A DESIGN OF A REINFORCED
CONCRETE BUILDING FOR A
SOY BEAN PROCESSING PLANT

Thoskfor tho Donn cl 8. 3.
mCHlG-AN STATE COLLEGE

W R._ Radcliff
1947

TH 5515

A Design of a Reinforced Concrete Building

for a Soy Bean Processing Plant

A Thesis Submitted to
The Faculty of
MICHIGAN STATE COLLEGE
of

AGRICULTURE AND APPLIED SCIENCE

By

Robert Redcliff

M

Candidate for the Degree of

Bachelor of Science

June 1947

THESiS

3.;

k7“

ACKNOHLEDGEMENTS

I wish to thank the members of the Civil Engineering Staff for their

assistance in the preparation of this_thesis. In particular,

Professor 0. A. Miller, for the time and effort spent with me to per-

fect the design both structurally and practicably.

189?8‘

BIBLIOGRAPHY

Underwriter's Laboratories, Inc. - Report on 8 in. Bearing Walls.

Engineering News Record", may 1, 1947.

Pulver - "Construction Estimates and Costs."

Parker - "Simplified Design of Reinforced Concrete"

Sutherland and Reese - "Reinforced Concrete Design"

O'Rourke - "General Engineering Handbook"

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

A.C.I. Building Code

T.E.C.O. Truss Manual

Joint Committee Report. 1940

Portland Cement Association - "Reinforced Concrete Design Handbook"

merriman and Wiggin - "American Civil Engineers' Handbook" - 5th Edition.

National Lumber Manufacturers Association - "National Design Specifications
for Stress - Grade Lumber and its Fastenings"

Bateman - "Highway Engineering

Portland Cement Association - "Design and Control of Concrete Mixtures"

CONTENTS

I. Purpose and ScOpe of Thesis

II. Computations and Sketches

PART I

PURPOSE AND SCOPE pg THESIS

 

The purpose of this thesis is to present a practical and economicd.
design of a building for use by the owner. The property is located in
the Central Ohio farm area. The objective of the owner is to have a
building large enough and strong enough to house a Soy Bean Processing
Plant. Mbst of the construction work will be done by the owner's
organization. The building will be constructed with materials avail-
able in that area, and a cost of materials estimate will be included.

The project offers several practical design problems for the author.
The methods used are those set forth in the respective design courses
as taught at Michigan State College. much more labor was applied to
this problem.than would have been necessary if it had been handled by
an experienced design engineer. Despite this fact, it is a worthwhile
project because it employs design fundamentals necessary to a good de-
signer as a basis for the tricks and short cuts of the trade.

The author will design the building in accordance with specifications
supplied by the owner. The design and materials estimate will constitute
the object of the thesis.

The owner plans to design and fabricate all plant equipment. While
this is being done the building will be used for farm.machinery repair
and as a farm.locker agency. By means of these Operations the owner hOpes
to establish himself in the area.

The scope of the thesis will include the following:

A. The design of a fireproof, durable building to
perform.the following functions:
1. Provide suitable space for Operating a
Soy Bean Processing Plant.

2. Provide suitable space for repairing farm

machinery.

3. Provide suitable space for Operating a farm

locker agency.

4. House the following equipment:

a.
b.
c.
d.
e.

f.

Metal wOrking tools.
Welding equipment.
Overhead crane.

work benches.
Woodworking machines.

Lubrication facilities.

B. To locate and design:

1. A concrete driveway from the rear of the build-

ing, running parallel to the side of the build?

ing and connecting to the existing road.

2. A three-ton overhead crane system.to be located

on the ground floor.

C. Estimate the cost of materials for construction.

The scape of the thesis is limited to the structural design of units

heretofore described.

No attempt has been made to perform.the functions

of an architect in that architectural features and details of the allied

trades such as plumbing, heating, or wiring have not been mentioned or

treated.

 

 

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BUILDING DETAILS

 

Drawing 1-2-3-4.

The building will be 120.0' long and 80.0' wide and will be located
on the property as shown in Drawing 1. Because the author was unable to
visit the building site, the data for the site map was obtained from.the
owner and from U.S. Geological Survey tOpographic maps of the area. The
ground floor walls will be concrete blocks. The roof system will be con-
structed of wood with wood sheeting and aSphalt roofing. The remainder
of the building will be constructed of reinforced concrete. The driveway
will be poured concrete with no tensile reinforcement. The drive will be
widened at the curve to facilitate entrance into the building. Due to the
heavy floor loads expected, the floor system will be designed for a live
load of 400 p.s.f.; column Spacing taken as 20.0'o.c. with T-Beams inter-

esting the girders at the 1/3 points.

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_- -v'.,_,,_‘,,-_-__

-—‘_‘_ -_~~_

ALLOWABLE UNIT STRESSES USED

 

Taken from A.C.I Building Code 1946.
Concrete

Using a concrete with fc'

3000 p.s.i.

n : 10
Compression fc : 1350 p.s.i.
Shear v
Beans

W/b web Reinforcement
wyb SA - 60 p.s.i.
w/ SA - 90 p.s.i.

'fi/ ‘Web Reinforcement
w/b SA - 180 p.s.i.

W/ SA - 360 P0801.

Footings - 75 p.s.i.
Bond u
Beams
Plain bars - 120 p.s.i.

Deformed bars - 150 p.s.i.
Two-way footings
Plain bars (hooked) - 135 p.s.i.
Deformed bars (hooked) - 168 p.s.i.
Bearing fc
0n full area - 750 p.s.i.
On one-third area - 1125 p.s.i.
Steel
Billet, hard grade fs - 20,000 p.s.i.
Lumber - Pine Southern Shortleaf (Dense Structural)
Extreme fiber in bending - 1900 p.s.i.

Horizontal shear - 120 p.s.i.

Compression perpendicular to grain - 455 p.s.i.

Compression parallel to grain -l450 p.s.i.

Nodulus of elasticity -1,600,000 p.s.i.
Soil Pressure - "American Civil Engineers' Handbook."

Dry Silt-Loam -3 Tons/ sq. ft.
Bearing on Masonry Wall

Portland Cement Association -800 p.s.i.

PART II

ROOF DES IGN

 

Sketch 1
This truss was selected from Timber Engineering Company Truss manual.

Fink Truss
80' - 00" Span
16' - 00" Rise
16' - 00" Spacing 0.0.
1' - 00" Eaves

Dead Load a 2470 a ~§§3§ FBM = 7410# Truss Weight

-% Roof Area.: 44.0 x 16 = 704 s.fg/truss

Wind Load 2 20 p.s.f. of vertical surface

P normal I 2 0'3” 20 .-. 13.1

Snow Load a 20 p.s.f. vertical
Maximum Expected Load
Dead / Wind / % Snow
Dead / % Wind 7/ Snow

‘Wooden sheathing 1.0" thick - 3 p.s.f.

Asphalt covering - 2 p.s.f.
Roof covering 5 p.s.f.
Truss = 7410 a 5.8
(W

 

10.8 Total D.L.
Dead / Wind / —%- Snow
10.8 / 13.1 7/ 10 a 88.9 p.s.f.
Dead/ 7% Wind ,1 Snow
10.8 / 6.6 ,1 20 -.- 57.4 p.s.f.

Therefore 40 p.s.f. is a safe value to use.

1

DESIGN OF

MEMBERS

 

Sketch 2

Top Chord

Length . 10' 9%" a 129.25"

Try 4.0" member

L s 129.25 : 35.7
'3 3 625

 

4
WOrking stress : C(l-l/S (REE) )
l. 2 .l
K = 0.64 (3.5.5.3? .-. 0.64 (2'5 x hggoiooof = 33.7

4
Working Stress = 1450 (1-1/3 (1.331%) ) = 845 p.s.i.

Area required = 60300 = 71.5 sq. in.

Two 4 x 12" timbers with a cross sectional area of 2(41.69) a

83.38 sq. in. will satisfy requirements.

Bottom Chord - Lo - L1 - L2

web

Area required 3 56000 : 29.6
'13RR5

Two 4 x 5" timbers with a cross sectional area of 2(16.77) x

33.54 sq. in. will satisfy requirements, but further investigation
shows that the joint between L1 and L2 requires a larger width.
Therefore two 4 x 8" timbers are used.

members - Compression

Select a 4.0" timber for web members in compression.

v2 length = 8' - 7 3/8" .-. 103.375"

L = 1030375 = 2805
'3

 

.l
K -_- 0.64 (W? . 21.3
3 = 0.2743 . 0.274 :5 1,60q,000 = 543 p.s.i.
A (28.5)“

(3)
Area required a 11,900 = 22.0 sq. in.
'7RE§"

A 4 x 8" member with a cross sectional area of 2719 sq. in.
will satisfy the requirements, but further investigation shows
that joint C requires a 12" width. Therefore a 4 x 12" member
is used.

v1 and v3 length .4 4.0' 3 11/16" -.- 51.688"

Use 4.0" member

L I 51.688 - 14.25
3.

K greater than L greater than 11.0, therefore the intermediate

column formula is used.

4
. _ 14.25 - . . . .
g 1450 (1 1/5 (2.1.3.) )- 966 p s i

Area required : 5950 : 6.16 sq. in.

’966'
A 4 x 2%" member with a cross sectional area of 7.70 will satisfy
the requirement. Four inch Split rings are to be used at the
joints and a minimum of 5%" width is required. Therefore a 4 x 6"

member is used.

Web Members - Tension

maximum tension load on D4

Area required.= 24000 = 12.6 sq. in.
'IUDU'

Two 2 x 6" members with a cross sectional area of 2(9.l4) : 18.28
will satisfy this requirement but in order to eliminate fillers

and to provide enough thickness for split rings, two 3 x 8" members
are used for D3 and D .

D1 and D2

Area required = 8000 : 4.22 sq. in.
I966

At least 6.0" members are required for 4.0" split rings, there-

fore two 2 x 6" members are used for D1 and D2.

DESIGN OF JOINTS

 

Sketch 3

Joint "A"
The load on 01 and 02 acts at an angle of 20.20 with the chord U1 -

U2. The allowable load on one ring : 4400#.

Therefore two rings are required, one in each member.

n

The standard edge distance = o [/16".

8000 . 91% of full load deveroped.
x 4 00
Therefore the edge distunoe can be reduced to 2 3/4".
The edge distance furnished by a 6.0" members = 2 13/16".
The lo:d on V2 acts an angle of 900 to U1 - U3.
The allowable load on one ring = 3750#.

11900 2 2.54
"4760

Therefore three rings are required.

There is not enough room for three rings, so two rings and four
3/4 x 15%" machine bolts are used.

The allowable load on bolts = 700 x 4 ; 2806}.

2800 /'2(4700) e 12200#.

Therefore four bolts and two rings are sufficient.

Minimum bolt spacing : 4d a 3.0"

Spacing between rows - 5d : 3 3 4".
Joint "B"
D4 - U3
Angle of load to grain.: 22o
Allowable load on one ring = 6200#
.Eégg%,= 3.87 rings are required - 6.0 rings used.

0 O of ca acit develo ed : 24000 3 64.5 9/9
/ ’ y p 6 x 6200

Standard Spacing parallel to grain in D4 = 9.0"

Reduced spacing : 5.0"

Spacing used = 7.0"

Edge distance required = 2 3/4" - 3 3/4" used
End distance required e 3 3/4" - 5%" used

U3 to splice plates

Angle of lord to grain = 22°

Allowable load on one ring 2 6200#

32000 = 5.2 rings required - 8.0 rings used
2

q/o of capacity developed

32000 . 64.5 o/b
8 x 6256

Standard spacing parallel to grain in D4 = 9.0"

Reduced spacing = 64'5 g 2 ' 100 = 36 o/b of capacity for one group.

50 o/b reduction allows 4 7/8" minimum

Use 7.0" spacing

Edge distance required : 2 3/4" - 2 3/4" used.

End distance required : 3 3/4" - 5%" used.
Joint "C"
72 to L1 - L2

Angle of load to grain : 57.5°

Allowable load on one ring = 4900#

11900 = 2.43 rings required - 4 rings used
"—4950

11900 = 61 e/o
4T49007

Reduced edge distance = 2 3/4"

o/b of capacity developed =

Spacing required parallel to grain : 5%"
End distance required : 3%"
D3 to L1 - L2
Angle of load to grain : 45o
Allowable load on one ring : 5350#
16900 = 2.98 rings required - 4 rings used

o/b of capacity developed : 16000 : 75 o/b

4T5355)

Reduced edge distance : 2 5/3"
Spacing required parallel to grain = 5%"
End distance required : 3%"

Length of Me bers

U0. U1. Uz. U3 10' - 9—

L0, L1 = 11' - 7 3/16"
L2 = 16' - 9 3/8"
71 = 4' - 3 11/16"
v? = 8' - 7 3/3"
v3 : 4'-1.@@"
D1 -.- 11' - 7 3/16"'
D2 = 10' - 7 7/16"
05 = 11' - 7 7/3"
D4 = 10' - 10 3/16"

NUTERIELS REQUIRED FOR TRUSSES

 

Building 120.0' long

Nine trusses @.16.0' C.C. required

Lumber
36 pieces 2" x 6" x 12'
36 pieces 2" x 6" x 14'
36 pieces 2" x 6" x 16'
72 pieces 3" x 8" x 14'
18 pieces 3" x 8" x 20'

9 pieces 4" x 8" x 12'

18 pieces 4" x 6" x 16'
54 pieces 4" x 8" x 14'
45 pieces 4" x 8" x 18'
18 pieces 3" x 12" x 16'

9 pieces 3" x 12" x 10'

18 pieces 4" x 12" x 14'

81 pieces 4" x 13" x 16'

27 pieces 4" x 1:" x 18'

18 pieces 10" x 12" x 18'

18 pieces 4" x 12" x 6'

Total = 22,230 FBM

liardware
2736
144
72

72

' Machine

Teco Split rings 4"

Taco shear plates 3 1/8"

Machine
Machine

Machine

machine
machine
machine

machine

machine bolts 3/4"

Steel plates

Steel plates

‘ Steel plates

Steel plates

bolts
bolts
bolts 3/4"
bolts
bolts
bolts
bolts o

bolts

1
3" I 12"

18"
23"

24"

3/4"
10"

-%-" x 3" x 10 7/8"
%" x 3" x 12"

" x 20" x 16"

(73?“

3 8" x 10" x 12"

Angles 4" x 7" x.3/8", 12" long

9 Threaded rods 17' - 3" long

144‘Washers

2a

2124 Plate washers 3" x 3" x 3/16"

cheek bearing on wall

Total roof load . 80 x 120 x 40 : 384,0t0#

well load = 192,000#

18 pieces 4" x 12" x 14'
81 pieces 4" x 13" x 16'
27 pieces 4" x 12" x 18'
18 pieces 10" x 12" x 18'
18 pieces 4" x 12" x 6'
Total : 23,230 FBM
IEbrdware
2736 Teco Split rings 4"
144 Teco shear plates 3 1/8"

72 machine bolts-%" x 12"

72 machine bolts 3/4" x 6%"
216 rachine bolts 3/4" x 12%"
126 machine bolts 3/4" x 15%"
396 machine bolts 3/4" x 17%"
144 Machine bolts 3/ " x 18"

36 Machine bolts 3/4" x 23"

36 machine bolts 3/4" x 24"
144 machine bolts 3/4" x 10"

9 Steel plates-%" x 3" x 10 7/8"
9 Steel plates-%" 1 3" x 12"

" x 20" x 16"

4‘0ng

36 Steel plates
36 Steel plates 3/8" x 10" x 12"
36 Angles 4" x 7" x 3/8", 12" long
9 Threaded rods 17' - 3" long
144 washers 2"
2124 Plate washers 3" x 3" x 3/16"
Check bearing on wall
Total roof load I 80 x 120 x 40 : 384,0h0#

wall load a 192,000#

18 pieces 4" x 12" x 14'
81 pieces 4" x 12" x 16'
27 pieces 4" x 12" x 16'
18 pieces 10" x 12" x 18'
18 pieces 4" x 12" x 6'
Total = 23,230 FBM
IEmrdware
2736 Teco split rings 4"
144 Teco shear plates 3 1/8"

72 machine bolts-%" x 12"

72 Machine bolts 3/4" x 6%"
216 Machinebolts 3/4" x 12%"
126 machine bolts 3/4" x 15%"
396 machine bolts 3/4" x 17%"
144 machine bolts 3/4" x 18"

36 machine bolts 3/4" x 23"

36 machine bolts 3/4" x 24"
144 Machine bolts 3/4" x 10"

9 Steel plates " x 3" x 10 7/8"

l‘Olb-J (Oh-J
:

9 Steel plates 1 3" x 12"

36 Steel plates " x 20" x 16"

as:

36 Steel plates 3/8" x 10" x 12"
36 Angles 4" x 7" x 3/8", 12" long
9 Threaded rods 17' - 3" long
144 washers 2"
2124 Plate washers 3" x 3" x 3/16"
Check bearing on wall
Total roof load 3 80 x 120 x 40 = 384,000#

wall load : 192,000#

Load er lineal foot : 1927000 = 1600 .f.
p "m— P
Gross area of 8" x 8" x 16" concrete block:: 15.75 x 7.75 :
122 sq“. in.
122 x.%§.= 91.5 sq. in. per lineal foot
1600 8 1705 p.801.
91.5

Therefore, the bearing on the top course is within the allowable.
Use 3/4" bearing plate over top course on E and W walls to
distribute truss load.

UPPER WALL DESIGN

 

'Weight of 8" x 8" x 16" heavy bearing blocks‘/ %" joints : 50#
14.0' wall height : 168"
158 = 21 courses
120.0' wall length = 1440"
1440 s 90 blocks
16 .

'Wall area = 1680 sq. ft. requires 1890 blocks

1.125 blocks required per square foot of wall

 

 

North‘Wall
80.0' long 1 - 12' x 14' door
2 - 10' x 20' Windows
South W811
80.0' long 1 - 12' x 14' door
East'Wall
120.0' long 3 - 8' x 10' windows
west W811
120.0' long 3 - 8' x 10"windows
N.W. S .W. E.W. W.W.
Area sq. ft. 552 952 1440 1440
No. Blocks 620 1080 1620 1620

Total Wt.# 33,000 63,000 94,500 94,500
(Assuming solid wall)

Unit ht.#/lin.t 787 787 787 787

Check for bearing on bottom course - Esst and best halls.
Live load - 1600

Dead load - 787

Total load - 2337fi/Lin. ft.

h:
I»:

 

hIRDON LINTEL DESIGN

East and'West'halls

Sketch 15
3 courses on top : 3 x 50 : 150%
150 x ‘1‘; = 1133%/lin. ft.
Live load leoofi/lin. ft.
Dead load 113. lin. ft.
Total lord l713#/lin. ft.

2 ~ -2
thimum.B.M. = W'l = 1713 x 100 = 14,3001W

 

 

‘12
Section modulus = M= 14,300 x 12 = 8.6 in.3
T 20,000

Use 2 angles 6 x 4 x 7/8"

section modulus = 2 x 7.2 : 14.4 in.3
North‘Wall
Sketch 16

3 courses on top
\. 160 x 1'3 : 113#/ft.

sax B.M. = wl2 1":

 

: 113 x (20)2 = 3770
*12

Section modulus = 5770 x 12 = 2.27 in.3

'1fififfifif“

1
Use 2 angles 4 x 4 x-g"

Section modulus : 2(2.0) = 4.0 in.5
Doors

Use 4 x 4 x-%" angles for door lintels.

DESIGN UPPER FLOOR SLAB

 

Sketches 4 and 5
Live load assumed 400 p.s.f. This figure used in order to satisfy
future requirements to which building Hwy be put.
Assume an 8.0" slab

L. L.

400 p.s.f.

D. L. : 100 p.s.f.

T. L. : 500 p.s.f. or for a 1.0' section 5007mm. ft.

Considering this a simply supported beam with a uniformly distributed

2
load, B0143. = _w_8_]_’__

Max B.M. = w 12 = 500 x 64 = 400011"
8""
V': "'1 - 500 x 8 : 200 ”
T ' ‘—2'— O"

From table #2 RCDH
d : 4%"
Change sltb thickness to 6.0"
TL 3 475 p.s.f.
BM .-.- 38001" d : 4%" v = 19005;é
From table #1 RCDH

a = 1.44 As a M .: 3800 : 0.587 sq. int/ft. required

Use %" circular 4.0" spacing
Check shear

- V _ 1900 _ 41 1 0 K
V— n . - .S. o o o
5.33 1?): 7/8 x 4.5 p ""

Check Bond

u . v 3 190,0 .1 106 p.s.i. 00K.
2013 4.7:: 7/8 x 4.5 ‘-

10

Placement of negative moment steel.

Sketch 6
Reinforcing steel of the same size and placing as that used
for tension will be placed over supports; the length to be
fi-clear span each side of support.

8000" "’ 1000" = 07000"

70-0 = 17.5 Use 18.0" each side.
"T

T - Beam Design
Sketch 7
Span 20.0' cc
Assume clear span 19.0'

LL

475 x 6.67 = 3170#
DL : weight of stem assumed a 239i
TL : 3400%/'
v'= W’l : 3400 x 19 2 3g 30 '

T "_"2'—_ , 074

v allowable with web reinforcement : 180 p.s.i.

v = FEE bd : 32,:00 : 205 sq. in.
Let b : 10.0"
d = 23.0"
d / 3 : 26.0"
26 - t = 20.0"

Check weight of stem

9'5 x 20 x 150 = 19
T 8#

Assume j : 0.875

BM _ w 1‘2= 3400 x (19)2 x 12 : 1,470,000"#
‘ ‘127' *10

 

BM .1 Tjd

T = 1.470.000 = 72,900#

11

As . 72,900 = 3.65 sq. ind/Tt. required

9
Use 4 - 1" square bars
Area : 4.0 sq in. two rows
Check_bond

- V 32 300 - 101 s i
u - ’ .. .. - O O 0

Check fiber stresses
Sketch 8

%-0f Span.: 20 I 12 = 60.0" Least

 

(8 x 6) 2 7’ 10 = 106.0"

(80'0" ‘ 10") I 3': 35"

35x2/10= 80"

moment NA = (60 x 6) (x - 3) = 40 (23 _ x)
X.= 6.25"

.25
mfc _, 0.04fo .. Z

23.0 - 6.25 = 20.92"
"3"

Compression C Arm. Mbment
C1 - 0.04fc x 6 x 60 : 14.4fc x 20 : 288fc
02 = é-x 0.96fc x 6 x 60 = 172.8fc x 20.92 : 3620fo

187.2fc 3908fc

01 2'02 = C

3908fc = 1,470,000

f0 = 377 p.s.i.

C

T = 377 x 187.2 = 70,500#

T Asfs

f3 = 70,500 = 17,600 p.s.i.

Check T beams at support

12

Sketch 9
(N - 1) As = 36 sq. in.
NAs : 40 sq. in.
10X : 10 x sq. in.
36 (x - 3) /10 1022): 40 (23 - x)
36x - 108 /'5x3 = 920 - 40x
51:2 7’ 76X : 1028
3.: 8.6"

jd = 23.0 - -_- 20.2

8.6
T
Co = .152 (10) (8.6) ;_- 43.0fc

Cs : gig. fc (36) = 37.6fc

43.0fc x 20.2 : 876fc
37.6fc x 20.0 :- 752fc
80.6fc (a) .-. 1628fc
a = 20.2

T = C = 1,470,000 : 72,700#
20.2

f0 3 72,700 = 901 P0301. 00K.

f8 = 72,7(X) : 18,200 p.s.i. 00K.

T Beam Web Reinforcement

Sketch 10
The maximum shear occurs at the ends when the entire span.is
loaded.

vnax : 329500 : 153 p.s.i.
x o 1

Center shear taken as 25 q/b of maximum.
vb : 153 x .25 = 38.3 p.s.i.
Shear takeniby stirrups :- w x 10 : 43,000#

Use-%" circular stirrups.

13

'-

.. 1,
1m 7
a .7
_
«I-

r——-——-—_-_._ q"‘_ _- __...~._.. _.

‘r-vv" ~ .
r

.1. ———-.-.—..w-__-

—__.._‘ -——

\1 am

"Ills-"1 d ‘1...“

a
4
51-1.! 17.1.»;

a a 7

I

m n

,1 ll:

 

E
I

 

GJPM‘

—_—..--.~——< _

, 3:

As = 0.1963 x 2 = .3926 sq. in.
16,000 x .3926 = 62807}E taken by each stirrup

45.000 = 6.85 Use 7.
'YEfifiT'

d. 23 = 1105"
'2 '—2

93 0
. - 80
1175- 1

Therefore 9 stirrups will be used from support to center.
205"
7@10"

GIRDER DES IGN

 

Sketch 11
Span 20.0'
Clear span = 18.5' assumed simply supported beam.
Assume stem weight : 5004/rt.
B.M. = 64,600 x 80 x 2/3 /’1/10 x 500 x (18.5)2 /’12

3,450,000 /'205,000

3,655,000"#

 

vmax. : 64,600 /'500 g 18-5

= 69,225#

Assume j : 0.875 at support
69 225

bd : V' = = 440 s . in.
180j 180 x .875 q

Use b .. 15.0"

d 32.0"

d‘/ 3

35-6 _ 29.0"

35.0"

Check stem weight

39 I 15 x 150 : 455' '
._4F@f_. ua/

Assume j : 0.92 between supports
B.M. .-.: Tjd

14

As : T = 124 200 = 6.21 Sp. in
'TE "26f606 ’

Use 1.0" circular bars

Use 8.0 bars - two rows .. 3%"

Spacing.
AS : 6.32"
o = 25.1"

Bond

'V - 69 225 '
u = - , . 9307 0801. 00K.

Review of Girder Design

Sketch 12

pp

span : 32.11.13 : 60.0"

8 1 thickness x 2 /'15 111.0"

Clear span : 18.5 x 12 222.0"

meent NA.= (60 x 6) (x - 3) = 63.2 (32 - x)

X = 7036"
Z : fc
036 7.36

Z = 0.0488fc

fc - Z = .95fc

cl = (60 x 6) (.osrc) = 18.0fc x 29 = 522fc
02 : (60 x 6) $942323) = 171fc x 30 : 5130fc
c ; cl /’cz : 18.9fc = 5652fc

f0 = w = 646 13.8.1. 00K.

T = c : 189.0 x 646 = 122,200#

fs =.1§§i§g2.= 19,500 p.s.i.

07..

Check girder over support
Sketch 13
meent 34.: Q§)(15X)1/'(56.9) (x - 2) a 63.2 (32 - x)

15

X = 10.7"

Co = £29 (15)(10.7) 80.3fc x 28.2 = 2270rc

c - 8.7 fc (56.9) 46.4fc x 30 = 1390rc
8-

iU.V 125.7?0 3600fc
a = 28.9"
T : C = 3,655,000 : 126 50 '
“—2‘8 7‘9 ’ 0;)

f0 = _126,500 3 1000 p.s.i.
126.7

fs : W = 20,000 p.s.i.
.0

Girder'fieb Reinforcement
Sketch 14
maximum shear occurs at the end with the entire span loaded.

VV : 69,225 = 157 p.s.i.
‘15 x 0.92 x 32

Shear at center taken as 25 q/o of mpx.
157 x 0025 3 39.2 P0801.

Sheer taken by stirrups

15 x 9.1.3.233 : 66,600#

66.600 - 10.6 stirru a re uired-
"'62—‘80 - p 9

Use 12 stirrups from support to center.
Spacing
4 @.5.0"
4 @ 8.0"
2 @ 10"
2 @ 12"
Eg§LL.BEAM DESIGN
Sketch 17
North and South wall beams will carry half the floor load of the

of the other cross beams and no roof 108d. Therefore the floor

16

w
¢-,-s.

, ,1
3A 0.1
‘. I
. . :41
1 u .,
: T ...I.s!i.ri-s!-lll
_
r u
v I ~
I 1 .
,/
/ .
. 2
,, _
, w
/
, V:
.. .
\. 2_
.A V.
w.
+1
<
y l f l .1411 I 4" in I 0‘
, ..
v. .
_
_/ _
.
.
/

r.
.v

‘9‘.

H
/ I
I
, _ .
v
A‘ f.-
/l
1 1t
H\ \ to
I , . n
1 1/ ’ls.

2.. xv

beam design will suffice for the North and South wall beams.
East and west Wall Beams

Assumed weight of stem.= 500#/'

 

Wall load =17102’Z'
2210#/'
Max. v = 32,300 / 221012: 18.5
V'= 52,800#

B.M. = 32,300 x 80 x 2/3 / 1/10 x 2210 '(18.5)2 x 12
= 2,707,000"#
Since the B.M. and shear for the East and West wall beams are
less than that used in the design of the interior girders, the
previous girder design will be used for the East West wall
beams.
COLUMN DESIGN
Sketches 18 and 19
A11 columns concentric - Axial Icaded.
Columns 1, 7, 29, 35.
Sketch 20
Maximum.load will be on columns 29 and 35 due to the crane on

the first floor.

Girder load 3 69,225#
Floor beam.load - 32,300#
Crane load = 3,000#
Column'Wt.(assumed) a SIOOQfi
N ‘ :107,525#

From R.C.D.H. table fi20

Use 10" x l4"-column.

Load taken by concrete 3 95,000#
Load taken by steel = 15,099fi

Mex. allowable load -110,000#

17

1F

'- I
r . ._
—
x ll ‘ _
, a a _
f a. H
v. M. i
do
I \n.h yr L
V II“ J 3‘
1 a , .. r . “1.:
x
1.1
VI A,
r .
. ..
,r _ x d .
r 1. .. .w
..
.
... .H
....\
A
\ ‘
,C
. t .1
T. .. . .
V
. m f \
J L _. -1 , . 3
‘ v I. _ , I p
! ullllliv 1211' i I l. r5111»! V i 05!... . ..
III‘ y :0 n!| I L

¢
‘
_
d
. _
v
.
IL »1
‘
V
.
a u :I . ll ‘ w y‘lr. 1-6.-..
_ r. . . .._
.. -. 2 11.13: 1.1
'
,,
V . H-
a
a
1
,n
.o, ._
”A; m
1-11-. -. - - - 1 L; 1- Q
. L . 1 . V
l a (I 1 . n.“ _
3 - r - .. .
a - , . -- . sq z I.
_ f
I . v.
.a

Use 4 - 7/8" circular bars.
Tie spacing - Use the least of the following conditions.

1. 16 bar diameters - 14.0"

2. 48 tie diameters 24.0"

3. least column dimension.= 10.0" Least
Use-%" ties @ 10.0.

Check column weight

10 x 14 x 14 x 150 - 210 ’ 0.K.
‘12?- ’ 0* --

Columns 2, 5, 4, 5, 6, 30, 51, 32, 33, 34.

Sketch 21
Maximum.load will be on columns 50, 51, 52, 55, and 54 due to
the crane on the first floor.
Girder load = 2 x 69,225 . 58,450#

32,300#

3,000#

I

Column wt.(assumed) a 3 50 ”

Floor beam.load

Crane load

N 177,250#

Use 14" x 16" column.

Load taken by concrete = 151,000#

Load taken by steel - 50,009fi

Max. allowable load = 181,000#

Use 6 - l" circular bars.

nee fin ties @ 14.0".

Check column weight.
131$=14x150=3270# _C_)_._K_.

Columns 8, 14, 15, 21, 22, 28.
Sketch 22
maximum.load will be on columns 22 and 28 due to the crane on

the first floor.

18

3 so

>VT'11 I.

V4“.
V

:3

 

a a a:

'9
D
1 $1

 

Girder load

69,225#

Floor beam 108d - 2 x 52,500 3 64,600#

Crane load - 5,000#

Column wt. (assumed) = 5,00

N 159,825#

Use 10" x 18" column.

Load taken by concrete = 122,000#

Load taken by steel = 19,009fi

Nbx. allowable load : l4l,000#

Use 4 - l" circular bars.

er-%" ties @ 10.0".

Check column weight.
_m__10118x14x150=2620# 2:};

Columns 9, 10, ll, 12, 15, 16, 17, 18, 19, 20, 25, 24, 25, 26, 2?.
Sketch 25
Nbximum load will be on columns 25, 24, 25, 26 and 27 due to
the crane on the first floor.
Girder load : 2 x 69,225 = 158,450#

Floor beam load = 2 x 52,500 ; 64,600#

Crane load : 5,000#
Column wt. (assumed) g 4:009fi
N = 210,050#
Use 16" x 18" column.

Load taken by concrete = 194,000#
Load taken by steel = 25,009fi
max. allowable load : 217,000#
Use 6 - 7/8" circular bars.

Use é” ties @ 14.0".

Check column weight.

Eifilé x 14 x 150 - 4200;“,4 Egg:

19

‘_.. ”AA...” . A

v-.. 4*-

: -

1

11
n-M"

, .\
fl
. .
. ..
A a
a.
1 .
\l...
A
A
- 1 a .,
2 a.” _
. ., _
to ._
. _,
w. A 2-!
q». A I .\
0’. .x
.L ,.
a,
11 -1 A ._
_

 

 

11.1 L

 

A A

II

 

COLUMN FOCTING DESIGN
Sketch 24
Column footing l.

Allowable soil pressure

6000 p.s.f.

Footing weight will be assumed 6 o/b of live load.

Hooked, deformed bars will be used in all footings.

Column size 10" x 14"

DL = 6,459&

TL1: 115,975#

Area =.£%§6%%§,= 19.0 sq. ft. required.

Use L: 4' - 6"

Net pressure : 107,525 :
‘20-'25-

A = 20025 sq. ft.

5320 p.s.f.

. _ . 14 22:: 22
3'1““ - 5330 ( x m)/ 5520 ((22)210.6 €32) : 30,000'#

d - M = 30,000 x 12 = 15.2

’K‘B 256x10

d 2 16.0" h 3 400

h/d

Use 16.0“

20.0"

Check weight. 20.25 x'gg 2 150 - 5060#

As 3, M I, 50,000 x 12

3;, 20,000 x .866 x 16

Use-%" circular bars.
A -.- 0.2 sq. in.

1.30 = 67’

Check Bond.

3 1030 sq. in.

Use 7 bars @.6.0" 0.0.

u = V‘ = Neg pressure (L2 - (q/Zd)zfll_x'%

262-13

053

u : 5C20 (20' 25 - 12' 2Q: 70 p. ..s is 00K.

11 x .866 x 16 x 4

Check shear.

V-

42, 560 = 1805 PoSoio

=4! a72dljd= ’168 x .866 x’16

20

Column footing 2.
Column size 14" x 16"
LL : 177,250#
DL.: 10,659fi
TL = 187,900#

Area : 187,900 : 51.5 sq. ft. required

4'
USe L: 5' " 8" A 8 32.2 sq. ft.

Net pressure = 177,250 : 5500 p.s.f.

13.1. .. ”500(‘11'21— x 21);! (Lag x x 17)‘ 56,100 #

d ; 56,100 x 12 .-_- 20.4" Use 21.0"

 

236 x‘14
d :- 21.0" h .-. 4.0" d/h = 25.0"
Check weight. 32.2 x 12,; x 150 = 10.10056 0.x.
As : 56,100 x 12 a 1.85 sq. in.

 

20,000 x .866 x‘ZI
Use 10 - %" circular bars @ 6.0" c.c.

Check bond.

u - 5500(32°2 - 21.8) = 50 p.s.i. O.K.O
"1527’x .866 x 21 x_4

Check shear

V = 5500 x 10.4 = 14.1 p.s.i. O.K.
‘224 x .865 x 21 —__-'

Column footing 8.
Column size 10" x 18"
11.: 159,825#
DL : 8, 40%
TL : 148,225#

Area = 148,225 a 24.7 sq. ft. required.
6,000

Use L = 5.0. A = 25.0 Sq. ft.

21

Net pressure : 159,825 : 5590 p.s.f.

, , 18 25 25 _ .- 2 _
8.1-1. = seam-Igr— x E) / 5590 (1.3.21 x 0.6 x g) = 48,500'#

d = 48,500 x 12 = 13.7" Use 14.0"
256 x’18

 

a = 14.0 h -.- 4.0" h / d = 18.0"

Check weight 25 x 18 x 150 = 5630;} O.K.
'12

As : 48,500 x 12 g 2.40 sq. in.
20,000 x .866 x‘I4

 

Use 13 --%" circular bars @,4.0" c.c.
Check bond.

11 .-. 5590(25 - 10.1) = 84.2 p.s.i. O.K.
2024 x .866 x ll’x 4

 

Check shear.

V = 5590 x 14.9 8 45.2 p.s.i. O.K.
38 zilfix .866 xfil4

 

Column footing 9.
Column size 16" x 18"
LL = 2l0,050#
DL = 12,609fi
TL 3 222,650#

Area : 222,650 g 57.1 sq. ft. required
Tum-

9

Use L8 6' - 2" A: 38.0 sq. ft.

Net pressure : 210,050 a 5520 p.s.f.
'1fl?77"

' 2
B.M. = 5520(EIZI‘ZPE x .32) 7’ 5520 ($12.24) x 0.6 x g) . 70,800:#

 

d : 70,800 x 12 = 22.5" Use 23.0"
256 x 16
a = 23.0" h = 4.0" h 7’ d = 27.0"

Check weight. 58 x 27 x 150 = 12,800# O.K.
' IE. '___-

22

"-7 —_—— _,_

\.

 

 

 

fil-l|l-l r - 1- n . |.1 i . |l| u
.
g
1
4
_ _
, V n ‘ll 11
. yr
4
_
~
_ 1"... 'laflflltlnla .J
_
A.
«I __
\fi 7
. n
_
O _
_
.l: _
. (L 1 p .L
7
7 4
_
_
.

-- Q - f- 5--.}--.
\ H
_
.\. _
_
m
.
_
.
m
i
i
. 1
.
,_ n
.
7
1,
. A

As : 70,800 x 12 = 2.15 sq. in.

 

Use 11 - g" circular bars e 6.0" c.c.
Check bond.

u a 5520(38 - 26.6) = 45.7 p.s.i. O.K.
173 x .866 x 237x 14’

 

Check shear

v = 5520 x 11.4 : 12.9 P0801. O.K.
2487x.366 x<Z3 -1

 

BLSELEMT FLOOR DESIGN

 

Due to the expected use of the building the basement floor will be
designed to carry heavier loads than the ground floor.

Lax. wheel load expected : 6000#

Bearing area assumed : 10 sq. in.

Unit pressure : 600 p.s.i. £Efi2_ 1350 allowable.
A subgrade of 6.0" of conpacted stone will be prepared to receive
the slab. A 6.0" plain concrete slab will be poured in alternate
sections 20.0' x 20.0'. A %" filler will be used at the junction
of the floor slab and all other members such as columns, walls, etc.,

and between the 20.0' sections.

RETEIHILG LILL DESIGN

 

Sketches 25, 26 and 27
Since the driveway is to be on the West side of the building the
basement wall on thrt side will be a retaining wall in order to
eliminate damage to the wall from driveway parking and traffic.
The wall will be made 16.0' high, and 2.0' of the wall will be
below grade as a precaution against frost damage. In order to
compute the earth presaure against the wall, it is necessary to
compute the equivalent surcharge for the maximum loads expected

on the driveway. The largest vehicles expected to use the drive-

23

 

ii!

I..si.|.ll'

vvq--

.f - -~~a4

‘ ‘gYu

1.

bin. .

L n.
'
/
p
O!
.
I
n .
/\
x
j
.
. r
V
.
o I
..
.
t
_
.
II“ III. 1.1.: .l-‘ 1w l. .11 [4.1 lfllllIi‘I
.. / . r /
I. . . , /
. (I / / / I .
II .1 I. ll .
I i
.1 . ..... v. . / / I. / . .
y
1
.
r
O V
v

A——'..-‘ v—

I.

-_._.--__.n._. J

k.. .

-

way tare semi-trailers as shown in Sketch 25. The maximum condition
expecrted is shown and the axle loads of 20,000# are considered to
act LLniformly on an area of 240 sq. ft.

.Area = 40 x 6 = 240 sq. ft.

Unit load : 80,000 : 355 p.s.f.
240

'Weight of earth assumed 120 #7bu.ft.

355 : 2.78' Use 3.0' surcharge.
'IEU

The magnitude and point of application of the earth pressure will be
determined by use of Rankine's Theory of Earth Pressure.
Sketch 26

Angle of repose for clay loam 36° 53'

P:%§wh(h/2h') l-Sinfl

l f'Sinlfi

 

 

P = % x 20 x 16(16/6) 1 — .602 = 530034;
1 7’ .602
h2 5h 1 (18)2/s 18 3
Y': = a} x = 6.75' from.base
5 h 3(18/6)

Mo _-, 5800 x 6.75 = 35.800'#

To determine stem thickness at base

M .-. Rbdz

d - ( M)% - 5300 7 2 % - 10.5" USS 12.0"
- a 4W) -

d : V : 5500 : 8.52"

 

'x'r'fi 12 x .8661: 60
Stem.thickness taken as 15.0". This will provide sufficient
cover for stem steel. No batter will be used on the stem.
Design of base slab

The following dimensions will be assumed:

Toe : 3.0'

Heel : 5.0'

Thickness : 2.0'

24

Heel

120 (5 x 16) 9,600 x 6.75 . 64,800

r
IA.
"

g?
u

1.25 x 18 x 150 - 8,880 x 5.68 = 12,200

g?
u

5 x 2 x 150 = 1,500 x 6.75

10,150

W3 : 3 x 2 x 150 : 900 x 1.5 1,350

RV ' = 15,380; 11 88,500»#

88,500 55,800 = 52,700v#

52.700 3.43' from toe

15,580

Therefore, the Resultant acts within the middle third of the
base.

4.63 - 3.43 = 102'

P (1 69 = 15 880 1 6 x 1.2

I’ Z 5") “§f§3‘ ( 4/'-g-25-)

6

S

: 2957 p.s.f. max.

367 p.s.f. min.
slab
Shear A-A Moment APA

9600 x 2.5 24,000

1500 X 2.5

_J{11,100& J/ 27,750!#=

4,590

5,750

367 x 5 = 1835 x 2.5

5,840

 

1400 x 2.5 - 3500 x 1.67
" 5L‘535# .. 10,4300#
Shear A-A : 11,100 - 5335 = 5765#

Moment AFA = 27,750 - 10,550 = 17,320»#

d : V' = 5765 = 9.25"
V35 86 x .866 x 12
%I 7 320%
M - 1 n
d : - Lg.) : 8.58
(F) 236

Base slab thickness changed to 16" (d a 12.0"). This will

provide sufficient cover for steel.

AS = 1 17,320 = 0.0832 sq. in./ino
20,000 x .866 x‘rz

25

 

Use 1" circular bars @-9.0" c.c. As = .0872 sq. in./in.

Steel placed in top of heel slab.

 

Toe slab

Stem

_ 5765 _ - -
v.12 x .866 x 12’- 46.2 p.s.i. O.K.
-46.2 x 9 - 132 i 0 K
u - p08. . o .
W
Shear B-B Moment B-B
/ 900 x 1.5 = ,1 1550'#
2117 x 3 = 6351 x 1.5 : 9540
840 x 1.5 = 1260 x 1.0 : 1260
.. 7611# -10,800'#

Shear B-B = 7611 - 900 = 6711#

Mement B-B a 10,800 - 1550 = 9450'#=

 

d = 6711 = 10.8"
60 X .866 1‘12

d = 9450 %-= 6.34"
(7%)

Toe slab will be mede the same thickness as the heel slab
16.0" (d = 12").

As . 9450 = 0.0454 sq. in./3n.
20,000 x .866 x‘Iz

 

Steel should be placed in the bottom of the toe slab. Stem.ateel
will be carried around for reinforcement in the toe slab. This
will provide sufficient reinforcement for the toe slab and it

will also provide enbedment length for the stem steel.

The stem will be 15.0" uniform thickness. ioments are computed

for every 2.0' of stem.from.which the cut off points for the stem

steel are determined.

 

E. E M'f’ As sq. in./ft.
2 80 55.5 0.00309
4 520 427 0.0205

6 720 1440 0.083

Rf _.-._

.«7

 

~41

In... AIL.

8 1280 3420 0.197

10 2000 ' 6670 0.585
12 2880 11,550 0.666
14 5920 18,400 1.06
16 5500 28,500 1.64

Use 4 - 5/4" circular bars a 5.0".

From this table the cutoff points can be determined. Sufficient
length will be added for embedment.

L 3 £32 : 18.0"

For points of cutoff see Sketch 27.

In addition to the reinforcement steel, temperature steel will be
placed in the stem, 0.15 0/0 of cross section.

As = 15 x 12 x .0015 : 0.27 sq. in-/Tt.

Use é” circular bars @716" c.c. front . 0.15

Use %" circular bars @ 16" c.c. back = 0.15

0730' sq. in./ft.
Factors of safety

Overturning = 88, 500 ; 2.47

I

Sliding = 15,580 x tan 25° = 1.56
5500

Crushing = 6000 _-,- 2.03
2957'

The retaining wall will be continuous for 60.0' sections. In
order that the wall will frame into the other members, the wall
will be poured flush with the outer face of the columns and wall
beams. This will eliminate modification of the wall beams and
columns along the West wall.

BASEL'ENT 1A_I_L_S_
The North, South and East basement walls will be poured concrete

walls. Temperature steel will. be placed in the walls, but no other

27

illll.

 

A
D. 1111lll'rlli on.-- urlllvllllrl.‘lll 1.. .. .I'- I1 a1
. I, V .4
. g

. ,\ v . .r
.l. U.

i I

I I ll 1 1 .1111 ill I

 

 

 

 

.‘II‘IIIIII. .. .1. .Dllllllill.,5|.'n [I'lll‘lt’iltL - ‘

. "p I

\
{CD

/.

a

L £lffi

. l
|

 

_—_-¢

reinforcement will be used. The walls will be independent of the
enclosing members such as the upper wall beams and columns. There-
fore, it is safe to assume that there will be no transfer of loads
from.these members to the walls. The only direct load on the walls
will be the earth pressure due to back fill.

The wall height will be 14.0' - 29" = 139" for the East wall
and 14.0' - 20" = 148" for the North and South walls.

The wall thickness will be taken as 8.0" in order to provide

“‘"i?

a water-tight wall and to resist any stress caused by the back fill. 5
The only Opening in the basement walls will be a door in the

South'wall. The door will be 14.0' high.and 12.0"wide. 'Wall Beam

-#12 will serve as the top of the door hence no lintel is required. 1

Wall temperature steel

8.0 x 12 x .0015 = 0.144 sq. in./ft.

Use-%" circular bars @ 16" c.c. in inside wall surface.

BASEMENT WALL FOOTINGS

 

For a 1.0' section of wall

Thickness a 8.0"

 

Height : 148.0"

Weight -.- 8 x 148 x 1 x 150 - 1255,5/rt.

Wall weight -_- 125#/£t.
Footing weight (assumed) : 200#/ft.
T.L. : 1455#/£t.

Area : 1435 : .239 sq.ft. required.
ififififi

Since such a small area is required, a practical size footing will
be used without further investigation. Use a footing 2.0"wide and
8.0" thick. No reinforcement is necessary, but temperature steel
will be used to prevent cracks from.0pening.

Use 73;" circular bars a 9.0" c.c.

28

 

Egg; POUR SCHEDULE
All columns and wall footings will be completed in one pour.
Dowels of the same diameter as the vertical steel in the joining
members will be placed in the footings. The dowels will extend
at least 25 bar diameters beyond the footings. KeywayS'will be
placed in the wall footings while the concrete is still in the
plaster state. This will provide added anchorage for the walls to
the footings. ' 57"
The basement walls will be poured in two lifts. This is deemed
necessary for proper puddling. The first lift willbe 8.0' high, and
a keyway will be placed on the top surface of the first lift. The
second lift will complete the wall to grade. Dowels will be used to
anchor the wall to the floor slab. These dowels should extend 25

diameters into the wall and into the slab. This will necessitate

 

bending the bars 90°. The temperature steel in the wall will extend
25 diameters into the joining columns in order to provide pr0per
anchorage with these members.

The stem of the West wall, the retaining wall, will be poured
in two liftS'with the keyways provided in the same manner as for the
other walls. The wall will be poured in two 60.0' sections with an
expansion joint between the two sections. The expansion joint will
be a keyway, doweled together, and with-%" filler between the two
sections. This type of joint should provide a flexible, water-tight con-

nection. The stem will also be doweled to the west wall columns and

wall beams.

STLIR DESIGN

Stairs located and dimensioned as shown in Sketch 28.

LL = 100 p.s.f.
DL .1 75 p.s.f.
TL = 175 P.8of.

29

 

Use 6.0" slab.

Total rise from top of floor to first lending - 7.0'.

Stairs with 10.0" run, 6" rise, and 1.0" nosing will be used.
horizontal projection of stairway : 10.0'

Landing platform 4.0' long, 8.0' wide, 6.0" thick.

The stairway slab and platform slab will be designed as one slab,

 

length ; 14.0'.
v : wl :- 175 x 14.0 ; 1225;;
‘2' ‘2

8.1.1. : wl2 ; 175 x (14.012 3 55507}
'10 10

d(for shear) : 1225 = 1.97"
125x .866 x 60

'If-E_236— N

 

d(moment) : 3.82"

Use d : 4.0"

A8 .. 54:50 I 12

- 0.594 s . in. ft
20,000 x .866 x 4.0 q ‘/ .

 

 

Use %" circular bars @ 4.0" c.c.
Check bond
u " 1225 ' 75.2 p.s.i. O.K.

' 4.7’: .866’x 4 -
The lower and upper stairway slabs are fixed to the landing slab
along the inside edge. The landing slab, 4.0' long, is fixed in the
East wall. The stairway slab reinforcement is continued through the
landing slab and anchored into the wall with hooks. This same rein-
forcement will be run along the 8.0' length of the landing slzb.
A beam will be designed to support the inside edge of the plzt-
for“ S] b. This beaUlwill be supported by a short colunm.at each end.

Stairway slab reaction 1225 g%/'

Platform 51:b reaction - 4 X 75 = 150i/"
'“‘Z“'

Total load on 15nding beam : 1575:fi/'
Assume stem weight : 50-#/'

30

 

T.L. : 1Q05,31/'

V': 1405 x 8 = 5520#

fiith web reinforcement

 

bd : 5620 : 35.6 sq. in. required
00 x .275
Let b : 6.0"
d : 9.0"
q/a.0 ; 11.0"

d-t : 5.0"
Check stem'weight

5 x 4

‘1117'x 150 = 21.0% O.K.
8.x. ; 1405 x 64 = 11,250”#
8
AS : 11,250 x 12 : 0.815 sq. in. required

 

20,000 x .92 x79
Use 5 --%" circular bars @ 2.0" c.c., two rows.

Check bond

11 : 5620 2 85.8 p.s.i. O.K.
7.9 x 0.92 x 9

Check shear

Max. shear at end of beanlz fig 582” = 113 p.s.i.
6 x .92 x 9

Web reinforcement is required.
Max. center shear will be taken as 25 0/0 of max. Shear.
Vc : 0.25 x 113 : 38.3 p.s.i.
Total shear taken by stirrUps
113 - 38.3 : 74.7 p.s.i.

115 - 60.0 = 55.0 p.s.i.

74.7 e 55 x = 54.0"
'75?’ '7:
v = 55 x 54 = 900#

2

Using %” circular stirrups A.= 0.05 Sq. in.

31

 

i

A

fl
_
i
A
_
_
.

W

it‘ll] ti ll til! .

lllllll 01+-

V---..,i

l“" -~.—-

3111.... C III. .

1 1.1.7.1,

1.31.3.

all

 

1‘. -H.‘

_.__ m*.1--.—- —

0.1 x 16,000 = 1600# taken by each stirrup
One stirrup required.

max. spacing : g_: 4.5"

Therefore 34 : 7/

Use 8.0 stirrups @ 4.5" c.c. at each end.

Landing Post Design
Two 3" x 2 3/8" - 5.7% standard steel columns will be need. he

Allowable concentric load for 7.0' height : 9900# l9;§; ;
A 3/4" - 4 x 5 bearing plate will cap each end of the column . 1
maximum beam reaction : 5620#
.424
566059

Column weight = 7 x 5.7

[wk—w

Total axial load

5660 3 283 p.s.i. compression stress upon floor. O.K.

1X5

Allowable compression stress = 1350 p.s.i.

 

Therefore no footing is required. The landing beam.posts will

 

be encased in concrete in order to provide a fireproof structure.

CRANE SUPPORT DESIGN

Sketch 29

A three ton overhead crane is to be placed on the first floor
along the East wall. A four wheel crane will be used, crane to be
designed by owner. The crane will operate on two rails which will
be supported by steel columns. The columns will be embeded 12" into

the concrete of the floor system, and the t0p ends of columns will

be tied together by lateral bracing.

Six inches clearance will be allowed for the crane wheels and
2.0' clearance will be allowed for the crane mechanism.

Maximum load on the column will occur when one crane wheel is

32

 

positioned over the column bracket. Due to insufficient data con-
cerning the crane wheel spacing, the steel columns will be designed
for the condition that one wheel is positioned over the bracket and
the resulting moment will be computed for the weak axis of the column.
The brackets will be placed so that the actual moment will be about
the strong axis of the column.

A 5 x 5" - 18.5# light weight column will be checked for maximum

moment about the weak axis.

Crane load : lSOOfi/wheel

Moment ; 1.5 x 6 : 9.0" Kips.

. '. ‘.__-..,-_-.

Equivalent direct load : 9.0 x 1.56 /'l.5 a 15.36 Kips.

fag/'fb shall not exceed unity.
'FI '55

“‘ .

l.‘ 144 : 112.5
r 1.28

FA.: 10.81

fa :§::

FB

20.0

ft §!: 9.0 : 0.91
9.9;

2.82 0.91_
1m XW 0.306 1633 than unity O.K.

Allowable axial load for unsupported length of 12.0' equals

49.0 Kips. Therefore the column selected is adequate.

 

Bracket Design
From.A.I.S.C. handbook
P = 1.5k
1 : 6.0"
3/1" rivets
3.0" pitch

Allowable stress on one rivet = 15.0 x .4418 = 6.63 Kips.

33

 

n
a
. .
.
.
‘ .
. .
.
.
I.
— x
.
II
.J.
.
.
.
A
A

 

. 1 l‘ui‘lll. I'll-w nil. Jenxr. ' . ,.‘...Qvu

I... (r
.u. .1.
, v
A, + . ,,
L
1 f
.. m _ - -
11 ..
a 1.
. . J
1.. . C .1
.u1 _. , W
s. T 1
£1 to. .1. . .1144:
P dv *0 ..‘..|aa|n!. 0. ~

5:

‘21

c : P ; 1.5 ; 0.103
233 216.63

Use 3 rivets in each of two rows.

Use-é" plate and 2 x 2 x-%" angles for bracket.

The columns will be 15.0' long, 1.0' embeded in the floor
system. The brackets will be placed 2.0' below the tops of the
columns. The tops of the columns will be braced as shown in
sketch 29. The columns will be tied together with 3 x 2 3/s" -

5.7# standard I bea s, and the top cross bracing will be 2 x Z x

‘L F v‘)... “I;

3/6" angles. Rail size and fastenings will be determined after
the crane has been assigned.

DESIGN 9: DRIVE

 

Sketch 30 H 111‘

A two lane concrete drive will be placed parallel to the West
side of the building. The drive will be $.0' from the building wall
and will extend from the existing city street to the rear entrance

of the building. The slab thickness will be determined by the use

 

of Older's Theory.

fluxural strength of concrete

wheel load

slab thickness
1

.l ..
(105W)2 = (105 x 10,000)2 = 4.47"

s
‘w
d
d 750

A.6.0" slab with thickened edges will be used. No reinforce-
ment willbe placed in the slab. A transverse expansion joint'will
be placed at a point midway along the drive. weakened plane
contraction joints will be placed at points midway between the
expansion joint and the ends of the slab. The total width of the
drive will be 16.0', laid in two 8.0" slabs connected by a doweled

longitudenal joint. .A gutter will be placed in the outside edge of

34

 

the left slab, in order to carry the run off from the slab away from the

basement wall. The drive will be widened to 26.0' at the curve in order

to facilitate entrance into the basement.

 

"_£€-' "_

 

55

 

 

! ‘Illit. I

1.. 1_ -_.

‘I :

 

a

 

b

-‘..a

. —vv~~~. ~pv»-.«"-‘_., -_.-

‘1 r' - 7y r'~.

 

+£",v‘ Jen“?

C. CH” f V 1.11.-

COST 25 MATERIALS ESTIMATE

 

The unit prices for the construction materials used in this
estimate were obtained from the May 1, 1947 issue of "9ngineering
News Record". The unit prices used are those listed therein for
the Central Ohio area. Referencewas also made to "Uonstruction
Estimates and Vests" by H.E. Pulver, published by MbGraw-Hill

Book Company in 1940.

Concrete Block Wall it":

4940 blocks G $0.17 = $3839.80 1;
Cement Mertar. 1:4 mix used
0.25 cu. yd./100 blocks

49.40 x 0.25 : 12.35 cu. yds.

 

Cement

12.35 x 6.75 @ $2.62/bbl. = e 54.70
Lime

12.55 x 70 o $30.00/Ton : 3 15.00
Sand

12.35 x 1.0 @ $1.15/Ton = 51 21.10
water

12.55 e $0.10/eu.yd. .-.- s 1.50
Total Wall Cost .-.- $929.90

Concrete

3000# concrete will be used for all structures. The
columns will be poured with a very plastic mix; 131 3/4: 2%
mix by volume, 6.0 gal./back. The remaining sturctures will

be poured with a 132:3 mix by volume, 6.0 gal/sack.

 

2.2; fit; Absolute Wt.
Cement 5.10 94#/eeek 62.4 x 5.10 . 194#
Sand 2.65 110#/cu.ft. 62.4 x 2.65 : 165#
Gravel 2.65 lOO#/cu.ft. 62.4 x 2.65 = 165#
Water 1 .00 62 .4#/ou .ft .

36

 

The yield will be computed for a 1.0 sack unit of each mix.
1:233 mix by volume.

Cement 94 a 0.484 cu. ft.
194

Sand 11%;}3: 1.333 cu. ft.

Gravel 100 x 3 = 1.818 cu. ft.
"T6?"

Water 6.0 : 0.802 cu. ft.
7.48

Yield in concrete = 4.437 cu. ft.

. _'_"o“(7"‘q
.
1

1.1 3/432% mix by volume. 6.0 gala/sack

Cement 94 g 0.484 cu.ft.
T91

 

Sand 110 x 1.75 : 1.167 cu. ft.
165

 

Gravel 110 x 2.5 3 1.515 cu. ft.
6

Water 6 .0

" 00802 cu. ft.

Yield in concrete - 3.968 cu. ft.

First floor slab. 4800 cu. ft.
T-Beams 2200 cu. ft.
Girders 1800 cu. ft.
Column footings 2400 cu. ft.

Basement floor slab 4800 cu. ft.

Basement walls 2200 cu. ft.
Retaining well 3900 cu. ft.
Wall footings 400 cu. ft.
Stairway 75 cu. ft.
Driveway 1300 cu. ft.

volume of 1:233 concrete = 23,875 cu. ft.

23:875 - 5"80 units
Tat-57"”

37

 

Cement, l x 5380 : 5380 sacks

Sand 1.333 x 5380 : 7170 cu. ft.

Gravel 10818 x 5380 = 9780 cu. ft.

Columns 800 cu. ft.

800 cu. ft. of 1:1 3/432% mix required

_8_0_9__ = 202 units
5.958

Cement 1 x 202 ; 202 lacks

Gravel 1.515 x 202 -

Total cement : 5582 sacks 5.5 2.62/bb1. : 8 5550.00
Total sand 3 7406 cu. ft. 5.8 1.15/Ton ; s 407.00
Total gravel = 10,085 cu. ft. 0,5 2.50/Ton ; 5 1260.00
Total water = 6.0 x 5582 @.$ 0.15/1000 gal. ; 5 5.00
Total concrete cost = 3 5333.00

Reinforcing Steel
Floor slab and stairs
T Beams
Stirrups
Girders
Stirrups

Columns

Column footings

Retaining wall

Basement wall

Wall footings

38

306 cu.

256 cu 0 ft.

50,000!

7,0001
2,520'
4,800'
6,000'
1,800'
1,500'
2,800'
5,500!
1,040'
4,550'
2,880'
5,550'

560'

ft.

-%" Circular bars

1" Square bars

~%" Circular bars
1" Circular bars
%fi Circular bars

7 8" Circular bars
1" Circular bars
Circular bars

" Circular bars
1" Circular bars
3/4" Circular bars
%" Circuler bars
%" Circular bars
%" Circular bars

in

 

 

%” Circular straight bars 30,000'
2,800!
2,880!
5,550!
550!
59,500!
59,500 x .668 = 25,500#
25,500 5.5 5.20/100#r= 5 850.00
5” Circular bent bars 2520!
6000!
5500!
15,820!
15,820 x .668 : 9250#
9250 0.5 5.60/100#=: 5 555.00
5/4" Circular bars 4550!
4550 x 1.502 5.3 5.00/100#== 5205.00
7/8" Circular bars 1800!
1800 x 2.04 5.3 5.00/100#:= 5110.00
1" Circular bars 4800'
1500!

1040'

 

'7540!

7540 x 2.57 = 19,600#

19,500 5.5 5.00/100#== 8 588.00
1" square bars 7000'

7000 x 5.4 0.5 5.00/#=s 5 714.00
Total reinforcing steel cost = 5 2802.00

Filler

100 sq. ft. @.$ 0.05/sq.£t. : 5 50.00

39

 

 

Structural steel and rivets

Steel 2755# o 5 4.52/100#~- 5 120.00

Rivets 500 8,8 5.25/100 _ 8 15.75

Total

5 155.75
Lumber _

22,250 ft. B.M; 8.5 100.00/1000 ft. B.M. - 5 2225.00
Roofing

21,120# 0.5 2.26/100# : $480.00

T
E

Hardware
9 x 8 10.00/Truss : 3 90.00

Total Estimate = s 12,045.55

 

 

 

From this estimate it is believed that all the materials required
for construction of the building, with the exception of doors, windows,

and miscellaneous fixtures, could be supplied for approximately 8 12,000.