- ~ - ---0N¢I¢v~rnq.ygwr1r \g’I'Tr w "w' 0"" “TV 'Y ' PY

A REINFORCED CONCRETE -

mow Deals-N

T Thai: for fin Degree of 3., S

. MlcflchNi-ST‘ATB..-COLLEGE

. Bill E. 'Hanelfi

jut-15:5

 

 

A Reinforced Concrete Stadium Design

A Thesis Submitted to

The Faculty of
MICHIGAN STATE COLLEGE
of

AGRICULTURE AND APPLIED SCIENCE

by

/

\

Bill Efiifianel

Candidate for the Degree of

Bachelor of Science

June 1943

THESIS

 

 

ACKNOWLEDGMENT

Going on the theory that "two heads are better than one", help
was secured throughout the working of the thesis from various professional
men about the college and the city of Lansing.

For their help, patience and suggestions, the author wishes to
express his sincerest thanks and appreciation to Professors C. L. Allen
\and C. A. Miller of the Civil Engineering Department of Michigan State
College and to Mr. C. Thornton, Assistant City Engineer of Lansing,

Michigan.

‘

f ‘5
b:- .I
(.0
{O
G)
x 3

INTRODUCTION

Because of the need for individualistic thinking, a thesis
is undertaken by each senior in his final term of school to put to
actual use some of the highly theoretical knowledge he has obtained
throughout his four year stay at the institution. It is in most
cases the students first "real" design problem. Each student picks
a problem that he thinks would be to his liking and at the same time
educational as far as learning something new in his field.

For my problem I have chosen the analysis of design of a
reinforced concrete football stadium. Because a new school has been
recently built in the City of Lansing proper, namely, Sexton high
School, it was thought to be timely to analyze a football stadium that
could be built at this school if and when the opportunity presents
itself after the war.

The first question that must be answered is: Of what type
and how large must this stadium be? The concrete stadium proper
should seat approximately six thousand persons. A stadium of this
kind then can be built along only one side of the playing field
without benefit of any horseshoe curves. The football stadium
at Pattengill Field answers these Qualifications nicely, so an
analysis of this design will be made accompanied by various design
changes.

The Pattengill Stadium was originally designed in 1922 by
Mr. J. N. Churchill.

Various constants were assumed in figuring the stadium

section line curve and will appear in the compiled computations.

1|
The concrete and steel used in this design will be taken as ZSOd‘and

2000le" respectively. Because of the highly erratic costs of concrete,
steel and labor in these war times, no attempt will be made to

compute the cost of building this stadium.

COMPUTATION S

a“

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(”/ba #63

 

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I- ._i l_-._ ”1/? /«._.l. __ To- ___,i_.._
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NOMENCLATURE FUR GJUL Vlad FOEEULas

a= horizontal distance from assumed sight point to
spectator in first row of seats.

nt to lowest

b-vertical distance from assumed sight poi
first spectator.

‘0
,.

floor elevation of s adium at feet ..

c: assumed vertical measurement froms

top of head or hat. also Lie verti
his focal ray to the eyes of LLB preceding spectator.

sp;ct atcr's Q)€S to
( #1 Cl guruflfic from

d: width of treads or sack to back distance of
assumed vertical measurement from floor to ezss J:
W " C.

e:
hqual to sitting height lea-

seated spectator.

f: assumed focal or sight point taken as origin of the
sight line curve.

o-origin of stadium section curve at oase of first riser
or e units below eyes of first spectator.

organ.

xy. co—ordinates of stadium section cueve wit” o a:

X u=.14X z' z=1.077z' y (ft.)
ft. ft. ft. ft. Computed Actual
2.08 .29 .50 .54 .83 .75
4.17 .58 .90 .97 1.55 1.50
6.25 .87 1.38 1.49 2.36 2.25
8.33 1.17 1.92 2.07 3.2' 3.00
10.42 1.46 2.40 2.59 4.05 3.83
12.50 1.75 2.31 3.03 4.78 4.67
14.58 2.04 3.20 3.45 5.49 5.50
16.67 2.33 3.92 4.22 6.55 6.33
18.75 2. 2 4.43 4.77 7.39 7.25
20.83 2.92 5.01 5.40 8.32 8.17
22.92 3.21 5.43 5.85 9.06 9.08
25.00 3.50 6.20 6.68 10.18 10.00
27.08 3.79 6.70 7.22 11.01 11.00
29.17 4.08 7.40 7.97 12.05 12.00
31.25 4.37 8.00 8.62 13.00 13.00
33.33 4.66 8.60 9.27 13.93 14.00
35.42 4.96 9.30 10.00 14.96 15.08
37.50 5.25 10.00 10.77 16.02 16.17
39.58 5.54 10.50 11.30 16.34 17.25
41.67 5.83 11.20 12.08 17.91 18.33
43.75 6.13 12.00 12.92 19.05 19.49
45.83 6.42 12.70 13.70 20.12 20.67
47.92 6.71 13.30 14.33 21.04 21.83
50.00 7.00 14.05 15.10 22.10 23.00

SIGHT LINE DATA:

Constants assumed:

a:
e=4 "
b =

50 feet
3 I!

c =0.5 feet

1.1:

d==25 inches or 2.08 feet

(b+ e)x ::
a

(

w

50

3+4)X=.14X

From table 111a p. 28 "Stadium"

\

Coefficient of z for values of c=*0.45', d==2.083
is 1.077

y=u+z

DESIGN OF TREAD SLABS:
Width of slab - 25 inches
Person is assumed to weigh 143# and have a seating width of 18 inches.
The impact load will be taken as 100? live load.
Live load:

Area occupied by one person :18 x 25::3.12 sq. ft.
1"

. 4%
Live load 140#
Impact load 140#

Total live and impact==280# distributed over 3.12 sq. ft.

or 280 = 89.8 lbs./sq. ft.

3.12
Dead load:

,. .‘ ~ ’1
Assume slab thickness to be 4" at front and 45" at rear for
drainage purposes.

Average thickness = 4.25"

Weight of slab = 1 x 4.25 x 150==53.2 1bs./sq. ft.
12

Assume width of riser stem==5n
Clear span = 25 - 5 =20"

.. - a” a . . , r, 2 .
Bending .anznent = on :51 =(89. 8153.29: 20 =476 in. lbs.

 

 

12 ‘~ 12
k = l = 1 = 0428
1+_£§_ 1+ 20003
j‘l-_1§=l-.143=.857 say.86
3
Bi‘vlzgdjkbd

A

4

476 =1_QQQ(.86)(.428)12d2
2

min. d = 76 33.51."
T1835

min. h = d+-clearance = .51:+1.5-P.25 a 2.26

For tOp protection use h = 4"
Note: hin. d was less than req'd. cover of 2".

Commercial d = 4 - 2 = 2"

M

Shear = gel-ell? x 20 —_-. 118.2 lbs.
2 2 x 12

bjd 12(.36)2
Bond: Use wire mesh

Steel req'd. A5 = 416 =.0139"" OK
zoooo(.36)2

DESIGN OF RISER BEATS:
Design of 9" risers:

Assume: Span = 19'
Inclined beams = 12" wide
Clear span = 18'
Riser stem = 9"): 5" assuming a 2 to 1 ratio as
most economical for h and b.

Load on riser:

Total slab load = 143.0 x 2.08 = 297.5
Riser weight :32 x 15’_ .-. 46.8
144 344.3

say w = 35032

Using: Concrete: 1:4 .30 /cu. ft.
Steel = .03 /lb.

Cost r = 5'20 2L9. : 98
30

Econ. d: “98 x_£_3_?fl___ +
20000 x b

M“- 534 = sf = 250 x_l§"= 11340” or 136030”
10 10

 

h) "5

Therefore, Econ. d = 10.22nwhen b = 10"
Corrected weight with b = 10"

Dead load and live load 8 297.5
Riser weight=l x 9 x 10 x 15Q=93.8
144 391.04 say W =

EM.“ 179:: 400 X 182: 1.2980“. or 1555001”?-
10 10

BMr‘.ElT= 400 X 181x 12 = 129800"”
12 12

 

Econ. d= #155500 x 98-+ 2 = 10.72"
20000 x 10

h = 10.72+-2.63==13.35" Too Large

Try b = 11"
Econ. d = 10.32"
h = 10.32 + 2.63 = 12.95” 0K
Commercial d =13.00 — 2.63 = 10.37"

0.92 at supports (negative)
0.86 between supports (positive)

Assume: j
3'

II II

Positive steel:

A, = 129800 e .727“
20000 x .86 x 10.37
Use l—l'th bar A = 1.0“"

Negative steel:
A, . 155500 = .815 °“

20000 x .92 x 10.37
Use 1-1" 0 bar A = 1.0““

 

Shear:
Vsll=400 x18 e 3600*
2 2
v e v e. 3600 - 36.7%.. 0K
bjd 11 x .86 x 10.37
Bond:
,a. v .. 3600 = 101*? OK

 

QM

..jd 4 x .86 x 10.37

Design of 10" riser:

Assume: b = 10"

 

h- t g 2 (max.)

b
11:1 _.
10 ‘ 1 OK
Dead load slab/ft. riser 111.0
Dead load riser 1 x 10‘3.10 x 150 104.0
144
Live and impact loads 18626

Total 401.67. Use w = 4027

 

Econ” d= 98 xh156200 + 2 =10.75"
20000 x 10

h =-10.75-+ 2.63 1: 13.38" 0K
Commercial d = 14.00 - 2.63 = 11.37"

130300“

a:
I52.
1
II
F.
a)
N
r4
‘N
H
to
u

3:11.: 402 x 13.3.1; = 156200 “*

Positive steel:

1?: _1_3_g}ppm_“n___ = .666 °"
20000 x .86 x 11.37

Use 1—1" <1» bar A = .73 “"1
Negative steel:

A“ —. 156200 = . 747 ““

 

 

2
v r. __3~6.8 ‘ = 37.01%” OK
10 x .86 x 11.3
Bond:
)4 = __.3_.€>_8. -,,__ _ = 1132,, 0K
3.14 x . 6 1.37

Design of 11" riser:
Assume: b = 11"

1‘; - 4 es 1.38 0K
8

Dead load slab/ft. riser
Dead load riser 1 x ll‘x 8 x 150
144

Live and impact loads

 

Total
Bm = 320 {18.3.1.2 =126300M
12
13:1“: 320 3C 133: 12451700....
10
Econ. 6.: 28 x -5, 700 + 2 = 11.63"
20000 x 8 '

h "‘ ll.63+ 2.63 7-14.26"
Commercial d = 15.00 - 2.63=:12.37"

Positive steel:

AP=___126300 = .593“

20030‘) X .86) \ 12. 37
Use l-l'Nb bar A = .73 Du

Negative steel:

1§1700 = .666 ““
20000 x .92 x 12.37
Shear
v —— 390 x 1 =35101r

_§
2

41.2%..

V: _3_510
8 x .86 x 12.37

 

Bond:

1
ll
‘03

\11
H
C)

I

— 10517111

 

314 x .86 x 12.37

111.0
91.8

186. ___6
389—: 4/. Use w -- 3907/,

OK

OK

Design of 12" riser:
b = 6"

16 - 5
6

Ass ne:

2 OK

Dead load slab/ft. riser
Dead load riser l x 12 x 6 x.150

144

Live and inpact loads

 

Total

132.1,: 373 3: 181412
12

120800"*

 

BM”: =73 $321342 = 145000”

 

10
Econ. d = Fegggggg + 2 = 12.87"
zoo

00 x 6
h = 12.87'+ 2.63 = 15.50" OK

A
“ =
J

Commercial d = 16.00 - 2. 13.37“
Positive steel:

Ar = “QIEQEEOO
20000 x .86 x 13.37

Use 1-7/8" cpbar -

I
11
Negative steel:

A u = _.112399_.

20000 x .

‘GO——'-'

92 X 13.37

 

Use 1-7/8" 4: bar

43.67,“

Bond:

435.7 = 1067...

x4
2.75 x .86 x 13.37

Note: Continue to use h - d

bar diameter has been decreased.

111.0
75.0

186.6

372.6”Z Use w =37372

OK

OK

2.63" even though reinforcing

Design of 13" riser:

Assume: b = 6.5"

6.5

Dead load slab/ft. riser

Dead load riser l x.12 x _;§ x 150
11.4

Live and impact loads

BMP==380 x_13zx l§_= 123030":
12

3.1“: 3_so x 18‘}: 12 .. 147800”
10

 

 

Econ. 01: ‘93 x14jggga- 2 = 12.53"

20000 x 6.5

h = 1.2053 + 2063 = 15016" OK

Commercial d = 17.00 - 2.63 = 14.37"

Positive steel:

1, = "123900 A - .49
zoo-00 x .36 x 14.37

Use 1-7/8" 4’ bar
Negative steel:

L:
/

Au 3 $2.893. ._ _ =
20000 X .92 X 14.37

Use 1-7/8" ¢ bar
Shear:
v=380x1§ = 32.20"
2

V- 343:) -.. = 42.
6.5 X .86 X 14.3

 

2.75 x .86 x 14.37

.3£9___A = 1000 57

A = 63 ““

OK

Design of 14" riser:

Assume: b = §;:_£ =3l§L:_é;= 7n
’7

n

I

‘- 5.

Dead load slab/ft. riser
Lead load riser l_; 12 x 7 x 150
144

Live and impact loads

 

Total

132.1,: 38g xlis‘x 1,3 = 124800”:
4'

Blur-3.85 x 18‘}: 12 = 150000”
10

 

 

Econ. d- 23;}.1‘10009 4.. 2 = 10.23"

20000 x 7

h 3 10.23‘+ 2.63 = 12.86"

111.0
87.4

186.6
385.0

OK

Commercial d‘= 18.00 - 2.63‘=-l5.37"

Positive steel:

AP =__ 124803 = .472un

200033736 1: 15.3?
Use l—7/8"4’bar A ‘ .60"“

Negative steel:

A N = ——-l59999_‘ = . 530 a“

23003 x .92 x 15.37
Use 1—7/8" 4’ bar
- Shear:

v = 385 xig. 3465*
2

V = “2&5 = 37.4%,,

7 x .86 x 15.37

M = 2465 — 95.2%

2.75 x .86‘E—15.37

OK

OK

Use w

385%

DESIGN OF INCLINED BEAMS:
Upper inclined beam:

Assume beam 12" x 30"

Clear span: ~14. 833' horizontal, 16. 88' inclined

Angle beam makes with horizontal==28° 17'

Eight riser beams carry load to inclined beam, 4 @ 385%

403W“
Uniform load carried to beam1=838§ x AI+§§79 x gala = 3590
16.17
Assume weight of inclined beam 32§

Note: w==vertical unit load of beam
N==unit load on beam normal to beam

= 3915 x .8806 = 3448?.

Maximum BM 1 _N_1‘ 342g x 16.852). 12 —-= 1175000"1t

 

 

 

10 10
Min. ch" 1175000 = 23.0"
.86 x .428 x 12 x 500

 

Econ. d=v 98 3: 1175000 = 21.9"
20000 x 12

D'= 23.0—F3.0.= 26.0" OK is less than designed 30"
Commercial d‘= 30.0 - 3.0 = 27.0"

Positive steel:

 

BMP= 3448 x 16.88;): 12 = 979000W
12

= 979000 = 2.12““
20000 x .86 x 27

Use 3-1"41 bars = 2.36""
Negative steel:

A”: 1175000 = 2.36
20000 x .92 x 27

Use 3-1" 0 bars

Minimum width of beam:

3 bars @ l" 3.0
2 spacings @ 1.5" 3.0
2 coverings @ 2" 4.0
2 stirrup bars @ 0.5" 1.0

Total

|._.o
F"
0

O
O
W

Shear:

v = 3448 x 16.88 = 29100“
2

v: 29100 = 104.4%“
12 x .86 x 27

Using ordinary anchorage and web steel v==.06fg
v = .06 x 2500 4502“ allowable

Bond:

Point of inflection at 0.1!. for EMF: 0
0.1 x 16.88 = 1.688' or 22.2"

v,“ = 29100 - 3448 x 1.688 = 23280“

 

..-. 23280 = 106.1?" OK Using deformed
3 x 3.14 x .86 x 27 bars 1252.15 allowable.
Stirrups:

Inflection point .215l.for M“: O
.215 x 16.88 X 12 = 43.6"

Allowable shear stress = .021} = 502
'u=uoaa';“

 

 

 

 

 

Shear stress diagram 1

Shear stress of v = 104.47%“ is less than allowabkefor web
steel, therefore the stirrups for maximum Spacing is:

s e 3/4d = 3/4 x 27 = 22.5"

Using em: stirrups A = .20“ rs= 160007...
v = vuv. = -02fét_(_a_f.)__
bs(sina9

118 = SO-r.2 x 2 x 16000
12 x s x 1

min. 8 = 9.8"
Stirrup Spacing: v = 50'+ 523
s

2-9" Spacings until passed x 10" which is 18" out.
l_l5 I! I! I! H x 24" I! ll 33 H N
1-21" n n u x 53" n n 54" n .

Middle inclined beam:

Assume beam 12" x 30" to conform with upper inclined beam.
Clear Span = 15.00' horizontal. 16.50'inclined

Angle beam makes with horizontal = 24340'

Eight riser beams carry load to inclined beam. 4 @ 390‘1and

 

 

 

 

 

403vt.
Assume weight of inclined beam 325
Uniform load carried to beam Ii§90 x 42 (373 x 4iI1.-= 3550
16.33 w = 387551
N’= 3875 x .90875 = 35205:normal to beam
Maximum BM = E17: 3520 x 16.5?x 12 = 1150000"*
10 10
Min. d= \{1150000 ‘5' =. 22.8" say 23"
.86 x .428 x 12 x 500
D=2300+ 300-12600" OK
Econ. d: 98 x 1150000 a 21.6“ 0K
' 20000 x 12
Use D = 30" Commercial d =30 - 3 = 27"

Positive steel:

Bmvzflz': Bizo X 1605zx 12 = 9580001!“
12 12

A,—— _958000 a 2.07““
20000 x .86 x 27

Negative steel:

AN=- 1150000 2 2.31“"
20000 x .92 x 27

Use 3—1"4> bars bent as in upper inclined beam.

 

Minimum width = 10 5/8" 0K allowable is 12"
Shear:
v = 3520 x 16.5 = 29050":
2
v = 29050 = 104.353" Use ordinary anchorage and
12 x .86 x 27 web steel. Allow.s150*/m.

Bond:

Point of inflection at 0.12. for MP=-O or
0.1 x 16.5 = 1.65' or 19.8"

v“: 31680 — 3520 x 1.65 = 25870“

= 25370 = 118.326" 0K
3 x 3.14 x .86 x 27

 

Stirrups:
Use «3% stirrups A .. .20“ r, = 160009....
Max. Spacing = 22.5"

V: .02 fc'*‘asf$2

bs sinx
104.3 = 50-k533
5
Min. 5 =53; -.-. 9.8"
54-3
v = 50 +.233
s

Stirrup Spacing:

Spacing Stress Dist. from column face

8 v x

9" ....... ._....

12" 94 10"

15" 86 17"

18" 80 23"

21" 75 28"

-— 40 61"
2—9" Spacings until passed 1 10" which is 18" out.
l__15" SpaCing I! I! X 23" H I! 33" I! .
2—21" Spacings n n X 61" n n 75" n .

Lower inclined beam:

Assume beam to conform with upper and middle inclined beams
12" X 30"

Clear Span = 14.833' horizontal, 15.92' inclined

Angle beam makes with horizontalat2lfl7.2'

Eight riser beams carry load to inclined beam, 4 @ 350§1and

4 @ 40232
Assume weight of inclined beam 325
Uniform load carried to beam 0 x 02 x 1 = 35§5

16.17 w = 3860‘4
N = 3860 x .93178 = 3600‘):

Max. BM = 111} .-. L600 x 15.59;): 12 -_ 1095000....
10 10

 

Note: Complete design of lower inclined beam will be the same
as for upper inclined beam since BM for design are

practically the same. Beam was designed for larger BM.

DESIGN OF INCLINED END SECTION BEAMS:

Clear span = 14. 833' horizontal

Assume beam 8" x 30"

Load from 8 risers carried to inclined beam

4 @ 385"? and 4 @ 379“/. 3056 x 19.5 = 184.0%

Beam.weight‘= 2175{

w = 2057 ‘7.

N = 1810?.

Max. BM = 618000"‘

Min. d = 20.5“ Use

Commercial d =-27"

Positive steel:
BMP=-515000"*
A..= 1.11“"

Negative steel:
BM": 618000“
A. = 1.25 ‘1“

Use 2-1" 4) bars

 

16.17 x 2

D = 30"

U“

A = 1.57

Minimum width of beam:

2 bars @ l"

2.00

2 coverings @ 2" 4.00
2 stirrups @ .375" .75 3/8"<b stirrUps
1 spacing @ 1.25" 1.25

Shear:
v = 15300“:
v =.- 82.3%“ OK

Bond:

8.00" OK

Place stirrUps same as in full inclined beams.

Point of inflection at 22.2"

vm = 12720“

m: 87.1%.. OK

Make all end section inclined beams 8" x 30". This beam was designed

for maximum moment.
J

DESIGN OF BACK COLUMN BRACE BEAM:

This beam serves only as a brace to the columns to decrease the
unsupported height of the column. It must be designed large enough to
resist buckling when end thrusts are placed upon it and to carry its own
weight. In view of this fact, a beam 12" x 24" will be chosen and checked
to see if it will resist its own weight.

w: l2x24x150= 300V.
144

 

 

 

 

Max. BM = 11‘... 300 x 17.672): 12 = 93700“
12 12
Min. d= 93700 = 6.5" OK
.86 x .428 x 12 x 500
Commercial d = 24.0 - 3.0 = 21.0"
Steel:
As= 93700 .. .26“
20000 x .86 x 21
Use 2—3/4"¢bars A = .88""
Shear:
v = 300 x 17.67 = 2650'
2
v e 2650 a 12.2%.. 0K

 

12 x .86 x 21

Note: No stirrups necessary, allowable V - 5022

Bond:

,4... = 2650 = 31.2%.. OK
4.71 x .86 x 21

DESIGN OF BACK COLUMN:

Inclined beam carries 34487. normal or 39157. vertical.
Clear span of beam = 14.833' horizontal or 16.88' inclined.

At.exterior SUpport:
v =- 3915 x 14.833 a 29100“
2

BM... 31‘: 3448 x 16.88%: 12 = 735000"
16 16

Assume: t= 16"
b -.- 32"

Weight of back wall:
3.0x2/3x1x150= 300

Weight of top brace beam: 2 300
Total w = 600%

Load brought to column = 600 x 18 = 10800“
Load

 

 

 

 

N = 29100 + 10800 = 39900“
M = 29100 x 8 + 735000 = 967000“*
e =

g s 967000 = 24.2"
N 39900

Eccentricity ratio== = 24.2 2.1.51 Case II

.2
t 16
h = 11.55b -ll.55 x 32 = 370"

With brace beam in place 10 feet up from grade, column acts as
a short column.

Assume: rga 6000‘/ou
Steel ratio - p°=’0.04

1?+3(1.51-.5)k’+6 x 5 x .04 x 1.511: = 3 x 5 x .04Ea(.3l2)’+ 1.51]
18+ 4.03k‘+1.8k 2 .60[.l95+l.5-l}= .6 x 1.705

k3+4.03k‘+1.8k = 1.02

k = 0.32

rc = 173191
bt‘ k' (3.21:) + 12np.(a/t)‘

 

res 12.x .32 x 967000 ,,__ 4754/“
162x 3; (.3213--64)+-12 x 5 x .04 x .312'

This would be OK but the steel ratio can be reduced. Allowable
fC = 6000 x .225 = 1350 Z“

Repeating using table on p. 450 of Peabody's "Reinforced Concrete
Structures".

d 2 020
t
Try— p.= .020

3: 16 = .66].
e 24.2

mpg: .02 x 5 =- .10
C,_= 11.3 k =- .27

fc = ix Ca = 267000 X 11.3 1335 0K Allow. f 1350
bt‘ 162x 32

f, = d - kt x ni; = 13- .27 x 16 {5 x 1335): 132.00%“
kt .27 x 16
an OK Allow. f5: 2000017,,"
A,= pbt = .02 x 16 x 32»: 10.25 of steel necessary.

Check on lower half of column below point of application of
intermediate brace beam load.

Weight brought to column by beam 2 300 x 18 = 5400‘"
New N = 5400 +39900 = 45300“

e = g .-_ @7000 = 21.3"

N 45300
Eccentricity ratio =Ԥ_= 21.3 s 1.34 Case 11
t 16

Try- p, = .020

$2216; =o75
e 21.3

npos .02 x 5 = .10

else 11.2 k = .28
r, 1320‘2u 0K Allow. re: 1350*...
f, = 125502,, 0K Allow. f3 = 2000070..

as: pbt = .02 x 16 x 32 = 10.25“" of steel necessary.

Use 8-1 1/8" ¢ bars A .3 10.12““

Use 1/'" ¢ stays 12" c-c

DESIGN OF INTERIOR COLUMNS:

Rear column:

N = 3915 x 14,833 + 3875 x 15 = 29000 +29000 = 58000"
2 2

No eccentricity

f): .1 -.-. .289b = .289 x 16 = 4.625
A

Limiting Length = 11.55b = 11.55 x 16 = 185" Short Column
Assume weight of column 2 150 x 15.1. x 16 x 16 -—-— 4100’“
144

Total N = 4100 +58000 = 62100“
Minimum area: Assuming p = .007

N =f¢A[l+(n—l)p]

62100 = 6000(.225)A[1+4(.007)]

a = 44.7“" 0K Allow. A = 256““
Longitudinal steel:

As: pi = .007 x 256 : 1.78”“

Use 4-3/4" Q bars A = 1.770“

Use 1/4"<} stays l2" c-c

Front column:
Load brought to column = "9000 + 3860 x 14.833 = 576001;
2 ,
Limiting height'= 11.55b =-1l.55 x 16 ==185" Short column
Assume weight of column = 150 x 7 x 16 x 16 = 1865W say 1900*
4* 144
= 57600-r 1900'” 59500
Minimum area: Assuming p = .007
59500 = 1350a[1+ 4( .0077_|

A = 22500 = 42.8w. OK Allow. A -.= 256°“
1387.8 .

Longitudinal steel:
Ag= pA =- .007 x 256 = 1.78cm
Use 4-3/ " cp bars a = 1.77“"

Use 1/4" o stays 12" c-c

DESIGN OF FRONT EXTERIOR COLUMN:

Slab load to column = 4 x 4.083 x 150 x = 103*/.

12

hflh‘

Beam load to column = 12- x 16 x 150 = 200‘“
144

Total beam and slab to column = (103+ 200)18 = 5450*

Inclined beam carries 3860U6 Clear span horizontal =~14.833'

at exterior support:

v = 3860 x 14.833 = 28600“

K.

BM = y: a 3860 x 14.83221; 12 = 637500”

 

16 16
assume: t = 16"
b = 32"

N = 28600 + 5450 = 34050“
N = 28600 x 8 + 637500 = 866500“?

14 =;8_6aiqg --- 25.4"
N 34050

m
1|

 

Eccentricity ratio = = 25.4 =.1.59 Case II

2.
t 16

Assume: f5 = 6000
p0 = .020

np° = .020 x 5 = .10

 

d}: _}_.= .187 say .20
t l6
1;: l6-_.63
25.4
CL=-1l.3 k‘= .275

rcs 866500 J: 11.3 —_-_ 1205?. 0K Allow. re: 13507....
32 x 161

rS s. 13 - .275 x 16 [1205 x 5] -.-. 117502.. 0K Allow. r.--— 200007...
.275 x 16

As"pbt ‘ ’02 x 16 x 36 =’10o25nuof steel necessary
Use 8‘1 1/8"¢ bars A = 10.12""

Use 1/4" 45 stays 12" c-c

DESIGN OF BOTTOM AISLE SLAB:

Assume: Live load 100
Impact load (100%) 100
Dead load 50 Assuming slab 4" thick

Total w =2 2507.

BM: 11}- 250 x 4‘5: 12 = 4000'":
12 12

 

Min. d3 4000 = 1.34"
500 x .86 x .428 x 12

t - d.='1.504-.25-= 1,75“
Min. to = 1.75 + 1.34 2:: 3.09" Say D = 4"
Commercial d = 4.00 - 1.75;: 2.25"

A, = 4000 = .103 “"
20000 x .86 x 2.25

Use wire mesh.

FOOTINGS:

Weight of column 12500
Load on column 45300
Assumed wt. of footing 2200
Total 60150“t
Unit soil pressure = 70002;. rge3ooo’7m
A = 60150 = 8.6‘“ say 9‘“ L=3'
7000

Net pressure =. 57800 =: 673052.
8.6

M (111%.) :3 fl:
2

MAQ=_E7%2__£_2_€[16+ 1.2 1: lg : 2620:13-
x 2

M“: 6730 x 102 [32+ 1.2 x 2] = 80400"

144 x 2
Min. ‘1; 1:]! 2620 _—_ .86"
kb 222 x 16
Min. (15‘80500 = 3.36" say 3.5"
222 x 32

 

h = 3-54—5 = 8.5"

Weight of footing = 3 x 8.5 x 150 = 956* OK
12

Shear:

v = 6730[ - 2 x 1 = 36700“
144

v = ~367OO = 11.77....“ column perimeter OK
2(16 x 32)3.5 x .875 No web steel necessary.

Diagonal Tension:

v: 6730‘: - 34 x 23 224700“
144

Allowable stress v 8 .03 f5: 90%“

Average d: V = 24700 -.=-. 2.8"
ij 2(34 23).86 x 90

Average h=2.8+ 5.0 = 7.8" This is safe.
Weight assumption 0K.

 

BM at column face AH = 80400”

A = 80400 = 1.34
20000 x .86 x 3.5

Use 6-3/4"¢ bars 6" c-c A=2.64n"
BM at column face AB:= 2620"» Steel not needed when double column
is used, but needed when single column is used. For this

web steel use 6-3/4'W bars, 6" c—c placed in footing both

directions.

BEARING CHECK ON FRONT WALL:

Weight of 1' length of wall = (8 x l x 1)150=1200"f or 1.200%.
Load of slab carried by wall 500 '-
Total load per sq. ft. 17007.. 0K

.2
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BIBLIOGRAPHY
"A Design of Reinforced Concrete Stadium"
by Fred M. Barron

"The Stadium"
by Myron W. Serby

"Reinforced Concrete Construction"
by George A. Hool

"Reinforced Concrete Structures”
by Dean Peabody, Jr.

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