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Full text of "A text-book on retaining walls and masonry dams"

^ 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 TEXT-BOOK 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 text-book." 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 TEXT-BOOK 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 pin-connected 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 TEXT-BOOK 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 TEXT-BOOK 



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 right-angled 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 = 



P-fN 



(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 -J-0) 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* 

P-fN 



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 P-fN<cl\ 



or if n be a number greater than unity, called the factor of 
security, 

n(P-fN) = 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(i-cos(0-0)) ...... 



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 



v-v'v' .v-u< 




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(P-fN)-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 = 



284-2 ^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 co-ordinate 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 
12-foot 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 t-he 




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 well-known 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 so-called 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 well-known 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(0-tf)/ 



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 four-place 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 unit-pressure 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 two-thirds its height. 
This may be proved by Figure 10, which gives a graphical 
representation of the pressure against the back of the wall, the 
unit-pressures 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 
two-thirds 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 unit-weights 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 


0-3333 


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 cross-section is made to scale, and 
its centre of gravity found : this is G in the figure. The area 
of this cross-section 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 first-class 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 cross-section 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 first-class 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 first-class 
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 unit-stress 5, at B will be equal to the 
unit-stress 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 unit-stress 
S t at B becomes smaller than in the first diagram, while the 
unit-stress S at C becomes greater, and the unit-stresses be- 
tween B and C are taken as varying proportionally. In the 
third diagram the distance MT is such that the unit-stress at 
B becomes zero ; this occurs when CT is one-third of CB 
(since the line of direction of R passes through the centre 
of gravity of the stress triangle) or when MT is one-third 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 cross-section 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 cross-section 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 cross-section 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 cross-section 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 cross-section, 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 lever-arm of P with 




N 



FIG. 17. 



reference to it be /. The lever-arm of V is b s t, and 
that of R is zero. Then the equation of moments is 



Pp =V (b-s- 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 pounds-feet. These inserted in (50) give 
t = 1.75 feet, which, being greater than one-third 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 
one-half 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 first-class 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= r-Pcos(0 + *X (53) 

in which P is the earth-pressure 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 unit-stress 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 unit-stress at 
(7, and 5", be the least unit-stress at B. Then the unit-stress at 
the middle of the base is equal to the average unit-stress, or 



and as N is applied opposite the centre of gravity of the 
stress-trapezoid, 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 unit-stress, 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 
unit-stress 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 unit-stress 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 one-third 
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 cross-section whose area is K 2 4~ 7-57)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 cross-section 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 i-J- 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 cross-section is rectangular. Compute the 
thickness of the base when the cross-section is triangular. 
Compare the two sections with respect to amount of material. 



ARTICLE 23. SECURITY AGAINST SLIDING. 
(p-n//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 cross-section 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 cross-sec- 
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 


9-5 


3-50 


67 


7-3 


2.77 


53 


2.0 


9-4 


3.80 


73 


7-1 


3-03 


58 


3-0 


9-5 


4.17 


80 


7-1 


3-37 


65 


4.0 


9.6 


4-57 


88 


7-1 


3-70 


71 


5-o 


9.9 


4-97 


96 


7-1 


4-03 


77i 


6.0 


10.2 


5-40 


104 


7.2 


4.40 


85 


7.0 


10.5 


5.83 


112 


7-5 


4.83 


93 


7.8 








7-8 


5.20 


IOO 


7-9 


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 


5-1 


1.70 


33 


1.0 


5-9 


2-30 


44 


4.2 


1-73 


33 


2.0 


5-4 


2.47 


47* 


3-4 


1. 80 


35 


2. 9 








2. 9 


1-93 


37 


3-0 


5-1 


2.70 


52 








4-0 


4-9 


2.97 


57 








4-9 


4.9 


3-30 


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 b-a 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 


1-77 


32 


1.38 


19 


1.0 


2.77 


21 


2.00 


28 


1.40 


J 9 


2.0 


3.10 


19 


2-30 


24 


1-45 


19 


3-0 


3-50 


16 


2.63 


22 






4.0 


3-97 


13 


3-00 


19 






5-o 


4-50 


9 


3-40 


15 






6.0 


5-07 


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 lever-arm 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 + 9-757- 
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 cross-section 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 lever-arm 
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 cross-section ABCD so that it is 
easy to find the weight V = 5240 pounds, and from (52) its 
lever-arm 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 lever-arm of Fwith respect to >, and let B l C l be denoted 
by b. With respect to the centre M the lever-arm 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 
cross-section so that its weight V is found to be 108.50 pounds, 
and the lever-arm 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 lever-arms 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 lever-arm 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 cross-section 
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 cross-section 
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 cross-section 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 one-third 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 cross-section 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 cross-section 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 weep-holes 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 waste-weir 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 " one-hoss 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 cross-section, 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 down-stream side, and for the second case it should 
not pass without it on the back or up-stream 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 cross-section, 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 lever-arm of P\ let / repre- 
sent the distance CT, and s the horizontal distance from B to 
the line of direction of V. The lever-arm of V is then 
b s /, and the equation of moments is 



Pp = V(b-s- 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 lever-arms 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 cross-section. Then 

P'p' -P"p"=.v(Ab-As-At\ . . . (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 4-U 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^^^f-i 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 one-sixteenth 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 one-third 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 one-third 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 
lever-armsandjg' are to be stated in terms of the otter y 
s : fcd*Jv **s*<<<* 



quantities, thus 









-. . . 

&*ZCL v r jjL ^^p^^^^uL^ '-t&t-c 

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 one-third of MB. 
Th distance BS can, however, be obtained by dividing (77) 
by (76), whence 



_ 

"*HhflT 

and then by the use of (58) the unit-stress 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 unit-stress 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 unit-stress at the toe C. The base thickness will hence be 
greater than that given by (78). 

Let 5 be the given limiting unit-stress 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 unit-stress 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 sub-section 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 sub-section 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 unit-stresses 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 sub-section will be found from the require- 
ment that the line of resistance for reservoir full cuts that base 
at one-third 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 sub-section, 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 sub-section, 

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 lever-arms with respect to 
Tbep' and/". Then the equation of moments is 

P'p' -P"p"^v(A^A^(b^-s-t\ . . (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 unit-stresses 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 unit-stresses 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 unit-stresses 
^ 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 sub-section is made a 
rectangle whose height is by (84) equal to a Vg. The follow- 
ing sub-sections 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 unit-stress 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 


34-7 


20.0 


834 


O 


6.7 


10. 


13031 


6516 


50 


26.2 


1187 


O 


8.7 


10.5 


14156 


ii 328 


70 


37-4 


1823 





12.5 


12.4 


15234 


15234 


90 


53-4 


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 ' 


35-1 


31-7 


16384 


18462 


150 


114.6 


7698 


0.170 


45-3 


40.1 


17078 


19930 


171 


137-4 


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 sub-trapezoid 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 up-stream when the reservoir is filled, the 
stability of the dam is increased ; if it blows down-stream, 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 unit-stresses. 

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. 




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