COMPLETE DESKSH
OF A DEPARTMENT STORE

{REINFORCED caucus?! STRUCTURE.
FLAT 5w TYPE}

Thai: {*or fit. Dunc cf M. 5.
MiC‘dIGAN STATE mum
Hang-Yuan Wong
1949

TH $513

This is to m-rtifl] that the

[In-sis entitled

COMPLETE DESIGN OF A DEPARTEENT STORE

(REINFORCED CONCRETE STRUCTURE~FLAT SLAB TYPE)

prcsmnml In]

Bong—Yuan Wong

has ln-cn amrelolvd luwm‘clx {llllilllnmn

ml Ilw requirements for

“.8. degrm: in_ (LE.

51:] i¢ vl‘ PFUICSMH‘

Hate May 25’_1_949 _

 

 

This is to certify that the

thesis entitled

COMPLETE DESIGN OF A DEPARTRENT STORE

(REINFORCED CONCRETE STRUCTURE - FLAT SLAB TYPE)

The complete archictural and engineering
design for a sample department store

presented by

HGNG-YUEN WONG
HAS BEEN ACCEPTED TOWARDS FULFILLMENT
OF THE REQUIREMENTS FOR
M.S. DEGREE IN CIVIL ENGINEERING

flflm.

 

May 21, 1949 Major professor

COMPLETE ESIGN OF A DEPARTUCNT STORE

(REINFORCED CONCRETE STRUCTURE - FLAT SLAB TYPE)

By
Eons-rum WONG

A THESIS

Submitted to the School of Graduate Studies of Michigan
State College of Agriculture and Applied Science
in partial fulfillment of the requirements

for the degree of

MASTER OF SCIENCE

Department of Civil Engineering

June 1949

DEDICATED TO

THE AUTHOR'S PARENTS

2180-15

ii

iii
ACKNOWLEDGMENT

The author expresses his thanks to the following
people and organization for their help and guidance in

the oparation of this thesis.

Professor Chester Allen
8080 Bios. 00E.
Head of Civil Engineering Department

Mflzhigan State College

C. A. Miller
Assistant Professor of
Civil Engineering Department

Michigan State College

PREFACE iv

Let us look backward into the dark days of our national
crises which is still continuing. Our nation used to say
that the economy of our country is based on agriculture. But,
in the world of today it is impossible for a country to stand
only on one power support. In other words, besides agricul-
ture, we shall have to take cognizance of educational in-
stitutions, industry, commerce, and the armed forces. Form-
erly it was thought that scholars were superior to the mer-
chant. Can we accept this archaic notion today? Surely all
the nations of today will give a negative answer. We should
lose no time in developing these five forces.

I am trying to use all that I have learned in Richigan
State College to make a complete design for a department
store. My minor, Engineering Drawing, will be used as back-
ground for my architectural design, my major, Structural
Engineering as background for my computations for the whole
store buildinc. Therefore, I write this thesis to furnish
a complete treatment of the subject of department store design
and as a guide for the architects and engineers of the future.
One department store is only a small unit in the vast web of
commerce, but is on these small units that great commerce
is built.

In the first section of this thesis I dealt with the
principlesof the good modern design of a department store;

in the second section I considered the economic aspects.

of construction, and in the third I gave a sample design.

In this last section I have narrowed my plan to three articles:
The general idea of the design of the sample department store,
The complete computations for the sample design, and the comp-
lete drawings for the sample design.

In conclusion, I want to remind once more the peeple of
our country that commerce is one of the five important el-
ements in the develOpment of a strongnation and must be
built up. However, it is a wise policy for our government
to encouragepeople to develOp all of the five vocations.

By so doing, our country may step ahead as one of the most

modern in the world.

 

 

 

mm @m

—Hong-Yuen Wong

 

Michigan State College

Civil Engineering Department
East Lansing

Michigan

United States of America
May 1949

quoo JO

TABLE OF CON ENPS vi
SECTION PAGE
I. DESIGN OF A DLPARTKENT SthE . . . . . . . . . . 1
II. PRINCIPLE OF FLAT SLAB CONSTRUCTION ..... . . . 9
III. A SALPLE DESIGN
ARTICLE 1.

GENERAL IDEA OF THE SAEPLE DEPARTMENT

ARTICLE 2
COHPUTATION . . . . . . . . . . . . . . . 21
ARTICLE 5
DRAu‘e'II-IGS................. '72
Plate 1: First floor plan of Kitty's
Department Store
Plate 2: Secong floor plan
Plate 5: Basement plan
Plate 4: Front and rear elevation
Plate 5: Band plan
Plate 6: Foundation plan
Plate 7: Detail of foundation
Plate 8: Detail of beams and schedules
Plate 9: Detail of column and partial
cross section

Plate 10: Detail of stairway

SECTION I

PRINCIPLS OF FLAT SLAB CONSTRUCTION

DESIGN OF A DEPARTJEET STOhE

To design a depirtment store one must face the problems
of store construction and layout. The store serves its real
purpose when it is so constructed and laid out that the mer-
chandise can be displayed, sold, and delivered advantage-
ously. Department stores are so planned that retail buyers
and sellers may meet for mutual exchange and so that cust-
omers may make selections under favorable circumstances. In
design everything should be worked out on the basis of ab-
solute simplicity. The sole purpose of the design should be
movement for customers first, and then for merchandise.
Besides the physical laws which have a bearing on the use-
fulness of these structures, artistic effect should also be
considered. Beauty, proportion, and symmetry are regarded
as very desirable attributes in a department store.

The store that is beautiful will be patronized in pre-
ference to one that is not. Architectural perfection is very
important in the stimulation of sales. The store must not
only be designed attractively, but the comfort of the cust-
omers should be taken into account. Sufficient heat, good
ventilation, and adequate lighting are factors which add to
the well-being of the shOppers.

One of the serious problems in designing a department

store must be the consideration of future additions or

extensions in store planning. The location and layout of

.,

 

 

 

Specific departments must all be carefully located before it
is possible to be determine whether the permanent addition
will take care of the selling and non-selling activities.

The designer who recognizes the necessities of store planning
will locate the various departments, both selling and non-
selling, according to that which is to be sold. The buying
potentialities of the community, city, and adjacent terri-
tory must also be considered. In case the store fails, the
building might be used for other purposes, such as an office
building, warehouse, and etc. Thus the interior design
should be such that it could be easily changed.

There are many stores that have been unable to purchase
the necessary ground for expansion, either adjacent to or at
a convenient ’istance from the store. In such cases it would
be necessary to add new floors. In the large city centers
it is necessary to secure the maximum value from the location.
It means building high buildings and using the tOp floor for
non-selling departments.

The type of architecture and the actual construction of
the store building are planned so that the store attracts
customers, and the windows display is a great factor in in-
‘icating what the store has to offer in quality and style.

The beauty and dignity of a building should enhance the
appearance of the city. A dignified style of architecure

usually identifies the sales place. Thus the construction

and design make the building individual.

 

 

 

 

 

n.

 

 

Specific departments must all be carefully located before it
is possible to be determine whether the permanent addition
will take care of the selling and non-selling activities.
The designer who recognizes the necessities of store planning
will IOCate the various departments, both selling and non-
selling, according to that which is to be sold. The buying
potentialities of the community, city, and adjacent terri-
tory must also be considered. In case the store fails, the
building might be used for other purposes, such as an office
building, warehouse, and etc. Thus the interior design
should be such that it could be easily changed.

There are many stores that have been unable to purchase
the necessary ground for expansion, either adjacent to or at
a convenient distance from the store. In such cases it would
be necessary to add new floors. In the large city centers
it is necessary to secure the maximum value from the location.
It means building high buildings and using the top floor for
non-selling departments.

The type of architecture and the actual construction of
the store building are planned so that the store attracts
customers, and the windows display is a great factor in in-
‘icating what the store has to offer in quality and style.

The beauty and dignity of a building should enhance the
appearance of the city. A dignified style of architecure
usually identifies the sales place. Thus the construction

and design make the building individual.

The styles of architecture will differ according to the
section of the country in which the store is located. A
great deal depends upon the style in mode at the time the
store is built. Some city planning restricts the style of
architecture to some extreme.

THE 'RYCUT OF INTERIOR ARRAN ELEE'

The correct place for the elevators and stairways must
be carefully designed. The most advantageous location for
the elevators would be toward the center of the store. When
the elevators are so located, the customers, as they enter
the store, all have an equal distance to walk in order to
reach their location.

Adequate arrangements must be made for the handling of
the merchandise. The freight elevators should be located in
a place convenient to the receiving department. It is es-
sential that facilities for the receipt of merchandise at the
store platform be made adequate.

LOCATION OF EHTRARCES AND ELITS

The entrances and exits should be located at a short
distance from the corner, so that peOple who happen to be
waiting on the corners of the main shopping throughfares can
spend their time examining the merchandise on display. When
the side of the store is on a side street, where the traf-
fic is negligible, a large corner display window might not be
advisible for display purposes,and this location could better

be used for the location of the entrance.

4
It is better to eliminate any obstacle that might deter

people from entering the store. It is better to eliminate
steps leading up to the entrances and to make entrances level
with the street if possible.

Smaller stores usually use the ordinary hinged door.
Larger stores usually use either the revolving doors or
double or triple sets of vestibule doors. Often, combined re-
volving and swinging vestibule doors are most satisfactory.

FUTURE NECESFITIES

In order to meet the varying circumstances that might
arise due to the expansion of the business, it may be advis-
able to move the non-selling department to new quarters in
order to get more selling space.

The store should be adequately illuminated to present a
pleasing appearance; also special lighting colors can help to
eliminate harshness. The lighting equipment must be artistic
and well arranged, so that it will produce a pleasing appear-
ance without drawing one's attention away from the merchandise.

The painting of the store should be such as to avoid a
clash with the merchandise on diSplay. By careful planning of
the lighting and the colors of paints used, the interior of a
store can be made very attractive andpleasing.

Each storahas a certain characteristic tone, and the
quality, style, color effects and lighting of the show cases,
should be carefully considered in order that they blend har-

moniously into the style of the store. The show cases should

.I ’14

 
 

 

 

also provide a convenient and attractive method for storing
and displaying the merchandise. Also,the other interior
features should give the mximun diSplay value and tone to
the entire store.
PLALEN INC

The placement of selling and non-selling departments,
the layout of the specific merchandise departments, and the
placement of special sales tables and counters,should be con-
sidered. The locations of various departments on the differ-
ent floors, when once made, is likely to become permanent.
If any of these are changed afterwards, it would not only be
expensive, but perhaps it would halt business in that de-
partment for a time. A move of this sort would not be wise
from the customer's viCWpoint. Customers tend to familiarize
themselves with the plan of a store, and frequent changings
may baffle some customers into going to a store which has a
more stable arrangement.

LOCATIONS FOR SELLING AND NON-SELLING DEPARTEERTS

For any kind of department stares, it is logical to put
the non-selling departments in a place not likely to be
valuable as a location for the selling of merchandise. The
upper floors, certain parts of the back of the store, and
other locations away from the main stream of customer traffic
are usually devoted to the non-selling activites. It is found
more profitable to use the ground floor Spacr for selling,
and any store services rendered to customers should be in

the rear of the building.

No two stores sell exactly the same merchandise, nor is
the same importance given to the same class of goods in all
stores. Also the placement of goods in a store is an arbi-
trary matter, and no set plan has been devised which is stand-
ard for all stores. The type of customer as well as the class-
ification of the merchandise, helps to decide this matter.
Certain kinds of goods have their greatest sales when locat-
ed in a place where the maximun number of customers pass.

The store should be carefully arranged so that the loca-
tion of the various departments can be adapted to sell items
most in demand in that locality. If, for instance, a store
is located in a college town, books and school equipment would
be placed on the ground floor, whereas luxuries would be
placed on the other floors. The number of entrances should
depend on the number of customers, and care should be ex-
ercised that a free circulation for the customers entering
and leaving the store is provided.

RELATION BETwEEN DEPARTMENTS

Merchandise departments should be in close proximity
to allied departments. For example, the stocking department
should be close to the shoe department, and the pattern and
dry goods department may well be placed near the dress de-
partment.

To obtain a unity of appearance, the height of the store
fixtures should be uniform, and the width of aisles should be

up .

the same, considering always that they are wide enough to

' 7
take care of the traffic. Sometimes on the ground floor the

merchandise needs a different arrangement, but the layout
should be logical and attractive. Some stores divide their
departments into sections. In the sports department there

will be many sections: Ball, sport shirts, skates, fishing

equipment, hunting equipment.

A

11' 1.3

nrue-SELLILEG DE APPLENTS LAYOUT

In many stores the officers and credit desks are hidden
so that customers have great difficulty in finding them. It
might be advisable to provide windows in the back of the main
floor for peOple who wish to pay their accounts, and to have
signs directing them to this point. If it is necessary to have
the credit windows on an upper floor it might be advisable
to have them facing the elevator or stairway or adequate
signs placed at these points directing the customers to the
locations of the credit windows.

THE SPECIAL LAYOUTS

The sales tables are better located on the traffic line
so that customers can easily find them. The most desirable
locations are the entrance, the elevators, thestairways and
the escalators. Because of the different natures of marchan-
dise, it is often necessary to arrange them in several locat-
ions. Swimming suits may be displayed in both the men's
store and in the sporting goods department.

The design, size, and arrangement of a department store

depends mostly upon the territory in which the store operates.

8

The arrangement of departments depends on the increase or
decrease in the demand for the goods sold in that department.
The space given to a special department should be so flexible
that it can be changed in order to meet seasonal sales.

In conclusion, it is necessary for the seller of today
and of the future, to be familiar with the real needs of his
territory, and to be aware of the comfort of his customers.

A fact of the greatest importance, however, is the location
of the store. The designer may be able to furnish the finest
functional design which science and architecture can provide,

but lhould the store be located poorly, it will not be a success.

SECTION II

PRINCIPLE OF FLAT SLAB CONSTRUCTION

FLAT SLAB CONSTRUCTION

The flat slab is a type of construction distinctive
to concrete. These slabs have no supporting beams or
girders except at the margins, only rest directly on
columns which usually built with enlarged 'capital'.

The portion about the column capital of the slab called

a 'drOp panel' is made thicker than the rest of the floor.
This is the most economical and advantageous reinforced
concrete construction, so adapted to build department
store often.

Flat slab constructions are suitable for certain
cases: 1. A building at least two and preferably three

bays wide.

2. The column spacing can be made fairly regukir.
(17 - 50 or more feet each way)

5. Live load of 100 #/ft?
Advantages:
1. Forms are much less expensive than in beam and
girder construction.
2. Saving steel and concrete.
5. Great saving building height.
4. Saving overall thickness.
5. Saving in columns, walls, and partitions.

6. Easily layout sprinklers and piping or shafting.

9.
10.
ll.

12.

ID
Easier artificial lighting, better day lighting
with winddws that extend to the ceiling.
Better ventilation because of the absence of
pockets in the ceiling.
No easy to catch fire - no sharp corners.
Less danger of sudden failure.
Easier for inspection.

Greater storage capacity.

DEFINATICN

Floor slab supported directly by column without using

beams and girders.

Classification

(1) According to reinforcement:

(a) iwo way
(b) Four way
(c) Three way

(d) Circumferential

(2) According to ceiling construction

(a) ‘f-J'ith drOp; for w >150 #/ft‘a

(b) ‘u'x'ithout amp; for w< 150 #/ft2

 

 

 

 

 

 

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J 8410 - 'u‘uidth >1/5 1, 1: panel length in that
direction.
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and shear.

J 641a C; 0.201 l=span of square panels or
average of rectangular panel.

C=Diameter at under side of
slab or drop panel no portion
of column capital should be
considered effective which
has outside of the largest
90° cone that can be included
within the outlines of the
column capital.

Cap - For architectural purpose (1 in. - 2 in.)
Bracket - For exterior column - To take the place
of capital.
J 84lb - Angle of leping face<145° with
vertical width of bracket g width
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WIDTH OF BRACKET

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Statistical Analysis of Flat Slab.

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- Variation of moment coeffiCient 6» steel in

north-south direction same as that in east-west direction

for a Square panel.

Thickness of slab J859.

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For interior panels

t1:

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Moment at support of column
strip.

Width of drop for slab with

drdp panel and full width of
column strip for slab without
drop panel.

Shear at a distant d out from
column capital.

0.05 f5 when at least 50%

of total negative reinforce-
ment of column strip pass
through the section.

0.025 r5 when 25% passes thru.
shear at a distance d out from
the edge of drop v=0. 03 f'
with at least 50% of steel

area placed within the strip
above the drop panel.

(b)

Order of
(l)
(2)
(3)

(4)

(5)

15

For shear no reduction d=tl - 1.5
For deflection
1. Floor slab with drOps, 4.5 in. min.
without drops, 5 in. min.
2. Roof slab with drop, 5.5 in. min.
without drOps, 4 in. min.
5. Floor slab r5 = 2500 #/in2
(a) End panel 0.05 l,
(b) Interior panel 0.02 l,
4. Roof slab fé = 2500 fi/inz.
(a) End panel 0.025 1,
(b) Interior panel 0.025 1,

For fé other than 2500 modify 5 and 4 with “215560

Design

Determine C and width of drOp.

Determine moment and shear.

Determine t1, and t2 from moment, shear and

deflection.
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Determine steel area AS
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Steel placement to cover bond and anchorage.

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SECTION III

A SAMPLE DESIGN

ARTICLE I

GENERAL IDEA OF THE SALPLE DEPARTEENT STORE

18
Canton is the oldest and the most important city in
Southern China. It has over one and a half million peOple.
It is situated on the delta formed by the meeting of three
important rivers, and all the major rdilroads junction at
this city.
Before the war there were only four big department

stores in this city; Sincere Company, SunnSun Company,
and two stores of The Sun Company. Now the commercial busin-
esses are growing up day by day. There used to be many small
stores selling merchandise. But many failures in these smaller
stores are traceable to the fact that the owners are not trained
merchandisers and do not recognize the need for having

specialized help, trained to do a certain task in the best
posible uy. The volumn of stock carried and the type of ‘

stock must be estimated in accord with some preconceived
plan. Stock turnover depends on this. Good buying is predicated
upon it. Therefore, the larger store has been studied in
order that the smaller merchant may anticipate his difficult-
ies and may know what type of problem is being investigated
by the leaders and how its solution will help him.

1 The smaller store owners are tending to cooperate to
start a department store. Therefore, the department store

designs and constructions are very ponular and growing up

reason I chose this subject of department store design for

19
my thesis, is to get more experience and knowledge to meet

some of the needs of my peOple. Also, in Canton there are
no women's department stores at all, and I think it is very
useful and necessary for women to shop for their goods pri-
vately. Therefore, I chose to design a women's department
store.

There is a large cement plant called, 'West Village
Cement Plant', in Canton. Their product is of very good
quality. The most economical material for building construc-
tion in Canton is reinforced concrete, and for public build-
ings, the flat slab floor construction is the most economical
method. Therefore, I adopted the reinforced concrete con-
struction in flat slab style for my design.

A SAMPLE DESIGN

The Kittys Department Store in Canton, China, sells

womens merchandise, and gives facilities to the women in

O
.

a private. The stores

(.9

shOpping for their daily necess'tie
supposed location is on the corner of a main street and a
secondary street.

The proposed new store is to be located on Waioi Road
and Dutching Road, in East Canton. It is to be two stories
high and have one basement. Frontage on the Waioi Road side
will be 111 ft. 8 in., and 67 ft. on the Dutching Road side.

The dimensions meet the specifications of the Canton
Municipal Building Code. The specification for computation

is almost the same as the 1940 J.C.R. specification.

20
The main entrances are on the south side or Waioi Road.

The dignity of the enterences are enhanced by setting the
doors back from the building line. These doors are of the
ginged construction type.

The Kitty's Department Store is built of reinforced con-
crete structures in flat slab type. hollow tile covered with
plaster is used for both exterior and interior walls. The
Openings are equipped with automatic fire-doors. Seperate
elevators and stairways are used by customers and employees.
A fire-escape on the back of the building insures added pro-
tectien.

4.1

A highly perfected ventilating system, removes the foul
air, and insures a constant supply of clean, fresh air through-
out the building. Power and ligh are furnished from the
Canton Lunicipal Power plant.

The elevators and stairways are located in the center
of the store, so it allows the same distance for the customer
to reach them from either entrance. Thus carrying out the
merchandising idea, that customers will look at the advert-

isin' displays while valking across a part of the main floor.

n
I

The incoming goods are sent to the receiving department
by the ally behind the building. The non-selling depart-
ments are located in the basement.

For the future expansion of business or to meet any
varyingpircumstanee, I did not put permenent partitions or
counters on the floor, so it could be changed to meet these

circumstances.

ARTICLE II

COMPUTATIONS

22

CALCULATIONS

 

 

 

r5 = 2500 y/sq.in. r0 = 0.45 x 2500 = 1125 fi/sq.in.
_ 15000 _ y _
r — 1125 — 16, n - 12
k = n = 12 = 0.425
n+r 12+15

 

 

5
n
p = ——~———— = 12 = 0.0154
2r(n+r) 12x15(12+15)
K = %fckj = % x 1125 x 0.425 x 0.557 = 207

PERIOD I

DESIGN OE ROOF SLAB

I. Thickness of slab

A. Corner bay Control - moment coefficient is greater.

-\

D. t={S +.§ — L] l QFIESOO
m 10 72 fic
= F2lxl€+221 x 12 _ 2x268]1 Qf2500
3 l . 10 ‘72 2500
= 25“ + 255 - 55.5 x 1 :48? 4
c o ] 72X 1 .__?§__
= 6.7 use 7
wt. of slab = 150x x.%§ = 87.5§/55.ft.
total load = 50 + 57.5 + 40 = 157.5 fi/sq.ft.
Where t thickness of slab
3 short Span
N total length of slab which continues with
adjacent slab
m ='§
1
C. l'loment for 1 ft. strip (middle)

 

+ a center = 0.057 w1§= 0.057 x 157.5 x( 2.55) x12
= 0.057x157.5 x §%2x12
= 0.057x157é5 x 449Q§g = 41400 f
- m discont. end = 0.025 x 157.5 x (22.35)2xl2
= 254002
- m cont. end = 0.045 x 157.5 x (22.55)2 x12

= 55000;

24

D. Effective depth

._ 11'- _ v O O _ ,2 __ r ,1”
d jfgg _(2%7%I2 — J2o.6 — o.15" use 52

Assuming %" bars

total depth 7 0.5.

II. Steel reinforcement:

A. Riddle strip - short Span
Center of Span -

As: h = 41400 = 0.54 in
rsjd 18000x0.557x5

fl'l

 

use l'hbars space a? c.c. fur'd =0.55 sq.in.

 

 

2 >0.54 sq.im.
At discont. end As: 28400 =O.568sq.in.
18000x0.557x5
. * A = 55000 . =O.712s .in.
At cont en“ 3 5000x02557x5 q
use .Jghbars at 4 c.c. <5t=5x7=2l"
' 1 . n 1 _ 1”
'>2 = .5x_.—1_
2 l 2 2 4
l
'71§

hvery 5rd bar are straight through the support,
and the rest are bent at both ends.

(fur'd at center) =0.59sq.in4‘>6.556sq.in.
(lur'd at disc. end) =0.67x0.59=0.594 >0.5683q.in.
(fur'd at cont. end) t0.45x1.55x2=0.788 >0.712sq.in.

5. Column strip

At center of span
As(req'd)=0.556x0.67=0.557 sq.in.

At discont. end
As(req d)=0.558x0.67=0.245 sq.in.

At cont. end
As(req'd)=0.712x0.67=0.475 sq.in.

use % sq. bars spaced 6 c.c.
As(fun'd) at center=0.25x%3 =0.5;>0.557 in.2

As(fur'd) at disc. end=0.25x2x0.57=0.555.70.245in.2

25

As(fur'd) at cont. end=0.25x2xl.55=0.57;.0.475in.2

C. Bending Schedule

11 =l2i=12 x % = 5“

or 11:1/20 L=1/20 x12 x22.55* 1/20 x 12 x.éZ

 

 

 

 

 

 

 

 

5
=15.4 use 14“
12 = raizsooo x % = 4500 _ 94
*u 4 x 125— 4 x 125
15 = fsi_213:2 x 9:13” u=0.05f¢ (deformed)
4a =125 #/sq.in.
% of straight bars (a) E: 2214' 1.4)
5/4 of straight bars (b) LE? |328" :t{|
22.55 x 5/5 l5'-8" VIT
Dent up bars (0)

22.55 - 15' - 5' 5'5"

(If, U-er_fi\\‘* ’/' .
’7 ‘I 1 ;| A 15"8” :L 8 ~8'
P l

 

 

l

 

—
rfi

J
‘

26

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

MNHA: _ml d. 1

lawmmivafi. a. b 5.15.5. r to. t.“— .. _

I _ _. _

as: . . _

— . . b

__: ___ 9...". ___:"nxnmnuu rfl _
__: : ___. _ __

___"=____ ___.___:___"= a. .
.55. :5. ___"? “flush... . ..
___. _“________ __ __"___"___.__._ .
=._ ___ ___... . ____.”_:____= +

E... ___... .
r ... .3 u
/1/vflflvpopqpoa m H
h.»
7

 

 

 

 

 

 

-:F__ (___-

PERIOD II
DESIGN OF ROOF 53AM

5:22' - 4: 22.55' Dead on slab = lS7.5§/sq.ft.

- 57 2
Total load on one beam =2ws“: 2 x 187.5 x(j§)
4 4

 

=46800 #

9' Or’ _, C? 7 .
Lquivalent uniform load=2%.§::2 X 197;5 fl UZ/U = 2790 a/:
v.)

Total load = 5060 #/ft.
2

-‘ I-: 2
hzl/lo w12=l/10 x 5050 x(%§)x 12:505 x (5;) x 4

=l,550,000 5"

End shear = %545500 * 219.2.22 = 25400+5000=25400#
‘ 2 X 5 (v=0.05fc'=l50#/in.)
b'dt V =°6400 =26400 = 200 sq.in.
bj 150x; 151

when b‘ =10 d(req'd)=50"
b' —12 d(req'd)=24'
use 12 x 24 =288 sq.in. 7'200 sq.in.

total depth = 2

depth of stem = 21

wt. of stem = 12 x 211:150 =262 y‘(270# (0.K.)
144

2.5teel reinforcement
Approx. jd=(24 - 7/2)=20.5

As: 1554000 4 4.95 sq.in.
' 15000x20?5

5.

28

I
use 89%gnbars furnished = 8x0.5g=5.52 >4.95 sq.in.(0.h.)
b<1/4=57" b-b'/2‘<8t=56m<b‘+16t=124'

 

 

 

b - b' =255 - 12 4-255: 125 b<268"
_ 2 2
use b: 67
p§§§= 5.52 = 0.0041
bd 51x24
t- 7 - 0 292 - - 07
ETEZ — . (from diag.II 1.55)
Kd=0.26 x 24 = 6.254<t=7 design as sq. beam.
(Table VII)
k=0.266 3:0.911
As(req'd)- 1554000 = 4.55 sq.in.<;5.52 sq.in.
15000 x 0.911 x 24

Investigation at support

d‘=4" 5:21" g'=‘3.=0.157

o 24
Ilp,_.:np_l2 x 5.52 =0.25
12 x 24
p=p'=.§a§&__ = 0.0192
12x24
diag 1v 2:0.405 j=0.55
r = —i- = "1334000 = 15100 4 .‘n.
S aeJd 5.5 x 0.55 x 24 ‘ T/Sq l

< 11153000 a/sq. in.

 

f0: fsk 2 15100 x 0.405=595 fi/sq.in.
n(l-k) 12(1 - 0.405) ,
<1125 fi/sq.in.

 

 

-1: ,4
ft: fs X kd a} = 15100 x 0-4-O-lqz
=16100 x 0'250 =6270 #/Sq.in.

 

0.6

T: o . 1
none stress and unit snear

u: v = 26400

=57.2 yin/s .in.<125 4' 's .in.
2534 5x .75x24x7/5 ’ q 7/ q

 

29

 

— V — 25400 , ,
v———— .— — L“ .1. . .
bid l2x24x778 _ 105 n/Sq.1n.<.150 t/sq.ln.

5.Bend up of bars
Assume pt. of inflection at a dist. of‘é

span from support.
lst 2 bars band up at 5 pt. at a dist.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Diag.l. \0.27 x 255 = 72. 4'- lxl/5 x 255 =44. 7'
0.17 x 255 = 45. 5" - l/4xl/5 x 255 =22. 5"
1 fag-l 35'6A4 » Mfwmz'e... 3.4.1
= 24‘ o 25me a 455 91" a." 4.. '°,o:‘f.-,°;, | «W131 ° n;
bgd? b???D'E‘OL.-+> ID fog-:0" ;}-’ 211.0; 1"!vcbb:o'- abp'lpldzoo +0.98
A L # J— [\ I I
- l
. :9‘hg : I \\ l I II |
‘9 V5: ' I I \ I 1 1
5 b. I ' I 4 I I
09 ll fig,>\
L..__...° 451....1 L.__I_.1 AI__11_..L
‘ -56 °b .
1? 9’ 5 A '“558” ’7 3‘*—1

 

 

 

 

 

S_A fs d :0 sé‘x 2 x 15000 x 24 x 7i5 I
7 x 26400-(50x12x24x7/B)

X 16000 X '7 Kg: {5.8"
300
U-stirrup

Xl=_]2- -V b.d =67 _ 5OX12X24X7=112 - 4 1_ 7 11
’ "‘5050"‘

1“" ;;;:__j‘:,t L_ T___{

" b-B sec:

 

 

 

 

 

28'
O 0.4.
01’

0

 

...___1_

 

 

 

 

 

 

 

 

 

1E?
...,
a
g,

 

 

 

 

....-. v
__o - «--
u
. , ..

v
,
I

 

 

 

-.._-"p“

0

50

use 12 for all

6.Diagonal torsion

chvbjd=50x12x24x7/e=12500 #

w =SOGO f/ft.

xl=1/2 - vbjd: 57x12._12500x12
w 5x2 5060

1:34 ”4:905
= 84.5"

 

I
use 3/4\stirrups at support.

S_Avf3jd:=2x0.1418x16000x24x7/8==21.5-

 

 

V 26400 - 12600
J
Allowable . s = 12' 5/2 = 12
Use s=l2 thru a dist. 85"from support

s=24" thru out from 85"to the other.

PERIOD III

DESIGN OF FLOOR SLAB

 

I. Column capital "
0:0.20 l=0.20x22.53=4.46‘=55.5

II. Drop '
b=1/5 x 22.55 = 7.5' = 90

III. t1 and t2; Assume wt. of slab l§O%/sq.ft.
Total 15ad=(210=150) x (22.55)“ =550 x 22.552
=40 x 573:150,000 #

Mo=0.09wl(l-2/5 x C/1)2 x 12
=0.09 x 150,000 (1-2x4.45/5 x 22.55) x12
=0.09 x 150,0 0 x 0.702 x 12 x 22.55
=5,040,000 #“

 

d: [0.5M_.0.5 x 5040000

V Kb 207 x 90I“=81°5=9~O5

 

for interior panel

d1=_Jl . 15 ..
d qfl.15 x 81.5 =J93.6 =9.7
for exterior panel

t1=l.2d + 1% =1.2 (9.7)+1.5=10.5=1.5=12'
t2=l2 x 2/5 = 5“ .

(52 min.=0.05 x 22.55 x 12 = 5)
T.L.=22.552 (210x150+ ) = 150 x 12-8x7.52

*

12 12

=22.552 (510)+150 X.é_x 7.5”
12
=154000 + 2500 = 155500<,150000 (CK)

IV. Check shear
a. V=(210 f_§ x 150) [22.552 - (72.55+.2 x 5.5)2]
12 ”*5 " :2

520005
.5)=4 x (7.5 + 1.05)=54.5'=412"

 

2 n p
=olO X 420

6
I3 51

v: V
bjd
b.V=156800

=155500
=155500

 

b=2fl x (2

52

152000 =32,4 5/3 .ft.<:75“/s .ft.
412 x 10.5 x 7/5 q W q

- (210+%% X 150) fl (2.23 + 10.5)?

- (210 + 150)n(2.25 + 0.575)
- 10900 = 145400 5

ll

 

.25+1$55)=19.5'=25.5

‘-

 

6.le - 05 = 10.5"
V: 145400 = 67 #/sq.ft.<L75 #/sq.ft.
255 x 7/5 x 10.5

V. Steel reinfor

5. Interior

(1)

(2)

(5)

b. Exterior

(1)

c ement
panel

Column strip "
-M=O.5Mo=0.5 x 3040000 = 1520000 #

 

As: ___E___ = 1520000 4_ =9 4 Sq ft
fsjd 15000 x 0.557 x 10.5

use 14 ~1Qbar
fur'd=ll SQoinu

Column strip .
+m=o.eomo=507000 5“

As: M 507000 = 5.05 sq.ft.
15000 x 0.557 x 5.5

use 8 - l"§'bar
fur'd = 6.3 sq.in.

Mid-strip "
+M 5 -M = 0°15Mo=456000 #
As: 66 456000 _ 4.

15000 x 0.557 x 5.5 "
use 6 -1 ‘bar
fur'd =4.7 sq.in.

55 SQoino

 

panel

Column strip
-m=1.15 (0,5 Mo): 175000 sq.in.

55

As=l.l5 x 9. 4 = 10.8 sq.ft.
use 14 - luibar
fur d = ll sq.in.

(2) Column strip
+Lfi=l. 15 (0. 21-10): 093000 sq.in.
use 9 - lear
fur'é= 7. 03 sq.in.
As=l.15 x 6.05:6.96 a“

(5) Column strip at wall
As=l.5 x 4.55 = 5.91 sq.in.
use lO-l" ‘8. bar
fur'd = 7.85 sq.in.

(4) Mid- strip +M & ~M
As: 1. 5 x 4. 55: 5. 91 sq in.
use 8 - l 5 bar
fur' d=6. 5 sq.in.

(5) Mid-strip at wall
As=0.8 x 4. 55 = 5. 64 sq.in.
use 5 - l'kbar
fur' d=5. 94 sq.in.

p= _ 3-95 , =0.0047<;0.0025 (OK)
80.5 x 10.5

 

VI. Bondstress

az=vtr2 a=2.25ffi=5.94
V=1w (assumed)=1 x 155500 = 59070 #
4 4
20:104.’ X 5014 = 404..
d=10.5
(allowable bond stress = 0.05fé x 1.5 x 0.75
=112 5/sq.in.)
v= ___..Y... ___.52079 = 99<112 #/sq.in.

 

z.jd, 44 X 0. 857.x lO. 5
VII. End anchorage
L1: l2i

E.A.= 10000 1 =22.4 i = 22.4\|
4 x 12

Total length = l2i + 22.4i= 54.4i = 54.4

VIII. Bending schedule

a. JLnterior panel

{are-44m—
COL. STmP

 

-——-

MID. 5T9")

 

 

 

COL. $TR| 9

{13"— _8__ J4 -_

 

 

0’ COL. 5mm

 

 

..oa.‘ -. 4 ‘. -
v
A
‘ ,

 

   

.
. I
- :-. ‘t . .
w. . 4-
C , , I I
—-- .-- . . . .-1 _ . _ ....-.- “ .. ,. . ,
. .

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

DUVQ vi
2m? ———————— - /
P /
/
¥_ _ _ ... _ ... _/
dtaLS mm «,3
..ol 13v HL‘DNB'I ‘WLOL
L A; 1
Ft»; T‘ .Z',|\==?€0 “gals qga'o
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L .L 711 J
F6"€T .Otw “3 3 ‘4 [Ts-’2']
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fir—i .979 ‘ ma ,y-g
,,O-,L| - H19Na w101 ‘
q IN?
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. JV‘79 .9‘“II* “97' ‘ 9'
(1 1L f _ I 141 J
r "2le-190 T ‘99 “92,0 yrs-two
/ I \
¥ 0 avg

.7.

b. Exterior panel

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

l ' L
- -_.-._L-
I
. 14 | 9 I y 10 col. strip
_4t_—_J ___l_.l__
l | .
8| 8 | 5 mid-strip
| I .
"1r'-“-P—'""‘—'r'-—+—'-'-'
14 | 9 ( 10 001. strip
-___LH____. - f
‘ “L - I wall

 

_1_

(1) col.

(551mb

 

strip

 

 

“"""‘("‘+"““

 

 

 

 

 

 

 

 

 

 

 

 

 

57

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

.P A! Ldl l» . l*|( ll
- _ 1 a I_ E _
.Il |LI$uLI l _ . _
_ d u #9; _ _ _
_m .... I 1 _
wI ....ml. \..Uuoll+l [.ilill.
In? Iiiinw w L _
m . .5 -II.1 .._.oIIhl I. ..II
m _ Im ..._. 1 _ mm _ I
M III. : - IL .I.I.. I. II
r
m I 1. fl _ _ 1
m _ .ca _ .
Q l _
.m. _ 5 2 p “1 _
m MXtIQ T ..fllw—LNWWIW. l_lll Ill") Ill
m l l-l.b a m _ . I I
ma . 5 _ ..II +IIII.
IL I. 19. I. I
A“. 0.53% LN—l l.l Llllll
o I I.L,.L.l l-l
u l M . # _

 

 

 

Q
L

P

[:11

0

OD IV

f—l

DESIGN 03 INTERIOR COLUXH

 

I. For axial load only

A.

2nd floor column

(1)

(2)

LJoad of roof slab and live load
= 157.5(22.55)2 = 95,500#

Load of beam 2x§%§%gx 22.55xl50 = 11,700#

Assume wt. of col. = 215035
Total =107,500#

 

Let rg= 0.02

 

0.225r5 925
0.225fé +ngS=552 + 0.02 x 15000 = 882

#854a.
Use D=161n. Ag fmr'd = 201 in8

Wt, of 001.: (144-28)X201X 150:2040#

M 8 H
1’2 25005
Slenderness ratio
illé = 7.25<ClO short col.
16
Safe load =O.225féAg+Agfs = 562x201+l6000As
= 176000#

7

The col. strength is made successful to
take care of bending.

Vertical bars
AsngAg=O.2 x 201:4.02 sq.in.

Use 10-5/4 in.‘n\ A(fur'd)=4.42 sq.in.
Space at about n(16-5) =3,45 in.

2.5 x 5/4 (on)

59
Actural rg = éaig = 0.022

201
(4)5pira1
From table 15 - 5/5 in 0 at 2 in. pitch
<31/6x16
<15 in.

:>1 5/ein.
3 Spacer bars are used

Check p'=0.45(-‘{5‘-E 'l)£§

 

AC fé
5. first floor column
(1)“oad of roof =105,500#

Wt. of 2nd floor column = 2,040;
Wt. of slab, drop and L.L. =156,800#
Assume wt. of col. capital = 5,0005
Total =259,140¥
Let Pg = 0.02
Ag(req'd)=3%%%lg = 295 in.sq.
Use ’= “O in. Ag(fur'd)=514 sq.in.

I . :. _ t I” 37
Mt. of caolta1<_lflg_(l.5ifl(44) I
l 1728 L 4 J)

=1/12 n x 22(442 + 202+44 x20)
= 150

1728
=155 ’

Wt. of col. =(l44-55.5)A81%§28—2960

:.Wt. of col. d capital =155O+2950=4310#<15000#

 

(2250+12720)

(2)51enderness ratio = 10265 = 5.45<:10 short col.

 

Safe load = 514 x 925 = 290,000#

(5)Vert. bars
As=0.02 x 514 = 6.28 sq.in.
Use 15 - 5/4 in. Q\As fur(d = 6.62 sq.in.

Spaced at abort fl(20'5) = 3,142 in;>2.5x5/4in.
l5

4O

Acturql Pezélég =0.021
“ 514

(4) Spiral

From fig. 15 - 5/8 in. e‘at 2.75 in. pitch.
Spacer bars are used.

C. Basement col.

(1)

(2)

Load from upper fls . =25a,450#
nt. of slab, drop and L.L. =l56,800§
Assume wt. col. = 5 000?
Total =45l,250fi
Let p=0.02

Ag=.fi§l§§9_ ; 457
925
Use D=26 in. Ag(fur'd)=55l sq.in.

. to 1
Wt. of ceeitai— 10“ 2250+ 1n x 22(442+262\
- (2' [ 'IZ
1"3 +44X26)J
_ 150
1725

Iw’t. of col. = (120-55.5)Agi%%%_=u420#

(2250+21500)=2070;

Wt. of col. and capital =2070+5420=5490$<6000#

Slenderness ratio = 8:°5 = 5.26<110 short col.
6

Safe load = 551 x 925 = 490,000#

 

Vert. bars
As=0.02x551=19.65 sq.in.
Use 24 - 5/4 in. d A(fur'd)=10.61 sq.in.

Spaced at fl(25'5):2.74 2.5.X5/4 (OK)
24

.1912; =0.02

Actural P8 ?
551

Spiral

From Table ~15
5/8 in. d at 2.5 in. pitch and also 5 spacer
bars.

41

 

 

 

 

 

 

 

 

 

II. Lap. All vertical reinforcement of these above col.
provide a lap.
For 2500? concrete, lap for deformed bar is 24in.x4/5
=52i and that for plain bars is 52i(1+.25)=40i
In this case, the lap = 40x 5/4in.=50in. or 2.5in.
III. Investigation, for bending
The unbalanced rogf load moment =1/4O W1.
=1/40 x 50(22.55) x12 = 200,000 in.lb.
The unbalanced moment for other floors
=1/40 x 210\22.55)Qxl2 = 698,000 in.1b.
The resultant load N on each col.
(1) The resultant load N on each col.
Load of roof aner.L. =105,500#
1/2 x 50 (22.55)‘ = 14,9504 K7
N = 90,440;
(2) fihe lst. f1. col.
Section at the bottom of capital
Load above to, that section = 265,490#
1/2(210)922.55) = 52’409ixf
N = 215,090#
(5) Basement wt.
N = 451,250-510-5420-524002574,0004
I
Col. Dia. 1g (n-1)Is _
Table 10 Table 11 I=Ig+(nel)iIs h h M(in.bl.)
in. (in.)4 (in.)4 (in.)
Top 16 5,217 757 5,954 116 54.1 2001000 Top
‘ 220,000'Bot.
475,000 Top
Inter 20 7,854 195 8,047 1085574.2
125,000 Bot.
575,000 Top
Base 26 22,452 6,440 28,872 84£5541
0 Bot.

 

 

 

 

 

 

 

42

 

 

 

 

 

 

 

Col. Point At N M I
sq.in. (effective) in.lb. =Min (I s npg K r=CE§ ) r0
--- 15. ‘N D '3 Hr” dia
top 105,550 200000 1.54 .115 1.5 555““‘575
Top 249.5 (159 .254
bot. 107,540 220000 2.05 .125 1.4 747 590
Ifite t0p 215,090 475000 2.25 .112 1,5 1020 920
555.5 0.75.255 .
bot. 255,450 125000 .47 .024 __ .95 521 750
top 575,550 575000 1.55 .059 1.15 512 556
Base 547.7 0.5 .24
, 450,740 0 0 0 0 0 0

 

 

 

 

 

 

 

 

 

°.The size J'ntermediate col. must be increased use D=22in.

 

A fur'd = 550 sq.in.
Wt. of capital = 150 +1 fixgg 443+223¢44x22
1725 2260 I2 ( )
= 150

(2250+19500)=1590#
1725

Wt. of col = 2950xéégt=5,550#
514
Safe laiding F08Ox9225551,000#
Wt. of col. and capital = 1,890+5,580=5,470#
As = 0.02 x 580 =7.6 sq.in.
Use 18- 5/81n.¢ As<fur'd)=7.95 sq.in.
Spaces at about fl(173) =

 

 

18 2.96 in.
“ctural Pg = 7.95 = 0.021
550
Spiral - From table 15 - use 5/5 in.5 at 5 in. pitch
Ig = 11,499 (Table 10)
(n-1)Is = 5,170 (Table 11)
I total = 14,559 in?

 

 

 

 

I/h M
200
Top 54.1 ‘ ’OOO
- 557
Int 155 ’000
° 195,000_
Base 541 500'008

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

- II ,
Col Pt. At (effect) 14 =%(in) _I__ _S__,
sq.in. 1b. in.lb. d d
Iep 105,550 200000 1.54 0.115
Top 24 9 . 5 0 .59“,
Bot 107,540 141000 1.52 0.125
iop 215,520 557000 2.51 0.119
Int 506.3 0.75
Bot 269L280 198000 0.?N5 0.055
TOp 575,550 500000 1.55 0.051
8as 647.7 0.8
Bot 452,170 0 0 0
~ NK
hP5 K £0<=fi§2) fc(diag. 9)
1.60 863 870
0.264
1.40 747 890
1.40 785 950
0.252
0.96 710 860
1.10 780 860
0.24
\
/
§
'5
2?
I
r:
.1
.42525

 

 

 

 

 

45

PERIOD V

925155 OF EXTEYIOR COLUKN

 

I. Second floor

1/2 x 22.55 x 157.5 + 12/12 x 150

x’g; X 22.55
12 2

= 45500 + 2950 = 49,550#

Parapet wall = 1 x 3.5 x 20.17 x 120 8,700#
14,900#

Roof load

Wt. of col. = 16 X-%% x %§ x 15" = 7,890#

lst floor

Wt. of floor = 1/2 x 155 ,500#
“all load: spamdral wall = 2. 5 x l x 120 x
(22. 55 x 2. 5): 5950#

wall beam 1 x2 x20.17 x150=5840#

window 8 x20.17x(12-2.5-2)
=1220#
wt. of bracket =O.5L26x18 18x _
12X12h12x 150 567#
_ 26 18
wt. of col. —l2x .__ =
3:21:12X 150 5850#
Basement
Wt. of floor 78400
wall load 15240

wt. of bracket: 567

3§§3§x150x29 =5250
144 12

wt. of floor

45

 

 

 

 

  
   
  

 

 

 

 

 

 

 

 

Counter moment (2nd.fr.top)=l4900x1=14900 in.lb.
=100925xl+13240x2=127405 in.lb.

(Roof) 1
5541

(lst.) 1 1
55

(Base)—;
55 wl

M(col.)

=1185100 x

=1072595 x

=1073595 x

(basement)

load *Amount 3125 of

Floo from. of load col. vert.5teel Po At=A+(n-1)A3
Roof 48,060 26x18 8-1 sq. 0.0171 555

2nd. wall 14,900 A: 458 As=8
col. 7,800
total 70,755
floor 78,400 26x18 8-lin.sq. 0.017 556
wall 15,240 4:455
. lSt. sq.inbAs=88qoing

bracket 367
col. 5,850
total 168,622
floor 78,400 26x20 8-lin.sq. 0.015 608
wall 15,240 A=52

Base sq. n. As=85q.in.
col 5,280
total 255,909

W = 7075"-1o 26 53 — 9 — i

.rt 9 XI§5f§x150—70750-4550—555504

th= l68600-4880=163720#

N t= 265900-8xgéxggx150=255900-455o=251570#

b 12

- Mc=3%x 95500 x22.55x12 -l4900=70k,1001n.lb.

l

' Mo=ggx 156800x22.53xl2 -l4900=1,185,lOOin.lb.

- Mc=—$x155500x22.55x12 —127405=1,072,5951n.1b.

35

=701,100 (2nd. top)
=1155100 x

296

156

H

2:2'615000(Base top)

140:560000(2nd.bat.)
295
1§§=527000(1st.top)

~415000(1st.bot.)

46

 

 

 

 

2nd ton e=.E = 701100 =10.7 .2 _ 10.7 _
* N "55550 a “‘—Ig- ‘O°5
2 d t. = M = 560000 =8.5 g = 8.53 =
n bo e ‘N ‘*55550’ 5 a "I8" 0.475
= M 527000 - e _ 5 22 _
s p N 155720 a 15
— E _ 415000 _ e _ _
t b t. e— — _ _ .66 _,_ 1.66 _ .
15 ° N 251570 1 a 7g- 0 092
= M — 615000 -
N 251570 2 a —-5" 0'122
a
Base bot.e=.L =.§Q§§QQ =1.06 .2 = 1.05 =0,053
N 255900 a 20
From diag. 24, diag. 27:
M M
2nd. T. - k=0.45 '——7— =0.l55 r = ___... =.191199_
fc ’ ° .15eba2 .155x25x152
=606lb/sq.in.
2nd. B. - k=0.54 =0.155 rc:§§9929 =4eelb .in.
' .155x25x152 /Sq
let. T. - k=0.90 =0.12, fc:§22999 =5151b/sq.in.
.l2x25x152
BC
Allowable rc = fa 1+'§”
1+C——;§
fa_o 5 0.224r¢+3fl2g
1+(m-lypg
C = fa
0.4551“c
R2 =.£
A
Allowable fc (2nd fr.)
fa=0.8,0.225 x 2500 + 15000 x 0.071
l+llx0.017l
=O-8X 565 + 508: 0.8 x 374 = 585
1+0.188 1.19

47

 

 

 

 

 

C = 585 = 0.52
0.45 X 2500
R2: l,=,l§229_= 54.7
A 468
(2nd top)
6 5
1+ _-__
rc = 555 x 854.7 2 555 X 1 + 0.1575
34.7
=585 x-§L%§ = 860#/sq.in.‘> 613#/sq.in. (OK)
(2nd. Bottom)
6 5
1+e—L——
f0 = 585 x 34:07 = 5.85 x l + 0.187 x 8
6.5 l + 0 098 X 8
+ ___-___ O
1 082654.7

: 585 X fl : 6257i;/sq.in. > 492#/sq.in. (OK)
1.08

(lst Top)

_ 1+0. 8 0‘
re — 555 x 1 7 x 5 85 = 555 x.12449
1 0 0.095 x 5.85 1.055

624#/sq.in.‘>4622 #/sq.in. (05)

(let Bottom)

f

= P 1 + 0.157 x 1.59 _
c 555 x — 555 x.1;22§ = 570;> 550#/sq.in.

1 + 0.095 x 1.59 1.014 (0K)

(Basement)

_ 0.225 x 2500 + 15000 x 0 154
r — 0.5 -
a x 1 + 11 x 0.154

= 0.5 x 555 + 275 _ 0 5 x 541

1 + 0.159 _ ° 1.159 = 075

 

C=fa.___ =_515 A :051
0.45f'c 0.45 X 2500

48

(Basement Top)

575 x l + e'géé7
o __ = 575 X l+O.1755

7 5 1+0.0895e
+ 0. e—L———
l 15 42.7

fc

 

1.289
1.148

575 x

 

:= 645#/sq.in.7>643#/sq.in. (OK)

(Basement Bottom)

 

1 + 0.175 x 0.515 = 575 x 1,17

fc=575 x
l + 0.09 x 0.815 1.08

Exterior Uolumn Value of.%

 

 

 

 

 

 

 

 

 

 

 

 

Columm][lc:gii (n-l)Is I=Ic+(n-l)Is h I
12 'E
in4 in
2nd 12500 5590 15290 11511 1400—-
lst ' 12500 5590 15290 104 155
Base 17500 4950 22250 90 247

 

 

 

Exterior Column

Computation of stress

49

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Counter 1/55 Wl-Mc H
N d'/a homent ‘ e/a ‘k M2 f0
71% in.lb. in.lb. in.lb. b8. f0
=—
65880 710500 0.5 0.470.155 515
0.159 57,000 710,500
70755 555000 (14440.55<1155 505
155720 550000 0.2140.51<1155 554
0.159 55,000 1,194,700
155500 555000 0.0770.9c 012 550
251570 451000 0.05 0.95c1115 550
0.125 455000 754,000
255900 215000 0.40
Nrt= 70125 - 9.5 x E; x 2 x 150 = 70125 -5150=54015#
th= 97557 - 5150 = 91597#
th= 97257 - 5150 = 911274

Roof 1/55 w1=1/55 x 95500 x 22.55 x 12 715000 in.lb.
Floor 1/55 wl=l/55 x 155500 x 22.55 x 12=12000001n.1b.

 

fc (allowable)= fa

 

fa: 0.5 x 0-225f”c+fcpg =0.83.0.225x2500+15000x0.0171

 

 

 

l+(n-1)pg l+llx0.017
1+0«187 1.19 '
C = fa = —El§———— = 0.515
0.451% 0.451"c
R = 4; R2 = I = 16290 = 34.7
A I’ 455
2nd floor

54.7

 

915 K ME = 1050 #/in.“

 

 

 

 

 

 

 

 

2.69
Bottom fc=9OO
lst floor tOp
bot.
: Basement top
9
It ..
w W
5 ”0'0" F. 0" 6 fits. herb—(no '0‘ 1:;
fi 9'76; f
1 0-0
L-1
1-17
4-fl :
“i" l'-1"‘T?
:l l-r (3 It
h 5~\‘R-E¥O"
TIES l2."c.c.
—i 22— YA‘Q
‘0
-& i.2"
it k"‘0‘5‘315‘23"“35a “0’";
N +

 

 

T;

6';

 

 

3-431 - 13’-d'
TIES ~12"c.c.

 

 

 

 

 

 

 

7'1 22" V4”!
2‘3 ..
“ r P2
I? 51:35 ~101 1.05:1":‘6‘ '. Q'OL O. rM'O 0‘
\N '3‘: h'CAP f
.11.. :_ LC"
L-L :
11-4 3 .. ..
f?- 9; 85—! 67 —— VII-O
'15.", "ED-4.— TlES __ lzllc.c ‘

 

     
  
 

 

 

O

 

 

212—924

 

fc=875
f0:750
fc=690

 

 

 

l8”

 

 

 

 

 

 

M¥§.Ufcmc.

2ND. 4 \ST. FL.

26'

 

 

 

20"

 

 

“Kb-’30 TIES
V4331 12"c..c-

fiASEthnrr

 

 

PERIOD VI

DESIGN OF INTERIOR AND EXTERIOR COLUMN FOOTIEG

I. Interior col. footing:

 

Column load =432,l70#
Assume wt. of footing =8% of col. load = 54,50Qfi
Total load =466,670#

Allowable soil pressure=2.7T/ft2=2.7x2000= 5400#ft3

1 _466670_ 8
A(req d) 5400 86.5ft ”%h_58..4‘°«

 

 

 

., I, ' I I DMSLS
Use 9'6x9'-6 :90. 25>>55.5 rt“(OK) L
P!=駧lZQ = 4790#/ft“

90. 25 "

 

 

Equivalent sq. sect. of 001. 40-585! -—I2’-2." 3.1“.
a 2
4 28
= 4552 = 25in. = 1.92'

M: 45.5 x114 x 4790 x 45.5
144 2

= 5,950,000 in.lb.
(n=l5) K=155

= [4. _ 5950000 _ —-_

use 221n. for d 751
total depth = 22 + 4 = 25 in.._i.

 

 

     

 

 

 

 

 

 

 
 
     

/ --s//
[III/III.

 

 

 

 

 

I4 1 9L6"
Check for shear r“ 1
Total V QOin. out from edge of 001.=479Qx(6i::14)r2: 5
=4129 25. 5 _ 4790 35.5 H
144 x 2 144 x 177 x “§“-= 752OQ#

v _ v = 75200
bjd 55x0.9x22

 

= 6O.l#/in2= 60#/ina(OK)

52
A (req'd)= 0.85H_ 0.55 x 5950000 = . 2
S fsxjxd lBOOOxO.9x22 9.561n

Use 40-5/5ln.¢ baf, As(fur‘d)= 40 x 0.5055=12.5-1n2
>9.55 in3 (OK) 2.. =40 x 1.954 = 75.5111.

 

Bond u=_z_ V=45.5xll4
2’7on 144

u_175000 _“
’75.5x0.9x22

 

X 4790: 175000

 

= lllfi/in2‘<.ll2f/in2(OK
Allowable u=l.5x(o.05x2000xO.75)=112 42/111a

Dowels 1=fsl= 15000x5/5 =
4u 4x100 281n°

55 - 5/81n. 0 dowels at 25V : 2 in.
55

As= 10.7 in.2:>10.5 in“(05)

0.0.

Since the fooring is not enough to extend the dowels,
a pedestal of 6in. x 50in. is used.

 

'd: = fb 3 _ 800 3
A(surface area) req ( 0.25fé ) A'—(25x2000 ) x550
=l.5 x 550 = 795 in2 50in. x 501n. (OK)

Length of bars

Ordinary extension = 121 = 12x5/8 = 7.5 in.

 

End anchorage =.£9991_ = 1000 X 5/8 = .
4u 4 x 100 7 5 in'

Total length = 25.1 in.

If r=8in.

Total length of hook = fl x 8 x 4i: 25.2 x 2.5
27.7in.7725.1in.

Total length = 2x28 +(ll4-2x4-2x5)=27.7ing725.1 in.
(OK) say 28 in.
= 55 + 90 = 145 in. = 12ft. 2in.

Wt. of footing and pedestal = 9.5 x 9.5 x I; x 150
= 29500+470 = 29770#<:54500# (on)

 

 

 

 

 

 

 
   
 

‘ 53
' .,. e.o.s|5":opss’
, , 55763! was II I
II. Exterlor colund footing +LE' . I 1 if“
A(req.d): 1.08x255900 = 53.2 fta ' 1 f “1r-
5405 ' ' 3' If"
. “J ...-t
Use 8 ft. 5 in. X 6 ft. (Sin. 25"”?c‘ L 3'14! ‘1' 1+“
fur'd': 55. 5>55. 2 ft2(C1{) gs—sg'h-At’c? ' '9' "
5‘ Bad
p'=33§999— = 0.815 in. = 0.068' ‘7?-
55.0
Mab;—§§§Z§<X8.5x4780xa2&§£§ -l- ¥
144 2 ‘o
= 1,870,000 in.lb. 9°

 

 

M =.22£§E§ , 29. 815
be 12 x8.554780x——§———

= 1,595,000 in.lb.

 

(1:!1870000 =j2_05= 14.4 use 20 in.for d.

J155 x 78
Check for shear
V ___ 8 '78 60 '78
ab470x§g+ 1.2+(‘2IXI2)] X%2.x%

24780 x(4.87+4.5) = 4780 x 9.17 = 45,800#

v =4780 x(§§:§é .2_x l _ 150
be 12 1.12 '2) _ 4780.xL—I§-—§Z)

=4780X4.7= 22.500#

Governed by Vab

v: 1V d. V _ _45800
bjd vjb 60x60x0.9

 

=15.45 in<f20 in.(OK)

In long direction.

As= 0.85M = 0.85 x 1595000
rsjd 18000 x 0.9 x 20

15 - S/Bin. 5 bar = 15 x 0.5054: 4.5 1n2<:4.451n2(ox)

= 4.45 in2

:a,= 15 x 1.95 = 29.4 in4

PERIOD VII

COT‘BIHEJD FOOTING HEAR THE PROPERTY LII. E

 

Interior column footing = 452,170# H
Soil pressure =2. 7T/rt2 = 54OOfi/ft2

Exterior column loading =265 ,900# (8: O. 815in.=0. 068ft)

Assume wt. of footing = O. l2(452l70+265900)= 85800#
Distance of c. g. of column from the center of exterior

     
     
 

 

column pseesgoa «211°!
_ 452170 x 21.5 _ ,, 2"”
— — 10.3 ft.

698070 2L

  

2(15.5 + 3.2.0)

Length of footing

 

= 28ft. 5in.
Area req'd = 6980220; BSSQQ =145 ft

Area furnished = 6 x 28.25 = 195 ft

  

pl: 698070 _ 2060735/in. EAR D'AG.

28.25x12

  
  

Edge of exterior column
Shear =( -265900+24700£ %)=-2247oo#
1.‘loment=- -}%_2§_O;202=2,250,000 in.lb.

Inner edge of interior column M’ssgooou'

Shear = -2050 x 59.5: -122,500# MOMENT DlAG.
Moment: 3—2-53 x 59.52=5,540,000

Outer edge of interior column

Shear =452l70 -(2060x82.5)=452lVO-175OOO=257,170#

55
Moment =.§g§9(82.5)2-452,170x11.5 = 7,000,000-4,950,000
= 2,040,000 in.lb.
For max. moment

Shear = 265900-206OX=O X=150in.

 

max. moment = -255900(150—10)+2360(1so)2
=-31’eoo,ooofl7,450,OOO=-l4,150,000in.lb.

 

 

5—14050000— 54.8 in. use d=55in. total deptn=55+5+.5
6xl2xl65 =58 5 in

wt. of footing = 6x28.25x§§§§x150=82,200#<’85,800#
As req‘d between the col.
=14,550,000
18,000x0.9x55

= 25.5 in2

 

pt. of inflection applies at Xa from the left end

of footing
2060

2 x8 - 452 ,170 (Xo-74)=0.

2050X8 - 864540X0+ 54,000,000=0 Xo=95.8in.
and 95.8 - 85.5 = 10.5in. from.the outer edge of
the interior col.

v for bond =257l70-10.5X2O6O=256970§

£0: 256170 __
l50x0.9x55

 

=48.6in. use 26 -lin. sq. bar
fur'd AS=26 1na
20:104in.

(deformed bar and anchared at both ends spacing
2.5 in. cc)

For the portion og the longitudinal beam which
projects beyond the interior col.

m=+5,540,000 in.lb. v=-l22,500#

A =5540000

—6 043 in2
ll2xO.9x55

 

55
25:4122500 = 54.7 in.

ll2xO.9x55

Use 25 - 5/8 in. d furnished AS=7.05 in27>6.45 in2
ffi=45.2 in. >540? in.

Design for shear

At edge of exterior col.

= 225000 _ a . 2'
v 6x12x0.9x55— 99°2:>6O#/ln
Use vertical stirrups from the exterior col.
toward center for
X; 225000-60x6xl2x0.9x 5 =.§2999

2060 2060

 

= 45.2 in.

Space req'd of 5/4 in. stirrup
S=6x0.4418x16OOOxO.9x55
89000

 

= 15 in. 9 stirrups space
at 5 in.cc.

At left edge of interior col.

X;__257l7O-6Ox6xl2xO.9x55._121170 =
2060 2060

S____6xO.4418xl6OOOxO.9x55
121170

58.8 in.

 

 

= 11 in. use 17 stirrups'
at 5.5 in.cc.

At right edge of int. col.

v:127500 _
5x12x0.9x55‘

.3 safe for shear

V=127500

 

56-#/in2< 6Of/in2

Design of transverse beam under int. 001.

M_452170 x.l(

24.5)2=1800000 in.lb.
6x12 2

Effective width of col. = l.5x25= 54.5

_J180
d-/___9999 = 17.5 in. Actual d=55-O.5=54.5 in.
V’54.5xl65

>17.8 in (OK)

57

 

As= 1800000 = 5.25 in.2
18000x0.9x54.5
At edge of col.
V=-駧lzg x 24.5 = 147000 #
6 x 12
, 147000
30: = 42.5 in.

 

112 x0.9x54.5

Use 11 - 1 in. sq. bar fur'd As=ll ins, 20:44 in.

The bars are Spreaded in a width of 92 in., spacing
of bars are about 9 in. 00.

Investigation for shear. v=l~47000 :== 51.5#/in2

92x0.9x54.5

 

<150#/ln”(on)

Design of transverse beam under exter. col.

.-255900 1
M4_______ x ..05 8= 977000 in.lb.
5 x 12 2(‘ )

Effective b: 20 in.

d= 577000 = (255‘: 17.2 in.< 54.5 in.(OK)

J155 x 20

Asf[977000 A—= 17.5 in.2
18000xO . 9x54 . 5 E

29'
Shear at the edge
255900

 

 

 

V: x 25 = 85000 #

 

6 x 12

0‘
ll

20+2x54.5= 89 in.

é _ 85000 4#_ _ 24 5 ,
‘ 112xo.9x54.5 ° ln'

 

Use 7 - l in. sq. deformed bar
furnishing As: 7 in“ at 5 in.cc
("0: 28 in.

58

V: 85000 147000 is safe for shear.
Extension of bar

For bar (A)

Extension of hook

2229'
5E...-

For bar (B),(C),(D).

fl x 111 + 41

 

 

58.517571

 

Ex tension = 571
length of hook = 58.51

(1 ‘3
Total length of bar
(A) 54’4- 22' + 58”: 28t-0"
(B) 120” + 48”: 14'-0”
(c) 44”+ 78”: 10'-2”

«v.1. (D) = (c) = 101-2”
' I
l

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

:5 fl 11--
Em4-k ______ 1 __________ ___.
1- \“dl . E
«r: z
I §
fikmf ) &
ID 3-
I l a?
4.....1...sa__._._.__fl___..._”_.a.....s_
' 7 26‘6"

 

 

 

"1"!"

PERIOD IX

DESIGN OF WALL BEAXS

 

2 I15"
22’4"

 

 

Slum
k

 

l
Pa 195-I“ ‘ 0" 2"
4 4214. 2- 4

The total load on the wall beam per linear ft. is:

Brick sill = 2.5 x 1 x 140 = 350#

 

Windows = 7.5 x 8 = 60%
Stem of beam (assumed) = 200#
F 155880 ' 4
100? load 4X22.53 ~1750fi
0§§§95r§

v: 2560x3§é§§_.= 25400 # v=0.5 x 2500 = 150

d(for shear)" 26400 = 16.8 in.

l50x7/8x12

 

”-1
Maia w12:%6kx2560x21.252x12= 1280000 in.lb.

'Effective flange width b=«%§x2l.25xl2+l2=55.25 in.

M=l/2fcbf(d-1/2t)
1280000 = 1/2xll25x55.25x8x(d-8/2)
£.Approx. d= 7.56+4=11.56 in.

Design as rect. beam.

d: II/1k280000‘ '“

=fl‘55‘“= 15.5 in.
\' 207x55025

 

60

Total height = 15.6 + 8 = 21.6 in.;>16.8 in.
Effective Bopth = 22-5= 19 in.
Design of Bl
As: H _ 1280000
fsjd 18000x0.9x19
Use 6 - l in. 5 bars fur'd 4.7 in.2;»4.2 in.2 (OK)

= 4.2 in.2

Design of 52
5: %5 x 2550 x 22.553 x 12 = 1175000 in.lb.

2

 

AS: 1175000 = 5 84 in
15000x0.9x19
Use 5 - 1 in. n/bar fur'd 4.7 in.2;75.84 in.2 (on)

Moment at support = %§w12= 1175000 in.lb.

As(req'd) = 5.84 in.2

5 bars from each beam are bent up.

d'_ 50 4 7
57* = 0.15 p=p'=__;11__ = 0.0205
EU 12x19

From diag. 4 k=O.402 j=O.86
fs= M = 1280000
ijd 4.7x0.86x20

f0: fsk = 15800x0.502
n(1-k) 12(1-01402)

15800 #/in2< 18000 #/in2(0K)

 

 

= 885 #/1n3<;1125 #/in2 (on)

Both of which are satisfactory.
Max. unit bond stress

u: V = 25400
553d 18.8x7/8x20

 

= 80 #/in2

In Bl the bars may be bent O.18x21.55 = 5.85' from the edge
of col.

In 82 = O.2lx22.55 = 4.68' from edge of col.

Assumed the point of inflection to be 0.251 from edge

of col.

0.25 x 21.25 = 5.52' in B1 0.6x5,52'=2.66' pt.of bent dn.

61

0.25X22.55=5.55' in 82 O.5x5.55'=2.77' pt.0f bent dn.
Design gor vert. R.

In B2 x1: 1 -.Eaglgz 22.55 _ 50xl2x7.8x1§
2 W 2 ‘__ 2550

 

=1l.155-4.2
=7 ft.
At support V'=V-Vc=26400-(50xl2x7/8x20)
= 25400-10000: 15400#
with 5/8 in. 5 U stirrup ft: 15000 #/in2

S: 2x0.5068xl6000x7/8x20
15400

 

=lO.7 in.

max. 3: O.5x20= 10 in. use 10 in. cc.

First one 2 in. from the edge of col.

Total =.§§;3 = 8.4

 

10
In Bl x: Elséé - 50K12X7/8xl2 = 10.55 - 4.2 = 5.45'
2 2550
At the support
V=2i53§ x 2550 = 25100.%

2
v'= v-vc= 25100-10000 = 15100 #

_gx0.5068x16000x7/8x29
15100

S:

 

= 11.8 in.

use 9 - 5/8 5 U-stirrup arranged as in B2

 

 

 

 

 

 

 

 

 

LENGTH OF BARS & STIRRUPS

 

 

 

62

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(N 3| Q‘d-E
6L
flan A 23.0. v .037 J. ‘mn !‘ 25.0" ‘4': 3,-‘0'
TL. 3 321d"
IN 52. 510' , K.“ 1.6“ M.

 

 

 

 

 

 

12L. 2m15"=ss’
LENGTH OF STIRRUPS

T.L. - 58": 4le

 

 

 

 

 

 

 

‘N L
*0 q 0
55’4- 1‘—
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PERIOD X

DESIGN OF STAIR SLAB AnD BEAM B3 FRAEING STAIR WALL B

<1 1
\\\\\\\\Jb

Use 8: 10.5 in. b= 7 .2 in.
15
axb = 74:>70

<175

5
10 at 7—— = 6 ft. 0 in.
16

9 at 10.5 = 7 ft. 10.5 in.

I. Slab stair

 

L.L.= 100 #/ft2
Tread w: 10.5 in. R: 71% in.

No. of tread = 18

L: 7.88) +5345.5=15.5 ft.
Assume wt. of slab =lOO#/ft2
T.L. on 1 ft. strip:

 

Treads =gle-5X7.187

 

2 x 144 x150: 548
slab = 15.5x100 = 1550
L.L. = 14.06xlOO = 1450
Total = 5504 #/ft.

M: éx5504x14.06xl2=70000 in.lb.

d=(19299 =j§éti = 5.5 in. use 6 in.
207x12

Total = 6+1 = 7 in.

 

54
wt. of slab =i%x150=82.5 #/rt2 100#/rt2 (ox)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

L 9@ 103-71403? 1 5'4"
1_r‘ 5“
CR
‘fo :9
r 8
l 9 La
‘\\11
. 52:6" .. 4?
Le we
As(re 'd) = 70000 5=— =0.755 in2 r ft.~ dth
q 18000x0.897x6 p6 W1

use 5/8 in. 5 bars 4 in. cc. furnishing 0.918 in2

' :>0.74 in2 (on)
1 - 5/8 in. e 5ft.Oin. long bar placed below each
tread perpendicular to main reinforcement is used
for construction.

2. Beam B5

wt. of partition 20x12: 240 #/ft.

wt. of beam (assumed) = 100 #/ft.
= 340 #/ft.

M:%x540x14.052x12= 100000 in.lb.

bda=lggggg = 484 in5 b=7in.

d=é§é = 8.5 in. use 9 in.

total d= 12 in. wt. =.1— x.lZ x 150= 87.5 #(OK)

 

 

1 2
L/D= 14.0: x 12 a 24.1 = 24
As = lOOOOQ— = 0.686 in2
18000x0.9x9

use 4 ~ 1/2 in. 5 bar 2 in. co. in 2 rows

65

 

v=0.5xl4.1x540 = 43%,3/1112

 

 

7x7/8x9
u_0.5Xl4.lX54O ___ 4:8 f):/in.2 (OK)
4xl.57x7/8x9
l-lBOOOXO‘SXO‘S— = 11.5 in. use standard hook
4x100

L= flx5x0.5+4 = 17.7 in.

Tetal length of bar = 17 ft. 0 in.
use 4 straight bar to take care of -M 1 ft. each.

PERIOD XI

DESIGN or BEAIS B4 AND B5

     

26mfl$

      

 

 

 

I 0 Beam B4 “l‘ u H I .. III/ullll/J'I/Illl /1
A . Loading R'l‘“7°0*
Partition wall 20x10 = 200 #/ft.
wt. of beam assumed = 450 #/ft.
1/4 panel load = 2020 #/ft. ( 6x22.5>>l/4
Total ' 2670 #7ft. total area)

load due to stairway 0.5x5504 : 1625 #/ft.
Concentrated load from B 2580 #/ft.

Design

51:2580xlll 2570x22-55 + 1650x113
22.55 2 22.55x12
= 1170 + 29800 + 4500 = 55500 #

Point of lero shear

X1.44700 = 44700
2670 + 1650 4520

= 10.4 ft.=10ft.5in.

t = 10.4x12 [44700 -,% (2570+1550)x10.4]

=10.4X12(44700-22400)= 2,790,000 inlb.

Allowable unit shear = 150 #/in.2
b’(assumed)= 12 in. E.W.(b)= 12+ %§x22.55x12=54.55in.

 

b'd=_éé199— = 540 d—54O— 28.5 in.
150x7/8 12

use d3: 50 in. Ta: 54 in.

wt. of beam = 12x(54't) x150 = 425 #(OK)
12x12

 

immens-v-4w-.<r a1
,

 

67

19.8 in<Z30 in.

 

dz/2700000‘ =,5§3
207x54.55

V

 

 

n2einformement

As“ 2790000 = 6 in2 use 6 - 1 in.sq. bars
18000x0.857x30

p = 6 =0.0058 E: _ 0.3
54.55x50 d 50

From diag. 2, neutral axis in flange.

Bond stress

 

—- 44700 4'5. 0 2
u— = 142 188 , 1n (0.075fc)
0.5x24x7/8x50 <1 . 7/
Def. bars end anchoraged

Bent 5 - l in. sq. bats to resist negative moment
at both end, use 5 - l in.sq. bars.

stirrups v3: 50 #/in.2

vc= 50x12x7/8z 15800 a

Left end 55500-15800 = 19700 = 7.4 in.
2670 2670
use 1/2 in 0 U-stirrups
Smax = O.5x50= 15 in.
V'= 55500-15800: 19700 #
822x0.196x16000x7/8x50_

 

 

 

 

= 8.35 in.
19790

use 8 in. through a distance of 3 ft. from
support.
Sz2xo.198x15000x7[8x50_ u o

19700-5870 14 use 6

Right end

X_44700-15800__2<3900 6.7 ft.

(2870+1850)"4520
use 1/2 in. 0 U-sttrrup

 

II. Beam B5

A.

68

V'= 14700-15800=29800 #

s: 2x0.l96x16000x7/8x50
52250

use 6 in. for a distance of 2 ft. and 4 ft.
for the rest.

= 5.8 in.

 

 

 

 

 

 

 

 

 

 

R. ‘ fin
Loading 1:]
wt. of partition = 20x11: 220 #/ft.
wt. of beam assumed =550 fi/ft:_
770 #/ft.
load due to floor = 2020 #/ft.
load due to B4 =44700 #
Design
BL = 44700Xl6.3 + 2790 x 22.55 = . L
w 22.55 2 65900 #
22.55 2
Point of zero shear x=§2§99 = 21 ft.
2790

nmax = 21x21(55500-l/2x21x2790)= 8,150,000 in.lb.

Assume b'= 16 in.
b'dz 59500 = 450
150x778

d: 450 = 28 in. total length = 52 in.

16
wt. of beam.= l§E§2x150 = 555 #<'550#

12x12
E.W. (b)= 16 + %5 x 22.55 x 16 = 45 in.

d: 5150000 = 635 = 26 in.< 28 in. (OK)
1(207 x 45 [

 

Reinforceme nt

 

AS: 8150000 _ , 2
18000x0.857x28 14 ln’

‘ “—- "—.r-o _qg-M‘... -.r—
.

 

69
use 9 - 1.25 in.sq. bars As: 14 in.2

14 = 0.011 t — 9 = 0.52

D: __
‘ 45x28 ‘3 28

 

from diag. 2 neutral axis is in flange.

59500 — 674/1n2<1150 #/in3

Bond stress u= _
40X7/8x28

 

Negative moment at both ends use 5 - 1 in.sq. bar

stirrups:

(1) Left and V0: 50x16x7/8x28= 19800 #
v'= 74500-19800: 54700 #
x: 6 ft. 0 in. use 5/8 in. 0 U-stirrups
q_2x0.507x18000x7/8x28
” 54700

’v

 

— 4.64 in.
use 8 in.

(2) Right end
x=59500-19800 = 59700
4810 4810

S_2x0.507x16000xZ/Q§g§ _240000
59500-50x16X7/8X28 59600

 

 

= 8.25 ft.

 

= 6.1 in.
(5) Bending schedule
L1: 12 in. or.3§;§§= -
20 12.8 1n.

L2= 9000x_1 1 = 18 in.
4 x 125

L5: 56 in.

(4) B4

.5,

7O

anon-6'

 

 

*0"

 

 

 

54-

 

r:

 

tag;

 

 

 

85

 

 

and

.1.

 

 

 

 

 

(b) DETAILS

 

 

 

 

 

 

 

 

 

 

 

 

22! "

 

 

 

 

1'-'.v

 

 

 

 

 

 

 

 

 

 

 

71

BOOKS

Urquhart, Leonard Church and Charles, Edward O'Rourke
Design of Reinforced Concrete Structure, New York

and London, McGraw-Hill book company, Inc., 1940

Doubman J. Russell, The Organization and Operation of
Department Stores, New York, John Willey and Sons,

Inc., 1927, Pp. 155-171

ARTICLE III

DRAWINGS

 

 

 

 

 

 

 

 

 

 

 

 

 

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