770 UC-NRI ENGIN. B 3 mi LIBRARY GIFT OF Mars ton Campbell, Jr li,ngmeermgf Library RETAINING -WALLS ; FQR EAUTH. THEORY AS DEVELOPED BY PROF. JACOB J. WEYRAUCH. EXPANDED AND SUPPLEMENTED BY PRACTICAL EXAMPLES, WITH NOTES ON LATER INVESTIGATIONS, BY MALVERD A. HOWE, C.E. NEW YORK : JOHN WILEY AND SONS, 15 ASTOR PLACE. 1886. Engineering Library Copyright, 1886, By JOHN WILEY AND SONS. NOTE. FOR the translation of Prof. Weyrauch's paper the writer is indebted to the labor of Prof. A. J. Du Bois, of the Sheffield Scientific School, .. Yale College, who had copies printed by the electric-pen process. However, only the leading equations of Prof. Weyrauch were given ; hence a great deal of labor has been devoted to expanding, verifying, and filling in the intermediate steps of the work, and this nucleus of the mathematical part alone has grown to about double the original quantity. For most of the historical notes acknowledgment is also due to Prof. Du Bois for his article in Journal of the Franklin Institute for December, 1879, in which he called the atten- tion of American engineers to the value of Weyrauch's theory. M. A. H. 834140 PREFACE. THE following theory of Prof. Weyrauch is presented for the use of the practical man, although, at the first glance, the array of mathematical formulae may cause him to smile at such a statement. However, a brief examination of the results and their practical applications should con- vince him that a "long- felt want" has been supplied, not only in the graphical constructions that can be performed anywhere with the simplest instruments, but by formulae so simple that substitutions can be made and solutions obtained in a very short time. The mathematical operations are somewhat tedious, and are now presented for the first time in English, fully ex- panded and verified, so as to be easily followed by those who are inclined to doubt the results or wish to satisfy themselves that they are correct. In fact, all that it is absolutely necessary for the prac- tical man to have is the contents of the Recapitulation of Formulae, in order to determine the earth-thrust and its direction for a wall not leaning backward. For walls leaning backward, Prof. Kankine's method has been combined with Weyrauch's. Walls having a curved profile and those with counter- forts have not met with the approval of American engin- vi PREFACE. cers, and vith good reason, as they can as yet be treated only by empirical formulae ; for these reasons they will not be considered. Some may question the accuracy of the theory for sur- face of earth inclined, i.e., for surcharged wall ; yet in any case Prof. Weyrauch's theory is to be preferred to any of those previously advanced. An attempt has been made to present the subject in a simple manner, and to show by a few examples the sim- plicity of the application of the formulae and construc- tions. The reader who does not care to follow the theory until he is persuaded of its practical value in application should turn at once to examples and discussions in Part II. THAYEK, SCHOOL OF CIVIL ENGINEERING, April, 1886. M. A. H. INTRODUCTION. OLDER THEORIES. RETAINING- WALLS were first treated in 1687, but until Coulomb's time no theories were advanced that are worthy of much notice, as they were for the most part founded upon mere assumptions, for which the reasons, if given at all, were statements unproved. In 1773 Coulomb founded a new school that assumed the earth-pressure to act normally to the wall, and to be induced by a prism of maximum thrust. As Coulomb's theory was the only one generally ac- cepted for some sixty or more years, a brief outline of the principal points will be given. According to Coulomb, the surface of rupture is a plane along which a prism of rupture tends to slide ; he also assumes the direction of the earth's thrust to be normal to the wall. The weight of the prism of thrust, G, and the reaction of the wall, E, are decomposed into forces respectively perpendicular and parallel to the surface of rupture. Then the difference of the horizontal components must represent the resistance to sliding of the prism of thrust. This resistance consists of friction, which is proportional to the normal pressure, and cohesion, which is propor- tional to the surface of sliding. Vlll INTRODUCTION. B C FIG. 0. Let /= the coefficient of friction. c = the coefficient of cohesion. y = the weight of a unit of volume of the earth. h = the height of the wall. x = the distance BO. E = the earth- thrust against the v all. The wall will be assumed as vertical, and the earth-sur- face as horizontal (as this is the only case discussed by Coulomb). Then from the above figure __ *(x +/A) Coulomb then reasons that there must be somewhere a surface of rupture corresponding to a prism that will ex- ert a maximum pressure against the wall; and he proceeds to differentiate the above expression with respect to x, and finds that E is a maximum when x = fh + Ji Vl +/"" 1 , INTROD TTCTION. ix an expression wholly independent of the coefficient of co- hesion. The point of application of E is shown to be at one third the height of the wall. These deductions are correct for this particular case, as Prof. Weyrauch's theory shows. But when either the earth or the wall is inclined, the direction of E is not normal to the wall, but makes an angle, (5, with the normal. The value of E, as given by Coulomb, depends directly upon the position of the surface of rupture, and changes in intensity but not in direction as the surface of rup- ture is assumed to change. In reality E is constant for any given wall and earth, and does not depend upon the position of the surface of rupture. See Recapitulation ; and notice that GO, the angle the surface of rupture makes with the vertical, does not occur in the equations for the value of E. Later writers have proved (?) Coulomb's prism of maxi- mum thrust to be limited by a plane which bisects the angle made by the natural slope of the earth and the vertical rear face of the wall. In the nineteenth century, Rankine, Levy, and Mohr have considered the conditions of the earth-particles, and arrive at their results by integration. Rebhahn (1871) and Winkler (1872) advanced theories founded on assumptions identical with those of Rankine. Rankine assumed the surface of rupture to be a plane, and that the direction of the earth's thrust is parallel to the top surface. As has been said before, Coulomb assumed the pressure X INTRODUCTION. of the earth to act normally to the wall. Since 1840 it has been customary to assume the earth-pressure to make an angle with the normal equal to the angle of repose. If this were true, a horizontal wall would be pressed, not ver- tically, but by forces acting at an angle with the vertical equal to the angle of repose, which is manifestly incor- rect. It will be seen that Prof. Weyrauch's theory is closely allied to Prof. Rankine's, but conclusively proves that the earth-pressure acts parallel to the top surface of the earth only in special cases. WEYRAUCH'S THEORY OF THE RETAINING -WALL/ PART FIRST. IN the following the earth is supposed without cohesion, and its pressure is determined independently of any arbi- trary assumptions as to direction of the earth-pressure, and with sole reference to the three necessary conditions of equilibrium. The single and only supposition, then, is as follows: That the forces upon any imaginary plane-section through the mass of earth have the same direction. This assumption lies afc the foundation of all theories of earth-pressure against retaining-walls. For those cases, therefore, to which the following discussion does not apply no complete or satisfactory theory is yet possible. In what follows, the ordinary assumption as to the direction of the earth-pressure will be proved to be incorrect, except for special cases. * Zeitsclirift fur Baukunde, Band I. Heft 2, 1878. 2 TUEOHY. OF THE RETAILING - WALL. I. GENERAL RELATIONS. Let the surface of the earth have any form, and the wall AB, Fig. 1, have any inclination. The earth-pres- sure makes any angle, tf, with the normal to the wall. Suppose through the point A the plane A C. Then the weight G of the prism ABCis held in equilibrium by the reaction of the wall, E, arid by the resultant, R, of all the forces acting upon A O. Now decompose E, G, and R into components parallel and normal to AC-, then for every unit in length of the wall, denoting by e, g, and r the lever-arms of E, G, and R respectively with reference to A, the sum of the forces parallel to A = 0, or (i) GENERAL RELATIONS. 3 the sum of the forces perpendicular to A C = 0, or the sum of moments about A = 0, or Gg + Ee-Jtr = (3) Equation (3) was first introduced by Prof. Weyrauch. Further, according to the theory of friction, if cp is the coefficient of friction for earth on earth, p / P P / -~ - tan (p or n * = tan (p. . . (4) Va V T~ Vi If now there is any plane for which P P l (Q + d) tan cp, . . . (5) P P this plane AC will be a plane of equilibrium, and jr^r-fr will be a maximum, or ? p p \ i /i \ . n j Jl-^ir^O '(*) This plane is designated as the "surface of rupture." From Fig. 1, for every position of A G, P = G cos GO, Q G sin GO, Substituting the above values of P, P l9 Q, and Q l in equation (5), it becomes G cos GO Usiu (GO -f- + <5) = [G sin 6? -J- ^cos (GO -f- <* + ^)] tan
Performing the differentiation indicated in the equation
(la), considering G and GO as the variables, it becomes
Sw sin wdw G J?cos(w+ a+S)<2wl [Cr sin a>4 Ecos(o>+ a+ fi)]
n u> + cos du> G - E sin (a> + a + g)cfto] [G cos ),
or
sin 2 (cp-{-G>) cos 3 (or
2 sin 2 (a +3) cos 2
+ sin 2 (a-{-d) cos 2 (^-j-r^) -f- cos 9
) cos 2 (a+d) cos 2
or sin 2 (^-J-cj) cos 9 (a-\-d) -{- cos 9 ( ) sin (a+ 6) ,
~Sos (a+d~ '
GENERAL RELATIONS.
which reduces to
COS (
and hence, according to equation (9),
G = Func. y = y A A CJ. .... (12)
Also, if AK is perpendicular to CJ,
CH Ic cos ((p-\-co) _ E
and if JL is made equal to JC, then, since the perpendicu-
lar from L upon CJ is equal to CH,
ACJL _ CH _E
ACJA~ AK~ G'
or E=yACJL (13)
If, finally, AM=AC,
AM. CH
or R = yAACM. (14)
All these geometrical results may be summed up as fol-
lows :
Draw from the highest point C of the surface of rupture
a line CJ, which makes with the normal CH to the natu-
ral slope the angle a -\- d, or the angle which the earth-
pressure makes with the horizontal ; then the A A CJ is
10 THEORY OF THE RETAINING-WALL.
equal in area to the A ABC, the prism of rupture. Then
lay off JL = JC and AM= AC and draw CL and CM ;
then for every unit in length of the wall the following
relations exist :
Weight of prism of rupture, G = yACAJ\ J
Earth-pressure upon wall, E = yACJL; >- (14a)
Reaction of the surface of rupture, R = yACAM. )
The first two relations were first made known by Bebhahu
in 1871, for tf = or cp.
Since, now, G : E : R = AJ : JC : CA, . . . (15)
it can be asserted that
The weight of the prism of rupture and the reactions of
the wall and of the surface of rupture are to each other as
the three sides of the A A CJ.
Thus far no assumption whatever has been made as to
the value of the angle tf. This is determined by equation
(3), which, in all theories following Coulomb's method,
does not occur.
PLANE EARTH-SURFACE INCLINED.
11
II.
PLANE EARTH-SURFACE INCLINED
ADOPT in this case the notation of Fig. 3, anct let E be
first determined for any value of d.
II =90 (e+6>)
111= 90 (a- e>
FIG. 3.
If A G is the surface of rupture, then AAEG AACJ\
or, since
AB _sin II
277 ~ shTlTP
In like manner, AJ = AC
sin II
sin III '
sin V
snTVl'
or
But since A ABC = J^ 6V,
^<7sin/ = A7. ^
sin /sin// sin /F sin V
sin ///
'. (16)
; . . . (IGa)
sin VI
THEORY OF THE RETAININO-WALL.
or, finally,
sin (ar-f-Gtf)cos
sin
cos
cos
cos (a: s).
Further, from Fig. 3, if BN is perpendicular to 'AD,
AADB = ZAAJC+AJDC,
or AD . BN= 2AJ . CH+JD . Off;
and since
and
BN _BO _ OP
~CH ^W' ~JD
AD . OD = JD (AJ+ AD),
AD (AD-AO) = (AD- AJ) (AJ + AD),
whence AO . AJ= AJ . AD. . . . (17)
D
B
FIG.
Upon this relation rests the well-known construction of
Poncelet for the earth-pressure. Draw (Fig. 3') BN per-
pendicular to the natural slope AD\ draw BO, making the
same angle with ^^that^makcs with the horizontal, and
PLANE EARTH-SURFACE INCLINED. 13
then determine the point /so that equation (17) is ful-
filled, that is, make A J a mean proportional between A
and AD ; then draw JC parallel to OB. Thus the surface
of rupture AC is found, and use can now be made of the
relations already deduced in I.
In order to determine J (A, 0, and D being given), there
are several methods, one of which is indicated in the
figure. In all these constructions 6 is assumed.
Now from equation (13), E % y TO* cos (a -f d),
but
CJ_ _ AD- AJ _ AD - V~AD~. AO _
BO ~ AD - AO ~ AD-AO ~~AO
~D
J0-, tnen UJ = - iU = -
From Fig. 3,
A ^ r
Let w = y __ then CJ = T -2?0 = ---
AD l n* .1 -f n
= ) AB _ sin (cp - t)
AB ~ cos (a: + 6)' AD ~ cos (a -~e) ;
and the multiplication of these equations gives
=
-- n ^
cos (a + 6) cos (a - f)'
^L=^/.
cos (a -\- 6) *
14 THEORY OF THE RETAINING -WALL.
and by substitution of SO and n in the value for CJ, and
of CJ in that for E,
rcos(-a)-| _ Zy_ r cos (0 - a) n ft Y
L n-f-l J 2 cos (a + 8) L(n + l) cosaJ 2cos(a+) - '
For the special Ciise of the earth-surface parallel to the
angle of repose, = cp, n = 0, and
cos
J
'^ '
These formulae hold good for any value of d. But the
angle d is determined by equation (3). In order to insert
e and r in this formula, the points of application of E and
R must be known. The angles d and G? are connected by
the relations in (16&), in which there are no other unknown
quantities. Since now d, according to the single assump-
tion of Prof. Weyrauch's theory, is independent of the
height, so also is co, and then for variable li equations (19)
and (11) become
E = Cl\ R = CJc\
dE = 2 CW, dR = 2 CJcdJc.
Let x and z equal the distance of the point of application
of E and R from A, respectively. Now considering the
top as the origin or centre of moments,
Z, R(k - x) = 2
and therefore x = $l and z $k.
Now O must act through the centre of gravity of the
A ABC, and it has been already proved that the points
PLANE EARTH-SURFACE INCLINED. 15
of application of E and R are at distances $1 and -J& re-
spectively above A ; hence (Fig. 3') ah = ed and hf=g =
bd ah = ^Jc sin GO ^l sin a.
Substituting these values in equation (3) and referring to
equation (15),
AB (CJ cos S AJ sin a) = AC (AC cos -j- G?)
-f cos 2 a: sin GO sin d cos ( )sinG?] sin#
p-\- GO] cos GO-\- cos 2 ** cos
cos 2 a cos ,
in the second to cos cp. If in the third cos 2 a = I sin 2 a,
and in the fourth sin 2 a = 1 cos 2 a, equation (22c) be-
comes
+ cos GO cos (cp + G?) cos d sin a: cos cos d sin 9)
sin a cos a sin # cos cp
+ sin # sin 2 sin (^-j-^) cos co+sin d sin a? cos (^>-|-CL))
sin 2 : sin GO sin (5 1 cos ( -)- GO)
cos 2 a: cos cp cos d -\- sin a: cos a cos ^? sin #
Reducing and dividing by cos #,
sin of cos a sin sin (cp -\- GO)
-\- cos 2 a cos GZ? cos ( cp -j- GJ
Since
cos GO sin (< 4- GO) sin GO cos
= 0.
18 THEORY OF THE RETAINING- WALL.
and
sin GO sin (cp -f- GO) -f- cos GO cos (cp -f GO) = cos ), it becomes
sin (a -j- GO) cos (a -f- $) cos a?
= sin ( ) sin sin
-j- cos (^ -f- <*>) sin a cos d
-f- cos (^> + a?) cos sin
20 THEORY OF THE RETAINING- WALL.
and equation (24#) reduces to
cos GO sin (a + G?) cos a cos $
cos oo sin (a -|- GO) sin a sin
cos 3 a cos ( ) sin (cp + G?) tan tf f ~
cos sin cos 2 ( ) cos 2 oo =
cos 2(q> -{- &?) cos 2
which equals
2 sin -J- [2&?
sn
-f (2(7? -f
__ 2 sin ( q>) sin (2&? -j- cp)
'
or
and
sin
also,
sn
cos
sin 9?,
cos
% sin 2(^> + GO) -J sin
sin 2r , 3 cos 2 t
sin a cos = - , and cos 3 a = 1- -J.
/ /v
Hence, after multiplying by 2, equation (24^) reduces to
sin 2a sin (2cc? -f- 00 s ^ u V
cos 2a -J sin 2(q> + a?) -j- cos 2 | sin 2 <
-J sin 2(<7? -f- GJ) -\- -J- sin 2G?
- tan 6^00820: cos 2 (^+c 6 9) cos 2 (-}-<) tan (J j, = 0. (24=e)
2 tan d sin 2 or cos 2 OD
-f- sin 2 sin ( ) tan d
sin 2a sin ca cos a> tan d
22 THEORY OF THE RETAINING-WALL.
Now
2 tan dsin 2 a cos 2 GO = [since sin 2 a 1 cos 2 a]
[cos 2 6? cos 2 a cos 2 ,
and
sin 2(q> -\- GO) sin 2&? _ sin 2(^> + co) sin 2oj
~~2~~ ~~~ ~2~
2 sin j- (2y + 2&? ^<^) c
2 '
= sin cp cos (2cj+ ^?),
and
tan d cos 2or cos 2 ((p -{- GJ) -\- 2 tan # cos 2 a cos 3 GO
L cos 2 . . \
= ( by making cos a = - --- 1- -J J
tan # cos 2a [cos 2 (q) -f <*>) cos 2 GO] + tan d cos 2 G?,
or tan d cos 2# sin 2&? -- < s ^
I
HORIZONTAL EARTH-SURFACE. 23
Also, cos 2 (cp -\- GO) tan d -\- tan d cos 2 GO
= tan # [cos 2 (cp -j- GO) cos 2 GO]
= sin cos GO tan (J
= tan d sin 2# [sin ( ) sin G? cos G?]
F sin2( cos (2G? -j- cp)
+ sin 9? [sin (2G? -j- 9?) cos 2ar
+ cos (2 G? -f- + ft)
= 2 cos 2 w -sin ft [sin (2w+ft) cos 2a+cos (2w+ft)"sin 2a] - sin ft sin (2w-f ft)'
By making sin 2a = 2 sin a cos a and cos 2a = I 2 sin 2 a
in the numerator, and cos 2 or = 2 cos or cos <* 1 and sin
2a = 2 sin cos # in the denominator, this becomes
tan 8 =
sinft [sin (2o)-fft) 2 sin a cos a cos (2a>-|-ft) + cos (2a>+ft) 2sin 2 a] - sin ft cos (2 cos a sin
sin ? sin %a
a)
siu (2o? -j- cp) sin 9? cos 2a '
Dividing by sin cp, clearing of fractions and dividing by
sin a, also transposing, this becomes
sin (2ft? + cp -j- a) sin (2oo -J- 9)
jj -\- cp -{- a) sin
. /rt
sin (%(& -\- cp -j- a') sin cp cos 2 -- cos GO
sin
cos (2ft? -f- cp) [sin
cp) sin
sn a
or
sin (Sea -}- cp -\- a) sin (2&? -j- <^)
sin cp cos 2^ sin (2a? -j- ^> -j-^) 2 cos or cos 2
-{- sin 9? 2 cos 2 a sin (2oi9 -}- cp -j- <*)
_ cos (2G? -f- 9?) [sin (2Gt>+ 9*) sin ^ cos
Since
sin a
2 cos 2 cos 2 -(-a?4-a)8in (co+cp) cos a cos 09 =0.
2 cos &?
The formulae for GO, d, and E can now be found in the
simplest manner. Equation (25) is satisfied for269-j-
and likewise for a height h
J t 2 jji j t a
(7 representing the constant quantity.
From equation (29*) E = C(H* 7i*); hence dE ' =
2CHdH2Ch dJi . Now let x equal the distance of the
centre of pressure below the top of the wall, then
Ex =20 f H H*dH- 2C f\*dk,
or C(H* - Ii *)x = %CH 3 - f 6V,
2 H 3 - h s
'
and if y = the distance from bottom,
Equation (30) holds good when the earth-surface is
loaded and the loading is equal to a distributed load of the
height h . Still, even then, h is often so small that can
o
be substituted for it just as for unloaded earth-surface.
In all cases d is determined by equation (28).
28 THEORY OF THE RETAINING-WALL.
Instead of using equations (28) and (29), the following
simple construction can be used :
a E B c v
FIG. 4.
Draw (Pig. 4) AC and A D vertically and horizontally,
each equal to h, also DF making the angle FDG = 45 ^
So
with the horizontal. Through the points D and ^describe
a circle whose centre lies in AD. Then draw GH parallel
to AB, and through A the straight line HJ. Then JO is
the direction of the earth-pressure upon the wall AB. If
^i^is made perpendicular to AB, and equal to AH, then
the AABK gives the intensity and distribution of the
earth-pressure, or
E = yAABK.
The proof of this construction is as follows : Conceive, in
Fig. 4, JD and FG drawn, then
A irn A P _ AG cos a
~ PH~ HG-[AGsm~a=PG]'
in which AP represents the perpendicular let fall from A
upon GH.
HORIZONTAL EARTH-SURFACE. 29
but AG : AF :: AF : AD =h,
AF
therefore AG = -T = li tan 2 _.
\ 2
Now
## GD sin or = ( J. G + AD) sin
/ \
= h sin a -\-li tan 2 (45 ^j sin a\
tan ^#(7 =
A tan 2 (45-^
h sin a + h tan 2 ^45 - |) sin a-Ji tan 8 ^45 - |) sin r
therefore
tan AHG = ^-^ tan 2 ^45 - ?} = cot tan 2 (45 - %}.
sin a \ 2/ \ Ji/
From Fig. 4,
f
cos (cp -|- Ce9) cos (^? -j" ^ + ^) si n ^ sin a:
I + cos (a cp) cos (a -}- tf) cos cp
= \ ~ cos (<* 9?) sin (9? -j- GO -f ) sin G;? cos 6
( cos (a cp) cos (^-j* GO -\- a) sin (J sin a? ;
ANGLE = ANGLE cp. 33
dividing by cos d and transposing,
cos (a cp) cos (a -\- d) cos cp
cos tf
4- cos (a cp) sin (cp 4- <& -\- a) sin &>
4~ cos (cp 4~ G&) cos ( _ _ cos (a 4-Ge9-|-a')cos(a' ) sin (cp -\- GO -f- a')
-j- sina cos(^>-|-c) cos(^>-)-G?-}-a')cos(: cp) sii\(cp-\-Go-{-a)
sin GO cos 3 (a cp) cos ( + <*)],
which becomes
sin GO sin (GO -j- a) cos (cp -f- GO).
Terms number 5 combined give
cos(0+a>)cos(4-w4-) [cos a cos (t/> + <>+) +sin a sin (< 4- + )]
wliich becomes
cos (cp + GO -j- a) cos (cp -|- GO) cos (cp + GO).
Terms number 4 combined give
-j- cos cp cos (a -{-<# -f
cos (a cp) sin GO cos
oo -f a) cos GO i
(CP + GO+ a}}
Substituting for &?, 90 9>, this becomes
cos cos (9? + 90 ^+a)sin( ^-|- 90-^)
-f cos a cos (cp+ 90 ^>) sin(^+ 90-^+ a)
[ + cos (-^) sin (90- ^) cos(^ + 90
As the second term reduces to zero, this becomes
[cos a sin a cos (a (p) cos a sin ) (cos a sm 9 cos 9? sm a) to,
or
Fsin 2a . \ / x~| 7
cos ( #>) sin (9? a) J&9
sin %a sin ^(<^ a)
],
ANGLE = ANGLE cp. 37
or
F 2 sin j(2 2a + 2a) cos |(2 2<* 2aQ "I
which equals sin cpcos(cp 2a) dca.
The differential of the denominator is
f 4- sin a cos ( ) cos (a
or [1 sin cp sin (cp
therefore
tan * =
1 sin 2a)
To find an expression for the sin d, clear equation (33)
38 THEORY OF THE RETAINING -WALL.
of fractions and deduce tan d tan d sin cp sin (cp
= sin cp cos (cp %a). Multiplying by cos tf,
sin 6 sin d sin cp sin (cp % as a radius ; draw GH parallel to BA from
IT through A draw HJ', then through a point one third
the length of AB above A draw E parallel to GJ, and this
will give the direction of the thrust, and the angle made
with the normal is found to be d = 27 13'.
Now make AK perpendicular to AB and equal to AH;
find the area of the triangle ABK, and multiply the re-
sult by y 170, and obtain the value ofE= 3037.5 pounds.
APPLICATIONS. 51
The next step is to find the centre of gravity of the wall ;
to do this, make Ex and By equal to six feet ; Dm and Ac
equal to two feet ; connect m with y and x with c, and the
point of intersection, g, is the centre of gravity of the cross-
section. The value of G = 15 X ^4"^ X 170 = 10200
/&
pounds. Constructing the parallelogram of forces, R is
found to cut the base within the middle third or q = 2.2
feet; therefore the wall is theoretically safe, so far as over-
turning is concerned.
In order that no one can say the wall has been favored
by taking a large coefficient of friction, it will be taken as
0.4 ; it is well known that for rough limestone it is nearer
0.75. Then the wall offers a resistance to sliding equal to
10200 X 0.4 = 4080. pounds ; as this is much greater than
E, it must be still greater than the horizontal component
of E, and hence there is no danger from sliding. The
plane upon which the wall is supposed to slide is at the
bottom of the wall and on top of the foundation.
To ascertain the values of E and d analytically :
Equation (26), GO = 45 - f = 26;
O 7*
tan a = ^- = .18333; .-. = 1023 f .
Equation (28), tan ( + *)= ;= 7704 ;
. . a + d = 37 36' and 3 = 27 13'.
Equation (89a), E = = 3 037. 5 Ibs
52 THEORY OF THE RETAINING-WALL.
These results are identical with those obtained by the
graphical method. To find the value of q the graphical
method used above is preferred, as being much simpler titan
the analytical method.
EXAMPLE 2. Determine the dimensions of a trapezoidal
wall built of dry, rough granite, having a vertical back and
being 20 feet high, to safely retain the sides of a sand cut,
the surface of the sand being level with the top of the wall.
W = 165 Ibs. cp = 33 40' y = 100 Ibs.
The graphical solution is given in Fig. 5.
Let AB represent the back face of the wall, 20 ft. in
height ; make AD = AB = h ; draw DP, making the angle
ADF= 28 10' = GO = 45 - ?; pass the arc DFK through
A
D and F, the arc having its centre in AD ; draw BK, then
the area of ABK X y E = 5740 Ibs. jE'acts normally to
the wall at one third the height.
The dimensions of the wall must now be determined by
the process commonly called "cut and try." In this
case suppose I' = 2 ft. and I = 8 ft., and therefore
n i Q
G = - - X 20 X 165 = 16500 Ibs. Now find the centre
&
of gravity g as in example 1, and draw the parallelogram of
forces. R is found to cut the base in the middle third,
and q about 2.8 ft. The coefficient of friction of granite
on granite is at least 0.5, and hence the wall resists sliding
at the base by 16500 X .5 = 8250 pounds, which is much
greater than the thrust E. The triangle ABK represents
the intensity of the thrust; and to find the thrust that
must be resisted by friction at any height, all one has to
APPLICATIONS. 53
do is to find the area of the triangle above the plane, and
multiply it by y. Thus, it is seen, it is a very simple matter
to find whether the wall will slip on any plane above the
base.
The above wall, then, is perfectly safe.
Analytical Solution. a =0, G? = 45 -^ = equation
(26), which gives a = 28 10'.
Equation (28), tan (a -f 6) = ; .-. 6" = 0, and
normally to the back face of the wall.
Equation (29d), fi= .2867 1Q = 5734. pounds,
which is the same as obtained graphically within 6 pounds.
The dimensions are obtained by the above graphical
method.
From Trautwine's "Engineer's Pocket-book," 1885, p.
690, the above wall would take the proportions H = 20 ft.,
I - 20 X .389 = 7.78 ft., and V = 20 X .096 = 1.92 ft.,
and hence G = 1600 Ibs. The wall to be built of cut stone.
For dry rubble, which the example calls for, his propor-
tions are : H = 20 ft., b = 20 x .528 = 10.56 ft., V = 20
X .236 = 4.72 ft, and hence G = 25212 Ibs., or 8712 Ibs.,
or over 50 cu. ft. more masonry per lineal foot than is nec-
essary. Such walls are no doubt safe, but involve a needless
waste of material. Bear in mind, only well-laid walls are
considered, and foundations are supposed to be immovable.
A wall, if built properly, will be so bonded as to leave
but few voids, and hence the average weight of the mate-
rial used in construction may be taken in all practical cases
with safety.
54 THEORY OF THE RETAINING-WALL.
We infer, then, that Trautwine's table used above, giving
the dimensions of retaining- walls, specifies much larger
quantities of material than are absolutely necessary, and
leads to waste of material.
EXAMPLE 3. The same as Example 2 with a = 8, or a
batter of 1.68 inches per foot in height. By the graphical
method of Fig. 4, assuming H = 20, b = 8, V 2, it is found
t\rdtE= 6328 Ibs., and that R cuts the base, making = 2.7
ft., or 0.1 ft. less than for a vertical back. The horizontal
component of E which tends to make the wall slip is 5200
Ibs., which is again less than for the vertical back. Hence
it is seen that the wall in this case is less stable, but less
liable to slip, and, besides, owing to the inclined back, it
will not be affected to so great extent by the action of frost.
The wall, exclusive of the foundation, exerts a pressure
of less than 14 Ibs. per square inch in Ex. 2, and less than
18 Ibs. per square inch in Ex. 3 upon the earth.
EXAMPLE 4. What must be the dimensions of a rubble
wall of large blocks of limestone, laid dry, to retain a sand
filling which supports two lines of standard-gauge railroad-
track ? '
H= 15 ft., JF=170 Ibs., a = 8, y = 33 40', y = 100 Ibs.
Graphical Solution (Fig. 6). Assuming the moving load
of the railroad to be 3000 Ibs. per lineal foot, the pressure
per square foot as distributed by rails and ties will be
about 400 Ibs., which will be considered to extend over
the whole surface of the fill to compensate for the shocks
due to moving trains, etc., or h = H -\- 4, or 19 ft.
= H+ 4; make AD= AC-, draw FD, making
APPLICATIONS. 55
an angle of 28 10' &? = 45 -^ with AD describe the
arc DJFGH, the centre being in AD ; draw GH parallel
to AB, then HJ is the direction of the earth- thrust E,
which acts at a point above A equal to one third AB ;
draw AK perpendicular to AB and equal to AH, then the
triangle BAK x y gives the intensity of E or 5760 Ibs.
It is evident that the wall could not be triangular, since
the thrust at the top is not zero, as the triangle ABK in-
dicates. To find the least allowable value of #', find the
thrust at a distance of, say, 1 ft. below the top : this is
about 375 Ibs., or horizontally about 300 Ibs. Then, in
order that the top stone will not slip off, it must have a
width of ' = 3.53 ft. (0.5 = coefficient of friction).
Hence V may safely equal 3.5 ft.
The resultant R will cut the base within the middle
third, if I = 8 ft. and q will equal 2.7 ft.
Analytical Solution.
Equation (26), GO = 45 - % = 28 10'.
.-
Equation (28), tan ( + <*) = ~ = .4901;
. . 26 7' = (a + ) and 3 = 18 7'.
Equation (29,), E = ' = 5758 Ibs.
Dimensions I' = 3.5 ft., b = 8 ft., obtained by graphical
method above.
56 THEORY OF THE RETAINING-WALL.
If the above wall had a vertical back and was not loaded
by the railroad, Mr. Trautwine's proportions would be (p.
690, 1885) about H 15 ft, I = 15 X 0.51 = 7.65 ft., V = 15
X 0.343 = 5.15 ft., and G = 16320 Ibs. ; and hence the wall
contains a few more cu. ft. per foot run, and is also more
stable (as a vertical-back wall is more stable than one with
a battered back, dimensions being the same), but the ad-
ditional weight of the railroad has not been considered at
all.
EXAMPLE 5. What must be the dimensions of a 20-ft.
wall, to retain the foot of a side of a deep sand cut ; mate-
rial to be rough blocks of limestone (case of surcharge) ?
W = 170 Ibs. ; H = 20 f t. ; y= 100 Ibs.
a = 8
Graphical Solution (Fig. 7). Let AB represent the
back face of the wall, BCthe natural slope and surface of
the cut ; draw GAD parallel to the surface BG 9 and, with
A as a centre and AC as & radius, describe the arc DCHG;
draw GH parallel to AB\ draw DP horizontal, and through
If and F draw HJ\ then JG is the direction of E.
Make AK perpendicular to AB and equal to HP, then
the area of ABK X y = E = 23600. Assuming V = 2 ft.
and ~b = 9 ft., R is found to cut the base within the middle
third, or q = 3 ft.
APPLICATIONS. 67
Analytical Solution.
Equation (32), GO = 90 - cp = 56 20'.
Equation (34),
~ .13917 + .55436 X .90133
tan < + *> = .831*8 x .90133 - = - 8517 '
.- . a + 6 = 40 25' and 6 = 32 25'.
.90133 (20) 2
Equation (20), - ~- ^ = * = 23600 Ibs.
The wall is proportioned by graphical method above.
These five examples illustrate the method of using the
graphical constructions and the formulae. The graphical
method seems to be the easier, and is fully accurate enough
for all practical purposes.
It has been noticed, no doubt, that cp has been taken
throughout, excepting in the first example, to equal 33 40',
or equivalent to a slope of about l to 1. Mr. Trau twine
says, on p. 690 of his " Engineer's Pocket-book/' 1885 : "For
practical purposes, ive may say that dry sand, gravel, and
earths slope at 30 41' or 1|- to 1, as abundant experience on
railroad embankments proves." This statement is reason-
able, and for the majority of earths the angle is too small ;
hence walls proportioned for cp = 33 40' will be on the
safe side.
In all that precedes it is supposed that there is no fric-
tion between the earth and the wall, or, in other words, tf
does not depend upon the structure of the wall for its value
in any respect.
58 THEORY OF THE RETAINING- WALL.
Now, it is plain that as soon as any movement of the
wall takes place the friction existing between the wall and
the earth has been overcome ; or if a coating of earth
sticks to the wall, as is usual, the friction overcome is that
of earth on earth ; if cp' represents the coefficient of fric-
tion of earth and walls, then the direction of E must make
an angle with the normal to the back face of the wall
equal to at least cp'. To introduce cp' into Professor Wey-
rauch's theory it is only necessary to find the value of d as
given by his formulae, and see if it is greater or less than
cp'; if it is less, use the value of
Average
Ibs. per
cu. ft.
Alabaster
4-1
144
Brick best pressed
40 to 300
R
150
" conirnon linrd . .
125
" soft inferior
100
Chalk
20 to 30
S
156
49 6 to 102
Flint
162
g
166
Granite
300 to 1200
&3
& C
170
Gneiss
^
168
Greenstone trnp
fe
187
Hornblende black
203
Limestones and Marbles, ordinary ....
Mortar hardened . .
250 to 1000
8|
a m
O> OJ
j 104.4
[168
103
Quartz common
!g
165
Sandstone
150 to 550
II
151
Shales .
3 QQ
162
Slate
400 to 800
175
Soapstone
H-
170
VALUES OF y AND +a, +. + ) 3 cos ( + a.)
+
which reduces to
o __ cos (