THE DESIGN OF A BUS TERMINAL FOR
LANSING, MICHIGAN

THESIS FOR THE DEGREE OF M. S.
W. B. Edwards
1932

 

 
 

SUPPLEMEiTARY
MATERiAL

3N BACK OF BQOK

The Design of a Bus Terminal

for Lansing, Michigan .

A Thesis submitted to
The Faculty of
MICHIGAN STATE COLLEGE
of

AGRICULTURE AND APPLIED SCIENCE

:”By‘:
W. a; (Egan
candidate for the Degree of
Master of science.

June 1932.

Acknowledgement.
The author of this thesis wishes to acknowledge
the valuable assistance of Professor C. L. Allen in the
preparation of the material presented.

W.B.E.

95049

 

 

 

Table of contents.

Introduction

I Description of the building and the materials used.

II Design
Roof beams
3rd floor beams
and floor beams
1st floor beams a sbutments

miscellaneous

III Drawings

conclusion.

 

 

 

Bibliography

Structural Engineers Handbook - Ketchum
Architectural construction -. voss
Structural Aluminum Handbook

A. I. S. C. Handbook

Building code - city of Lansing
Articles on welding - mngr. news - necord
Reinforced concrete construction - Hool
Theory of structures - apofford

unit Prices - nngr. News Record

Handbook of cost Data - Gillette
Estimating Building costs - Dingman

Estimating Building costs - sarnes

 

 

 

 

Introduction

 

 

 

The thesis herewith presented was chosen for
two reasons. rirstly, it presents two features of civil
Engineering construction; namely a building and a bridge,
and secondly, it offers a structure which has long been

needed in the city of Lansing.

 

 

 

Part One
A Description of the building

and the materials used.

 

 

 

 

The structure, the design of which is presented
in this thesis, is to be used as a bus terminal and office
building. The lower floor will serve as a bus garage and
loading station and as a waiting room for the bus passengers.
The upper two floors may be used as offices for the various
bus companies occupying the building, club rooms for the
bus drivers or offices for companies other than the bus
operating concerns.

The building is situated on a bridge spanning
the crand niver at the foot of mast Ottawa Street. It is
one hundred and eighty feet long, seventy feet wide and
three stories high.

The design was undertaken with the intention of
securing a light weight, fire-proof structure.

The roof of the building is composed of tar and
asbestos built up roofing on corrugated aluminum sheets
supported on lightweight steel trusses which are in turn
supported on steel 1 beams. The side panels are fifteen
feet by fifteen feet with the trusses spaced at five foot
intervals. The center panels are ten feet by fifteen
feet with trusses fixed at the middle of the ten foot dis-
tance. A plaster ceiling supported on metal lath is hung
from the underside of the roof frame to provide a dead
air space for insulation and to provide fire protection
for the roof itself.

The floors are of the American Institute of
Steel construction Battledeck floor construction "T"
sections, composed of steel plates welded to small I

beams spaced twenty-four inches center to center. These

 

 

 

 

 

 

r,‘

sections are welded to the floor frame and are protected
from fire on the under side by patented fireproof blocks
which are applied after the floor is laid. The surfaces of
the upper floors and of the waiting room are covered with
linoleum, laid directly upon the steel plates. The floor
in the garage sectidn of the first floor is covered with
two inches of asphalt which serves as a cushion for the
heavy loads and also deadens the noise.

The exterior walls of the building are of hollow-
tile faced with limestone with steel sash windows and
aluminum spandrels.

The interior walls are of hollow gypsum.block,
plastered on one or both sides as necessary.

The columns, like the rest of the frame, are of
rolled steel sections being fastened by means of welded
Joints.

The piers, footings and abutments are of rein-
forced concrets, the west abutment having two Openings
through it to allow for outlets for storm sewers which
are laid down Ottawa Street.

The elevator is a one thousand pound steel cab
type with an overhead drive using a one to one wrap traction
system. The machinery and cab are supported by means of
steel columns, and the shaft is of six inch terra cotta
tile blocks.

The stairs are of steel with tile covered treads.

While a heating system for the building has not
been designed, the use of aluminum spandrels provides con-

venient space for the new convector type heaters in the

 

 

 

 

offices and waiting room, while unit heaters would probably
be the most satisfactory in the garage section. Since the
building has no basement, the most desirable method of
heating would be to purchase heat. This could be conven-
iently done, as the city heating plant which supplies steam
to downtown buildings is on the adjoining property.

Part Two

Design

 

Roof beams

 

The first step in the actual design of the build-
ing is the design of the roof system. The roof itself is
composed of tar and asbestos roofing on corrigated aluminum
sheets. The panels are 15 feet by 15 feet with light steel
trusses 5 feet 0. to c.

The load on the roof is as follows:

Weight of roofing - 2 pounds per sq. ft.
Weight of sheet aluminum - 1 pound " ” "
Live load (Lansing Bldg. code) ~_§2;Pounda n u w

The total load is 35 pounds per sq. ft.

The weight of the suspended ceiling is 10 pounds
per square foot making a total load upon the roof system
of 43 pounds per square foot.

The allowable load on s a S 16 gage corrugated
aluminum supported at 5 foot intervals is 39 pounds per
square foot which makes this type of roof safe to use.

The framing plan and the design of the roof beams

follows.

note: The number of inches of 3/8" fillet weld required
for any Joint equals the reaction divided by 3000.

 

 

kflN kON

 

DESIGN OF BEAM TYPE R.B.N.
Load a each truss point
15 x 5 x 33 z 2475# 15 x 5 x 10 : 750?
ungx : 3225 x 5 x 12 a 193,500 from a to b
Assume wt. of beam 2 6# /ft.

Mb 2 6 x 15 x 15 x 12 - 2025"#
‘8

Total M - 195,525
3 = 195E525 = 10.9 use 7" 17.5? std. I
s - 11.11
DESIGN OF BEAM TXPE R.B.E.
Ceiling = 10 x 5 =‘ 50
Load = 33 x 5 a 165? per lin. ft.
Assume wt. of beam 5? per lin. ft.

Total load -220f per lin. ft.

 

 

1&8 3
M = i x 15 x 15 x 12 - 74,500"#
I
8 = 74 500 = 4.15 use 5", 10? std. I
mime
s = 4.84 ‘
DESIGN OF BEAM TXPE R.B.M.
15 x 5 x 10 “ ‘750 0
con Load = 15 x 5 x 33: 2975
Mnax - 3225 x 10 x 12 = 96,750
1"
Mb = 5 x 10 x 10 x 12 a 750 use 6" 12.5? Std. I
"""'5‘
577555 a = 7.27

s - 97 500 - 5.42

‘ mimic

 

DESIGN OF BEAM TYPE 3.3.5. assume wt. = 4%
Load - 47 x 2.5 - 117.5; /ft.
M’

117.5 x 11 x 11 x 12 - 21,3oo"#

 

 

*s
3.: §5fggg - 1.19 use 3", 5.7“ std. I.
s - 1.67
DESIGN OF BEAM TYPE 3.3.1. ass. wt. - 4"
Load a 47 x 5 - 235 r /ft
M - 235 x 158x 15 x 12 - 79,300
3 - 79,300 I 441 use 5", 10% std. 1.
s = 4.84
DESIGN OF BEAM TYPE R.B.J. assume wt. I 5%
Load - 45 x 5 - 240; /ft.
M = 240 x 11 x 11 x 12 - 43,500 "#
= 1:388 = 2.42 use 4" - 7.7 std. I.
s - 3.0
DESIGN OF BEAM TYPE R.B.Q. assume wt. 5" /ft

Load - 45 x 5%.: 254 f /ft.
m = 254 x 11 x 11 x 12 - 4sooo"# use 4"x m7§ std. I.

 

s
8.5e0
s - 45,000 - 2.57
DESIGN or BEAM TXPE R.B.V. Assume wt. = 5% /ft

Load - 48 x 5 - 2405 /ft.

M 240 x 11 x 11 x 12 8 43,500 use 4", 7.7f std. I.
B .

s - 3.

i
(0
II

43500 n 2.42
18000

 

DESIGN OF BEAM TYPE R.B.R. assume wt. 3 6#
195,525 5 = 10.9 use 8" - 18.4? ltd. I.

Gone Load

M..:

B ' 4.22

DESIGN OF BEAM TYPE R.B.T.
cone Loads I 3225 wt. of beam = 8?
moons I 3225 x 2 x 12 = 387,000

 

Mb I 8 x 208x 20 x 12 I 4,800
Mmax ' 391,800 use 10" 25.4? std. I.
S - 391500 = 21.7 s = 24.42
“15000
DESIGN OF BEAM TYPE 5.3.3. assume wt. of beam 55 /ft

cone Loads I (assume 2' supp. by beam)
-43 x 2 x 5 = 430?

MC: 430 x 5 x 12 - 25,800

Mb I 5x16 x 16 x 12 = 1 920

 

s """'
"max I 27,720 use 3" 5.7f std. I.
s - 27720 . 1.54 s . 1.57
15050 '
DESIGN OF BEAM TYPE E.B.U. ass. wt. . 5# /ft

cone Load (assume 6' supported by beam)

- 43 x 5 x 5 - 1,290

 

M = 1290 x 5 x 12 - 77,400
Mb I 5 x 16 g 16 x 12 : 1,920 .
Mug; 3 79,320 use 5", 10? std. I.

S I 79320 I 4.42 s I 4.84
10000

 

DESIGN OF BEAN TYPE R.B.F. assume wt. of beam - 5# /ft
Uniform Load 33 x 2.54-10 x 2.54-5 - 112.5? /ft
M I 112.5 x 15 x 15 x 12,= 38,100

 

8
S I 38 100 I 2.12 use 4", 7.7? std. I
‘Isfooc'
8-3.0
DESIGN OF BEAN TYPE R.B.w.' assume wt. of beam 7? /ft

Cone. Loads I (10x7.5x5)+(33x7.5x5)+(10x915)+33(9x5) I 3543
Moon I 3545 x 5 x 12 I 212,700"#
Mb 4 7 x 15 x 15 x 12 - 2,351"#

 

8
“max I 215,081 "#
3 3 215,081 : 12.0 use 8”, 18.4? std. I
’ s I 14.22 '
DESIGN OF BEAM TYPE R.B.K. assume wt. of beam 5? per ft.

cone. loads I (10 33) (5) (9) I 1935
Moon I 1935'x 5 x 12 = 116,100
Mb 8 5 x 15 x 15 3,12 8 1,690
Ev Nmax - 117,790
3 I 117 790 I 6.55

“IstUU use 5", 12.5? std. I
s I 7.27
DESIGN OF BEAM TYPE R.B.L. assume wt. of beam 53 /ft

conc. loads 43 x 6.5 x 5 I 1400
Moon = 1400 x 5 x 12 I 84,000

Nb‘= 5 x 15 x 15 x 12 - 1,590

N = 85,690
3 I 85 690 I 4.76 use 5f, 10? std. I

Isfooo

' - 4e84

 

DESIGN or BEAM TIPE N.B.p. assume wt. of been I 6f /ft.

Gone. loads = 43 x 6.5 x 5 I 1400

n,on,= (2100 x 10 - 1400 x 5112 = 155,000 "r

Nib-6x20120xl2I
8

“max
8 = 171500'a 9.55
‘18000

3,600

171,600 “I
use 7“ 15.3? std. 1.
s = 10.34

DESIGN OF 52AM TIPE H.B.A. assume wt. of been I 5? /ft

uniform load I43 x 2.5 I 108

wt. of beam I 5

M = 113 x 18 x 18 x 12 = 55,000 use 4" 8.5f std. i

s
S = 55000 I 5.05
10000
DESIGN 0r BEAM TxPE N.B.B.

Uniform load I 43 x 5 I 215f

 

 

wt. of beam ' 5f
220
M I 220 x 18 x 18 x 12 I 107,000
8
3 = 107 000 I 6.5
“rel-005
988100 or BEAN TIER 8.5.0.
uniform load I 43 x 25 = 108
wt. of beam = 5?
m = 113 x 13 x 13 x 12 = 50,700

 

8
S I 50700 I 2.82
10000

B 8 3.15

assume wt. 5% /ft.

use 6", 12.5? std. I.

s = 7.27

assume wt. 5f /ft

use 4", 7.7? std. 1

a = 3.0

 

DESIGN OF BEAM TXPE h.s.0. assume wt. = 10#

suitors load = 220s

 

‘M = 220 x 13 x 15 x 12 I 89,600 use 5", 12.75 std. 1
1T
8 I .9600 = 4.97 8 I 5.40
18000
DESIGN OF BEAM’TYPE R.B.Y. assume wt. = 5? /ft

Assume beam supports 10'
Cone. load = 43 x 5 x 10 I 2150
Moon . 2150 x 5 x 12 = 129,000 "#
Mb I 5 x 15 x 15 x 12 I 1,690

 

B.
mmax = 130,690 Use 6", 12.5w std. I
S = 150 590 = 7.14 s . 7.27
~15655-
DESIGN OF BEAN TYPE 2.5.2. assume wt. =-5# /ft

Assume Beam supports 13'
Cone. load I 43 x 5 x 13 I 2800
'Mcon = 2800 x 12 x 5 I 168,000"#
Mb I 6 x 15 x 15 x 12 = 2,025"#

 

1T
mmax 170,025 086 7", 15e3# Btde I
S . 170025 - 9.5 s = 10.54
'10000
DESIGN OF BEAN TIPE 3.2.1. assume wt. - 5} /ft

Assume beam supports 7'

Con. load = 43 x 7 x 5 = 1510

 

 

Econ, - 1510 x 5 x 12 - 90500
my a 5 x 11 x 11 x 12 I 910
s

um.x 91510 use 5", 12.255 std. I

 

 

S I 91510 3 5.1

DESIGN or BEAN TYPE R.B.G.
Load = 5 (45) a 240 9 /ft
M’

240 x 5.5 x 5.5 x 12 I
S a 10900 I .61
‘18000

DESIGN OF BEAM TYPE RB.0.

 

8 I 5.40

assume wt I 5% /ft

10900"#

use 3" 5.7# std. I.
B 8 1e67

assume wt. I 5% /ft

, P,P , R . 40 x 5 x 7.5 = 15002
4- M 5
I 1 . P I 43 x 6.5 x 5 I 1400#
Q
5, R, = 1500'2 4 x 1400 - 1310
2 10

M I 1310 x 5 x 12 I 78,600

Md.“ 5 x 10 x 10 x 12- 750
F

M ., 79,350
S I 79350 = 4.41
15000

use 5", 10.0# Std. b
B .4e84.

 

Third floor beams

 

The next step to be considered in the design is

that of the third floor system. The type floor used is the

A.I.S.c. Battledeck floor with linoleum covering and

patented fireproof ceiling.

There are two types of floor required, one a 3

inch beam with 3/16 inch plate where the Span is fifteen

feet and the other a 4 inch beam with 3/16 inch plate where

the span is eighteen feet.

span.

span

The load upon the section having the fifteen foot

is as follows:

weight of floor 10.5 pounds per square foot
Weight of fireproofing 13.0 " " " 7
Live load (Lane. Bldg. Code)_§Q;_’ " 7 I "
Total load 73.5 pounds per square foot
Tabulate safe load 85.0 " " N N

The load upon the section having the eighteen foot

is as follows:

 

weight of floor 11.5 pounds per square foot
weight of fireproofing 13.0 " " " N
Live load 50 u n u n
Total load 74.5 pounds per square foot
Tabulated safe load 100 " " " I

A check upon the safety of the 3 inch floor

follows, the section used being one “T" section 24 inches

wide

Max.

and 3 3/16 inches deep.
L.L. = 100# /lineal foot
L.L. moment :IE'I 100 x 15 x 15 x 12 I 22,500"#

Mix. D.L. moment I l x 23.5 x 15 x 15 x 12 I 5,300"#

'12

 

Max. Total moment : 27,800")?

IQ of plate - 1 x 24 x ( 3)3 = .00152
eBe I! (IE)

A 01 plate 3 24 x 3/16 I 4.5 square inches
Ic.g. Of'I beam = 2.5 A : 1.54 n n

'1? of plate = 3.09375

.Y. of I been I 1.5

4.5 x 3.094 1.64 x 1.5 I 6.14 x c
2.668

0
.425

A
I = .00132 4.5 (.425)2 2.5 1.54 (1.168)2
I = 5.55

E, = 27 800 x 2.668 = 13,480 r /sq. in.

fins——

Allowable :, = 18,000

The framing plan and the design of the third floor
beams follows.

Note: The number of inches of 3/8" fillet weld re-
.quired for any Joint equals the reaction divided by 3000.
Floor Beam Type N.B.3N.

Total load /ft I 73.5? on floor
load /ft on frame - 73.5 x 15 = lloop
assume wt. of beam I 20? /ft

wt. of wall (4" hollow Gypswm 12' high with pégg7;{)s

 

 

 

 

 

 

 

 

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M = 1 x 1460 x 15 x 15 x 12 I 328,500"# 1.3.2

'12
:3 = 328,500 x 1%,: 27.4 ~ use 10" 35# std. I been
' ~s = 29.15

Floor Beam Type F331!
Total load /ft on floor - 73.54 load on frame - 1100f
Assume wt. of beam = 151' /ft '
Total load 11155 /ft.

s 3 1115 x 10 x 10 x 12 I 6.2 x 12 I 9.3
12 x 18000 ‘1!

use 7, 15.3? std. I beams = 10-34

Floor Beam Type FB3W
Floor load /ft. on frame = (74.5 x 9) (73.5 x 7.5) = 1230
Assume wt. of beams = 25? [ft
Wt. of wall = 3354 /ft
M = l 1590 x 15 x 15 x 12 = 358000"? I 1218
12s 3 358 000 x 12 2 28.4 use 10' 35 std. I beam

Term—5

8 : 29e16

rloor Beam Type NB3E1
3" been as reqd. by floor
Floor Beam Type NB3G
3" beam as reqd. by floor (see also FB3Gl)
Floor beam Type rB3D
3" Beam as reqd. by floor
Floor beam Type FB3B
4" Beam as reqd. by floor

 

 

 

 

 

 

Floor Beam Type F832 assume wt. 10f /ft.
wall load I 336? /ft
1M =1; x 350 x 15 x 15 x 12 I 74,200"# x 12/8

use 6" 12.57 std. I beam
3 2 74200 I 4.12-x 12 = 6.17 s I 7.27
10000 '8'

Floor Beam Type rB3N

Floor load /ft ‘ (73.5 x 6.5) (74.5 x 9) I ll48f/'

Assume wt. of beam = 25%
Wt. of wall I 336 f/'
Total load 1510 f/'

S I 1510 x 15 x 15 x 12 x,1§ I 28.4 use 10" 35#/' std. beam
12 x 18000

a = 29.16
Floor Beam Type NB3Q
assume wt. of beam = 1011:t /'
wall load = 335#/' > A
S 2 350 x 11 x 11 x 12 = 3.52 use 5", 10% std. I beam

 

8 x 18000
a I 4.84
Floor Beam Type NB3I
same as £332 6" 12.5% std. I beam

Floor Beam Type PB3!
Assume wt. of beam 3 15% /ft
Assume 12' of floor'carried by beam (max)
s I 900 x 12 x 12 x 12 I 10.8 use 7" 17.5? std. I beam

Tm .00
s 3 11.11

Computations for load on wall supporting beam

wt. of wall 6_x 10 x 32 I 1920f

wt. of windows 3 10 x 9 x 5.5 = 495#

 

 

Wt. of spandrel 9 x 3.5 x'1 x 170 = 450g
12

a; wt. or stone facing .31? x 6 x 10 x 2 x 160 I 400193
' 12

'Wt. of plaster = 6 x 10 x 5 I 300?

 

 

 

 

 

 

 

 

 

 

 

 

d i 9' 3.
t. .
437 / 76555 1‘437/4
R 3 1785
M I (1785 x 7.5 - 437 x 3 x 6 - 105 x 4.5 x 4.5) 12
K
. F.B.3F. M 3 53,800 5 8 531800 3 3 use 4", 8.5? ltd. I
' s = 3.45
F33A o’ 9' 5,
.437 ’/' fl 1051/. J 4371/.
31 I 2680# R, - 2200
MBA! 3 (2680 x 6.57 - 437) x 6 x 3.57) 12 3 99,500
S I 99 500 I 5.53 use 6%, 12.5f'std. I
m

s I 7.27
FBSC load = 437? /ft '

 

M I 437 x 13 x 13 x 12 I 74,000"#xl%§

‘12
S I 74 000 I 6.15 use 7" 12.5# std. I
mimo-
s I 6.88

Load on FB3L Assume wt. of beam I 15? /ft.
Same as FB3F‘r73.5 x 5.5? /ft I
M 3 53,800+_% x 405 x 15 x 15 x 12 = 188,800"#

M I 188,000"# use 7" 17.5? std. I
8 I 10.4 s I 11.11
NB3J

Use 10%, 15.3# std. channel for stairs
a . 13.38

 

FB3G1
Use 10" 15.3? std. channel for stairs
s I 13.38
r333“ wall load I 437f /tt. assume wt. = 15? /ft
M I 452 x 11 x ll'x 12 I 54600"# x 12/8

3 I 54600 x 12 I 4.59 use 5", 10# std. I
10000" ‘0'
s I 4.84
pm ' assume wt. = 15? /£t.

Wall load I same as E83! assume wt. 15# /ft.
M W811 108d I 53,800

 

m I (7.5 x 89.5) x 15 x 15 x 12 = 151,500 x‘lg - 227,500"#

 

 

‘12
Mmax I 281300"# use 9" 21.8f std. I
S = 281 500 . 1506 B ' 18087
151000
FB3x . .
3 5. 3
1 437’” 10:51". ‘43747. assume wt. of beam I 155I /ft
M I (15 +(7.5 x 73.6)) x 11 x 11 x 12 I 66,600 "#
floor 12

 

M wall I (1574 x 5.5 - 1311 x 4.0 - 263 x 1.25)12 I 37,000“#

'M I 103,600 "# use 6" 12.5? I
S I 103 000 I 5.75 s I 7.27
151000
F335 assume wt I 15? /ft.

Wall 1oad I 437? /£t.
Floor load assume 220% /ft

M - 682 x 15 x 15 x 12 - 172,000 "# x 12
. 12 ‘8

use 8" 2005? Std. I

S I 172000 x 12 I 14.35 8 I 15.05
'10000' '0'

F330 assume wt I 15f /ft
Wall load I 437 # /£t
Floor load 588% /ft

M I 746 x 16 x 16 x 12 I 191,000"? x %_2_ use 8" 23? std. I

 

12
S I 191000 x 12 I 15.9 s I 16.05
“10000' ”0'
FB3Z assume wt. 15% /ft

Floor 1oad I 73.5 x 11.5 I 845 # /ft
m = 860 x 15 x 15 x 12 I 194,ooo"# x‘lg

 

12
S I 194000 x 12 I 16.2 use 9" 21.87 std. I
'10000__—-0'
s I 18.87
XBSA assume wt. I 10? fit.

wt. of wall 280* /ft
M I 1 x 290 x 13 x 13 x 12 3 49000" /ft x 12
'12 "0
3 I 49000 x 12 I 4.18 use 5", 10? I
10000 0—

B = 4.84

wt. of wall 2802 /ft assume wt. 10? /ft
M - 1 290 x 5 x 5 x 12 = 10500 "I x 12
IE '8 '
use 3", 507? Std. I

S 3 10 500 3 .6 x 12 I .9 s I 1.67

 

 

 

F33? * 82w». a au-
?! 6' _ 7'

efldii‘

3% Eh
I 1595 ' R stair - 2950

 

 

assume 30" /ft

 

 

 

Rxb3a
RI. -.-.- 8700 +13 1: 1595+ 7 x 2950 I 10770
20' 20

.8, I 11,175

.Max moment occurs 10.54' from ”I

 

 

M‘I (10,770 x 10.54 - 1595 x 3.54 - 870 x 10.54 x 5.27)12 I
718,200“?
3 I 718 200 = 40.5
"IEfUUU use 12" std. 1 beam 40.8? /ft.
8 I 44.82
FB3'1‘ 153mm» , 2950 floor load left or]: 1148f/i’t
7 1 ‘7 l 7' w .. .. -
145077 . . loom“ right of 1 - 670#/ft
5 “wall load I 280 #/ft

 

 

‘1 . 'fiH .
r. Vt. of beam I 50% /ft
RL I 13.5 x 13 x 1480+ 13 x 5985.. 7 x 2950+ 3.5 x 7000 I 18.850
20 '20 '20 20

233Ai-XB3B I 5985?
R, = 15325

Max Moment occurs 8.68' from “L

M I (18,850 x 8.68 - 5985 x 1.68 - 1480 x 8.68 x 4.34)12 I
1,282,000"?

S I 1 282 000 I 71.4

“18060"

use 14" - 58" 0 Beam

s I 78.25

eno'
;xasc length 13' [ Assume wt. I 15? /ft
1

, 5'
6” 25055
.5566”

 

 

 

 

 

 

KL : 4954 13' a, I 5806
8, . 5/13 x 870+ 9 (280 x 8)+ 13 x 503 = 5805
1'3 ““2""

L I 5806 I 6.7' from rt. end

Mmax (5805 x 5.7 - 858 x 5.7 x 5.7112 I 232,500"#

use 8" 18.4? std. I

 

 

 

 

S I 232500 I.12.9
"10000 e I 14.22
1338 length 18' assume wt. I 20? /ft
5506
12'
250% 6
5
RL I 4390 18' R, = 5170

KL . 180+%(5806)+§(12 x 280) =

I (5170 x 5 - 120 x 3)12= 357,92074

 

Mmax‘
s I 20.4 use 9" 25# std I
s I 20.31
1083!! assume wt. I 1073* /:ft
I
4254
.1 5' x e
1
1 R1. = 2755 r Rr I 2315!
33 I 5.5 x 10.1; x 4954 . 2755
R, = 5.5 x 10%; x 4954 = 2315

 

FBBV’ (cont.)
M . (2755 x 5 - 50 x 2.5)12 = 155,000

 

 

 

 

 

s I 155000 I 9.15 use 7" 15.35 std I

'18000

a I 10.34
5530 2315 9. Assume wt I 205 /ft
- .- I ‘7
I
770% ' 9551/.

RL 2 5 x 5 x 770..1 ; 955 x 5..1 (2315) I 5250 100 - 535

1’ ‘Z '2
8,. I § x 5 x 955.; x 5 x 770+%(2315) - 5710 100 = 5810

mm.‘ = (5350 x 5 - 790 x 5 x 2.5)12 = 203,000"?

 

 

 

 

S I 203 000"# I 11.3 use 7" -20§ std. I
151055
a I 11.97
5IlO
58351
I , :
1060“ [3'25 / assume wt 25?} /l't
‘BL = 11500 Br I 13425

 

 

BL I 7 x 5710+.3.5 x 7 x 1525+.11 x 8 x 1080 I 11,600
15' 15 15'

M I [11600 x 8 - 8640 x 4) 12 I 700.000"#
8 I 700 000 I 38.8 use 12" 40.87 std. I

183555

a I 44.82

 

Second floor beams

 

The loading and type of floor used on the second
floor is the same as on the third floor with the exception of
the concentrated load at the midpoint of the transverse beams
caused by the column loads applied at these points. This
load is taken as 50,000 pounds, computations for which are
shown in the section on columns.

The framing plan and design of the second floor

beams follows:

Note: The number of inches of 3/8" fillet weld required for
any Joint equals the reaction divided by 3000.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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FLOOR BEAMS
FBZN length 30' . 2ND FLOOR
assume wt 150? /ft
Floor load I 15; 73.5? /ft
wall load I 280 5 /ft
Wt. of beam I 150? /ft
Cone. load 57 Mid pt. I 50.00051

 

030%

 

 

 

“mg: (86.000 x 15+ 1 x 1530 x 50 x song =5.555,5oo
3' ,

SI 6 865 500 I 381

 

 

 

 

"lHddd"
use 0 27 "C" I beam 145# /ft
R I 47,950 s I 408.05
5322 length 30' wt. 1507 /ft
Floor load I 9 x 74.5 I 670,5f /ft
wt. of beam 150.05 /ft
Wall load , .
3' . . 6 . » 3 .
w 9 1—
_ . . ‘
cone. load I 2 x 7850 (5833) . 157207
2 x 1990 (883) I 39801
Hoof column 1385
1 x 1440 (F533) I 1440f
1 x 1980 (EBB) I 1980?
3rd floor column 588

Ilse 25,000 as design load

 

.EBZK (cont.)

3 a 5 1: 437+ 9 x 105 +15 1: 820+; x 25,000 a 28,367?

"max 3 (28,367 x 15 - (437 x'6't105 x 9+'820 x 15) 7.5) x 12 =
3,680,000"#

S 3 3680 000 = 204.2

I8,UUU use 21" 987 0 Beam

3 = 209.24
FBZL Length 55' wt. 150»v /1’t

2 ol‘ 4f50’ ’
15' GPT 13' 7' I

[066' "/’

 

 

 

l

 

Location of windows uncertain-whole beam designed to carry

full well load.

RL = 20 x 25000,. 1065 x 35" '7); 2950 = 33,515
35 2 ‘35

Hr = 51711
Mmax - (33513 x 15 - 1065 x 15 x 7.5) 12 = 4,500c000
S 3 4 500 000 = 250 use 27" - 101$ 0 Beam

W

s = 264.72

FBZR Length 35' 6%“; . 1505‘ /ft 21:50:

 

%

 

 

 

 

L 1586”!
4
HI. = 20 x 50000. 7 x 2950135 1: 1586 = 55,910
3'5 . 35 2
R, - 51540

3 - _¢_55910 x 15 - 1585 x 15 x 7.5) 12 = 449.9
18060

use 27” - 160? 0 Beam
8 = 450.13

F523 25000#
AS' 25
667‘? wt . 1002t /ft

floor load 350# /ft

I

 

 

 

 

M = (25000 1 32,1 x 880 x 32 x 52) 12 = 3,752,400 "a?
1" 8’ '
S 3 3752400 = 209.0 use 24” - 945 0 Beam
‘IUUUU'

s 2 225.02
R = 26,580#

FBZY' use a concentrated load of
37,500#=@ mid point
wall load 3 280? /£t. assume wt = 100$
let floor load a 735% /ft (10 sq. ft. of floor)
M = (37500 I 32.4%11115 x 30 x 30) 12 II 4, 878,000

 

 

 

8 = 4878000 3 271 use 24,110fi 0 Beam
'IFUUU'
8 8 276.83
19132111 wt. 1501i /1’t span 30'
Cone. load 8 50,000 d mid point . 57103t 8' from Rr
8’
I
I5, 1 7 [will
Laooi%

 

 

 

.RL 2 25,000+ 4 x 8 x 16504.19 x 22 x 1200 = 43840
30 30

M max = (43840 x 15 - 18000 x 7.5) 12 = 6,271,200
S I 6 271 200 = 348 use 27" 137$ 0 Beam

157000-

8 - 358.73

 

1821

F828

F820

F82D

F828

F828

F820

F828

8821

F82J

EBZMC

F820

FBZGL

FBav

Same as

I!

'3

883A
F838
10380
5335
F833
may
use.
FB3H
beI
255.:
F83M
F830
1:550

FBSV

 

11821

;X82A

11328

.XBZU

1828

Same as

883x

183A

X838

1830

X83D

 

First floor beams '

(including girders)

 

The design of the first floor system varies some-

‘Ihat from that of the other floors in that the live loads are

greater, and the longitudinal members are girders which trans-

mit the stress to the abutments and the pier.

The loads on the north section are as follows:

15 foot span.

L. L.
2" Asphalt
Fireproofing

Weight of floor
Total load
using 5', 10f I's with
240 pounds per square foot. .
18
Load

150 pounds per square foot.

20 I! I! II N
1 5 u n n It
15 u u n n
300 n n n n

1/4" plate the safe load is

foot span

200 pounds per square foot.

Using 5" 14.75? 15 with 1/4~ plate the safe load 1:

220 pounds per square foot.

The loads on the south and middle sections are as

fiallows:
15
.L. L.
Fireproofing
Weight of floor

Total load

foot span
100 pounds per square foot
15 W a u u

15 7' 5' IV 7'

 

130 pounds per square foot

Using 4" 7.77 1'8 with 3/16 plate the safe load is

150 pounds per square foot.

 

18 foot span
Load 130 pounds per square foot
Using 4" 10.5? 1'8 with 3/16 plate the safe load
is 135 pounds per square foot.
The framing plan with the design of the members

follows.

Note: The number of inches of 3/8" fillet weld
"Wired for any Joint equals the reaction divided by 3000.

 

 

 

 

AP‘UTNIEN T

g g or P452

 

4* AEU rMENT

QIV!

b.

 

wiiN’Vi,

379°,

p.71

A 193 " F"

“"0 ;
.

Lina 2,95,117#

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5U: T? QMINAI
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. HMSINU

UNION

Brid1

 

Load

Loa

iBridge floor beams (North side)

8518 Span 30' -wt. 1005 /ft
Load = (15 x 200)+-100 a 51005 /ft
M 2% 5100 x 50 x 50 x 12 . 4,185,000

8 = 4 185 000 8 232

4157050-

use 21" 0 Beam wt 104? /ft e = 235.74
neaotion = 3100 x,15 = 46,500#

881W span 50' wt. 1505 /ft
Load = (15.5 x 200)+ 150 = 54504 /ft
m - 115 x 5450 x 50 x 50 x 12 = 4,555,000!»'

S - 4 655 000 = 259

use 21" 0 Beam wt. 120% [ft 8 t 2.72"
Reaction 2 3450 x 15 8 51.750?

251! Span 50' wt. 1002 /ft
Load =-(15 x 200)+ 100 = moon/rt
m - '5 x 2700 x 50 x 50 x 12 = 5,550,000?

S = 3 650 000 3 203

150015——

Use 21" 98? 0 Beam 8 209.24

neaotion '-' 2700 x 15 a. 40,500?

\.....(r 3..

 

 

 

Bridge floor beams

8'le Span 30 '
5!

 

’2' r 0;
’00?” [6’6 /’

(North side cont.)

wt. 150% /ft ,
(5

 

 

Floor load 8 9% x 200 = 1900 7? /ft

Bead load = 1501?

Total uniform load = 20504 /ft

8}; 8 3 x 1618 x 28.5+12 30100 x 21+ 15 x 1618 x 7.5 +2050x30x15

8L 3 138,500 + 25200 + 182,000 + 920,000

1T

8L 3 42,190

Mmax':

Hr' : 49,650

(49650 x 15 - 182,000) 12 = 6,753,000”;t

,s = 6 753 000 = 375

W

use 24" 150? "0" Beam 5 - 384.94

8813 Span 52' wt. 1504 /ft

12’

r

max floor load 1' 5.5 x 200? / ft 0 R and

I7

I

 

(N)

since this beam rests directly on the abutment the area need

only be large enough to withstand the compressive stress.

Max load /ft =

1100 150 1618 = 286875: whish load

would require only a fraction of an inch of web thickness to

Provide for joints and floor connections.

"I”.

use 8" - 18.4% std.

 

 

 

 

 

 

Bloor Beams (middle).

palm Span 10' wt. 25? /ft.
L. L. a 1005 /eq. ft.
Load 3 (130 x 15)+ 25 = 1975? /ft
Partition load - (15' high) 5.4.22? /ft
motel load 2395
u = % x 2395 x 10 x 10 x 12 = 360,000"f

3 3 360000 : 20
‘18000

use 9" 257 std. “I" s = 20.31
neaction = 2395 x 5 = 12,000f

rBlv Span 11' wt. 107 /ft
Load = 4207 /ft
8 = 1 x 430 x 11 x 11 x 12 = 4.35
8 18000

use 5" 10? std. 1 s = 5.84
neaction = 420 x 5.5 8 23103

1810 span 10' wt. 50%
Load = (150 x 15; 5o - 2150 420 = 25507; /ft
m = 2510 x 5 x 12. % x 2550 x 10 x 10 x 12 - 525.100";

5 = 526 100 = 29.3

wiped

use 12" 31.8" std. I s = 35.97

8811
See F818
Use 8" 18.4? Std. "I"

 

 

Bridge floor beams (Middle)

F810 Span 15' Wt. 504' /ft
wall load = 1918? /ft
‘wt. __591 /ft
Total load = 1968# /ft

 

S a 1968 x 13 x 13 x 12 = 27.8 use 10" 30# C Beam
851—18000—
s 3 31.91
1315 Span 4' Wt. 205 /ft

wall load = 1918# /ft
_Floor load 4 6.5 x 130 = 845
Total load 3 {/ft

s = 2778 x 4 x 4 x 12 = 5.71 use 8" 18.45 std. 1*
5 x 13000

s = 14.22

* for floor connections

 

 

 

(South Side)

BRIDGE FLOOR BEAMS RBlL It. 100* /ft
partition load - 450# /ft
Floor load - 150 x 15 = 19507 /ft
Wt. of beam . - _1291 /ft
Total load 2500# /ft

M = % x 2500 x 30 x 30 x 12 3 3,375,000 "#

S 3 3 375 000 = 187.5 use - 27"0" beam

"TB-.300" .
85? /ft 8 = 215.20
Reactions = 2500 x 15 = 37,500?

F812 Span 30' Wt. 100# /ft
rloor load = 15 x 150 = 1590; /ft
Partition load = 4505 /ft
Wt. = 1005 /ft
22404 /ft
m = % x 2240 x 50 x 50 x 12 : 5024,000
S = 5024 = 158.5 use 21" 80# "0" Beam
Reactions = 15 x 2240 a 33,600# a = 170.9
F818 See F818
8" - 18.4# Std "I"
5518 Span 50' Wt. 120# /ft

Floor load = 17 x 130 = 2210
Partition load = 450

w ht 38
eig 1 /ft

 

 

s : 27800 x 50 x 50 x 12 a 208.5 nos 27" - 85% "e" Beam
”58000

8 x 1
Reactions = 15 x 2780 = 41,700# a - 215.20
FBlT Length 50' Wt. 100% /ft

‘ Floor load = 150 x 15 = 19505 /ft
Partition load = 450# /ft

st. of team = _199fi /ft
Total load = 25004 /ft

M = 5,575,000"# s = 187.5

use 18" - 100# 0 Beam s = 19557
Reactions I 37,500#

 

 

5515 Length 50' Wt. 150* /ft
Floor load a 150 x 5.5 - 7155 / ft
Wt. 15gf / ft
Total uniform load = 5554 / ft
6‘ Io' ’4'
I (007/: ’6’5’fl’ ] 'VALL 107405
(’4’ 7
5L = 1518 x 5 x 27-«10 x 100 x 19 +14 x 1518 x 7-+555 x 50 x 15

 

5L ' 24,520 5r - 28,690
514. from rt = 12(28.59o x 14 - 2285 x 7): 452,500"#

3 = 462 500 3 25.7 use 10" "0" Beam

"15500-

Wt. 255 /ft s = 27.55

 

 

 

.8563 .. 263% 4 0853 58.381.33.65 35.3 . mm
322» u an

lobl
.2. K onbnmc +89" M oom.0m.C +:.om N oora. + “2.0m. N wow. H mm

 

a: 2:.an . 2.5.3.35: 0862 -

3.01:. +5 +3 +3 5.: 3 a Sod .

3.2. +23 2.3+ 5.223123; 2 ans. - Hm om

 

 

000.0! ooh 0! 00.0.0: BR 8

 

 

I III

 

GIRDER 0N1 (Cont.)

Mix shear 2 355,581# a right egg
The max moment will occur 0 pt. 48' from the left end
M (48' from rt. end)
12 (322891 x 48 - 29,124 x 30 - 127,760 x 30 - 120,500 x 15 -
231 x 27 x 22.5 - 1618 x 3 x 1.5) = 106,050,816"#
D = 1 span I 1 x 90 x 12 = 90" use 96"
IE I!

use 96 x %" Web A = 48 sq. in.
Using iob atlrrners {_mny - 10,000 A = 355 581 = 35.56"

m
Assume wt. 2 370* /It

Dead M (48- from L end) = (16,650 x 48 - 48 x 570 x 24)12 -
- 4.470.ooo"#

A - 110 520 800 - % x 96 x % - 65.3" - 6.00 sq. in. = 59.5 sq."

W94

Use 1 channel 15" - 55f = 16.11 sq. in.
1 P1 16" x 1" a 16.00 sq. in.
1 P1 17 x 1" = 17.00 sq. in.
1 P1 18 x 1/2" = 9.00 sq. intoomp. flange only)
1 P1 19 x 1/2" - 9.50 :3. 1n.

68011 'q. in.
Allowable comp. = 18,000 - 170 x 180 - 15,2oo# /Iq. In.

H 3 96 2 (1.63 - .82) = 97.62
Ac I 110 520 800 - 6.00 2 64.0 sq. in.

arm

AT - 110 520 800 - 6.00 = 57 sq. in.

mm

 

4';

 

Han.oon . mm
«He.nn¢ - oomdaa..oam.me..nmn.oa..oon.moo H mm

IJBJI

«He.nne . mam «no on . am

in.” .Hm

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GIRDER GN 2
1—
L.—

   

3 a m
s
093.:

 

83 00.0 a!
en... 00.». ea cube! _ 3.0 3‘

 

 

.5“. a?“

 

   

GIRDER 0N2 Mont.)
Mex shear 3 433.7141} 0 left and

Max moment will occur 0 e pt. 48' from the left end.
D = 1 x 90 x 12 - 90" use 96" web
TE

Using web stiffners %’t 10,000

Req'd Ares - 433714 = 43.3714 sq. In. use 96"x§"'web
IU'UUU

9

Assume wt. - 400# /£t
x 30 - 231 x 27 x 22.5 - 3 x 1618 x 1.6)12“
(460 x 45 x 48 - 400 x 48 x 24)12 =
105,424,854. 4,502,000 - 110,925,854"#

110 926 864 ’.% x 96 x'% - 59.7 sq. in.

Approx. A =
15,000 x 94

use 1 channel 15" - 66" 16.11 sq. in.

1 P1 16" x 1" 16.00
1 P1. 17" x 1" 17.00
1 P1 18" x i” 9.00 (comp. flange only)
1 P1 19 11% _g;§g

68.11 sq. in.
Allowable comp. = 15,000 - 170 x 150 = 15200 f /sq. in.

u a 93.35
A0 = 110 925 854 - 5.00 . 54.1 sq. in.

EVth‘i‘IETeee

AT = 1100926 824 - 6000 3 57e9 BQO ine

977521413000

 

 

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GM 1 (Cont.)

M31. shear - 606,060? 0 left end

Max. moment is 0 middle

Web Area = 506 060 - 50.606 sq. in.
107600-

use 96" x 9" web

assume wt. - 400? /ft.

M (max.) = (405,120 x 45 - 530 x 45 x 22.5 - 255000x22.5)12
M = 139,750,300"#

use 1 channel 16"-65f 16.11 sq. in.

1 P1. 15 x 1 - 15.00
1 P1. 17 x l - 17.00
1 P1. 15 x 1 - 15.00
1 P1. 19 x 11/15 13.05*

80.17 sq. in.
Allowable compression 2 18,000 - 170 x 180 = 16,200

Required Area

A (Comp.) = 139 750 300 -1% x 54 8 79.45 sq. in.

1572007100"

A. (Ten) 139 750 300 - 6.75 = 70.76 sq. in.

We

* use 19"x t" on tensile flange

making tensile area = 76.61 sq. in.

v a I- 93., hr: 317, le

 

 

 

GIRDER 6M2

03.3.» n 38.2: 2m. 3 u 89 u mm ... Ax

 

 

 

. .5 2 a . .0. a . .2 .
00602 on“ .NQ 00.9.»?

.n..\

 

 

.-O

gnr

 

 

GM 2 Wont.)
Max Shear = 540,160 0 end

Max Moment is 0 middle
Heq'd. area = 340 150 I 34.016 sq. in.

“£01000

use 96" x 3/8" web
assume wt. a 40015 /ft

M max. = (198,750 x 45+% 1820 x 90 x 90) 12 = 117,288,000"#

117 288 000 - 4.5 3 63.6 sq. in.

Approx. A '-"
95 X 18,000

Use 1 channel 16" - 66f - 16.11 sq. in.

1 P1 16 x 1" ' 16.00 sq. in.
1 P1 17 x 1" = 17.00 sq. in.
1 P1 18 x :11" 8 15.60 sq. in.
1 P1 19 x i” - 9_.52 " (comp.)
72.11 sq. in.

Allowable comp. = 18,000 - 170 180 = 16,200

A0 = 117 288 000 - 4.5 S 69,00 sq. in

991151-200

AT = 117 288 000 - 4.5 - 51.30 sq. in.
x .

m

 

GIRDER G M 3

 

. 005.000 u xx

005.009 . 0:

 

. 0m + . pm. . Db. Db.
000 00H x HH:.00_.H¢ 005 00H x 00_.005 00 u so +000 u 00 u 00

. obs. . haw. Iohl.u A
+02. 00H N 3. con #9 N >0 000m N n N 0 mm m

 

 

06.0. «.3

, 1
$ OMV ,, ‘0‘ l0§~ ‘Nzlvfi ms &
. .10

 

 

 

. 005.0! 02.00. 02.0! 000...»
8.0. fine 00% an.

 

 

GM 3 Mont.)

Max. shear = 404.7603I 0 left end
nsxu moment is e A point 49' from the left end.
web area 2 404 760 I 40.476 square inches

101-000
use 96" x i" web

Assume weight = 400? /ft
M (max) = (386,730 x 41 - 125,500 I 45 - 830 x 41 x 20.5)12

m = 103,217,150";
use 1 channel 15" - 55? 15.11 sq. in

1 P1 15": 1" 15.00
1 P1 17" x 1" 17.00
1 P1 15 x 3/4“ 13.50

62.61 sq. in
Allowable compression = 16,100? /sq. in

Required flange areas
A (oomp.) = 103 217 160 - 6.0 = 68.9 sq. in

98.76 x 16,100

A (Tens) 2 103 217 160 - 6.0 = 62.0 sq. in

mtg-.000

 

000.00» u 02

 

 

GIRDER G M 4

 

00H.0He u 0:
AHH. 000m0HH.2100..H0..00. He. 00. 000unma..0ne 2 me u an
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bx \ \
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on» a“ 20th 09.002 one»! so... .0!

3.35

 

 

 

 

0 ll 4 (cont.)

Max shear '-’ 418,140 6 left and

Hex Moment is o 49' from left and

web area = 418 140 2 41.814 sq. in.

107000-

use 96 x 1/2" web

assume wt. = 400 :5 /ft

I! (max) = (357600 x 41 - 123,500 x 15 - 119,600 x 30 - 430

x 41 x 20.6) +(400 x 45

use 1 channel 15“

1 P1

1 P1

1 Pl
allowsole comp. = 16,100

15"
177
15"

x 41 - 400 x 41 x 20.5) 12 a

103,368 , 730"?
55:»' 16 .11
1" 16 .00
1“ 17 .00
3321220

52.51 sq. in.

As 3 103 368 780 - 6.0 it 59.2 sq. in.

957757615300

At - 103 368 780 - 6.0 3 52.2 sq. in

Mo

03.000 n m

m
0.3. Bus n Am
00 ..0e 2 000.00

ed M 02.6.3“... Anoton +0.: 0005:”... 32: n N 0.34.1431. on+mH. o N mama

.10.00+0.s0+0.m~+ 0.: 0 x 2.0+ 0.5 s 0H u 33.. 00 s «A s 0000 u 02
svnxusl
+\

 

 

 

 

 

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0s - ‘

 

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s

at“. 3‘

GIRDER GS].

 

 

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GB 1 (Cont.)

M x shear = 319,415 0 left end
Max. moment is @ the middle
V 3 10,000
I Req'd. area - 319416 3 31.9416 sq. in.

use 96 x 3/8" web
assume wt. = 400# /ft

M : (292,725 x 45 - 223,000 x 22.5 - 273 x 27 x 22.5 - 1618

x 18 x 22.5) 124-% x 400 x 90 x 90 x 12 3 92,486,500 "f

92 486 500 - 1 x 96 x 3 I 49.5 sq. in.
x 8 8

Approx. A 3 5 ,

use 1 channel 15" 66# 16.11 sq. in.

1 P1 16" x 1" 16.00 sq. in.
1 P1 17" x 1" 17.00 sq. in
1 P1 18 x 1/2" 9.00 sq. in

58e11 sq. in.

Allowable comp. = 18,000 - 170 x 150 - 15,100? /sq. in.

A0 3 92455500 - 4.5 = 54.5 sq. in.
. x 5100

A T 3 92 486 500 - 405 = 5707 sq. in.

mug-.050

00H.¢Hn u 250.45 K m.m.+.nnm K but 93." N 0.: u mm .0. HM

 

 

G82

 

 

 

G 3 8 (Cent.

Max. sheer =- 314,105? 0 right and
Max mement is @ the middle

D = 1 x 90 x 12 - 90" use 96" web
12’

using web etifi’nere % - 10,000

heq'd. Area = 314 106 - 31.4105 sq. in.

use 96" x 3/8" web
Assume wt. 3 400 f/ft

M = (569,855 x 22.5 +% x 400 x 90 x 90) 12 3 88,260,000 "1?

Approx. A = 88 260 000 - 1 x 96 x 3 = 46.5 sq. in.

mime-#93 e a

use 1 channel 15"

66% 16.11 sq. in.

1 P1 16" x 1" 16.00
1 P1 17" x 1" 17.00
1" P1 18" x ‘1‘" 4.50

63.61 sq. in (comp.)

Allmweble comp. : 18,000 - 170 x 180 = 16,100f/oq. in.

A0 = 88 260 000 - 405 . 5200 HQ. in.

9506 x 16.156
AT 2 88 260 000 - 4.5 - 45.5 sq. in.

93f0‘i‘18000

columns

 

 

 

      

we provide equal resistant!0 to ‘'11!!! ltreaaes in

gmels, all columns on each floor are designed to

ex}.
“9.901% the maximum load oocuring on any column on that

floor 0
These computations with the size of columns chosen

£011 ow:

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ROOF COLUMNS
max Lead on Roof Column
REE = 7.6 x 220 = 1660
1980

BBB 3 9 x 220
2 RBI 8 2 (3545,4'15 x 20 3 7390

110205 which would require an
area 1 sq. in,.Iut for Joints
a larger beam must be used.

use 6" 12.6% std. "I"
Total load transmitted to column below a
11,020 11 x 12.5 = 11,155#
for design load use 11,600

Check stress for partial loading

RBE 3 (75 2.5 x 30) 7.5 1125

1373

355 a (77.5 2.5 x 30) 9
RBWL,D = 5545 7.5 x 20 = 3595

snub - 18 x 75 225 = 1575
s = 4455,,2120 x 5 x 5 = 2115# /sq. in.
“3751 .
RBB = 1950 555 = 75 x 7.5 = 5405

e - 5515 1340 x .115 x .115 = 1770f/aq. in.
*7 1.1

2 555% a 5595

 

 

 

 

a “‘1 interior oolumnfor 3 rd floor.

39:;
2 (FB3W) = 1690 x 16 8 23,860
533E1 = 73.6 x 2 x 7.6= 1,100
F3313 = 74.5 x 2 x 9 a 1,330
load from roof . 11,600

47,780
Assume 8" 18.4% A = 6.34 sq. in.

check for eccentric load

a = 15075 ,10775 x? 4 x 4. = 5050
'6'31' _-"'66—§""

Use cone. load of 60,000#

 

 

-1
? so an!

 

   

1.0
EX (£3215) :: 25,000 22,900 = 47.9007;

'23 (332]?) I 2(1786) '8 3,670
RE. (FBSN) - 11,000
ZR (FBZF) - 2 (1786) = 3,670
RI. (BEN) = 3,367
25 (REF) = __1_,_7_25_

71,1225

Wt. of Roof column 8 1385

" " 3rd Floor Column = 388i
Lead on Column 8 71,648

Assume an 8" Bath. 8 001. 14' long 275

 

s = 71,545+47,900 x 4 x 4 - 9,120 5,540 = 17,6601: /sq. in.

Total conc. load on bridge) = 71,848 27 x 14 = 72,240?
from above .
use 74,000;

Uniform load

Wt. of windows = 10 x 5.5 - 555: /ft Wt. of well - 445# /ft

wt. of 5' of wall : 52 x 5: 961‘ /ft wt. of 14' of facing-37,01/ft.

wt. of facing = git/rt {-wt. of zo'or tacingfiggt/tt.
251;: /ft 15159/rt.

 

 

 

 

Piers and Abutments

 

 

 

 

Design of Pier 5 Abutmenta.

According to a survey made by 11. J. Olver and
m: flYgant for a thesis on flood control, the elevation
of Ottawa street is 123.12. This corresponds with 93.0
on the map included in this thesis. High water is at
121.0 or 90.9. This would necessitate the bottoms of the
girders being at an elevation of 92.0 giving a clearance
of approximately 12 feet over the flow as of December 1,
1931 which is also the present condition (April 1932).

The heights of the girders are 8 feet 8 inches
making the elevation of the first floor 100.67 feet or 7.67
feet above the present level of Ottawa street and 13.32 feet
above the road step the east bank.

The only available data as to bearing conditions
as obtained from the city sanitary engineer shows a stratum
of blue clay at a depth of twelve feet below the center of
the road on the east bank. It is assumed that this soil
extends down for several feet so that the base of the pier

and the abutments will bear upon it.

West Abutment.

The loads upon the west abutment are as follows,
from north to south; 410,640#, 624,060#, 422,760#, and
337,416#.

If six inch rollers are used on this end of each
girder, and each roller is 24 inches long the number of
- rollers required is as follows:

Under northermost girder (0N1)
410 640 . 4.8 use 6 rollers

555-7572:

 

   

\hndlcur (0M1)
-- 624 060 .-. 6.06 use 6 rollers

E950 x 6 x 21
\hnler (Gl3)

4.9 use 6 rollers

422,750

Under (931)
337 415 = 3.9 use 4 rollers

5571-55—

The allowable bearing stress for steel on con-

 

crete is 600 pounds per square inch. The required area

under the maximum load would be 624 060 - 873.4 square

'15:)"

inches or at 24 inches wide, this would be 36.4 inches ,

long which is the minimum width at the tOp of the abutment.
The allowable bearing stress on blue clay is four

tons per square foot. The total load on the abutment is

1,694,766 pounds. This would require a base area of 1 694 7663

"5555‘-

222 square feet. If the abutment is equal in length to the -
building (70ffeet) the required width is 222 or 3.17 feet,

but since the required tcp width is 36.4 inches the base area
must be larger than required.

Since all the buildings along the west bank have
retaining walls along the river's edge, there will be little
danger of the water washing behind the abutment; therefore,
a straight type abutment will be satisfactory.

The height of the abutment from the base to the
seat of the girders is taken as twenty feet.

With such a large load being produced by the
bridge, the weight of the abutment itself and the earth
Pressure against the back of the abutment will be of minor

-.I'1'..

 

'I—‘l' '

-.....--'

      

ortance, therefore if the resultant of the girder loads

ass?

‘9‘6593 ‘through the center of the base the abutment will
Qto‘bfibly be safe. A check of this is shown in the accompany-
ing data and drawing.
Bast Abutment.

The concentrated loads on the east abutment are,
from north to south, 414,700?, 419,400#, 436,400? and
387,866#. There is also a wall load which is a distributed
load along the t0p of the abutment. The maximum wall load
is 1618 pounds per foot.

Using six inch rollers 24 inches wide under each
girder, the number of rollers required is as follows:

Under (0N2)
414 700 = 4.8 use 6 rollers

m

Under (GM2)
419 400 s 4.9 use 6 rollers

‘3

Under (0M4)
436 400 : 6.03 use 6 rollers

3

Under (032)
387 866 a 4.6 use 6 rollers

'3

The required area under the maximum load is 436 400-

156—“

726 square inches or at 24 inches wide, 30.26 inches which
is the length of the bearing surface which will be required.
r5. required area of the base is 1 770 515 = 225.5

”"5555—

square feet. The required width at 70 feet long is 3.22 ft.

To provide against any danger of the dirt being ’,

 

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washed away from behind the abutment, wings will be extended

back on both sides of the abutment.

The design of the abutment is shown on the abutment
drawings, .

Middle Pier.

The pier at the center of the bridge supports the
1'1de ends of all the girders.

The maximum sum of the loads on the pier at any
Point is 898,770? at the reaction of 0111 and (.112. If a bear-
ing plate 24 inches wide is used, the width of the top of the

Pier, as determined by this load is 595 770 ; 52.5 inches.
566 x 21

The total load on the pier is 3,304,266 pounds. The
baae are required is 3304266 . 41303 square feet or at 70 feet
10113 this would require a width of 5.9 feet.

The design of the pier is shown on the drawing of

the pier and abutments.

m“ 1;. has: -.

 

 

Elevator

 

DESiGm 0F ELEVATOR

Since the building is very low, the smallest
elevator will be large enough.
The floor area used will be 26 square feet.fbr
an area of 26 square feet:
width - 6‘ - 0“ Depth - 4' - 3“
G - gate Opening = 3' - 9"
The design load for this size elevator - 76f / square foot.

Live load ' 76 x 26 I 1876 use 2000#
Assume wt. of frame = 1000?
Wt. of cab = 16? /sq. ft. of floor - 400%
Wt. of platform 3 12f / sq. ft. - 300f

Total wt. of car 1700?

counter weight‘: 170041.40 x 2000 - 2600f

Assumef wt. of cables = 160?

Max. live load I 1700 +2000+ 2600+160 I 836%

L.L.+ 1007. impact = 5550 5550 = 15,700#

Max "Rope Pull" (load on cables) = 3,860?

Lead on supports =~16,700f

Assumef wt. of machine - 3,000?

Total load on supports - 19,700? (This checks very closely
with the load of 19200? specified by westinghouse K a n

elevators.)

‘ counterweight a wt. of car 40% live load

f Assumptions made after following similar design in “Architec-

ural construction" by voss a varhey-

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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heads 0 A, B, a o are proportioned as follows:

A = 10,900 B = 6,100 0 = 2,700
A s c = 19,700 which is the total load.
Assume distance from L of supporting beams to edge of
hatchway = 3"
Length of sheave beams 2 car widtht-front clearance'firear
clearsnce+ 6"
Length = 4' - 3'+ 6&"4 9&'-06" = 6' - 0'
Beam R - R.
g : 12 (10900 i 4.084 x 1.916) : 14.20

 

Use 9" 21.83? I S = 18.87
Beam 8 - 8.
M a 12 x 6100 x 4.08 x 1.916 a 7.95
3 6 x 12,000

Use 9" 13.4# S = 10.61

Beam T - T.

M a 12 x 2700 x 4.084 x 1.916 2 3.52
3 6'x31250

Use 9" 13.4# 3 3 10.51

 

R 3 10 900 x 4.084 = 7,430

—J___6.____

51 = 10 900 x 1.915 = 5,470

.—l___6.___
s = 5,100 x 4,054 - 4,150

51 = 5,100 x 1,915 = 1.950

T : 2 700 x 4 054 = 1,540

_l—_TJ___

T1 = 2 700 x 1 915 : 550

__l____6___L__.

Front supporting beam

 

 

eflo ma>aw
i, (5:24151
I 7,-du j
7.16
RL = 3126 RR = 2166
due to elevator and machinery
RL = RR 273 x 3 x 7.16 = 2936# due to floor load

Total reaction 3 EL = 5050, RR - 5100
s . (2955 x 7.15.,5125 x 5.555 - 5470 x .557)12 - 14.14
‘4" 12000

Use 9" - 21.8#I s = 18.87

 

 

Rear supporting beams.

Load due to machinery.

RL - 6660 RR : 4610
KL 3 RR = 2936 due to floor load

Total reactions

Rn = 9595 RR = 7545

3 = 5.25918.90 = 24.15
use 10" 85s“ I B . 24e42

 

Penthouse Design.

Floor area

7' - 2" x 11' - 0"

Live load I 260# /eq. ft.

Wt. of floor) 23# /sq. ft
and fireproofing

' total 273# /sq. ft.
Span 7' - 2"
Use 3" beams with 3/16" plate.

Supporting beams
Leads
980 = 3.68 x 273 #/ft from floors
224 3 7 x 32? /ft from wall
187 9.1%: 7 x 1601)t /ft from stone facing

126 = 3.68 x 35? /ft from roof
1616# /ft total load assume wt. 44# /ft
1476 a 7(32*.% x 160) x 3.68 a each end from walls

—I 1.550% I 5507/ I /560 YAJ

216” T 6’ T 2,6”

 

RI. = RR 3 2.5 x 1560+ 14754-3 x 580 = 7,115

M (center) = 7115 x 3.00 - 1560 x 5.5 x 2.75 -1476 x 5.5

 

M (+) = None

M (-) I 1475 x 2.5 *1560 x 2.5 x 1.25
M (-) 3 103,756"#

8 = M 8 103 756"# 5.76 ‘

1570557152500— ‘~

max 7 a 7,116?

A 6" 12.6# I Beam having an S of 7.27 and an
allowable shear of 16,600 will be used.

 

 

Elevator columns

N. W. Column

Leads

1. From rear supporting beam - 9695

2. From pent house beam 7116

3. From roof beam HBK 2100

4. From floor beam 533K 6816

6. From floor beam FB£K 28367
63992

Total load on tap = 18,810#

The distance between supports is not greater‘ than 12'
Choose an 8" Bethlehem Girder Beam

29.5? /ft

3 = 15 810 + 9595 x .145 x .145 . 2,1755 / sq. ft.

 

 

‘8969' 28.4
s a first floor a 55 992,_25557 x 5.94 x 5.94 - 10,500
"5f59' 100.7
s allowable = 12,000 5" Col. OK

Approx. wt. of column 3 40 x 29.6 = 1180?
Total lead on bridge = 66,172 66,000?

‘* Supports for guide rails must occur at least every 12

:feet.

 

Maul-l

 

 

 

 

Elevator Columns (Continued)

N. E. Column
Leads

1. From Rear Supporting Beam - 7646

2. " Penthouse Beam - 7116
" Roof Beam BBQ - 1300

4 " Floor Beam E330 1760
6 " Floor Beam rB2Q 1760
19460

The smallest column allowable for an unsupported

length of 12' is a 6" B.S. section, 16.65 /ft.

The greatest possible bending stress would occur
if the elevator load were applied without any loads on the

other side of the column.

s = 14660, 7545 x .12 x..12 I 3172f /sq. in.
1‘61—-TT9—_

Full Load stress

3 = 19450, 2745 x .12 x .12 a 4254i / sq. In.
4.51 9.19

Allowable stress 3 12000? /sq. in.
Approx. wt. of column - 40 x 16.6 = 620#
Total load on bridge - 20,080
use 21.000#

 

 

Elevator Columns (Cont inued)

S. E. Column

Leads

1. From Front Supporting Beam - 2166

2.
3.
4.
5.

Use 5" 5.3. Section 15.55: /ft

'1

fl

Penthouse Beam -
Roof Beam RBV -
Floor Beam B837 -
Floor Beam 532V -

Total

Total load on bridge - 16,730?

use 18,000#

8. W. Column

Leads

1.
2.
3.
4.
5.
e.
7.
8.

from Front Supporting Beam -

ll

'7

Penthouse Beam -
Roof Beam BBT 4
Roof Beam BBC -
Floor Beam 833T -
Floor Beam 8830
Floor Beam EBZR
Floor Beam F820
Total

7115
1300
2765

2765

16110

3125
7115
3725
750
16325
2840
51540
2840

 

88260

 

 

Elevator columns {Cont inued .)

S. ‘7. Column (Cont.)

Choose A 35# 8" B. H. Section
Max stress 2

88 260 51 540 x 4 x 4 a 14,420

man-W

.kpprdx. wt. of Column = 40 x 35 I 1400?
Total load on bridge 3 89,660

use 90,000¥

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Hatchway enclosure.

For the hatchway enclosure 6" hollow tile as
allowed by the Lansing Building code will be used.
Wt. of 1' or 6 " hollow tile 40' high £ 24 x 40 - 950%
Enris wt. is carried to support beams by a beam having 3 -

950 x 7 x 7 x 12 - 51.3
"‘1nr1xnr““-"

.
use a 12" 31.8? std. I s = 35.97
Support Beams
West Beam (331;)
Length - 10'

cone. Load 4' from RL

1. from column 90,000
2. from front of hatchway - 960 x 3% - 3.360
3. from F310 3 1968 x 6.5 12.700
106.060?
Uniform Load (Rt. 6 Ft.)
1 from well = 1968? /:t
Uniform Load (left 4 ft.)
1 from floor = 2500? /ft
(see 331T)

RL 3 6 x 106,060. 8 x 4 x 25001-3 x 6 x 1968
16 IU 15

RL 3 75,176 RR ' 52,692 (Wt. not included)

8 :(75,176 x 4 - 10,000 x 2) 12 = 280,704
412,000

Use 27" 112# "c" Beam
s = 293.17

 

Support Beams (cont.)
East Beam (EBlB)

Length 10 ft. Assume wt. = 100#

Loads

cone. load 4' from BL

1. Column load - 18000?
2. Front of hatchway - 3360
3. Front FBlV - 2310

23670

Uniform load (Right 6 ft.)

1. Floor - (130 x 9) - 1170? /ft.
2. Hatchway - 960? /ft
2130? /ft

Uniform load (left 4 ft)
1. Floor (130 x 5) 3 650? /ft

R '6123,670 814x750+316x2230*
1‘ 1‘6 ‘ro 10'
BL - 20,622 KB - 19,428

8 = (20622 x 4 - 750 x 4 x 2) 12 = 74,488
“F’ 12000

use a 16", 50? "C" Beam 3 = 81.95

*Weight included

 

 

 

Wind Stresses,

Although the building is low, and wind bracing is
unnecessary; to comply with the Lansing Building code the
stress produced by the wind stress must be checked. Accord-
ing to the building made, the sum of the dead plus live plus
wind stresses must not exceed the allowable stress (18000
pounds per square inch) by more than one-third. If the dead
plus live stress were 18000 pounds per square inch, the
allowable wind stress would be 6000 pounds per square inch.

Using a panel length of fifteen feet and a height
of three feet on the parapet wall, ten feet on each of the
upper stories and fourteen feet on the first floor, the wind
panel loads, at twenty pounds per square foot, are 2400
pounds on the roof, 3000 pounds at the third floor and 3600
pounds at the second floor.

The maximum total windstress in a column support-
ing'the second floor, assuming the point of contraflexure at
the mid-point of the column is equal to twice the stress in
an exterior column on the same floor. The stress in an ex-
terior column is found by dividing the moment of the wind
panel loads by the width of the building. This gives a
,maximum stress of 2010 pounds over the entire section. The
maximum stress in an interior column would be 4020 pounds or
4020 e 7.89 3 5.10 pounds per square inch.

since the maximum possible windstresa on an en-
tire section is less than the allowable stress per square inch,

it is evident that wind bracing is unnecessary.

 

 

 

All plans'are included in the pocket in the back

cover.

 

 

 

 

conclusion

 

\
Conclusion.

An accurate estimate of the cost of this structure
is impossible with the rapidly changing costs of construction
and the uncertainty of the footing conditions of the location.
To complete this project, however, an approximation of the
cost has been made using a combination of several methods
of estimating.

The cost as estimated is :

Steel frame 280,620# 0 $0.03 $ 8,418.60

Steel erection - 2,348.85

Roof - Aluminum sheets 12,600? 0 30.43 5,418.00

Roofing surface 25,200? a 30.024 604.80

Installation 12,600 sq. ft. @ 80.0125 157.50

Floors - in place 37,800 sq. ft 0 $0.75 28,350.00
(Covering and fireproofing included)

Walls - Limestone 5080 Cu. Ft. 0 $1.60 8,128.00

Laying limestone 5080 cu. Ft. @ $0.71 3,606.80

8" Hollow tile 30480 0 20.14 4,267.20

6" " n 1040 0 $0.11 ' 114.40

Laying tile 31,520 0 $0.065 2,048.80

Windows (garage) 918 sq. ft. 0 $0.76 697.68

" (ventilating) 11,260 sq. ft 0 $1.14 12,836.40

Aluminum spandrels 17,000 f @ $0650 8,750.00

Plaster 7758 sq. yds. @ $0.25 1,939.50

Ceiling (Third floor)*
Metal lath 1400 sq. yds. @ $0.114 159.60

Partitions - Gypsum blocks
24240 sq. ft. 0 $0.17 4,120.80

 

 

Elevator - Installed 9 5,000.00

Piers and Abutmentsi’ 24,000.00
Heating, Plumbing 6. Lighting 500,0000.F.e60.06 26,000.00

 

145,964.13
Add 10% for ommissions, profit, etc. ~#14,596.4‘1
Estimated cost 3 160,560.54

note: All items not followed by a charge for in-
stellation are given in place.

Although the above cost does not seem high as com-
pared with the cost of other office buildings of equal volume,
it must be remembered that prices are now (1932) approximately
1/3 to 1/4 lower than a few years ago, and since there is no
present need for more office space in the City of Lansing,
and since the bus companies of the city, since the starting
of this thesis, have remodeled an existing building to suit
their needs for several years in the future, the undertaking

from a commercial standpoint would be unwise.

* Plaster for third story ceiling included under plaster.
# Estimated from data in "Gillette” with preportionate prices.

 

 

 

 

 

 

 

 

 

 

~r—Jf.

 

 

  

Iain _

 

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