quantities.
To investigate a wall, the factor of security n is to be de
termined. From formula (45),
n =
MC
MT ' \b 
(48)
^x
62 INVESTIGATION OF RETAINING WALLS. [CHAP. IIL
and n will be known as soon as / is found. To do this the
value of/ is expressed in terms of /, thus :
and this being inserted in (47) there is deduced
/tf cos(0 + *)
sn
In this formula F is the weight of one unit in length of the
wall, or, for a trapezoid,
(51) 
and Vs is the moment of that weight with respect to the
inner edge B of the base. By considering the trapezoid
ABCD as the difference between the rectangle AaCc and the
two triangles AaB and CcD, this is found, by the help of the
principles of statics, to be
Vs = \vh((t + 0& + P (2a + b)k cot 0), . (52)
and, dividing by F, the distance s can be determined if it
should be required. To investigate a wall, formulas (52) and
(51) are first used, then (50), and lastly (45).
ART. 20.] COMPRESSIVE STRESSES OF THE MASONRY. 63
As an example, let the following data be taken : A sand
stone wall retaining a level bank of earth ; 034 degrees,
w = 100 pounds per cubic foot, h 18 feet, a = 2 feet, b = 5
feet, 6 =. 80 degrees, v = 140 pounds per cubic foot. The
value of P, from Article 8, is 3570 pounds, z being zero. The
weight V is found by (51) to be 8820 pounds. Vs is found by
(52) to be 4405 poundsfeet. These inserted in (50) give
t = 1.75 feet, which, being greater than onethird the base,
shows proper stability; and lastly, from (48), the factor of
security is n = 3.3.
It will be interesting to test the same wall by the pressure
theory of Article 9, where, z being 34 degrees, P is 2590
pounds. All other data being the same as before, there is
found from (50) the value / = 3.23 feet, which is more than
onehalf the base, so that T in Figure 17 lies between M and
13, and the tendency to rotation about B is greater than that
about C.
Problem 19. Compute the factor of security against rota
tion for the data given in Problem 18.
ARTICLE 20. COMPRESSIVE STRESSES ON THE MASONRY.
As a general rule, the working compressive stress upon the
base of masonry walls should not exceed 150 pounds per
square inch in firstclass work. A tower 150 feet in height
will produce this pressure on its base if the masonry weighs
144 pounds per cubic foot.
64 INVESTIGATION OF RETAINING WALLS. [CHAP. III.
The total normal pressure N upon the horizontal base of a
retaining wall will be given by (44), making a = o, or
N= rPcos(0 + *X (53)
in which P is the earthpressure acting at an angle z with the
normal to the back of the wall, V the weight of the wall, and
the angle which its back makes with the horizontal. If P is
computed by Article 8, the angle z is zero ; if by Article 9, its,
value is 0, the angle of repose of the earth. For any ordinary
case cos (0 + z) is a small fraction, and in most cases it is
a sufficient practical approximation to regard N as equal to V.
The compressive stresses upon the base BC (Figure 17) will
be regarded as caused by the vertical pressure N alone. N is
\ evidently the vertical component of the resultant R. The
horizontal component of R produces shearing stresses along
the base which are supposed not to increase the compressive
stresses. The distribution of the compression over the base
will then be similar to that shown in the diagrams of Figure
15, and will depend upon the position of the point in which R
cuts the base.
If the resultant cuts the base at its middle point (as in the
first diagram of Figure 15), the compression due to N is uni
form over the area b X I square feet, and
S =J^ (54)
s**V
is the compressive stress in pounds per square foot.
ART. 20.] COMPRESSIVE STRESSES OF THE MASONRY. 65
If the resultant is applied at the limit of the middle third
(as in the third diagram of Figure 15), the unitstress at the
edge B is zero, that at the middle is the average value given
by (54), and the greatest stress_at the tQ_6js double this aver
age value, or
S=2y ........ (55)
If__the resultant is applied without the middle third at a
distance / from the edge C (as in the last diagram of Figure
15), the compression is distributed only over the distance 3;,
so that
(s6)
gives the stress in jpmmdsjpei^square foot.
The case where R cuts the base within the middle third
at a distance / from C (as in the second diagram of Figure 15)
remains to be considered. Let 5 be the greatest unitstress at
(7, and 5", be the least unitstress at B. Then the unitstress at
the middle of the base is equal to the average unitstress, or
and as N is applied opposite the centre of gravity of the
stresstrapezoid, the value of t is
_
~s+s, v
fa**.
:
66 INVESTIGATION OF RETAINING WALLS. [CHAP. III.
Now eliminating ^ from these two equations, there result
v
. .
which is the greatest unitstress, namely, that at the toe C.^ h
As N is in pounds for one foot in length of wall, and b is in
feet, these formulas give compressive stresses in pounds per
square foot, and dividing by 144 the values in pounds per
square inch are found.
^ example, take the wall of the last article, where h 18
feet, a = 2 feet, b 5 feet, 6 = 80 degrees, z = o, P= 3570
pounds, F=882O pounds, and t 1.75 feet. In (44) the
value of a is o, and N is found to be 8665 pounds. Then
from (57) the greatest compression is 22.9 pounds per square
inch, which is a low value even for inferior work.
For ordinary walls a sufficiently exact computation of the
unitstress 5 may be made by taking V for the value of N.
Thus for the above case V= 8820, and from (57) 5= 23.3
pounds per square inch. When z = o and a = o, formula
(44) gives N = V P cos 0, which differs but little from
N = Fwhen is near 90.
If there be no pressure behind a wall, the point T coincides
with 5 (Figure 17). Then the normal pressure V produces the
greatest unitstress at B, whose value is given by one of the
formulas,
' ' (58)
ART. 20.] COMPRESSIVE STRESSES OF THE MASONRY. 6?
according as the distance s is less or greater than onethird
of A
According to the theory here presented, the vertical com
ponent of R alone produces compression on the base of a
retaining wall, while the horizontal component is exerted in
producing a shearing stress. This theory has defects ; but
upon it has been based the design of structures more impor
tant than retaining walls.
Problem 20. Let the back of the wall be inclined forward
at a batter of 2 inches per foot, and let the normal pressure of
the earth be P= 22,760 pounds. Let its height be 36 feet,
the top thickness 6 feet, the base thickness 18 feet, and the
weight per cubic foot 150 pounds. Compute the greatest
compressive stress on the base.
68 DESIGN OF RETAINING WALLS. [CHAP. IV.
CHAPTER IV.
DESIGN OF RETAINING WALLS.
ARTICLE 21. DATA AND GENERAL CONSIDERATIONS.
When a retaining wall is to be designed for a particular
location the character of the earth to be supported is known
and also the height of the wall. The data then are : w the
weight per cubic foot of the earth, <p its angle of repose, 6 the
angle of inclination of its surface, and h the height of the
proposed wall.
The thickness of the top of the wall, a, is first assumed.
In doing this practical considerations will generally govern
rather than theoretical ones. Theory indicates, as will be
seen in Article 24, that the thinner the top of the wall the
less is the quantity of material required ; but theory supposes
the earth to be homogeneous and takes no cognizance of
the action of frost. Experience, however, teaches that the
freezing earth near the top of the wall exerts a marked lateral
pressure which can only be counteracted by a substantial
thickness. To possess proper stability against the action of
ART. 21.] DATA AND GENERAL CON SID ERA TIONS. 69
frost and the weather a wall should not have a top thick
ness less than two feet. Usually when the height of a wall
varies, as in a railroad cut, the top has the same thickness
throughout. If a wall be only a foot high, its thickness
should not be less than two feet, else in a few years the frost
will push it over. Even in latitudes where frost is rare this
rule is a good one to follow.
The engineer will next decide upon the batter of the back
of the wall, or upon the value of 6, the angle between the back
and the horizontal. In doing this he must have regard to the
batter which the front of the wall will have, and to the theory
of economy of material set forth in the following articles. In
construction the back of the wall will be left rough or built in
a series of steps, so that 6 need be taken only to the nearest
degree of the average inclination.
The pressure of the earth can now be computed by the
proper formula of Chapter II. The theory of Article 8 which
supposes its direction to be normal to the back of the wall is,
in general, to be preferred, because in practice the earth is
tamped against the wall so that there can be little tendency
to slide along it. Article 8 demands a heavier wall than
Article 9, and is thus on the side of safety. In our opinion
Article 8 gives the pressure of earth against a wall which
stands firmly with a high degree of stability, and Article 9
gives the pressure of the earth when motion or failure is about
to begin. As walls are designed to stand and not to fail, the
engineer should be careful in erring on the side of safety.
70 DESIGN OF RETAINING WALLS. [CHAP. IV.
Therefore in this chapter the lateral pressure P will usually be
taken normal to the back of the wall, so that in all previous
formulas the angle z is zero.
The next procedure is to determine the thickness of the
base so that the wall may have proper security against rota
tion ; how this is done the following Article will show. If the
front of the wall thus designed does not have the desired
batter, a change can be made in the value of 6 and the work
be repeated. Then it will be well to test the work by deter
mining graphically (Article 18) the factor of security against
rotation. Lastly, the question of sliding must be considered
and proper security against it be provided (Article 23). The
practical points regarding the coping, the frost batter near the
top of the back of the wall, the weep holes, the foundation,
the drainage ditches, the quality of the masonry, and the
details of construction will, of course, receive full attention
and be fully set forth in the drawings and specifications.
Problem 21. If ft be the angle which the front of the
wall makes with the horizontal, prove that b a equals
h (cot ft cot 0). Find the batter of both back and front
in inches per foot when b = 5 feet, a 2 feet, h = 18 feet, and
6 = 80 degrees.
ART. 22.] COMPUTATION OF THICKNESS. 71
ARTICLE 22. COMPUTATION OF THICKNESS, y
,
The discussion in Article 19 furnishes the following funda
mental equation for the stability of any wall against rotation :
To apply this to the determination of the thickness of the
base of a trapezoidal wall the values of /, V and Vs are
inserted from (49), (51) and (52) and t is made \b, thus giving
a factor of security of 3.0 against rotation (Article 17). The
value of the angle z is taken as zero because the earth pressure
is computed from Article 8 under that supposition. Then
results
b cos B + = &h(P + *" + bh cot0+ 2*cot0), (59)
and the solution of tljis equation with respect to b gives
b =  A + V + A\ (60)
in which A and B have the values
_ _ 2/>, cos 9
vh '
B =  + a*  2ah cot B,
v sin u
7 2 DESIGN OF RETAINING WALLS. [CHAP. IV.
from which the base thickness can be computed for any
given data.
When the value of is 90 degrees this takes the simple
form
(61)
which is the formula for the b***e thickness of a wall with
vertical back.
In these formulas P l is the earth pressure computed by
Article 8, h the vertical height of the wall, a the thickness of
its top, 6 the angle at which the back is inclined to the hori
zontal, v the weight of a cubic foot of masonry, and b is the
thickness of the base which gives the wall a factor of security
of 3.0 against rotation, the resultant R then cutting the base
at the limit of the middle third. For all joints above the base
the factor of security will then be greater than 3.0.
For example, let a wall with vertical back be 20 feet high,
sustaining a level bank of sand which weighs 100 pounds per
cubic foot and has a natural slope of 34 degrees. Let the
masonry be 165 pounds per cubic foot and the top of the wall
be 2 feet in thickness. It is required to find the thickness of
the base BC (Figure 16). From formula (26) the pressure of
the earth is 5650 pounds. Then from (61)
ART. 22.] COMPUTATION OF THICKNESS. 73
which gives a crosssection whose area is K 2 4~ 757)x or
95.7 square feet.
As a second example take the same wall except that the
back is inclined backward so that 9 is 80 degrees. Here the
value of P l is found from (25) to be 4410 pounds. Then
whence from (60) A = 229, B = 44.2, and b = 4.45 feet,
which gives a crosssection whose area is 64.5 square feet.
The advantage of inclining the wall backward is here plainly
indicated, the vertical wall requiring nearly 50 per cent more
material than the inclined one.
If the wall be of uniform thickness throughout, a equals b,
and the solution of (59) gives
(62)
in which C has the value
r _ 3/z cot 6 2/>, cosfl
2 ~~vh '
If in this 6 be 90 degrees, it becomes
which is the proper thickness for a vertical rectangular wall.
As an illustration take the same bank of sand as in the last ex
ample ; then for 8 80, C ' = 4.81 and the required thickness
is ='4.0 feet. If, however, 6 = 90 degrees, there is found
74
DESIGN OF RETAINING WALLS.
[CHAP.
b = 8.3 feet. Here again the great advantage of inclining
the wall is seen.
Sometimes it may be desirable to assume the inclination fi
of the front of the wall, and then to compute both b and a.
For this case Figure 17 gives
a b //(cot /? cot 0), .... (63)
and inserting this in (59) and solving for b there is found
..... (64)
in which D and E are determined from
D = /Kcot 6 + j cot ft)  
B
vh
p
z>sm
For example, take the same bank of sand as before and let the
back be vertical, or 6 = 90, and h = 2O feet. Then P l = 5650
pounds per linear foot of wall. Now let the front of the wall
have the batter of iJ inches per foot, or /3 = 82 52', and cot ft
= 0.1250 (Article 13). Then Z> = 1.25 and = 68.5 and
from (64) the base thickness is b = 7.12 feet ; lastly from (63)
the top thickness is a 5.87 feet.
The formulas above given can only be used when the earth
^pressure P l has a direction normal to the back of the wall.
Those who believe in the theory of earth pressure set forth in
Article 9 are referred to the latter part of Article 24 for a
formula by which they should compute the thickness.
ART. 23.] SECURITY AGAINST SLIDING. 75
Problem 22. A wall weighing 140 pounds per cubic foot
has a vertical back, is 18 feet high, and the horizontal earth
pressure on it is 4580 pounds. Compute the thickness of the
base when the crosssection is rectangular. Compute the
thickness of the base when the crosssection is triangular.
Compare the two sections with respect to amount of material.
ARTICLE 23. SECURITY AGAINST SLIDING.
(pn//C^~a jLl&iftr^l
The base thickness b computed in the last Article provides
proper security against the rotation of the wall under the lat
eral pressure of the earth. The crosssection thus determined
should now be investigated and full security against sliding be
provided. This can be done in three ways.
First : the masonry "may be laid with random courses so
that no through joints will exist. If the stones are of suffi
cient size this checks very effectually all liability to sliding.
Second : all through joints may be inclined backward at an
angle a (Fig. 13) so that the resultant R shall be as nearly nor
mal to them as possible. This will occur when Fin formula
(44) is zero, or when
/>, sin (# + )= Fsin a,
and this reduces to
cot a = V  cot 0, .... (65)
P. sm 6
7 6 DESIGN OF RETAINING WALLS. [CHAP. IV.
from which a can be computed for any joint, V being the
weight of the wall above that joint, P l the earth pressure above
it, and the inclination to the horizontal of the back of the
wall. As b is computed for a horizontal base, the value of Fis
a little less than \vh (a f b). For example, take the wall de
signed above where B = 90 degrees, // = 20 feet, P^ = 5650
pounds, ^=165 pounds per cubic foot, a 5.9 feet, and
b= 7.1 feet. Then Fis a little less than 21 450 pounds, say
21 ooo pounds, and cot a = 2.7, which gives a =. 20 degrees
nearly. This backward inclination should be made less for
joints above the base, becoming nearly zero for those near the
top of the wall.
Third : for cases where a through horizontal joint cannot
be avoided, as when a wall is built on a platform, the thickness
of the base which will give a required factor of security against
sliding can be computed from (43). To do this make both z
and a equal to zero in (44), and substitute the values of F and
TV from (43), giving
*/> 1 sin0=./(F/>cos0).
Now in this let the value of V be inserted, namely,
V= kvh(a+ b\
and the equation be solved for b, thus :
*2f*l, ... (66)
ART. 24.] ECONOMIC PROPORTIONS. 77
in which a is the top thickness, h the vertical height, the in
clination of the back of the wall to the horizontal, P^ the nor
mal pressure of the earth, v the weight of the masonry per
cubic unit, /the coefficient of friction of the masonry on the
through horizontal joint, and b the base thickness for a factor
of security of n against sliding, It would be desirable that n
should be about 3.0, but to secure this the wall must be thicker
than is required for rotation. Accordingly, this method of
obtaining security against sliding should be used only when all
other methods are impracticable. Thus in the last article a
vertical rectangular wall is determined to be 8.34 feet thick
when k=. 20 feet and P l = 5650 pounds, v = 165 pounds per
cubic foot; now, if n =3.0 and /= 0.5, formula (66) gives
a = b 10.3 feet.
Problem 23. Compute the proper inclination of the joints
in the rectangular wall of Problem 22 at distances of 6, 12 and
1 8 feet from the top.
ARTICLE 24. ECONOMIC PROPORTIONS.
By the help of the formulas of Article 22 the thicknesses
of several trapezoidal walls will now be computed in order to
compare the quantities of masonry required, and thus obtain
knowledge regarding the most economical forms of crosssec
tion. All the walls will be 18 feet in vertical height, and
sustain a level bank of earth whose angle of repose is 34 de
grees and which weighs 100 pounds per cubic foot. The
7% DESIGN OF RETAINING WALLS. [CHAP. IV.
weight of the masonry will be taken as 150 pounds per cubic
foot.
Case I. Let the back of the wall be inclined forward at a
batter of two inches per foot, or 6= 99 28' (Fig. 18). From
formula (25) the normal earth pressure P l is found to be 5690
pounds. Then assuming the top thickness a at o.o, i.o, 2.0
feet, etc., the proper base thickness for each is computed from
formula (60) and given in the table below.
Case II. Let the back of the wall be vertical, as in Fig. 19,
or 6 = 90. From formula (26) the earth pressure P l is found
FIG. 18.
to be 4580 pounds. Then assuming thicknesses of the top, the
corresponding base thicknesses are computed and inserted in
the following table.
In this table the column headed "cubic yards" gives the
quantity of masonry in one linear foot of the wall, and it is
seen that in each case this is least for the wall with the thin
nest top. It is also seen that the vertical walls require less
masonry than the corresponding ones with forward batters.
The columns headed " per cent " show these facts more clearly,
the standard of comparison being the vertical rectangular wall
which is taken as 100.
ART. 24.]
ECONOMIC PROPORTIONS.
79
Assumed
T*/\f\
Case I. 9 = 99 28'.
Case II. 9 = 90.
1 Op
Thickness.
a.
Base
Thickness.
*.
Cubic
Yards.
Per cent.
Base
Thickness.
b
Cubic
Yards.
Per cent.
Feet.
Feet.
Feet.
0.0
9.6
3.20
62
7.8
2.60
50
I.O
95
350
67
73
2.77
53
2.0
94
3.80
73
71
303
58
30
95
4.17
80
71
337
65
4.0
9.6
457
88
71
370
71
5o
9.9
497
96
71
403
77i
6.0
10.2
540
104
7.2
4.40
85
7.0
10.5
5.83
112
75
4.83
93
7.8
78
5.20
IOO
79
lO.g
6.27
120
Case III. Let the back of the wall be inclined backward
at a batter of ij inches per foot, or = 82 53' (Fig. 20).
Here the earth pressure P l is found to be 3850 pounds. Then
FIG. 20.
FIG. ax.
assuming values of the top thickness a, the corresponding
values of the base thickness b are computed from (60) and
given below in the tabulation.
Case IV. Lastly, let the back of the wall be inclined still
more backward, the batter being 3 inches per foot, or B =?$ 58',
as in Fig. 21. Then the earth pressure is found to be 3200
8o
DESIGN OF RETAINING WALLS.
[CHAP. IV.
pounds, and as before values of b are computed for assumed
values of a.
The following table gives the results of these computations
for Cases III and IV, the columns "cubic yards" and "per
cent " having the same signification as before. It is seen that
the general laws of economy of material are the same, namely,,
Assumed
Case III. = 82 53'.
Case IV. = 75 58'.
Top
Thickness,
a.
Base
Thickness.
b.
Cubic
Yards.
Per cent.
Base
Thickness.
b
Cubic
Yards.
Per cent.
Feet.
Feet.
Feet.
0.0
6.6
2.20
42
51
1.70
33
1.0
59
230
44
4.2
173
33
2.0
54
2.47
47*
34
1. 80
35
2. 9
2. 9
193
37
30
51
2.70
52
40
49
2.97
57
49
4.9
330
63
the thinner the top and the greater the backward batter of
the wall the less is the quantity of masonry. The considera
tion of these principles in connection with the local circum
stances of an actual case will hence tend toward economy of
construction. Chief among these local circumstances is the
price of land, and where this is very high a wall with a verti
cal front and a forward batter of back is often used, al
though this requires more masonry than any other form, for
the saving in cost of the land may more than balance the
extra expense for masonry. In all cases of design the first
consideration is stability, and the second economy not econ
ART. 24.] ECONOMIC PROPORTIONS. 8 1
omy in the cost of material, but in the total expenditure of
money.
Those who believe in the theory of earth pressure set forth
in Article 9 may ask if its use would lead to the same conclu
sions regarding economic proportions. To decide this it *is
necessary to deduce a formula for the thickness of a trape
zoidal wall under such pressure, and then to make the same
computations for the four cases with the same data.
The fundamental formula (47) is good for all cases. In
this let the values of/, F, and Vs be inserted from (49), (51)
and (52), making z = and t = \b. Then results
P a [2& cos (9+ 0) + ^ ] = \vh(P + a ba i + bh cot f 2ah cot 6),
\ sin 6 /
arid solving this with respect to b there is found
b = A+ VB + A*, (67)
in which the values of A and B are
2P~ COS . ,
B = H jr + # 2 2^ cot 0,
z/ sin
and from this the base thickness b can be computed for any
values of the given data, namely, the angle of repose of the
earth 0, its inclined pressure P a as found by Article 9, the
82
DESIGN OF RETAINING WALLS.
[CHAP. IV.
angle of inclination of the back of the wall 6, the top thick
ness a, the vertical height 7z, and its weight per cubic unit v.
Using the same data, the inclined pressure P^ has been
computed for each case, and the base thicknesses found from
formula (67) for the same assumed top thicknesses. The
cubic yards in one linear foot of wall are next obtained, and
an inspection of these shows that the same general laws hold
as before, namely, the thinner the wall and the less the angle
6 the less is the quantity of masonry required.
The subjoined table gives the quantities of masonry for
Case I, Case II, and Case IV, and by comparing them with
Assumed
Top
Thickness.
a.
Case I. = 99 28'.
Case II. = 90.
Case IV. = 75 58'.
Cubic
Yards.
Per cent
Difference.
Cubic
Yards.
Per cent
Difference.
Cubic
Yards.
Per cent.
Difference.
Feet,
o.o
2.43
24
177
32
1.38
19
1.0
2.77
21
2.00
28
1.40
J 9
2.0
3.10
19
230
24
145
19
30
350
16
2.63
22
4.0
397
13
300
19
5o
450
9
340
15
6.0
507
6
those previously deduced it is seen that they are all less, the
difference being greatest for the triangular walls and least for
those of uniform thickness. The column "per cent difference"
shows in each case the percentage of material which the walls
designed under inclined pressure are less than the correspond
ART. 25.] THE LINE OF RESISTANCE. 83*
ing ones designed under normal pressure. As in practice
walls are not built with a top thickness less than two feet, it
may be said as a rough rule that the hypothesis of inclined
earth pressure (Article 9) gives a wall from 10 to 20 per cent
less in size than that of normal earth pressure (Article 8).
Problem 24. Deduce a formula for the thickness of a wall
under inclined earth pressure when a = b. Compute the
thickness and quantity of material of such a wall for Case I,
for Case II, and for Case IV.
ARTICLE 25. THE LINE OF RESISTANCE.
Let a be the top thickness and b the base thickness of a
trapezoidal wall whose height is h. Then the thickness b 1 at a
vertical distance y below the top is
a), ..... (68)
and this is represented by B'C' in Figure 22. Let P be the
* a
FIG
pressure of the earth, and V the weight of the wall above
B'C. Let T' be the point where the resultant of P and V
DESIGN OF RETAINING WALLS.
[CHAP.
cuts B 'C '; as y varies T' describes a curve called the line of
resistance. When y is zero T' coincides with the middle of the
top. When y equals h the point T' coincides with T as de
termined by (50).
The line of resistance is the locus of the point of intersec
tion of the resultant of the forces above any horizontal joint
with the plane of that joint. This is a general definition
applicable to triangular and curved sections as well as to trape
zodial ones.
For a rectangular vertical wall under normal earth pressure
the line of resistance is the common parabola. To prove this
let the origin of coordinates be taken at the corner A in Figure
24, and let AB' =y and B'T' = x. Now P' = cy\ in which c
FIG. 23.
is a function of w, <f> and d (Article 8), and its leverarm with
respect to T' is \y. The value of Fis vby, and its arm with
respect to T' is x \b. Then the equation of moments is
or
ART. 25.] THE LINE OF RESISTANCE. 8$
which represents a parabola with its vertex at the middle of
the top of the wall.
For a triangular wall with a vertical back the line of resist
ance is a straight line drawn from the top to the point where
the resultant cuts the base. The proof of this is purposely
omitted in order that it may be worked out by the student.
For a trapezoidal section the position of the line of resist
ance can be computed from (50), (51) and (52), making z = o
for normal earth pressure, putting P= rj/ a , h = y and b = b f .
For example, take a wall for which = 34, d = o, w = 100,
h 1 8, 9 '80, v 140, a = 2 and b 5 feet. Here from
formula (25) P is found to be 11.027". From (68)
and this inserted in (51) and (52) gives the values of Kand Vs
in terms of/. Then substituting all in (50) there is found
280 + 67.487 2.393?'
28o + 9757
From this equation the curve is now easily constructed, thus*
y = o, t i.oo, and b' = 2.00
y= 6, =1.77, and ' = 3.00
y ~ 12, 7=1.88, and ^ = 4.00
718, *=i.8i, and ' = 5.00
and it is seen that the line, while lying always within the mid
dle third, departs most widely from the middle at the base of
the wall.
86 DESIGN OF RETAINING WALLS. [CHAP. IV.
Whatever be the form of crosssection the line of resist
ance can always be located by first determining the earth
pressure and the weight of the wall for several values of y and
then for each making a graphical construction as in Figure 12.
The curve joining the points thus found on the several hori
zontal joints will be the line of resistance, and to insure proper
stability against rotation it should lie within the middle third
of the wall (Article 17).
Problem 25. Locate graphically the line of resistance in
one of the walls of Case II, Article 24, determining points at
depths of 6, 12 and 18 feet below the top.
ARTICLE 26. DESIGN OF A POLYGONAL SECTION.
Retaining walls with curved front are now and then built.
The advantages claimed for such a profile are, first, finer
architectural effect, and second, that the line of resistance
may be made to run nearly parallel to the central line of the
wall, thus making it a form of uniform strength and insuring
economy of material.
The determination of the equation of a curved profile to
fulfil the condition that the line of resistance shall cut every
joint at the same fractional part of its length from the edge is
of very great mathematical difficulty, if not impossibility, be
cause the weight of the wall above any joint and its leverarm
are unknown functions of the coordinates of the unknown
curve. By considering the curve to be made up of a number
ART. 2C.] DESIGN OF A POLYGONAL SECTION. 87
of straight lines, however, it is easy to arrange a profile to sat
isfy the imposed conditions which will not practically differ
from the theoretical curve. The method of doing this will
now be illustrated by a numerical example.
A wall 30 feet in vertical height is to be designed to sup
port a level bank of earth whose angle of natural slope is 34
degrees and which weighs 100 pounds per cubic foot. The
back of the wall is to be plane and to have an inclination of 80
degrees. The top of the wall is to be 2 feet thick, and the
weight of the masonry is to be 165 pounds per cubic foot. It
A p
FIG. 24.
is required to design the wall so that the line of resistance
shall cut the base B* at its middle point, and also cut the
lines Bi and BC at their middle points, B 1 C 1 being 20 feet
and BC 10 feet from the top. This insures a factor of infinity
against rotation (Article 17) which is a greater degree of sta
bility than is usually required in practice, but the method em
ployed is general, and the example will serve to show how a
wall may be designed to satisfy any imposed condition
DESIGN OF RETAINING WALLS. [CHAP. IV.
First, take the upper part ABCD and consider it as a sim
ple trapezoidal wall, upon which the normal earth pressure is
found by (25) to be 1410 pounds. In the general formula (47)
the values of /, V and Vs are now to be substituted from (49),
(51) and (52), making z = o and putting / equal to \b. This
gives an equation in which all quantities but b are known, and
by its solution there is found the value b = 4.47 feet. This
completely determines the crosssection ABCD so that it is
easy to find the weight V = 5240 pounds, and from (52) its
leverarm s = 1.55 feet.
Second, take the trapezoid BCC^B^ and consider it as acted
upon by four forces, the weight of the upper part 5240 pounds,
its own weight V, the normal pressure of 20 feet of earth
which is 5650 pounds acting at 6f feet vertically above B l ,
and the reaction R of the wall below it which by the hypoth
esis passes through M, the middle point of B r Let s be
the leverarm of Fwith respect to >, and let B l C l be denoted
by b. With respect to the centre M the leverarm of V is
ART. 26.] DESIGN OF A POLYGONAL SECTION. 89
\b s, that of the 5240 pounds is \b + 0.21, and that of the
earth pressure is 6.77 f 0.087$. Then the equation of mo
ments is
5650(6.77 + 0.087$) = V(kb  s) + 524o(i$ + 0.21).
Inse/ting in this the values of Fand Vs in terms of b, and
then solving, there is found b 8.70 feet. This determines the
crosssection so that its weight V is found to be 108.50 pounds,
and the leverarm of this with respect to B l to be 2.62 feet.
Lastly, the trapezoid B^Cf^JB^ is treated in a similar man
ner, as acted upon by five forces, the weights 5240 and 10850
pounds, the pressure of 30 feet of earth which is 12 720 pounds
applied at B l , its own unknown weight F, and the reaction R
which passes through the middle of Bf v The leverarms of
the known forces with respect to that centre being found, the
equation of moments is
12720(10.15 + 0.087^) = V(\b s) + 5240^ + 1.97) } 10850(1^ 0.86),
in which b is the base JB 9 C 99 and s is the leverarm of Fwith re
spect to B v From (51) and (52) the values of Fand Vs are to
be expressed in terms of b and inserted ; then by solution
there is found b 13.6 feet.
The points C, C l and C^ in the profile of the crosssection
are now known, and a curve may be drawn through them, or
the front may be built with straight lines. The economy of
the curved profile is indicated by the fact that the crosssection
as determined is 209 square feet, whereas a trapezoidal section,
90 DESIGN OF RETAINING WALLS. [CHAP. IV.
designed under the same conditions has a base thickness of
15.2 feet and a crosssection of 257 square feet.
Problem 26. Design a curved wall for the same data as
above, but under the condition that the line of resistance shall
cut each of the bases BC, B^C^Bf^ at onethird its length from
the outer edge.
ARTICLE 27. DESIGN AND CONSTRUCTION.
When a retaining wall is to be designed its vertical height
will be given. The inclination of its back and the thickness of
its top are to be assumed, in accordance with the principles of
Article 24, so as to result in the least total expenditure for
land, labor and material. The form of section selected will be
usually trapezoidal.
The normal earth pressure is now computed by the proper
formula of Article 8.
The thickness at the base is then computed by formula (60),.
and thus the crosssection of a trapezoidal wall is determined.
The batter of the front of the wall is known by (63), and if
this proves to be greater or less than is thought advisable new
proportions are assumed and another crosssection determined.
By the help of formula (65) the approximate inclination of
a few of the joints should next be found so that the wall may
be built with full security against sliding. It is not always
ART. 27.] DESIGN AND CONSTRUCTION. gi
necessary to give the joints the full inclination thus computed,
however, since this implies a factor of security of infinity.
As a check on the computations it is well to make a graph
ical investigation of the proposed wall and determine the fac
tors of security at the base against rotation (Article 18) and
also against sliding (Article 15). These will, in general, be
less for the base than for any joint above the base.
Lastly, the maximum pressure per square inch at the edge
of the base joint may be computed (Article 20). If this is less
than the allowable working strength of the stone, the wall is
safe against crushing. Only for very high walls will this com
putation be necessary.
The computed thickness b is the horizontal thickness of the
wall at the top of the foundation, as BC in Figure 22. This
foundation should be built with care, not only to bear the
weight of the wall and prevent it from sliding, but also to pro
92 DESIGN OF RETAINING WALLS. [CHAP. IV.
tect it from the action of the rain and frost. Provision should
be made for the drainage of the bank by longitudinal ditches
and by weepholes through the wall, so that water may not col
lect and increase the pressure.
It is good practice to batter the back of the wall slightly
forward for about two feet near the top, in order that the frost
may lift the earth upward without exerting lateral pressure
against the wall.
Whether the wall be built with dry rubble or with cut stone
in hydraulic mortar, great attention should be paid to details
of workmanship and construction, all of which should be clearly
set forth in the specifications. The earth must be thrown
loosely against the wall or be dumped against it from above,
but should be carefully packed in layers which slope upward
toward the back.
Problem 27. Let = 38 degrees, 6 10 degrees, w= 100
pounds per cubic foot, >:= 150 pounds per cubic foot, #2
feet, Q 80 degrees. Compare the quantities of material re
quired for two walls, one 9 feet high and the other 18 feet
high.
ART. 28.] THE PRESSURE OF WA TER. 93
CHAPTER V.
MASONRY DAMS.
ARTICLE 28. THE PRESSURE OF WATER.
All doubts regarding the direction and intensity of the
lateral pressure against walls vanish when the earth is replaced
by water. For since water has no angle of repose, = o and
6 = 0, and all the formulas of Chapter II reduce to (35),
which gives the normal water pressure against a wall of height
h when the depth of the water is also h.
The principles of hydrostatics show that the direction of
water pressure is always normal to a submerged plane ; also
that the total normal pressure on such a surface is obtained by
multiplying together the weight of a cubic unit of water, the
area of the surface and the depth of its centre of gravity below
the water level.
The water level is usually lower than the top of the dam,
as shown in Figure 27. Let d be the vertical depth of the water
above the base of a trapezoidal dam, 6 the angle which the
back makes with the horizontal, and w the weight of a cubic
94 MASONRY DAMS. [CHAP. V.
unit of water. Then the surface submerged is  X I, the
depth of its centre of gravity below the water level is \d t and
hence the normal pressure is
P=\wd* v sin 0,
which agrees with (35). The centre of pressure, or the point
at which the resultant pressure must be applied to balance the
actual presssures, is on the back of the dam at a vertical height
FIG. 27.
of \d above the base ; this is known by a theorem of hydro
statics and likewise by Article 12.
The angle 6 is never less than a right angle for masonry
dams, and hence it will be convenient to use instead of it the
angle ip which the plane of the back makes with the vertical.
Then 6 = 90 f ip, and the normal pressure is
P = $wd a sec i/>, ...... (69;
>jue, $f~
and for a vertical wall, where i/> = o, this becomes P=
ART. 29.] PRINCIPLES AND METHODS, 95
The normal pressure P may be decomposed into a hori
zontal component P ' and a vertical component P " , whose values
are expressed by
P 1 = Pcos $ = %wd\ P" = Psin $ = %u>d* tan ^ ; (70)
and if ^ be a small angle, as is usually the case, the horizon
tal component \wd* is sometimes taken as the actual water
pressure. This is an error on the side of safety, since the ver
tical component, acting downward, increases the stability of
the dam, unless the water penetrates under the base BC, which
is an element of danger that ought not to be allowed.
Problem 28. For a wasteweir dam the water level may be
higher than AD by an amount d,. Prove that the normal
pressure is
cosifr, .... (71)
and that the centre of pressure is at a vertical distance^ above
B, whose value is given by the formula
h d
ARTICLE 29. PRINCIPLES AND METHODS.
The fundamental requirements concerning the design of
masonry dams are the same as those governing all engineering
work ; first, stability, and second, economy. The first requires
that the structure be built so that all its parts shall have
g6 MASONRY DAMS. [CHAP. V.
proper strength, and the second that this shall be done with
the least total expenditure of money. This expenditure con
sists of two parts, that for material and that for labor, and
economy will result if material can be saved without increasing
the labor. Hence all parts of a structure ought to be of
equal strength (like the " onehoss shay "), provided that the
cost of the material thus saved is greater than the cost of the
extra labor required ; for if one part exceeds the others
in strength it has an excess of material which might have
been saved.
For ordinary retaining walls and for low masonry dams the
trapezoidal form is the only practicable crosssection, since
curved faces do not save sufficient material to balance the cost
of the extra expense of construction. But for high masonry
dams, and as such may be classed those over 80 or 100 feet
high, it not only pays to deviate from the trapezoidal section,
but it is often absolutely necessary to do so in order to reduce
the pressure on the base to allowable limits. The section
adopted in such cases is therefore an approximation to that of
a form of uniform strength.
The general principles of stability of retaining walls set
forth in the preceding pages apply to all masonry structures,
but it will be well to state them briefly again, with especial
reference to dams.
First, there must be proper stability against sliding at
every joint and at every imaginary horizontal section. This
can be done either by bonding the masonry with random
ART. 29.] PRINCIPLES AND METHODS. 97
courses so that no through joints exist, or by inclining such
joints at the proper backward slope (Article 23). The first
method is alone applicable to a dam, and by the use of
hydraulic mortar the whole structure should be made
monolithic.
Second, there must be proper stability against rotation at
every horizontal section of the dam. This will be secured
when the resultant of all the forces above that imaginary base
cuts it within the middle third (Article 17) or at the most at
the limit of the middle third. In a dam there will be two
cases to be considered : (a) when the reservoir is full of water,
and () when the reservoir is empty. For the first case the
line of resistance should not pass without the middle third on
the front or downstream side, and for the second case it should
not pass without it on the back or upstream side.
Third, there must be proper security against crushing at
every point within the masonry. As a general rule this de
mands that the compressive stress per square inch shall not
exceed 150 pounds, although in a few cases higher values have
been allowed.
It will be found in designing a high dam that the second
principle will determine the thicknesses for about 100 feet
below the top. For greater heights the third principle must
generally be used, and the formulas of Article 20 be applied.
It is indeed doubtful whether these formulas correctly repre
sent the actual distribution of stress on the base of a high
dam with a polygonal crosssection, for it would naturally be
98 MASONRY DAMS, [CHAP. V.
thought that greater stresses would obtain near the middle
rather than near the edge of the base. If such is the case,
however, the application of the formulas can only err on the
side of safety.
Problem 29. A masonry dam 36 feet high and 24 feet
wide weighs 150 pounds per cubic foot. Find the point
where the resultant cuts the base when the water is 33 feet
deep above the base.
ARTICLE 30. INVESTIGATION OF A TRAPEZOIDAL DAM.
1 6i ?4,*C *^ (^^^^CL^j^
The given data will furnish the dimensions of the dam, and
the normal water pressure on its back will be computed by
(69). Then by the method of Article 18 a graphical investiga
tion for rotation may be made and the factor of security be
determined for any joint BC. Through joints should not
exist in a masonry dam, and hence BC will be taken as hori
zontal in the construction, or even if they do exist BC may be
an imaginary horizontal joint.
The factor of security against rotation may be computed
by the formulas of Article 19, first making z = o, and 9 =
90 + ^. Then from the given datq^P is found by (69), Fand
Vs by (51) and (52), and / is computed by (50), in which h is
to be put equal to */, whence finally n is derived by (48). It
will however be more satisfactory for a student to make an
analysis directly from first principles rather than to arbitrarily
use formulas for mere computation. This will be now done
for a particular example.
ART. 30.] INVESTIGATION OF A TRAPEZOIDAL DAM,
99
The largest trapezoidal dam is that at San Mateo, CalK
fornia. The top thickness is 20 feet, the base thickness is 176
feet, the vertical height is ijh feet, the batter of the back is i
to 4, and the masonry is concrete, which probably weighs about
150 pounds per cubic foot. It is required to investigate itc
stability when the water is 165 feet deep above the base.
Let P be the normal water pressure on the back, and V
the weight of dam, both per foot of length. Let BC be the
base, and T the point where the resultant of P and V cuts it.
. &
,D
C
FIG. 2 8.
f(/U*^
Q t
Now with respect to this point the moment of Pwill equal
the moment of V. Let/ be the leverarm of P\ let / repre
sent the distance CT, and s the horizontal distance from B to
the line of direction of V. The leverarm of V is then
b s /, and the equation of moments is
Pp = V(bs f) or
Vs Vt.
(73)
The first member of this equation may be replaced by
Pp' P"p", in which P' and P" are the horizontal and verti
cal components of P, and p' and p" are their leverarms with
IOO MASONRY DAMS. [CHAP. V.
respect to the point T. Also V may be replaced by vA r
where v denotes the weight of the masonry per cubic foot and
A is the area of the crosssection. Then
P'p' P"p"=.v(AbAsAt\ . . . (74)
which is a formula better adapted to numerical operations.
To apply this to the San Mateo dam the data are d = 165
feet, tan = 0.25, a = 20 feet, b = 176 feet, h = 170 feet, and
v = 150 pounds per cubic foot. Then from (70)
P' = 850 780 pounds, P" = o.2$P = 212 700 pounds,
4 r 4U G
and from the figure,
/ = & = 55 feet, /' = 176  0.25 X 55  '
Also the area of the trapezoid is
A = i X 170(176 f 20) = 1 6 660 square feet,
and the moment As is computed by regarding A as the sum of
the triangles AaB and DdC and the rectangle AadD (Figure
28), thus ; ^ /^P^&ttf ^C*/ >^UrZ^L4^f>>^
6_ t>~ t UK < :'. r i ^Jfjy & AA+**++/ ty 4stsL4Usvi*C't*t
As=AaBx %Ba + AadD(Ba + Ja^jf + DdC(BC \dC\
t^i^^^fi cAc*~^.^i^ /^^i^wc^rr
whence As = I 248 820 feet cube. Inserting now all values in
(74) and solving for t there is found / = 88.6 feet. The result
ant therefore cuts the base very near the middle, so that
the factor of security against rotation is practically infinity
(Article 17).
ART. 30.] INVESTIGATION OF A TRAPEZOIDAL DAM. IOI
It is the custom of some engineers to neglect the vertical
component of the water pressure, and regard only the horizontal
component. Testing the San Mateo dam under this supposi
tion, P" equals zero, and, all other quantities being the same
as before, there is found t = 80.2 feet, whence the factor of
security against rotation is
88
which shows that the degree of stability is ample.
A masonry dam should be investigated not only for the
case when the reservoir is filled with water, but also for the
case when the reservoir is emptied. Here the tendency to
rotation, or overturning, is usually backward instead of for
ward. Let 5 be the point where the direction of the weight V
cuts the base, and let M be the middle of the base. Then the
factor of security is the ratio of MB to MS, or
in which the distance s is computed by dividing the value of
As by that of A. Now, for the San Mateo dam,
I 248 820
: J = 75.0 feet,
16660
and then n is found to have a value of 6.8.
IO2 MASONRY DAMS. [CHAP. V.
No through joints exist in this dam, and the method of
construction of the base is such as to preclude all possibility of
sliding. Moreover by the use of (43) the coefficient of friction
which will allow sliding to occur on the base is
850780
~~ I50XI6660" ' 34 '
a value which would be very low for an imperfect construction.
The compressive stresses on the base may next be investi
gated by the method of Article 20. When the water in the
reservoir is 165 feet deep the resultant R cuts the base so near
the middle that the compression can be regarded as uniformly
distributed. The pressure normal to the base is V\ P" t and
hence the stress per square inch is
150 X 16660 + 212700
S = 2 = 107 pounds,
144 X 176
which is probably less than onesixteenth of the ultimate
strength of good concrete when one year old.
If the reservoir should be empty the greatest stress would
come at the heel B, and as V is applied at 75 feet from B, that
stress in pounds per square inch is, from formula (58),
2X 150 X 16660 / 3 X 75\
O ^^ ~^ " I ~ ? j == 14.2.
144 X 176 176 /
It will also be found that the stress at the middle of the base is
99 pounds per square inch, that at the toe C is 142 99 = 43
pounds per square inch.
ART. si.] DESIGN OF A LOW TRAPEZOIDAL SECTION. IO3
Problem 30. Investigate the security of the San Mateo
dam for a horizontal section 100 feet below its top (a) when
the water is 95 feet deep above that section ; (b) when the
reservoir is empty.
ARTICLE 31. DESIGN OF A Low TRAPEZOIDAL SECTION.
When a trapezoidal dam is to be designed its height h will
be given, and also the depth d of the water behind it. The
weight per cubic foot of the masonry v will be known, at least
approximately. The thickness of the top, #, will be assumed ;
usually this will serve for a roadway or footway and hence
cannot be less than 8 or 10 feet. The batter of the back, or
tan 0, is next assumed, and usually this will be taken small in
order that the weight of the wall V may fall as far away from
the toe C as possible. Let M be the middle of the base BC ; let
5 be the point where the direction of Fcuts it, and T^the point
where the direction of the resultant R cuts it. It is plain that
MS will always be less than onethird of MB for any trapezoid
whose back leans forward, and that it becomes equal to one
third of MB only when AD is zero and AB is vertical.
' '
IO4
MASONRY DAMS.
[CHAP. V.
Let b be the length of the base BC, and let t be the distance
CT. It is required to findj so that MT shall be onethird of
MC, or, what is the same thing, that t shall equal %b. Full
security against rotation will then exist both for reservoir full
Formula (74) is a funamental one applicable to any sec
tion. To aoply it tathe problem in hand, the values of the
leverarmsandjg' are to be stated in terms of the otter y
s : fcd*Jv **s*<<<*
quantities, thus
. . .
&*ZCL v r jjL ^^p^^^^uL^ 't&tc
Also the area A is expressed by )
' 0Kt$M6& J
. ..... (76)
and by the method of the last Article the value of the
moment As is found to be
. (77)
As = u" + ab + P + h(2a + b) tan
Inserting, now, all these quantities in (74), and making / = %b,
there is found a quadratic equation in b whose solution gives
b F+VF'
in which F and G have the values
(78)
G = 2 (P' + P" tan 0) + a* + 2ah tan t/>,
ART. 31.] DESIGN OF A LOW TRAPEZOIDAL SECTION. IO5
and from these the proper base thickness can be found, P' and
P" being first computed by (70), or if desired the expressions
for F and G can be written
tan*;
gh
in which g is the ratio of v to w, or the specific gravity of the
masonry.
If ty = o, the formula (78) takes the simple form
(79)
which gives the proper base thickness of a trapezoidal dam
with a vertical back.
~f o g:
The compressive stress at C in pounds per square inch is
now found from (5*5), or
// 'an.
144^
and if this is less than the specified limiting value, no further
investigation will be necessary ; but if greater, then the above
formulas for thickness will not apply and those of the next Ar
ticle must be used. The limiting value of 5 is often taken
at 150 pounds per square inch.
The compression at the inner edge B when the reservoir is
empty is less than that at C when it is full, for in any trape
IO6 MASONRY DAMS. [CHAP. V.
zoid where tan fy is positive MS is less than onethird of MB.
Th distance BS can, however, be obtained by dividing (77)
by (76), whence
_
"*HhflT
and then by the use of (58) the unitstress at .# is computed.
In order to show the application of the formulas and at the
same time study the question of economic proportions, let the
following data be taken : h = 60 feet, d = 57 feet, a 9 feet,
v = 150 pounds per cubic foot or^= 2.4. Let three designs
be made for which the back has different batters, namely,
tan # = J, tan $ = fa, and tan $ o. Using the formula (78),
the base is first found, and then by (76) the area of each
trapezoid ; thus :
tan i/> = J, b = 36.5 feet, A 1365 sq. ft., = 109 per cent
tan i/>=^, b = 34.4 feet, A = 1302 sq. ft., = 104 per cent
tan i/} = o, b 32.75 feet, A = 1253 sq. ft., = 100 per cent
From which it is seen that the most advantageous section is
the one with the vertical back. This conclusion might also be
inferred from the discussion in Article 24.
It is the custom of some engineers to neglect the vertical
component of the water pressure. Formula (78) may be
adapted to this hypothesis by making P" equal to zero in the
quantities F and G, which then become
73
G =  1 c? + 2ak tan ib.
gh
ART. 32.] DESIGN OF A HIGH TRAPEZOIDAL SECTION. IO?
The thickness of the dam computed under this hypothesis is
greater than before. Thus, for the above example,
tan tp = , b = 39.8 feet, A 1466 sq. ft., =117 per cent
tan $ =: 1 i g ., b = 36.2 feet, A 1356 sq. ft, = 108 per cent
tan ^ o, b 32.75 feet, A = 1253 sq. ft, = 100 per cent
Problem 31. Find the compressive unitstress at B and C for
one of the cases of the above numerical example.
ARTICLE 32. DESIGN OF A HIGH TRAPEZOIDAL SECTION.
When the value of h is so great that the formula for thick
ness deduced in the last article cannot be used the dam is said
to be " high." For such cases the condition / = \b cannot be
applied, but / must be made greater than \b so as to reduce
the unitstress at the toe C. The base thickness will hence be
greater than that given by (78).
Let 5 be the given limiting unitstress in pounds per square
foot. The corresponding value of / is, from (57),^s. Q
Trr
in which vA is the equivalent of the weight V. Inserting this
in (74), and also the values for / ', / " , A and As, there is de
duced a quadratic in b whose solution gives
108 MASONRY DAMS. [CHAP. V.
in which K and Z, have the values
K=(P" i^ 2 tan^)i
" tan $) + 2/% 2 + 2^ tan #).
If in these P" = o, the vertical component of the water press
ure is neglected ; and if tan fi = o, the back of the trapezoid is
vertical.
In using these formulas the given data are a, k, d, tan ^, v
and 5. Then b is computed, taking the water pressures P' and
P" from (75). When b is found, s should be determined by
(81), and then by (58) the stress at B when the reservoir is
empty is computed.
For an example take a 20 feet, h = 170 feet, d= 165
feet, tan i(> = o.2,v= 1 50 pounds per cubic foot and 5 = 2 1 ooo
pounds per cubic foot. Let it be required to find b, neglect
ing the vertical component P" of the water pressure. From
Article 28 the value of P' is 850 780 pounds, and by hypothesis
P" o. Then inserting all values, K 20.64, L = 15 506.6,
whence b = 145.2 feet. This gives for the area of the section
A 14 042 square feet, and from (82) t =. 0.425 b, which locates
the point where the resultant pierces the base when the reser
voir is full. From (81) there is found s = 61.9 feet = 0.426$,
which gives the point where the line of action of Fcuts the
base, and when the reservoir is empty the unitstress at the
.back edge of the base is, by (58),
5, = ^ 150X14042 (2 _ 3 x
145.2
ART. 33.] ECONOMIC SECTIONS FOR HIGH DAMS. IOO,
so that the compression at B for reservoir empty is about the
same as that at C for reservoir full.
Problem 32. Discuss the above example without neglecting
the vertical component of the water pressure.
ARTICLE 33. ECONOMIC SECTIONS FOR HIGH DAMS.
A high trapezoidal dam designed so as to give proper se
curity against crushing on the base has an excess of stability
in its upper part. Accordingly if the section be polygonal, or
bounded by curved lines, both in front and back, these may
be arranged so as to save material in the upper parts, thus less
ening the weight that comes on the base, and hence reducing
its width from that which a trapezoidal section would require.
Such a structure will be approximately one of uniform security
against rotation in its upper portions, and of uniform security
against crushing in its lower portions. The method of design
ing the upper part will be similar to that used in Article 26 for
the retaining wall.
Local and practical considerations will determine the thick
ness of the top AD. From the principles deduced in Articles
24 and 31 it is plain that to secure the greatest economy of
material the back should be vertical for some distance below
the top. If the upper subsection AA'D'D be rectangular, the
line of resistance for the case of reservoir empty will cut the
middle of A'D' ; and if the height be properly chosen, the line
of resistance for reservoir full will cut it at the front edge of
no
MASONRY DAMS.
[CHAP. V.
the middle third. To find what this height should be let a be
the thickness, h' the height A A ', and d the depth of water
FIG. 30.
above A'. Then the equation of moments with reference to a
point in the base distant \a from D' is
%wd* X^ = vati X \a.
Now if d be taken equal to h', as it may be in an extreme case,
the solution of this gives
(84)
in which g is the ratio of v to w y or the specific gravity of the
masonry.
The next subsection should be a trapezoid, and the entire
section in fact may be considered as made up of trapezoids,
the widths of these being so determined as to secure economy
and stability. The former requires that the back should be
vertical or that its batter should be as small as possible, and
"
fa
ART, 33.] ECONOMIC SECTIONS FOR HIGH DAMS. Ill
the latter requires that the lines of resistance for reservoir full
and reservoir empty shall not pass without the middle third,
while the resulting unitstresses are kept within the specified
limit.
In the upper part of the dam the question of the com
pression of the masonry need not be considered, and the width
of the base of each subsection will be found from the require
ment that the line of resistance for reservoir full cuts that base
at onethird the length from the front edge.
In the lower part of the dam the widths are to be de
termined by regarding the compressive stresses. Owing to
uncertainties concerning the theor)^ of distribution of these
stresses, and to differences of opinion concerning the manner
in which it should be applied, engineers have not agreed upon
a uniform method of design. The general form of section,
however, is that shown in Figure 30, the back being battered
below a certain depth in order to keep the line of resistance
for reservoir empty well within each base, while the batter of
the front increases downward. The views of different authori
ties are fully set forth in WEGMANN's Design and Construc
tion of Masonry Dams (second edition, New York, 1889),
where also are given sectional drawings of all existing high
dams.
Problem 33. Prove that a triangular section is one of uni
form stability against rotation when the water level is at the
vertex of the triangle.
112 MASONRY DAMS, [CHAP. V.
ARTICLE 34. INVESTIGATION OF A POLYGONAL SECTION.
The graphical investigation of the stability and security of
a polygonal section like Figure 30 is so simple in theory that
space need not here be taken to set it forth in detail. The
general method of Article 18 is to be followed for the base of
each subsection, and the only difficulty that need to occur will
be in connection with determining the positions of the centres
of gravity of the areas above the successive bases. These may
be best computed by the method explained bdow. When the
points 5 and T have been found for each base the factor of
security against rotation is known by Article 17, both for reser
voir full and reservoir empty, and then the maximum com
pressiye stresses are determined as in Article 20.
IV ^ "* ^T^K^tZ
The analytical investigation begins with the top subsection,
which is either a rectangle or a trapezoid (Figure 30), and finds
as in Article 30, or by the formulas of Article 19, the degree of
security for its base A'D'. Thus is determined the area A^
the corresponding weight V^ and the horizontal distance s l
from its point of application to its back edge. Now let
A' BCD' be the next trapezoid, let h be its vertical height, a
its top width, b its base width, ip the angle of inclination of the
back to the vertical, A^ its area, v the weight of the masonry
per cubic unit, V^ its weight vA 9 which is applied at a horizon
tal distance 5 2 from the back edge B. The sum A l \A 9 is
ART. 34.] INVESTIGATION OF A POLYGONAL SECTION. 113
the total area A whose weight is vA = F, and the line of action
of this cuts the base at S, whose horizontal distance from the
back edge B is called s. The value of s can be obtained by
taking moments about B, thus :
which is the formula for locating the line of resistance when
the reservoir is empty. The values of A^ and A 2 s^ are found/''
from the given quantities a, b, //, tan fy by the help of (76)
and (77). ^ = fcfcfi V^*$ VtfA**^2
When the reservoir is full let P f be the horizontal compo
nent of the water pressure on the entire back above B, and P"
the vertical component. Let their leverarms with respect to
Tbep' and/". Then the equation of moments is
P'p' P"p"^v(A^A^(b^st\ . . (86)
In this the value of P' is %wd*, and that of p' is %d. If the
batter of the back be uniform from the top to B, the values of
P" and p" are known by (70) and (75). If, however, the dif
ferent trapezoids have different batters, values for P" and p"
are not easily expressed. Hence it is often customary to neg
lect P", and then the distance CT is
...... (87)
in which g is the specific gravity of the masonry. From this
the line of resistance can be located when the reservoir is full.
The factor of security against rotation can now be found,
if desired, by (45) both for the case of reservoir empty and
that of reservoir full. The degree of security against crushing
will be deduced by computing the unitstresses at B and 7 by
the help of the formulas of Article 20 and then comparing
these with allowable and with ultimate values. The degree
of security against sliding could be easily determined if the
coefficient of friction were known, but as the base is not a
real joint, it will be sufficient to use formula (39), and deduce
the value of f which would allow motion if a joint actually
existed.
The above formulas can be applied to each trapezoid in
succession, A l being taken as all the area above its top, and
thus the lines of resistance can be traced throughout the en
tire section.
As a numerical example let it be required to test the fourth
trapezoid of the theoretical section of the Quaker Bridge Dam
given in Article 35. The data are A l = 1823 square feet, s l =
12.4 feet, h = 20 feet, tan ^ = 0.115, a = 37.4 feet, b = 53.4
ART. 35.] DESIGN OF A HIGH ECONOMIC SECTION. 115
feet, d = go feet, and g = 2 J; and it is required to find s
w
and / with the unitstresses 5, and 5. First the area of the
given trapezoid is 908 square feet, and its moment A^s^is
21 808 feet cube. Then from (85) the value of s is 17.8 feet,
and inserting this in (87) there is found t 17.8 feet. The
lines of resistance here cut the base at the ends of the middle
third so that the factors of security for reservoir full and for
reservoir empty are each 3.0 (Article 17). The unitstresses
^ and 5 are also equal, and each will be found to be 1 1 1
pounds per square inch. Lastly, from (i) or (39) the coeffi
cient of friction necessary for equilibrium is 0.59, a value
which cannot be approached in a monolithic structure.
Some authors use the term " factor of safety " as meaning
the ratio of the horizontal water pressure which would cause
overturning to the actual existing horizontal water pressure.
This should not be confounded with the factor of security
used in this book.
Problem 34. Given #, #, h, V l and s l for any trapezoid
(Figure 31). Deduce the value of tan ^ so that s shall
equal %b.
ARTICLE 35. DESIGN OF A HIGH ECONOMIC SECTION.
The application of formula (86) will in general lead to com
plicated equations, unless the vertical component of the water
pressure P" is neglected. This is an error on the side of safety
and is hence often allowable, particularly when tan $ is small.
Il6 MASONRY DAMS. [CHAP. V.
The following method is essentially like that devised by
WEGMANN for the design of the Quaker Bridge Dam, and is
here given because of all the different methods it appears to
be best adapted to the comprehension of students.
Using the same notation as in the last article, the top width
a is first assumed, and the uppermost subsection is made a
rectangle whose height is by (84) equal to a Vg. The follow
ing subsections will be trapezoids with vertical backs, each
base being determined so that / = ^b. To find b for any
trapezoid there will be given A t and s l from the preceding
trapezoids, its upper base a, its height h, the total depth of
water d, and the specific gravity g, while tan ip equals zero.
First A^ and A^ are expressed in terms of a, b and ^, by (76)
and (77), and these are inserted in (85). Then the resulting
expression for s is put into (86) and / made equal to \b. Thus
is obtained a quadratic, whose solution gives
b K+VK*+L, ..... (88)
in which K and L have the values
If, in these, A equals zero, the formula reduces to (79), which
should be the case, as the whole section above the base then
becomes a single trapezoid.
ART. 35] DESIGN OF A HIGH ECONOMIC SECTION. 1 1/
After having found the base of a trapezoid by (88) the
value of s should be computed by (85), taking tan ^ = o. This
will be at first greater than J#, but in descending lower (usually
before d becomes 100 feet) a trapezoid will be found where ^
exceeds \b. As soon as this occurs formula (88) ceases to be
applicable, for the section has not a sufficient degree of sta
bility when the reservoir is empty. The back must now be
battered so that s shall equal ^b, at the same time keeping
t = ^b. Introducing these two conditions into (86) and solv
ing for b there results
which gives the base of the trapezoid, and thus A^ becomes
known. The amount of batter required is now found by in
serting in (85) the value of s 9 from (81), and solving for tan ip r
namely :
(9 o)
in which s is to be taken as ^b. Thus the trapezoid is fully
determined, and the next one can be designed, taking A^\ A t
as the new A l , s as the new s 1 , and b as the new a.
After having found the base of a trapezoid by (89), the com
pressive unitstress at the ends of said base should be com
puted by (80). The value of this will be at first less than
1 1 8 MA SONR Y DAMS. [CHAP. V.
the allowable limit, but in descending lower (usually before d
becomes 150 feet) a trapezoid is reached where it is greater.
As soon as this occurs formula (89) ceases to be applicable, for
the base of the section has not sufficient security against
crushing.
The next value of b is to be derived by taking t as given
by (82) and making s = %b. These introduced into (86) pro
duce a quadratic in b, and this will be used until the com
pressive stress at B reaches the allowable limit. When this
occurs s must be made greater than \b by expressing its value
from (58) in a manner analogous to (82). The two values of s
and t are thus stated in terms of S^ and S, the limiting unit
stresses at B and (7, and inserting them in (86) and solving for
b a quadratic is found from which all the remaining trapezoids
are computed. As soon as any b is found A z is known, and
then s is derived by (85), taking A l f A^ as A. Lastly, using
this value of s, the batter tan ip is derived by (90).
This method is open to the objection that the formulas of
Article 20 do not probably give the correct law of distribution
of stress on the base of polygonal sections, and also to the
objection that the water pressure is always taken as horizontal
in direction. On the other hand, it has the advantage of being
simple in use, whereas other methods but little, if any, more
accurate in principle lead to equations of high degree whose
solution can only be effected by tentative processes.
By the help of this method the engineers of the Aqueduct
Commission of the city of New York deduced an economic
ART. 35.] DESIGN OF A HIGH ECONOMIC SECTION.
section for the proposed Quaker Bridge Dam. The top thick
ness was taken at 20 feet and the specific gravity of the
masonry at 2.5. The following are results for the theoretical
section to a depth of 171 feet (see Table II in Report of the
Aqueduct Commission, 1889).
d
b
A
tan ^
t
s t
s
Si
347
20.0
834
O
6.7
10.
13031
6516
50
26.2
1187
O
8.7
10.5
14156
ii 328
70
374
1823
12.5
12.4
15234
15234
90
534
2731
0.115
17.8
17.8
15984
15984
no
71.2
3977
0.100
25.2
23.7
16391
17453
130
92.9
5618
0.170 '
351
317
16384
18462
150
114.6
7698
0.170
453
40.1
17078
19930
171
1374
10339
0.171
56.1
49.1
18 219
21 822
In this table the first column contains the depth of the water
in feet, the second the base of each subtrapezoid in feet, the
third the total area above that base in square feet, the fourth
the batter of the back, the fifth and sixth the distances in feet
from the front and back edges of the base to the lines of
resistance, and the seventh and eighth the stresses at those
edges in pounds per square foot. It will be seen that the San
Mateo dam, 170 feet high (Article 30), has about 61 per cent
more material than this economic section of 171 feet height.
Problem 35. Design an economic section, taking the top
thickness as 30 feet and the specific gravity of the masonry
as 2\.
I2O MASONRY DAMS. [CHAP. V.
ARTICLE 36. ADDITIONAL DATA AND METHODS.
There has now been given snch a presentation of the
theory of masonry dams, adapted to the needs of students, as
will serve to exemplify the principles which govern their
design. A few concluding remarks concerning data, princi
ples, and methods will now be made.
The force of the wind has not been considered in the data.
If the wind blows upstream when the reservoir is filled, the
stability of the dam is increased ; if it blows downstream, its
effect will be to produce waves rather than to add to the
water pressure on the back.
The pressure due to the impulse of waves may be inferred
from the fact that the highest pressure observed by STEVEN
SON in his experiments was 6100 pounds per square foot. The
maximum horizontal pressure per linear foot on the top of a
dam from wave action can, therefore, probably not exceed this
value acting over three or four feet of vertical depth, and this
only when the reservoir is of wide extent.
The horizontal pressure at the water line due to the thrust
of ice should be taken, in the opinion of a board of experts
on the Quaker Bridge Dam, to be 43 ooo pounds per linear
foot. (Report of the Aqueduct Commission, 1889.)
Let H be the horizontal force at the water line due to ice
thrust, or wave action. Its moment will be Hd, and this is to
ART. 36.] ADDITIONAL DATA AND METHODS. 121
be added to the moment of the water pressure. In all the
preceding formulas, therefore, the quantity should be re
placed by 1 fn order to include the effect of this hori
hg vh
zontal force in the computations. For instance, if the example
in Article 32 is to include the effect of the ice thrust, formula
(84) must be modified as stated, taking /^= 43000 pounds.
Then b will be found to be 156.3 feet instead of 146.8, and the
.area of the trapezoid will be about 5f per cent greater than
before.
When the computations extend below a permanent water
level on the front of the dam the effect of the back pressure
can easily be introduced into the formulas by substituting
d* d*, for d\ where d is the depth of the water on the back
of the dam, and */, that on the front.
When the back and front of the dam are covered with
earth or gravel below a certain level its action may be approxi
mately estimated by computing the earth pressures according
to the method of Article 8, and then adding the moments of
these to the other external moments. Such computations
however, will always be liable to more or less uncertainty, and
hence should be made with caution.
It is not probable that the theory of Article 20 gives the
correct distribution of stress on the wide base of a polygonal
section, and it seems more likely that in such cases the unit
pressures at the ends of the base are less than those near the
122 MASONRY DAMS. [CHAP. V. ART. 36.}
middle. If this is the case, the formulas probably err on the
side of safety, even though they neglect the influence of the
shearing stress due to the horizontal pressures. It is known
(see Mechanics of Materials, Article 75) that a shear combines
with a compression normal to it and produces in another
direction a greater compression. But the application of this
principle to stresses in masonry can scarcely be made until
experimental evidence is afforded concerning the laws of dis
tribution of the unitstresses.
The theory of a dam which is curved in plan and which
acts more or less like an arch has not been considered here.
It may be stated as the general consensus of opinion, that a
section which resists water pressure by gravity alone, like
those designed in these pages, will not usually be rendered
stronger by being curved in plan. A curve, however, is pleas
ing to the eye and impresses the observer with an idea of
strength, so that it is often advisable to employ it, even if the
length of the dam be slightly increased.
AN INITIAL FINE OF 25 CENTS
~TCB~TS19?4
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W_H^494
,'UN 2 5 1953
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