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An Investigation of the Structural Design of
the Michigan State College
Brick Apartment Buildings

A Thesis Submitted to

The Faculty of
MICHIGAN STATE COLLEGE
of
AGRICULTURE AND APPLIED SCIENCE

by

Robert T. Chuck
‘ Inn—II.-

Candidate for the Degree of

Bachelor of Science

June 1948

THESIS

 

 

 

 

(oh/+8

-

“T:

5:

INTRODUCTION

The purpose of this thesis is to check the structural
design of the Brick Apartment Buildings on the Michigan

State College Campus.

From.the standpoint of structural design, the different
Brick Apartment Buildings do not vary from.each other. The
main difference in the various buildings is their size and
location of basements. With this in mind, the structural
design of one of the buildings was completely analysed and

taken as representative of the whole group.

These Brick Apartment Buildings have been constructed
and are now being occupied by the faculty and students of
Michigan State College. They were built on preperty owned by
Michigan State College, south of Shaw Lane and east of
Harrison Road, east of the Michigan State Police (East Lansing
Post), and just off the Michigan State College Campus at East

Lansing, Michigan .

These apartment buildings are of reinforced concrete
footings and concrete block foundation walls. The frame is
composed of formed strip steel framing members with.a self
formed nailing strip to facilitate attachment of exterior
sheathing and interior base for wall and ceiling finish.
Slabs are 2%" thick concrete on #26 guage corrugated iron

sheets supported on the steel Joists. The roof is textured

20731.6

asbestos shingles on wood roof boards supported on steel

rafters.

The exterior is face brick with cut stone sills. Sash
and entrances are wood. The interior partitions are steel
studs with gypsum.lath and plaster to make an overall thick-
ness of 6", for all bearing partitions and dividing par-
titions between apartments, and 4" overall finished thick-
ness for all others. Floors are asphalt tile, terrazzo,

cement or other finishes as indicated.

In examining the structural design of these buildings,
several text books and pamphlets were consulted and the
methods and procedures as taught in the courses at Michigan
State College, were followed. Hence, more time may have
been spent on this investigation than would have been in
ordinary practice. In.many cases, however, the computed
design was checked by following the methods an actual

designer would employ, using handbooks, graphs and tables.

The specifications and plans used in this investigation
were thm 6 actually used in the construction of these
buildings. Whenever there was a question of interpretation
of the plans or of actual construction procedure the firm

of 0. J} Munson was consulted.

Investigation of Design of Roof Truss

Weight of roof covering:

wooden sheathing 1" thick 5 lbs./sq. ft.

Asbestos shingles and felt 4 lbx./sq. ft.
Snow load:

Minimum value of 25 lbs./sq. ft.

for all slopes up to 20°. Load

may be reduced 1 lb. for each

degree of slope above 20°.

Hence:

Slope 2 to l or 26.50

Use: 20 lbs./sq. ft.

Iind pressure: (by Duchemin)
Ph : the horizontal pressure

per sq. ft. on a vertical
surface

'13
II

the normal pressure per
sq. ft. of sloping surface

p.
u

the angle of inclination
of the sloping surface

2 sin 26
O( n265 5 )

(I ; (.4467?)

.892
(1.199)

n : 22.55 1bs./sq.th. 22.35 lbs./sq. ft.

Weight of truss:
w= Ll
ISO/51,1251
3
IV = weight of truss in lbs./sq. ft. of
horizontal covered surface.

I = span of truss in ft.

8 = the distance center to center of
trusses in ft.

P load in lbs./sq. ft.

Combinations of loads:

(1) Dead and snow load
20 ,1 7 = 27 lbs./sq. rt.

(2) Dead and wind load
7 K 22.4 - 29.4 lbs./sq. rt.

(3) Dead, one-half snow, and wind loads
7/ 10 ,1 22. 4 - 39. 4 lbs./sq. rt.

1! = : (40) (38)
150 [51E [pg 150 /(5)(58)/§4o )2

S
II

7.42 lbs./sq. ft. of horizontal covered surface.

Length and areas of panel:

Length of upper chord 8.21.1 ft.
Length of panel = 10.55 ft.

Total load per panel:

 

Asbestos Shingle (5.7)(2)(1o.55) = 77.8 lbs.
Roof felt (0.5)(2)(10.55) = 6.5 lbs.
‘Wood sheathing (5)(2)(10.55) = 65.0 lbs.
Truss, per panel (7.42)(2)(58)(%) = 44.0 lbs.
Dead panel load 288.1 lbs.
Snow panel load (20)(2)(10.55) = 420 lbs.
Wind panel load (22.3)(2)(10.55) = 469 lbs.
Ceiling load (§i%)(58)(2)(150) = 2575 lbs.

Determination of stresses:

= 5100
R = 2550 lbs.

(see drawings for sketch of roof truss and joint and
member numbers). «

Joint (1):

1-2 sin 26.60 / 205 / 255 cos 26.60
1-2 (.447) / 205 / 255 (.894)

1-2 (.447) s 2155

Member 1-2 = 4770 lbs.

2550
2550

1-4 4770 (003 26.6) - 235 (sin 26.6)
4260 - 105

Member 1-4 = 4155 lbs. tension
Joint (4):

 

_1187 - 0
Member 2-4 ' cos 26°4
Member 2-4 - 1187

.894

Member 2-4 2 1400 lbs. tension

4-5 = 4155 / 1400 (sin 26.6)
= 4155 X 625
Member 4-5 = 4780 lbs. tension

Joint (2):

2-5 = 4770 - 410 (.447)
2-5 = 4770 - 184
Member 2-5 = 4586 lbs.

Design of tension members:
Maximum stress = 4780 lbs.
Using allowable unit stress of 18,0001bs./sq. in.
Net area required:

37380 = 0.266 sq. in.
’

8" Jbists used:
area of section”: 2.90 0.K.

6" Joists used for member 2-4
area of section = 2.50 0.K.

The 8" and'g" joists used are both large enough to
support the required load, hence design is 0.K.

Design of upper chord:
Since direct compression in the end panel 1-2 is
greater than the other panel of the top chord, the top
chord section will be fixed by the stresses in flhis
panel 1-2.

At center total direct compression is equal to maximum
compressive strength in.member : 4800 lbs.

Allowable compressive unit stress:

g . 18
r i_7_9%g____

lBOOOr2
l = 126"
'r = 2.43
f1 = 18000

 

l /' ¥(l26)3
18000 x (2.45)2

fl = 15,700 lbs.

Moment:
Where a member is continuous over panel

points, twenty—five percent reduction in the
bending moment is allowed for simple beams.

Weight per running foot on.a panel:

1177 3
16755 111.5 lbs./ running foot.

Moment at support:

M

1177x 10.55 x 12 x 5/4 = 56,000 in.1b.
”‘2‘”

p.
OI

N fLMe
f1 f252

p.
u

effective area of the cross-section.
I

Z
I

' total direct stress.

- unit direct stress.

H
H
I

f2 : critical extreme fiber stress.

3
u

bending moment .

(D
II

distance from the neutral axis to the
extreme fiber.

8
u

radius of gyration.

A = 4600 / 56000 2.94
15705' 18000 2.45

A 2 1.86 sq. inches.

Area furnished by a 6" joist = 2.50 sq. in.,
hence design is 0.K.

All joints are to be welded:
(From.Lansings Building and Safety Code - Amended 2-16-48)

All welding shall be done by skilled workmen and the
Commissioner may refuse to accept work done by any
workman who cannot give satisfactory proof of his skill
and ability in welding. All welding metal shall be in

 

 

shear as far as possible. The shearing stress shall
not exceed five thousand (5,000) pounds per square
inch based on the minimum section of welding metal
and welding metal in tension shall not be stressed
beyond six.thousand (6,000) pounds per square inch.
All working shall be done in a neat, accurate and
workmanlike fashion. Connections showing unsound
metal or*important defects shall be removed and
replaced.

Before approving any type of welded connection, the
Commissioner may require one or more full sized
specimens to be constructed and tested to deStruction,
The designed load to be allowed on such connection
shall not be more than one-sixth of the minimum test
load resisted.

Investigation of Room Slab Design

Specifications:

1'6 800 1bs./sq. in.

f3 = 16,000 1bs./sq. in.
V 3 40 1bs./sq. in.
M = 1/12 W12
U = 100 1bs./sq. in.
n = 15
Loads:

(From Lansings Building and Safety Code - Amended 2-16-48)
Minimum live load for an Apartment is 40 p.s.f.

Live load = 50 p.s.f.
Asphalt tile and asphalt base 8 5 p.s.f.
2" slab (150)(§§)(1)(1) : 25 p.s.f.
W’ = 80 p.s.f.
f0 3 800
f
a - 19999 -
n - 15 - 1067
§_: 800
d 1067 800
I = .429 d
jd:d-§_=d-.429d

5
id = d - .145d = .857d

Clear span a 1 foot 10 inches

Moment:

IH by

”max 12

' 12 (80)(1.85)2(l2)
267 in.-lbs.

Depth (d)

M = ijd
=(§%9)(12)(.6576)(.429d)
d2 = 267
400 12 .4 9 .657
d = .59 inches 0.K.

A 2“ slab is used, hence, 0.K.

End shear:
V = (unit load)(span)
2

: 8O 2
2
80 lbs./Tt. wide strip of slab.

Unit shear:
v - V - 80

- de - Iglo85752 : 5'9 ibs/Bq. in.

Well within allowable 40 lbs./sq. in.

For first and second floor s1abs,concrete with no
reinforcement is used on No. 26 guage corrugated sheet iron
forming on steel joists.

Investigation of Corridor Slab Design

Loads
Live load 50 lbs./sq. ft.
Terrazzo 1%” l9 lbs./sq. ft.
2%" slab _§1 1bs./sq. ft.

100 1bs./sq. ft.

Clear span = 22 inches.

Moment:
2
M :. 1151
I2
: lgg(l.65)2(l2)
: 554 in.-1bs.
Depth (d)
M I ijd
554 :(§%Q)(12)(.429d)(.857d)
d2 - 554

' T4667T1271722§TTT§57T
d = .454 inches '

We have 2.5 inches, hence 0.K.

End shear:
v = (1001(2)
2

= 100 lbs./ft. wide of slab.

v = .1.
b3
: 100
12 .657 .5
= 5.69 lbs./sq. in. Well within allowable

4O lbs./sq. in.

The thickness of 2%” for the corridor slabs is found to be
safe.

Investigation of Basement Slab Design

Loads
Live load 50 lbs./sq. ft.

4" slab 50 lbs./sq. ft.
100 lbs./sq. ft.

Clear Span = 14 ft. 8 in.

Moment:
M = 1/12 wl2
= 1/12 (100)(l4.67)2(l2)
m = 21,500 in.-lbs.
Depth (d)
M = ijd
21'500 ‘ (§%QI(12)(.429d)(.857d)
21,500 = new2
d2 = 12.15
d = 5.48 inches

4 inches is 0.K.

Steel area:

 

fsjd
: 21 500
16000 .857 4
: 21,500
54,800
A6 = 0.592 sq. in/ft. width required.

6"x6" N0. 10 wire guage:
1 Thickness of one wire a .130 inches
In one sq. ft- 4 x .150 = .520 sq. inches 0.K.

Hence, the 4' concrete slab with.6"x6" No. 10 mesh is
structurally safe.

Investigation of the Design of Beams

The complete design of a steel beam.may include the
consideration of a number of items such as deflection, shear,
flexure, etc. However, the basic principle underlying the
design of a beam is that the beam cross section shall have
a resisting:moment equal to or greater than the bending

moment. Thus:

M = fS
or S : M
f

in which.M is the bending moment and the quantity fS is the
resisting moment .

Here are the steps that will be followed in the design
of a beam:

1. Compute the loads the beam will be required to
support.

2. Compute M, the maximum bending moment in inch-lbs.

5. Find S by S : M, the flexure formula.

4. Refer to the tables giving the properties of beams
and select a beam.having a section.modulus equal

to or greater than the required section modulus
found in step 5.

5. Investigate the beam for shear and deflection.

Corridor Joists:
Loads:

Joists spaced every 2'-0".
W.= 100 Tbs./sq. ft. for corridor.

(2)(1oo) = 200 lbs./ running foot evenly distri-
. buted.

Maximum Bending Moment (Span 5.67 ft.)
M : W12
8

(2001(5é67)(5-57) : 600 ft.-lbs.

or 9600 in.-1bs.

Section.modulus:

S : M : 9600

16000
s = .60 in.3
2"x8" Joists have S = 4.6 in.5 0.K.

The weight of the beam was ignored in computing loads but
a beam having a larger section modulus was chosen, hence
there is additional strength to provide for the weight of
the beam.

weight of 2"x8" joists.
(6.5 lbs./Tt.)(5.67) : 56.6 lbs.

56.6 / 1150

Total weight
" " 1169 lbs.

Shear:

V : 1%22 = 565 lbs.

Depth of section = 8 inches
Thickness (t) = 0.15 inch

To find the actual shearing unit stress:

: V - 5 -

The shearing unit stress is found to be 565 lbs./sq. in.
and gs the allowable stress is 15,000 1bs./sq. in; the
beam is acceptable for shear.

Deflection of Beam:

The allowable deflection is limited to 1/560 of the
span.

D = 5'57 12 = .169 inches

560

For uniformly distributed load simple supports, the
actual deflection is:

384::

5 (1169)(68)5
EEZ'T30,000,000)(18:7T'. .0085 inches

.0085 inches actual deflection

As the actual deflection is less than.the
allowable .189 inches, the deflection is not
excessive.

Room.Joists:

Loads:

Joists are also spaced 2'-0" apart.
V = 80 lbs/sq. ft. for the rooms.

2x80 s 160 lbs./ running ft. evenly distributed.
Maximum Bending Moment: (Span 15'-0")

For simply supported beam, uniformly distributed load.
2
M : El.

8
'(160)(15)(15)
8

4500 ft.-1bs.
54,000 in.-1bs.

Section.Modulus:

s a N

54,000
16,000

5.56 in.3

2"x9" joists were used and hence from the tables
3 = 5.8 in.5

Total weight:

Weight of 2"x9" joists s (7.5)(15) a 112.5 lbs.
I = (l60)(l5) = 2400 lbs.

2515 lbs.

Shear:

R1 = R2 and therefore
v : §§%§,= 1257 lbs.

Depth of section (d) = 9.

Thickness, t = 0.145
To find the actual shearing unit stress:

_ 7
rs 'dt
rs = 1257
(9)(145)
fs = 965 1bs./sq. in.

As the allowable stress is 15,000 1bs./sq. in.,
the beam.is acceptable for shear.

Deflection:

The allowable deflection is limited to 1/560 of the
span.

D = 155602 = 0.50 inches

For uniformly distributed load simple supports, the
actual deflection is:

D a 57115
_584E' '1 "'
_ 5(2515) (160)3
’ 584(50,000,000)(26.2)

- 0.55 inches

The actual deflection is less than the
allowable, hence the deflection is not excessive.

Design of Girder

Maximum bending moment:

Maximum moments for the concentrated loads and
uniformly distributed load will be found separately.

31 = 32 = 5526 lbs.

Mmax

(5526)(5) - [(1842)(5) ,1 (1842)(5) / (1642)(1)]
11,052 ft.-1bs. or 152,624 in.-1bs.

Section modulus :

s g u = 152,624 g 5
5' 16,000 8'3 1n°

Beam weight:

Maximum moment due to beam weight.

2 , 2 -
u . w; _ 17(132 1121 - 2550 in.-1bs.

-M-
S - f" §%%%55-= 0.1595 in.3

Section modulus required for concentrated load and the
uniformlyvgistributed load is 6.5 x .16 = 8.46 in.3.
The 8"WF1 has a section modulus of 14.1 in.5, hence
is acceptable.

Design of Stairways

Stairs are designed as simply supported slabs with a
span equal to the horizontal distance between the floor
beam and the landing beame In computing the dead weight
per horizontal foot for use in design the inclined part
should be considered to have the total vertical thickness.

Loads:

Live load

80 lbs./sq. ft.

9" slhb 9/12 x 150 115 1bs./sq. ft.

195 lbs./sq. ft.

Span: (8 ft.)

2
‘12
(195)(8)2(12)
(12)

12,550

Depth (d):

M = ijd
12,550 - (2%Q)(12)(.429d)(d)
d2 = 12,550
1765
d2 = 7
d = 2.65"

There is an effective depth of 6", hence 0.K.

End shear:
V = (unit load)(span)
2

v : (1952(8)
2
V = 772
v : .2.
bjd
: 772
12 .85 6

V = 12.5 1bs./sq. in.
‘Within.allowable 40 1bs./sq. in.

Steel:

A : M
S fsjd
12 350

(16,000)(.857)(6)

0.150 sq. in./sq. ft.

%Ӣ'@.6" o.c. have a steel area of .40 sq. in./sq. ft.,
hence 0.K.

Bond check:

- N
U - ond
2 772
(3.1)(.857)(6)

= 46.5 1bs./sq. in. ‘
0.K., within 100 1bs./sq. in. allowable.

As to further check the table on.page 85 of the 'Simplified
Design of Concrete Floor Systems" by the Portland Cement
Association was referred to:

‘Iith a span of 8'-O' the thickness should be 5 inches
and the reinforcing steel should be %”¢¥7%“ o.c.

This desi has an effective depth of 6 inches and reinforcing
steel of "5L6" o.c., hence 0.K.

Investigation of Column.Design

In the design of the steel columns the following
allowable unit stresses obtained from.the Lansing Building
and Safety Code - amended 2-16-48, were used:

With values of %.not greater than 120:

r = 17,000 - 0.465 13
r2

With values of l.greater than 180:
r .

f = 18,000
2
l/ 1
16,000:-2
In which.1 = the unbraced length of the column, in

inches, and r = the least radius of gyration of the
section, in inches.

Attic columns:

Load:
Total load per panel 1177 lbs.

weight of 8“WF l7#ridge 17 lbs.
1194 lbs.

Computation of allowable axial load:

For 4"x4' H @ 7.5#
Unbraced length of 7 ft.
Slenderness ratio:

1%; $111.13.). 8 39,5

.94

It is seen.that l.does not exceed 120 and

therefore
by the

f

HH"!
II I. ll

r
the allowable unit stress is found

equation:

17,000 - 0.485 (192

1‘

17,000 - 0.465 (69.5)2
17,000 - 5055
15,945 lbs./sq. in.

Area of section 5. 2.22 sq. in.

(f)(A)

(13.945)(2.22) = 50,950 lbs.

= P = allowable axial load.

0.K.

Steel studs (supporting roof truss):

Load or steel studs I 2550 lbs.

Computation of allowable axial load:

Unbraced length of 9 inches.

1‘

"3

fill-J fill-J

- I

A
= bd3
12‘

: (5.65)(5.65)3
" 12"““

= 14.4 in.4
14.4

“V (5.65)2
= 1.045 inches

The slenderness ratio is less than 120, hence:

r = 17,000 - 0.465 (%92
= 17,000 - 0.485 (6.6)2
r = 16,964 1bs./sq. in.
P = f(A)

(l6,964)(5.65)2
224,000 lbs. Allowable axial load. 0.K.

Investigation of Loads on steel studs adjacent to door
opening (maximum.stress condition)
weight of Brick masonry:

45 lbs./bn. ft.
(45)(4.67)(8)(4/12)

560 lbs.

Weight of plaster 200 lbs.

Weight of a 5-5/6" stud per ft.

(490)(5.63)2(1) a 45 lbs.
144

Reactions:

2550 / (4.67)(45)
2760 lbs. 2 R2

(2760) / (2500) / (500) /'(210)
5970 lbs. = R4 '

w
(A
II n

Computation.of allowable axial load:

r =“ll
.A

 

I = bd3
12
; (5.65)(5.65)2
12
I = 14.4 in.4

ref—13%;?

r = 1.045 inches

Maximum.unbraced length of 8 ft. 5 inches.

r 1.045
1.2.96.7
I'
l
g'less than 120, therefore:
r = 17,000 - 0.465 (1)2
r a 17,000 - 0.465 (96.7)2
r = 17,000 - 4550
f = 12,460 lbs./sq. indh.
P = fo
P = (12,460)(5.65)2
P = 164,000 lbs. allowable axial load.
0.K. to use 5-5/8' steel studs.

Investigation of design of steel studs (ordinary case).

All exterior walls and corridor partitions have
5-5/8" steel studs spaced at 24 inches.

Loads:
From.roof and roof truss 2550 lbs.
Brick masonry (45)(2)(16)(4/12) 460 lbs.
Plaster (5)(2)(16) 160 lbs.
Floor and joist reaction 1500 lbs.
‘Ieight of stud (45)(18) _§ll_lbs.

5501 lbs.

Slenderness Ratio:

.1. z: 6.42 12
r - 1.045
'% = 96.7 1666 than 120, hence:
f = 17,000 - 0.485(%)2
r = 12,460 lbs./sq. in.
P = fo
= (l2,469)(5.65)2
P = 164,000 lbs. allowable axial load.

5-5/8' steel studs are 0.K.

Investigation of Desigm.of Steel studs in cross
partitions (2-5/16" steel studs spaced.@ 24 inches).

Loads:
Weight of 2-5/16" steel stud 405 lbs.
Weight of 24" of steel joist 20 lbs.
Reaction from floor and joist 1500 lbs.
1725 lbs.
ll“: 96.7
I‘
r = 12,460 1bs./sq. in.
P = (f)(A)
= (12,460)(2,512)2
P = 66,000 lbs.

The allowable is more than the actual load,
hence tteel stud is 0.K.

Investigation of Design of Columns in Unexcavated
Part (Interior Column)

Loads:
From.roof truss and roof 1500 lbs.
Steel stud (45)(8.45)(2)(5) 2280 lbs.
Wall Partitions (14)(10.17)(9)(2) 2560 lbs.
From Second floor joists (1642)(5) 5526 lbs.

Weight of 6" we 1719‘ - 175 lbs.
uniformly distributed:

Reaction (1842)(5) /’65 5611 lbs.
1"7',47"'7 lbs .

From”Lansing Building & Safety Code"- amended 2-16-48:

Compression in Bulk Masonry equals 90 lbs. per square
inch for concrete blocks and the bearing pressure for
concentration loads equals~115 lbs./sq. in.
Hence:
For a 16" x.l6” masonry column.
, (16)2(9o) = 25,000 lbs.

The allowable is more than the actual load, hence
design is 0.K.

Investigation of Design of Masonry Walls.

Exterior basement wall:

Analysis will be made for section of one foot thickness.

Loads:
From roof and roof truss 2550 lbs.
From steel studs(double)(45)(16)(2) 1440 lbs.
From brick masonry 500 lbs.
From plaster 200 lbs.
From.second floor slab, live load,
partitions and joist 1500 lbs.
From first floor slab, live load,
partitions and joist 1500 lbs.
7290 lbs.

From."Lansing Building & Safety Code" - amended 2-16-48:
Compression for Bulk Masonry equals 90 lbs./sq. inch
for Concrete Blocks and the Bearing Pressure for
Concentrated Loads equals 115 Ibs./sq. inch.

Area of 144 sq. inches

Hence (l44)(90) = 12,960 lbs.
The Code also states that no masonry wall shall have
a heighth between horizontal lateral supports of more
than twenty-two times its thickness:

(22)(l) = 22 ft. allowable
Design.has 10 ft., hence 0.K.

Wall is twelve inches thick and hence, conforms to the
Lansing Code which states that the minimum masonry
wall thickness shall be twelve inches.

Investigation of Design of Twelve Inch Concrete
Wall at Stair Section.

This wall rests on a continuous footing and resists earth

pressure increasing with the depth. The span is ten feet.

Using w = 100 lbs/.sq. ft.
P=cwh
= (O.50)(100)(10)
P = 500 lbs.
RA /'RB = 300 lbs.
(Rs)(1o): (500)6.67
Ra : 200.0 lbs.
R3 = 100 lbs.

Maximum bending moment :

M (6.6)(200)(12)
M 15,850 in.-lbs.

Area of steel:
A M

; 15 650 1
fst (16,000)(.657TT127

A5 = .0964 sq. in.

5’9“? 12" o.c. = 0.20 sq. in. Hence 0.K.

Depth:

M = ijd

62 3 15,650 = 90

1768

d 2 5" Design has 12" 0.K.

Shear:
12 .857 12)

V Z 2.42 lbs./sq. in.

Bond:

$3
I

_ _2' - (2.4)(12)
20 ' 1.6

16 1bs./sq. in. 0.K.

:2
ll

Investigation of Design of Footings

Wall Footings:

Wall footings are continuous and.can be designed as
cantilever beams one foot wide When the wall loads are
constalt.

Loads:

7290 lbs./ft.

Wall load

167 lbs./ft.

Footing load(§%)(1)(l)(m50)
~ 7457 lbs./ft.

Using an allowable bearing capacity of 4500 1bs./sq.ft.
(Allowable used by 0. J. Munson.Firm in designing
footings).

Area of footing gggg : 1.6 sq. ft./ft. length.

Footing is 20“ x 12", hence area is just enough.

Bending moment:

The footing is 1 ft. 8 inches long and projects 4 inches
on each side of the wall, forming cantilever beams.

not pressure p - ¥$§§7 = 4500 lbs./sq. ft.

Take a width of 1 ft. of this footing as a unit

 

beame
2
Miximum moment a '1
'2?’
a (4500)(4)2
~ 1
”max = 5,000 in.-1bs.
d = 1‘;
Kb
: $53001 A depth of 12" is being
used, hence safe.

d = 1.2 inches

Shear stress:

V3—L
bjd
g 4500 4
12 .857 12

V = 12.1 lbs./sq. in.
within allowable shear stress of 120 lbs./sq. in.

Steel:
A:M‘
s TEE
: 5000
,000 .85
A; = .0182 sq. inches

An area of 0.12 sq. inches is Obtained from the
5-%’¢ continuous bars used, hence is ample.

Column footings:

Loads:
Column load 3 18,477
Weight of footing (2.55)2(1.67)(150) = 1 360
19,657

Area of base:

19 657 -
W - 4041 sq. ft.

Design has (2.55)(2.55) = 5.41 sq. ft., hence
safe.

Net pressure:

p = 19.837
5.41

p = 5660 lbs./sq. ft.

Moment :

M : 11.2.
2

5660 26 6 2
(127112Y127L'

 

M : 12,600 in.-lbs.
Depth:
d :\r%5
- ‘ 12 800
- (157H28)
d 2 1.71 inches 0.K.
Steel:
A = .___n
s fjd
_ 12 800
115T00031.657YT167
43 = .0585 sq. inches

Che ck for bond:

- v
5° :35
= 5660 26 6
144 140 .857 16
20 = 2.25 in.

Steel used:

6-%"g{ @ 4w o.c.

As .60 sq. in. 0.K.
$0 4.7 in. 0.K.

Summary

Structurally, the design of the Brick Apartment
Buildings is very safe. In most instances there is some
overdesigning. It is not so much that large structural
members were chosen, as it is that the spans were very short
and the members were not stressed under a very large maximum
moment. There is only one feature that this author found
to be bordering on the unsafe side and that is the design
of the wall footings.

The size of the wall footings are 20" x 12" Which
gives an area of 1.67 sq. ft. per lineal foot. The
allowable bearing soil pressure used by the designing firm
is 4500 lbs. per sq. ft. Thus with the computed foundation
load of 7450 lbs. per lineal foot on the wall footings, the
area of 1.67 sq. ft. is just barely enough to withstand the
load. It is this authors opinion that the width of the
wall footings slould have been enlarged to an even 2 ft. to
safeguard against excessive detrimental settlements. Other
than this one aspect, the apartments are definitely
structurally safe.

Bibliography

"Stresses in Simple Structures" by Urquhart and O'Rourke.
Copyright 1932 McGraw-Hill Book Co., Inc.

"Design of Steel Structures" by Urquhart and O'Rourke.
Copyright 1930 McGraw-Hill Book Co;,, Inc.

"Steel Construction" by Burt and Sandberg.
Copyright 1958 American Technical Society.

"Simplified Design of Structural Steel" by B. Parker.
Copyright 1945.

"Steel Construction.Maaual" - American Institute of Steel
Construction. Copyright 1941.

"Reinforced Concrete Design” by Sutherland & Reese.
Copyright 1945 Wiley and Sons.

“Reinforced Concrete Structures' by D. Peabody, Jr.
Copyright 1946 Wiley and Sons.

'.The ACI Building Code” ASA A89.1-1948.

"Reinforced Concrete Design Handbook" American Concrete
Institute.

"Building and Safety Code", City of Lansing.
me 11de 2‘16-48 o

'Jbint Committee Report for Recommdnded Practice and
Standard Specifications for Concrete and Reinforced
Concrete: American Concrete Institute?

Copyright 1940.

"Continuity in Concrete Building Frames" - Portland Cement
Association.

"Simplified Design of Concrete Floor Systems" - Portland
Cement Association.

"Soil Mechanics and Foundations" by Plummer and Dore.

 

 

 

 

 

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