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/ re boaryeis of a Reinforced Concrete
Factory Building /
/

A Thesis Submitted to

The Faculty of

KWICHIGAN AGRICULTURAL COLLEGE

by

‘

\

P, R: (Von Sprecken) J. F. Gibbs
/

Cundidates for the degree of

Bachelor of Science

June , 1920
THESIS
[31
(29 O
“THES
Index.

Introduction
References
Specifications
Moments

Loading

Dead weights

Loading on roof slab
Loading on beams
Loading on girders
Loading on columns
Analysis of roof slab
Analysis of beams
Analysis of columns
Analysia of saw tooth

Conchision

 

AO4GiO

pg. l-=-2

10
11
13
23
28
35
58
42
44

S92

257
LIST OF PLATES,

Plate 1, Location Diagram of Column Footings.

» Framing Plan of Hanger Bemms and Flat Roof.
» Plan Showing Saw Tooth Construction.

» typical Section.

» Bent Rib Bar Diagram,

» Beam Sizes and Reinforcing.

2
3
4
" 5, Plan of Typical Slab.
6
7
8
9, Girder Sizes and Reinforecut.::.
" 10, Column Sizes and Reinforcing.
* 11, Roof Slab Stresses.
* 12, Beam Anatysis.
" 13, Girder Analysis.
2 14, Omitted.
" 15, Plan of Blower Room.

" 16, Column Stresses,
 

 
INTRODUCTION.

The Sheet Metal Building of the Olds Motor Works con-
gists of a one story reinforced concrete structure of the
beam and girder type, 180 feet wide by 480 feet long. The
roof is entirely of reinforced concrete and is a combination
of flat slab and saw tooth construction. The walls are of
prick to a height of about three feet. Truscon steel win-
dows are placed above this wali and extend to the roof,
A1:2:4 mixture of concrete was used,

The building is used as a light machine shop. The
beams which run longitudinally are provided with Truscon
slotted inserts cast in tne concrete to which the shaft
hangers are attached. No provision is made for attaching
machinery to the girders which are transverse of the building.

The building was designed by the H. G&G. Christman Co.
and erected by them in the summer of 1919. The steel was
furnished by the Truscon Steel Co. and consists of Kahn
bars and Kahn rib bars.

It was our purpose in choosing this as the subject
for a thesis to continue the study of reinforced concrete
as applied to a practical design and to study the unusual

problems presented in this type of a wvuilding.

The writers desire to acknowledge their indebted-
ness to Prof. C. L. Allen for the valuable assistance
and advice given, and to Mr. Conrad of the H. S. Christ-

man Co. for allowing free access to the building at all
oo —

 

 
Le

INTRODUCTION.

times, for furnishing complete plans, data, and information,

and for the practical advice so kindly given.
REFERENCES.

Hool, Reinforced Concrete Construction, Vol. 1 & 2.

Hool and Johnson, Concrete Engineers! Handbook.

Ketchum, Structural Engineers' Handbook.

Malcolm,Graphic Statics

Michigan State Highway Department, General Specifications
for Steel and Concrete Highway Bridges.

Taylor an& Thompson, Concrete Plain and Reinforced,
SPECIFICATIONS.

4,

Te following working stresses have been recommended

by the Special Committee on Concrete and Reinforced Con-

crete of the American Society of Civil Engineers present-

ed before the society Jan. 17, 1917.

Concrete Engineers' Handbook, page 845-6.

Per cent of
compressive

strength.

1,Structural steel in tension

Ze

Se

4.

Concrete in compression

where the surface is at

least twice the loaded area 35.90
Concrete for concentric com-
pression on a plain concrete
column or pier, the length of
which does not exceed 4 dia-
MOTEGrG wenn -oceewnnweees=-= 22.5
Compression on columns with
longitudinal reinforcement

to the extent of not less than -

1 % and not more than 4 %,

and with lateral ties of not

less than + inch in diameter,

12 inohes apart, nor more

than 16 diameters of the
longitudinal bar <------------ 22.5

Hool and Johnson,

Loe. per

Ss Qe inch e

16 ,000

700

450

450
Se

Per cent of Lbs. per

Compressive

strength.

5, Compression on columns

6.

reinforced with not less

than 1 3

4 % of longitudinal bars

and not more than

and with circular hoops or

spirals not less than 1 %

of the volume of the con-

crete, the clear spacing of

the hooping to be not great-

er than one- sixth of the
diameter of the encased

column and preferably not greater
than one-tenth, and in no case
more than 2-4 inches, where the
ratio of unsupported lenzth of
column to diameter of the

hooped core is not more

than 10 -<«c«e<<«cc-eee- ~<o 34.875
Compression on extreme fibre

of a beam, calculated for
constant modulus of elasticity
(stresses adjacent of the sup-
port of continuous beams may

be 15 % higher) ----------- 32.5

8q.

inch.

697.5

650

Se
Te

9.
10.

il.

Per cent of
compressive
Strength.
Shear in beams with horizontal
reinforcement or without
reinforcement -----<-<------- 2
Shear in beams thoroughly reinforoed
with web reinforcement (the
web reinforcement exclusive of
bent-up bars to be designed to
resist two-thirds the external

shear) --------- aneweennnn -- 6
Punching shear,only ------- - 6
Bond stress between concrete
and plain reinforcing bars 4
Bond stress between concrete

and drawn wire -<«<----<<<--- 2

Lbs. per

8q. inch.

40

120
120

80

40

For the above a 1:2:4 mixture (Portland Cement Concrete)

was used as a basis, having a strength of 2000 lbs. per

Ssyusre inch. (Compressive).

12.

MOMENTS .

All steel to ve allowed a working stress of 16,000

los. per aq. inch.

General Specifications for Steel and Concrete Highway

Bridges ,- Fourth Edition 1920,- Michigan State High-

way Dep't.,- Spec. 179% 180,- page 14.

When the beam or slab is continuous over its supports,

reinforcement shall be fully provided at points of

negative moment, and the following stresses shall not
Te

be exceeded; (See Speco. 237 & 238, page 17) for a 1:2:4
mixture with ultimate compressive strength per square
inch of 2000 lbs. an extreme fiber stress of 650lbs.
per square inch will be allowed, and adjacent to the
support of continuous beams, stresses 15 % (or 747.5)
higher may be used. (Agrees with Spec. 6 above).

In computing the positive and negative moments in
beams and slabs continuous over several supports due

to uniformly distributed loads the following rules

shall be followed.
(a). That for floor slabs the bending moments at center

and at supports shall be taken as wl for both
dead and live loads, where w represents the load
per lineal foot and 1 the span.

(ov) That for beams the bending moment at the center
and at supports for interior spans shall be taken
as wl , and for end spans wl for center and adjourn-
ing supporte, for both dead and live loads.

(c).In case of beams and slabs continuous for two spans
only, the vending moment at the center support shall
be taken as BL and near the middle of the span as ma

(d). At the ends of continuous beams the amount of neg-
ative bending moment will be left to the judgement
of the designer, but it must be provided for.

(e). Continuous beams and slabs designed for concentrated
loads shall have their moments calculated as if
they were simply supported and the resulting moment
shall then be multiplied by the factor eight-twelfths
or eight-tenths to give the designing moment: the

factor eight-twelfths shall be used where the co-
13.

14.

15.

8.
efficient of wl for uniform loading is one-twelfth,
and the factor eight-tenths shall be used where the

coefficient of wl’ for uniform loading is one-tenth.

LOADING.
Live load on floors. Hool - Vol. 2 = page 144.
Toilet room (same as public buildings) 100 lbs. per sq.ftn.
Live load on beams. 500 lbs. per lineal foot of beam.
Note: Mr. Conrad of H. G. Chr isman Company advised us
that this was the load assumed in designing the build-
ing. It includes the weight of the shaft hangers, shaft-
ing, pull of the machinry, and impact. |
Live load on columns. Schneider's Spec. 8. H. B. pg. 74.
Use specified uniform live load per square foot with
minimum of 20,000 lbs. per column.
Loads on foundations. Sohneider's Spec. S.H.B. pg. 75.
Live loads on columns shall be assumed to be the same
as for the footings of columns. The areas of the bases
of columns shall be proportioned for the dead load
only. That foundation which receives the largest ratio
of live to dead load shall be selected and proportion-
ed for the combined live and dead loads. The dead load
on the foundation shall be divided by the area thus
found and this reduced pressure per square foot shall
be permissable working pressure to be used for the dead
load for all foundations.
Permissable pressure on foundations. S.H.B. pg. 75.

2 tons per square foot.
De

16. Wind load. Malcolm's Graphic Statics, pe. 75.
30 lbs. per square foot of vertical surface, the normal
pressure to be the largest as determined by Duchenin's,
Hutton's, or the Straight Line formula.

17. Snow on roof. Hool and Johnson, Concrete Engineers’
Handbook pg. 512.

50 lbs. per square foot of horizontal surface.
DIMENSIONS.
18. The minimum width of web in beams and girders shall
not be less than 1/24 of the span. Spec. 189 - pg.15
8S. & C. H. Bridges.
10.

DEAD WEIGHTS.

The following weights were used in figuring the dead

loads.
Concrete - ------+-+--+-+-+-- 150 lbs./ou. ft.
Brick ---+-+-+-+-+-+-+-+-+-+--- 120 lbs./eu. ft.
Plaster ---+--+-+-+-+-2-+e-¢-- 5 lrs./eq. ft
Roofing, (3 or 4 ply) - ------ 5 lbs./sq. ft.
Windows (Estimated) - -----+-+-- 15 lbs./sq. ft.

Hollow tile Hool, Vol. 2, pg. 69
6-12-12 22 lbs. each.
4-12-12 16 lbs... each.
ll.

LIVE and DEAD LOADS ON ROOF SLABS,

Flat Roof,
Snow load -----+-- 2-2-2. 2.2.2. LL. 50.0 lhe. per sq.ft,
Roofing, 4 ply tar and @ravel ---+-.-.-22. 5.0
Weight of slab 12X12x3x150 = 57,5
144x12
Total load 72.5 lbs. per sq.ft.

Roof over saw tooth,
The roof slab as scaled from the blue print makes an
angle of 2835 with the horizontal.
Wind Load, Malcolm's Graphic Statics, page 173,
P=pressure per sq, ft. ona vertical surface,
Pr pressure per sq. ft. normal to the roof surface,
A = angle the roof makes with the horizontal in degrees,

Duchemin's formula, -

P= p 28in*A = $0X 2x%,47844 =23.4 lbe. per aq. ft,
n 1+ sin’A 1+ (.47844F

Rutton's formula, -
P, . P sin al.842cos A-l _ 30x(.47844 )1-842x. 87812-1
Phz 21 lbs. per sq. ft.

Straight Line formula, -

P, x pA = 30(26.58) = 19.1 lbs, per sq. Fb,

Use the largest of these or 23.4 lbs, per sq. ft,
Dead Load + Maximum Snow Load,

Max. snow load, 30 x cos 28 35 26.34 lbs./ sq. ft.

Roofing, 3 ply tar and gravel = §,.00
Weight of slab, 2x 12 x 12 x 150 = 25,00
12 x 144

Total vertical load = 56.34 lbs./ sq.ft.
Total load normal to roof = 49.47 lbe./ sq.ft.
12,

Dead Load + Min. Snow + Wind.

Min. snow, 15 x cos 28 35 = 13.17 lbe./ sq. ft.
Roofing, 3 ply tar and gravel, = §.00
Weight of slab 2 x 12 x 12 x 150 = 25,00

12 x 144
Total vertical load

43.17 1bs./sq.ft.

Normal to roof,-

43.17 x cos 28° 35' 37.89 lte./sq.ft.

Wind = 25,40 .
Total load normal to roof = 61.29 lbs./sq.ft.

Use the larger of these two or 61.29 lbs. per sq. ft. as

the normal load on the roof over the saw tooth,
13.

LOADING ON HEAMS.

Beam A.

Roof (live + dead) 0.0

Weight of beam 8x16x150_ _ 133.3

Live load ( machinery) 500.0
Total 633.3 lbs./lin. ft.

Beam 5
Roof (live + dead) 0.0
Weight of beam 6 x 16 x 150 133.3
144 =

Live load ( machinery) §00,0

Total 633.3 lbs./lin, ft.
Beam C,

Roof (live + dead) 17.5 x 72.5 = 543.75

Weight of stem 19.5 x 8 x 150 . 162,50

Live load (machinery) = _§00,00
Total 1206.25 lbs./lin. ft.

| Beam D,

Roof (live + dead) 17.5 x 72,5 = 643.75

Weight of stem _8 x 22 x 150 . 183.33

Live load (machinery) ue =  §00,00

Total 1227.08 los./lin, ft.
L4e

LOADING ON HEAMS.

| Beam B,
Roof (live + dead) 7.6 x 72.5 = 543.75
Weight of stem 24.5 4 Sxls0 - 204.00
Live load (machinery) “ = §00.00

Total

Beam F,
Roof (live + dead) 3e5 x 72.5

Weight of beam _12 X 30 x 150
144

Live load (machinery )
Total

Beam G,.

1247.75 lbs./lin. ft.

= 254.00
= 375.00
= 800,00

1129.00 lbs./lin. ft.

The amount of load on this beam from the slab in the saw

tooth is indeterminate. Ve have assumed that one-fourth
of a slab load will cover this load.

Roof (live + dead), flat roof, 3.75x 72.5 = 271.9

Roof, saw tooth, ¢ x 3.75 x 72.5

Weight of stem 12% 27 x 150
144

Live load ( machinery)
Total

=135.9
=338.0
=500,0
1245.8 lbs/lin.ft.
15.
LOADING ON HEAMS.

Beam H.
Roof (live + dead) 17.5 x 72.5 = 543.75
Weight of stem 8 x 19,5 x 150 = 162.60
Live load (machinery) ue = 500.00 |
Total 1206.25 lts./ lin. ft.
Beam I,

Roof (live + dead) 7.5x 72.5 = 643.75
Weight of stem 8 x 22 x 150 = 183.33

144

Live load (machinery ) = —600,00
Total 1227.08 lbs./ lin. ft.
Beam Je

Roof(live + dead) 17.5 x 72.58 =z 543.768

Weight of stem 24.5 x 8 x 150 =z 204.00
144

Live load (machinery) = 500,00
Total 1247.75 lbs./lin. ft.
Beam K,

Under the flat roof at the east end of the building,
Roof (live + dead ) 7.5 x 72.5 2 543.75

Weight of stem le x 27 x 150 =z 337.00
144
Live load (machinery ) = _§00,00

Total 1380.75
——-~L

a

LOADING ON BEAMS,

Beam K,

16.

Under the saw tooth. Assume + slab load from the slab

on the saw tooth,

Roof (live + dead) #x 7.5 x 72.5

Weight of stem 12x 27_x 150

144
Live load (machinery )

Total

Beam L,
Roof (live + dead) 22x 7.5 x 72.5
Weight of stem 21.5 X 27 x 150
144
Live load (machinery)
Total

= 407.81
= 324,00

= §00,00
1231.81 lbs./lin. ft,

= 407.81
* 324,00

= 500 .00
12.31.81 lbs./lin. ft.

In the above $ a slab load was assumed to cover the

load which might come on the beam from the slab over the

saw tooth.

Beam KM,
Roof (live + dead) 7.5 x 72.5

Weight of stem 8 x 24,5 x 150

144
Live load (machinery)

Total

= 543.78
=z 204.00

= $00,900

1247.75 lbe./lin. ft.
17.

LOADING ON HEAMS,

Beam N,
Roof (live + dead) 7.5 x 72.5 = $43.75
Stem 50 = 1835.33
Live load (machinery) | = 500,00
Total 1227.08 lbs./lin, ft.
Beam 0.
Roof (live + dead) ¢x 7.5 x 72.5 2271.87

Weight of 3 inch slab on the monitor

2.75 X 1 x 3 X 150 = 103.10
12
Weight of sash § x 15 =z 175,00

Estimated weight of slab extending under

the monitor 2,5 x 3 x 150 = 94,00
Weight of stem 8 x 19.5 x_ 150 = 162.70
Live load (machinery) = 500,00
Total 1206.67 lbs./ lin. ft.
Beam P,
Roof (live + dead) +x 7.5 x 72.5 = 408,00
Weight of stem 12 x 27 x 150 = 338.00
Live load (machinery ) uae = 500,00
Total 1246.00 lbe,/lin.ft.

The above is on the assumption that the load on the beam
from the saw tooth slab is equal to one-half a slab load.
18.

LOADING ON BEAMS.
Beam Q.
Beam supporting the toilet room located next to the stairs.

6 in. tile wall 5,5 X 20 X 22 = 77.0
20

Weight of plaster both sides

o,9 X 20 X 2x § =z 35.0
20

5 in. floor slab + 2¢ in. granolithic

surface 7 x 7.5 x 150 2704. 0
12

Live load (fixtures + people) 7.5x1002 1750.0

Live load (machinery ) = §00.0
Weight of stem 8 x 16 x 150 = 132,0
144
Total 2198.0 lbe./lin. ft.
Beam Q.

Beam at center of the toilet room span.

Weight of cuttain wall located 2 4 feet from center line
of the beam. (4 inch tile, plastered both sides).
Weight of wall 7 x 20 x18 #*® 2520

Weight of plaster 10x7x20 = 1400

5920 lbs.
Weight of wall on beam 5x3920 = 2610.0
Weight of slab 7.5 x 20 x 7.5 x 150 =14080.0
Live load on floor a 20 x 100 #15000.0
Live load (machinery) 20 x 500 #10000.0
Weight of beam 8 x Ao 4 ee x _ 150 = 2670.0

44560.0 lbs.

44,360,0 = 2218.0 lbs. /lin. ft.
20
LOADING ON BEAMS.
Beam Q,
Located at the back of the toilet room floor,

Curtain wall 2,5 X 1960 =z 655.0
Floor slab 10 x 3.18 x 7,5 x 150 = 3520.0
Live load 3.75 x 10° 100 = 3750.0
Live load (machinery) 10 x 600 = 5000.0
Weight of beam 8 x 16 x 150 x 20 = 1355.0

nae Total 14260.0 lbe.
14.260,0 = 1426.0 lbs. per lin ft.

Beam Q.

Supporting heating coils ( See plate /° ),.

Weight of motor + x 1500 = 7500.0
Weight of fan + x 3000 =1500.0
Floor slab 5§ x 7,5 x 20 x 350 x i2 = 9300.0
Weight of beam 22 xe x 20 x 350 # 3670.0
Live load 20 x. 5008 #10000 ,0

Total j§§ 265220.0 lts./
25 220.0 = 1261.0 lvs. per lin. ft.

Beam(Qx)' (See plate /9 )

Weight of heater + x 32688 = 16,344.0
Floor slab 6,5 x 3.76 x 20 x 150 = 6,090.0
Weight of beam je x on x 20 x 150 =z 5,280.0
Live load (machinery ) “0 x 500 = 10,000,090
Total 37,714.0 lbs.

37,714.90 = 1885.0 lbs. per lin. ft.
20

19.
LOADING ON BEAMS,
Beam Qx.(See plate /°5 )

Van $ x 3000 = 1500.0
Weight of heater 4 x 32688 = 16344.0
6 "floor slab 6,5 x 3.75 x 20 x 150 = 6090.0
5° floor slab 5 x 5.15 x 20 x 150 = 4800.0
Live load ( machinery) 20 x 500 = 10000.0
Weight of beam dak oe © 100 2 20 = 5280.0

Total 44014,0 lbs.

44,014 = 2201 lbs. per lin. ft.
20

Beam R, (See plate /° )

Located under slab supporting heaters,

Weight of motor + x 1500 = 1750.0
Weight of floor slab4x5x17,5 x20 = 4650.0
Weight of 8" wall (8215x2021 500 - (5x15x150x8 )
12 12

222500 .0
Weight of windows 5 x 15 x 15 z 1125.0
Weight of stem 12 x 28 x 20 x 150 = 7000.0
Live load (machinery) aa 20 x 500 = 10009,0

Total 46025.0 lbs.

46,925_ = 2301.2 lbs. per lin. ft.
20

Beam R,

Same as beam G - - = « = = = = = 1245.8 lbs. per lin. ft.

20.
21.

LOADING ON BEAMS,

Bea T,
Roof (live + dead) 17.5 x 72.5 =z 543.75
Weight of stem x x 150 =z 225.00
Live load (machinery) “ae = §00,00
Total 1268.75 1be./ lin. ft.
| Beam U.
Roof (live + dead) 7.5 x 72,5 = 543.75
Weight of stem § x 24,5 x 150 = 204.00
Live load (machinery) “ae = §00,00
Total 1247.75 lbs./lin. ft.
Beam V,.

Roof (live + dead) 7.5 x 72.5 = 543.75

Weight of stem 8 x 22 x 150 =183.353
144
Live load (machinery ) =500,00
Total 1227.08 lbs./lin, ft.
Beam W,.

Roof (live + dead ) 7.5 x72.5 = 543.75

Weight of stem 1205 Xo 20 =z 163.00
Live load (machinery) = §00,00

Total 1206.75 lbs./lin. ft.
22,
LOADINGS ON BEAMS,

Beam <X,
Roof (live + dead) 7.5 x 72.5 = 543.75
Weight of stem 12 x 27 x 150 = 337.50
Live load (machinery) = 500,00
Total 1381.25 1bs./lin.ft.
Beam Z.
Roof (live + dead) 3.5 x 72.5 = 254.0
Weight of beam 12 x 30 x 150 = 375.0
144
Live load =_ 500.0

Total 1129.0 lbs./lin.ft.
LOADING ON GIRDERS.

| Girder AA.
Roof (live + dead) 71.5 x 72.5
Weight of girder 11,5 x 20 x 150
144
Live load (machinery)
Total

Girder B B,
Roof (live + dead )

Woight of girder 1],5 x 20 x 150
144

Live load (machinery)

236

= 0.0
= 240.0
=  §500,0

740.0 lbs./lin.ft.

= 0.0
= 240.0
= _500,0_

740.0 lbs./lin.ft.

Total
| Girder CC,
Load from beam F $ $(20 x 1129.0) = 11290.0
* # © © 20 x 1206.25 = 24135.0
8 * * D 20 x 1227.08 = 24541.6
" re © B 20 x 1247.08 = 24955.0
e # * 6 $(20 X 1245.81) = 12458.1
Weight of girdet $x 17 x 27 x 15.5 x30x150__
144 =_10656,0_
Total 108,035.7 lbs.

108 ,035,7 =z 3601.2 lbs. per lin. ft.

30
Girder D D.
Load from beam G $(20x1245.810
3 beam A 3x20x635.5
beam L $x20x1231.81

Weight of girder $x13x27x15.5x30x150
144

Total

72,5907,2 *= 2417.9 lbs. per lin. ft.

50

12,458.1
37,998.0
12,381.1

9,700,0
72,537.2 lbs.
LOADING ON GIRDERS,
Girder EB. &.

Load from beam L $x20x1231.81 = 12,318.1
peam M 20x1247.75 = 24,955.0
beam N 20x1227.08 = 24,541.6
veam O 20x1206.75 =z 241 35.0
beam AA $x20x1283.75 = 12,837.5

Weight of girder $x17x27x15,.5x30x150 = _10,656,0

“ae Total 109 ,443.2 lbs.

109 443.2 = 3648.1 lbd. per lin. ft.

50
Girder F F.

Due to the unusual loading the moment was figured on
the basis of concentrated loads.
Load at 1/4 point,

Beam Q 10x2198.0 21980.0
Beam A 10x638.0 6380.9

28,310.00 lbs,
Load at 1/8 point,

Beam Q 10x2218.0 22,180.0
Beam A 10x633.3 6 .333,0 28,513.0 lbs.

Load at 3/4 point,

Beam Q 10x 1426.0 14,260.0
Beam A 10x633.3 6 .333,0 20 ,593.0 lbs.
Weight of beam 16 x 30 x 150 | = 500#/lin.ft.
144

Weight of wall 6x10.5x4x10.5x$x18 = tile = 1512

6x10.5x4x10.5x4x10 = plaster = 840
eooe lbs.
25,
LOADING ON GIRDERS.
Girder G G.
Load at 14 point,-
Beam (Qx)' 37,714
Beam A 10 x 633.3

18, 857.0

6,333.0 25,190.0 lbs.
Load at 1/2 point,-

Beam (Qx) 44,014 = 22,007.0
2
Beam A 10 x 633.3 = _6,333,0 28,340.0 lbe.

Load at 3/4 point,-

Beam Q 25 , 220 = 12,610.0

Beam A 10 x 633.3 = 6 .555,0 18,943.0 lbs.
Weight of wall 26,5 x 11,5 x (22+ 10) = 4,873.5 lbe.
Weight of girder 16x 20 50 x150 = 15,000,0 lbs,

Total uniform load 4,873 #215.,000 = 662.4 lbs./ lin. ft.
3

Girder H H.
Load from Beam G +x 20 x 1245.8 = 12, 458.0
Beam A 3x% x 20 x 633.3 = 18,999.0
Beam L $x 20 x1231.81 = 12,318.1
Beam K 1x 20 x 1380.75 = 27 615.0
Beam T 3x4 x 20 x 1268.75 = 38 ,062.5
Weight of girder 27 x 15.5 x 30 x 150 # 13,100.0
Weight of wall 210 x 6 x 150 = 15 ,750,0
‘ 138 ,302.6 lbs.
158,502.6 = 4610.9 lbs. per lin. ft.

350
266

LOADING ON GIRDERS.

Girder I I.

Loading from Beam Z +x 20 x 1129 5645.0

Beam H #$ x 20 x 1206.25 = 12062.5

Beam I 4 x 20 x 12270.8 12270.8

Beam J +¢ x 20 x 1247.75 12477.5

Beam K +x 20 x 1380.75 = 6903.7
Weight of girder 27 x 15.5 x 150 x 30 2#_13100,00

144
62 ,459.5 lbs.
62,459,5 = 2081.98 lbs. per lin. ft.
D
Girder J J.

Loading from Beam K +x 20 x 1380.75 = 6903.75
Beam T 3 x4 xX 20 x 1268.75=27615.00
Beam K +x 20 x 1380.95 = 6903.75
Weight of girder 27 x 15,5 x 30 x 150 =_13100.00

144
64,522.50 lbs.
4,522.50. = 1817.4 lbs. per lin. ft.

Girder K K,
Loading from Beam Z +x 20 x 1129 = 5,645.0
Beam H #4 x 20 x 1206.25 #212,062.5
Beam I + x 20 x 1227.08 #® 12,270.8
Load from the Enameling Building, -

Floor slab 10x20x8.5 x150 =
12
Live load 10x20x150 z= 51,200.0
LOADING ON GIRDERS.
Girder K K Cont'd.

Wall load 7,0x20x3,75x120
12
Sash §x15x20

12,250.8

1,500,0

Total 94,928.3 lbs.
24,928.38 =z 4746.4 lbs. per lin. ft.

Girder L L.
Load from Beam J $x20x1247.75 = 12,477.5
Load from Beam K %¢ x20x1380.75 = 13,807.5
Load from Beam B 4$x20x633.3 z 6,333.0
Loads from the enameling building, -
Floor slab ( See girder K XK) = 51,200.0
Wall load 7x160x120 =z 11,200.0
Sash 15 x 120 = 1,800.0

Total 96,818.90 lbs,

96,818,0 = 4840.9 lbs. per lin. ft.

20
28.
LOADING ON COLUMNS,
Columns T 6 = 28 and Z 6 - 28,

 

 

 

 

 

 

 

 

 

= =
Beam F-. : Beam F
a =
Girder |
CC
ea
Beam F 20 x 1129.0 = 22,580
Girder C C
Beam C 24,135
+ Beam D 12,270
4 weight of girder 5,328 = 41,733
Weight of column 12 x 16 x 14 x 150 = 2,800
144
Total 67,113 lbs.
Columns U 17 - 18 and Y 17 - 18
a
Girder
Cc
Beam C 4 Beam
R } G
a wo
Girder
GG
Beam G 10 x 1246.8 =z 12,468
Beam R 10 x 1246.8 = 12,468
Girder C C (See page 29) = 42,553

Girder G G (See page 25) 4$x92346. = 46,173

Strut 136,800 cos 30 S1 , 869

Weight of column 5,1416x64x14x150 = 2,916
144

 

Total 145,964 lbs.
LOADING ON COLUMNS.

Columns U9 -16 &19 = 25
Y¥9-e16 &19 = 25

Beam G 20 x 1246.8

Girder C C
Beam EB 20x1247.75 = 24,955
+ Beam D = 12,270

4 weight of girder = _§,328
Girder D D
Beam A 20x633.3

12,666

+ Beam A 10x633.3 63333

+ weight of girder =__4,850
Strut I 36,800 cos 30

Weight of column  3,1416x64x14x150
144

Total

29.

24,936

42,553

= 23,849
= 31,869
= 2,916 _

166,123 lbs.
LOADING ON COLUMNS,
Columns U 28 and Y 28,
For typical columns seepages 28 and 29.

Beam G = 12,468
Beam K = 13,808
Girder C C =z 42,553
Girder H H =z 40,331
Strut I = 18,153
Weight of column = 2,916
Total 130,229 lbs.

Columns U 7 - 8 and 26 - 27
Y 7 - 8 and 26 ~- 2

Beam G = 12,468
Beam P = 12,460
Girder C C = 42,5535
Girder F F @ 47,907
Strut I = 31,869
Weight of column = 2,916
Total 150,173 lbs.

Columns U 8 -16 and 19 -25
Y 8 ~ 16 and 19 - 25,
Assuming maximum load from girder CC, No machinery load
from girder DD.

Beam G = 24,936
Girder C C = 42,553
Girder D D = 8,849
Strut I = 31,869
Weight of column = 2,916

Toasl 111,123 lbs.
LOADING ON COLUMNS.
Columns V 7 - 8 and 26 ~ 2%,
x 7 - 8 and 26 o27.

Bean L = 12,318
Beam P = 12,460
Girder EB E = 42,554
Girder F F : = 37,279
Beam H (Saw tooth) 52707 cos 30 = 45,644
Weight of column = 2,916
Total 153,171 lbs,
Columns V 9 - 16 and 19 < 25.
X9- 16 and 19 - 25,
Beam L = 24,636
Girder E E = 42,554
Girder D D = 23,849
Beam H (Saw tooth) = 45,644
Weight of column = 2,916
Total 139,599 lbs.
Columns V 17 - 18
X11? -18
Bean L = 12,318
Beam R = 28,512
Girder E E = 42,554
Girder G G = 45,739
Beam H (Saw tooth) = 45,644
Weight of column = 2,916

Total 177,683 lbs.

Sl.
S26

LOADING ON COLUMNS.
Columns V 28 and X 28.

Beam L = 12,318
Beam K = 13,808
Girder E £E z 42,554
Girder H H = 35,080
Beam H (Saw tooth) =z 45,644
Weight of Column = 2,916
Total 152,520 lbs.

Columns V 9 - 16 and 19 = 25.
X9-16 and 19 = 25.
Assuming maximum load on girder E E and no machinery load
on girder D D.

Beam L = 24,6356
Girder EE =z 42,554
Girder D D =z 6,849
Beam H (Saw tooth) = 45,644
Weight of column =z 2,916
Total 124,599 lbs.
Columns W7 = 27.
Girder AA = 14,800
Girder E E =z 85,108
Weight of column = 2,916

Total 102,824 lbe.
S56

LOADING ON COLUMNS,
Columns VW 7 - 27.
Assume a maximum load on one girder E E and no machinery

load on the other girder E E.

Girder AA = 14,800
Girder E E = 42,554
Girder E E = 27 554
Weight of column Total =_ 2,916

Total 87,814 lbs.

Columns T 29 and Z 29.

Beam Z =z 11,290
Girder I I = 24,748
Weight of column 16x16x14x150 = 3,735

“As Total 39,773 lbs.

Columns U 29 and Y 29.

Beam K =z 13,807.5
Girder I I z 26,162.9
Girder J J = 25,581.2
Weight of column =  2,916,0
Total 67,467.6 lbs.
Columns V 29 and X 29.
Beam K = 13,807.5
Girder J J = §4,522.5
Weight of colum = 2,916.9

Total 71,246.0 lbs.

 
| 34.
LOADING ON COLUMNS.

Column W 29.
Beam X = 13,812.5
Girder J J = §4,522.5
Weight of column = 2,916,9

Total 71,251.90 lbs,
556

ANALYSIS OF ROOF SLAB.

In the analysis of the roof slab the formulas for rect-

angular beams given in Hool's Reinforced Concrete Construction

Vol. 1, pages 112 and 113 were used. The entire load was

assumed to be carried the short way of the slab. This is

in accordance with the recommendation of Hool (Vol. 1, pg.140)

that when the ratio of the length to breadth is more than

1.5,the load is carried entirely by the reinforcement the

short way of the span.

NOTATION.

f= unit compressive stress in outside fiber of concrete.

fg = unit tensile stress in steel.

n

ratio of modulus of elasticity of steel in tension to

modulus of elasticity of concrete in compression.

&@s = area of cross-section of steel.

b
d

¢q¢4 &% Rr SS &

breadth of beam.

distance from compression surface to axis of reinforce-
ment.

bending moment.

steel ratio.

ratio of depth of neutral axis to depth of steel.

ratio of lever arm of resisting couple to depth of steel.

total shear.
unit shear.

v' = unit working shear.

u = unit bond.

o = circumference of one bar.

w = uniform load in pounds per foot.

l1 = span of beam in feet.
36,
ANALYSIS OF ROOF SLAB.

Reinforcing = 7/32 inch bars spaced 5" c. toc.

Span = 1 =7.5 ft.

 

 

 

 

 

 

Loading = w = 72.5 lbs.
d= 2.5
Mo= wi°/l2 = 72.5 x 7.5 x 7.5 x 12/12 = 4080 in. lbs.
p=% = ,03758 = .003
ba 5 x 2.59
k = YSpn * (pn)~ - pn = V2 x .008 x 15 + (.008 x 15)e
e 003 x 15
k = £2583
j=1i-8= 1- .2583 = .9139
3 3
a = 44 x .03758 = .08267
5 20
f, = é = 4080 = 21,600 lbs. per sq. in
asja 208267 x .9139 x 2.5
f.= 2X = 2 x 4080 _ = 461 lbs./sq. in.
a 79139 X 02583 x le xle.5)* |
vzV = 3075 xX 72,5 = 9.9 lbs. per sq. in.
bid 12 x .9139 x 2.5
u= V = 3.75 x 72,5 = 5.76 lbs. per sy. in.
jdsumo 09139 x 2.5 x 1.515

sum o = 3.1416 x 7/32 x 2.2 = 1.815

Over the Support.
Reinrorceing consists of alternate bars bent up over the

support, or 7/32 in bers spaced 10" c toc.

 

 

 

p= as = 403758 = 0015
bd 10 x 2.5
k = V2pn + (pn)© - pn =\2x,00l5xl5 + (.0015x15)~ -,0015x15
k = .179
j=1-k/3 =1 - .179/3 = .9403
rf, = = 4080. = 42,000 lbs. per sq. in

a jd. .0414 x .9405 x 2.5
57.
ANALYSIS OF ROOF SLAB.

 

 

 

 

fo * eae ~ & 79403 ore KSB x 12
f. = 646 lbs. per sq. in,
u=_V _ = ____ 3,75 x 72,5
j ad sum of o 09403 x 2.5 x 3.1416 x 2.2 x 7/32
u = 78.2 lbs. ver sy in.

End panels,
Reinforceing,- 7/32 in. bers spaced 5" c toc.
12 - 7/32 in. bars spaced equally.

 

 

Mo= wl2/10 = 72.5 x 7.5 x 7.5 x 12/10 = 4890 in. lbs.
a = 52 x .03758 = .0977
5 20
p= as = ,0977 = 00327
bd 12 x 2.9
k = .267
j =.911
f= YM = 4890 = 21,950 lbs. per sq. in.
8 a,jd 40977 x .911 x 2.5
f =2y = 2 x 4890 _ = 536 1bs./sq/in,.
Cc jkba 91] x.267 x 12 x (2.5)*
v =V_ = 3,75 x 72,5 = 9.94 lbs. per sq. in.
vjd 0911 x 12 x 2.5
u= V = 3.75 x 72,5 = 66.5 lbs. per sq. in.

jd sum o e911 x 2.5 x 1.79
oe,

ANALYSIS OF HEAMS.

Beam C - Typical Tee-beam.

Reinforcing,- 1 - 3/4 x 2 3/16 kahn bar 0.79 sq. in.
1 - 5/8 rib bar 0,3906 *
Total area 1.1806 sq. in.

Loading = 1206.25 lbs. per lin. ft.

span = 20 ft.

According to the recommendations of the joint committee; -

(1). b = + span of beam = + x 20 = 5 ft.

(2). b= six times the thickness of the slab on each side
of the stem = 2 x(3 x 6) + 74 = 43% inches.

(3). b = distance between beams = 7 ¢ ft.

Use the least of these three conditions, or b = 43 } in.

kd = 2nda. + bt” = 2 x15x21x1,1806 + 43,5 x 9
ena, + 2vot 2 x15x1.1806 + 2x43,.5x3

kd = 3,83
Therefore the neutral axis falls in the stem, and the

formulas for case 2 apply.

p=a. =1,1806 .001293

 

 

 

 

 

43. 5x21
j = ca oli + off} « ff fey" (Ben)
6-3 (a)
yn 6 - 6h) (30 ¢ By) 5x LOIS I)
© (St)
j = .945

M = wi°/12 = 1206.25 x 20 x 20 x 12/12 = 482,500 in. lte.
f =M = 482,500 = 20,600 lbe./ sq. in.

* aljd 1.1806 x .945 x 21

 
59.6

ANALYSIS OF HEANS.,
Beam C - Tee-beam.

 

f= u = _ 482,500
ec Ti - t_)bvtjd (1 -__3 ) 43. 5x5x.945x21
2kd 2x3.83

te = 306 lbs. per sy. in.

v=V = 10 x 1206.25 = 80.9 lbs./ sq. in.
b' jd 7.5 x .945 x 21
u= $V = 10 x 1206,25 = 78.8 1lbs./ sq. in.
jd sum o 0945 x 21x 7.7
u =f, «= .126 x 16000 = 103 lbs. per eq. in. = bond
O1! 2x.81 x 12,05

in the web reinforcing.

1'§ = ,6d -4 4 4+=- 24) = 12.05 = effective length of
2707 «(3707 )

web reinforcing when inclined at an angle of 45.

x=] - y'bid = ge - 40 x 7.5 x ,945 x21 = 5.07 ft.
' 2 w 1206.25

= distance from support to where web reinforcement is needed,

sg = bagi gle = 3X,126x16000x,945x2l = 7.12 in.= required
° 10x 1206.25 x1.4

spacing of the web reinforcement.

f = 2/3 x_0,7Vsa___ ™_2x0,7x10x1206, 25x] 2
agJjd SX, 126X, 945x211

i 2? ,000 lbs. per sq. in. = stress on the web bars from
diagonal tension.
40.
ANALYSIS OF HBRAMS,
Beam A- Rectangular beam.

Reinforcement, -

1 - 3/4 x 2 3/16 kahn bar

1 - 1/2 rib tar
Span 20 ft.
Loading - 633 lbs. per lin. ft.
@, = 1.04 sq. in.

p= "3 = 1.04 °+§$%}j«jT ,00956
ba 7.5 x 14.5

k = (Zon + (pny? - pn
k =( 2x.00956x15 + (.00956x15)2 = .00956x15

 

 

k = .4112
J=1-k/3 = 1 - .4112/3 = .863

 

; in.
M = wi7/12 = 633.3 x 20 x 20 x12/12 = 253,320 1be./ sq
f= Mx = MM = 53,320 .
" pjbaé aJd 1,04 x .663 x 14.5
f. = 19,400 lbs. per sq in.
f = aM = 2 X 253,920
© xjpde 7863 X .411 x 705 KX 14,5"

- = 907 lbs. per sq. in.

v= V = 10 x 633.3 = 67 lbs. sq. in.
bjd 7.5 KX .863 x 14.5

uz YY = 10 x 633,3 = 70 lbs. per sq. in.
jd sum 0 e865 xX 14.5 x 7.2

1' = 15.8 in. = effective lenght of the web reinforceing.

u= &f, «= 2126 x 16000 = 79.2 lbs per sq. in.
01! 1.62 x 15.8

xel-y'ibjd = 20 -_40 x 7,5 x ,863 x 14,5
w 2

‘ 2 633.3
x = 4.05 ft. = distance from the support t@ the point
g

where web reinforcement is needed,
41.

ANALYSIS OF BEAMS.

Beam A continued,

s = 38sfsjd = 3x,126x16000x,863x14,5 = 8.6 in. = required
1.4 V 1.4x10x633.3 :

 

spacing of the wing bars.

f. = 2/3 0.7 x Vs =_2x0,7x10x633,3x12 = 23,020 lbs. per
, a,Jja . 126X.863x14.5

sq. in. = tensile stress in the web reinforcement.
The negative moments for both the beams and girders were

computed as above for a rectangular beam.
42.

ANALYSIS OF COLUMNS.
Columns U and Y 17 and 18.
Reinforceing,- 4 - 5/8 in. square bars.
Length - 14 ft. |
Gross diameter - 16 inches.
Diameter inside reinforceing - 13 inches.
Total load - 145,964 lbe.

p = 1.5624 = .01178
132.7

 

f= P = 5 964
o “AQ + (15 = 158) Et + (15 -1) x .01178)
f, = 945 lbs. per sq. in.

f. = 15 x 945 = 14200 lbs. per sq. in.

Considering Eccentricity.
Columns W 7 = 27.
The method of analysis outlined in Hool's Reinforced

Concrete Construction, pages 395-413, of Vol. 2 was
used. The formulas for beams continuous over three
spans were used, This is not exactly correct but Hool states
that this method may be used in the case of beams continuous
over four or more spans without any great error.
B = ratio of moment of inertia to length for a beam,
Cz ratio of moment of inertia to lenght for a column,
“% = the slope or angle made by é@he neutral surface at the end
of a member with its original position.
E = modulus of elasticity.
K = constant in equation M © KCE x
I, = moment of inertia of beam.

I. = moment of inertia of column.
435.

ANALYSIS OF COLUMNS.
Column W 7 - al.

B= ,1) x (28)9 x 15,5 = 73.5 (See Fig. 317, pg 405

30 x 12
Vol.2 of Hool. )

C = 4125 = 24.5

12 x 14
x = 1° 1,5 w) - wo * (30)* (12)” 1,5 x 3468 - 2135

l2k Kc + 5B 12xE 4 x24.5 + 5 x 73.5
YY *_7100

E

M= 48CX = 48 1g 7100 = 1691
12x14 «x c

stress = Mx = 169lc x 4.5 = 760 lbs. per sq. in.
Io
 

 
~*~ et”

44,

ANALYSIS OF THE SAW TOOTH.
2 inch roof slab.
Reinforceing - none.
Span - 12 inches,
Load - 61.29 lbs. per sq. ft.

Shear = 4 x 61.29 = 1.28 lbs. per 8q. in,

2x12

=wl°/ 12 = 61,29 x 1x 12 = 61.29 in. lbe.
12

M
f, = aa ~ $2.29 x4 = 7.65 lbs per sq. in.
ANALYSIS OF THE JOIST.
Reinforceing - 1 = 3/8" rib bar
1 = 1/2" rib bar
Span - 20 ft. Size 4" x 8"
Loading normal to the joist -
Roof slab(live + dead load) 16/12x61.29 = 81.72
Tile - 12x12x6 = 22.0
Plaster = 5.0
Weight of joist = 29,0
92.0 cos 28 35 = _ 45.66
Total . 127.38 lbs.
b= (2x 6)2+ 4 = 16"/

kd = 2nd®s + bt© = _2x15x7x,3906 + 16 x4 = 1.93
ena, + 2bt 2X15x.5906 + 2x16x4

Therefore the neutral axis is in the flange and the formulas

for rectangular beams apply.
45.

ANALYSIS OF SAW TOOTH.

p =8s = .3906 = .003495

 

 

 

bd 16x17
x = 2pn + (pn) - pn = [2x.003495x15 +(.005495x15)~ = .00349x15
k = ,2755
j= 1-k/3 = 1 -,2755/3 = .908
M = wl"/12 = 127.38 x 20 x 20 x 12/12 = 50,952 in. lbs.
f= Mw = 50.952 = 20,500 44, lbs./ eq. in.
a, ja 00906 x .9082 x 1
fo aphe Seis aa POF 8a: An.

v= V #10 x 127,38 = 50.1 lbs. per sq. in.
vj 7x .908 x 4

a
u= iV = 10 x 127,38 = 57.5 lbs. per sq. in.
jd sum 0 908 x 7 x 3.5
Stresses in joist due to tangential force
of the load.
Reinforceing - for tension 1 - 1/2" rib bar
for compression 1 - 3/88 rib bar

Loading,- snow + roofing + slab = 16 43.17 = 57.56

Tile - 1 - 6x12x12 @ = 22.00
Plaster 16/12 x 5.00 = $.00
Weight of joist 4x6 x 150 = 25,00
a 111.56
111.56 x cos 28 35 563.37 lbs./ lin. ft.

Mo= wi"/l2 = 53.37 x 400 x 12/12 = 21,348 in. lbs.

p = 88 = .25 = .01
bd 6x4

p’ = .1406 = .00585
6x4
46.
ANALYSIS OF SAW TOOTH.

x =\2n(p + pigy + ne(p #pry - nlp + pi)

 

 

k = Y2x15(.01 + .00586x1/3) + 225 x.00159 = 15x.01585
k = ,3872

Z =_1/3k54 +2p'nd'$k -d'/a)
x? +2p'n(k - d'/d

z = 1/3(.2378)°x3 +30x,00585(.3872-1/3) = .0095
EIT + Dxi5(.3872 So 445872-1/5)

jd =d -2=3 - .0095 = 2.99

f

 

© pd“(3k -k° + ein (k= d'7a)(i - d'7a)

f, = 640 lbs. per sq. in.

f= YX = 39,600 lbs. per sq. in.

 

f' ex2nf -d' =1345 lbs. per sq in. = compression in
steel. :

Considering the stresses from both the normal and
tangential loads,-

Maximum stress in the concrete = 640 + 520 = 1160 lbs./ sq. in.
Maximum stress in the steel = 39,600 + 20500= 60100 ** *® *
This, the worst possible condition that could exist
and the loading as assumed probably never would act in this

way. The roof slab would take at least a part of the tang-
ential load and it is possible that it would take the entire
load. It is beligved that it would be necessary to consider

only the normal load, in the design of the joist.
~ af

47.

ANALYSIS OF THE SAW TOOTH.

Alternate joist,
Reinforceing - 1 = 3/8" rib bar

1 = 3/8" rib bar
The computations are the Same as given for the previous case,
and the following results were obtained, -

Due the normal load.
f= 27,950 lbs. per 8y. in.
f. = 587 lbs. per sq. in,

v= V_ #= 49.4 lbs, per sq. in.
bjd

u = 65.8 lbs. per sq. in.

Due to the tangential load.
f, = 56,700 lbs. per sq. in.
tf, = 643 lbs. per sq. in,

BEAM H.

Reinforceing,- 2 - 14 x 2t kahn bars

2 - 7/8" rib bars
Span 24.5 ft. Size - 12 x 20
Loading,-(roof slab + snow + wind)=24,5x20x61.29 #30 ,032.1
Joist 19x20x25 = 9,500
tile 19 x20x22 = 8,360
Plaster24.5x20x5 = 2,450

beaml2x18x150x24.5= 5.512
144

25,822 x cos 28 35 22,674,8
Total 52,706.9 lbs.

52,106.9 = 2150 lbs. per S4//MKL/ lin. ft.
4.5

Computing as a tee-beam with a 2" concrete Slab the following

results were obtained, -
ee /
—

 

 
™

48,

ANALYSIS OF SAW TOOTH,
Ream H continued.
tf. = 20,200 lbe. per sq. in.
t. = 1595 lbs. per sq. in.
v = 130.5 lbs. per sq. in.
u = 78 lbs. per sy. in.
In the web reinforceing,-
u = 74.8 los. per sq. in.
r= 11,950 lbs. per sq. in.
ty
Considering beam H as a strut supporting the
tangential load.

P

Ff =
° Al + (mn - lp)

P = 25,996 lbs.

p = “s = 01812

ft, = 96 86.4 lbs. per sq. in.
L2x20(1 - (15 - 1).01812)

Considering both the normal and tangential loads, the

maximum sfress in the concrete is found to be 1481.4 lbs./sq.in.

Beam D.
Considering beam D in the same manner, the following results
are obtained,-
Reinforcéing, -2 - 3/4 x 2 3/lo kahn bars
2 - 3/4" rib bars

Span - 24.5 ft. Size - 10 x 20
ant

49,

ANALYSIS OF THE SAW TOOTH.
Beam D,
Loading,- slab + snow + wind 15 ,016.0
joist 19x 10 x 25 24750
tile 19 x10x 22 «= 4180

Plaster 24.5x10x5 = 1225
beam 12x18x150x24.5 =_2756

144 12,911 cos 28 35 11,.337.4

26 ,353.4
26.555.4 = 2330 lbs. per lin .ft.
24.5
Computing as a rectangular beam.

p =.0154
k = ,484
Jj = .839

Mo= wi7/10 = 959,000 in, lbs.
f= 22,850 lbs. ver ey. in.
f, = 1454 lbs. per sy. in.
v = 110.4 lbs. per sq. in.

u = 95.6 lbs. per sq. in.
Stresses in the wing bars.

u = 92.3 lbs. per sq. in.
f = 70,000 lbs. per sy. in.

° Considering beam D,as a strut supporting the
tangential load.
Loading = 13,871 lbs.
Pp =,0135
ty = 28.6 lbs. per sq. in.

Maximum stress in the concrete considering both loads
50.

ANALYSIS OF THE SAW TOOTH.
Strut I.

In the analysis of the struta the maximum stresses were
found to exist as a result of the dead weight of the roof
and snow load a¥tove and at the Same time the wind blowing
against the sash produceing a bending moment. Hence only

these results will be given.

Loading, -
Roof slab + snow = 30 x 20 x 37.89 = 22,734
Weight of joist, tile, etc = 25 ,248

Only one-half of the roof load comes on the strut, the
balance resting on the girder,

1/2 x(25,248 + 22,734) = 23,982
Weight of strut and bracket = 2,634

ll

Weight of sash 5,160
Total load = 29,785 lbs.

Reinforceing,- 4 - 1/2" rib bars.

Size,- 12"°x 12" within reinforceing bars - 9"x9"

Length,- 11 ft.

p= %g = 01235
A

f = P = 29 7.85
c “RU #(m - 1)p) 81(1 * 14 x .01235)

f, = 15 x 315.5 = 3291 lbs. per sy. in.

313.5 lbs./sq.in,

 

 

Considering the wind pressure against the sash,
Wind pressure against the upper sash = 6 * 20 x 30 = 1800
Normal pressure = 1800 x cos 30 = 1559

Angle of lower sash with horizontal is 60 degrees.

P_ = 30 x 2x ,866
n Sa S777 = 29.7 lbs. per sq. ft. of surface.
Sl.

ANALYSIS OF THE SAW TOOTH.

Strut I.
Pressure on lower sash= 6 x 10 x 29.7 = 1782 lbs.
1/2 x (1559 + 1782) = pressure at mid point = 1670
Weight of sash at same point = _430_
Total load = 2100 lbs.
The reaction = 6/11 x 2100 = 1145
M = 10/12 x 6x 1145 x12 = 68,700 in. lbs.
f, = 6 x 68,700 = 355 lbs. per sq in.
1160
{, = 68,700 = 14,400 lbs. per sy. in.

 

.00397 x .91 x 12 x (10.5)*
Maximum stresses,-
f = 355 + 313.5 = 668.5 lbs. per sy. in.
C = 14,400 - 15 x 313.5 = 14,071 lbs. per say. in.
Strut J.

Since there is no beam resting on the top of strut J
none of the foof load will be considered as coming on it.
Reinforceing - 4 - 1/2" rib bars.

Size - 12" x 12" Size of core 9" x 9",

Length - le ft.

Loading due to the wind pressure against the sash,-
2100 lbs. (See strut I).

Reaction = 1/2 x 2100 = 1050 lbs.

M = 6 x 1050 x 12 x 10/12 = 63,000 in. lds.

f = 326 lbs. per sy. in.

f = 63 .000 _. = 13,200 lbs. per sy.in.
8 (00397 x .91x l2 x (10.5)*
aw

52.

CONCLUSION.

At the time the analysis was started very little
machinery was placed, Hence it was necessary to adopt the
loading for which the building was said to be designed.
This live load of 500 lbs. per lineal foot of beam which
was adopted is believed to be extremely high. Probably
200 lbs. per lineal foot will cover the total load exist-
ing on the beams and provide an impact coefficient of 100
per cent. If the loads are reduced in this ratio practically
all the apparent over stresses shown on the stress sheet
will disappear.

According to some authorities the snow load of 30 lbs.
per ayuare foot of horizontal surface might ve reduced to
20 lbs. per square foot. The construction and location of
the building makes it possible for a large amount of snow to
accumulate and be held on the roof and it is thought best
to use this higher load.

In the roof slab the concrete was found to be within
the allowable working stress. The steel at the center of
the slab is over stressed. Over the support the stress was
found to be 42,000 lbs. per ay. in. Only every alternate
bar was bent up and hence the stress was nearly doubled.
According to the Joint Committee the span length may be
taken as the clear distance between the faces of the
support for slabs and beams built monolithic with the sup-
ports.(H.& J. pg. 318, Art. 44),If this rule were followed

as the designers state was done, the span length would
535 6

CONCLUSION.
be reduced to 6 1/2 feet instead of 7 1/2 feet thus reducing
the stresses,

In the beama the compression on the concrete was low
except in the beams A, B, R, and W. Beams R and W were
analized as rectangular beams due the their irregular shape.
It may be permissable as was done in designing the beams,
to consider these as tee-beams. If this were done the stresses
would be reduced to a considerable extent. However, as the
roof slab on one side is at an angle of 30 or 60 degrees, it
is not thought advisable to do this. The steel in all the
beams shows a slight over stress, This may be account:*for
by the fact that the span length used in design was the
clear space between faces of the supports and also by the
fact that the live load is high. The stresses in the steel
due to the negative bending moment at tne supports are not
correct as additional steel not shown on the steel plans
was placed in the beams at these points. This was not
known until the analysis was completed and the amount of
steel is not known.

In the girders the stresses were found to be hiffher.

In desizning the buildinz the live load on the girders was
reduced 20 %, In the analysis the full live load was used
based on Hool, Vol. 2 page 211 where he states that the
full live load should be used on the columns for a factory
building,

The stress in the concrete over the supports is high.
54,

CONCLUSION.
The girders may have been designed as doubly reinforced.
This is not permissable as the steel at the bottom of the
girders does not continue through the center of support,
leaving a space of about three inches where there is no
steel. The negative moment is a maximum at the center of
support and decreases rapidly toward the edge of the column.
Hence there will only be a small space which will be under
high stress.

Assuming that a kahn bar is a deformed bar the bond
stress at the center of the span was found to be safe for
both the beams ana girders. The values found for bond over
the supports apparently shdow over stress. The hook on the
end of the bars may be counted on to reduce this to the
allowable value.

The Joint Committee recomment an allowable shear value
of 120 lbs. per sq. in. when the beam is thoroughly rein-
forced for diagonal tension. These beams and girders are
not fully reinforced since the wing tars are spaced too
far apart to take the required 2/3 of the diagonal tension.
It can only be said that the shear value is higher than
recommended,

The columns were analized first, considering only
the total load, second, considering the possible eccent-
ricty. The over stresses found when the load was considered

as centrally applied are undoubtly due to the high machinery
356

CONCLUSION.
load and to the fact that the building was designed on the
basis of a 20 % reduction of live load in the girders and
a 10 @ further reduction in this live load to the columns,
In a building of this type the effect of eccentricity
shoud not ve overlooked. The load on the column from girder
C C may be double that from D D. This produces large stresses
which are equal to if not greater than those due to the dead
weight alone.

In the saw tooth the steel,in the joist was found to
be over stressed somewhat due to the normal load alone.
Whdle it is possible that the joist would be balled on to
bear the tangential load it is not at all likely. The roof
slab is a part of the flat roof slab and may be depended
upon to take considerable of this load in compression.

The steel in the beams D and H in the saw tooth was
found to be over atressed. The conerte in the same beams
was found to be highly over stressed, beam H having a
stress of 1595 lbs. per 3 3. in. due to the normal comp-
onent alone.

In the struts the maximum stresses were found to be
produced by the wind blowing against the sah which are fasten-
ed to the struts near the center and the dead load of the
roof and snow above. The stresses in both the steel and
concrete were found to be well within the allowable work-

ing values,
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CONTENTS OF POCKET.

Plate 1, Location Diagram of Column Footings. 7

2, Braming Plan of Hanger Beams & Flat Roof.
" 5, Plan Showing Saw Tooth Construction,
4

» Typical Section.
"5, Plan of Typical Slab.

6, Bent Rib Bar Diagram,

7, Beam Sizes and Reinforcing.
e 8g a n n "

9, Girder Sizes and Reinforcing.
"10, Column Sizes and Reinforcing,
"11, Roof slab atresses,

"12, Beam Analysjs.,
" 13, Girder Analysis.

" 15, Plan of Blower Room,

 

 

" 16, Colum Stresses,

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