A DESIGN OF A REINFORCED
CONCRETE BUILDING AND
SERVICE UTILITIES FOR A

CHERRY FARM

Them in: the Degree of E S.
MICHIGAN STATE COLLEGE

L. E. Sobkowski
I947

TH ESIS

. .E: II:..1||1|1-

.14.‘
(Dual-Wwiufip‘ .‘
L

 

‘f.

.1.....‘;;....14.'!r. i

. ‘1

111.1,! 3‘13! “til: 3 film.“ ‘0»

r. 1r .r

. 1511...“... .
. :IIII » . . I ‘4! . tightlivriiflllfw 141.. z. ,.
up} . . ....I..fl1uIJ.V 11.. 4 .1 a . .. . .

.
. . u I. x. .. v‘_1xu5&_—. ‘3‘ III...=~.«:¢. . e

A Design of a Reinforced Concrete Building
and Service Utilities for a Cherry Farm

A Thesis Submitted to
The Faculty of
MICHIGAN STATE COLLEGE
of
AGRICULTURE AND.APPLIED SCIENCE

By

L. E. Sobkoweki
Candidate for the Degree of

Bachelor of Science

march 1947

3/1! [H7

(- ._
“JWT

DED ICAT ION

To my sons, Stevie and Joe

18.33 '30

ACKNOWLEDGMENTS

I wish to thank the members of the Civil Engineering Staff
for their assistance in this thesis, in particular, Pro-
fessor C. A. Miller, for the time and effort spent with me
to perfect the design both structurally and practicably.

BIBLIOGRAPHY

Sutherland and Reese - "Reinforced Concrete Design"

O'Rourke - "General Engineering Handbook" I -

H001 and Johnson - "Concrete Engineers Handbook"

A.I.S.C. - "Manual of Steel Construction

Trautwine - "Civil Engineer's Reference Book"

A.C.I. Building Code

T.E.C.O. Truss manual

Joint Committee Report. 1940

Reinforced Concrete Design Handbook - Portland Cement
Association

Specifications for Design of Highway Structures -
Ohio State Department of Highways, 1946

NI‘BNTS
I Purpose and Scope of Thesis
II Computations and Sketches

 

  

‘__-._.._~-. .... -.. - -.~.'_.- -.\.._‘_ ._._—— - , _~ H -~ ._._.- “,7 -..—...,

a - 7706/27!“
b «SprayEr

(2-7 Corn shredder

 

I!
to

A

 

‘I/eh/c/es as shown

 
 
   

:,"‘P:¥' “ , I
I 43¢“ . ,7
tiff- “a, ‘ ' ‘ l

 

 

 

 

ll

g
“i

 

I

m z, ~52; 'ir: .

:U
I
1‘."

Mimi»

; w;
Sb

1
3

swmwvé

‘1‘-

Si‘orage Tank

 

 

  
   

PROPOSED Bumowg

 

Dwg. No.1

 

 

 

 

 

 

 

    

 

 

 

 

PART I
Purpose and Scope of Thesi .

The purpose of this thesis is to present a practical
and economical design of a building and service facilities
for use by the owner, a fruit grower in Grand Traverse Coun-
ty. The owner plans an eXpansion in property and operations
and thus I have taken the Opportunity to gain some eXperi-
ence in practical design. Design methods used by the author
are those set forth in the respective design courses as
taught at Michigan.State College. Thus it may be stated
that much more labor had to be applied to this problem than
would be necessary if handled by an eXperienced design en-
gineer. Despite this fact, it is a worthwhile project in
that the application of fundamentals is necessary before
acquiring the tricks and short cuts of the trade.

The owner operates about 150 acres of cherry and apple
orchards. This holding is to be increased in the near fu-
ture and the required additions and changes to the present
set-up constitute the object of this thesis.

A. To locate and design a fireproof, durable build-

ing to perform the following functions:
1. House the following listed tools and machin-
ery:
2 tractors
l pick-up truck
1 flat trailer

2.

3.

4.

5.

1 corn shredder

1 Orchard sprayer unit

WOrk benches

Welding set

Woodworking machines

Lubrication facilities

Provide means for maintenance and repair of tools
and machinery.

Stock all necessary spray materials for orchard
pest and disease control.

Provide an inepection, weighing, and crating cen-
ter for all seasonal fruits.

Provide a separate section of this building for
extended storage of fruit. This section will be
a part of the main building but of conventional
storage design.

The owner desires this building to be located at the edge

of an orchard, at the crest of a slight hill leading to the

East boundary of this orchard.

B.

C.

(See Contour lap - Drawing no. EL)

To locate and design a well to be driven adjacent to

this building.
Select the most efficient and economical method of

pumping water from:this well for the following pur~

poses:

1.

water for orchard spray. Under the present plan
of operation, a 400 gallon tank sprayer is being

 
 

i

5057‘ Orchard

 
   
 

From Orchards

  
 

A — Preposed b/dg. /oco7‘/0n

B— We// and pump

 
 
 
 
 

   

C — Sforage funk

  
  

0 " New road

(6"
44

   

BMr/00.00’

   
   
    
 

 

 

,.. q)
C .
. r ‘\ Top of foundaf/on) SW
3 5’ corner of house (mar/red)
x Q) 5‘ Q
(U S ,. ‘0 Q
E [\1 &
Q
5* z: r" Q 2
K I
K I
Q '.
S
:“I/li"
OI
\0 E
x Da/ry Barns
1 /' I
g/ . 1 70 Houses ;
[ll/i" .' I E
f A I
'. I I
E I
v . / ,
7‘0 H13 no. way 350 I
Y _ §
. , __§_0N TQU R WaMgIE M A P E
E ~ E . E; _f [057‘ Prop. [.1779 I
z -- E
I

   
 

Dwng.No. 2

 

 

 

 

used and at the height of the spraying season,
water must be supplied at the rate of 400 gal-
lons per hour.

2. Water for dairy stock and other farm animals.
Predicted future needs are 100 gallons per day.

3. Fire protection needs. A main of suitable size
installed to make available adequate supplies of
water at a sufficient head.

4. Garden irrigation. Fresh vegetables comprise an
important item.of sale in the owner's farm marnet.
In order to assure high quality and early appear-
ance on the market, the owner requires a pipe
line installed to the gardens and some means of
effective distribution designed.

D. To locate and design:

1. .An approach branch and turnaround from the exist-
ing orchard road to the upper story of the build-
ing.

2. A road from.the lower story of the building run-
ning adjacent to the farm yard and houses, join-
ing the highway at a point other than the exist—
ing drive. This will eliminate all heavy traffic
through the farm and near the houses.

The sc0pe of the thesis is limited to the structural de-
sign of units heretorore described. No attempt has been made
to perform the functions of an architect in that architectural
features and details for the allied trades such as plumbing,

heating, or wiring have not been mentioned or treated.

lgLLOWABLE UNIT STRESSES U§__
From.ACI Building Code 1946
Using a concrete with f5 = 3000 p.s.i.
n = 10
Steel - Billet, Hard Grade

rs = 20,000 p.s.i.

 

COMPRESSION fC 1350 p.s.i.
SHEAR v
(Beams)
W/o Web Reinf. -
60 p.s.i.
W/o S.A. -
W/o'Web Reinf. -
90 p.301.
‘Nith 80A. "'
180 p.s.i.
W/O SsAs '-
360 psSsis
With S.A. -
One way footings ’ - 75 p.s.i.
B0 u _
(2-way footings beams)
Plain Bars - 120 p.s.i.
Deformed Bars - 150 p.s.i.
Plain Bars - 135 p.s.i.

(Hooked)

BEARING fc
On full area - ' 750 13.5.1.
1/3 Area or Less - 1125 p.s.i.

O'ROURKE
SOIL PRESSURE
3 tons per sq. ft.
EA.S.T.M. C 90-39
Bearing on masonry block - 1000 p.s.i.

PART II
Loading Platform.Design -
Sketchl#l
Class E concrete

1:61} Vol. mix

’2
H

3000 p.s.i.
20,000 p.s.i.
1350 p.s.i.
n = 10

H
to
H

H
H

Computing Min. Live Load
Apples to occupy a space 6' x 16' x 8' = 768 cu. ft.
w/cu. ft. apples = 35# (Bldg. Code Com. of the National
Bureau of Standards November 1924)
768 cu. ft. x 35#/cu. ft. = 26,880# total live load. '
Total Area = 18' x 6' = 108 p.s.f.

Unit L.L. eggagag = 250 p.s.f.

Floor Load Recommended - 300 p.s.f.
Assume D.L. = 50 p.s.f.
Total Load = 350 p.s.f.

For a 1' section:

V = w = 350 x 6 = 1050#/Lin. ft.
51...?—

BM = .712 = 350 x 62 = 1575'#
__§._.

From table 2, RCDH, d

°°|

3" RM”: 2.12' K
1.44

m
H

.575 3 .364 sq. in./ft.

11"“)! I‘ll "IIIIIJIIOII. 1.0.‘1‘;.‘IIOIIsI'IIII'Iv’qu‘IIIl‘II. 1.7. .I'Irn.lllllll‘lvi . .l. I u Inc .I 15 VI. I?v1’sl!ll-)\.v,1.wl 'a..l. .‘l.'.sn . I I. .1 .i....l.s:A<

 

 

J
F
u
5
I . . ,. ,, . 0. 5
.. -. II- 4 . m I: ., amid: - .. . .- I/., ..>,_. g “0 w
a . ... T
E
K
6

n . I! .

a e i -vr / x
m C l . . - ..

_ .I

, e. . . a ,- /

I
. /
f-
/

|
N.

1......- -........
)—
/
/
/
9
‘-

En
ET
I
I

 

. . {n .

. x .1 a. J .-\
. . In I - All \ t . i."

I“. a: I u I: I . \ e. a...
b I .Vw K 1&4... _ x x. E , Iv
Ks . x . .x b .. . i x . .[r M
. . . (k
. x C a \
x x .

 

-Z-

20 = V = 1050 = 2.67"
3du 773 x 3 x 155 .

Table 3 RCDH
USe 3/8" 0 3%" c.c. Spacing. To = 4.0
Use 1" cover, depth = d i c = 3" / 1" = 4"
Check.weight - 4" x 1' x 1' x 150' = 50 p.s.f. 9g

 

3:2."
SHEAR
v ='V = 1050 = 33.4 p.s.i. 95
b3d 12 x 7/8 x 3
BQND '
u "V = 1050 = 100.2 p.s.i. QE

4 x 778 x 3
Check Bearing on Support - Sketch 2

II

R slab = 1050# Allowable Bearing Stress = 750 p.s.i.

ACI T 305
A 3 égsgg = 1.4 sq. in required
0 p03. 0
A = 12 x , x = 1.4 - .1165" thickness required
12

Use 4" thick support wall.
Check Earth Bearing at Base of Support Wall - Sketch 3
Allowable Bearing Stress - 8000 p.s.f.
(Ref. - O'Rourke Hdbk. - Compact, coarse sand or
stiff gravel)
R = 1050#
W2 = 1 2/3 x 1/3 x 150 = 125# (Support)
Total Vert. Ld. = 1175# ‘

Area in Bearing = 2" x 4" = .33 sq. ft.
44

117 : 3530 p.s.f. 9K.

 

 

  
   
   
  
  
  
  
 

 

 

\0
an F
N u
w /,
Q} 5 .0
W) b B
2’05” wP/czfes
0 6-1. 3X3"; /@/8’—0” Raga“,
F V F , C 2 é) 6f" 0” ”’
A I A . ~ . 0—- L 3x25- 2@ 49 0 ’“
- In E, I A I I I / / l/ //
K5 1‘ 3@ 4 W ‘3. G..L 3K 3XZ 2 @ // H9 ‘
HOG/(8
\ VI I ' E E - Lace/ed as mar/(ed F
\ I i I /u l/

2- ¢/2o52/—”/Pood., 5:221— 93/“

   

ed as mar/(ed J
:44— ¢ 2@4§"Reqd.) 5:2” 3:23—-

Al

 
   

  
 
  

     

2- 0A D LN£-E?LA If .9 BMW as T I L 3

”-9-?7” -m

 

2’— 4”
5667‘. 8’5

    
 

 

Sketch 4

ROOF DESIGN

Fink,Truss

40'
8'
16'
1t

- O" Span
- 0" Rise
" O" COCO

- O Eaves

This truss was selected from'Timber Engineering Co. Truss

Manual.

Total load 2 4O p.s.f.

Loading will be checked to see that this is not exceeded.
i Roof Area : (21.5{; 1.0') x (48' x 2) = 1125 s.f.

e = 21° 50'
Sin e 8 = .3721
EITB
C08 9 20' 3 .9302
élTB
LIVE LQADS
Wind Load, p1 = so p.s.f. (HOriz.)
P ”137%.
= 30 x 2 x .3721
i'7'7575i
P = 19.6 2.9.1. Say 20 p.s.f.
P" = P cos 9
= 20 x .9302
= 18.6 p.s.f.
Snow Load - 25 p.s.f. (Vertical)

It is unreasonable to eXpect a roof to be subjected to max.

snow load and max. wind load simultaneously.

Therefore, the following load combination is judged adequate:
Dead Load { §~wind load on one side / full snow load on
both sides

T.L. = D.L. ; 18.6 x 25
‘2‘”
D.L. / 34.3 p.s.f.

DEAD LOADS

From.Truss Design:

501.FBM»@ 2580 #/1000 FEM = 1290 # Truss Wt.
Roofing - Asbestos Shingles 200#/100 sq. ft.

200 x 11.25 = §§§Q#
Felt 25#/100 sq. ft.

25 x 11.25 280#
Total Dead Load ~ 1290 g 2250 / 280 = 3175# (For é-Roof)
‘2" .

é Horizontal Area = 21 x 50 = 1050 sq. ft.

Unit Dead Load = 317§fi = 3.02 p.s.f.
10 0 sq. ft.

Total Load = 34.3 I 3.02 = 37.32 p.s.f.

Use 40 p.s.f.

Reaction = T.L. x Area = 40 x 1050 = 42,000#
2 -

USe 4 trusses 16' - 0" c. to c.

42,200 3 10,500# Each wall truss reaction

Length 3 48'
22:899.: 875#/L1n. ft. of wall - Unit Live Load

Gross Area of 8 x 8 x 16 Concrete Block
15 3/4" x 7 3/4" = 122 sq. in.
1000 p.s.i. Allowable Bearing
ASTM C90-39

-0-

122 sq. in x 1% = 91.5 sq. in. per lin. ft.
1

875 = 9.6 p.s.i. Bearing on Top Course 93

91.5
‘UPPER WALL DEAD LOAD
8 x 8 x 16 Concrete Block
Heavy load bearing
i" Joint - 50#

10' wall Height 3 120"

120 = 15 courses
8

40' wall length = 480"
488: 30 blocks

48' wall length = 576"

576 = 36 blocks
15

450 block per 400 sq. ft. of wall area

450 = 1.125 block per sq. ft. of wall
405

North Wall

40' Long 2 - 3 x 5 windows
South Wall

40' Long No windows
East wall

48' Long 2 - 3 x 5 windows
West Wall

48' Long 1 - 3 x 5 window

1 - 7 x 8 door

N.W. s.w. E.W. w.w.
Area Sq. ft. 470 500 450 409'
No. Block 530 563 506 460
Tot. Wt. # 26,500 28,150 25,300 23,000
Unit Wt-#/Lin.' 663 704 .527 480

Check East Wall for Bearing on Bottom Course
D.L. - 527 #/Lin. Ft. 91.5 sq. in. per lin. ft.
L.L. -__8_'Z_5__ Area of Block
T.L. -1402 #/L1n. Ft.

1402 #/ft. .= 15.3 p.s.i. 91;
91.5 sq. in./ft.

Check South Wall
D.L. " 704 #/L1no Ft.

L.L. - 0
TOLO " 704 #mno Ft.
7 lfto 3 707 p.801. 9;;

.5 sq. in./ft.
WINDOW LINTEL DESIGN ,
East Wall Windows Sketch 5
3 courses at top
3 x 50 = 150#

150 x 12 = 113#/Lin. rt.
18’

8" = .67' width

1' x .67' 3 .67 sq. ft.
L.L. - 875#/Lin. Ft.
D.L. -_;;§;
T.L. - 988#/Lin. Ft.

W 3 988#/Lin. Ft. 3 1480 p.s.f.
737 sq. ft.

 

.. ‘a'AAV .munmw

”v—

>-d v,” “We-s;

' ‘Wfiufli 9"‘O

- ~

_ -‘—~'

- V -.o‘>-Aawg—n&

. u; . “' -wm'H-Qv-‘r

“-a -.-.¢H7.v ‘v—wn x,‘cv

Nth.

.F'"

w~um _ A
I A4 “AHA-m‘”. .guu- lflM:~h.§-t-“‘ >‘- "

"‘0' "-K

 

 

 

 

11-1-1111

d ccuwsts
7“?“

1

3x15

 

 

L.

5'600R565

- .. .- u.” -...... ._.._fi

#
5 W/NDow

 

‘ ’*.-V ‘r‘va-p-‘v—mm—Oc'cb ---qcu—-r.---¢‘u _. 1“" -...-.~--H-"-rwlewWfi-.

'4 Roar TRUSS.

LL = 675%7‘
A

/010”

 

 

 

 

 

‘fi
8 _ LINTEL. DES/6N

 

_.___-L_ 1 11.1.1. .1 .5...

415“couqst$

M“ “~*¢.‘ “w-.-”

'7xa i

 

" 7 TR/AN 601. AR
1. GA DING

5K5 TCHES

13M?! 5PACIR

”4-5

-‘--—I~v~.~-

Sketch 7
Max. B.M; = wL? = 1480 x (3.2523 = 2120'#
- 24 .

24

S ='M = 2120 x 12 = 1.275 in.3
f 20,000

Use 2 - 4 x 4 x i L's s = 1.1 in.3 25 = 2.2 1n.3
Sketch 8 '
This design will suffice for all windows in the upper
story.
DOOR LINTEL DESIGN
‘West wall Upper Story -

Sketch 6
4.5 x 50 = 225#
225 x 12 = 169#/Lin. Ft. D.L.
16

D.L. = 169#/Lin. Ft.
L.L. = 875
T-L. =1044#/Lin. Ft.
w = 1044 = 1560 p.s.f.

767'
max. B.M. = EL? = 1560 x (8.3313 = 39,200'#

24 24
s a 39,200 x 12 = 23.5 in.3
20,000
Use 2 - 8 x.4 x 7/8 L's s = 12.5 28 = 25 1n.d

Sketch 8

UPPER FLOOR SLAB -
Sketches 9 & 10
Live Load on Storage Section

Area 16' x 40' = 640 sq. ft.

Vol. of storage, considering apples in bulk 8 ft. high.
15' x 38' x 8' = 4550 cu. ft.

Weight - V01. x unit Wt. of apples
W = 4550 x 35 = 160,000#'

160,000 - 250 p.s.f.
640

Floor Load Recommended - 300 p.s.f.
Assume a 6" thickness

Wt. = 1' x .5' x 1' x 150 = 75 p.s.f.

L.L. = 300 p.s.f.

D.L. = .ZQ

T.L. = 375 p.s.f. or for 1' ft. section 375#/Lin._§t.
Considering this a simply supported, uniformly distributed
loaded beam:

Max. B.M. = 1L2 = 375 x 64 = 3000'#
8 “"8‘"’ _

V's g; = 375 x 8 = 15084
2 "‘2“‘

From Table 2 R.C.D.H.
d = 3%"
Changing thicxness of slab to 5"

B.M. = 2900'# d = 34" v = 1450

AS = ‘M a = 1.44
ad

3 2.9

Af_ 0575 Sq. in./ft. RGQdo
1.44 x 3.5

 

Table 3 R.C.D.H.

Use Q" d 4" Spacing ' 4.7"

“1
o
I

S7nAR

w

v = V = 1450
335 12 x 7/8 x 3%

40 p.s.i. pg

BOND

u = V’ = 1450 = 101 p.s.i. 93
2333 4.7 x 7/8 x 3%

T - BEAM DESIGN
Sketch 11
Span 20' C. to C.

Assume 19' Clear Span

L.L. = 375 p.s.f. x 8 ft. = 3000#/ft.
D.L. 8 Assume Stem Wt. - 200#/ft.
T.L. = 3200#/ft.

<
u
2
H
II

3200 x 19 = 30,400#
2 2

Allowable v .06 f5 = 180 p.s.i. with web reinforcement

v = V b'd = 30,400 = 193 sq. in.
b'd} 180 x 7/8
Let b' = 10" d / 3 = 25"
d = 22" d - t = 22 - 5 = 17"

Checking for weight of stem

w = 10 x 18 x 150 = 188 f . g;
“144“

Assume J = .92

 

 

BM'=‘E12 = 3200 x 19 x 19 x 12 = 1,390,000"#
10 10
BM = Tjd
T = 1,390,000 = 68,700#
.92 x 22
As = T = 3.43 sq. in. per ft. Reqd.

H:

s
Use 4 - l" ¢'Bars, 2 rows Area = 4.00 sq.in.

E3
E?

I

u = V’ = 30,400 3 94 2.5.1. 95
zojd TX 47 .92 x 22

- 10 -

120 p.s.i. All. Art. 306 AC1

REVIEW OF T-BEAM DESIGN

Sketch 12

To select flange width "b", the least of these three con-

ditions is chosen:

1. Span = 20 x 12 = 29" *LEAST
4 4

2. 8 1 Thickness x 2 / 10" : 23"

3. (8'-0") - (0'-10") 3 7'-2" clear Span

2(3'-7") { 10 3 96"
ZIMna : 0
(50 x 4) (x - 3) 3 40 (22 - x)

 

x = 6.17"
.3_=.fc
2.17 6717

2 3 .352 fc
fc-Z : .648

m- , .2 41:1

cl = .352 2, x 4 x 60 = 84.5 to x 20

cg = g-x .648 re x 4 x 60 = 77.8 re x 20.67
3300 fc ' 1,390,ooon#

 

fc : 42030301. g
c = 162.3 x 420 = 68,000 #
c = 'r =' Agra

f8 3 68 000 3 17,000 p.s.i. Q;

1690 re
1610 to

 

3300 re

 

 

3 *——- -- .v“.4‘.m v "—1

a- u-vu nu ----¢---— “anon---o..~-t‘d~_>-‘

O~—a

*mw'. ‘- -7-. ”.-,~.

 

2

A

N

6
I!

L

 

I

c -

F-r'
{' 66C TION

__ .7,
I
I

‘
-w ~M‘w‘—~ 9*”--.

IT} A

' rv
(J 4
q '
but

1‘,

 

!
:77f‘.’
.5" a
1
i
1

 

I: Illu‘l'ltlllllllllll 1.!2‘IP" III
..
4
p
o

a
,
.

A

T. -.. I; In- [t 3.1 1.9:.

.Iu‘illlul

._

A. \mmwnewi 0km...
. _ .113ka .

 

 

 

 

”/0

'9

__

w

~ . r I

~aillttlcllinfl
a

 

 

 

it.
,7

 

4:- r.

.3
4.... P - .b.
L
2 : H
[’1‘

 

t b.' i ’6: .l
_ I . a
. . 4
1'. A“! ’1 2.“:
.k.'r/ _ .
. “ e. .I. .
. / 1 m :1. J A. ./,._
. . .
. . _ w .u. 1. Ir #
I. . . . 9..
.. . k . . .,
l u .
§ . s. ,
..N a . .n a} e - 7r ..
_ __ n a ".1 a
L .62. .
. m . . 7 A t I...
v . I
n M (F P J 7 N‘ a
.i. i . _ m@ u
.r u If: . ‘ a
_ - .L . a _._ I. .11 J
_ _ "v .a _ Fm C 0 ./

 

 

. .14-‘.|¢I:.Lu|."1’..

I

I.

1.0...

.
a

.1!
.f1
11”. l‘s...’. .- .r ).I {slit} . 0?...

CHECK OF T-BEAM.RT SUPPORTS

To locate neutral axis: Sketch 13

EEM, = 0
Al (x - 2) / 12 (x/2) = A3 (22 .x)
x = 8.2"
2 3 f 3
°373
= .476 fc

f ‘2:0524

 

c8 = 36 x .476 fc = 17.1 fc
Cc = §~x .524 fc x 10 x 8.3 = 21.8 fc
T = cS / cc 2 38.9 re
08 x 20 = 17.1 re x 20 = 342 to
00 x 20.67 = 21.8 2c x 20.67 = 552 re
38.9 f 894 f
c c
a = 23" MOment Arm of Resultant C
T = c = 1311 = 1,390,000 = 60,400#
. a 23
f = 60 400 = 1330 p.s.i. 95
c “38?9" "“'
rs = 60,400 = 15,100 p.s.i. g;

4
T - BEAM WEB REINFORCEFENT
Sketch 14
The maximum Shear, 30,400# occurs at the ends when

the entire Span is loaded with the uniformly distri-
buted load - 3000 #/ft.

 

vm = 30,400 3 150 p.s.i.
10 x .92 x 22

The center shear will not be computed but will be

calculated as 25% of maximum shear stress.

vc 3 150 x .25 = 37.5 p.s.i.
x = 114" x 90 = 91"
112.5
V taken by Stirrups 90 x 91 x 10 - 41,000#

2
Use %" d Stirrups A 3 .1963 sq. in. x 2 3 .3926 sq."
16,000 p.s.i. for web reinf. (J.C. Spec.)
.3926 x 16,000 3 6280# taken by each Stirrup
41,000 = 6.5 Use 7

6280
121" = 13" Spacing
7
but 3 11", therefore use 9 Stirrups

MID:

Spacing:
From end to center
1{@ 5"
8t? 10"
PLACEMENT 0F NEGAT IVE MOMENT STEEL
Sketch 15
Reinforcing steel, same size and Spacing as that
used for tension, will be placed over supports.
Length to be governed by general rule:
& clear Span on each Side of support.

b 3 10"

x

M
[K M‘ ~__ A.... . 4‘..-_,. __.......-”__,

 

f ;“ r11 '01
a" 'U‘
i “'1’ 3! O

— - ..-~.......... .

 

——— Cunt.“

 

.H.
....__.
\\

n .

 

...--.... -....-n—. -. .
l

-9
i-‘ - -..—.—

,-_ ,..,-u-- O-r" can. . .- .

.
3 A.
w- 'r‘..- _- a -.._n- WWI .~,~.- -._,q....
{ o. dog-v --O -.‘_.. D-ru I!“
—— H—‘ -4. —.—~-- ’. —
‘--’ “1.. ~ .n n..-‘ ‘...- —-

 

15 HEAR

.,... ..
' 1“~; i
} ‘ n ’
I . .

I

. I ‘\,
t, ' .
I 0" ,l
7" r. "
f
s, l l
I
, 1
II
_... . ... .—

.4.— -.n-n— n- —-—-\. M‘~-.~

TAKEN

 

\
I f
5 l
I
I
. 5
\}_ 1
_"‘l 4,- s I
s—co- -—— .¢. .-.. no .._--L«..-.:h.._ ; ‘
“"7"" .._-- " *7 x. . .acz 5' -i;

"walnut c.c..- . hr .sp-

pun—r..- b~o~

55y CON’C. .‘

ps/
i
1

._1-. .. -4‘

0-4

I c "1' '
AsurfihV ,
l
- w ..--.......-....u—‘

Cu“- ‘ n- '0” W’ """‘V
n“

m

m““*“

 

 

mafia. -u‘ ru-

fl

- 'W-- w~~r .4th 03". , ,,. .- _-.. .‘v

['1'2. 5p$I

 

 

 

A..~.. -‘..

 

- 13 -

supports - 8'-O” c.c.
Clear Span — (8'-0")-(0'-10")
- 72-2"

7'-2" 3 22" Length
4

GIRDER DESIGN

Sketch 16

Considering this beam simply supported
BM at center = 60,800# x 8' = 486,400'# = 1/8 wl2
but BM’3 1/10 W12 . 3

.°. Actual BM at center 3 486,400 x 8 x 12 3
i0

4,670,000"#
Weight of Stem, assumed - 400 #/ft.
l7,640#'
211,680,"

BM’3 1/10 w12 = 1/10 x 400 x 21 x 21

211,600 4 4,670,000

Total BMT@ center line
4,881,680 #"

v = 60,800 4 w; = 65,000#

(0

b'd 3 V 3 65,000 3 413 sq. in.
' 180 x j 180 x 7/8
Use b' = 14" d l 3 = 33"

d = 30" D - 4 = 29"

Checking weight of stem

29 x 14 x 150 = 420#/' pg
144“

Assume j 3 .92

BM = Tjd

- 14 -

T 3 4,881,680 3 177,000#

.92 x 30
As 3 1 3 177 000 3 8.85 sq. in.
fs 20,000

Try 1%" d A 3 1.56 sq. in.
8.85 3 5 / Bars
1.56

USe 6 - 1%" d 2 rows 3%" Spacing
A39.36 sq. in. £036x5330"

u 3 'V 3 65000 3 78.5 p.s.i. Q;
zojd 30 x .92 x 30 -

REVIEW OF T-BEAM DESIGN - GIRDER

Sketch 17
To select flange width "b"
l. Span 3 24 x 12 3 72" * Least

4 4
2. 8 x thickness x 2 % l4 3 78"

3. Clear Span 3 21' x 12 3 252"

To find neutral axis:

25m, 3 0
(72 x 4)(x - 2) = (93.60) x (30 - x)
x : gégn
cl = (72 x 4) .55 re: 158.5 re x 28" = 4430 re
02 3 % x .45 fc x 72 x 4 3 64.75 fc x 28.67" 3 1855 f3
cl 4 C2 = c = 223.25 fc 6285 re
Rm = 4,881,680 "#

f 4 881 680 = 777 2.6.1. 95
c 336285

- 15 -

T = c = 223.25 x 777 = 173,500 #
r = 3,: 173 500 = 18,050 p.s.i. (g;
5 As 339763” -
_ CHECK 0F GIRDER OVER SUPPORTS
Sketch 18

To locate neutral axis:

:zmg 3 0

 

(§)(14 x) / (84.24)(x - 2) = (93.6) (30 - x)
2 -
x 3 11.5"
RM=C1‘11"C2d2
3 (.65 fc x 72 x 4) 28 / (fi-x .35 fc x 72 x 4) 28.67
= 5240 re 4 1450 fc = 6690 re

f = 4 881 680 = 730 p.s.i. 93
° "6650—"

c = cl / 02 = 187 fc ; 50.4 fc - 237.4 re
0 = 237.4 x 730': 173,500# = T
f8

3 z 3 173 500 3 181500£.S.i. _0__Ig
As 9.30

WEB REINFORCEMENT
Sketch 19

3 168.5 p.s.i. max. at ends.

V 3 V' 3 65 000 -
m. 533 IE x..92 x 30

v 3 v x .25 = 42.1 Say 42 p.s.i. at center.

c m
Shear taken by Stirrups = 108 x 108.5 x 14 3 82,000#
2 -

3/4" 0 Bars A 3 .44 sq. in. x 2 3 .88 sq. in.
16,000 p.s.i. for web reinforcement.

16,000 x .88 3 14,100# each Stirrup

82,000 3 5 / Use 6
14,100

.- '.-.. v.' 4.5-- .__.

.‘ .- .— u..- -. -1» -~ . ~ . _. _ l .. H ‘9' .. r .7... .- w ..
1 fl -

: 5280.9 65.“..- '

I _.

O ' g

-. ‘ as .. 8’ 5 :3 ,_ ,
5 G/F'Z DER J
5 24‘ cc.
2/’CA£AR SPAN

”"5

 

»» __ .,_.~'. 0*

i ' ”/5

. -.- a-«-——~~.—c
3 .5'—“w“: ‘

. I .

'3 2

’ ZS” £545.75,
5

I

5

 

*- q~ ‘ -_,~,.. c...» m... ~v*-H-.““m I"m‘fl n».—— >4...—

\
I
4

a..—
\
4‘.
5‘.
I

y " . ,I 5 “
‘
. . \ I / I} ““.‘-.‘ . I
.. /' . 7‘\- i
V 60 V) '44 / lJ ' ‘ " ‘~ .
1 - Y —- .- - —--‘—-.--—‘"--v--..-- ‘-

‘ I
---~..... .-A...._. .. ........‘

Q “\~ 1
Av ~-.,,_

. . x . .. a... ,. ,
4! ‘1 4"ps'l

 

SHix‘vQ TA/TEA/ 5y CaNCQEr‘d‘

’U_,.,._.‘..,.. ',_‘_..,.__,..-.-...A-.. 0’ "

5
5
5
5
1
5
5
g

a...- «aw .
v awn“

l/

1 3.. . ~ /26

«Vas- xv‘ - - ‘

5’93““

«,1
.ch

 

5K5 TLC/1’55 576453 i

 

 

. o v .. ~ -1 .v» -. ,- -» 3'93‘ . '7. an— ~.. - N- fl-w-o.o'--~“h--'

- 16 _

Minimum Spacing - g : 15"

108 3 7.2 Reqd. Use 8
15

Spacing:

From end to center.
1 3'3"
7 € 15"

EAST &:WEST WALL BEAM'DESIGN - Sketch 2O

M.B.M. @ Center 2 30,400# x 8' : 243,200'#(06up1e)
M.B.M. = 1/8 w12 ‘
but B.M. 1/10 wl2 may here be used.

Therefore 243,200 x 8 x 12 : 2,335,000"#
' 10'

Assumed weight of Stem 3 300#/ft.
1/10 w12 = 1/10(300 x 480)(21)2 = 33,200"#
398,000"#

B.M.

B'M.
TOtal BoMo at Center : 2,733 ,000"#
VI: 30,400 / El 3 30,400 K 780 X 21 3 38,600#
2 ' 2

b'd =‘v = 38,600 = 245 sq. in.
53— 180 x 7/8

It is evident that this wall beam.will be smaller than the
girders, therefore computations will be discontinued and the
girder design will stand for the wall beam design.

All girders and wall beams 2, 3, l2 and 13 - Same design.

NORTH & SOUTH WALL BEAM’DESIGN 3 Sketch 21
LL (Masonry) =- 704#/ft.

704 x 8 = 470 p.s.f.
12‘

"J ---.

1“

VV—o»
.

g.” o— a
i .
.’
;--—w—.-m—+~.—‘

1".

(“'6‘
EV

 

-,/-' /I

 

w. a- .——---.—_,.— .
C r.
46‘ ‘
IN“

I

-.,.~'
-‘u

,0

“.

 

T

r
‘
3
a...
i
!

Z’f‘
5

_.,~_-,_ —. -_ , c

.3

-r
.v——-- M—r-+-—--~—--—-—- ----- -—-j’
." * a
:19 .
L

l
i
!
L—aov‘
0
q
4
‘.
l
I
i
4
I
.' P n
a.) b
'r" r
l
d ('3 r3
6‘
w fi-vl

5
3 . .. i
4. _ 1.... m
f» A W

uh
.‘
7. .. {I

o "' i r.
2.4
MI
‘/

‘

.5 . _ 2..
r

‘22

‘

..-v.5\flvmu XV §m....- i... v...‘ fin

.
a .0; '1'}

 

 

. i . . . -1... . .
-1... . c . 5 1.551. w. 6835:.
. .. 5 .13.... ¢ 31.111.31-13- all..."

C ’ '
WW...
M
5%

 

 

T -

1 !
L—J———L—

7" 1..

f - ..__.. .__--_ 1-..- -
i 1'
x
L...”.....

“4’
1

 

 

.0
2

'5 K5 77. H

 

.mh- ‘r.

 

 

 

. .. . 1.0).... . | I .I'I. ..

11. (Floor slab) "-'- 351.150 3 50 p.s.f.

1
TOto 1.1L 3 52° po‘efe 3 520 x 4' 3 ZOBW/fie
D.L. = Assumed wt. of Stem = goggrt.
T.L. ‘

228M/ft.

Vsz'ZZBOX 9'21 600#
2... .74. .

Allowable v (Web Reinf.) - 180 p.s.i.
Assume J - 7/8

b'd = v = 21 600 = 137.5 sq. in.
' 9T 1331 778

3.11. = 112 = 2280 x (19)? x 12 .-.-. 937,000-#
10 10 “’3‘

Since B.M. and V are both less than used in design of
the floor beams, that same design will be adequate for
n11 beans 1, 4, 5, and 11.

Sketch 22 - COLUMN DESIGN
All Columns - Concentric, Axial Loading
Col. 1 Sketch 23

f'c = 2000 p.s.i., n? 15

N = 103.6 kips
In order for the Joining members to frame into this
column, the dinensione of Col. 1 are set at 10' in. x 14 in.
Area = 140 sq. in.

to e 2000 p.s.i.

N/wach [1! (n- 1) 13]

103,600 I 140 x 150 x 111 . .225 x 2000 x 140 [1 I 141)]
m 12"

105,100 3 63 ,000 I 880,0001)
p . 0048

-18-

As =AXP=140x .048=6.7 sq. in.
Use 1 1/8 A.‘ 1.27 sq. in. x 6 8 7.64 sq. in.
iFor tie spacing, the least of these three conditions:
é” o ties
1. met over 16 bar diameters 16 x 1.125"'-'18"
2. 48 tie diameters 48 1: § . 24"!
3. Least dimension of column 10" *Least
Reference 2. 1104b ACL 318-41 ‘
This design.will apply to Cole. 1, 3, 7 and 9.
Col. 2 Sketch 24

N32X65kf38o6/W

Let Col. dimensions be 14" x 16.5"
Area I 231 sq. in.
p'- .04
‘As a 231 x .04 I 9.24 sq. in.
USe 1i" A.= 1.56 sq. in. x 6 = 9.36 sq. in.
§" 5 ties V
For tie spacing, the least of these three conditions:
1. 16 x 1.25 - 20"
2. 48 x .3" . 24"
3. Least dimension of column I 14" *Least
This Design will apply to Gals. 2 and 8

 

.- ~- ...,....1 “‘0‘... —~ ..

 

t.‘§.¥.l I'.III

..._ 7/
m...\./ w... A 3 J ..... .1 x
J ,1 7-”... . Ix. . AA 5 1.. a 9f“ I. ..
P v 4.4. o adv
C 7.1 cu, _ "(w r r . ,0 _ /
r ,., / . . ... a
F 7 H u r . . . «tannin. XIII... 1.ij

[ I!
v
yrx:
I
I
l
C
I
1

 

r.

 

-'63
t.

V
. -...-.__.. ..__

\

._._ '

82'

..

 

 

 

./ {ED-bill“. IO! .Ilvlf'l .’o! ‘0‘ LII‘AO‘IL.
.

,. .../.

n H. . . M.
.9 m 64

V
- .. .I c J TiltiakL

///”

I 7
“WU/I\).\IJ VIU‘t‘IlIt-\ .11. 4.:
.H. .. .. .,,.
.1. I W. . r r . ,
a. 7v H x... .. c
. 74 al.24—
A .
I

 

10.....- y} e n I- ..IO.|'0|

 

.44/E / 3:?”

If
. TK/{Y HIV-'13

IL‘ c'

1
l

l,

I
D

a.

31:71)". ' . '
1

I

c

1

.
'0 C
.

. .
-_.--i.

p
w"
j
W —"
;, E
“as-4
p

 

5 NA“ TcH --

J J «5 H"? 3.} 7-

\

a/..'

41’

‘-T'

‘ 'u.‘
5 4

ll

// .. ,,
Z ;,c7/£/*f

 

21‘ el
1‘ III!" III . 0.1T?!
I. 1‘ Io‘ldno'i. -.-.(II| I
‘1 PE‘. ill-t
.}\.A-.r|.l~ a ‘Iv ‘I "II
I'll"
1.1.

ll‘-4l.l 0

Col. 4 Sketch 25
N'8 77.2 I 65k I 142.2
Let the column dimensions be 14" x 14"
Area I 196 sq. in.
142,000 I ggg x 150 x 111 = 450 x 196 (1 I 14p)

144 12
p I .0458 '
A3 3 196 x .0452 = 8.85 sq. in.
Use 6 - 1%" Area I 9.35 sq. in.
%" 0 ties

Ties Spaced 14"..
This design will apply to 0016. 4 and 6.
001. 5 Sketch 26 h I
N'I 2(65 I 38.6) = 207.2 kips
Try 18" x 17" column.

w'- 306 x 150 x 111 I 2950
144 12‘

210,150 a 450 x 306 (1 I 14p)
1g46§9,s p - .0377
1,930,000
A3 I 306 x .0377 I 11.50 sq. in.
‘USe 8 - 1%” .A, I 12.5 sq. in.
5" 5 ties Spaced 17"
COLUMN“FO0TINGS
Col. Footing 1 Sketch 27
001. Size 10" x 14" p I 105.1 kips

-20-

.Allowable Soil Pressure = 6000 p.s.f.
LL I 105,100#
DL M62! (6% LL)
TL I 111,406#

A 3 1113406 3 18.6 sq. ft. Required
4'5" x 4'5" 3 19.36 sq. ft.
L 3 4.4'

Net Press. w = 105 100 = 5430 p.s.f.
, 19.30

B.M. = 5430 (1.5 x 1.672 x .6 I 14 x 202 = 23,400'#
144 x 24 .
Min. d .'%0 Let b = least dimension of columm.= 10"

d I 233409 x 12 = 10.9" Say 11"
x

Le0.h = 4" (Cover)
hIdI_1_§_" Weight a4.42x1§x150'3_2§2#.91<
MsAsgjd-

As '7' M =' 23 400 x 12 3'- l.475 sq. in.
?sjd 20,000 x .836 x 11

Use %" 0 bars .A = .1963 sq. in.
1.475 I 7 I bars. ‘Use 8
1503

Spacing 6" c.c.

 

BO_NQ §TRESS
2
u 211%. . 5430 (4.42 -G-°—ll;:.42§ ) x .25 = 120 p.s.i.
o d

5.1.57 x 8 x 7866 x 11 ,9;
Allowable bond stress I I; x .056 I 168 p.s.i.

Deformed bars, hooked.

- 21 -

Check for Diagonal Tension

Col.

BOND

 

v = w L? - a‘I 2612 = 47.2 p.s.i. '95
a / 4d) .

Footing 2 Sketch 28
C01. 5126 = 14" x 16.5" A = 231 sq. in. P = 168,600#
168,600#
10,100#
T.L. 178,700#

A = 178,700 2 29.8 sq. ft. Reqd.
6000

L.L.

D.L.

L 3 5.5'
1:28 30.3 sq. ft.

Net pressure w = 1685600 : 5630 p.s.f.

B.M. z 5630 (24 x 131 x .6 x 2 I 14 x 24 x 2 x .5 )
: 28,000'#

Min. d g, 28 000 x 12 x 9.5" Say 10"
236 x 14

Let h : 5" (Cover)
h I d g 15"

s = 28 000 x 12 _ g 1.94 sq.in.
,000 x~.8 x 10

Use 3/8" a A ; .11 sq.in.

A

$122.: 17 / USe 18 bars. Spacing 3" c.c.
ll -

STRES§| 2
14 20~

u .-.- 5630g5.52- x .25 z 168 p.s.in. 9};
8 X 101 x .8 X 10

Use deformed, hooked bars.

 

. -‘l ‘I“ u

 

A“

1
1
. ac. 1| IV. (\L
D. A
_ Ru ..\r\ \ 0 u
1...|l|'.l.|!l.‘ [ ‘!- 1
... 7 1
1 , 1
I _
_. / w
.1 ..
Ti -iii-.i.l,_ 1
. .
1 / 1
_ /
.nIll -u' l. 4. I .|\
4 .5... . , . 1 1.
.. / / I ~
‘1! ... .. , . / 1 L
N. sf .4.‘ .. I
1

ll '5' 1! 1 'y‘ ‘.'O1llllo

.‘ * —_ “fig.-- * -I.

 

 

 

 

hm...“ -

 

-
. V . v.1. In I. .0 4 .
J. (I Y if q . ICI-

Smu r.

I. . 1".sctll'i t! It‘ll-.- 'F‘ ‘0. It... i"0:-‘i {llg‘

kw.

rt}; 9...» .1

 

 

 

 

M
1

5K5 73st "Z 7'2 8

 

 

 

 

“WW ..~—.-—-.;n.—-—.:——-_ ‘ .r, ...

 

DIAGONAL.TENSION

Col.

v : 98 p.s.i. Too high
Therefore, d must be increased.
let d g 12"
Then v z 68 p.s.i.‘Q§
AllowingAS to remain unchanged is on the coservative

side and decreases the bond stress.

Footing 4 Sketch 29

Col. size 14 x 14 p : 142,2oo#
L.L. - 142,2oo#
D.L. - 8154Q#
T.L. - 150.74o#

A = 150 740 = 2500‘ egofto RGQdo
6000

Use L = 500‘

Not pressure - w - 142,200 a 5670 p.s.f.

 

25

B.M. : 5670(1.792 x .6 x 1.79 ; 1.17 x 1.792 x .5)
= 30.000'# . '

Min. d =fso§ogo x12 3 10.5 Say 11"

h g 4" Cover

h I d - 15"

30 000 x.12 = 1.89 sq.in.
20,000 x .866 x 11

USe 3/8" d A = .11 sq.in.

A .-
S

1.89 = 17 / USe 18 bars -Spacing 2%" c.c.
.11

 

 

 

 

BOND STRESS 2
(14 l 22)
18 x 1.18 x .866 x 11 ,
Use deformed hooked bars.
DIAGONAL TENSION

v = 5505 p.s.i. 9E.

Col. footing 5 Sketch 3O
Col. size = 18" x 17" P = 207,200#
L.L. - 207,2oo#
D.L. - 12,4oo#
T.L. - 219,600#

A z 219,600/6000 = 35.5 sq.ft.

L = 6.0'

Net pressure w 3 207,200/36 . 5830 p.s.f.

B.M. : 5830(2.1253 x .5 ; 1.5 x .5 x 2.1252) = 39,800*#

Min. d = 39 800 x 12 = 10.6" Say 11"
236 x 18

h z 4." COVGI’

h/dsl5"

As 839 800 x 12 ' 3 2051 sq.in.

Ems é" d'bars A 3 .196 sq.in.
2.51/.196 : 12 / UBe 13 bars Spacing 4%" c.c.

 

BOND STRESS

13 x 1.57 x .866 x 11

 

Use deformed, hooked hers.

 

I .ku‘; ltzn\.0c_lv .I ..'I\..

 

.1... Q .0 V
.
1: \
.“n‘ «1'... {I} If ...., t1 .3...
\1
1
. .
1
1
.
1
.1... Ill-l III. III... tall-I III I...
/

o
o

 

A. .,. 0 V1
wrwék\ xmd\ I.
\\

I .
.I p A
I
. I...» .. .I *3! ill Wyl1t ‘Ivvr‘n'cllal- .
. x .

e;

»‘ “|.|.|I’Ii.7i‘1‘(l.3

 

1
I;
I
f
L

 

 

‘~-_--_ _. -.~‘~—¢ ...-—- -

 

 

 

 

 

 

SKETCHES

 

 

on... 1.3.1.1.. islii.‘ Iql.

 

 

1... .- Ww-w‘ ow“... — —---

Diagonal Tension

v 2 87 p.s.i. Too high

Increase d Let d = 12"

v 3 72 p.s.i. OK
FLOOR DESIGN -
I?ron1an.analysis of all equipment to be stored in this build-
imug, the max. wheel loading to be eXpected is 13523.
[Being a 3000# concrete, the max. extreme fiber stress in com-

pression, 1?C 3 1350 p.s.i.

'Fhe bearing area of a Ford tractor tire may be taken as 19
sq. in.
1250 ‘ 125 p.s.i. Q;

10

.A sub grade of a 6 inch layer of compacted cinders will be

prepared to receive the slab.

.A 6" unreinforced concrete slab will be poured in alternate

sections as illustrated in Sketch 31.

A %" filler will be used at the junction of the floor slab
and all other members such as columns, walls, etc.
DESIGN’OF RETAINIVC WALL -

As this retaining wall is to serve as the back wall of
the first floor, it must be at least 12 feet in height. As
.a further precaution against frost action the wall arm will
be made 14 feet with 2 feet of the wall above the footing be-
low grade. '

In preparation for computing the total earth pressure

 

_. m— o *g-p- -nh.--.._.-.—~

lo .0)" Ic’lyul‘i’ ‘ A .I .l.:.

 

 

v
.x :4 . .
IN P {P1P f_r .
p p . .
/ o
-- 1 1 -. l .. I 5 d .
1 .1 1 1 .11 1 I). . t 1,
... s u. . 12.111! .1... 0 I . -4--. filial-:1: -n. all .1. 1V 1 ._
.. 1 ._ 1 17 . a 1 1 1 1 - . x. W ,1 1.
r! ._ _ ~ VJ _ _ w! I! A— .1/1 . * f’: o
1 . c| [I I II .n’ , m .
_ rll I 1 ~71?! t , .fo/I t r . M...\. .ls,\ .nh~.\ 3 .. // f .

4.1-.. _
n —. .
p.111
1: 1

by u i

'.‘\ N /-“\’v'-)

f .

‘1 If

I] t

.5 1 3., 1 1.1.1 v

u u.
,IL' . l‘ul.llL..HH1lk I

/2

)\\\.<..\u J. nu U

/ .. N -

R
- - x . . . 1...
L x .

N1

r

M/ . , ._ .. A. /2

“ lliO‘ 0,...0 . I»

’1’
N

. ../
1 uoll 'll \ - 4.
III} .. . m
u H. r Mull !
n .m M llllll 1‘? 5.: Il‘ II A m —
_ 1. _ . {fr 1
”I! 7 . .lll ll 6L

,
-w‘

1.,1

_ . _
. .(prlrb .5? 10“; .. h '. I.....r.!1l P Iii, i .
_
1' no

.Iuvhfr.- ,1.

I g! $.1- ll.

.\
1\ __.

lg

1

--O —-L - ..b—~——-'~—--‘ An ,

.
.
q .
. I
. ..
1
y
.
.

\. _
a

x 7

_

s
. 1" L

’_
L ‘h—‘n... «mvfl.

/C(:7-/Ns.

04.

/
L.

,.vl.ll.ll

’
1 f
\A
1 1 .
\\ 1 .
.I‘..
c.\ . .
.‘I
v
‘ _
I
. .
1
_
1 . \k
\ l. 1
\‘ n
“\ I
\
A
Q .
.
u I
1.
“l. \
K I
1 iii- .I In“. ltll'ublt'ii 1.: III“ I01.

..-.|.‘\|..O|.I. ‘ u 'yI.’ .. wt

‘li'tl .I‘l“- "Ltlo

.| (Ill.

3/

,‘i’f 7 C H

<
x.)

\w"“-

,.-..- - 75-:

Gin-t 1.4-.1.

agyainst the wall, the equivalent surcharge of earth must be
ccnnputed for all loads eXpected to be impressed on the drive

Emissing by the west wall and at the loading platform.

’Tlue largest vehicle expected to park next to the loading
jplatform is a semi-trailer as illustrated in Sketch g2. Its
‘Nheel.load of 10,000# is considered to act uniformly on an

.area.6' x 84', the dimensions of the tractor and trailer.

Area 204 sq. ft.

L.L. 4,800 / 2 x 20,000 = 44,800#

Unit L.L. = 44,800 = 220 p.s.f.
204

220 = 1.83 feet equivalent earth surcharge
120

Total Platform Load - 1175# per Lin. Ft. of support.
Total L.L. = 1175 (18 x 2) = 42,300# '
This load, too, is considered distributed over an area 6' x 18',

the dimensions of the platform.

.'. Unit L.L. = 42 300 = 332 p.s.f.
G x l§

392 = 3.27' Equivalent earth surcharge
120

Total surcharge = 3.27' l 1.83 = 5.10' Say 5'
It is considered that an economy in the wall design out weighs
the added cost of modified wall beam and column design if the
loads imposed on.WB2 and W33 were transferred directly to the

retaining wall as vertical loads.

This therefore will alter the design of Wall Beams 2 and 3

and also Column 2.

-26-

'Tlue thickness of the wall at the t0p will be 14", and there

will be no batter.

Eur use of RANKIVE'S Theory of Earth Pressure the magnitude
axui point of application of the earth pressure may now be
calculated.

3 15H(2h / H) 2 6250#

= l6t

25‘1“”:
n
()1

/ H) = 6.35' distance from base
/*H)

B.M. at base 2 6250 x (6.35 - 2.00) = 27,200'#

M
n
A
[‘3‘ h)
‘33"

.‘V = P = 6250#
To determine "d" required at base section x - x,

(1) M6 = kbd2

from which

d = 27.200 x 1é* = 10.7" Reqd.
“236 x 12

Say 11"
(2) v = _y_
bjd
from which
d 3 6250 3 19:9" Reqd.

 

12 x .266 x 60
The selected thickness of 14" provides ample cover of the

maximum required "d".

 

As 3 M = 27.200 x 12 3 1.71 sq. in. Reqd.
ijd 20,000 x .866 x 11

From Table 3 RCDF

Use a" 0 Spacing 3" c.c. A = 1.76 sq. in.

 

«at..-

7-a .1"~ 'IQ-r-vr

 

o'- ‘1‘ “firm. ".‘N' n‘f- v“ rfip—o—WJ‘H—_.. ”-

. .0 ..4......

- —--- v- u—- c..-4

‘m—r“.—‘pfl-‘ --.¢—;
—

I. ll .I'I Ill. i ‘01:: it“s‘li

It’sllllv‘lrii n.-. .[l’luhlll .4iv10?’ A (t. '17 l . Vi i . .it. t. n o 1|»? .I . . . . .7.

—- 8,...“— .~__ _.-_.
;
I

:‘J
'/«4’
W

A

. ~ .- . Oil- 1‘
‘ r
. Y) . - _ ._ ..
fl... ,1 .1 .1 . D 1 z ....
,.\ a . I. ,
* f" o .,.. _ .o, ,
1 - d .
If“
w
. f
_ - .
«x _ .. A '5. r) v .. x
.. _
I -
n. 1:: III- 1;! n l v]: 1
_ _ f . . ., I“
a. I a .

r _ ‘ I ‘ 1 . .5‘ .1.”
.l i’t- :l-I‘ slit 6": 1
A \
J l . .. u
'11 ID- \0 .I-.-..l -..l... . I-..L||l4ll|’|t. . .- -lu . -1Iluol o .. . Olh _
a. x H 1 .4
. ,u m ’ .rn it. o...
s . . o
.1 .C .. . 1
1 . 5 . 1 1 _ .
. o . I.
a . . . a L
. P‘ a! .—' “M L t V will. ’1 o
l 1 . 1.
u m .. * XI
— . ’74 1 “ L
1 . _
1 a 1 . , f v
a. A . r - .III III -. l.—
w . . b
n ‘ w .~\‘
.

. ti 0
it u;‘ll"¢"!‘l’. .l‘.x§. \In'l‘ 0"?)

.r . {lit-0n

 

IF‘II". 1" {'3'

 

 

 

a

 

--. ~. —— “A'—

--oo‘

I

“o~4«—~ 4 ”~w'V‘h-‘M h noun-- .. - 1‘

'u = 5250 : 70 p.s.i. pg 160 p.s.i. All.
9.4 x .205 x 11

Ixength of emoedment required.

1 = :5 D = 20.000 x .75 = 24" Reqd.
4u 4 x4160'

Simme there is but 24" in the base slab, this vertical tensile
steel can be bent and extended into the base toward the toe '

in order to gain the required length. See Sketch.§§

Temperature steel will be used in both the front and back
faces of the wall as illustrated in.DRAWING NO. ‘4 .

DESIGN OF BASE 8 AB - HEEL
With the base slab dimensions selected as follows:
Thickness - 2'
Toe - 3‘ Sketch 33a
Heel - 5'
The resultant passes through the base at a point 3.5 feet
from the toe, within the middle third.

Factor of Safety against overturning - n 3 l 3 9.16 3 4.25
l-Za 90 16-7 -_

P - earth pressure 3 6250#
F - vertical component of Resultant 3 22,428#
E - Resultant 3 24,000# See Sketch §§

To solve for foundation pressures Sketch gg

P max. 3 (41 - 6a) F 3 4170 p.s.f. QE 6000 p.s.f.

H

All.

P min. 3 (6a - 21) 720 p.s.f.

F 2
l2

 

o---. .m..._

n 7...--

'-" Mr. -1.-

"b"’ -< ”h- u.- .M-uuo-- m. - am...- .~--- ..~..-

-—.-~‘—r—_-a._...~------ .—-—...-. .
.

p...

[V :‘I _'

.’ fr"

>Iullb’llr' -ill’ln“- .

0t ’!'.i| ht.
\w“il»‘ 0".

\I .I‘Irll .A

w

’I'tl'lr.

r 1......)l"
ML.

 

I' [70.30

m“. —.—-—.. “——

'\

 

 

Q
‘I 1" '00—. -v.--—... mac-0,

. .C
‘ 'r
it ill. a I ll .1 a
a . M W r...
‘
R.
w bk)»
* ‘ ‘
1
a 1 .
1
1 .1
1 Pl». .
1 . ..
/ _ .
_ w 30 w
; Q

~
5._ «yu—gfl n.-~-—..o-. —

. . 1 w. .1
. 7. .

.t '21 ’I-‘|ll‘.f.|lllo.|l'l t’l..‘|'l‘ll' '¥.I‘;§LF .
N . m
a

‘ ‘
1/.,1.. ..
1 4.1. u
.v 1.7: .1

w/

 

ii- - 1 i l y, 4 _ m /

.1 ,1 m 1.1

W 1... u p. a r 1

z 3 ‘ L 7.1 1a .

a M 1eflxxomu

.. 1 _ m .6. f M ...

. / .

N 1 m. M N r w .1

w h .. J / m

A w LiL .fl +1! a” N S.

w. c . N U i. a)“ w

r .... . ... .11.. 5. I .

-17 . 1 Ga wM / 1

1 . 51.. A a

t I. It if...£.......1l c... ill.“ .5 ul 253?...0195

f

11,400 x 2.5 / 1500 x 2.5 - (2700 g 720) x 5 x 2.9
2 ‘

10,ssot#

11,400 1 1500 - 1710 = 11,190#

d (for moment) 3F5850 x 12 = 6.8" Reqd.
2&3 x 12

d (for shear) 3 11 190 3 18" Regd.
. 35 x..836 x 12

.A.slab thickness of 24 inches was tentatively used and thus

5

 

 

it is evident that it will more than suffice.

As 2 10 850 x 12 = .42 sq. in. Reqd.
561mm
From Table 3 RCDH
Use 3/8" 5 Bars Spacing 3" c.c.
A 3 .44 sq. in.

QOND STRESS
u = 11 190 = 153 p.s.i.
Z.$ x .836 1 IE
For sufficient embedment length:

1 = 20 000 x.3/8 = 11.7"
4 x 130

- TOE DESIGN -

 

M33 = (4170 3000) 2.2 - (3 x 2 x 150) 1.5 = 6537'#
V33 = 3585 - 1350 = 2235# -
d (for moment) = '6537';tl2 = 5.75" Beg .

253 1.12
d (for shear) = 2235 _= 3.58" fieg .
30 x..§33 x 12

A slab thickness of 24" was used for the heel, therefore the

toe will be the same. .

A d = 18" was selected.

3 Q537 X 12 = .253 sq. in. Reqd.
AS 20,000.x‘T§€6‘§‘I§

No tension steel will be designed for the toe section as the
vertical steel of the stem willbe bent in the slab toward the
toe in order to obtain sufficient embedment length. This
steel, a" d 3" c.c. will more than provide the required steel
area in the toe.

- FACTORS W SAFETY -

ggarturning

F.S. = 4.25
Sliding

v.3. = 22g428 x.tan 25° = 1.68
22211123

F.S. = 6000 = 1.44

Zl76
- POINTS OF CUT-(FF FOR STEM VERTICAL STEEL -
Bending moments will be computed for 1' increments of height
of the stem. Area of steel required for each bending moment
will be computed and it will then be a simple matter to de-
termine points at which steel may be cut-off.

 

 

figight from top $133.11. A 339d.
1 80 Ft. Lbs. .005 Sq. In.
2 340 ’ 0.022 '
3 810 0.051
4 1520 0.096
5 2500 0.157
6 3780 0.238
7 5390 0.339
8 7360 0.463
9 9720 0.610
10 12,500 0.785
11 15,730 0.990
12 19,440 1.220
13 23,660‘ 1.490
14 27.200 l=Z;Q_
As = ¥:§3%gv rs = 20,000 p.s.i.
- 3 a .866
" Betas . = 11"

4 - i" 15 Bars 3" Spacing
Add Embedment Length - L

L 3 fc x D 3 lg"
Zu
For points of cut-off see Sketch 34.

.i‘u.‘ b . ’y o t. . v. 1: . t . . l "1 {I‘ll c [I u . 4
l 4' 3’0 .4 y I .60 . . .. . I- .I n . .7! .ID . 1.1 {I'll '35 . .Iu}.l.1t'l. ' |.>‘: Irxrl
4 1. . II. '1 ‘r 0...,
i..

 

 

 

 

 

Ii. II! 1“ l
\l. \
.
\.. ' ’
I
#1 [III IIEI’I‘... '1 I»
-
\.. (\ II“.
I."
I I {II 1.1!»..IIIIIII] I [I’ll l. .I'OI - -

 

_ ,
1 I: . .. \ l..\ I us\

3?

- s

A-~._ -- 5

, . .4' o.“ _.

-n. m---v_-- .- q‘fiso—g “—_.—o-‘-.

. _~_-,__', -

r- .3... “~-

?
-....4L-
.3

.L. o‘L‘lI):.’l I.|1¢l‘| ‘.. 3|

.{;.:}

.M-vm aR —O

a...‘ -

. _
p... t,
r.“ 7
a- a.
I
z!
e\
a .
. — pt

~ a

 

I. - .. . ta 0"..l‘01... li'ii‘. 1.1.0'1 (CI-IV! !.t‘.l." 1,1“? 1*|zl'tl“. 1+, {It'll}. ‘- .|on..l'. . Ln.lvu-..lltlill'ul‘lalltl : 1"

“q-
A
J

3:4 s— cums-ts .

14 K .... i4 .... \ 4.14144 .4 m.

.6“...
--- M p~—.4._.__--A4- .A-«n-o—r «Ac.— M -,....-v “14—-

 

: I .0 i. I‘I.«.lu ll 1...! ll
IE.»II}‘III§.I’I-§ilg s 4
til}...

i ‘11! ‘l

-31...

WALL POUR SCHEDULE
All column and wall footings will be completed in one
pour. Dowels will be placed in these footings of the same
ldiameter as the vertical steel in these reapective joining
members. Dowels will project 25 bar diameters beyond the

footings. L 3 18"

A key way will be formed in the t0p surface of the wall
by placement of a 2 x 6 board on the centerline while the
concrete is still in the plastic state. Before pouring the
stem, the footing surfaces will be thoroughly cleaned, scour—
ed, and a é-inch layer of neat cement paste applied before

the wall pour begins.

The wall proper, excluding the columns will be poured
in two lifts. This is deemed necessary for proper puddling.
Form.work will be placed for the front or inside face of the
wall for the full height - 14; Form work for the outside
face will first be placed up to 7' for the first lift. Here
again, a keyway will be placed on the t0p surface of the
first lift. The second lift will complete the wall to grade,
the bottom of the floor slab.

To provide anchorage for these two members, dowels bent

90° will be placed so that 25 Die. extend into both the wall

and floor slab. Use E" d bars. L = 4'.

To provide anchorage of the wall and the columns, the
longitudinal temperature steel of the wall will be extended

25 D13. into these columns. L = 16".

 

 

 

 

 

Re 2‘0 I'm'ng. We //

586/. A~A

 

 

 

 

 

F/Oor 43/35

3.1”

. 35 “r?
I ‘.
7') l [#3
C: ' * 1”"
..- _.
J ‘32- 33
N; "
\r’
-4
n (’7'
.1 ’1’!" I/ - f/
“A / ,‘ "Y
/

FRAMlNG PLAN

Dwg, No.1

SECT/O/VS

  
    
   

 

 

 

 

 

-32-

Upon reference to "Reinforced Concrete Design" by Sutherland
and Reese, it is decided to build this wall in one section.
The author prescribes a maximum.wall section length of 60 ft.

without putting in eXpansion joints.

- STAIR DESIGN -
Stairs located and dimensioned as on.Sketch.§§
L.L. - 100 p.s.f. Use 6 inch Slab
D.L. - _Z§ p.s.f.
T.L. - 175 p.s.f.
Total rise from top of floor to first landing - 6 feet.
Stairs with 10" run, 6" rise and l" nosing will be used.
Horizontal projection of stairway = 9.16'
Platform.slab 4' length, 8' width
The stairway slab and platform slab will be designed as one
slab, length = 13.16'.

v = 11. = 175 x 13.16 = 1150#
2 M2“"""

3.1!. = 112 = 175 x 13.162 = 303054
10 "‘"1'0' "" .

 

 

 

d (for shear) = 1150 = 1.85"
x . x 0
d (for moment) = 33030 x 12’ = 3.6"
412 x 235
Use d = 4"
AS = 3030 x 12 = .525 sq. in. per ft.

20,000 x .866 x 4.6

USe 3/8" 5 bars Spacing 2%” c.c.
BOND " L1 ‘: 1150 = 5802 peeoio

e X o X

The lower and upper stairway slabs are fixed to the landing

.ili‘tiii ’Illll‘lslt In- II 0". a '5 -‘Il

 

I 19"]. I.I.. .Icl‘

 

 

 

 

 

 

, -ngv- ..- ,2,

 

 

 

u“- —~.. w

—.—-..-- . ..

11 {it} I.O.I'3|a% .1 A

 

 

‘tlbl‘ll'iv't‘hl. ‘-“\. In... ‘1‘... n I l x n

,/

,c J.
a

. L
’l
..
. 1 .. :
(713.11.- L.
. 4. ll...’

\IL \ rusl..
x...
\

Jl

u
r

 

 

1‘i I‘Is...‘ (I14

-J

1._ v

4’ ".4

I I
1
.-J ;- -d-—.._——-J

but.

9/5

 

 

-33..

slab along the inside edge. The landing slab, 4 ft. long
is fixed in the west wall. The stairway slab reinforcement
is continued through the landing slab and anchored into the
wall a length:

l 3 20000 x 3/8 3 l6"
4 x 120

This same reinforcement will be run along the 8 ft. length
of the landing slab.

A beam will be designed to support the east or inside edge
of the landing also. The north end of this beam will be
fixed in the north wall and the south end of this 8 ft. beam

will be supported by a short column.

pesign of landing slab beam -
Stairway slab reaction 3 ll5Q#/ft.
6 inch landing slab 3 75 #/ft.
R 3 4 x 75 3 l50#/ft.
Landing slab reaction 3 150#/ft.
Total load on landing beam 3 l300#/ft.
Assume weight of stem 3 20#/ft.
T.L. 3 1320#/ft.
V 3 1320 x 8 3 5280#
___§___.
v 3 ‘V , b'd 3 5280 3 33.6 sq. in. reqd.
5753- EEG—x 7/8
Web reinforcement needed - stirrups
Let b' 3 4" then d 3 9"
d / 2" 3 ll"
/ 3 6"

- 34 -

 

ll - 6 3 5"
5 x 4 x 150 3 20.4#/' Weight 0K
144
B.M. 3 1320 x 64 3 10,560'#
8
j = .92

B.M. = Tjd

T 3 10,560 x 12 3 15,300#
.92 x 9

As 3 15,300 3 .765 sq. in. reqd.
20,000

Use 2 - g" 3 bars AS 3 .88 sq. in.
BOND

u 3 5280 3 136 p.s.i. Use deformed bars.
4.7 x .92 x 9

 

Maximum shear stress at ends:

V1 3 5280 3 160 p.s.i.
4 x .92 x 9

max. shear stress at half Span 3 v1 x 25% 3 40 p.s.i.
Total shear taken by stirrups:

v = 100 x 3.33' x 4 x 12 = 8000#
2

Using i" 3 stirrups, A 3 .05 sq. in. 2A 3 .1 sq. in.
.1 x 16,000 p.s.i. 3 1600# taken by each stirrup

8000 3 5 stirrups Reqd.
1600

Minimum stirrup Spacing : g : 4.5"
2

3.33' x 12" 3 9 stirrups @ 4.5" c.c. at each end.
4.5 '

 

Landing Post Design -

Let A 3 4 x 12 3 48 sq. in. Gross area.

- 35 -

p 3 .02
As= 48 x .02 3 .96 sq. in.
Use 4 - 4}"$bars A = 1.00 sq. in.

Wt. = 48 x 515 x 150 = 275#
1214.

Beam reaction 3 5280#
Total axial load at base 3 5555#

5555 3 1155 p.s.i. Compression stress upon floor.
48

Allowable = 1350 p.s.i., therefore no footing is re-

quired for this post.

PARTITION’WALLS
The Herth, South and East lower floor walls will be
concrete partition.walls.
A tentative thickness of 6" was arrived at by the use

of / = 10' x 12: 4"
30

Assuming a wind load of 30 p.s.f. as before in the
roof truss loading, this thickness is investigated.

This wall will be independent of the inclosing mambers
such as the upper wall beams and wall columns. Therefore it
‘will be assumed that there will be no transfer of loads from
these members to the wall.

The maximum wall area under action.of wind loads will
be taken as fi'of the East wall minus the door area, the col-
umn area, and the wall beam area.

.A 3 12 48 - (14 1.12 l 3 3 1.24 I 12 x 10)
%(X)12 1‘2 _

A 3 90 sq. ft.
Total wind load 3 30 x 90 3 2700#

Unit wind load 3 2700# 3 208#/Lin. Ft.
. 13'

This force may be considered to act at the third point from

the base. Height 3 12 feet - 33 inches 3 111"

 

B.M. 3 208 x 111 3 770 in. lbs.
’2

U

 

 

 

AS 3 770 3 .011 sq. in. per ft.
20,000 x .866 x 4
Use %" a 5%" c.c. Spacing
Shear 3 v 3 208 3 5 p.s.i. OK
12x .866 x4
BOVD
u 3 208 3 38 p.s.i. OK
2 x .79 x .866 x 4
Temperature Steel .002 A

C

AS 3 111 x 6 x .002 3 .0093 sq. in. Reqd.
144

Use i" 0 l2" c.c. Spacing
Door Lintel
There is no appreciable load of partition wall above the 10'
doors, therefore lintel is deemed unnecessary at these loca-

tions.

Special reinforcement will however be placed at the top cor-

ners of the door openings in the wall.

PARTITION’WALL FOOTIVTS
For a 1 ft. section of wall, 6" thickness, height 117"

W 3 .5 X l x 117 x 150 3 730#/lin. ft.
12

Assume wt. of footing 3 150#/1in. ft.

 

T.L.: 880#/lin. ft.

280 3 .147 sq. ft. bearing area Reqd.
6000

Use a 2' width 6" thickness
No reinforcement needed but temperature steel will be
used.
Use i" o bars 15" c.c. Spacing
For Drawings -
Girder Stirrups -

%n 5 l (2) 3n 7 3 15"

(1)

(2)

(3)

- 38 -

WATER DEMANDS
irrigation
3 acres @-§ in. - acre/week
Each acre to be irrigated once every 6 days

43,560 x .gg 3 967 cu. ft./acre/2 days
12

Q 3 967 x 7.48 3 7240 gal./2 days
This quanity of water to be pumped in about 10 ~ 15
hours and stored for 2 days.
Tank must have min. capacity of 7240 gal.
Pump and well must be able to supply - 7240 3 724 Gal.
10 Hr.
or

7240 3 485 Gal.

 

15 Hr.
Livestock and Toultry Reqm.
Cattle - 20 Ed. @ 10 g.p.d. - 200 g.p.d.
Pigs - 10 @ 2 - 20
Poultry 200 @ 5g/100 - __;Q__
Total - 230 g.p.d.

Spraying Regm.
The Spraying equipment, capacity 400 gal., is used in the

Spring, summer and fall. Spraying services are required
mostly in the Spring and fall, therefore any demand for
this purpose will not be added to the other demand for
the total figure.

The max. Spraying accomplished in one 24 hr. period is
5 loads 3 5 x 400 3 2000 gals.

- 39 -

It may be stated that max. demand conditions will occur in
the summer when irrigation will be carried on at almost a
constant rate. To this irrigation demand will be added the
constant poultry and livestock demand.
Qt 3 7240 l (2 x 230) 3 7700 gal. (Every 2 days)
22%9 3 770 gal./hr. Rate for a 10 hr. pumping period.

Assuming the water table at 75' below Gd. Surface.

A 4" well casing used
Using these conditions, different type pumps will be selec-
ted from "Deming Pump Catalog" and a final choice made.

 

 

 

l pipegiet 2 pipe_1et Force Pump

Unit No. S-660 Unit No. C-660 10150

15m: 15m: 1§HP

Capa. - 845 g/hr. 845 g/hr. 855 g/hr.

2299 3 9.1 hr. of 2299 = 9.1 hr. 2299 3 9 hr.

845 pumping 845 855

4" Casing But, a 5" casing 3%" casing
must be used. --

Unit Cost $266.00 $262.00 22229200

The 1 pipe Jet pump will be installed because of its low
initial cost and size of well casing required.

- STORAGE TANK DESIGN -
The pump unit selected was capable of pumping against a
20 p.s.i. tank pressure. Since a pressure tank is not be-
ing used , the maximum allowable lift from the pump to the
storage tank is 20 x 2.31 3 46.2 ft.

- 4o -

The minimum tank capacity was computed as 7700 gal.

In anticipation of future expansion and peak demands, a
required tank capacity of 8500 gal.

8500 3 1136 cu. ft.
7.48

Assuming a tank 10' in.height.
.A 3 113.6 sq. ft.

 

\

Dia. =F13.6 x 4 = 12.1 ft. Say 12 ft.
”‘IT'

The flow of 100 gallons will cause a decrease of 100 3 13.35 cu.‘
7.38 ,

13.35 3 13.35 3 .118 ft.
A 1:3

1 ft. = 847.5 gal.
The discharge of each 100 gallons from.this tank will lower
the water level only .12 ft. This of course is true when the
sides of the tank are vertical or if all horizontal sections
are identical. This feature is favorable, that is, a constant

static head under discharge provides uniform.flow.

Rather than use pressure switches to govern the action of the
pump - a float switch arrangement will be installed. This
arrangement will operate so that pumping will begin.when the
tank water level is lower than 5.00 feet and shut off when
the water level reaches 9.75 feet.

9.75 - 5.00 3 4.75 foot range

4.75 x 847.5 3 4030 gal.

4030 3 4.75 hour pumping period.

335
The outlet of the tank will be installed at least fi-ft. from

the bottom of the tank in order to prevent flow of settled
material in the line.

The advantage of static head gained in elevating this stor-
age tank is deemed insufficient as compared to the cost of
constructing a suitable support structure. Therefore the
tank will simply rest upon a Spread footing. Therefore the
gross static head available at the tank site is 9.75 feet.

The well will be located adjacent to the tank and the pump
will be mounted in the building, in the compartment on the
first floor under the stairway landing. The distance from
the pump to the well and tank is 20 feet.

Ground elevation at tank - 130 ft.

Length of pipe line to the point of irrigation distribution 3
163 feet.

Elevation - 108 ft.
Difference of elevation - 22 ft.

Total gross static head - 9.75 f 22 3 31.75 ft.

Ground SlOpegjalong pipewline)

L1 - Water tank ,4 4s' - (130 - 116') - 14 ft. Diff. Elev.
L2 - Next 51' - (116 - 110) - 6 ft. Diff. Elev.
19 - Next 69' - (110 - 108) - 2 ft. Diff. Elev.

The pipe line will be laid to a minimum depth of 2 feet.
No account of frost protection will be taken as the water

facilities will not be used in freezing weather. PrOV1sions

iliit‘} 1" $10!-."5‘13 ’ 2.7.! .l”\ .0544.

.m--—s -w'q— .“2‘. .

\<0\.k v. v V Wh.r..>\\
Q. \\<\5Q\

v- .. ”we”: pvr A—mn

tun P 1. i.

s... w .

.0.”

.\u_

if.

u:

_ i.

1:.

.10.,

.1:

1
4K. .: .

 

 

r ~§.l-.'.-. ‘11.] I141

 

'.Ilp 'nll' I

\. .l

a» llk. '95.-.l.~"...4 I... .l. ....ll.0.ln

ell-1’. i .(I‘i

- .o‘L..blwI-’o,vl

~.‘L "Ihr

 

xi,

 

 

'ow-Q-u -- ~ ,

+-

\\i ;.0|I'-l' 3-,| Ill..." III). 'Qll‘vtillg.

-42-

will be made to drain the line.
Slope L 3 l4 3 32.6%
1 as

= 6 3 11.75%
L233:

3 2 3 2.9 %
1369

- Pipe Size Selection and Friction Losses -
By the use of a diagram based on Hezen - Williams Formula

with C 3 100 for C.I. pipe after 20 years service.

Selecting a 4" diameter pipe, and a desired Q 3 100 gal./min.

Head loss for L1 3 1.25 x .43 3 0.54'
Head loss for L2 3 1.25 x .51 3 0.64'
Head loss for L3 3 1.25 x .69 3.9299'

Total Head Loss 3 2.04'

V 8 2.6 f.p.s.
A total head loss of l velocity head will be assumed sufficient
to cover bend and nozzle losses:

hL = 2.62 = .105 ft.
2 x 32.2

Total Head loss .5 this point:

2.04 / 0.105 3 2.145 ft.
The next point of distribution and the end of the line for
the present period of design is centrally located in the farm
yard 110 ft. from the first point.

Elevation - 101'
Difference in elev. - (108-101) 3 7'

Head loss in L4 3 1.25 x 1.1 3 1.375'

7"1“!
-'Il“}.l‘lflui luv.

 

 

 

new EL 6:

. ”\U\\¥ “\(\nn.\\\fi.nu.\ V“?

 

 

m

(\(Nh

 

Werngm

 

 

 

 

\<v\k \Wxfivv

 

u“...— M o..-- - - -o. - .—_-

 

 

ril'ntI" 1!-.1"..Ulv|‘" ...u" l | I, It- I 0'! n...‘ '. ‘I‘Q

 

-43-

Head loss for entire length:
2.145 ¥ 1.375 3 3.320'
Net Available head ' (130 - 101) l 9.75 - 3.32

35.43'
Pressure 3 35.43' x .433 3 15.3 p.s.i.
This pressure is available for fire protection for the barns

and residence.

W
Road width - 12'
The material for this road will be taken from the gravel out

crOppings on.the farm. maximum aggregate Size will be 1%".

Those sections of the road with the crown on the centerline
will have:

Thickness at crown - 6"

Thickness at edge - 3"
This type of road is an adaptation of the feather edge type.
12 ft. was deemed an adequate width and since most of the
travel will occur along the edges, these were made 3" thick

instead of 0".

The surface of the ground along the read line will be cleared
of trees, stumps, rocky and other objectional material before

placing the pit run gravel.

After placing the gravel and compacting, a binder of
native clay will be Spread on the surface and allowed to
‘weather and drop among the aggregate. The proportion of clay

.-.__-.-'-‘-~‘_ ._ -..—».4——-~_.»..._.. .. fiv- m-.. .

__ ~~—o

,_ ”‘_y- a" H... .—._

.4 Roll: J
\/\ \qu his...“

0 Temp. Gr. ex»... .38
..,.. 2 t. .9. yum“
.b

.,... HVVVV_~._ - ..

. . 0px umux. E 0
I Q\.\_3uc.v.>.kok M

1.15.. {00¢ Nb

comfy

I

.w

a A.

'1‘.-
"i

I, [.1 ,lc. In . I 'u. ,. tin: ,‘ u I Ii :1 . .I u .I!A.III .lll‘lll.‘ -.II.|'III

M. MO

0, .u -MN QXJ.
new . Q 0

.9936, . 00 x V

Mei u .00
. v
5.6.69 ... UQXV

\

h

.6: 00
ed... “manta .m

\1 HI; . . . e. u

0.. ..\/ -... I. .>... .X
\o {I fix \ . x
(n 1

Ni .. v.3 - law»... :3. ed Y1...
av. .0...“sz - \\<
.w... VLU .r..lw.w\ev.w\.v (A X

‘II. I II! .I IV Ill...

.30.,...K The -V

\

M Sci . 0m

«0.3.x \ 2%.... (NR - V\

7“.

-‘~_ .%3-._fi.— 1",,

 

 

 

..II|.I bil- ,

to be thus applied will be 20% or 5 cu. ft. per 100 cu.

ft. of gravel.

_- FINIS -

 

RSI?" Ll

7'1“ 31:11:.

SHIFTS

‘ s
‘ |
l