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Design of a Counterfort Retaining Wall
along Red Cedar River, East

Lansing, Michigan.

A Thesis Submitted to

The Faculty of
Michigan State College
of

Agriculture and Applied Soience
BY
Philip H. ggderson

Candidate for the Degree of

Bachelor of Science

June 1930.

m was a. fflI‘I‘iun ,.
o.‘ I _ ', ' . «‘Q‘ fir g' f '. ' 5' a g .. - s y
.4 ‘_ '1'! a .fi > .
J ‘0; i 5 9*" an. N , ' (
Sam ;

The writer wishes to acknowledge his indebted-
ness to Mr. C. A. Miller, professor of Civil Engineer-
ing at Michigan State College, for his able assist-
ance and advice thruout this work, and to Michigan

State Highway Department for material and valuable

information.

Philip H. Anderson.

9377'?

BIBLIOGPlPHY

American Highway Engineers Handbook

Arthur Blanchard author

Reinforced Concrete Construction

George A. Hoel author

Cyclopedia of Civil Engineering

Vol. VI American Technical Society

The protection of property from damage by floods has

always been one of primary importance.

The site chosen for this design is located on the out-

-skirts of East Lansing at the intersection of the Red Cedar

River and Harrison Road.

In the mid summer this land is comparitively dry, but

in the early spring the Red Cedar rises above its banks

and floods the area.

The difference in the extreme High

Water level and the normal water level is approximately 10

feet. This difference provides an excellent opportunity

for study of the variations of the resultant pressures and

the subsequent reversal of stresses.

In this particular case the following sketch shows

the variation in pressures.

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This design was made adequate to provide safety in both

cases and is found necessary to deViate from the minimum

allowable dimensions in several cases to produce a satis-

factory design.

The site was surveyed and countours run.

A log of

Borings was obtained from the State Highway Department and

are shown together with a Topographical map on Sheet No. l

in the back folder of this book.

Profiles were taien at intervals of important changes
in contour. From this data a height of 21 feet was con-
sidered necessary as determined on-the profile map marked
Sheet No. 2.

The counterforted type of wall was chosen, since the
height exceeded 20 feet and results show that this type
is more economical when the height is greater than 20 feet.

A value of‘t equals .42 was selected with an assumed
weight of 35 andha design was worked out. The Resultant
pressure failed to pass within the middle tnird and these.'
values were discarded. After several trials value of
t equals 10' h equals 21' were found to satisfy the
condition and were used.

A value of 30“ was assumed for the thickness of the
footing, which was later found to be high and a correction

was made in the next trial.

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49'- a' L ’9,
Verticle Wall

W due to earth equals 25#
W due to earth & water equals (62.4 - 25) equals 37.4#
Verticle.flall
The pressure at a depth of 16.5' is wh. equals 37.4 x 18.5
wh equals 730#

The counterforts are spread 8' apart 0 - c with a

thickness of 16“. Clear Span equals 8-1.35 equals 6.67'
M equals 1/10 wlav equals 750/10 1 (6.67?” equals
3242# per ft. of width
d equals \fi7bi equals V32427107T7 equals 5.71
For Shear--—
V equals 730 (4 - .6?) equals 2430
d equals V/vbj equals 2430/(40)(.874).(12) equals 5.75 or
5”
The minimum allowable d to allow for plaCing of the
steel is 12”. It was found advisable to use 14' here.
Assume 1/2" round bars spaced 4' on centers.
A equals p d b
A, equals (.0077) (12) (5.78) equals .535
u equals V/eoad equals 2450/12t4 x (1.571)
X (.874) (5.78)
equals 102#
Since a bond stress of 102# is too high the spacngg

was decreased to 5'

u equals 2430
l 1.5 .8 .78
equals 76#
v equals V equals 2430 equals 40#
533 IZI-873(5-587

For Outer Steel:
wh equals 25 X 18.5 equals 487#
M equals 1 x 48? 1:46.07)? equals 213236;E

10
d equals1‘2132 equals 4.45
157.7

A, equals p d b equals (.0077) 12 (4.45) equals .4l
Use 1/2' round bars Spaced 5 1/2" c - c

Spac1ng of Steel:
Inner Wall,

at Quarter Point equals 4 x 4/3 equals 5.33"

at one-half point equals 4 X 4/2 equals 8.“
Outer Wall

at Quarter Pt. equals 5 l/2 X 4/3 equals 7.53"

at One-half Pt. equals 5 1/2 x 2 equals ll"

Dealgn 9;.gqggygg; (High Water)
To determine total pressure on Base.
wt. of Earth squall (18.5 x 5.09 x 100) equals 9425
' ” Stem " (1.165 X 18.5 X 150) ” 3230
' “ Footing " (10 x 2.50 x 150) n 3750

. . Water " (62.4 x 15.5 x 2.75) I 2660
15075“

r equals (9425 x 2.54) plus£3230 i 6.67) plus (3750 x 5)
plus 2660 g 8.63)
ldU'I'D

r equals 85250 equals 4.72
18075

To Find where Resultant Pressure Acts:
P (earth) equals 112 wdpequals l/2.25.(21)yequals 5520
P (water) equals 1/2 whrequals 1/2 62.4 X Sgrequals 10100#
Pr equals 10100 equals 5520 equals 4580#

1 equals 5520 X 7 - lOlUQfiX 6 ecuals 5.22
r 450v

1 equals 4580 equals 1.3
5.22 18575

X0 equals (5.00 ~ 4.72) plus 1.3 equals 1.58

P equals 18075 (1 plus ‘6) (1.58;

equals 1807 plus 1720 equals 3527#
Pb equals 1807 - 1720 equals 87#

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Safety Factors:

Sliding equals 18075 X .5 equals 1.9
4580

Overturning equals 4.72 equals 2.5
4.72:3.42

Pressures for Normal Water level:

Wt. of Earth equals (18.5 x 5.09 x 100) equals 9425
“ ' Stem ' (1.165 X 18.5 X 150) ” 3230
" " Footing . (10 x 2.50 x 150) N 3750

” ” Water ” (2.75 X 62.4 X 5.5) “ 945
17550

r equals (Q425X2L54)p1us(3230X8.67)plu§(3750§5)plus(945X8.65)

17350
r equals 70500 equals 4.05
17350

To Find where Resultant Pressure Acts:
P (earth) equals 3 . 25 X (21)*Equals 5520
P (water) equals 3 . 62.4 X 8yequals 2000
R equals 5520 - 2000 equals 3520
Xr equals 5520 X 7 - 2000 X 2.66 equals 9.45

 

3520
1 equals 3520
9.45 17350

1 equals 1.92

Xo equals (4.05 plus 1.92) - 5.00 equals .97

p equals 17350 $1 plus .97 x 6;
a 15 lo
equals 1735 - 1010 equals 725#

Pb equals 1735 plus 1010 equals 2745#

 

 

 

 

 

 

 

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Pressures on Base in both Cases.
H1 gh Water.
.1“ .
I
. x ‘37
3527

 

 

Normal Water.

41
725

 

.27vs”‘

 

Inner Floor Slab.

The decrease in load per foot is

725 plus 931 equals 828#

Uniform load on end Step equals 1850 plus 375 - 828#

equals 1397*
Max. Bending Moment equals (1397) x (6.67f”equale 6220#

10
d equals V6220 equals 5.8“
157 7

For Shear.

v equals 1397 (9.00 - 67) equals 4700#

d equals 470 equals lb“ plus or 16“

Total depth equals 16 plus 3 equals 19"

X equals 1? equals 6220 equals 27.6
de 15

P equals 27.6 equals .00194
' Xea

A8 equals (.00194)(l2)(15) enuals .342 sq. in.
Use & in. round bars spaced 12' center to center

SpaCing over Supports equals (80)(12)(2.36)(.87)(l5)
4700

8 equals 6.3 or 6.5
decrease in load per foot equals 2745 -725 equals.202#
At 2 feet from End: 10
Top Spacing equals 6 Xfil§97 equals 6.3 or 6.5"
Bottom Spacing equalslééX 6.5 equals 13'
At 4 feet from end:

TOP SpaCing equals 6 X 1397 equals 12.2 or 12.5”
6§8

Bottom Spacing equals 12.5 X 2 equals 25“ - 18“ equal Max.
P equals .44 equals .0049, 3 equals .894

x
:8 equals feazogsiz) r equals [Adam
.0 4 .5 28-)

74600 equals 6320#
1182

2; equals 6320 X .75 equals 14.8 or 15"
4n 4 X 80

 

length of Embedment equals 15"

v equals 4700 equals 4700 equals
(12)(.894)(15) 161
29.2#

Outer Cantilever:

Wt. of Wall equals 2.75 X 150 X 2.5 equals 1030#

" ' Water “ 945#
Total 1975*

distance
to c. g. equals 2.75 (2.(2189) plus 2745) equals 1.43

4934

M equals (2.467)(2.75)(1.43) (2.75)(2.5)(150)p1us 94511.375
d equals V7100 equals 8' plus
157.?

Sheer equals (3075)(3.75)-(2.75 X 2.5) X 150
5700 - 568 equals 5132

d equals 813 equals 16"
(12)(30)(.87)

Total equals 16 plus 3 equals 19”

Use &' round bars spaced 12' center to center.

P equals :44 equals .0040 3 equals .999
6 X 15 '
u equals 132 equals 76*
_§, 2.3 .899 6
6

F8 equals 7100 X 12 equals 7700*
1004 .699 .256 12 .

Length of embedment of Steel:

§;_equals 7700 75 equals 18'
u 4.80

Design of Counterfort.
Horizontal force transmitted equals fi(25)(18.5?¢8 equals 34200#
M equals 34,200 218.5; 12 equals 2,540,000'#
3

I equals 2 540 0 0 equals 34.6
16(5.65)¢.(12)"‘

P equals .0024?
AB equals (I.65)(12)(16)(.00247) equals 2.68

Use 4— 1 inch buss Spaced 3% c - c
Moment at 14' below top
4(25)(i4)Vx 6 equals 19600
M equals 19600 (ls/3) x 6 equals 1,090,000 in lbs.
Moment at 7' below tau
%(25)(49)(S) equals 4900
M equals 4900 (8/3) 12 equals 15700 in lbs.

At 14' below top

.5: .

K equals equals 25.8

 

Rigid P equals .0018
At 7' below top

K equals 15700 equals 6.90
16(3.13)V(12)

 

Rigid P equals .0005
2 rods carried above 14' point

2 rods carried above 7' point

 

 

Pl equals 1°7851 .2 equals .0020
15(4.28)(12

P2 equals _L-7851 .2 equals .0025
‘15(3.13)(12)

These rods may be stopped as clanned.

Horizontal Steel:

 

Shear on 1 ft. in height.
V equals 25 X (18.5)(6.67) equals 3080
assume 1/2" round bars.

No. per one it. ht. equals 3080 equals .97 or 1
16000 X .196

Spacing at Foating equals (122(2) equals 24"
l

Spacing at 10' point
V equals (25)(10)(6.67) equals 1670

No. per one ft. ht. equals 1670 equals .55
16000 X .196

 

Spacing equals 12 X 2 equals 43"
035

Max. spacing is 24"
Verticle Steel:
V equals (l397)(6.67) equals 9320#
Use 5/8" round bars

No. per ft. ht. equals 9320 equals 1.9 or 2
16000 . .306

 

0’)
O
on
O
H s
:3
r 1o

a equals 12 X 2 equals 12"
2

329 °trip
v equals 9320 - (3 x 828) equals 6836

No. per ft. ht. equals 6836 equals 1.3
T6000 . 306
Spacing equals 12.2 equals 18"
1.3

 

412 Strip
v equals 9320 - (4 x 828) equals 6000

No. equals 6008 equals 1.2
16000 X 306

 

Spacing equals 24 equals 20"
1.2

532 Strip
V equals 9320 - (5 X 828) equals 5180

No. equals 5180 equals 1.05
16000 X 306

9135101113 .. ~ 24-- r: 22.4"
1.05

Amount of steel for one

Bay & Counterfort.

Footing.
1r — 2” round bars - 8' a 188 lin. ft.
16 — 3" " " -3'1o"g_62 lin. ft,

.Z-

Wall — Horizontal

74 - 1" round bars - 8'0"
2
Verticle
3 - 1" round bars —20'l"
9 2
q _ an H H w "1”"
E
Counterfort
2 — 1" round bars —25'
2 fl 1" H H “20.
Verticle
2 -§" round bars -5'
2 -§" H H "8.6"
2 _n n u n “1396"
2 -H H H H “18'
Horizontal
2 - 1" round bars ~6'2"
2 ‘- ICEH II N “5'8"
2 a "H H H “5'0"

190 X 1.5

~592 lln.

62

-41

an.

40

H

. I
= 285*

ft.X.67= 396"

" Xl.5= 62?

_§5‘X 2.67 = 240%

~27

76

__z._ .
90 X 1.04 : 94a

R) R) [0 f0

[‘0

PO

oon't.

1" round bars

H H

H H

H H

-4'3"
—4'2"
~3'6"
-2'8"
-2'3"
—1'6"

~1'o"

H

6‘

I")

 

74 X .67 3 59%

Total

3 1169#

‘

Sne 3t No. III shows a detailed drawing of one
bay and coun tcrf art. This drawing shows the spacing

and arrang3ment of the steel reinforcement.

*3

ications used in design were 2003?

ml,
Ine speci

concrete, n: 15 , f: 16000 , f: 650 , k: 107.7

.

In assuming 3 = .42 a reference was made to
Trantwine's tables vhich was the basis of this
assumntmon, however HGV ral trials showed that
this ratio voild not hold in this case due to

the effect of the water pressure on the outside

0

of the vm all. The method of finding the corre t
ratio th at vould make the resul ant pressures
pass thru the third point was therefore one of

trial and error and considerable time was spent

in this manner.

 

 

ICHIGRN STRTE UNIV. LIBRRRIES

 

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