Simplified Methods on Building Construction

b. Solve for the minimum area of the vertical re-

inforcement.

.01 x 750 = 7.5 sq. em.
{.01 x 120 = 1.2 sq. in.)

c. Convert this area to the size and number of steel

bars by the aid of Table 5 -I. 9
Area of 4 pes. No. 5 (16 mm) bar = 8.04 sq. em.
(Area of 4 pes No. 5 (5/8") bar = 1.24 sq . in.

8 Maximlilt Reinforcement:

(English) a. .08 x 750 = 60 sq. em.
Metric: (.08 x 120 = 9.6 sq. in.)

b. Table 5-2 shows that:

10 pes No. 9 bars gross area= 10.0 sq. in
8 pes. No. 10 bars gross area= 10.12 sq. in.

10 pes. 28 mm gross area = 61.6 sq. em.

8 pes. 32 mm grons area = 64.3 sq. em.

From the result of the above illustration, it appears that the
minimum steel bars that could be placed in a 25 x 30 em. column
are 4 pes 16 mm steel bars. Likewise, the maximum reinforcing
bars that could be placed therein are ei'ther 10 pes 28 mm or 8
pes 32 mm diameter. The above example shows how to determine
the least and the most number of bars that could be placed in a
tied column.

Bundled Bars - Difficulties had been encountered in placing
concrete inside the forms congested with steel bars. A column
that is heavily loaded with reinforcement has this serious problem
when large nu mber of steel bars are positioned and held indiv idual-
ly by lateral t ies. Bundled bars are sometimes employed consisting
of 2 to 4 bars tied in direct contact with each other to serve or act
as one unit reinforcement placed at the corner of the lateral ties.

••u·~•DLID

Figure 8 · 6

140

TABLEB -2 ALLOWABLE LOAD ON A TIED COLUMN

+Pari 1. P {ki~) - {0.18/',A, 0.8/,A,) + 1000

G...,. Loed ou B&l'll x-d on Concrete 0.18J'.A1 + 1000
Al"e& Mlu.: 0.008/,Al + 1000 !'.
A, Mas.: 0.032/, ' + 1000

Colultlll Size

' · - 16,000 J.- 20,000

Mill. Mu. Jdlu. Jdu. 2000 2600 3000 3760 11000

1a2 120 16 Ill 19 77 43 M 85 81 108
140 18 72 22 00 72 711 ~ 1211
16 160 20 82 .26 102 50 81 88 108 1«
92 29 115 58 111 1112
ea10 65

us18 180 23

12 144 18 74 23 92 .s2 65 '78 07 130
14 168 22 'Z7 108 60 76 91 113 151
12 16 192 25 86 31 123 69 86 1(K 173
216 28 93 36 138 tao
-18 Ul 38 154 78 '¥7 117 146 194
20 240 31 123
86 108 130 162 2111

14 196 25 100 31 L25 71 88 1011 132 176
16 224 29 115 86 143 81 101 121 151 202
H 18 252 32 129
20 280 36 143 40 161 91 113 136 170 2'Z7
- - --- - - 22 308 39 158 45 ' 179 101 151 189 252
1.26
49 197 111 139 100 208 27'7

16 256 83 131 41 164 92 115 138 173 230
18 37 H7 46 184 104 130 156 194 .259
16 20 288 41 164 51 206 115 144 173 216 288
320 56 225 1'Z7 158 190 238 317
22 352 45 180 61 246 138 173 207 259 346
24 384 49 197

18 324 41 166 52 207 117 146 175 219 29.2
20 360 46 184
18 22 396 51 203 58 230 130 i62 194 243 3.24
•zt63 253 143 178 214 267 356
24 432 65 221 276 194 .233 292 389
26 468 60 240 69 300 156 211 316
75 168 ~

~

... 20 400 61 205 M 25& loU 1110 2111 270 8410
),)
lt2 oUO &6 226 70 282 158 198 238 207 39e
ao 81 :u6 77 307 178 216 259 824 482
24 480 ~83 ll38 187 8.51 468
26 620 67 2M a28m1 878 $06
.t8 6GO 72 287 g() 866 ~

·~21
22 484 82 248 77 8 10 174 218 :161 327
24 628 tiS 270 84 338 190 238 285 3S6 47.5
20 672 203 92 366 JOII 2~7 3011 336 615
73 99 3M 222 277 333 418 664
• -J8
28
28 816 79 815

24 67e 74 205 $2 31111 10'1 2511 311 88D 518
-28 28

80 80
auM 218 824 80 319 100 226 281 aa7 421 ~
.t8 672 811 108 430 :M2 302 . 863 454 6().~

676 87 346 108 Q3 ~ BOt 366 4541 6011
728 ga 373 116 828 893 481 666
2U

184 100 401 28:Z 868 423 620
900 116 <l8 1 324 405 608
- - ..1000 125 5m02 file 706
811 82 1024 181 loU 810
N 84 1156 148 624 164 B~ 691 922
682 185 4n16o5 38lt 461 780 1040
4 UI S20 624

128 612 160 &tO {,'j() 640 676 900

TABLE 8-3 LOAD ON BARS ~ - ON'..«. + tooo

Number of &r. N.-..~a...

Bar 8ise I 64 8 !If10 l2 1•6 18 20122 Bar Sise I • I • III·41 16 18 20 22
110 112

Ill~~ Grade: / o - 16,000 1lail or Bard G.-.de: /o - 20,000

16 24 32 40 .a 66 63 71 79 87 IS 10 · 80 40 00 60 ee 79 8lll ~ 100
~!ro " •11
18 84 t 5 56 118 N 00 101 113 124

,., IM81 Ill 77 lin 108 113 138 154 leG
16 28 42 66 70 8S 99 ll3 127 141 165
38 58 77 115 134 173 lin 211
40 61 81 101 121 142 182 1M .202 223 .51 711 101 120 15.3 177 202 227 2~ 278
77 10! 128 164 lN 205 230 256 282 18 64 Q6 128 1150 192 Z24 mm256 288 320 361
61 I& 81 163 203 244 2.15 366 447
66 98 130 163 195 llll8 260 .293 326 368 110 100 122 200 300 848 449 407 M9
80 120 160 .200 240 280 820 360 400 oUO Ill 160 260 49\l

-

1leproduaocl f rom the Amerioaa Concrete l..n.itute IJNVore«/. CMV:'Tfl.l Duip H~.

Allowable Load on a Tied Column - All parts of building
structures are. designed to carry load or resist forces classified ac·

tocording the manner how it was designed. Tied column design

could either be under designed, over designed or standard designed
which connotes unsafe, costly or safe respectively. The design of a
column shall be sufficiently strong to carry a super imposed load
which is referred to as the allowable load.

These tables are presented with the end view that it could be
of help in some ways to the reader in determining or checking the
column size and the steel bars required to support a given load.
The use of these tables will shorten the time and lessen the efforts
to be exerted _on the mathematical processes involved using various
formula.

The special features offered by the tables are:

1. How to determine the size of the tied column and the
quantity of the steel ban required to carry a given load.

2. To check the strength of a tied column if its size and rein·
forcements are either adequate, less or excessive to what is needed.

The principal consideration involved in the design of structure
are: cost and strength. The term cost is academic and easily under·
stood because anything that involves money be it in the form of
income or expenses is everyone's concern and it is where human
interest comes in.

· Failure of tied Column -Tied column failure is by crushing
and shearing outward along an enclined plane where vertical bars
fail by buckling outward between lateral ties. The failure of a tied
column is said to be abrupt and complete and is considered to be

more disastrous than the failure of a single beam or girder in the

same floor.

PAII.UR! 01' A TIf D COLIIIIIN

Figure 8-7

The design of a structure should be strong and safe to both life
and property but economical in the sense that the sizes and ma-
terials specif ied are just enough to resist all kinds of stresses im-
posed on it.

In using these tables, the following illustrations are presented.

PROBLEM:

Determine the size of a tied column having an unsupported
length of 9 feet and the reinforcement required to support an
axial load of lOO,OQO pounds (100 kips) with the following speci-
fications: .

F'c = 3,000 psi
Fs = 20,000 psi

SOLUTION:

1. Assume a column size, say 10" x 12'' having· a cross sec-

tionai area of 120 sq. inches.
2. Referring to Table 8-2, the load carried by the concrete

under the column F'c 3,000 is 65 kips.
3. Substract: 100 kips less 65 kips= 35 kips.

35 kips is the excess load to be carried by the concrete,
which is then to be carried by the steel bars. With the aid
of Table 8-3,
4. Under the column of Fs = 20,000 psi, 19 kips and 77 kips
are 1he values of minimum and maximum load of bars that
are allowed on a 10" x 12" cross sectional dimension of
tied column.
5. It will be noted that since 35 kips fall within the limit of
19 and 77 kips, the assumed column size is acceptable.
6. Referring to Table 8-3 under the column of Fs 20,000 psi
it shows that:

a. 4 pes. No. 7 bars could support 38 kips or

b.. 8 pes No. 5 bars could support 40 kips

Either of these arrangement will be acceptable being slightly
greater than 35 kips. However, the limitation for bar spacing as
explained in. Chapter 3 shall be observed. In this particular case,
values found on (b) is preferred.

I

1.44

Figure 8-8

Con~ion to Metric MaaiUre = Table 8-2 and 8-3 were repro-

duced from ACI Reinforced Concrete Design Handbook. Values
are of the old English measure including the computation of the

example problem.

The valtAes from the table together with the illustration could

be easily converted to the new Sl system of measure with the aid
of the conversion factor presented below. Consequently, it was
not changed abruptly - specially at this time of transition from
English to Metric system because, it would be difficult for one to
adjust if the figures were completely changed with a new one he

Is not so familiar with.

'Problem :

Convert to Metric equivalent the values on Table 8·2 and 8-3
as used in the illustration presented and the result with the aid of
the following conversion factor :

Multiply by to get

pounds per square inch (psi) x 0.704 kg.jsquare em.

pounds per square inch (psi) x 6.895 kilopascals

pounds of force x 4.448 newtons

pounds x .4545 kilograms (kg.)

.inch X 2.54 centimeters

kips x 454.5 kilograms

145

Construction Method of 1 Tied Column:

There are three methods presented in the construction of a

tied column for a small and medium reinforced concrete cons-

truction.

1. Block laying of walls after the concreting of the columns
2. Concreting of the column before the block laying of the

walls.

3. Simultaneous concreting of the columns and walls.

Tied column vertical reinforcements are anchored on the
footing by means of steel dowels tied to the footing reinforce-
ments or, the main reinforcements Itself attached to the footing
reinforcement followed by the pouring of concrete.

Sometimes the concreting of the footing is simultaneous with
the pouring of the column, depending upon the specifications and
methods being adopted by the Engineer or construction supervi-
sor. The construction of a tied column under the first method of

"block laying after the concreting of the column" shall be as

follows:

. Step 1 = construct the scaffoldings that will support the

column reinforcement to its vertical position. Usually there are
4 pes. of lumber vertically installed around the column provided
by horizontal braces spaced at 1.00 m elevation.

Step 2 = lransfer the markings and reference line of the build·

ing from the batter board to the lower and upper horizontal mem-
ber of the scaffolding. Check the vertical projection of these mark-
ings by the use of the plumb bob.

Step 3 = .Provide a temporary horizontal wood brace above
and below the scaffolding inserting it across the reinforcement to
hold the bars to its vertical position. The idea of inserting the
brace across the reinforcement is to give way to the installation of
the column forms.

=Step 4 Ascertain the vertical position of the reinforcement

in the row of several column In both directions, then install the
small sides of the forms in opposite direction and insure its vertical
position.

1.46

Figure8-9

=Step 5 Do not cover the forms until after the following

accessories have been verified from the plan and installed if there is:

a. Downspout

b. Electrical conduit & utility boxe~

c. Standpipe or fire hydrant

d. Plu.mbing and water line
e. Telephone line
f. Burglar alarm line
g. Intercom and door bell Iine
h. Steel dowels for wall doors etc.

Step 6 = In the final covering of the forms. see to it that the

wider cover is provided with charcoal mark and nails to serve as
guide in ascertaining the column size and in fixing the form to its
vertical position. Remove all dirt and debris before closing the
form.

Step 7 = Do not leave the column forms until it is firmly set

and completely supported. Most of the bulging failure of forms are
due to negligence and the inherited manana attitude.

Step 8 = Before concreting have the work inspected by the

authorized inspector or supervisors. Usually this is done before the
closing of the forms giving the inspector the access to see the sizes

and arrangement of the reinforcing bars.

The construction of columns under the second condition of

"Concreting the Columns after the Blod< Laying of the Walls" are

as follows:

=Step 1 The wall footing construction includes the installa-

tion of the vertical reinforcement of the wall. Block laying follows

immediately the concreting of the wall footing to save cement

mortar.

Step 2 =The space altoted for the column reinforcement is left

vacant in the process of block laying.

Step 3 = Install the pipes for downspout, conduits, utility

boxes and others.

=Step 4 Clear the column space with sawdust, earth, dlrts,

debris and wash thoroughly before installing the column forms.

Step 5 =Install the forms enclosing the column reinforcement,

check the allignment and vertical position, have it properly braced

or cross-tied with galvanized wire or machine bolts then pouring of
concrete mixture could follow.

Figure8-10

Comments: COitCitiT••• 01' c:o1.11•" '""Ill 1\.0c:ll

LAYIU

1. This type of construction requires only two pieces of forms

to cover each column, the, reinforcement being flanked on two

sides by the hollow block walls.

2. The bond between the wall and the column will be strong.

er, unlike when it was connected by mortar in the process of block

laying. Cracks between this joint will be unlikely to appear on the

surface.

148

3. Horizontal bars used in the block laying were laid conti·
nuous across the column reinforcement. This process minimizes
the horizontal overlapping splices and consequently, eliminate the
use of horizontal dowels supposed to be inserted across the

column in preparation for the wall construction if column con·

creting is ahead of the block laying.
4. The columns will not be much affected by shocks or

vibrations caused by removing the forms because the column is
laterally supported by the hollow block walls. Likewise, the work
is easy, fast and economical less the destruction of the forms, lum-
ber braces, waste of nails and labor aside from the handy handling
of transferring and re-installing of the forms.

5. Not all columns fall under this condition, because there are
also independent columns that are free from the wall layout of
which the previous methods discussed shall apply.

The methods of construction under the third condition of
simultaneous pouring of column and walls in one setting of mixing.
could only be made possible if the concrete mixture for both
columns and walls are of the same proportions. On the other hand,
if the proportion of concrete differs from one another, one must
be ahead of the other and it is preferred to give the column such
priority which in effect the method falls under the first condition.

8- 5 SPIRAL COLUMN

- Spiral column is the term given where a circular concrete core
is enclosed by spirals with vertical or longitudinal bars. The verti-
cal reinforcement is provided with evenly spaced continuous
spiral held firmly in position by at least three vertial bar spacers.
The column reinforcement is also protected by a concrete cover-
ing cast monolithically with the core. Comparatively, this type of
column is stronger than the tied column and is preferred for a
slender (long) column in carrying heavy load.

When .a load is imposed on a cylindrical column, a lateral
pressure is exerted at the confining materials which eventually
causes hoop tension in the spiral, a closely spaced spiral confining
the concrete and vertical bars counteracts the lateral expansion
while the concrete in the core increases its carrying capacity: The
sign of failure of a spiral column is advanced by the shell (pro-
tective covering} spall off due to excessive load, but failure of the
column occurs only when the spirals yield or burst. Unlike the tied

149

column that fails abruptly, the spiral column with heavy spirals
shows a gradual and ductile failure.

-=.:::::.··

~ 3 ~........Spiral-

.':'== e

. ~ J: ,,-......,........._._

Figure 8-11

Spiral Reinforcement Umltatlon and Spacing = For cast in

place construction, spiral reinforcement shall have a minimum dia-
meter of 10 mm. and that the dear spacing between the spirals
shall not be more than 7.5 em. or less than 2.5 em. The longi-
tudinal reinforcement area to the gross column area shall not be
less than .01 nor more than .08 and that the minimum number of
vertical bars shall not be less than 6 pes. of 16 mm bar diameter.

*Section 7.12.2 of the ACI Building Code specifies "Spi-
ral reinforcement for compression members shall consist of
evenly spaced continuous spiral held flrm~y in place and true
to line by vertical spacers. At least two spacers shall be used
for spirals less than .50 m. diameter, three for spirals .50 to
.75 meter in diameter and four spirals for more than .75 m
diameter. When bigger size of .steel bar is used for spiral such as
16 mm or larger, three spacers shall be used for a spiral having
.60 m or less in diameter and four spacers to a spiral having
more than .60 m diameter .•. The spirals shall be protected
from distortion due to h.andling and placing from the designed
dimension."
*note: conversion of measures from English to Metric were

supplied.

Spiral Anchorage and Splicing= ..The anchorage of spiral re-
inforcement shall be provided by one and a half extra turn of
spiral bar or wire at each end of the spiral unit. When splicers
are necessary for special bars it shall be tension lap splices
with 48 bar diameters as minimum but in no case shall be less
than 30 em. or weld.

l.SO

The reinforcing spiral shalt extend from the floor level in
any story or from the top of the footing to the level of the
lowe.st horizontal reinforcement in the slab, drop panel or
beam above. Where beams or brackets are not present on all
sides of the column, ties shall extend above the terminal of the
spiral to the bottom of the slab or drop panel. In a column

with a capital, the spiral shall extend to a plane at which the

diameter or width of the capital is twice that of the column."

Problem:

Determine the size of a short spiral column and the steel re-
inforcement required to carry an axial load of 200,000 pounds
when fc = 3,000 psi; fs "" 20,000 psi using cold drawn wire for t~e
spiral reinforcement and there will be - 1112 inches concrete pro-
tection.

. Solution:

1. Assume a circular column say 15 inches diameter .
2. The column load is 200,000 pounds or 200 kips.

3. Table 8-4 under round columns; load on concrete fc =
3,000 shows that a 15 inches diameter concrete carries
119 kips_

4. Subtracting 119 from 200 k ips, the excess load on con-
crete is= 81 kips to be carried by the steel bars.

5. Referring to Table 8-4 the load on bars under fs = 20,000

psi are 35 kips minimum and 187 kips maximum since the

excess load is 81 k ips which falls between the minimum

and maximum value, the assumed column size of 15 inches

is acceptable. '

6. Referring to Table 8-5 under "Rail or Hard Rail" fs ==

20,000 psi, seven pieces of No. 7 bars carries 84 kips load.

7. Table 8-6 .. shows that 11 inches core diameter column
could accomodate 8 pieces No. 7 steel bars; therefore, the
7 pieces of No. 7 found on step 6 is satisfactory.

8. Referring to Table 8-7 under ·•cold drawn 1 1/2., concrete
protect ion" and 15 inches column size; 3/8" spiral shatl be
spaced at 2 inches pitch.

15 t

TABLE 8 - 4 SPIRAL COlUMNS. LOAD OH GROSS SECTION
P O:ipe) • (o.216f.A, + I.A.> + 1000

8quano eoa-. Round ColumM

t.o.d OD Ban t.o.d c.. Cooerete Load OQ Ban~ oLo.madro.o-.C. o+oe1re0t0.e0

CCII. Grolt 1.- ts.ooo 1.-to.ooo o..=r~, + 1000
8iM
M..e...r. G. . - !.-1&,000 1.- 20,000
A.na
.x..t.... x!a. ..... ...,M1... ~ A or
• -!t f. Mtm. t t r.
Mln. Mf.K.
•11 lAM 1.-1 tal) JnO 100& Mn
- - - - - - - - - - -- -L<.d. Load Load Load 2000 2500 3000 3750 5000
-- -..11 IN 31 122 lU 1m10 m I. 22 1 154 i--
3a 110 t5 117 101 S5a 177 2S 122 3a1s 152 69 ffT 104 130 173
1'1 w 1111 :2u8 lSO 181 1111
17l Zit az.s m2101 150 80 ~ 138 1~ 1~
lN Is:!
18 ·c&e1 110 11 117 116 lU 36 1711 40 117 Ill 113 170 u22s8
176 68 218 130 181 4$ 218 102 128 192
-'"-Ill

20

-..21
324 52 200 M Z$0 14& 1112 Si t 2M 41 200 61 250 114 143 172 215 286
361 c6u8. 200 12 250 1&2 203 244 3015 4.5 200 !)7 250 128 1W 1!11 239 319
4.00 226 80 281 180 270 -1.50 2M 50 225 63 281 141 177 . 212 354
441 71 225 88 281 Its 225 2118 m337 S14 265
248 346 $$ 22S 119 281 166 196 244 21)2 390

12 484 'I? 260 97 312 218 212 3Z7 408 564 380 6 1 2SO 7sa6 312 17l 214 257 321 428
28 629 86 276 106 343 357 4411 415 66 276 343 187 2M 280 467
k 578 92 276 116 343 238 288 388 .all ~ 452 72 275 90 3~ 204. 2M 303 36() 509
26 626 100 800 126 374 422 SZ7 ~ till
259 324 456 510 382
281 862 492 816 703 1131 79 300 98 3'M 221 276 831 414 M2
· -~~ 5111 573
2e ne 108 3.24 135 40«1 ao. 880 0f07 710 780 816 8:) 82ol toe 406 239 299 3M +lS 1597
27 i ~~ 117 82ol 14& 406 328 410 eo8 760 120 661 322 387 483
,Sf 12& 349 157 487 161481818111M1 92 824 115 406 :2r5n8 3C6 4111 619 644
!28 I 185 3411 188 437 353 . 444713 735 9111 882 7ffl 98 3411 123 437 312 66'1
S41 878 Me 756 106 3411 132 437 297 446 693
.29 !,..l,O;;Ot 743
ao 10 13 8(M
tu 374 180 468 405 608 1081 113 314 Itt 488 318 31}8 477 598 796
SJ 154 mSliD 192 499 433 MO 1151 8M 121 3911 161 499 340 424 510 637 849
12 1024 499 4.61 67t 122& 129 399 161 499 362 M3 678 906
I88 1088 1M 624 :2u0s6 631 490 613 137 424 171 631 385 4~ &77 962
1n 722
tal

from Rart/rwutl. CoJOCrCt• Dui~rt H..Ubo<>l:, -C"o1

V'l

TABLE 8 - 5 SPIRAl COLUMNS. LOADS ON lARS

r-d on Bart, A. {kipe) •/..&0 + 1000 (Mai. ..t• - 0.08A1)

Number ol Ban

B.r 9 1101111121 18 Ia• lu It& In l1a llG I 20 I ZI 1211Z812~juJH

I I8iae
6
I7 8

Intetmediat. Grade: I.- UI,OOO

•te 42 411 56 63 70 17 M 92 99 106 113 120 127 114 1U 14.1 LIS 1Q 1811 176 Ul8
f/j 30 86 40 45 60 66 00 M 69 74 711 84 88 IN lOt IOD ·u• 111 124 129
.. -fll lliO 175 200 22$ 260 275 300 a:u· kll 314 liQ& 424 «II 414
11 68 67 71 86 116 106 115 126 lM 144 154 163 178 182 19% 2lln 211 221 230 240 260
f8 16 88 101 114 126 139 162 164 177 1110 202 215 228 240 m JM 278 ~1 303 316 329
nefll 116 112 128 144 160 176 192 208 224 240 258 272 2811 304 320
&52 368 384 400

f lO 122 IU 163 183 203 224 244 2M 28& 805 826 146 866 3811 400 U"T 447 467 488 508 521
m624 549 67~
824 640

Rail or Hard Grade: 1. - 20,000

371 I

15 43 liO 56 112 68 74 81 f¥1 113 99 105 112 118 124 130 138 143 149 15.5 161
16 63 62 70 711 88 97 106 IH 123 182 HI lli() 158 167 176 185 194 202 211 220 229
11 72 IH 116 108 120 132 144 156 168 180 192 204. 216 223 240 252 264 276 288 300 112
18 95 Ill 128 142 168 174 190 205 221 237 263 269 284 300 316 332 848 364 379 305 411

Ill ~ HO HIO 180 200 220 240 2M 280 300 320 340 360 380 400 ~20 440 460 480 600 520
110 1$2 178 203 229 25.1 Z79 305 830 356 381 406 432 457 483 508 634 569 384 610 685 G60

"-' fll 1fn 218 250 281 812 3-t3 37-l 4.06 481 468 499 631 562 5113 624 66.'1 686 718 740 780 811
· - ~- - --- -- ---

w

TABLE 8 - 6 Manmum Number of S.ra io Out.~: ~line. 0, aod in 11111« RiDe. I

S.r Ring ~~~" .10 d:~o(Core
SiH
, _ -;ao-17 18 1$ :10 21 22 23 24 26 26 21 28 29
11

15 0 9 10 u 1a u u 1 tf 1':' 18 HI 21 22 23 24 25 26 27 29 30 81 12
u ,I 4 $ 7 • 9 10 • 11
13 15 16 1T 18 19 20 21 23 24 25 26 27
- ---
a16 0 ;--

9 10 11 12 II 14 16 11 18 19 20 21 22 .23 24 25 26 21 28
- - - - --- - - - - - - - - - - - - - - - -"-u1 $ 8 1 8 9 10 11 u 13 14 IS 18 17 18 19 20 21 22 23
17 0 7 8 9 to 11 u 13 14 13 16 17 18 lll 20 21 22 23 24 2! 26 27
--- - - - --- - --- - ,_ --l -
a4 5 8 7 9 10 tl 11 12 13 14 IS 16 17 18 19 20 21 :.!2

- -18 0 7 8 9 9 10 11 12 13 14 15 HI 17 18 18 19 20 21 22 ll3 24 211
I 4 5 6 7 8 8 9 10 II 12 13 14 lli 16 17 17 18 19 20
- - -- ---- ------ -- --- - - - ----1-
-- 1---

- - -6 7 8 \) 9 10 11 12 13 13 14 15 16 16 17 18 19 20 20 21 22
6 7 7 8 9 9 10 ll 12 13 13 14 Ill 15 16
19 0

. - - - - - - - - -- -- - --- I
- - - - - -- - - - - - - -f---·-
s6 f-

--
, ,0 0

- -I

--- ~-_I-~- 1-=- --- - -------- -- --,11
7 8 8 9 10 10 11 12 18 13 14 15 15 16 17 17 18 19 19

- - li 6 7 8 8 9 9 10 u 12 12 13 13 H
- - · - - - --!-:--- - - ·-

6 7 7 8 9 9 10 u 11 12 12 13 14 14 16 16 18 17 18

- - · 6 6 6 7 7 8 9 9 10 11 11 12 12

-- - - -- .-lot\

c-.uAmMican Concrete l.o.atitute {rom Rftrt/tNUtl DuiiJI Ha~.

TABLE 8 - 7 SPII~.L COLUMNS. $IZ! AND PITCH Of SNAl$

ICoh•mn Con 1-~---S,qU&(-'l Co-luiZl.~ll ---1·I RoiUI<l Colu-
I I I ISlae Diameter 2000
2500 3ooo 3750 5000 2000 ·I2300 I 3000 11000

8760

Hot-Rollecl 1 :K-In. Cooeret.e ~n

•14 1t • •• •••
• • •16 12 }i-2 Ji-2 ~2 Uo-2 ~~·2"' ~~2I" ~~"'
• u-2 *•"'~2 Ji-2 )f-S
18 13 }i-2 }i-2 ~-.2 J+-2 H-1
' ~·"' }i-2
17 14 'i-2 !{ Jf-2!{ Jt-23( Jt-2~ ••• ~·"'~%)( H-2
K-2U)f-2)( K-l"' J+-2)( K-2
u-zu1815 K-2J.SK-2)( K-2 Ji-2J.S
Ji-2J.S Jf-2!{ ~2J.S Jt-2U J+-2H Ji-2
u-zuIQ 18
:.0 K-2 ~-2U ~-2.1-t • Jf-2"'
17 H-2"' Ji-2
Ji-2~ Uo-2)( Jt-2!{ H-2 K-2U Jf-2

u-zu21 18 }i-2 Jt-2" H-2 •~ H-2~ H-2
u--zu22 K-2U H-2
,.,...223 • H-3
19 :K--2 H-2"' H-2 •~ )f-2)( )f-2 K-2"' K-1

20 }i-2J.S .li--2 ~ -2J.S H-2)( ,.,...2 H-2"' )f-2 .

24 21 H- 2J.S .li--2 H-2J.S H-2 • K-3 Jf-2)( ~2 K-2"' )f-2

25 22 K-2J.S K-2 Ji-2J.S H-2 • ~3 *2K ~2 K-2
• ~3 H-214:
.2$ 23 .J.S-2)( ~ J.S-2J.S ~-2 H-2 K-2"' .H-2
)f-2"'

27 2' :K--2J.1 U-3 ,.,...2 • H-3
• •26
K-2K H-2J.S •• )f-2)( K-2 )t-2J( }f-2

H-2"' H-2 ~, ~ ~2)( H-2 .J.S-2" . J.S-2
•lit 26· K-2J.1 ,....2"' ~-2 }! H-2" H-2
•10 Z7 J+-3 H-2)( H-2
H-2~ H-2 }~ • *2H-3
Jt-2" H-2~ . H-2" H-2U
•32
u-zu ••aa
Sl 28 H-2K Jt-2U H-2U •.. u-zu u-2"~ )f-2 Ji-2)(
20 H-2
ao K-2 )f-2" ~2)( H-3 Jf-2)( )f-2· Ji-.2" Ji-23(
H-2~ • Jr3
H-2J.1 K-2 ' ~ Ji-2)(

Ut
CJt

SPIRAl COLUMNS. SIZE AND PITCH Of SPIRALS (CCNtfillueJ)

IColwnn Core

~ I ~ I ~ I I ~ I_~__I ~ l_&m I ~ I!De Diam9ter
Square Column Jlouad Column &000
3750
2 60

Hot-Rolled 2-In. Conttete l'lol.ection

14 10 • •

. .. ••• • • • •••• •••• •16 11
• • •• •• • ••

•11 12 ~~ K-2 ~2 ~1~

•17 13 K-1 K-2 *2 K-2 ~2 ~2 ~2 ~2
• K-2 )f-2 K-2 )f-2 ~~

11 1. ~2~ K-2~ Ji-2~ •••• • K-2 ~~" K-2~ ~2 ~2~
~~" *"2~
li 16 ~2K ~2~ ~2X • ~2 K-2 K-2~

:10 UJ . ~~K ~2~ ~2~ • -'T-2 ~~" )f-2~ K-2 *2X
• ~-2
21 17 ~2 *2" -'T-2~ ~-l" ~2~ *"2 Jf-.2~

22 18 K-2 ~2~ K-2~ •• • -'i-2 Jf-1~ )f-2~ ).f-2 ~~~
~2~ • Ji-2 Jf-1" K-23i )f-2 K-2~
aa 19 K-2 -'T-2

u 10 ~~ ~2~ K-2 H-2 H-1" *"2~ ~~ K-2~
-'i-2~ K-2 ~~ ~)i
• •26 :n ~~ ~2 ~X ).f-2~ K-1~
•• •ll6 22 K-2 K-23i H-2~ K--3~ ~23i
K-2 H-2~ *"2~ H-1" Jf-2~
• •• "'""'"rr 23 K-1~HU
K-2 K-2~

• •28 u K-2 K-2U K-2 K-2" JT--3~ *"2H ~~ H-2U
• •29 25 ~~ H-1~ H-2H
K-2K K-2 ~2X J+-aX ~2" ~X Ji-2)i
• •80 26 H--3 H-2X K-2 ~2~
• •31 'Z1 H-3 Jf-2X H-2 Ji-2)( ~~ ~~-"sx H-2H
• • •83 28 K-2U K-IX K-2" K-2H
• • •33 29 K-2" K-2X K-2~ K-SX K-2" ~" -oon
)i--2" *2H
Jof-2X K-SX
H-3}(
~~~

Oold·Drawn I H-~- Coacrete Protectioo

• • •u11
K-19( Jt-19( ;(-19( K-19( Jt-19( K-lJ( Ji-lj(
15 12 Jt-2 ;(-1" K-2 K-2 Jf-IJ(
16 13 K-2 K-1" JT-:1 JT-2 K-2 ;(-1" K-2 K-2 H-2
K-2 K-2 ~lj(
Jt-19( *-2 ~2

17u. K-2;( H-1" *-2~ ).f-2;( Jf-2;( U-1" K-2;( H-2~ Jf-2.14 Ji-1"
H-23( H-l" ).f-2,!i Jt-2.K J(-2 Ji-2U Jf-2,!i K-1"
,.....,.11litK-2;(Ji-1;(K-2.K H-2.K ,!i-2 K-2U K-2U Jt-1"
K-2.K K-2 Ji-2.K }(-2 Ji-2.!i
t•~1&Ji-2.14 ).f-2 ft-2" Jt-2U H-1"
Ji-2" Ji-2J( ~2;(
17

11 K-221 Ji-1" H-2U ).f-2 K-2.K ~-2 ~ ~ K-2K K-1"
11 Jt-2 Ji-2;( ~ ft-2;(
110 H-2,......22H-1" *-2U H-2 Ji-2;( K-2 Jt-a;( ~ H-1"
II H-2 H-3 *-2U H-2;( ;(-2 ~ 'f-2;( .K-3.14
"'"'"2S .K-2U K-2 3(-2 H-3;( 'f-2;(
22 Ji-2 H-3 3f-2U ).f-2 Ji-2;( ;(-2 J+-3.14
u ).f-2;( Ji-a;( H-2J(
23 H-2 H-3 K-2 H-3.14
~ U·2;( ~;(
~ )of-a;(
*2" ;(-2 "'"'*2G
u"'"'*27
,U-2 *2" H--2.14 ~ Ji-2 K-2 H--3;( H-3 )f-2;( H-3;(
~u-su *2"21 K-2" Jt-2 ;(-2 H--3;(
H--3.!i H-2" H-2;( Ji-2J( ,...2 ;(-2 H-3!1: Jt-a H-2}( H--IU
262$ ).f-3;( Ji-2;( H-2 H-3;( }S-3;(
27 H-3;( Jt-2" Jt-2;( 2U ;(-2 H-3 H-2.14
28:t.2" "'"'*"u-.auaoH-2U Ji-2 )f-3;( H-3 u-au
)f-2U Uo-2;( H-2" K-2 H-3-;( H--2~ . JHK
a11 Ji-2)i H-3
K-2U H-2 ;(-2 Jt-2;( H-SK
80'*2"u
).f-2 ;(-2 K-2;(
*'" "u"-'a3u-su""'238 Ji-2.)(

U.....l

Ul
CD

Cold-Dra•"Yl 2-ln. Cone~t. Proteetion
.. ..14 10
..16 11 ~1"• •• •• •• •• •• •• ••••
~~·2"
~2 ~2 ~-2 ~~" ~~" )f-2 J.t-2
J.t-2 J.t-2
~2 ~2
16 12

"'"2 "'"21713
~~" ~2 J.t-2 *2 * 2 ~2
'f-IH
J.t-2

18 14 ~~" u-zu ~2)4 u-zJ4 *2 "'"2 !4 "'"2)4 'f-2 J+-2X H---2)(
"'"1"~·"10 15 J.f-2~ J.i-2)4 Jt--2~ ~2 'f-2
~2~ J.f-23\ u-z~ ~2~ 'f-2~ 'f-2 ~-2 ~ H---2~
20 16 'f-2 ~2~ )+-2 ~
21 17 'f-1" J.t-2~ J.i-2 'f-2 H---2~ ~~ *2 J.i--2)4
u-2~ J.f-2" H---2~
u-z"

22 18
*2"'33
)f-1" J.i-2~ ~2 Jt--2 Jt--3 'f-2~ ~- 2 J.i-3 u-zu
~-2~ ~- 2
19 'f-1" J.f-2~ J.t-2 'f-2~ 'f-2 JT-3 'f-2~ 'f--3 J.i-2)4
'f-2~ *-2
24 20 'f-IH J.f-2~ J.i--2 'f-2~ •• 'f-3)4 J.i-3}( J.t-2~
u-z~ ~2"' 'f-2
~ 21 J.t-3 J.i--2 ~ ~-2 •• ~-a x *2 ~J4 J.i-2)4
26 22 "'-3 'f-2~ 'f-2 1{ ~)4
)+-2 )4 J.i-2 'f-2"' 'f-3 )4 'f--3)4 ~2U
27 23 *3 *2"' "'-2"'~-2 ~ "'-2\'
J.f-2U J.t-2 ~-3 U u-au H-23\
'-'-2"' ~-2 1{
28 24 )i-8 J.t-2 ;,. J.i-2 ~-2~ •••••• ~-3}1 J.t-3J4 H-23i
~-2~ '-'-2v. H-23i
29 26 'f--3 J.t-2U 'f-3 *2X "'-3 J4 Ju.t--3aJx4 J+-23i
30 Jt--2"' "'-21{ H-2 V.
~-3)4 H-•H Jf-2 X J.t-3 V.
"'-231 ~-8~ ~2)4
26 J.i-2H J.f-2}J4( 'f-3 *"21{ H-2}( - H-3~ ~2~
27 H-2H 'f-23i 'f-a3i
JT-3
32"'"2"8828 ~3}(
J.i-2" J.t-ll~ Ji-2~

29 J.t-2" ~2V. Ji-2" Ji-2~

Ex«eise Probl.-n:

Solve the above problem as illustrated in Metric Measure {SI)
using the following conversion factor:

Multiply by to get

psi X 0.704 kg/cm 2
psi )( 6.895 kPa
pounds )( .4545 kg.
pounds of force X 4.448 N
inch )( 2.54 em
kips X 454.5 kg.

*Note: for more conversion factor see appendices.

The preparation of the spiral reinforcement is very much dif·
ferent from that of the tied column reinforcement because the
former requires the skill and technique of making the spiral in
accurate measurement to a required diameter. It is suggested that
the spiral are bent continuously around acircular pattern disregard-
ing momentarily the pitch. The spiral will just be adjusted to the
specified distance or pitch by stretching the spring gradually
upward during the tying or assembling stage.

8- 6 COMPOSITE COLUMN

Composite column is another type of column where structural
steel column is embedded into the concrete core of a spiral col-
umn.

The work involved under this type of column is similar to that

of a spiral column after the structural steel have been set to its
position.

8- 7 COMBINED COLUMN

A column with structural steel encased in concrete of at least
7 em. thick reinforced with wire mess surrounding the column at
a distance of 3 centimeters inside the outer surface of the concrete
covering.

159

<.\\ IT

-· ~.

Figure 8-12

COMS\JilEtl COLUMN

Figure 8-13

The construction processes of a combined cofumn calls for the
installation of the structural steel as the main reinforcement,
followed by the attachment of the wire mess covering. The wire
mess serves as the holder ribs of the encased concrete. Usually the
wire mess is attached to the structural steel by weld. The form
makes no different with that of the previous methods discussed
for tied and spiral column.
160

8- 8 LALLY COLUMN
lafly column is a fabricated post made of steel pipe provided

with a plain flat steel bars or plate which hold a girder, girts or
beam. The steel pipe is sometimes filled with grout ·or concrete
for additional strength and protection from rust or corrosion.

IJI)HM-- B•am

- e o\t

J/61Jli-- W·I · S-tr41p

,.....J:~~i!:J,.t:~l!.ese 'P14lte
'f'ootin.9

Figure 8-14

161

9CHAPTER

PLATFORM- FLOOR STRUCTURE

9- 1 WOOD FLOOR SYSTEM

Floor framing rs that platform structure of the building sus-

pended by posts, columns, walls and beams. Wood, being the basic

construction materials, with the development of machineries and

sawmills advanced the knowledge and methods of construction

that skeleton frame type was introduced taking full advantages of

the different sizes of lumber that could be interchangeably made

into framing purposes .

The design of a platform - floor system depends upon the

following considerations:

1. Live Load 3. Types of materials to be used
2. Dead Load 4. Sizes of the structural members.

5. Spacing of the structural members

6. Span of the supports.

Live Load - Refers to those movable loads imposed on the

floor such as people, furniture and the like.

. Dead Load - Refers to the static load such as the weight of
the construction materials which generally carry the live toad.

Types of Materials to be used- The choice from the various
construction materials available such as lumber, concrete, steel

etc.

The sizes and spacing of the structwal members depends upon

its strength and capability to carry the load at a certain spacing.

Span of the supports- pertains to the distances between the

posts, columns or supporting walls~

The platform-floor framing structure is classified into the
following types:

a) The Plank and beam floor type
b) The panefized - floor system
c) The conventional floor framing system.

162

•) I'LAMI( en.. 8f:AN

Figure 9 -1

~.

Among the three different types of floor framing system, the
conventional type is the most popular and widely used because of
economy, simplicity and ease of work.

The different parts of a platform floor system are:

1. Girder 6. Trimmer

2. Sill 7. Tail Beam

3. Floor Joists 8. Ledger Strip

4. Bridging 9. Draftstop Plate

5. Header 10. Floor'ing

Girder: -Is a principal beam extending from wall-to wall of a

building supporting the floor joists or floor beams. Others define

girders as the major horizontal support members upon which the

floor system is laid. Girders may either be:.

a) Solid

b) Built-up

163

_....

.... _

Figure 9-2
Sill: - That part of the side of a house that rests horizontally
upon the foundation . Sill is further defined as those wood mem~
bers fastened with anchor bolts to the foundation walls.

Floor Joists: -Are those parts of the _floor system placed on
the girders where the floor boards are fastened. Joists are usually
nailed on the girders at a distance from 30 to 35 em. on center
rigidly secured by bridging to prevent from wagging sideways.

Figure 9 - 4
Tail ben, Ledger strip, Dr.tbtop Plate

Figure 9-5
164

Header end Trimmer: - Header is a short transverse j oist that

supports the end of the cut·off joist at a stair well hole.Trimmer

is a supporting joist which carries an end portion of a header.

Figure 9-6

Flooring: - The Tongue and Groove wh ich are popu larly

known as T & -G is generally specif ied for wood f looring. The

T & G board thickness is either t t \ 2 em. or (1") 2.5 em. w ith

varying width. that ranges from 7 em. to 15 em. {3 - 6") and the
length from (8. to 20') 2.50 to 6.00 m. long.

Figure 9-7

"REINFORCED CONCRETE FLOOR SYSTEM:"
9 - 2 BEAM

Beam is a structural member that supports the t ransverse load
which usually rest on supports at its end.

Girder - is the term appl ied t o a beam that supports one or
more smaller beam.

Beams are clasified as:
a} Simple Beam
b) Continuous Beam
c) Sem i-Conti nuous Beam

165

Simple Beam: Refers to the beam having a single span support-
ed at its end without a restraint at the support. Simple beam is

sometimes called as simply supported beam. Restraint means a

rigid connection or anchorage at the support.
Continuous Beam:- Is a term applied to a beam that r&st on

more than two supports.

lltettrolnt Uof.\-tcUtCt •rr•••t"• "
•ncttara ••

tlfTIR ICHI II'AJI I Nfl~ tOft SPAN
COIITIIIUOUS I I Alo

Figure 9 -a

Semi-Continous Beam: - Refers to a beam with two spans
with or without restraint at the two extreme ends.

':~:.':::;··

b-------...~Jj~...,_______,u.-------.-.~TI

$111PI.f 8£AIII Oft liM• 111111 COIITIIIUOVI IIAM

"W suP,.o•no Figure 9-9

UAIII

Cantilever Beam:- Is supported on one end and t he other end
projecting 'beyond the support or wall.

~------ ··- - --

Figure 9 -10
T - Beam: -.When floor slabs and beams.are' poured simul-
. taneously producing a monolithic structure where the portion of
the slab at both sides Of the beam serves as flanges of the T-Beam.
The beam below the slab serves as the web member and is some·
tim~ called stem.

166

_FionQ"--, •"" ~ ~I
_,!.
~ •
I~"

W•t> or S?•tn

~•

ill

INTEO!tA'tfl> I>ESIO!t BASIS 01' TORTIONAL SEC-
OF T•BEAM
TION PROPER'tiES AND
Figure 9 ·11 TYPICA\. 1\E INI'ORCEMENT

Shear: - Is the effect of external forces that acts upon the

structure causi_ng the adjacent sections of a member to slip at each ·

other.

Strength- Is the cohesive power of the materials that resist an

attempt to pull it apart in the direction of its fiber.

Ultimate Strength:- Is the maximum unit of stress developed

at any time before rupture.

Moment:- Is the tendency of a force to cause rotations about

a certain point or axis. ,

Strain: ~ Is a kind of alteration or deformation produced by

the stresses.

Stress: - Is an internal action set up between the adjacent

molecule of the body when acted upon by forces. or combination

of forces, which produces strain. Stress refers to the pressure of

load, weight and some other adverse forces or influences.

9-3 RELATION BETWEEN THE MATERIALS AND STRUC·
TURE

Building structure has to be distinguished from building ma·
terials. The combination of different building materi.als that make
it into' a building part is called building structure. The building
material in its raw form or unit has nothing to do with the strength
or participation in supporting nor resisting the load unless utilized
to be a member of the structure. The utilization of the different
materials in the structure has their own purpose ot service in
counteracting the different forces affecting the structure. '

167

Thaf is where design comes in to determine their sizes, quan·
tity, quality, spacing, proportions, mixture etc.

Although the subjer.t matter in dealing with stresses, moments,
compression torsion and the like are beyond the scope of this
subject, it Is considered important to discuss the topicbrieflyto
orient the reader and the beginner builders of the rudimentary
knowledge on how these terms influence the principle of designing
structure. Likewise, the reacting b'ehaviour of the structure when
different forces are applied on it are relevant in the knowledge of
building construction. ·

The DIFFERENT KINOS OF STRESSES THAT MAY ACT
ON THE STRUCTURE ARE:

1. Compressive stresses
2. Tension (Tensile) Stress
3. Shear Stress and Strain
4. Torsional Stress and Strain

Figure 9-12

Stresses on structures are usually brought about by load whictr
are classified into three categories:

a) Dead Load:-

Dead LOads are those loads that are distributed or concen-
trated, which are fixed in position throughout the lifetime of the
structure such as the weight of the structure itself.

168

The dead load on a beam are also categorized Into two:

1. Concentrated Load
2. Distributed Load

b) Live Load: - ·•

Live toad refers to the occupancy load which is either partially

or fully in place or may not be present at all.

c) Environmental Load:-

Environmental load consist of wind pressure and suctions,
earthquake loads rainwater on flat roof, snow and forces caused
by temperature differentials.

Figure 9-13

.9-4 BEHA. VIOR Of BEAM UNDER THE INFLUENCE OF.
LOAD

A homogeneous concrete beam even if free from carrying live
or concentrated loads has to carry its own weight classified as a

distributed load. The gravitational effect of its own weight will

cause the structure to sag or bend downward between its support

as shown on the following illustrations:

. u ..·-·._TIr:r------------~-~·-~-n·=u--

~----91- __._--1 r··------- 1 r

Figure 9-14

169

Bending Moment: - Moment is the tendency of a force to
cause rotation about a certain point or axis. Bending moment are
· of two different types. the Positive bending and the Negative
bending. The positive bending exists when the beam bends down·
ward between its supports where the upper portion of a beam
. above the neutral axis is compressed while the lower portion is
stretched at the opposite directions. The Negative bending mo·
ments exist when the beam is bending above the supports com·
pressing the lower part of the beam below the neutral axis and
stretching the upper portion of the structure.

......... .....ft.

Figure 9· 15

9- 5 REINFORCEMENT OF CONCRETE BEAM
It could be clearly seen from the behavior of concrete beam

under the influence of load that the structure reacts correspond-
ingly with the kind of interacting forces applied on it such as, the
positive and negative bending which may cause its failure or
collapse. It is under this principle that concrete beam has to be
provided with reinforcement in order to prevent rupture of the
fibers under stress.
170

1...om;r ~::te as a homogeneous material is said to be strong in
supporting compression load but weak in resisting tension forces.
Steel on the otherhand, possesses the strength qual ity to resist
both compression and tension forces. The combination of con-
crete and steel producing "Reinforced Concrete" offers the solu-
tion to the problem. The principle behind the design of reinforced
concrete is to avail of the strength of concrete in its capacity to
carry the com·•pression load and the steel to resist tension loads or
forces. When the area of the concrete and steel are just enough to
carry the compression and tension forces simultaneously, the de-
sign is ca lled "Balance Reinforcement or Balance Beam". The
.building Cod e on balanced reinforcement so prov ides that · the

cross sectional area of steel reinforcement shall be equal to .005

times the cross sectional product of the w idth and the depth of
the beam. Thus -

"Find the cross sectional area of steel bars required for a
beam having a cross sectional dimension of 25 x 40 em. in
order t o be considered .as a balan.ced beam.

As = .005 X 25 X 40

= 5 sq. em

This is the minimum required area of steel bars in a 25 x 40

conc·ret e beam to be considered as "Balanced Beam"

Figure 9· 16 .. , ·

9 - 6 THE COMPRESSION AND TENSION IN A BEAM

From Figure 9-15 the depth of the beam is divided at the

center by a horizontal line called the Neutral Axis (NA). The

portion above the axis at the support or column is under tension
while the lower part is under compression.

171

Likewise, the lower portion of the beam that tends to bend down-
ward between the support is under tension while the upper part is
·under compression. With the principle that concrete is to carry the
compression load while the steel is to resist the tension forces,
steel bars are placed in the portion of the beam where tension
stresses developed.

For positive bending the steel bars are placed at the lower
portion of the beam. Whereas, in those areas where negative
moment occurs the reinforcements are placed on the upper
portion. To do these, there are two methods that may be em-
ployed. ·

[.[..].] ill

•••• •••t,.,,,.. A•••tl'4•"•"••"'•"' •• ••••• ..•teel btrt 4Yfell~e•e.t fO ••vN4rMt fiJI' .......¥. ·"· ••tat Ia ......
e4oot ~u.,..,, *'141'• a ...-

Figure 9 -~17

1. Bent Reinforcing Bars: Reinforcing bars are bent up on or
near the inflection points and are extended at the top of the beam
across the support towards the adjacent span. lnfltction points
refers to the porticm of a beam where bending moment changes

tfrom positive to negative. This is usually located at a distance of

about to t length of beam from the face of the support.

2. No Bent Ban: -When bars are not bent, an additional

straight reinforcing bars are placed on the top of the beam across

the supports extended to the required length usually a distance

about i the beam span length from the face of the support,

other straight additional bars are also placed at the bottom cen-

ter of the beam span where positive moment deveJops.

17~

Under the first method, the advantage of the bend bars is its
function to resist the diagonal tension and shear which are usually
counteracted by the stirrups or web reinforcement. On the other
hand, the second method offers ease in the fabrication and install·
ation of reinforcing bars unlike the former that inconvenience are
usually encountered in the fabrication of bent bars and the diffi~

culties of repair when cut or bent Incorrectly.

9-7 SPACING OF REINFORCING BARS IN BEAM:

Reinforcing bars are. placed accurately and properly secured in

position with the use of concrete or metal chairs, spacers, or
bolsters. If the beam design calls for a bent up bars, it is desirable

to use an even number of bars for the main reinforcement. The
idea is when other bars are bent at the inflection points of span,
there will be remaining straight bars at the bottom continued at
the supports where stirrups are tied up to their designed positions.
The minimum clear distance between the main reinforcing bars

1should not be less than (1"} 2.5 em. nor less than 1 times the

maximum size of the gravel.

TABLE 9 -1 MAXIMUM NUMBER AND SIZES OF BARS

IN BEAM

2-*11 3-'*11
3-*9 4-f9
4-*6 . 5-'*6 .

6-..4

Figure 9-18

173

The measurement given under this table has considered the
allowance of 4 em. ( llk"} protective covering of steel bars from
outside of the reinforcement on both sides of the beam including

the allowance for 10 mm ( i ) stirrups. The table also shows

the maximum sizes of bars for ct given beam width. When two or

more layers are required. the dear distance between layers of
bars shall not be less than 3 em. placing the uppper layer directly
above those at the bot~om layer.

9- 8 SPLICING, HOOKS AND BENDS

The ACI Code on splicing, hooks and bends of reinforcement
states, "Splice of reinforcement shall be made only as required or
permitted on the design drawing or in the specifications or as

authorized by the Engineer".

1. Lap splices shall not be used for bars larger than No. 11 or

35 mm bars ( ll p).

. 2. Lap splices of bundled bars shall. be based on the lap splice
length required for individual bars of the same size as the bar
spliced and such individual splicing within the bundle shall not
overlap each other.

3. Welded splices er other positive connection may be used.
A full welded splice is one in which the bars are butted and
welded to develop tension or compression of at least 125 per
cent of th:e specified yield strength of the bars.

4. If the splices of joints under maximum stress could not be
avoided. it should be staggered.

Hook and bend refers to "Standard Hook" accomplished by a
semicircular plus an extension of at least four bar diameters but
not less than (2lfz") 6.5 em. at the free end of the bar or a 90
degrees turn plus an extension of at least 12 bar diameters at the
free end of the bar.

The maximum IJend diameter (other than strirrups and tie
hooks} should not be less than the value given on Table 5·7.
Stirrups and hook bend shall not be less than 4 em. for No. 3
bars; 5 em. for No. 4 bars and 6.5 em. for No. 5 bars.

174

Bars shall be bent cold, unless otherwise permitted by the
Engineer. No bars partially embeoded in concrete shall be field
bend, except as shown on the plans, ~pecified or permitted by the
Engineer.

J.),_-_____,

Figure 9-19

9-9 STEEL BARS CUT OFF AND BEND POINT

It is a common practice to cut off bars where they are no
longer required to resist tension stresses or in the case of a con-
tinuous beam to bend-up some of the bottom steel bars usually
at 45 degrees to provide tension reinforcement at. the top of the
beam over the supports.

The ACI code so provides; ;;Every bar should be continued at

least a distance equal to the.effective depth of the beam or 12 bar

diameter which ever is larger beyond the point at which it is

theorftically no longer required to resist streS$. The Code further

states: uAt least 1 of the positive moment steeel 7' in conti-.

nuous span must be continued uninterrupted along the same face

of the beam with a distance of at least (6") 15 em into the sup-

ports. At least l of the total reinforcement provided for nega·

tive moment at the suppor~ must be extended beyond the ex·

treme position of the point of inflection, with a distance not /eS$
h.than
of the clear span or depth of the beam or 12 bar dia-

meter whichever is greater. "

175

r; 1 ....l l IL I. fl__
3
\I) ~\::J

7
J 1_£, i. It .....J....

).. 4 I.I

St"l bart. arranttment to cou"Wid th• ~1ft •nd nt9dive
moment in bottm. ~ ~~ adopt d5ffMenl errtf\ttmtnt as

.nown on flg~~re !>-4.

Figure 9-20

9- 10 BEAMS REINFORCED FOR COMPRESSION

When Architectural conditions limit the cross sectional di~

mension of the beam, it might be possible that the area of the

concrete that will resist the comJ'ression load becomes smaller ·

and insufficient. Under this situat•on, steel reinforcement is

substituted in place of the concrete area deficiency to supple·
ment the ~oncrete in counteracting compression stresses. This

type of beam is called ..Double Reinforced Beam" where stir-

rups or ties are used to hold the reinforcement together in posi-

tion spaced not further apart than 16 times bar diameter or 48

tie diameter. ·

If compression bars are used in a flexural member, care should

be exercised to ensure these bars from buckling outward spalling

off the outer concrete when under load. The reinforcing bars

should be properly anchored in the same manner as the compres-

sive bar·s in column are anchored by lateral ties. Such ties must be

used throughout the distance where the compression reinforce-

ment is required.

Tf.- \t•·• d

+

•r!-•

l--6=-l

Double Rtlnforcement

Figure ~21

176

9- 11 WEB REINFORCEMENT

Web reinforcement Is the same as the strirrups used in the
beam to hold the reinforcement in its designed position. The web
reinforcement is not only intended to hold the reinforcement and
provide lateral support but also serves to resist- diagonal tension
and counteract the shear action on the structure. The vertical
stirrups should encircle the main reinforcement and hook bent

w ith a diameter not less than 5 times the diameter of the stirrups

at its end and secured prop.erly to prevent slipping of the main
reinforcement in ·ihe ~oncrete.

U-stirrups Closed stirrups

Figure 9-22

9- 12 TORSION IN REINFORCED CONCRETE MEMBER

To resist torsion, the structure must consist of longitudinal
reinforcing bars provided with closely spaced stirrups. The U-
Strirrups commonly used for transverse shear reinforcement are
not suitable for torsional reinforcement, instead, a lateral ties
used in column is being employed as stirrups which is effective In
counteracting torsional stresses. Good anchorage is by hooking the
stirrups bar end around the longitudinal or main reinforcement.

If flanges of a T·Beam are included in the computation of
torsional strength, a supplementary slab reinforcement should be
provided. The main reinforcement should be well-distributed
around the perimeter of the cross-section to control cracking.
Spacing must not exceed (12") 30 em apart. Bars should not be
less than No. 3 in size and at least one bar must be placed in each
corner of the stirrups.

177

~oo1<ed ~nd

Figure 9-23

9- 13 T-BEAM DESIGN & LIMITATION

The ACI Code on T·Beam design specifies that:

1. The effective flange width shall not exceed {. the span

length of the beam.
2. The overhang width on either side of the web shall not ·

exceeed 8 times the thickness of the slab or 112 the clear distance of

the next beam.
3. For beams with only one flange at the side, the effective

overhang flange width shall not exceed ! of the span length of

the beam or 6 times the th ickness of the slab or! the dear distance
to the next beam.

4. The principal reinforcement in the slab (T·Beam flange)
is parallel with the beam; transverse reinforcement is necessary
for the slab. The reinforcement spacing shall not exc'eed 5 times

the thickness of the slab nor (18") 45 em. This is not applicable

to a rib in a ribbed f loor construction.

9- 14 OTHER CAUSES OF BEAM FAILURE:

The fail ure of a beam is not only due to shear, the positive
or negative ~nding which was alreadyexplained but also includes
bond. Failure in bond means the slipping of the steel bar rein.·
forcement inside the concrete when load is applied on the struct·
ure. lt is due to this problem that deformed steel bars were manu·
factured in order to give a strong bond or contact between the
steel and concrete.

178

COMMENTS AND OBSERVATION

The use of a relatively high or low strength concrete or steel
depends upon the cost, availability of materials, importlf1ct of
special · requirements such as minimum sizes of the members
structure and concern for deflection and crack width~

High strength concrete is attained by increasing the amount of
cement in a mixture. Cement nowadays is considered expensive
aside from several ingredient to be mixed such as, sand and gravel
which in some areas, prices are so high and prohibitive that the
cost of concrete increases substantially with the desire. to attain
high strength concrete. On the otherhand, high strength steel are
produced either metallurgically or by cold working available at a
slight increase of cost. The present trend of building construction
is to use reinforcements having an increased strength of (60,000
psi) 413,700 kilopascal while concrete on the otherhand will not
likely change from the present allowable strength of (3,000 psi to
5,000 psi) 20,680 to 34,4 75 kPa. Consequently. labor plays an im·
portant role in the cost of the building construction wherein the
work for ·concreting should be compared with the cost of the work
for the fabrication and instailation of steel bars. Records show
that concreting including its preparation cost is substantially h igh-
er than that of steel construction.

9 -15 REINFORCED CONCRETE SLAB:

Reinforced concrete floor slabs are classified into the follow-
ing types:

1. One way solid slab and beam
2. Two-way. solid slab beam
3. Ribbed floors
4. Flat slab or girderless floors solid or ribbed

Each type of the floor system has its own advan·tages in appl i·
cation depending upon the following factors:

1. Spacing of the columns
2. The magnitude. of the loads to be supported
3. Length of the span
4. The cost of the construction

\79

One way sllb: - One way slab is the common type of rein-

forced concrete floor system made of solid slab supported by two

parallel beams. The floor slab is known as one way solid slab, be-

cause the reinforcements runs only at one direction, that is from

beam to beam. The one way slab is comparatively economical for

a medium and heavy live loads on short spans ranging from 2.00

to 3.50 meters long. Although the reinforcement is said to be
running in one direction, additional reinforcements are also

placed in the slab parallel with the beams perpendicular -with the

main reinforcements called "temperature. reinforcement". Usually
No. 3 steel bar is used to counteract the effect of shrinkage and

changes in temperature. It also distributes possible concentration

of loads over a larger area.
Unlike beams and girders, floor slab needs no web reinforce-

ment or stirrups. In the case of heavy load where the shearing

stresses maybe greater than the allowable values, the depth of the

stab is increased.

Plan

One way s!ab reinforcement

Figure 9-24

TABLE 9-2 MINIMUM SLAB THICKNESS

Simply supported 1/ 20
One End continuous 1/ 24
Both Ends Continuous
Cantilever 1/ 28
1/ 10

180

Illustration;

A fully continuous slab is supported by a beam spaced at 12'

or 3.60 meters. Determine the minimum thickness of the slab.

Solution: (English)

1,2 ft. = 144 inches

t = .1M_= 5 inches

28

Figure 9 -25'

Metric Sl:

Span of the slab = 360 em.

t = 3..6.0.. = 12.8 em.

28

Temperature and Shrinkage Reinforcement: - One way floor
and roof slab are reinforced for shrinkage and temperature bars
installed at right angle with the main reinforcements. The Code so
provides; "that in no case shall these reinforcements be plaCed
farther apart than 5 times the slab thickness or more than 18" or

45cm.

Table 9-3 SHRINKAGE AND TEMPERATURE REINFORCE·
' MENT

Mintmum Ratio of Retnforcement Areas to Concrete Areas
Slabs where plain bars are used ....... . ........ 0.0025
Slabs where deformed bars are used ....... ... : . 0.0020
Slabs where wire fabric is used

having welded intersections not
farther apart in the direction of
stress than 30 em. . ..... . .. ...•.•..•.... 0.0018

In using this table, the following illustration is presented:

181

Problem:
. A concrete floor slab having a thickness of (4") lO·cm. is to be

p.ovided w ith No. 3 deformed bars for shrinkage and temperature
reinforcement. Determine the spacing· required.

Solution:
1. Find the cross sectional area of a {12") 30 em. strip of the
slab (used in designing slab)

10 x 30 = 300 sq. em.

2. Referring to Table 9-5 using deformed bars, the value is
0.0020 x 300 = .6 sq. em. This is the required area of
• steel bars per strip of slab.

3. From Table 5-:9 the area of No.3 steel bars or 10 mm dia-
meter is .7854 sq. em. or 78.54 mm2
.7854 x 30 em.

:::: 39.27 say 39 em.
Therefore: No. 3 bars or 10 mm diameter will be used as

temperature bars spaced at 39 em. on center.

One way concrete slab is designed by making .an imaginary
strip of 12 inches or 30 em. wide perpendicular with the beam
that supports the floor. This imaginary strip is considered as a
beam, hence, the design steps and method for rectangu lar be~m is
applied where the width is equal to 30 em. and the depth is the
thickness of the slab. The depth of the fl9or is purely dependent
upon the span length and the magnitude of the superimposed load.

Plan
(lookil!( up)

Seam

Figure 9 ·- 26

182

Placement of Ban in One way Slab- The bending moment at

the center of a fully cont inuous slab is equal. Therefore, there
should be the same quantity of steel reinforcements at each point.

In attaining the same amount of steel bars that will resist posi-
tive and negative bending of the slab, steel reinforcement are bent-
up alternately at the inflection point equal to ~ point of the

span from the face of the beam extended over the sup.port to t

distan~e of the adjacent spans.
The remaining unbent bars are placed at the bottom of the

slab extended at least 15 em . into the slab support or continued
for several spans. For an end-span, the slab is considered as semi-
continuous and that the bending moment is greater. Some designs
provide an add itionaL200Jo reinforcement placed between bent bars
across the supporting beam. The reinforcing bars are then hooked
at the top of the termination end.

Figure 9-27

T¥fO Way Slab - Slabs which · are supported on four sides

where the f loc;>r panel is nearly square is generalfy economical to
employ the two directions of reinforcing bars placed at right
angle with each other. This type of reinforcement will transmit the
loads to the four sides supporting beams or walls.

The code specifies that .thickness of the slab shall not be less
than 4 inches or 10 em. nor less than the perimeter of the slab
divided by 180. The spacing of the reinforcement shall not be
more than 3 times the slab thickness and the ratio of reinforce-
ment shall be at least .0025.

183

Construction Joints: The ACI Code on construction joints
so provides:

1. Joints not indicated on the plans shall be so made and
located as not to impair significantly the strength of the structure.
Where a joint is to be made, the surface of the concrete shall be
thoroughly cleaned and all laitance and standing water removed.
Vertical joints shall also be thoroughly wetted and coated with
neat cement grout immediately before placing of new concrete.

2. A delay of at least until the concrete ir columns and walls
is no longer plastic must occur before casting or erecting beams,
girders, or slabs supported thereon. Beams, girders, brackets.
column capitals, and haunches shall be considered as part of the
floor system and shall be placed monolithically therewith.

3. Construction joints in floors' shall be located near the

middle of the spans of the slabs, beams, or girders, unless a beam
intersects a girder at this point, in which case the joints in the

girders shall be offset a distance equal to twice the width of the.
beam. Provision shall be made for .transfer of shear and other
forces ~hrough the construction joints.

Placement of Steel Bars - Where no ~ent bars are used in the

slab reinforcements, straight bars are used for both the top and

the bottom reinforcements. The bottom bars are extended at least

15 em. into the supporting beams or walls. The top bars are ex-
ttended to
point of the adjacent panels. Top bars for discon-

tinuous floor edges shall be hooked. (See Table 9 · 5).

._...jl Ill 'I \I II -·· I1-L--

- ·-- li ·-I· .......... f--

A :- ...

L - -- -

-- ···--.

-

=IL1 '

;! It

Figure 9 • 28

184

TABLE 9 -4 MINIMUM LENGTH OF SLAB

... ... ~ti;t>~~-~: PlifiilNlIMCCUNIIf~ WITHOUT DfiOf' ""HELS WITH DROP ""NELS

ATSECTIOH

~e~ ... 50 ~.d-l

~...~ 1-o-o-..l
... *e~ :I! Aetlloillder I>
fll!
'

:z: 3"Mox--f /rMox.OIZ$1 .
Mol. 0.125!-..,..,
50 ; tr-24bor 4io.GI'I2"t.li~. ~~
~d- on lxlrs
i-o-6"
I _, .J:;<kt--: ---
-i - ofd
1"'-b-
I"'~ "' 5 llemoilldet I

I *... ... ~i Remai11e1et ~--·-
3 1--b..,j
8 .... 50 l-b-
0 t-e-\..i

I \.z...
~ 1~

CD
... ... ;·"' ~~--~- 14-9--\l
2 I ~~~·0.125/
0 3"Max.- • 6':.0

Remoindel 2.4 bCif dia. or 12.' Min.

~: l--6" i:.~~~~
I... ... ,.e 100 .I 1~1-c---i
~----~-- ' 1--c--f ~-~-~I

1r•

*.. l......~......X
~ ~:t 50
8 3"Yo•. ,, :flotol.

2 I
0.. i RM>Oindef ~6" MOl 0.15/ Mo•OJS.f
6-=.::;
.... so loc !Gil ~orsJo.l 1--c- I· leo--!
1-c{ollb«tl• ·
"'Cll: .0... Flemoindlr ~o• I ~e~
d
'/ I
d
e '~ so
.z... .. ·-;lt-·_j~ 6'~ iJ.
.Re"'oinder
7
..t-6"
3 Moa. 3 Max.
.r
looc ~CitGt spon-1. t Cleor span -.tn fl

e

foee of support foce ol support

* Bent lxlrs at utttior supports ~
mcrr bt UHdif o .-wol
onol71is IS.I!IGde 1 BAR LENGTH FI!OM fACE OF $l$'PORT

MINIMUM LENGTH MAXIIIUM U:HGTH

•tilARK o. I b I e I ~ I f Ia
lENGTH OJ4/nlozo/n lo22l0 lo.30/nlo~ln 02o.t.Jo2t1/.,

Figure 9-29

185

9 -16 RIBBED FLOOR SLAB - Ribbed floor slab is generally

an economical type of floor construction but is applicable only to
medium span length with light or medium load unlike the one way
or two way slabs that could carry heav~ loads.

A ribbed floor slab consists of small adjacent T-Beam wherein
the open spaces between the ribs are filled by clay tiles, gypsum

t iles or steel forms. The t iles are generally 30 x 90 em. with depth
of 10, 15, 20r 25, 30and 40 em. placed at 40 em. on center making
the ribs 10 em. wide.

The concrete surface layer placed on top of the tiles ranges
from 5 to 6.5 em. think. The reinforcement of a ribbed floor sys-
tem consist of two bars placed at the lower part of the rib where
one is bent and the other rem'ained straight, or sometimes, straight
hars are placed at the top and bottom of the rib, Temperature bars
are either No. 2 bars or 6 mm. or wire mesh which runs at right an
. gle with the ribs.

Figure 9-30

Gypsum Tile Filler - This is a lightweight material for the
floor which also provides a flush ceiling finish · The common width
is (9") 23 em alt hough some other sizes are available. Gypsum t iles
are placed at .60 m. on center forming a ri b or web of 13 em.
wide. On the r;ontrary if 30 em. blocks are used, they are also
installed at 40 em. from the center same as that of clay tiles, pro-

viding 10 em. wide ribs. The Code specifies that the maximum

span of ribbed floor slab should not be more than 24 times the
total depth of slab and rib combined. ·

186

B

Section A-A

Section B-B Plan

Figure 9-31

Metal Tile Filler: - Are generally in the form of domes en-
closed on four sides, this is sometimes called as "tin pan" cons-
truction. The metal forms are .90 Ill. long with various depth from
.15, .20, .25 and .35m depth placed at the center to make a rib
from 10 to 17 em. wide at the layout portion. The form widths
are either .50 m or .75 m. The .50 m forms are placed at.•63 m. on
center making a rib of 13 em. at the bottom.

The metal forms maybe removed or left in place. The layer of
concrete placed on top of the metal forms ranges from 5 to 7 em.
thick.

Steel pen form~

~··,jV'\Er·:·~·.f

Section

Figure 9 • 32

Flat Slab- Flat slab floor is a rectangular slab directly support-

ed by columns without beams or girders. The slab is either uni·
for'm in thickness or provided with. square symmetrical area di-
rectly above the column reinforced with bars running in two
directions. The increased area directly above the column is called
drop panet or simply drop. On the other hand, a flared head is
· employed in the construction of a flat-slab floor making a capital
of the column .

187

When the column design is not provided with capitals, a straight
flat underneath is provided in the slab throughout the system,
which is called flat plata construction.

Section

r---------------------l

Plan

L-~===~-~-~~1= 1J;~-;=. ~;=I
.w .W=Sectlon==;-A-A
Aan

Figure 9-33

The flat slab floor system is generally economical not only in
terms of materials as well as labor and is even the most suitable
type of construction for industrial buildings having a wider live
load and also for building in which the use of capitals are not otr
jectionable. ·

The advantages of the flat floor slab are:
1. Simplified formwork
2. Better light in the absence of beam and girder.
3. Advantage in height for a clear story heights.
4. Uniform surface for suspended water sprinkler system.
5. Piping and shafting
6. Absence of sharp cor·ners
7. Better resistance to fire.

188

9-17 THE ACI ON CONCRETE JOIST FLOOR
CONSTRUCTION

The American Concrete Institute on concrete joist floor cons-
truction so provides:

1. The joist ribs shall be at least 4 in. or 10 em. wide, spaced

not more than .75 m clear, an~ a depth not more than 3i times

their minimum width.
2. Ribbed slab construction shall conform to the limitations

as provided for by the above thickness and spacing and the arrange-
ment to span in one direction or two orthogopal directions. Other-
wise, it shall be designed as slabs and beams.

3. When permanent burned clay or concrete tile fillers of ma-

terial having a unit compressive strength at least equal to that spe-

cified strength of the concrete in the joists are used, the vertical

shells of the fillers in contact with the joists may be included in

the calculations involving shear or negative bending moment. No

other portion of the fillers may be included in the design calcula·

tions.

4. The thickness of the concrete slab over the permanent

fillers shall not be less than 4 em.. nor Less than rl- o'f the clear

distance between joists. In one-way system reinforcement shall be

provided in the slab at right angles to the joists.

5. Where removable forms or fillers not complying with the

provision· of No. 3 as stated above are used, the thickness of the
-frzconcrete slab shall not be less than
of the clear distance

between joists and in no case be less than 5 centimeters. Such

slab shall be reinforced at right angles to the joists with at least

the amount of reinforcement required for f lexure.

6. Where the slab contains conduits or pipes, the thickness

shall not be less than 2.5 em. plus the total overall depth .of

such conduits or pipes at any point. Such conduits or pipes shall

be so located as not to impair significantly the strength .of the

construction.

189