145
L 405
THS

A DESIGN AND COST COMPARISON
OF RETAINING WALLS FOR COAL
STORAGE AT THE NEW MICHIGAN

STATE COLLEGE POWER PLANT

Thuisforthobcgnoofls.
MICHIGAN STATE COLLEGE
D. J. Modes-D. N. Worfol
1948

THESIS

[ll-I.) [III

 

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PLACE IN RETURN Box to remove this checkout from your record.
To AVOID FINES return on or before date due.
MAY BE RECALLED with earlier due date if requested.

 

 

A Design and Cost Comparison of
Retaining'Walls for Coal Storage at
The New Michigan State College

Fewer Plant

A Thesis Submitted to
The Faculty of
MICE. GAR STATE COLLEGE
of

AGRICULTURE AND APPLIED SCIENCE

by

D. J. Morfee D. m.‘Worfel
M

Candidates for the Degree of

Bachelor of Science

Jane 19h8

I’Hfib‘lfi

c,”l

PRE FACE}

333318113

While the primary purpose of this thesis is to fulfill the require-
ments of the Michigan State College school of Engineering, it is also
presented as a practical solution to an existing problem.

The authors have endeavored to include sufficient computations and
working drawings to give the reader a clear concept of the procedures
involved.

Investigation was made of unit weight, allowable soil pressures,
angle of friction under various conditions of saturation and height of
the filling material (coal), etc. Much discussion with persons familiar
with the problem was necessary to more successfully and accurately produce
a practical solution.

The authors wish to express their appreciation to those who have aided
in this work. They stand fully indebted to Mr.'William Bradley, faculty
advisor and Evelyn Morfee, typist, who gave freely of their ideas and time.

June l9h8 D.J.M.
D.N.W}

CONTENTS

Page
Introduction . . . . . . . . . . . . . . . . 1
Maps . . . . . . . . . . . . . . . . . . . . a
Volume Requirements of the Enclosure . . . . 5

A Study of the Possibilities of Using
a Cantilever Back Wall Design . . . . . . o 9

Actual Design 0 o o o o o o o o o o o o o o o 23

A Study of the Possibilities of Using
a Counterfort Back‘Wall Design . . . . . . j8

Front wall Design 0 o o o o o o o o o o o o o 57
Side wall Design 0 o o o o o o o o o o o o o b“

008% Comparison 0 o o o o o o o o o o o o o o 76

INTRODUCTION

Due to the large increases in enrollment and the huge building
program at Michigan State College since l9h5, it was necessary for the
college to build a new Power Plant. This plant is located south of the
Red Cedar River and is to eventually substitute completely the existing
plant.

In line with the expansion program, it seems reasonable to assume
that all present planning should be done with utilization of space as
the primary goal.

The storage of coal for the plant would consume a large area if
merely piled on the ground with out the use of retaining walls. The
coal would be spread out and would be carried to the h0ppers from the
pile with a great deal of inefficiency. An enclosure would also add
beauty to the storage area.

With these aspects of the problem clearly in mind the required
size is the next consideration.

As previously stated, the new plant will substitute the existing
one. It will Operate three boilers and have a maximum capacity of 360
tons per day. In this thesis, 270 tons per day and h2 days supply are
the basis for design -- 270 being the maximum expected demand.

'We will treat the back wall as a cantilever and then as a counter-
fort wall. The front wall will be semi-gravity and the side wall will
be of three elevations. The front twenty feet of the side wall on the
north side will be semi-gravity and of the same design as the front wall.
The next twenty feet will be cantilever and the back twenty feet will be
cantilever, and of the same design as the cantilever back wall. The side
wall on the south side has only the front eighteen feet, the same design

as the front wall. The rest of that side is the same as the north side.

Counterforts with steel angles on the edges will be used on the
back wall. It is assumed that the shovel will not do damage with this
precaution taken because it will be traveling vertically at sixty feet
from the cab. Counterforts on the side wall, however, will not be used.

Expansion joints will be used at fifty foot intervals on the front
and back walls. There will be no expansion joints in the side wall.

In determining the required capacity of the enclosure, it was found
that many small piles of coal present less fire hazard than a few large
piles. With large piles the fine and coarse particles separate and pro-
duce air channels and drafts which aid fire. It is suggeSted that the
piles be kept of fairly equal height.

A ten foot portion of the wall Should be omitted to permit the
passage of a bulldozer. The bulldozer will allow a mixing of the coal
when deemed necessary.

The back wall was designed at a height of twenty feet, overall, in
both cantilever and counterfort design to allow a good cost comparison.
The side and front walls will cost the same because their design is the
same for both cases.

As a final pre-design consideration, let us trace the paths that the
coal takes as it goes to the boilers.

It is brought into the area along the north-south tracks, east of
the stadium and then onto the tracks close to the west side of the plant.
As mucn coal as possible is dumped in a pit at the northdwest corner of
the plant and it is carried to overhead heppers by a bucket escalator.
The surplus is put in the storage area by a crane to be carried back to

the pit when needed.

 

 

 
 

 

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VOU‘TFE RE QU IPEPUS N '1‘ S
of the

ENCLOSURE

In order to design a wall of sufficient strength, economical and
still fill the desired requirements several factors had to be considered.
First of all the volume was taken into consideration. The consumption
of coal was placed at approximately 275 tons per day and a b2 day storage
0 was required. It was found that 56 pounds per cubic foot was the best
average value of the unit weight of bituminous coal.

In the book, "Walls, Bins, and Grain Elevators" by'Milo S. Ketcnum,
an average value for the angle of repose was assumed as 35°. The volume

then equals 275 x 2000 x A2 : L12,500 cubic feet of coal.
56

 

The cross sectional views are found for the 57 foot and the 58 f00t
widths on the following page. Also in this thesis are shown location
and plan views which may be referred to.

The computations for the volumes were as follows:

IA. Cross Section Area for 58' Width.

(1) 12.3 x 25 2307.5 Sq. Ft.

(2) 19.5 x 15.2 = 296.11

(3) 15 x 21 ' 2315.0

(b) 30 x h ==120.0
W. Ft.

Excav. 3' x 57 22171.0

 

1,209.9 sq. Ft.

B. Cross Section Area for a 57' Width.

(1) 11.3 x 25 = 282.5 Sq. Ft.
(2) 19.5 x 15.2 =29o.)4
(3) 15 x 21 2315.0
(11) 30 x L; :2 120.0

 

T301309 Sq. Ft.

 

 

 

 

 

 

 

 

 

 

 

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F/F 7)/ ~ £7009 7 F00 7‘ 5567/07

F/F7V~J[Véw F007~ 55.6770”

 

 

 

 

 

 

Computations continued:

1,013.9 sq. Ft.

Excav. 3' X 56 2 168.0
1,181.9 Sq. Ft.

 

Volumes:
A. 1,209.9 x 300 : 362,970

Bo 1,181.9 x 100 ::118,190
h81,160 Cu. Ft.

 

This gives a value of 68,660 cubic feet margin which is sufficient
to allow for angle of repose (35°) on each end.

It was found that a wall 20 feet high in back, 10 feet high in
front, and 3 feet of excavation throughout would give the required
volume. Allowance was made for the loss in volume due to the slope
of the side wall. It was determined that about 25 feet above track
level was the maximum convenient piling level. This factor limited
the volume.

The wall is far enough below frost line, throughout, to be free

from frost heave.

A STUDY OF THE POSSIBILITIES
OF USING A

CANT I IL‘VER BACK WALL Db S I GN

10

INTRODUCTION

 

After the height was determined to give sufficient volume, the
type of design was the next consideration. The height of 20 feet
eliminated the gravity wall because experience showed that greater
cost was incurred. That left the other two main types, the cantilever
and the counterfort. The first design was the cantilever and the second
design was the counterfort.

Retaining walls have been designed for many, many years by many,
many people with the result that there are several theories too numerous
to mention. 'Used in this thesis were theories of Sutherland and Reese
as found in their book, "Reinforced Concrete Design" with variations
as noted. We attempted to use our own logical thinking to determine
methods on controversial subjects.

From the "Joint Committee - Concrete and Reinforced Concrete" -

l9h0, the following values were taken:

Ultimate concrete stress ::'fc':: 2000 p.s.i.
Allowable concrete stress : fc Z. 800 p.s.i.= .14 fc'
Steel stress :: £8: 18000 p.s.i.
Ratio E3 : n: 15
Ec
'R-_- 139
Unit Shear ::'v:: .02 fc'
Unit bond stress 2211:: 100 p.s.i. (less than 150 if anchored)

A soil report for the neW'power plant was found in the Michigan
State College engineering office and with this in mind and consideration
for the depth required a unit force of somewhere between h000 and h250

pounds per square foot seemed like a good value for allowable soil pressure.

11

It should be noted that if this construction is contemplated a complete
soil investigation should be carried out.

With the height in mind the wall was proportioned accordingly.
Several dimensions were tried and only the final results will be found
in the computations.

Three cases were decided upon, the worst of which form the basis
for our calculations. Case one involved Rankine's design using the
inert prism and omitting back wall friction.

Coulomb's theory with the Culmann construction considering an
inert wedge with full back wall friction was the second case taken
into consideration.

The third case was the same as the second, but an inert prism
was used in place of an inert wedge.

It may be noted that there are, perhaps, more logical possibilities,

but the three cases used should give approximately the same results.

 

 

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13

CASE I

 

With reference to the preceeding drawing the following compu-
tations can be easily followed. It can be seen that the resultant
of the earth pressure acted in the middle one-third and that the soil
pressures were within bounds of the assumed value. The safety factor
against sliding was well below the required safety factor of two;
therefore, a cut-off wall on the base was required.

Computations:

 

=cos e costs/c0826 - cosem

cos -9-:t\/cos?—e- .. (”32¢

 

 

c=.819

p bottom: th
==.819 x 56 x 2h.9
==l,lh2 p.s.f.

P=l/2 x 214.9 x 11142: 114,218#

v=8,161#

Ph=11.6LIS#

x=l/3 x 2n.9= 8.30'

Base Pressure: Moment About Toe.

W1: 11 x 2 x 150 = 3.300% x 5.5! =18,150'#
W2: 1 x 18 x 150 = 2,700 x 3.5 = 9,LI50
W3:- 1/2 x 18 x 150 = 1,350 x 14.31, = 5,859
Wu: 1/2 x 18 x 56 = 50h x 14.57 22,351,
W5 2 6 x 18 x 56 =6,0h8 x 8.00 =u8.38h
We: 7 1 11-9 x 1/2 x 56 = 960 x 8.67 = 8,323

 

 

111,862# 925.20%

Computations continued:

114.8627? 92.520'#

 

 

 

13"; 8,161 x 11 = 89,771
23,023# 182.291'#
Ph(d): - 116145 3: 8.3 = ~96.65h

’ 85.637'#

x: 85637 = 3072'
25023

5050 "' 3072 = 1078' .1: e

.05 within middle one-third

I=§( figs)
~ 23023 1i 6 x 1.78

= 2093 (11.97)
f toe: £1123 p.s.f.

f heel 2 63 Pas-f.
g. 23023 x .LI99
S.F.-—. 1161.15 .

.99 Therefore must anchor.

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CASE II

Again the computations are easily followed with reference to the
preceding drawing of the Culmann construction. The results again
show that the resultant of the earth pressure fell within the middle
one-third. This is an important factor because when the resultant
acts at a point outside the middle one-third the cost of construction
is greatly increased. This time the soil pressure at the toe was
well below the maximum and the safety factor against sliding was higher
but still required the cut off wall.

Base Pressure: Moment About Toe.

 

 

 

 

 

W1 = 11 x 2 x 150 23300-54 x 5.5 =18,150'#
W2 = 1 x 18 x 150 22,700 x 3.5 2 9,150
N; = 1/2 x 18 x 150 =1,350 x 14.314 2 5,859
W7 = 1/2 x 18 x 5.3 x 56 5.2.671 x 6L8 =17.175

10,021# 50.63hh#
Pv: 7:14-10 I 8009 25999147

17:731# 1100581”?
Ph(d): .- 51.,53 x 8.3 = 45,260

65.32109L

..65321 6
x—-__._._._ 2 0 8b
17731 3

5050 - 3068: 1082': 9
.017 within middle one-third
P 6e
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f heel==1612 x .01 == 16 p.s.f.

Computations continued:

3.33: 17731 x 4'99
5&53

 

:: 1.6 Therefore must anchor.

17

 

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CASE III

19

Case three was constructed similar to Case two with the exception

that the inert wedge was replaced by the inert prism.

This time the

resultant on the base moved closer to the center giving a more even

earth pressure which, of course, was well below the h250 pounds per

square foot maximum. The sliding factor was higher than the previous

two cases but still not the required two.
Computations:

Base Pressure: Moment About Toe.

 

 

W1 :11 x 2 x 150 23,3001 3:
W2 =1 3: 18 x 150 =2.7oo x
113 :1/2 x 18 x 150 =1,350 x
Wu 21/2 x 18 x 56 = 501. x
115 =6 x 18 x 56 =6,0h8 x
W6=7xh.9xl/2x56=__960M x
114.8621E
P. = 3.559 x
18.h21#

5-5'
3-5

h-Bh
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8.00

8.67

11

Ph(d):: - 5078 x 8.3

289522: .86'
x TB'IIE'I ‘*

5.50 - 11.86:: .6119: e

1.19 within middle one-third

(us—2)
22111.21 (1:.35)

f 1306 : 1675 X 1.35: 2261 p.s.f.

f heel==l675 x .65===1089 p.s.f.

: 18,150'#
= 9.1150
=3 5.859
=2.35u

2148:3814
=2 8,323

 

92.52%:

=39. U49
131,669*#

 

:— “Ll-291147

89,522 '#

Computations continued:

S-F._l%h21 ”9:999 _. 1.81
5078

 

Therefore must anchor.

20

21

SUI? “AR Y

It was necessary to design the wall for the worst case in order
to be assured that the wall would not fail. From the chart that
follows it can be noted immediately that Case one was by far the
worst case. The sliding factor was very low, high toe pressure and
a large horizontal component of thrust were all indications that

Case one should be designed.

22

 

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25

Next, was a closer study of the cantilever wall. If the dimen-
sions that have been assumed are correct the design is a simple
process. In this design the authors followed through the computations
numerous times until the following computations were derived with the
dimensions as noted on the previous page. The wall design was in
sections: toe, heel, cut-off, etc., always correlating the results
to assure uniformity.

Again, it can be said that the computations contain no involved
mathematics and can be easily followed by merely referring to the

drawings of the wall of Case one.

26

HEEL DESIGN

 

The heel was the first consideration. From the book, "Concrete,
Plain and Reinforced", Vol. 1, by Taylor, SmulsKi and Thompson, it
was noted that the entire pressure triangle is considered and that
'the force acts through the centroid of the triangle, parallel to the
surface and thus to the wall where its components are taken. This
reasoning seemed very practical and was used in our designs of the
cantilever walls.

Shear and moments were taken at the section noted and the results
used to compute the depths. The formulas used are those accepted by
the 19h0 "Joint Committee on Concrete and Reinforced Concrete".

The results obtained were reasonable and conformed to our already
accepted values and dimensions. note the fact that special anchorage
on the reinforcing bars was not required.

Computations:

 

x :11L12 x 18'7 x .5714: 1492-511: S .Ft.
1 211.9 /q.

x __11h2 I 22'9 x . 7 =2602. S .Ft.
2_ 211.9 EA 9#/q

Psz‘ge'3 2 602‘9 x 6:.- 3285.6#

(192.3 2 602.9

 

 

__6
3“? 1095.2 )= 3'“

27

C omputati ons continued :

iaxinum Shear and Moment on Section 11-11.

 

 

 

 

Pvz- 3,285.6# x 3.1' =-lO,185'—,1f

W5=- 6,0148 x 3.0 =-18,1L;.L;

W6=' 960 x [4.0 =- 3,8110
-10.293-5# “32.169'#

Earth? —_— 5,501.0 x 1.79 2+9,h89
V — u,992.6# M =22,680*#

Therefore, approximately 16" is required but
keep base 211" due to expected large shear on toe.

V = £19326

 

V: = o o 0..

Kid W 22 6 P S 1

no special anchorage required.

R = M ___ 22680 2 51 <59

bd2 212

Therefore fc is very low.
A =M 22680 069 ° '

5 _—_- __ _-_-_ o SQOIHQ per 111.

Use 3/1; inch square bars @ 6" c/c.
Gives .073 sq. inch.

Vb 22.6 x b ‘ ‘
u: .__= :2 1.15.2 P9301. <100 p.301.
20 3
no special anchorage required.

its». 11.4.1.1...) -J

L.

28

TOE DESIGN

By taking a section B-B the toe was very readily attacked and
in much the same manner the values were obtained. The earth above
the toe was not considered, thus giving a larger shear and a safer
design.

The values obtained are of good nature and conform again to the
accepted values. One difference may be noted in the fact that the
bars are anchored.

Computations:

P:

 

( 3823 2 2716) =9809#

x: 3 (2716 3823 x 2)
3 6559

\N

I: 10362 = 1.59,

"3579
Maximum Shear and Moment on Section B-B.
'V== 9809#
M: 9809 x 1-59 =15.596'#
v _ 9809

= 15 o 6" OK

 

:v-fi-b__ 60x.875xl2

The steel must be anchored.

...—g 41;? =

v: _\_r___ 9809
bjd l2x.875x21

=LIAOS p.801.

 

Therefore special anchorage is required.

 

15596 Z: 35 (f0 is very low)

29
Computations continued:
A8: L=i5596 : .0147 sq.in. per in. Bottom
fsjd 18000x.875x2l
Use 5/8" square bars @ 6-1/2" c/c.
Gives .0h77 sq. inch.

11: £935 3: 605 —_— 97 13.3.1. <100
5

30
CUT-OFF'WALL (Or Anchorage)

 

The purpose of the cut-off wall was to bring the factor of
safety against sliding up to a universally accepted value of two.
There are several places that the cut-off wall could have been
located, one of which is locating it under the stem. It was located
four feet from the toe, giving the following calculations and results.

Computations:

C= 1 + 83:.11 350 = 3.7
l - Sln 55°

 

@ C-C, wh= 261.17 p.s.f.
Possible resistance 2: 3.7 x 26h7==»979h p.s.f.
To make sliding factor 2:

R== 2 x 116u5 - .h99 x 23023

R = 11,802# required.

 

min. h== 11802 ==1,2 Make h ==l'-3" for convenience.
9791:
V = 11,8027]:

L1: 7.5 X 11802: 88,515 lboino

11802 = 28.1..

 

d=V=
ij
Use 2'-6" for convenience.
If no tension reinforcement:

12x60

31

STEM DESIGN

 

By using the same line of reasoning as used in the heel design,
the stem was designed. The assumed widthsof 12" at the top and 2h"
at the bottom were confirmed, leaving only the steel left to be found.
The stem schedule that follows the calculations was attained in
order to find the necessary steel at each three feet of the stem height.
By'using these values the curve was plotted and by the use of same,
it was then easily found where the steel could be cut off and still
give the desired strength.
Computations:
P: 1/2 x 857.7 x 18.7 = 8,0195%
V: 8019.5 x .819: 6,568#

Moment at Bottom of Stem.
M: 6558XI8‘7XI2 ___ 11,91,286 in. lbs.
3

d 17.. 6568

_
n

ij l2x.875xh0

d __ :_' 1491286 2 17.2"
VRb V 139x12

Make stem 12" @ top and 2h" @ bottom.

15.6"

 

-—-
~

ll

 

I

9 any leve l:

.. 2
v}:— 22.93 h

11,: = 7.6L; 113

52

Computations continued:

 

. I h —'
Embedment _~_ L=fS D: 18000 145"
Ht? [0:100

Hook Bottom

Temperature Steel:
(.002 of concrete area)

.002 x 12 x 18;: .hh Sq.in. per ft. of height.

{1/2" 00 9" c/c} + {1/2" 0 0 12" 0/0} = .116 Sq. in.

Front Back

Key:

A general formla of 2 l
2 [A

could be used here. Generally, keys are designed up to 9" and this,

=;K f'c where K ranges from .02 to .12

with the steel and concrete friction, take the shear present.

33

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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women. an been .XoN \on Rx 0.8%. ..Q
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hummus, Much unfit. $8M pk no \,\\ .h\ k toms .0
putts. is @108» it XQ but .0
0\\QQ. \\ 09% x\ X\\ DON in,
XII. MQQQX . eX ,mX 3\ \N h \_ a

 

 

N V\» QLuxxbh \ka m,

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

//[/6//7

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5'

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CU777N€ 0/7.
57EfL

 

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34

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?

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A 6.54 5 725222 {2‘ ’7' J

 

 

 

55
JOINTS

Coefficient expansion for reinforced concrete equals .000006
I
for 1° F. Therefore, maximum eXpected expansion for a 50 ft. length
equals .21". (Assume 60° F. change of temperature). In this

design a .5" expansion is allowed.

 

 

At each joint use 1" round

steel dowels encased in ‘;\\
l"¢ 2'-O“ dowels @ 18" c/c.

tubes to hold wall in align-

 

 

me nt .

Space dowels as shown.

 

Encase dowels in 1-1/8" round

tUbOSo X
___..,_.__1/2" joint @ 501-0" c/c.
Joint will be of l/'" pre- Premolded Bit. Felt.

 

molded bituminous felt.

 

 

 

 

 

 

 

 

'7

Expansion Joint - Construction Suggestions:
1. St0p steel at joint form.
2. Place dowell tubes before concrete hardens.
3. Copper plating can be used to prevent water flow through

joint. This will do away with wall staining from bituminous
material.

 

‘fiu 5.12:: at

 

 

36

“ CONCIUSION

 

 

The finished cantilever wall shown on the following page was
properly designed and is safe against destruction under the condi-
tions previously mentioned. The economy is good.

Before definite acceptance or rejection is in order the whole

picture must be drawn. Thus, we move on to the other conditions.

 

F
I

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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m5 J ‘y" 6 @62C(

 

 

 

 

 

 

 

 

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1
z

4’ ll ‘

._ i _6 g

7 ’16”: ‘ , ‘ ‘7 575M 57222
// . - ...; PLAC/A/G

 

 

..
—_..-.___._... -__.____._..__..._..-. . .._._..__ ._

5041.5 / ’4 476 ”

37

...“ - ... _~.___~__-_—A~m——__.. __ “...—.4» 7 ...—-— 4———__.———1

CANT/LEVER WALL

 

 

 

 

 

STUDY OF TfiE POSSIBILITIES

OF USING A

COUNTERFORT BACKWALL

DESIGN

39

 

‘3 3.33333 3333 333

 

33:.13i33‘llfp
ll ill“!
i‘33ll‘313
‘llllltll

 

b ‘3

\.

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”“3"

 

 

 

 

 

 

 

 

 

 

 

 

 

33 33 37,. , AM,
3 3 3 3 / W32 3. 33. ,
-3 3-3 33 C M I, 2,, ... ,.
3\ 3, _ P // 2
1&33 3333 3.3% W313 3 33 33 3 3 3 OWN 332.4 3 / 52333-3 3 M39 3 2,,
/,.,Alx. Mo! WNW-1333 3.333 V: C/ Pr ./ . 3
((333.33 V3331! /@ 2 m
.32... W W4 / , _
/. a C, 33- 7.
/ . j -33\33.3\\ \ j /,. W
/~ 3/3J \\ W333: . a v,
3234. ,4 3V
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WWW WZ-|33III 3333333333 ,/_v
Ir _ .3
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C W 3 j _
A 0 W : #3331
5 C .13. .V. _ 3 N
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JC/ME/’=4’

 

 

 

 

CC!

3.3L.

3»! u

ho

With only slight variation the preliminary investigation of
the counterfort wall was the same as for the cantilever wall.
Sliding, overturning, and excessive soil pressures would cause
failure as before, but also it was determined that failure could
occur through tearing of the counterfort from the stem, tearing of
the counterfort from the base, and bending of the stem both parallel
and perpendicular to the base.

The wall was made 20 feet high as before (see diagram). This
repitition allowed for a better cost comparison. The base was again
placed far enough below ground to defeat frost action. The base was
made ll-l/é feet long, the stem was made 12 inches thick, counterforts
12 inches thick (1/20 of the height) and spaced at 10 feet. These all
conform to general practice.

To determine whether the wall would fail as a unit, Rankine's
theories were used. To be consistent with the previous design, varia-
tions as set forth by Taylor, Smulski and Thompson were employed where
necessary.

The resultant of the pressures on the base fell within the middle
one-third (required in retaining wall design) and the soil pressure at
the toe was below the determined maximum of h,250 p.s.f. The sliding
factor of safety did not equal two and thus anchorage was required.

Expansion joints were employed as before.

Again, if actual construction is contemplated (to the present
knowledge of the authors it is not) more thorough study of the subsoil

should be conducted.

Computations:

C = .819

lamp: .819 x 31.9 x

56 = 221407 poSofo

pbOt‘: .819 I 21.109 x 56 = 1;].14200 p.s.f.

 

 

P = 1/2 x 233.9 x 11332 = lh,218#
Pb: .819 I 11218 = 11,6b5#
P3,: .5713 3: 11218 = 8,161#
x = 1/3 x 2L3. 28.30'
Base Pressure: Moment About Toe:
W1: 11.5 x 2 x 150 x 10 = 331,500.93 x 5.75'
We: 1 'x 18 x 150 x 10 2 27,000 x 13.00
W3 = 3.5 x 18 x 150 x l = 9.1.150 x 6.8314
“54:: 7 x 18 x 563x 9 :: 63,50h x 8.000
W5: 3.5 x 18 x 56 x 1 = 3,528 x 9.167
W6: 3.5 x 13.9 x 56 x 10 = 9.6033 x 9.107
1H7.586#
Pv x 10 :: 81,610 x 11.5
229,196#

Ph(d)= -116135 J: 10 x 8.3

x: mm?” = 13-2131

5075 - 14.214. 3:. 1051' = e
.141 within middle one—third

L11

= 198.575 “13*
:1 108,000
= 633,581
= 508,032

: 32,3141

== 88,0h0
999:569'#

 

= 938,515
1.937.8813'#

:2 " 9661535

 

971.314???

. I ’1‘“. ‘1

“cu—rim“ Lacy.-
'I

Computations continued:

(we

1.1053(10 1105

r = 1993 (11.788)
f toe = 35633 p.s.f.

f heel 2..- )423 p.s.f.

Overturning Factor of Safety:

19378813
F080 2 :
995535 2

Sliding Factor of Safety:

F.S. : 229196 at .1199

. .98
1181350

 

Therefore must anchor.

L33

VERTICLE SLAB

 

As explained in the cantilever backwall design, the unit pressure
on any section of the wall caused by the backfill was found by the
pressure triangle. The pressure is represented by a force acting through
the centroid of the triangle parallel to the backfill surface at a point
common to the backwall and the line of action of the force.

The horizontal component of this unit pressure equals w and the
moment designed for at any interior panel is given by the formula,

M== 1/12 wL2. For a panel bordering an expansion joint, or an end
panel M== 1/10 WL2. L is the clear span. To be consiStent the wall
was designed for its entire length as an exterior panel, thus given
additional strength.

By correlating the results obtained in the following chart the
steel design was accomplished.

Computations:

M _—= 1/10 wL2

133 = 1/10 x 676.1 x : 51475.1;311 or 65,717*‘#
V: £32 676.: x 9 .-= 3.03.2333

d_ V - 30132 :7.2"

 

“ 7333 z horn-875x12

A __ 133 _ 65717

.. __ _-_: .L151 sq.in. per ft. Assumed ‘:.9
8 fsjd 18000x.9x9 J

Use 5/8 inch round bars @ 8" c/c.
Gives .h65 sq. inches.

Computations continued;
......" 30132 .
u._ __ -_ . . . .
_ 2:,Jd _. Tr x.625xl.5x.9x9 — 127 5 p s 1

Therefore must anchor.

Check j:

 

9”

 

 

x (12);;9 - x («65 (15) I

6x2: 9 - 6.9753:

x2 + 1.1631: = 9
(“5815)") -- 9-338
X+o5815 = 3.055
x=2.337

i=9 " 1/3; 2'L‘7 = .909

OK Assumed 09

 

 

 

 

 

 

135

 

 

 

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wk M ibfihfimfifi. Quake. 3,
m a j 3303-33 .. ,c
3 /
tacit u w .

 

 

 

 

136
HEEL SLAB

 

The moments and shear on the heel were found by the same formula
as used for the verticle wall. The shear at the extremity of the heel
and at the back wall are different and were found to be of Opposite
direction. Note that there is a point where this shear is zero. From
the graph it was found that the steel should be placed as shown by the
diagram that follows. The heel design was conducted as shown in the
computations.

Computations:

One foot on right end of heel:

M: 1/10 wL2
M = 1/10 x 1616 x 92:: 13.09013: or 157.080 "373-

V= g5: T1616 " 9 = 7.272%

3 157080 "
”fist: 34:5;‘=9'7

V 2 2
a: W=W7 7" 5x = 17.3" 0K

335 _ 157080
1:38“ 18000x.9x21

 

AS: 2.1.162 sqoino per ft.

Assumed j: .9

Use 1/2 inch square bars @ 6-1/2" c/c.
Gives .hoE sq. inches.

One foot on left end of heel:

3331 = 1/10 wL2

M = 1/10 x L305 1! 92 _-= 3,281?# or 39,572"#

L 0 .
v=§. = 9—321—9— = 1,8254%

Computations continued:

__ ' __ 39372 _
8172—-”

V __ 1823 __ n
d: Bat-W * “'3

5“ ngd'-I§OOOX.9KEI .—

Assumed j: .9

Use 1/2 inch square bars @ 25" c/c.
Gives .120 sq. inches.

Check 3:

x (12);.—_21 - x (.1361 (15) )
6x2: 21 - .6915x

x2+ 1.153 = 3.5

(x+-5765)2= 3.5 .3. .3332
I +05765 =3 1096
X :2 0,461

3:21-0h61 = .97
~ OK Assumed .9

u V 7272

.=' 2 2: 10802 c 0.0
21:33 2x1.8146x.875x2l p s 1

 

Therefore must anchor.

—-.1157 sq.in. per ft.

137

 

5755A ABE/J JQ/A/

 

HEEL

35 7&2 Z- P; /4 ([fl/lf/V 7

u T 3 3 3 ‘ 3 “‘ T" ‘ ’m‘
f 3 I ‘ i 3 3
1 i 3 ‘3 f 3 ‘
3 3 3 3 1 3 4‘92
f ' 3 3 i i
3 3 '
3

g .
i i / 3
3 /
3 ', /
3' 3
3 3 /’ .
3 3 3 / 7 ' .
J f ,’I
3 / 3
E 4 3 /I
Z : 7 ‘ i x, . 20 7 1
3 1 3 ~
. 3 /

‘ 3 g
: 3 , - f3
3 I 2 | 1

\‘\ : // ' g
\)\-\04‘// 6‘53“ 3
0 \/ i

2’ ‘5’ A4/ ‘5.’ 3 6r ___- ”’7’
0/57/4/1/6 A“ f't’O/W 152406 0F 344411

 

3. 3
O /’
l
3

 

 

139

TOE SLAB

The shear and moment on the toe were figured neglecting the effect
of the earth above the toe. This earth will not be present during con—
struction and therefore was not taken into account. Note that special
anchorage of the steel was required. In.the base design a value of j
was assumed and then was checked and found to be all right.

Computations:

v =(3263‘ 2 2307) 3.5 = 9.71393E

 

M 2 9719 x 1.85 = 18.036'#

d: €33: mic—ZEXBO _: 15.5" Therefore mst anchor.

. '8'038 ,,

133 18 036

As: T333 =18000x.§::21

z: .053 sq.in. per in.

 

Assumed j:- .9

Use 5/8 inch round bars @,5-1/2" c/c.
Gives .056u sq. inches.

V"

_ 97m» _ .
““fl=1m3m21 - 12“ P's'l‘

Therefore must anchor.

Check 3:

x (12):}:21 - x (.670 (15))
6x2: 21 - 10.1132:

x2+ 1.69x:3.5

(I+-8135)2= 3.5+7-1L3

1: #30313

_ 21 "' 0340314 _.
3.. 21d __.98 OK Assumed .9

50
COUNTERFORTS

 

The tendency of the counterfort to tear was the next consideration.
This tendency was checked by steel bars vertically placed in the counter-
fort 3 inches from the back and running from the top of the wall to the
back of the heel. These bars were anchored where necessary. The d used
in figuring the area of the steel was measured from the front of the
wall at the base.

Similar computations were carried out at various sections and by
correlating the results into the graph shown, the required steel at any
point and the steel cut-off was determined.

It was also necessary to add steel to tie the counterfort to the
verticle and to the heel slabs. This was done as shown in the computations
taking care of the embeddedment required.

To assist in.the protection of the counterforts against the coal
scoop it was decided that u" x h" x 1/h x l9'-3" angles (or their
equivalent) should be placed on.the back of the counterforts.

Computations:

= 1/2 x 676 x 18 x 10: 60,8h0#

M: 1/3 x 18 3: 608140: 565,0ho'3e

dz/‘T =/3650340= —-52.2" OK 86.3" available.

A_ M 3650130 :3: 12
8:" fsjd=18000x.9x86.5

 

= 3012 sq. in.

Use 1-1 inch round bar
and h-7/B inch round bars, spaced as shown on final drawing.

Gives 3.185 sq. inches.

51
Computati ons cont inu ed :
This spacing decreases d by a small amount. Excess of As 13
sufficient, however, to take care of this. This is shown below. (Also

see final drawing of counterfort wall).

 

. _1.98515+l.2x5.7 __ . n . _ . =8."
New d _ 3.185 _ 34 017 89 5 L3 02 5 5
New A r 3650340 .: 3.165 sq.in.

 

V 608130 .
u== r = h = 6 o o S. 10K
:0 JG xnwc.875xr)5.5 L‘ 7 p

Steel Cut-Off (6' from top of base):

v= 12 1 1° " “51 = 27.0603é
2

m= 27060 x u= lOB,2L.0'#

(1.11/33: /1082—To_ = 28" OK 60.14" available.

15 __ 1082h0 x 12

As: ”83 —TW . . = 1.33 sq.in.

Use 2 bars 1 inch round, spaced at 6" c/c.
Gives 1.57 sq. inches.

u_ V 27060
‘ 3'3: 23: . 875x60mx TT"

230

 

: 8105 p.301. OK

Steel Cut-Off (12' from top of base):
V='1/2 x 275 x 6 x lo= 8.250%
1.1: 8250 x 2: 16,5oor-5é

d: 3; =ngggg= 10.8" 0K 3h.3" available.

16500 x 12

A- ._
S fmsjd:18000x.9xoO.L3

 

 

: .2 8Qoin.

 

COO/17217“ 63(97- 5/2257.

  
 
  

 

8‘52“” (4’ 6‘ ”(#9

 

 

~ ¢ ,, l
2 g7 @ 6 745 ;‘/~/ ‘éo (fix/229?

 

 

 

57554 AEEA {.747 m//

 

 

Jffé'é 7)"//V6‘ (7/:- 70 5/455
555 (“CA/PU 7/4 7/0/1/5.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

- A
~ . NV” ' 3"“?
3 "\\ 3.
x,“ \ ‘Q -—3 /' L—n
w \ 3V
30 ‘4“,
'\ v v
a 3x.-
w “it 3
3 (“t-L 3 “3“?
N! 3 E 2L“ +2 J
A
// If” A ‘A

 

 

53'

Computations continued:
Steel Tying Vertical Slab to Counterfort:

V: 676 x 9: 6,0931%

A __ cosh
S“'I8005‘

Assume hook on every bar.

= .358 sq.in. per ft.

A: WEBBXX 8 = .113 sq.in.

Use 5/8 in.sq. bars.
Gives .114 sq. inches.

fsd __18000 x 3/8
liu — 11:: 100

Steel Tying Counterfort to Base:

L = 3: 1608"

 

 

At Heel:
1 r
A8: 1471-16 = .808 sq.in.
12 x .5
A: = 083
5-5
Use 2-1/2 in.sq. bars on every horizontal bar.
f d 18000 x 1/2 . n
L: S = =22.
Eu ’h_x‘100 5
Key:
5 V

Again the general formula 2. A: K f'c could be used. A 9 inch

key will be used to take the shear that is present.

5h

CUT-OFF WALL (or Arghorage)

 

 

Since the safety factor was too small (must equal two) a cut-off
wall was required (see drawing). This involved the passive pressure

given by the formula:

szh l+sin Vb)
I-Sln w

where wh equals the soil pressure at the face of the wall and Q equals
the internal angle of friction of the coal. The required dimensions

of the wall were found by the method shown in the computations.

Computations:

__ l+-sin ¢ __
PM Wh (m)—- 307 Wh

= 2607 X 3.7 = 9,6)46 p.s.f.
To make S.F.== 2
R =.— 2 x 10 x 11615 - .1199 2: 229196

R = 113.531.:

118531
“If???“ one

At front of'wall:

Min..h:: =2 1.22' ‘Use l'-3" for convenience.

V = 113,531:%

d_v _ 118531
'. ij ‘IZx.875x60

 

= 19"

55

SU 11.3.31}? Y

By the previous computations the assumed dimensions were checked
and found to be suitable. The back wall was designed cantilever and
now counterfort. A cost comparison could now be made but it would be
of little significance. It was decided that the front and side walls
should be included and then a cost comparison be computed. The

variables, however, have been designed and the rest will be constant

in the comparison.

 

 

COU/VTé'f/‘UPT WALL

50445 /”=4’

13
5’:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

CF 5755/.
in # ‘ \ I,
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FRONT WALL DE SI GR

(Semi-Gravity)

58

 

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FPO/V7" WALL

 

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59

To allow the shell to easily clear the front wall it was, of
necessity, designed so that the top of the front wall was well
below the crane. The wall, being ten feet high with four feet
above the ground, could have been designed either gravity or canti-
lever.

It was decided that a semi-gravity wall would be the most
practical solution. This wall must be stable in the same respects
as all other walls. Excessive soil pressures, sliding, overturning,
and shear or tearing of members would cause failure as before.

Expansion joints were placed every fifty feet as in the previous
designs. The only steel required in this semi-gravity wall was
that shown in the stem. This steel eliminates the tendency of the
stem to tear away from the base.

The toe and heel did not need investigation because of the
limited dimensions involved. The remainder of the design was
repitition of the foregoing design and it is believed that the

reader can follow the computations without further discussion.

Computations:

C —-= .819
ptop== .819 wh
=.819 x 56 x 1.9
= 871$E
pbot ==.819 x 56 x 11.9
= $116,415
P = 1/2 x 11.9 x 51.6 -—- 3,2119%
P11: .819 x 3219 = 2,6611%
Pv = 5711 x 32m = 1.865%E

Base Pressures: Moment About Toe:

W1: 5.5 x 2 x 150 =1,650;% x 2.75! =h,538'#
W2= 3 x 8 x 150 =3,600 x r 2.30 =8,280
W3: 1/2 x 5.5 x 8 x 56 = 7814 2: L163 =3.552
“1.: 1/2 x 2.75 x 1.93 x 56 = 1149 x [1.58 = 682
6,18% 17,052t#
Pv = 1,865 x 5.50 = 10,257
8.0148369 27.309'#
16,665'#
x: 16665: .
80h8 2 07'

2075 " 2.07: 068' = 6

.2h' within middle one-third

60

Computations continued:

Soil Pressures:

P 66
fZI (lit—5)

__801+8 (1+ 6 x .68)
‘3'}? ‘7‘"

f toezlh63 x 1.68 = 2,1158 p.s.f.

 

f heel=lh63 x 032 :3 I468 p.s.f.

Sliding Factor of Safety:

F030: 801.18 x 0,499 = 1.51
2661

Therefore must anchor.

 

 

Cut-Off
Passive P: 3.7 wh
@ face wh = 1,372 p.s.f.
Possible R: 1372 x 3.7:: 5,076%
To make S.F...—.. 2'
R=2 x 2661 - .1199 x 80148

:: 1,297?)E

Min. h— 1297 — 26' Flake h— 6" for o v '
... €673 - . . __ C n 9111911090

On face of cut-off:

M '2' 1297 I 3 = 3,891.59"

611 /6 x a 1 '
d" Ff = TZ—xiég" = 5.7" Make d = 9" for convenience.

S hear of Cut-off :

23.21191 -18 .
2A“2x12x9" p's°1‘

..4

61

Computations continued:

Stem Steel Design:

 

d: 113.8"
Shear at base of stem:
p: .819 x 56 x 9.11
=h31 p.s.f.
V=1/2 3: L131 3: 9.11
:=2,026#

M: 2026 x 2.7 = 5,).170 ftoleO 0r 65,6140 inoleo

 

 

d— PCB- : “139236137; =6.3" 113.8" available.
M 656110 .
A — = =
8—.fsjd 1800 .3 h5-8 .09 sq.in. per ft.

Use 3/8 inch round bars @ lu-l/ " c/c.

Bond:

v T: 2026. : 51407 P050210
'7 J 775—“. x maxim m“; .
Anchorage:

_ f d __ 18000::5/8 _ ,,

Anchor 1?" as shown in sketch.

‘1: OK

Temperature steel:

Use 1" sq. bars in the corners to prevent cracking.
This is the only steel required in the front wall.
This design will be used for the front 15' of the north side

wall and the front 13' on the south side wall.

 

 

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SIDE ”NAIL DESIGN

p—————.———

 

 

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66

The design of the side walls involved a decision as to the best
method to change the elevation of the wall above the ground from In feet
at the back to h feet at the front. A three elevation system was the
outcome. The north side (being 60 feet in length) has the front 20 feet
of semi-gravity design.the same as the front wall. The back section of
20 feet was made cantilever and the design was the same as the cantilever
back'wall.

The south side (being 58 feet in length) has a similar arrangement,
but the front section as described above was made only 18 feet.

It was the 20 foot long center section that had to be investigated.
The design was carried out according to Rankine's principles with varia—
tions as needed. Expansion joints were not used. The height was made
15 feet making a step, both up and down, of 5 feet. This height made
a gravity design uneconomical. Counterforts would give poor design
because of probable damage to them by the coal scoop. The only remaining
and economical solution was the cantilever wall, and, using constants
and knowledge gained from'the previous cantilever design, the side wall
design continued. The following is a discussion of the principles and

computations involved in each portion of the wall.

Computations:
p= .819 x 56 x 18.68 = 856.7 p.s.f.
P: 1/2 x 856.7 x 18.68 =8002§£
Ph = 6.551495é
Pv= 11.59319
x = 1/3 x 18.68 =6.25"

Base Pressures: Moments About Toe:

67

111: 8.5 x 1.5 x 150 =1,912.5# x n.25v =8,128'#
W2: 13.5 x 1.25 x 150 =2.531.3 x 2-88 ="7’29O
113 = 13.5 x 5.0 x 56 =3,780.0 x 6.0 =22,68o
W14=1/2 x 5.25 x 3.68 x 56 == 51.1.0 x 6.75 = 3,652
8,7614.8# h1.750'#
P. = 14593.0 x 8.5 =39.ou1
13,357.8# 80.791'#
Ph(d)== - 65511 x 6.23 =-L10.851
39.96055L
x: % = 2.99'
14.25 - 2.99 =1.26 = e
.16 within middle one-third.
r=§ (as)
= 13358 ( 1+6 x 1.26)
we“ “at“
r toe = 2970 p.s.f.
1‘ heel: 175 p.s.f.
3.11.: 13358 x $99 .=-. 1.02 Therefore must anchor.

 

6551

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a
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11“ l, E

HEEL DESIGN

 

Shear and moments were taken at the section noted and the results
used to compute the depth required and the steel needed. In many cases
the size and steel used was larger than needed to make the actual con-
struction convenient. In this design the reinforcing bars did not have
to be anchored. The unit forces acting on the heel form a trapezoid
(as shown) and were found as in the heel design of the cantilever back
wall.

Computations:

X1 = 355-14

 

 

 

 

X2 = 152-3
__ 11.6 (h52.3+710.8)
x_ : 202'
T 807-7
Maximm Shear and Moment on Section A-A:
W3 2 -3,780#- x 3.25' ~ :7 —12,285'#-
WL‘: - 5111 x 3.52 : " 1.9014-
P7: '1’918 x 2055 = " 14’891
Earth P = +3,)478 x 1.53 =+5,321
-2361} 43.759’1"
V 2761 "
d=w=W - 6'6
M 13759 n
V: 2761 = 17 no special anchorage.
X. 9X
11 __ 13759 < , .
R =1)? -— 152 ::: 61.1 139 fc is low.
1.1 _ 15759 _ - .. .. . .
AS: 1'53“ 18000x375x15 ... .0583 Sq 11‘ per in

Use 3/14 inch round bars @ 7—1/2" c/c.
Gives .0587 Sq. inches.

C omputati on 8 continued :

u _ 113. = W = 514 <100 no special anchorage.

Zo .

69

7O
’TOE DESIGN

 

Werking about the section shown, the toe design was accomplished.
The assumed values for the dimensions were below the required value in

each case. It may be noted that anchorage of the bars was required.

Computations:

133(271'5 E 2005) I 2.25 2 5.0314111

32:32.82 (2005 + 2 x 27851.. 1.18'
3 1.75? I"

Maximum Shear and Moment on Section B-B:

 

V: 509141-11
M1: 53hh x 1.18':: 6,306'# per ft. or "# per in.

Keep toe 1.5' thick to match heel.

V .
Vzm = W = 33.9 P0801.
M 6506
12:13-32: -—-i—§2-= 28 (f0 very 10W)
M 6506 .
A = .7. ’ ' :2 .0267 Sq. 1n.
8 Is ia o 3.

Use 3/8 inch round bars @ 14" c/c.
Gives .0275 Sq. inches.

vb 3h x 7 .
u_—. __ -_— = 11 100 S ecial ancnora e.
,0 vac-172 5 > P g

71

CUT-OFF'WALL

 

The method used to design the cut-off wall for the side wall did
not deviate from the method used for the cantilever and counterfort
backwall and the front wall. The purpose of this wall was to increase
the sliding factor of safety by decreasing the tendency of the wall to
move over the ground.

The location of this wall is usually designed as shown but may be
elsewhere under the base.

Computations:
Passive Earth Pressure: wh 1+8?“ 550 - 3.7 wh
1 -sin 350

At C-C, wh = 1983.6

Possible Res.= 3.7 x 1983.6 = 7,339 p. s.f.

To make sliding factor== 2

R: 2 I 65514 - .1499 I 13358 = 61114211
Min. h:: 6bh2 :2 .878' Use 1' for convenience.
7339
On Section C-C:
V== 6,hh2
M: 61412 x 6: 38,652"-,‘f
d: FVfi-e-Wfi‘m = 15-3"

If no tension reinforcement:

d— E x 38652 =17.,9,,

-—1 12 x 60

Make 1'-6" for convenience.

72
STEM DESIGN

 

To discuss in detail the procedures used for the stem would be
mere repitition. Any inquiry developing out of the following compu-
tations should be referred to the stem design of the cantilever back
wall. The results, of course, were different due to the variations
in dimensioning. As a result, anchorage and area of steel differed.
Computations:
maximum Shear and Moment on Section D-D:

172 1/2 x 15.85 x 507 25.511211

_V __ 3511 _ '1:
“WW—8°36 OK

M ==35ll x 1/5 x 13.85 ==16,210'% per ft. or ”# per in.

16210

d: = 1008" OK
‘Izv‘
Keep stem:
12" top
18" bottom

@ any level:

_ 2
Va: — 18.78 h

 

Mx= 6.26 1.3
b 21 x 6.5 .
u: ._._=' = 8 o 0 o
2'. 771:? 5 P S 1
Embedment
__ r d___ 18000 x 3/11 __ n
L‘Ifi‘“ 1411100 _' 53'75
Hook Bottom

Temperature Steel:

.002 x 12 x 15 ==.36 Sq.in. per ft. of height.

1/2" 0 @ 12" c/c} + 1/2" 0 @ 15" c/c : .36 Sq.in.
Front Back

73

 

 

 

 

 

 

 

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COST COMPARISON

77

The following cost comparison was derived from prices quoted in

the "Engineering News Record" magazine of February 5, l9h8, and is given

as a comparison of the two designs heretofore presented.

The results obtained should not be taken as the cost of the con-

struction, but merely as a comparison.

Only the three main items were

considered here: (1) Steel in Place, (2) Concrete in Place, and, (3)

Excavation.

Location
Back‘Wall
Back‘Wall
Front'Wall

Side wall (north)

Side wall (South)

Location
Back'Wall
Back'Wall

Front wall
Side'Wall (north)

Side well (South)

Area

 

2,655 Sq. Yds.

Concrete in Place

 

 

Volumn
Design Cu.Yds;
Cantilever 772.2
Counterfort 737.7
Semi-Gravity 550.6
Combination 87.87
Combination 85.25

Steel in Place

 

 

 

 

 

Design EEigEE
Cantilever 7h,39h.l#
Counterfort 5o,556.9
Semi-Gravity 5,733.6
Combination 5,237.9
Combination 5,208.9

Excavation
Depth m
1 Yd. 2,655 cu.yds.

Unit Price

 

SA0.00
h0.00
u0.00
L0.00

M0.00

Unit Price

 

$.10
.10
.10
.10

.10

Unit Price

 

3.90

Total

 

830,888
29.508
21,22h

3.515
Bahlo

Total

 

t 7.h39oh1
5,055.69
573-36
523-79
520.89

,Total

 

3 2,589.50

Location

Back'Wall
Side wall
Front'Wall

Excavation

Location
Back'Wall
Side'Wall

Front wall
Counterfort Angles

Excavation

Cost of Cantilever Design

 

Steel

 

$7oh39-h0
1,0hh.68

573-35

 

e30,888.00
6.925.00
21,22h.00

GRAND TOTAL

Cost of Counterfort Design

 

 

Steel

 

$5.035-69
1,0uh.68
573-55
1,056.00

$29,508.00
6.925.00
21,22u.00

GRAND TOTAL

78

Total

 

$58.327-h0
7.969.68
21,797.56
2.389.50

 

$70.h83.9h

Total

 

$3h.5L3-69
7.969.68
21,797.36
1,056.00
2,389.50
$67.756023

 

79

DISCUSSION

 

The reader should not be falsely lead to believe that the results
just obtained in the cost comparison are all conclusive.

To illustrate this let us briefly follow through the assumptions
and results obtained.

As previously stated, the unit prices are four months old and if
construction is contemplated, new prices must be used. The foregoing is
merely a comparison of the major variables and constants involved in design
and construction. The variables are the volume of concrete and steel
needed for the cantilever and the counterfort back wall. The form area
for concrete will vary. It is important to note that this is not brought
out by the comparison, and will tend to increase the counterfort cost.

The side walls, front wall, and labor for the side and front walls,
remains relatively constant. These constants, however, were included to
give a better picture of the magnitude of the work and, also, the varia-
tion in the magnitude effects the unit prices.

The final prices given in the comparison are surprisingly close.

It has been an accepted theory that twenty feet is the dividing line between
cantilever and counterfort design and this design seems to substantiate

that theory. The volumes of concrete in the two back walls was practically
equal and the steel required for the cantilever wall was slightly greater.

The counterfort wall will require more labor and as the labor cost
fluctuates, so will a comparison of the costs of construction vary. so
conclusion has been made here as each design has its own place. This com-

parison concludes the study undertaken in this thesis of retaining walls.

I725
- M846 Morfee
! c.1

II I IIII IIIIIIII I

93 50028 1908