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wall is vertical, and the foundations not more than 3 ft.

Rule. When the backing is deposited loosely, being dumped
from carts, barrows, etc. , wall of cut stone or first-class large
ranged rubble in mortar, base C D equals .35 of the vertical
height D B; wall of good common scabbled mortar rubble, or
brick, base C D equals .4 of the vertical height D B; wall of
well-scabblcd dry rubble, base CD equals .5 of the vertical
height D B.

When the backing is deposited in layers and well rammed,
these dimensions may be somewhat reduced, but there is no
fixed rule. In general, the additional cost of spreading and
ramming will quite equal the saving in masonry.

In Fig. 397, the height B D is 6 feet. The wall, supposed
to be of dry rubble, has a base of 3 feet 4 inches. The
foundation is laid in a trench about 1 foot in depth, with a
footing or offset F G 6 inches in width. The face is
battered 1 inch to the foot, which gives the wall a more sub-
stantial appearance, though clearly adding nothing to its

Earth and sand are the materials commonly used for
backing. When broken stone,
gravel, boulders, or clay are
to be used, additional weight
must be given to the wall.

By inclining the base A B
of the wall (see Fig. 398), the
friction of the wall against the
foundation is increased and
the danger of overturning
lessened. As was stated in
Art. 1478, the rough bat-
tered back of the wall also FIG. 398.



increases the friction of the backing, tending to prevent
overturning. The batter of the face should not exceed 1
inches to the foot. Any increase is liable to catch water
running down the face and carry it into the wall. This
danger is increased where the joints of the masonry are in-
clined backwards, as in Fig. 398. To obviate this danger,
the face stones are sometimes laid in mortar.

1 48 1 . Guarding Against Frost. Where deep freez-
ing occurs, the back of the wall should be sloped
forwards, as shown in Fig. 399 at $, and smoothly
finished to lessen the hold of the frost, which

might otherwise displace the masonry. The foot
of the slope b should be at the frost line, usually
three or four feet below the surface a.

1482. Bulging. Where walls are too thin, FIG. 399.
they usually first manifest their weakness by bulging out-
wards at about one-third of their height above the ground,
as at a, Fig. 400. This effect is sometimes owing to the

yielding of fresh mortar,
and if not more than inch
for each foot in thickness
of wall at a, it need not
cause apprehension.

Sometimes retaining walls
fail on account of the com-
pression of the backing,
causing settlement and in-
creased pressure against the
wall. This is especially fre-
quent where the backing
supports railway tracks car-
rying heavy and rapidly
moving trains. In design-
'/$ ing walls for such situa-
FiG.4oo. tions, this heavy additional

weight must be provided for by additional weight in the




1483. Offsetted Back. Having proportioned a re-
taining wall abdc in Fig. 401, by the foregoing rule, we
can, by offsetting the back, as shown
in the figure, considerably increase its
stability without adding to the volume
of the masonry.

The offsets are determined as fol-
lows: Through^', the middle point of
the back, draw any line f g. From
/"erect the perpendicular///. Divide
g It into any even number of parts, in
this instance 4, and draw through
these points of division lines parallel
to/ //. Then divide/// into 1 great-
er number of equal parts than^V/, and through these points
of division draw lines at right angles to///, forming the
offsets as shown in the figure. By increasing the thickness

FIG. 402.

FIG. 403.

of the wall at the base, the center of gravity is lowered and
the stability consequently increased. The backing included
by the lines g h and ///exerts only vertical pressure against
the offsets, which tends greatly to prevent the overturning
of the wall.



1484. Surcharged Walls. When the backing is
higher than the top of the wall and slopes upwards from its
inner edge a, at the natural slope a b of \\ to 1 (see Fig.
402), the dimensions given in Art. 148O will be inadequate
for the increased pressure. The following table prepared
by Trautwine gives dimensions of walls for all probable
heights of backing:




!_T ^


' j3





O3 tfl

55 Jg

2 rf\

*"h In

m w

tfl *~

3 ^

*-hi ^

O rt

1 o

- 1



O j_

- 1

*- Pi ^

<+-! <5

Tl ^


jj Pi t^

vw fS>

*rt 4^


J3 6

o _,

O ^

^ .


_o c

o 2


.* O "Q

13 ^

o -

3 Q

.S 5 ^ c *o

'a "


13 Q

K * 5




ffi ^ 5



3 c 3 '!

Thickness of Wall in


Thickness of Wall in

g '~

Parts of Height.


Parts of Height.


































































or more




When the slope a b of the backing starts at the front a
of the top of the wall (see Fig. 403),. additional thickness is
required. The triangle a c d showing section of earth
above top of wall exerts only vertical pressure against the
top of wall, and, hence, increases its stability. When the
backing reaches above the top of the wall, as in Figs*. 402
and 403, the wall is surcharged. The following table by
Poncelet gives thickness of walls surcharged with dry sand
from the outer edge a, Fig. 403:




%'l 8







tal Height of
ng as Compare*
Height of Wall

Wall of Cut
in Mort


'5 O

"rt kp *J

Wall of Cul
in Mori


Thickness of Wall in

Thickness of Wall in


Parts of Height.


Parts of Height.













































1. 320










The table is applied as follows : If the height of the
backing is 20 feet and the retaining wall 10 feet, the tabular
height, of backing is given as 2, and the thickness of the re-
taining wall, if of cut stone, should be 10 X .707 = 7.07 feet.

1485. To Prevent Sliding* A retaining wall may
slide from its foundation without losing its vertical position.
Where the wall is built on a timber platform or a smooth
rock surface, the danger of sliding is great, oAving to insuf-
ficient friction between the wall and foundation. To pre-
vent this, strong projecting beams are built into a timber
platform running at right angles to the direction in which
the wall would slide, as shown in Fig. 404. - On wet clay the
friction is about the weight of the wall; on dry earth, from
% to |, and on sand or gravel, from f to f the weight of the



The friction of masonry on a timber platform is about -fc
of its weight if dry and of its weight if wet, i. e. , a retain-

ing wall under the above given conditions will not slide
under a pressure of ^-, f, f, etc., of its total weight.

1486. On the Theory of Retaining Walls. Let

a b d c, Fig. 405, be a retaining wall with battered face and
vertical back. The
top b e of the back-
ing is level with the
top of the wall.
Let d e represent
the natural slope of
the material com-
posing the filling,
viz , \\ horizontal FIG. 405.

to 1 vertical, which is the average of materials used for back

It is assumed that the wall a b d c is heavy enough to re-
sist sliding along its base, and that it can fail only by over-
turning, i. e., rotating about its toe c. Now, if the angle


ode between the vertical line o d drawn from the inner bot-
tom edge of the wall and the natural slope ^/rbe divided by
the line a '/into two equal angles o <^/and / ' d e, the angle
o d f is called the angle and the line </_/~the slope of maxi-
mum pressure. The triangular prism of earth o d f is called
the prism of maximum pressure because, if considered as a
wedge acting against the back of the wall, it would exert a
greater pressure against it than would the entire triangle
o d e of earth considered as a single wedge. For though the
last is more than double the weight of the former, yet it re-
ceives much greater support from the underlying earth. It
has been proved by experiment that if the triangle of earth
o d e be divided by any line d f into wedges, the wedge that
will press most against the wall is that formed when the line
d f divides the angle ode into two equal parts.

The angle o d h formed by the vertical o d and the hori-
zontal ^ //is 90. The angle of natural slope hde is 33
41'; hence, the angle o d f 'of maximum pressure is equal to
90 - 33 41' -5- 2 = 28 09'.

In making calculations, only one foot of the length of wall
and of the backing is taken, so that all that is necessary is to
take the area of the section of the wall and backing. The
material composing the backing is supposed to be perfectly
dry and possessing no cohesive power, which is practically
true of pure sand.

If we conceive the wall a b d c, Fig. 405, to be suddenly
removed, the triangle b d f oi sand included between the
line of maximum pressure ^/"and the vertical back /; d of the
wall would slide downwards impelled by a force ;/ P, acting
in a direction n Pat right angles to the side b d oi the tri-
angle, i. e., at right angles to the vertical back b d of the
wall, its center of force being at P distant way between b
and (f, measured from the bottom of the wall d. The amount
of this force is expressed by the following formula:

Perpendicular _ weigJit of triangle cf earth bdfx of .
pressure n P ~ vertical depth od~ ''

This formula not only applies to walls with vertical backs,


as in Fig. 405, but to those with inclined backs, as in Fig.
406, for inclinations as high as G inches horizontal to 1 foot
vertical, which is rarely met with and never exceeded.

1487. Friction Caused by Pressure of Back-
ing. If all the backing material contained between the
line of natural slope and
the back of the wall were
unconfined, it would slide,
producing motion, but
confined by the retaining
wall, the force is converted
into pressure of earth
against the back of the
wall, resisted by the/r/V-
tion between the compressed earth and the wall.

If the wall were to begin to overturn about its toe c
(Figs. 405 and 400) as a fulcrum, its back b d would rise,
producing friction against the backing. So long as the wall
does not move, the friction of the backing acts constantly,
and must, therefore, be one of the forces which prevent
overturning. We ascertain the amount and effect of this
friction as follows: Let a b d c, Fig. 407, be a retaining
wall, and let n P represent to some scale the perpendicular
pressure against the back of the wall calculated by formula

perpendicular _ weigJit of triangle b df x of
pressure n P vertical depth o d

Make the angle n P h equal to the angle of wall friction,
viz., that at which a plane of masonry must be inclined in
order that dry sand and earth may slide freely over it, and
taken at 33 41' with the horizontal. Draw ;/ // perpendicu-
lar to ;/ Pand complete the parallelogram n h k P. Then,
k P will represent to the same scale the amount of friction
against the back of the wall. As the friction acts in ttfe di-
rection of the back b d of the wall, it may be considered as act-
ing at any point Poi the line of the back, and we will have two
forces, viz., the perpendicular pressure Pand the friction



/' P acting at P. By composition and resolution of forces,
the diagonal h P measured to the same scale will give us the
amount of their resultant, which is approximately the single

x q x ft o e t Jicoretical

force both in
amount and di-
rection which
the wall has to
resist. This
force includes
the wall fric-
tion. The force

// P is always equal to the perpendicular
force n P divided by the cosine of the
angle of wall friction. The cosine of the
angle of wall friction is .832, and the value
of the force h P may be expressed in the
following formula :

Approximate tJicoretical pressure h P =
weigJit of triangle b df X of
FIG. 407. vertical height o d x . 832

When the back of the wall does not incline forwards more
than 6 inches horizontal to 1 foot vertical, equal to an angle
of about 26 34', the following formula by Trautwine is
used, viz. :

Approximate theoretical pressure hP =
weight of triangle bdfx . 643, ( 1 08.)

which includes friction of earth against the back of the wall.
When the back of the wall is offsetted, as in Fig. 401, the
direction of the pressure of the earth will be the same as
though the wall had the batter f g.

1488. To Find the Overturning and Resisting
Forces. To find the overturning tendency of the earth pres-
stire and the resistance of the wall against being overturned
about its toe c, as a fulcrum, see Fig. 407. Find the center
of gravity g of the wall, and through g draw the vertical



line g i. Produce the line of pressure h P, and draw c v at
right angles to this line. To any convenient scale, lay off
/ / equal to the weight of the wall and to the same scale / m
equal to the pressure h P. Complete the parallelogram
/ m s t. The diagonal / s will be the resultant of the pres-
sure and the weight of the wall. The stability of the wall
will be the greater as the distance c r, from the toe to the
point where the resultant / s cuts the base, increases. To
insure stability, c r must be greater than \ c d.

The pressure k P y if multiplied by its leverage c i\ will
give the moment of the pressure about c, and the weight of
the wall // multiplied by its leverage c r' will give the
moment of the wall. The wall is secure against overturn-
ing in proportion as its moment exceeds that of the pressure.

For example, let the height of the wall a b d r, in Fig. 407,
be 9 ft. ; the thickness at the base c d, 4.5 ft., and at the
top a b, 2 ft., and the batter of a c be 1 in. to the ft. The
triangle of earth b d f has a base b f '= 6.57 ft. and altitude
d o= 9 ft. Taking the section as 1 foot in thickness, Art.
1486, we have the contents equal to 6.57 X 9 -=- 2 = 29.56
cu. ft. Assuming the material to weigh 120 Ib. per cu. ft.,
the weight of the triangle b d f is 29.56 X 120 = 3,547 Ib.,
0/=4.81ft. 3,547X4.81 = 17,061. 17,061 -i- 0</= 1,895.7
Ib. = the perpendicular pressure ;/ P. Lay off on a line per-
pendicular to the back of the wall at P, to a scale of 2,000 Ib.
= 1 in., n P= 1,895.7 -f- 2,000 = .948 in., the perpendicular
pressure. Draw P //, making the angle n P h-= 33 41'.
Draw // /; intersecting h P in h ; then will ;/ h to the same
scale equal the friction of the earth against the back of the
wall. Completing this parallelogram, ;/ Ji k P, the diagonal
h P 1.139 in., which, to a scale of 2,000 Ib. to the inch,
amounts to 2,278 Ib., and is the resultant of the pressure
and the friction.

Produce the resultant Ji P to ;/. We next find the center
of gravity g of the wall a b d c. The section of the tfall is
a trapezoid, and the center of gravity g is readily found as
follows: Produce the upper base of the section to .1% making
a x = c d 4.5 feet. Produce the lower base in the opposite



direction to /, making d y = a b 2 f t. Join x and y. Find
the middle points x' and y' of the upper and lower bases of
the section. Join these points. The intersection g of the
lines x y and x' y' is the center of gravity of the trapezoid
a b d c.

The volume of the section of wall a b d c is readily found.
The sum of top and bottom widths = 2.0 -f 4.5 = G. 5 ft.
6.5 -=-2 = 3.25 ft. 3.25 X 9 = 29.25 cu. ft. 29.25x154 =
4,504 Ib. (the weight per cubic foot of good mortar rubble

= 154 Ib.) = the weight of the section a b d c. Draw
through g a vertical line g z, and lay off in it to a scale of
2,000 Ib. to the inch from the point /, where the line of
gravity intersects the prolongation of the line of pressure
// I\ the length / / equal to 4,504 Ib., the weight of the wall.
Lay off from / on the prolongation of // P, / m equal to
2,278 Ib. to the same scale. Complete the parallelogram
/ ;// s t. The diagonal / s represents the resultant of the
pressure and of the weight of the wall. The distance c r
from the toe c to the intersection of the resultant / s with
the base c d is more than $ of the width of the base, which
insures ample stability.


1489. Pressure of the Backing on Surcharged
Walls. In Fig. 408, the surcharge of backing m b o slopes
from b at its natural slope and attains its maximum pressure
where the slope of maximum pressure d k intersects the
natural slope b m at /. Any additional height of surcharge
does not increase this pressure. If the surcharge slopes
from a, as shown by the line a p, or from any point between
a and b, then the slope of maximum pressure must be ex-
tended intersecting the slope from a in the point k. The
prism of maximum pressure will then be d i k. The triangle
of earth a b i on the top of the wall exerts no pressure
against the back of the wall, but adds to its stability.

Having found the weight of the triangle b d f, we have, by
formula 1O8, Art. 1487,

approximate pressure = weight of triangle b d f X .643,

which includes the pressure of the backing and the friction
of the earth against the back of the wall.

Draw P n perpendicular to the back of the wall and draw
Ji P, making the angle n P h = 33 41', the angle of wall fric-
tion. Then, h P will be the direction of the pressure. The
point of application of this pressure will not always beat/ 3 ,
of the height of b d measured from d, but above P, as at r,
where a line drawn from the center of gravity g of the
prism of maximum pressure d i k (omitting any earth rest-
ing directly upon the top of the wall) and parallel to the
line d k of maximum pressure cuts the back b d of the wall.
The center of pressure P. will be at the height of the wall
when the sustained earth d b s or d b f forms a complete tri-
angle, one of whose angles is at /;, the inner top edge of the
wall. For all other surcharges, the point of pressure will
be above P.

1490. General Directions and Precautions.

The batter of the face of the wall should not (unless circum-
stances require it) be greater than 1 inches to the foot.
When the batter is greater, the joints catch more or less of
the water running down the face, which is carried into the
joints, tending to weaken the mortar. All mortar used in



retaining walls should have some admixture of cement, es-
pecially in the lower courses of masonry. Common lime
mortar will not set so long as it is kept wet by the moisture
which comes from the foundation and the moist earth back-
ing. If allowed to remain long in this condition, it becomes
worthless. When the batter is considerably greater than 1
inches to the foot, the water may be excluded from the joints
by careful pointing.


1491. Earth Work. Earth work embraces the exca-
vation of all earthy material included within the section of
the roadway above the grade line and the transporting and
depositing of it upon those sections of the roadway below
the grade line, forming the embankment. It also includes
all excavations for foundations above the water-line, the
cutting of ditches, changing of watercourses, and all excava-
tion ordinarily required in the formation and protection of
the roadway. Those parts of the roadway formed by ex-
cavation are called cuts and, according to their various sec-
tions, are called through cuts, those in which the entire
width of the roadway is in excavation, and side cuts, in
which only a part of the roadway is in excavation and a part
in embankment.

Ordinarily the side slopes in excavation are made at an

FIG. 409.

inclination of 1 horizontal to 1 vertical, and the side slopes
of embankments at an inclination of 1% horizontal to 1 verti-
cal. The slopes for cuts in the alluvial soils of our western



prairies are commonly made 1 horizontal to 1 vertical, on
account of their small resistance to frost and water. Fig.
409 represents a section of a through cut, Fig. 410 a com-

plete section of embankment, and Fig. 411 a side cut, in
which part is excavation and part embankment.

Methods in handling earth vary widely, depending upon
the character of material, the situation, the magnitude of

the work, and very greatly upon the contractor. The sec-
tion of roadway shown in Fig. 411 admits of very eco-
nomical construction. The material is loosened with the
plow or pick, and cast by hand directly from the cut to the
adjacent embankment.

1492. The Use of a Road Machine. A more ex-
peditious and economical practice is to handle the material
with a road machine. A road machine of great strength,
especially designed for railroad use, is much used for work
of this character. The blade of the machine cuts and
scrapes the material from the higher points and carries it


along, depositing it in the low places. An experienced fore-
man, with a good and well-manned machine, can almost
complete the work on a side hill line, leaving only a little
trimming and ditching to be done by hand. Before using
the road machine, the ground is broken with a plow. Earth
under favorable conditions can be handled by road machines
at from 10 to 12 cents per cubic yard. A plow especially
designed for loosening earthy material is called a "railroad
plow." It is amply strong enough to stand the draft of

FIG. 412.

three heavy teams, and is an important part of a grading
outfit. Fig. 412 shows a good form of a railroad plow.

1493. Wheelbarrow Work. The transportation of
earth by wheelbarrows has of late years been practically
abandoned by the more progressive contractors. There are,
however, situations where they may be used to advantage.
When the haul is short and the work difficult of access to
teams, the pick, shovel, and wheelbarrow, on account of their
portability, are brought into use.

It frequently happens, especially after protracted rains,
that teams can not be used on account of miring. Under
such conditions, the work can be carried on with wheel-
barrows. A runway of planks under all circumstances is
necessary to secure a firm and even tread for the wheels.

In wheelbarrow work, each man loads his own wheelbar-
row. A sufficient number of pickers must be employed to
keep a constant supply of loosened material. The gangway
planks must be kept smooth, in order that there may be no
impediment to wheeling. A wheelbarrow carries y 1 ^ of a
cubic yard of ordinary material. The men in wheeling move


at the rate of about 200 feet per minute, or 2 miles per hour,
which is equivalent to 100 feet each way per minute, techni-
cally called 100 feet of lead. The length of time required in
making a round trip from the pit (as the place of excavation
is called) to the dump will be as many minutes as there are
100 feet of lead; to which must be added 1^ minutes for
loading and dumping. Delays of various kinds are met with,
which will consume about T V of the time; so that in calcu-
lating the number of trips which each man will make in a
day the total number of minutes in a working day of 10
hours, viz., 600 minutes, must be reduced by GO minutes,
leaving 540 minutes for actual work. We will, therefore,

the number of trips for each man per day =


1.25 + the number of 100 feet lengths of lead'

EXAMPLE. Allowing $1.20 per day as wages for men, what will be
the cost to the contractor for handling earth with wheelbarrows, the

Online LibraryInternational Correspondence SchoolsThe elements of railroad engineering (Volume 2) → online text (page 21 of 35)