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Design of A Service Garage
And Office Building

A The“. Submittod to

1310 Faculty of
IIGHIGAN STATE 001.180!
of
AGRICULTURE AND APPLIED SCIENCE

by
Frank Grove Foster, Jr.
Candidate for th. Dogréo of

Batcholor of Science

June 1946:

 

THESIS

a./

ACE]! OWLEDGIENTS

I wish.to thank the members of the Civil Engin-
eering staff for the assistance they have given.me in the
design features or this thesis. I also appreciate the
willingness of Ir. Grannan.or the Portland Cement Assoc-
iation in.making available the services or his organiza-
tion. Practical advice trcn.lr. Trumpower of the college

service garage was also helpful.

A
d

x p"l') I)
, . '€
‘. .‘ . y

L/v' . \. e

p.

BIBLIOGRAPHY

Pulver - "Construction Estimates and Costs".
Sutherland 3: Reese - "Reinforced Concrete Design".
Ecol a Johnson - ”Concrete Engineers Handbook".
Eshback - "Handbook of Engineering Fundamentals".
A.I.8.C. - "flannel of Steel Construction".
Parker - ”Simplified Design of Structural Steel".
'Ihe A.C.I. Building Code.
Lansing Building and Safety Code - (hereinafter re-
ferred to as L.B.C.)
Publications of the Portland Cement Association:
"Simplified Design of Concrete Floor 8y: tens"
(hereinafter referred to as 8.0.0.158.)
"Reinforced Concrete Design Handbook” - (herein-
after referred to as R.C.D.H.)
Bulletin 81' 51 - I'Concrete Floors on Ground".

CONTENTS

Purpose and Scope of 'Ihesis
Brief Description of Computations
Computations and Sketches
Drawings

 

PURPOSE AND SCOPE

OF THESIS

THE PURPOSE of this thesis is to present a
practical design of a building suitable for use by the
college for a service garage with separate office space
provided in the second story. The inadequacy of the exp
isting garage facilities make this a timely subject for
thesis and the variety of design features entailed make
it an excellent student problem.as well as one'which a

practicing engineer mdght be called upon to solve.

The existing service garage space could.be cone
veniently utilized for the parking of the trucks_operated
by the Buildings and Grounds Department whose offices are
in the same building. This would also serve to relieve
the parking problem in this congested area as well as pro-
tech the equipment.

The.college operates at East Lansing, including
automobiles, trucks, buses and trailers, about 125 vehicles;
this figure being liable to increases in the future. 0c-
casionally the college garage is called upon to service as
many as 25 of these in one day. The proposed building,
with.its more than 5,000 sq. ft. of floor space available,
would be fully capable of providing the facilities required
to cope with this situation.

The location of the proposed building, as shown
by the sketch, provides two convenient entrances to the

garage and does not in any way impair the:functions of the

existing'buildings in the vicinity. The entrance to the
east is directly in line with the exit from the main.campus
drive, while the entrance to the rear would.be used for the

necessary dead storage within the building.

The front (east end) of the proposed building is
to be two stories high. Ihe ground floor of this portion
of the building contains a garage office and a parts stor-
age room. The second story contains three offices with
dimensions 19'x17', 19*‘113', and lSi'xlS' for use by the

Stores Department.

TEE SCOPE of the thesis is naturally limited by
the problmm itself and‘by the time and space available for
its development. All problems of engineering design which
will be encountered in the evolution of this building have
been.adequately dealt with. No attempt has been made to
invade the sphere of the architect, however, in computing
dead loads certain basic architectural features have been
assumed which.should'be adhered to by the architect. These
assumptions, evident from.the computations, tend to give
the proposed building an architecture similar to that of
the existing Stores building to which it is attached.

Simdlarly no space is given to plumbing, heating or‘wiring.

Included in the thesis are all,computations,
sketches and drawings necessary to draw up complete arch-

itects and mechanics prints and form.the bill of materials

-4-

for the proposed building. Specifications should be made
with reference to the unit stresses used in design as

stated in the computations.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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:6-

BRIEF EXPLANATION OF
COIPUTATI 0N3

The Lansing Building and Safety Code classifies
all buildings according to occupancy and type of construc-
tion. The service garage portion of this building is class-
ified as ”sub-class B-l”, with types 1, 2, 5, 4 or 5 con-
struction allowable. I have chosen type 5 or skeleton con—
struction for the service garage. The second.floor portion
of the building containing offices comes under "class 0'
buildings, with types 1, 2, 5, 4 or 6 construction allow-
able. I have chosen type 3 or protected construction for
this portion of the'building. Because of the class B and
C occupancies in the same'building a complete fire separa-
tion is necessary, otherwise both.parts of the building
would.be subject to the strictest requirements of either
classification. The fire separation is Obtained through
the design herein presented. All unit stresses and design
methods are those prescribed by the L.B.C., A.C.I. Code
and A.I.S.G.

TIMBER AND STRUCTURAL STEEL DESIGN

GARAGE ROOF DESIGP:

Using type 5 or skeleton construction for the
service garage, unprotected steel and timber members may.
be employed. Using 5-ply felt, 5 coats of asphalt and 1”
sheathing, (From Pulver - ”Construction Estimates and Costs")
we obtain a dead roof load of 6# per sq. ft. The L.B.C.
specifies a live load of 40# per sq. ft. for flat (less
than l/6 pitch) roofs, giving a total of 46# per sq. ft.

-6-

Using unit stresses Obtained from L.B.C. for common struc-
tural fir or pine, timber rafters spaced 18' o.c. and pur-

lins spaced 5' c.c. were designed.

TO SUPPm‘I‘ ROOF SISTER:

Instead of using four 60' trusses, it was decid-
ed in the interests of economy to use eight identical 30'
trusses. This necessitates a row of columns at the center
of the garage, however, the economy of the shorter span

will outweigh any inconvenience involved.

has 30' truss is designed with 6 panels 0 5'.
Panel loads are computed at 4,740# per panel. Using a
horizontal lower chord and an upper chord with 1/6 pitch
the truss was analysed for total depths or 45". 6P and e}
ft. lbs lightest angle used in practice as truss members
is the 2" x 25" x 5/16" angle. With a total depth of truss
of 6% ft., it was found that two 2" x 2}" x 6/16" angles -
3/8" back to back - short legs outstanding could be used
for all members throughout truss, therefore this design was
used. its methods of analysis used were those of moments,
shears, Joints and summations of forces in H and V direction.
lhe final design was checked by the methods of graphic statics.
Shop details of truss were obtained graphically by drawing
each panel to 1/2 actual size and scaling off dimensions.

Rivet spacings used are those recommended by A.I.S.C.

TO SUPPORT TRUSS:

Using unit stresses in masonry prescribed by
L.B.C. no pilasters are required in the south wall to
support truss if made of brick 12" thick with portland
cement mortar. For columns #9, #10, #ll, #12 to support
trusses at center of garage rolled sections were selected

from A.I.S.O. tables computed for column fornmla -
r 3 18 000

W

as required by 1.3.0. for l/r ratio between 60 and 120.

At north side of building a cement block pier is used -
ratio of height to least dimension being less than 12 and
compressive stresses within the L.B.G. requirements. To
provide lateral support for trusses and facilitate the add-
ition of a moveable 2000# chainfall, a double angle section
was designed to be run from column to column, welded at
each end to truss bearing plate; the first and last member

to be anchored in 12" wall.

LINTEL DESIGN:

Although architectural details are not included
in this thesis it is assumed that window openings will be
similar to those of the existing adjacent building, thus at
the north and south ends of the building 11 ft. wide window
openings will be used and at the south side 13 ft. wide
window openings. Garage door Openings are 14 ft. wide and
office entrance is 3*} ft. wide. Lintels are designed to

support a 12" brick wall with surface area shaped as an

equilateral triangular with.base equal to width.of Opening.
Lintels are composed of "I" beams and angles welded to 12"

plates.

JOISTS TO SUPPORT OFFICE ROOF:

The character of the roof slab over the offices
necessitates a rolled section or small truss for its support.
I am.using Bethlehem.rolled steel joists spaced 2} ft. o.c.

for this purpose.

BEARING PLATES: .

Bearing plate areas are computed.from.allowable
L.B.C. unit compressive stresses for material carrying load.
Thicknesses are computed.by A.I.S.C. method. Bearing plates
were necessitated.by bearing of truss on outside wall, bear-
ing of truss on columns #9, #10, #11, #12 and‘bearing of
truss on cement block piers, also at base of central columns
and at base and top of column "A”. No‘bearing plates were
required for bearing of lustels on 12' wall, bearing of
office roof joists on 12" wall, and‘bearins of Sect. 1 on
12” wall or Sect. 2 on 4“ partition.

REINFORCED CONCRETE DESIGN
All unit stresses and requirements for reinfor-
cement‘bending, placing, anchorage, stirrups spacing and
etc. used in the design of reinforced concrete members are
those prescribed by the A.C.I. Code as the sections of the

-11-

L.B.C. covering reinforced concrete were repealed in 1940

and the A.C.I. Code substituted.

OFFICE ROOF DESIGN:

Ilhe office roof was designed as a 2%" concrete
slab with expanded metal reinforcement supported by rolled
steel Joists 2} ft. 0.0. The live load of 40# per sq. ft.

as given in the L.B.C. was used.

OFFICE FLOOR DESIGN:

The L.B.C. recommended live load of 50# per sq.
ft. for office space was used in the design‘of this floor
system. The design was based on a location of partitions
as shown on sheet 2 instead of increasing the unit design
load for partitions as is sometimes done. file greater part
of the slab is composed of a divided joist slab using tap-
ered metal pans. Between beams #4 and #5 a one way flat

slab is used.

REINFORCED CONCRETE BEANS, GIRDERS AND JOISTS:

Total loads supported are computed with no re-
ductions of live load and moments are figured by method of
moment coefficients. Spans are measured between faces of
supporting members. Actual design was accomplished by the
methods of “Sutherland 3: Reese” using A.C.I. specifications.

STAIR DESIGN:
The two flight stairs with 7}“ riser and 10"

tread meets the class 0 building requirements of one separate

exit from the office space. Stairs are designed for 10039
per sq. ft. of horizontal span with a one way flat slab
landing. Rolled sections #1 and #2, column "A“ and beam
#6 are designed to support stairs and landing.

REINFORCED CONCRETE COLUINS:

. Leeds on columns supporting continuous girders
were computed from the three moment theorem and then in-
creased fer ratio of I'height of column" to 'least dimension"
greater than 10, by A.C.I. formula. Rectangular tied col-
umns are used throughout. Reinforcement and lateral ties
are computed by A.C.I. methods and formulas.

FOOTINGS:

Due to proximity of proposed building to existing
Stores building, footings #1 and #4 were designed as rect-
angular beams with concentrated load and uniformly distri-
buted soil reaction. Spread footings were designed byuse
of table 35 of B.G.D.H. Computations show that no reinfor-
cement was required to resist bending or shearing stresses
for wall footings and footing "A”, therefore plain concrete
with temperature steel only is used.

Dowels were used to transfer stress of longitud-
inal reinforcement to footings and in some cases pedestals

were required.

FOUNDATION WALL:

Foundation walls were designed to bring the wall
footings below the frost line. The weight supported by
foundation wall produced unit stresses which required only
the minimum A.C.I. reinforcement which was provided.

GARAGE FLOCB:

Garage floor was designed by the methods set forth
in the Portland Cement Association bulletin, ST 51, "Con-
crete Floors (h Ground“. Assuming maximum wheel loads of
5000# a 6" slab is used with welded mesh reinforcement and
dowels, expansion and contraction Joints provided where

necessary.

C OIPUTATI OHS

-15-

DESIGN OF STRUCTURAL STEEL
AND TIIBER MEMBERS

Allowable Unit Stresses From.Lansing Building And Safety
Code.
TIMBER
Common Structural Fir or Pine

Stress in extreme fibre in‘bending.........1400 P.S.I.
Horizontal shearing stress................. 125 P.S.I.
Compression.parallel to grain..............1000 P.S.I.
Compression perpendicular to grain......... 250 P.S.I.
Extreme bearing stress.....................1700 P.S.I.

STRUCTURAL STEEL

Extreme fibre stress in bending............18,000 P.S.I.
Shearing stress............................12,000 P.S.I.
Compressive stress.........................15,000 P.S.I.
Rivet Bearing

Single shear..........................24,000 P.S.I.

Double shear..........................50,000 P.S.I.

5.11m

Extreme fibre stress in'bending............18,000 P.S.I.
Shearing stress............................15,5OO P.S.I.
Rivet bearing

Single shear..........................24,000 P.S.I.

Double shear..........................30,000 P.S.I.

 

Standard notation is used throughout this thesis except
that for any Joist,‘beam, girder, rafter, purlin, lintel or

other horizontal member supporting a uniformly distributed

verticle load:

"1

Total superimposed load in # per. lin. ft.

'2 In Total weight of member in # per lin. ft.

'1 pl“. '2

SAFE LOADS FOR MASONRY

Prom - Lansing Building a Safety Code

Lime and
Portland Cement
lortar

Cosmlon Brickwork

Clam Tile
Lead on Gross Area

Concrete Blocks
Load on Gross Area

Compression
In Bulk of
Masonry

150 P0801.

90 P.S.I.

90 P.S.I.

Bearing Press-
ure for concen-
trated loads.

175 P.S.I.
115 P.S.I.

115 P.S.I.

-17-

 

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-13-

RAFTER DESIGN

5 ply felt at 5“ per sq. = 150# per sq.
5 coats asphalt G 25# per sq. 8 1251? per sq.
sheeting -"' 5% per sq,

Total = 575# per sq.
. Say 6# per sq. ft.

D.L. 6# per sq. ft. -

D.L. : M e u a
Total roof load 8 46# " " fl

 

Rafters 15 ft. long supported by purlins every 5 ft. and

spaced 18" o.c.

Assume 2" x 6" nominal sise
1-5/8" x 5-5/8“ dressed size

'1 = 45 x 1.5 = SW/ft/

I2 I 2.5#/ft.

w : 71.5w“.

lax. llom. = .10 w12 (See Fig. l)

= .l x 71.5 x 52
179'# or 2150'#
Allowable Iom. = f8
I g 24e10 .3
3 e WI " 3'57

f3 8 1200 x 8.57 8 10,300'# (0.1:. - overdesigned
for rigidity).

PURLIN DESIGN
Purlins 18' long - simply supported - spaced 5 ft. o.c.
Use 4" x 14" timbers - dressed sise - 5-5/8” x 15-1/2”

~-

 

 

 

-19-

 

 

 

 

 

 

 

 

 

 

 

740
4740 a’ 740
4740 c é ‘9 4 ,r 4740
£570 A ’ M; {.0 5 0/ f 9 2570
I I o 9’" é '06
agar: 4 \ e 7 ‘ 5-06"
‘ chgémg,@t\9‘~//.
Q 6”? :0 3:5 ‘6 s_ ‘s. Ar“ as .33 a k0 2
Q9 0's * 5 ‘ \n ‘9 4 . d h
V‘ 1 3 <3 /0 /Z V
I. 4-
szo 4 @ 500' = 30’ #220
THEORETICAL; TRUSS A
73/63 3 DES/6N 5.12.5: /"= 5'
B
2
C
4
L:
7 3,/O
5:53} .4: Z
6
E
9
F
1!
{1}
H
GRAPH/c
F/a, #4 ANALYSIS 5041.5: /"—' @030”

I1 = 71.5 x 5 x 12/18 8 25% per ft.

'2 . 3 I” per ft.
' . 251.” ”1‘ ft.
lax. lam. . I? (See Fig. 2) = $1.341? a 10,150'#

or 122,000'#
Allowable Ion. 3 Sf
I 745 2:5
3-..s_3.:758 110.11

:3 = 110.11 x 1200 8 132,000"# (0.1.)

Check for horizontal shear:

V=SV3251x18x5
'51 I 1353

x 8 69.6 P.S.I. (0K)

LOADING OP TRUSS
It. Roof per panel I 251 x 18 = 4520!!
It. Truss per panel (assume) I 220#

Total panel load = 4,74.an
ANALYSIS 0]? 151038 ’
See Fig. 5 for theoretical truss proportions.
Taking free body - Fig. 5

sinOirggé.

d = 29 x 12.66.. .’ 18.15'
Summation IO = 0

-11,850 x 24 / X(ls.ls) s o
x a 15,690

Stress In (1-2) 8 15,6991:
Taking free body - Fig. 6
... - - as
g 4 85
d. 3‘ x m . 23e61
Sumtion Io . O

-21-

 

 

 

I

A )
P.)
b

F/G. 7

7/9. 3

 

F/G. 7

 

 

-11,S5O (24) ,I 4740 (29) / x(2s.el) . o
' x I 5,220
Stress In (5-4) = 61%!
Taking free body - Fig. 7

sin e I 31.3%
d I 59 x 3.1% = 29.20
Summation to = o
:11,950(24) / 4740(29) / 4740(54; / 2(29.2) - o
' x u: 479
Stress In (5-6) I 4—7319.
Taking Free Body - Fig. 8
Summation N I 0
(1-2)}: . 15590 (3725) = 12,250 ‘3 «5-2)n
15-2) I 12,250 (-54%?- ) - 12,400 ’
Stress In ib-2) «13 12.4%
Taking Free Body - Fig. 9
Summation V = 0
12-5) r’ 15-2), - 11550 I o
(5-2)v I 12,400 (3.1%) = 2,050
(2-5) = 9,920
as... In (2.5) I 9,92%
Taking Free Body - Fig. 10
Summation R I 0
(Io-21h - (o-uh / (5-411, :- 0
12,250 / ”20(37331 = te-4)h
\c-4)h I 16720
(04) I 16,720 (9-7-32) I 16,900
Stress In (0-4) I 16.9%

 

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-24-

Taking Free Body - r13. 11
Summation H I 0
(5-1) I (1-2)h I 12,250
Stress In (5-1) = 12,259fi2
Taking Free Body - Fig. 15
Summation B I 0
(c-4)h - (5-5)}, I (5-5),1
15,720 - wear-33») = (d-6)h
(d-6)h .-. 15,400
(d-6) = 15,400(§§°-£) = 15,520
Stress In (d-5) I 15,529fl
Taking Free Body - Fig. 12
Summation V'I 0
(4-5) = (3-4)'
52209—53): 4550
Stress In 4-5 8 4%
Taking Free Body - Fig. 12
Summation.fl I 0
(5-1) I (5-1) / (3-01;
(5-1) I 12250 / 4470 I 15,720
Stress In (5-1) I 16.72%T
mung Free Body - Fig. 14
Summation V I 0
(5-7) I -2(d-5)v / 4,740
(5-7) I -2(15520 x 52.3%) ,1 4740 I -710
Stress In (5-7) I 1151!
Taking Free Body - 5'13. 5
At Joint a-l-i-a

-25-

Summation H I 0

Stress In (a-l) I 14,220fl
Summation V I 0

Stress In (1-1) I Qt

TRUSS ANALYSIS
512353 SUMMARY

!2!22£. 222222
a-l 14, 220#0
5-2 12,4oo#0
0-4 16, 900%
d-6 15,520#c

1-1 0#
1-5 12,250#r
1-5 15,720#T
1-2 15,59o#r
5-4 5,220#r
5-5 479#0
2-5 9,920#c
4-5 4,55o#c

6-7 7109)”!

Selection of lembers for Truss
laximum tension 8 l6,’720# (Iember i-5)
Roq'd. Ares = §§§§g I .95 so. In. effective
A.I.S.C. effective area in tension 3 Back - Rivet ,1 1/2
‘ outstanding leg.
Using 2 - 2}" x 2' x 5/16" angles back to back with
L A short legs outstanding

f“: m:-

 

-25-

Eff. Area I 1.51 - (5/16 x 7/8) - l-ll/l6 x 1/2 x 5/16
3 .73
.75 x 2 I 1.46 Sq. In. 93
Compression Iembers
For l/r Ratio greater than 60, L.B.C. requires use of

column formula, f = 18000

I ; 1:71am (I7r)z

From A.I.S.C. Table,

2 angles - 2}" x 2" x 5/15' - 5/9' back to back
will withstand compressive force of from 29,000# to 59,000#
for unsupported lengths of from 4 to '7 ft. l/r ratio in
all cases less than 120, complying with L.B.C.

merefore 2 - 2}” x 2" x 5/16" angles - 5/8" back
to back with short legs outstanding will withstand maximum
tensile and compressive forces in truss, and being the min-
imum size angle ilich will take a 5/4' rivet.

his angle will be used throughout.

Rivets Regired
Use 5/4" Shop rivets.
One rivet in shear (double)

2
2 P1 (12 x r = 2 3 1416 7 x 15,500 = 11,9204)e

One rivet in bearing (single)
5/9 x 5/4 x 50,000 I 5,440!
Bearing stress will govern.

 

 

lenber Stress Rivets
IIIowa'SIe beariilgjer five? Req'd.

1:235

 

-27-

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

[ember Stress Rivets
Wearfgg pgr RIveE Red'dL—
... Its-91% 2
‘ 8.44

8 0

1-1 0 2

311?};-

1-3' 12 250 2
"61176

1-5 16 720 2
‘5"; 40

1-2 15.690 2

3-4 6 220 2
» 8.430

5-6 479 2

LEE: a
2-3 9.820 2
4-5 4 350 2
m _
6-7 710 2

For shop details of truss members see Figs. 15, 16 and
1'7. Dinensions were obtained by laying out truss to 1/2 full
scale and scaling off. Rivet spacing is in compliance with
L.B.C. and A.I.S.C. Upper and lower chord masters are to be
fabricated as shown in Figs. 16 and 1'7.

Values of "L' for web msnbers (Fig. 15) are as follows:

~28-

- W55 Mama-5R5

776 /5’

 

 

 

 

 

 

 

m 3% nos-«No- eke-C kwkom (tom-k

 

 

 

 

..a\\
II J...
mulllllunb -
. "a..- -..- LE- -T 1E..- L- I: \
~Illl-l-IL_. _ ____ . _._ _ __
EL): IILWon-rllln-fi “-1IL. \Nr ‘- -
liemfi. 1n \- gh ..\ .. \Nfix

523252.550 new-w ...-«0th.
{9(an 09.me mu Nun-SQQ A‘s-no?

we. sR-u-ww- N -mc-Q-Q tho-03 - 3.9K

Q
..wx
\e.e
. -

 

-29-

 

£92222. "I." in Inches
a-l 54-5/4
2-5 44-5/4
4-5 54-3/ 4
6-7 64-3/ 4
.1-2 50-15/15
5-4 57-11/15
5-5 74-7/15

Check Wt. of Truss
It. 2}“ x 2" x 5/16" angles I 4.5# Per Ft.

5.05 x 5 I 50.55 5.40
5 x 5 I 50.00 5.95
5.252712% 1.59,
97.11 20.91

21-5-3

97°11
Lineal Ft. of Double Angle = 155f53'rt.

159.95 x 4.5 x 2 I 1260 # Wt. Angles

24 at. 0.2. (Approximately)
50 # Wt. Truss

1500

(Assumed 220# per panel)

Truss BearflPlate in Isl;

From L.B.C. - Allowable compressive stress under concentrated
loads in common brickwork = 1'75 P.S.I.

Load = 14,220#

Ii.;§2 I 51.5 Sq. In. Req'd.

-30-

Use 8' x 12' plates.
8 x 12 I 96 sq. in.

’55—“ 220 = 149 P.S.I. - m1. 1: 15.3 than 150 P.S.I.
in'bulk of masonry - complying with.L.B.O. and
making pilasters unnecessary.

Bearing Plate Thickness (A.I.S.C.)

t2 :- 5.21:5 = 1% = .555
t i .6" - Say 5/8'
Use 8' x 12'' x 5/8" Bearing Plate In v.11.

Truss Bearing Plates 0n Piers #15, #14, #15, #16.
From,L;B.C. - Allowable compressive stress in cement block
masonry = 115 P.S.I.
- 4 -
Required Area - l—i§§9 - 124 sq. in.
Use 12" x 12' Plates.

p = 1422° a 95.7 P.S.I.

144
t2 8 9: 7 6 3 .53
t = .72”

Use 12' x 12" x 5/4" Bearing Plate on Piers.
Piers #15, #14, #15, #15 - 2' x 2' x 15' high - nude or
concrete blocks, ratio of height to least dimension.mnst
be 1ess than 12 (L.B.C.)

lé""3ar'7.5 0:

Oclulns #9, #10,1#11, #12 - See Fig. 18.
man Load a 28,440# ,4 4,000 = 52,440#
Unsupported Height I 15 ft.

For %. ratio less than 120 as stipulated‘by L.B.C. we reduce

/ [+155

BEAR/N3 PLATE-
5ux5'1x z/zH—\

i AL 3

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

X
I :1 .. .. ,.
\-- I---- --2—5’-5/’2x//z by)
BA SK 779 ’52.; ~'.T,:\’ —
COLUMN 3/1953 7‘ L535
L035" I 20‘3- OUTSTANDING
7/6. /<5
“WL
/\
/ \
WALL. AREA
jlllHHIHJB
a UPPOF? TED ‘ V a
$0 I I I I l l I ...—60
l AIM/TEL | I \
r: 7 7 I I l I LL 7 ,
7%“ "I." #:654
iro— I BEAM
r —1 J" FLA 7,5- /,2"x;;,r”x'z."+1'

 

FIG. M L/NTEL CRoss- Star/0N

"r" to r : 18000

1 1 1718000 117142

From A.I.S.C. Tables A 6" x 6"H 20# column is required.
Footing Bearing Plates -
1-
fo-2000
égaégg I 16.22 sq. in. req'd.
Use 8” x 8” x 5/4' plates welded to top and bottom

or columns.

Tb support truss lateralli and to facilitate the addition of
movable chainfall, angle sections will be run from column to
column and welded at each end to bearing plate. The first
and last member to be anchored in.the wall. See Fig. 16.
Ihese angles to support a maximum.concentrated load of

2000#.

lax. Mom. - 2000 x 9 - 1000 x 9 8 9000'# or 108,000"#.

I 108000
3 I .I I 6.0"3
f 1"“8,00"0
Use 2 - 5' x 3}" x-i' angles‘back to'back dhort legs
outstanding.
Check shear -
1000 g
m 200 P. 3.1. 2;
LINTEL DESIGN
It. brickwork I 125# per cubic rt.
lhll = 13' thick “L“ C 11 ft. (see fig. 19).

Area supported I 5.5 x 5.5 xW/S = 52.5 sq. ft.
52.5 x 15/12 x 125 = 7,100#
lo a %— «.- 11.03.312.11. = 15,000'# or 155,000'#

 

-53-

I 156000 - 03
s a s - 8.57
13,000

Use 5" I 17.25# «1550 to 12" x 2" x 12' Plate.

Check shear
Valli-9985550 1:5.02

v . 23.5%2 : 705 2.5.1. 93

Check bearing

5%.; = 59.2 2.5.1. 95
(Allowable 175 2.5.1. - L.B.c.)
1.111551 for "L" s 15' (213. 19)
Area supported 8 6.5 x 6.5 x /3 8 73.2 sq. ft.
75.2 1: 15/12 x 125 :- 9,900#

'3
I :- 1'33: :- Wig-iii a 21,450'# or 255,000"!

a - n I “253380 - 14.5

Use 5" I 15.4# welded to 12" x i" x 14' 21:50.

Check Shear -
723-299-85950 4-5.54
v = $23- : 954 2.3.1:. 01:

check Bearing -

Egg-g— = 82.5 P.S.I. 95
(175 P.S.I. 1115.351. - L.B.C.)
Lintel for 'L" = 14' (Fig. 19)
Area supported I 7 x 7 x /5 8 84.7 sq. ft.
54.7 x 15/12 x 125 = 11,450#
I 2 13—15. = wk)- = 26,800'# . or 521,000'#

In 321000 '3
“flaw-”7'3

Use 8" IF 21! welded to 12" x i“ x 15' Plate.

 

..

-34-

Check Shear -
v s $11429 I 5,725 A = 5.15
v a gag - 925 2.5.1. 2;
Check Bearing -

{-33-- 95.4 2.5.1. 2;

Joists to Support Office Roof (see fig. 20)
' Spaced 2&2 0.0. Span - 15'
I1 I 71.2 x 2‘} 3 1785‘ per ft.
'2 I 8.5# per ft. (assume)
I I 186.5# per ft.
I = 1;. = 132.235.1913 . 7,550'# or 90,700'#
“5.....- 555-5504

, 00
Use 6" x 4" - 8.5# per ft. Bethlehem Rolled Steel
:01..th
Check Shear -
78W =1575# 4:2.50 sq. In.

7 I «Egg = 570 2.5.1. 215
Check Bearing -
Load = 1678# Flange Width 3 4"

1%? = 70 2.5.1. 95
(175 P.S.I. Allowable - L.B.O.)

e .

REINFORCED CONCRETE D8310]
A.O.I. ALLOIIBLE'UNIT STRESSES
Beams, Girders, Columns, Poctings, Ialls :-
23 = 2500 2.5.1. n = 12 r. s 1125 2.5.1.
1'; . 20,000 P.S.I. '6 3 50 P.S.I. u ‘ 125 P.S.I.

Stairs and office floor slab :-
fé :. 200° P.S.I. n = 15 to . 900 P.S.I.
‘- ‘ 20,000 PeseIe '3 ' ‘0 P.SeIe u = 1.00

Garage Floor Slab :-
2; = 5000 2.5.1. 2. = 25,000 2.5.1.

ZI/z Con/c. 51.21.57

 

 

 

 
  
  

Exp/220050
. ” 5TEEL I
4.”; 4” ROLLED dot-'57: ~ 2 -6 o.c. F5247: T

.5 Rm». MALL

 

 

 

 

a), = Wig/Fr.
# L
w, = 306 /FT.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

)—
7‘)
$~
it—

 

 

F/G . 21 - 54¢2H\

 

page of 02:15” Roof 51.5 (See 215. 20)
Use 2}" Slab.
'1 s 1.1.. s 40# per sq. ft. (1.5.0.)
'2 I 1 x fig x 150 3 $51.25* per ft.
w I 71.24 per ft.

2 2
a . 1%- . W = 55.7'# or 557'?

kd : 1125 .
2:5 27:72 ““1001
2/5 kd = 1.01 2.2/5 I .57" 2.5-1.01 - 1.49

0 Isl—1535.11.01 :12 = 5,520#
5520 x .57 . 4,570":
a, x 20,000 x 1.49 a 4,570“!

‘2 ' 231355—115 "' '1“ "1° 1“
D x . p.’ ft.

Use Steel-Tex or comparable expanded steel reinforce-
ment having an area of .154 sq. in. per ft. width.

Check Shear -
71 2 x 2 5 . 89 .
-J-.T-+ - 89# m 8 2.9 P.S.I.
.95
Temp. Steel for Roof Slab -(A.O.I.)
lax. Spacing I 5t or 18"
gem : .0022 A. = .055 Sq. In. per ft.

5tI5x2.5Il2"
Use i" round bars 0 9" 0.0.

It. of Office Partitions
2" x 4" Studs - Plastered Both sides.
16# per sq. ft. 19x1 I 144} per ft.

 

 

 

 

 

 

 

 

 

 

 

 

 

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a, /// J? ,. d dI/
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C_J_|_L.lh.l___ _ _ _/_ __J

 

l/xz;

 

 

 

 

 

M

 

 

/
"H 'I ++++
Z
FM; 2.3 j/MFLE JP~W
a
I-———II
:—_-_ E‘H‘K ,g"

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

342' | fl ;l H
-, .1. rd
[70 52.4\/ F4, 24
a
——_l" " _.L,_
l —= ’ 77‘. if If 1. I
#4L'gt‘lJ—_———_L._ l ‘
, /
’42 III 042 I /Z | “/42
Ll ‘ ——~
Li’ffi-Jk CPA 7'}; 2 5‘

 

-59-

univ. Uniformly Distributed Load on Floor Joists -
-a=144x2xl§_:—§x5.5=500'#

2
T'1 =500

w 5005 a 22.1,? per ft.

22.1 a 11# per sq. ft. Equiv. partition load supported

by Jcists.

DESIGN OIFICE FLOOR
Using 20" metal tapered pans and concrete Joists.
Clear span I‘ 17'
Live Load 3 50# per sq. ft. (L.B.C.)
Pinish’lccrIl/24xlxlxl50I6i6‘per sq. ft.
Bquiv. Part. Load I 11# per sq. ft.
Total superimposed load 3 67# per sq. ft.

From "Simplified Deeigg of Concrete Floor
lstems - b e ,

6" Pans - 2}” Top - 4" Joists - 24" 0.0.

 

. It. Slab I 50# per sq. ft.
1‘ = 21-532 x 1.95 = .79 sq. in.
Use 2 - 3/4” round bars -2" 0.0.
Bend as mown Fig. 26
Check bond -
V . 1&31 I . x . "- 57.5 2.5.1. 2;:

Using Standard Hooks -

L : fuaD = ZOaOOOEE'IS) g 25!!

Tranverse Steel Req'd. =- .O49 Sq. In. per ft. at 5f 0.0.

11:

Use 1*" round bars - 12 o.c.

 

 

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I. .[3
I
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II II
II II
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.I

I
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____.___-l__ __
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1:11

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li'l‘l'l'lll'll

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

F'-
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F/G. 48$

 

-41-

03}? x 21 = .049 sq. In. 21;,

Dead Load on Sean #1
It. Roof -
Concrete 3 21% x l x 1 x 150 = 51.2# Per Sq. Ft.
Joiste I 8.5 x 1/2.5 I 5.4# Per sq. ft.
Plaster and suspended ceiling 3 10.0# Per sq. ft.
TOTAL I 44.6! Per sq. ft.

17x601144.6=
2x35

. 579# Per ft. of beam from roof.
it. lhll -
5/12 x 125 x 12 x 50 = 50,000#
Less Window Openings
11 x 7.5 x 5/12 x 125 x 5 s 20,5001II
Remaining -
50,000 - 20,500 a 59,400#

25.9.3399- : 5571‘ per ft. due to brick work.

5 x 9 I 45:? per ft. due to wall plaster.

(6 ,l 11 f 50) x 9 I 605# per ft. due to floor.
Total dead load on beam #1 = 579 ,l 557 ,1 45+ 505 =

16 r It.
Live Load 0n Beam #1

From Roof I 40 x 17/2 I 5402? per sq. ft.
.From Floor I 50 1: 17/2 I 4255‘ per sq. ft.

Total live load I 765# per sq. ft.

Design Bean # l
'1 s 75.5,! 1554 :- 25495I per rt.

 

'2 I 3007? per ft. (Assam)
I I 274315‘ per ft.
Span I 18' - 8"

From Fig. 1

2
la. non. = .1 '12 = 2"” " 18°67 - 95,700'#

or 1,148,000”
Fran fig. 22 Assume b I 12"
0 . } tobkd = l-l522(12)(.4050)d - 27255
27250 x .15 8 1,114,500
52 a 455
d I 22.0"
22"4 2"0over I 24" Bean
Check '5. Beam - I

zdllexlxlsoIsoofi‘perft. OK

I =._1114.500
“‘13?! W ”2°90 "1° 1“-

Use 4-1" round bars spaced 2%“ 0.5. and bend as
‘ sham in F135. 24 and 25.

Check Bond -
V 2749 x 18,6712 3
u C m- . . X . 6 x 106 P0301. .9}.
Check Shear

‘1 ' fir m 3 11° P-S-I-

Stirrups -
7° I 50 P.S.I. Allowsble (A.G.I.)
v' I 110 - 50 = 60 P.S.I.
P.0d. Iethod - From S.D.0.F.8., Fig. 1'7.
v'b I 720

-43-

-1 1'_6011807- ..- .-..
“21v “W's-m 5'1

lax. Allowable Spacing I 22/2 I 11"
Use 5/5' round stirrups 5 - 5 5 5", 1 a 5", 5 5 11"

Using Standard Hooks -

L , ran , 201000?) g 33.3.

Dead Load on Bean # 2

From Boat I 379;? per ft.
From Wall I 657# per ft.
From Wall Plaster I. 45;? per ft.
Prom Floor = _6_9_:_5_# per 1:.

TOTAL 168% per ft.

Live Load on Beam # 2
From Office Roof I 3403? per ft.
From 01:15. Plow I 32% per rt.
TOTAL 755# per :5.
Dead Load and Live Load from Garage Root -
£5522 x 12

T I 452# per ft.
'1 . 452-} 755+ 1554 I 2901:? per :5.
'2 I 350# per ft. (mums)
w I $261153 per ft.
Design Bean # 2
w I 325113é per rt. Span I 18.67'
I max. I .10 wl2 (see fig. 1)

_ 2
a 3251 13 6'7 : 1l:5,lOO'# or 1,358,000'#

 

 

From Fig. 22 Assume b I 12"
0 I i fobkd . 1126/2 (12)(.4050) a x 27250
27255 x Jd - 1,555,000
255052- 1,555,000
52 . 575
d I 24.0"
24"} 2" cover I 26" Beam
Check It. Bean -

§%x1x1x1160I325#perft. 21;,

I 358 000
‘a 'na W‘Hs’q- 1’"
Use 3-13" round bars o6" o.c.

Bend as shown in Figs. #24 an! # 25.

Check Bond -

v , 5251 x 15.57

“' 3.x.86'7x

- 105 2.5.1. g;

Check Shear -
v I 533 a 12 x .86 x ' 12° P‘S'I'

Stirrups -
'0 I 60 P.S.I. A.G.I.
v' I 120 - 50 I 70 P.S.I.
From 8.D.C.F.8. Fig. 17 -

V'b370x123840

a =%;- = We 6" x 7° 8 5.44! . 5'-5"

lax. 8 I 24/2 I 12"
1150 3/8" round stirrups, S I 1 Q 3", 4 0 6",
1 o 9", 5 5 12"

-45-

Using Standard Hooks -

L , r35~, 20,000 x 1-115 . 37.5.

DESIGI‘BEAI # 5 -
It. Partition I 144# per ft.
Floor L.L. I 50 x 1.5 I 75# per rt.
Floor D.L. I 5/12 x 150 x 2 I_Z§# per ft.
' w1 I 294# per ft.
w? : _§§# per ft. (Assume)
w I 379# per rt.

Span I 17'

575 17 3
I I 1/8 I].2 I -'—é—-l* I 15,700'# or 164,500“#
Assume b I 6" See fig. 22

c = 5 ram . l-1-52-§(5)(.4050)(0) = 15500
15500 x 10 a 154,500
115002 154,500
02 = 155
0 = 11.75"
11.75" .1 1.75" cover . 15%" Beam
Check It. Beam - ' '

15 5 x gz,‘ 1 x 150 : 54.5# per rt. .25

. mm = .784 “to in.

Use 2-3/4' round'bars -2" 0.0.
Bond ‘3 6110'!) Fig. 23a

Check Bond -

“‘fiz'MR-TWHH-SJ- 2'!

-46-

Check Shear -

v = 5220

' ' m = 51-2 P-S-I.

vc I 50 P.S.I. (A.G.I.) Therefore no web reinforce-
ment required.

Using Standard Hooks -
D

.t D U
L-t-zs

DESIGN ONE WAY RECTANGULAR SLAB -
13'-6" x 5'-O"
Live Load I 50# per sq. ft.
Wt. Slab I 38# per sq. ft.
Fran 15515 7 - S.D.0.F.S.
6" Slab Allowable Load I 140# per sq. ft.
As I .19 x.§%gé;I§§'I .10 sq. in.per ft.
Use 5/8" round bars spaced 8.5" 0.0.
Every other bar bent up at supports as shown
1n.fig. 26.

Check P -

5525 - ,,
Pz'fiéI‘FLE‘TF—1 .57 5--.7""5)"°9 25-

M. 315001 "

A
3-5;. 3 e002 , A. = .072 8Qe inc per ft.

lax. Spacing I 5t I 15' (A.O.I.)

Use 5/8' round'bars - 15” 0.0.

4‘7-

DESIGN BEAN # 4
Span I 13'-O"

Floor Load (Dead 5: Live) I 88 x 2.5 I 220# per ft.
Partition Load I 13% per ft.
'1 I 3645‘ per ft.
w2 (Assume) I fl per ft.
w I 434# per ft.
I = 1/5 '12 = 2§2é1§13, = 9,170'# or 110,000"#
From Fig. 22 Assume b I 6

0 a g r5520 3 (5)(.405)(0) I 15550
15550 x 30 . 110,000'#
115002 ' 110,000

(12

1125
T

95.5
d I 9.65 Say 9.75
9.757,! 1.5" cover a 11.25" Beam
Check It. Bean- I

g-é- :fi-EEE x 150 - 70# per ft. 21;

‘ 110 000 _
s fT‘Jd 'ms) .66 sq. in.

Use 2-5/4" round bars - spaced 2" o.c..
Bend as shown in fig. 25.

Check Bond -

uIv

4 .
‘ WW7: 70.2 p.01. 2;

Check Shear -

3 v 2520
V m 86.6 .5 =55PeseIe

V0 : 50 P0391. 7' = 55 - 50 . 5 PeseIe
From seDecePese H e 1"
over?

-43-

Prom S.D.G.F.S. Fig. 17
v'd = 5 x 5 a 50
"%"%"%§;§3%I .59' 87"
lax. Spacing a 5.75/2 = 4.5"
Use 24" round stirrups 5 4"
Using Standard Hooks -

L a 51% 8 20 OOO(.75 . 25..

DESIGN STAIRS
Lower 3115113 - “-6" Wide '- Horizontal span I 11I-5"
hon 5.5.0.5.5. - 7.51. 25
Thickness of slab I 7"
Steel Required I 5/8” round.bars - 7' 0.0.
For Bending and Placing details see Pig. 23
Wt . Lower Flight «-

zfgxg-é-x4.5x150x.5Il58#

E x {3- x 4.5 x 150 . 323%
4%
w I 605# per horisontal ft. of stairs.

Upper Illflt - 3'-6" Wild. - Horizontal Span I 5'-5'
From S.D.0.P.S. - Table 23
Ih11¢5kness of slab I 4"
Steel Required I 5/8" round bars - 5.5' 0.0.
For Bending and Placing details see Fig. 28.

v - ——_—-_____ V _

-49-

PART/T/O/f—j

 

 

 

   
 

 

£- FRONT WALL.

    
 

 

 

 

 

 

 

 

 

M / Maw/232; /”(‘/ear
IITF/oor] /;. Tofa/ dcpfh ofs/w
__‘=_—_=‘/
“' J l I H ' n 4"
[#1 2": -1 1569’=//-3 4 3-6» - _

 

 

Fm. Z7 LOWER FLIGHT

   

 

 

 

 

 

 

 

 

 

"f BEAM ”5‘

f _5/D£ WALL \ /

7% “I . \ /

‘. /"., 05/1107 , 1 ,,

.9 :1 7 / / [fear

I 1mm 1911:5125 0x551
é 55:7 '2 4

If,” I

6’5?" 4‘49" 7@ 9"=5’-3"

 

FIG. 26 UPPER 'FLIGHT

 

 

-50-

"to Upper Plight "

Zi-g-xg-gxs.5x1sox.5=125#
fix§§x3.5xl50 2131,53

254.3164

__:_§._..254 3 x 12 s 339.5! per horizontal ft. or spam.

loteMSteel of upper flight to be continued
across landing forming a one way slab «-
5.5" thick, capable of withstanding load

of 29“ p” “1' ft' 51515 7 - 5.5.0.5.5.
515110201101. 51551. 50 5529052 55115
Section # 2 - See Fig. 28
Dead Load «-
i upper £11355 5 559 x 5.25 x .5 I 5505‘
funding-44x5.5x4.5x.5 I_5_21£
1177#
Live Load -
1} uppor £11351; 8 100 x 5.25 x 5.5 x .5 a 915#
2 14mm; x 100 x 4.5 x 5.5 x .5 = 155i
1705#
1705544 11775! I 2555.?
Span I 3.5 ft. w I 2%}:- I 8245‘ per ft.
7 I 33-2-3- I, 1441

. 2 I
I = 1/5 an.2 I 824 3 5 = 1255'# or 15.200"#

3 5'15555'-85"
-7-1441 ,
“rm-dumm-

Use 5" I 1055‘ - 5'-lO" long - with Standard A.I.S.C.
A-l connection to column ”A"

Section I - See Fig. 27
Dead Load -
i lower £11355 = 11.25 x 505 x .5 = 5410#
Idve Load -
i lover £11355 :- 100 x 11.25 x 4.5 x .5 = 2550#
5410144 2550# . 5940#
3 9339'! 1,320# per ft.

2
I 8 1/3 '12 s 1320 4 5 8 3360'#' or 40,250'#
3=I=§g§§g.2,2y3

9

C” F‘
3

v = = 29705G

A: :%=.245 sq. in.

Use 5" I 10# -5 ft. long with standard A.I.S.C. been

..."!

connection A-l to column "A”.

Bearing of Sect. I on 12" I511. ,-
v = 2970,! 10 x 4.5 x .5 = 2995#
Roq'd- A ' {r = %$§ = 17.1 sq. in.
lb have 6' x 3' o 18 sq. in. Therefore no bearing

plate is required.

Bearing of Sect. II on 4' Clay Tile.

Partition -
V314411‘1013.51.5=l459#
ROQ'deA=?81§'§§= 12.7 Sq. in.

Nb have 4' x 3" 8 12 sq. in. therefore no‘bearing 9135.

is required.

DESIGN 00me "A”
Dead 5: 1.1" Load from floor = 454 x 5.5 a 2520#
' 7-3.55. 1 - 299519
V-Sect. II " M
Total mu Load: 7274#

351355 = 15'

Reducing f by A.I.S.C. column formula and using
A.I.S.C. table for l/r ratio less than 120, a 6" x 6" H
20# is required.

Bearing Plates -

P - 7274
"r-ww

c .
Use 8" x 8" x 3/4“ plates welded at top and

I 3.6 sq. in. Req'd.

bottom of colum.

0551511 55“ fl See Fig. 21

Span ’- 1'7 'oOO"
Floor Load (Dead 5: Live) 8 88 x 3.5 3 SW per ft.
Stair Load (Dead 5 Live) 8 88 x l x 824 = 912# per ft.
Find Point of laximum Moment -

505 x 17 x .5 = 2500964

912 x 5.5 = 5190#

5190 x 1.75/17 :- 528#

51 a 2500 / 529 a 2929#, 32 : 2500-,1 2552 = 5452:?

2925 - 5051: a o

x = 9.5'
lax. lam. . 2925 x 9.5 - 505 x 9.5 x 9.5 x .5
11 = 14,000'# or 155,000'#
'2 8 100# per ft. (Assume)

1'7 x. 100 x .5 = 850#

Additional Moment-
! = 550 x 9.5 - 100 x 9.5 x 9.5 x .5 =

5550'#~ or 42,500"#
Total n a 155,000.; 42,500 = 210,500"£
Pros: Fig. 22 Assume b = 8"

0 = 5:05:55 8 l%§§(5)(.405)d = 19205

15205 x :5 ' 210,500'#
157552 a 210,500‘#

52 = 155.7
d = 11.5"
11.5"-/ 1.6" cover 8 13" bean.
check Wt. Beam -

fixgxuosiow 21;,

‘a 'g-na-‘Wfim: 1.05 sq. in.
Use 2-7/8' round bars -.‘5" 0.5.
. Bend as show in Fig. 23.
Check Bond «-

V 5462 >
u I a :
m WW . . 98.7 P.S.I.
Check Shear -

V - 5462 ..
vgm-W-68 P.S.I.
7° ' 5° AeceIe

Prom S.D.GQF.8. F180 17

IE

v'-v-vc‘68-5o=38 P.S.I.
v'b='59x5=504

1 1'31'71138
“5H W'4"9"

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lax. Spacing = 13/2 3 6.5“
Use i" round stirrups S = 10 0 8"
Using Standard Hooks -

Ingfizzoooo 8'75 325':

DESIGN COLUIN f 6 85 i '7

From Fig. 1 - w 0 2750 1 1'- 18.67
P = 1.1 '1 = 2750(15.57)(1.1) = 55,457#
hon F180 29 '-

ast Dimension ' 1 " 14'25

Increase Load for 13‘- ratio greater than 10 (A.0.I.)

 

P' 3 P(1.5 - .osg)
P . figzm :5 64,900#

P = Ag(.225fg-/ raps).8
54,900 a 100(.225 x 2500-,4 20,000 p8).8

p8 ' .0125 1111s is between .01 and .04 therefore
0K

Ins-f; , .0125=f3-5

A, 8 1.25 sq. in.

Use 4 - 5/8" round bars as shown in Fig. 24.
lateral Ties

s : 15 x 5/5 a 10" Use 1" round ties as shown in
Fig. 29.

DESIGN COLUMN f 5 8: i 8

From Fig. 1 w = 2750 1 = 18.67

 

P I .4 w1 3 .4(2750)(18.67) 8 20,534#
his value is less than that of columns # 6 and '7,
therefore the same design will be used as this is the min--

I.

-55-

imum (A.G.I.) amount of steel allowable.

DESIGN COLUIH i 2 and fl

From Fig. 1 - w 8 3251 1 = 18.67
P = 1.1 '1 = 1.1(5251)(15.57) = 55,754#

PromFig. 29 -§-=¥f§§s14.25

Increase P for g greater than 10

1H : P(l.3 - .05§)
55 754

P " W‘ 76.60““

r . Ag(.225 1;,1 r.p3).5
75,500# - 100(.225 x 2500-/ 20,000 x p8).8
p8 ‘ .0198 This is between .01 and .04 - 95
P3 3 x‘: , A. 8 peas 8 100(.0198) 3 1.98 sq. in.
Use 4 - 7/8" round bars as shown in Fig. 29.
Lateral Ties «-
5:15:7/5s'14'
3848:}31211 Usei'roundtiesasshownin
.7 Fig. 29.
DESIGN COLUIN l a 4
From Fig. l - w 3 3251 1 3 18.67
P = .4 wl s .4(5251)(15.57) = 24,274#
his value is less than that of columns #6 and #7,
therefore the same design will be used as this is the mini-

mum (1.0.1.) amount of steel allowable.

POOTIHG DESIGN
Soil pressure 3 20009? per sq. ft.

I D
,_ I.
I
..
0 O O
r
.-

.. -..

-57..

2500 P.S.I.
3 12

a
0
II

1125 2.5.1.
1, a 20,000 2.3.1.
75 2.5.1.
141 2.3.1.
Column Footing £2, #10I #11, #12
Gal. Load = 25,4404 2000+ 15(20) = 50,7001?
From Table 35 - 3.0.13.3.

4
u

S
I!

Use Footing 4'-9" square - 15" deep
9-3/8" round bars each way with standard hooks
D placed as shown in Fig. 29.
Column Footing g 2, g 5 '
col. Load 3 55,7544 1 11 1 x 150 x 15 = 55,704#
From Table 35 - R.C.D.H.
Use Footing 6'-8" square - 15" deep.
17-3/8" round bars each way - standard hooks -
- placed as shown in Fig. 29.
column Foot 1
Col. Load a 24,274 ,1 1950 = 25,2241!E
From Table 35 - R.G.D.H.
Use Footing 4'-9" square - 15" deep
9-3/8" round bars each way - standard hooks -
placed as shown in Fig. 29.
Column Footi 6 and 7
ca. Load a 55,457,! 1950 = 55,417#
From Table 35 - R.C.D.H.

-8-

Use 5'-9' square footing - 13“ deep
11-3/8" round.bars each.way - standard hook: -
placed as shown in Fig. 29.
Column Pooti 8
051. Load = 20,554,! 1950 I 22,484#
From Table 35 - R.C.D.H.
Use Footing 4'-9' square - 15" deep
9-3/8' round bars each way - standard hooks -
placed as shown in.ng. 29.
Pier Foot # l3I f 14I £15. £16
Pier Load I 14,220,! 2 x 2 x 15 x 120 = 20,450#
From Table 35 - R.C.D.H.

 

Use Footing 3'-4' square - 15' deep .
5-5/5' round bars each ... - standard hooks -
placed as shown in Fig. 29.
Column Footing # l, i 4
Col. Load I 26,224#
qu'd. Area I 13-33%?— I 15.112 sq. ft.
Width Footing I 20" I 1.57'
I9:151:11 ". ‘3 LE-f—g-g = 7.5' I 7'-9"
I I 15,112 x i x 5/2 I 25,2249!I = 515,000'#
Assume total depth 3 16” d- 12' 3” clear.

 

From Fig. 22
c = i fobkd : i(1125)(20)(.4050)(12) = 54,500#
0 x Jd I 54,500(.5557)(12) = 557,000"# 93.
1 x Jd I A,(20,000)(.5557)(12) I 515,000"#

A. ‘ Wm = 1-515 “1° 1m
5 0 ‘

Use 5-5/8" round bars - 3" o.c. - Standard.hooks -
as shown Fig. 29 except only one way.

Check Shear -
13 112

"Té‘fis‘54-7 21S.

Check Bond -
V . 13112

u I 3 126 P.S.I. ‘QE

Column Footing "A"
001. Load I 7,274#
Req'd. Area I £31753 = 5.55 sq. ft.
Use20" x 20" Plain Concrete Footing 12" deep.
A.C.I. - Allowible for stresses for Plain Concrete ..
Footings -
T = .03 f; 8 .03(2500) I 75 P.S.I.
w I .02 :3 I .02(2500) I 50 2.5.1.
Using A.C.I. method we take amount about section half-

way between column and edge of base plate.

w = 1%? I 114 P.S.I.
n = 2000 (5/12) (20/12)(4/12) - 2(114)(5)(1/12)

I 559 '# or 7055'#
From Fig. 22
1 = c = 1% (5)(20) I 4500#
Allowable I 8 4500 1.8 = 36,000”# [fig
V (at edge of base plate) 8 §§-(%%) (2000) 8 555#

555 _
25";‘15 - 2-3 9.3.1. .95.

Therefore, no steel required in this footing.

~60-

Use Dowels to transmit stress of longitudinal column
reinforcement to footings.

Columns with 5/8" round bars - Use 3/4" round bowels.

Extension into footing-

20,000 x .51 = 5200# = 2.55 x L x 125

L = 21"
Use 1-42' -3/4" round dowel - 8" pedestals will
be necessary for 15" footings.

Columns with 5.4" round bars - Use 7/5" round dowels.

Extension into footings -

20,000 x .60 = 12,000# 3 3.14 x L x 125

L = 50"
Use 2 30"-7/8" round dowels - no pedestal req'd.

DESIGN FOUNDATION WALL
Maximum weight per foot of outside wall will occur
under south wall. With 12" x 3.5 ft. foundation wall this

weight '111 be -

5'0" Rm!" Lit-fig)- I 790# per ft.

From Brickwork - 125‘27‘1g'11‘13 = 2550# per ft.

From Windows - W : 40# per ft.

From Foundation - 3.5 x l x 150 I 525# per ft.
50111. - W per ft.
Allowable (1.0.1.) working stress for minimum reinfor-
cement = .25 f; I .25(2500) I 625 P.S.I.
625 x 144 I 90,000# per ft. morefore, use minimum
(A.C.I.) steel in foundation wall.

.0025 x 12 x 12 = .36 sq. in. per ft. each way
required.
Use 5/8' round bars spaced 18” o.c. vertically 2"
from outside face.
Use 5/8" round bars spaced 18" o.c. horizontally 2"

from inside face.

DESIGN WALL FOOTINGS
It. wall 3 3735# per ft.

Wt. Fboting I 265i per ft. (Assume)

TOtIl . 4000' per ft.

4000
2000

qu 2' wide footing - 12" deep - plain concrete.

Area req'd. per ft. I

 

I 2 sq. ft.

Allowable 1 I .05 1"; I .05(2500) I 75 P.S.I.
Thking lament at face of wa11 -
u I 2000 1 g5 11 g- I 1000'# I 12,000"#
CIT=75/2x6x12I2700#
a I 2700 x 8 I 21,500"# Allowable .95
Check Shear -
.02 r; I 50 P.S.I. Allowable
Punching Shear I 2000 'x g? x i, s 1000#

139-315 ' '7 IRS-I. 91!.

Therefore, no steel required for Shear or bending.
For Tbmperature steel - use 4-3/8' round.bars - 5' o.c.
running longitudinally to take care of cracks due to vary-
ing soil bearing and contraction.

DESIGN FLOOR 0F GARAGE
Using Portland Cement Association Bulletin No. ST 51,

"Concrete Floors on Ground".

lax. Wheel Load 3 5000#

Area Loaded = 30 sq. in.

Impact I 25% 5000 x 1.25 I 5250#
From Fig. 2 (ST 51)

Using 5000# concrete - modulus of rupture = 675 P.S.I.
From.Pig. 3 (ST 51)

A safety factor of 2 willfibe used due to frequent stress

repititions. Therefore the design modulus will be -

935 I 555 2.3.1.

Fig. 1 (ST 51) is based on modulus I 300, therefore -
300 -
6250 x 350 5360#
Entering Fig. 1 (ST 51) with load = 5360# and loaded
area of 30 sq. in. a 5.25" slab is required. Use 6'

sldb.

Steel to Prevent Large Cracks

5::
A: we x1x150I76#
a: i
L I 18'
r-2
1. I 20,000

75 18 2 -
‘ 3 2 25,000 as .54 sq. in. per fte

U50 any welded wire fabric having this cross sectional
area. A fabric with longitudinal wires, gauge I 00, and

spaced 2' 0.0. is suggested.

-55-

 

 

 

 

 

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3/4" expansion Joints as shown in Fig. 30 should be placed
on column lines as shown on sheet 1. Protected expansion
Joints as shown in Fig. 31 should be used at both exits.
1/2' expansion Joints should be placed at all walls, columns

and piers as shown in Fig. 32.

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