Beton Bertulang J Thambah Sembiring Gurki.

2

M

'

•

pb = ~ d 0,85f'c) x fy

600 600+ fy

= 0 85 [ 0,85*28] x[ 600 ' fy 600+400

Jx10_

3

= 0,030345 •

Diambil p = 0,5% ;

•

Sebut

= J.t_ = 0,0035

400 p =0,5 * 0,030345 = 0,0 1522

I

Rn

I=

w = px fy

f'c

1 --

•

pm in

-

-

-

0,0152x400 = 0 21714 28 '

---'

= w * fc ( 1 - 0,59 * w) = 0,2 1714 * 2~ ( 1 - 0,59 = 5,69 N/mm-~

* d2

·

Mn = 0,5Rn

3

b

= Rn * b = 0,5 d

d

=

b= 0,5

* 518

Mn

Bab l_Balok

3

* 0,1 0857) '

' 6

395,6* 10 0,5 * 5,69

=518 mm

= 259 mm 19

Cek Mn yang ada dengan persamaan

:EM =O) Mn )Mn kop :EH =O "TRIAL" I Coba Ukuran Coba d

•

•

=700mm

b

= 350 mm ;

b/d

=0,5

Berat sendiri,

qdl

= 24 * 750 * 350 * 10 - 6 =6,3 K.Nim' sama dengan asumsi diatas (OK) qdl

jadi Mu Mn •

As = p

= 316,48 KN m = 395,6 KN m.

* b * d =0,0076 * 350 * 700 = 1862 mm2

Pakai tulangan D 25 mm nt =

1862

~IT(25)

2

== 4 bh

4

Garis netral a= _A_s·....::i..::...Y_ 0,85 f' cb 89,4

x=--

0,8

•

Mn

=As

=

1862·400 = 89 4 mm 0,85 * 28 * 350 '

= '111 ,8 mm

* fy [d - a/2)

= 1862 *400 (700- 89,4/2) * 10- 6 Mn ada

20

= 488 KN- m > Mn

perlu

= 395,6 KN - m

Seton Bertulang

!@fti.@iil 1. Gambarkan penampang balok (lengkap dengan dimensi) dan dilengkapi dengan diagram tegangan dan regangan. Dari diagram tegangan dan regangan:

£s'

- 80 * 0 003 =0,002 192 = 297297 '

£y

400 = 2x10 5 =0,002

£s' > £y berarti tulangan tekan le1eh.

As'

= As'(fy- 0,55 f'c) = 707 xlO' = 1909,5 mm2

T

= (T 1 + T 2) = Cc + Cs

Jadi Cs

(400 - 0,85 * 35)

= As2

X

T2 =As' . fy

fy

As

= As, + As2 =As 1 + As'

T

= 2474 + 707 = 3181 KN 3

As perlu

= 3 181x 10 =7952,5 mm2 400

Contoh Soal- 5 Diketahui:

d

= 660mm

d' = 80 mm fy = 400 Mpa

b

= 350 mm

f'c = 35 Mpa Mn = 1450KN-m

Pertanyaan: Rencanakan penampang (b dan d) serta As dan As' jika ditetapkan garis netral

X = 0,375 Xb. Periksa kekuatan penampang Bab l _ Balok

2 1

Solusi: a) Perencanaan Penampang Balok X

= 0,375 * 396

= 1485 mm

a

= 0,8

*

= J 18,8 mm

Cc

= 0,85

148,5

* 35 * 350 * 118,8

= 1237 KN

=0 =Cc= 1237 KN = 123 7 (660 -

118 8 • ) 2

=743,3 1 KN 1237 x 103 = - - - 400

=3092,5 Mn2

.

=Mu - Mn 1 = 1450-743,31

Mn2

=706,7 KN -

Cs

= T2 = 7067700 = 1218 45 KN 660 -80 '

m

Kontro l regan gan bagi baja tekan; As'= As 2

c:s'

5 80 = 148148,5 * 0 003 • •

c:s' < c:y

=0,002

= 0 00 138 '

berarti tulangan tekan belum Jeleh.

sehingga fs'

= c:s' . Es

= 0,001 38 * 200000

= 276 MPa

22

Seton Bertulang

3

=As?= 1218,45x 10 - (276 - 0,85*35)

As'

= 4948 mm2 = As' - 1218,45 10.1- 3046 2 - (400) x ) mm = A s 1 + Asz

As

= 3046 + 4948

= 7994 mm2 Garis netral tidak ditentukan clulu , X

X

d

b) Memeriksa Kekuatan Penampang ts'

Cc

=

X * 0,0030 X -80

f'c

b

fy

35

350

400

Cs

T

1579.356 16 !R.243 3 197.6

Bab l _Balok

T - Cs

a

1579.356 !57 .CJ735

d

As

As'

rs

700

7994

4948

356./:l

148.5

23

Penjelasan Dengan " TRAIL AND ERROR" dicari nilai f's yang memenuhi syarat, melaiui persamaan- persamaan berikut: Dicoba harga f' s (f's sembarang) ..... (*) Cc

= 0,85 f'c. b. a = 0,85*35*350*a

Cs

= As' (f's- 0,85 f'c)

Cs

= 4948 (f's- 0,85 * 35) = As . fs = 7994 * fs

Cc

T = Cc+ Cs T :EM=O :EH=O

..... (1) ..... (2) ..... (3) ..... (4) ..... (5) ..... (6)

Melalui persamaan (1) s.d. (4) didapat harga a Maka: X = _!!____ 0,8 Kemudian dari diagram regangan dihitung harga c:s' dan f' s dengan

=200000 Mpa Maka harga f' s =c:s' * Es dapat dihitung

Es

Sebut f 's (**) sampai dengan f' s<*>di atas (sampai konvergen), seperti di atas diambil f's <*> = 356,8 Mpa f's<**) = 356,92 Mpa

24

Seton Bertulang

Sudah konvergen Mn dihitung = Mn

trial stop. OK. =3850 KNrn > Mn perlu = 1450 KNrn

Contoh Soal- 6

D iketahui: f'c

Es

= 30Mpa 5

eCU

= 2. 10 Mpa Ey

Beban mati merata Beban hidup merata Beban hidup terpusat

=0,003 = 0,00175

= 8 KN/m' = 10 KN!m' = 50 KN

..

300

...

Pertanyaan: a. Periksa kekuatan balok dengan penulangan 5D22 b. Apabila tulangan pokok salah pasang sehingga kedalaman efektifnya menjadi d = 500 mm, Periksa apakah balok rnasih kuat?

Solusi A Memeriksa Kekuatan Balok dengan Penulangan SD 22 Step 1. Analisa Pembebanan Berat sendiri balok = 24 KN/m3 * 300. 600 . .10 _6 Beban mati mcrata qo = 8 KN/m' qo = 12,32 KN/m' Bab l _Balok

=4,32 K.Nim'

25

= 10 KN/m' =50 KN/m'

Beban hidup merata ql Beban hidup terpusat pi qu

;;;; 1,2 (12,32) + 1,6( 10) = 30,784 KN/m'

pu

=1,6 (50) = ((1/8)

Mu

=80 KN

* qu *

2 L)

2

= (1/8 * 30,784 * 6

Mu

Mn perlu

= Mu= ci>

* L) * 80 * 6) = 258,528 KN/m'

+ ((1/4) * Pu )

+(lA

258 528 = 323,16 KN/m' • 0,80

Step 2. Analisis Kekuatan Balok

. 2 +-A!--..-'

c =0,85 f"c . a. b

Z =d-a/2

600

5D22 -

1r • • L-L----

~

•••l

300

t so

Ns

eS > Ey

Ill>

:EH= OHk.Hook Es = fy t:y

Ne

= N fy

= Es . e.y

=2.

0,00 175

10 5· .

0.85 f'c. a. b =As. fy = 350 Mpa

0.85f'c.a . b =p. b.d.fy

26

Seton Bertulang

p balance

_ 0,85f'c·a fy·d

p balance

= ~ 0,85.f'c . ( c'cu ) 1 fy c'cu+£ y

_A

-pi

f'c

= 30 Mpa > 28 Mpa

~I

= 0 85 - 0 05 ( ,. '

pb

= 0,85 0,85 * 30 (

0,85f'c . ..:.. fy d

30 28 ) = 0 8355 6,9 '

350

0,0032 ) = 0,03975 0,0032 + 0,00175

pmax = 0,75 p B = 0,029514 =0004 Pmin = ~ 350 , p ada

= As(5D22) = 1900,6635 = 0 01152 b.d

300 * 550

'

dapat dipastikan tulangan baja tarik sudah leleh jadi: a=

As.fy = 1900,6635 * 350 = 86 96 mm' 0,85/'c·a 0,85 * 30 *300 '

a= ~I·

X

x

= !!__= /31

86 96 • = 102 306 mm 0,85 '

= d- a/2 = 550Mo

86 96 • = 506 521 mm'

2

'

=T . Z = As . fy . Z = 1900,6635 * 350 * 506,521 . 10- 6 KNm'

Mo .... = 337 KNm' > M 0 perlu = 323,16 KNm' memenuhi kriteria design.

Bab l _ Balok

•

27

Periksa Regangan

s'cu ss - = -x d- x

cs

-

= E'cu(d -x) X

_ 0,0030(550-104,08) 104,08

cS = 0,01371 > £y = 0,004 Scbelum beton mcncapai regangan c'cu = 0,0030 maka regangan baja rarik lebih du1 u lelch j ika asumsi diatas bcnar.

Solusi B Apabila tulangan pokok salah pasang, sehingga kedalaman cfektif menjadi d = 500 mm, apakah balok masih kuat? Periksa p ada

=

As(5D22)

b·d

6635 = 1900 · = 0 '01267 > p min = 0 ,004 300 * 500 p ada

a tau

=0,01267 < p max = 0,029514

maka dapat dipasti kan tul angan baja tarik sudah Jelch. Z = d- a/2 = 500 Mn kapasitas

86 96 • = 456,52 mm' 2

=T . Z ==

As (5022) . fy . Z

= 1900,6635 * 350 * 456,52. w-6 Mn kapasitas

KNm

=304 KNm' < M perlu = 323, 16 KN m'

Tul angan pokok tarik yang salah pasang dengan kedalaman d tidak kuat menahan beban.

28

= 500

mm,

Seton Bertulang

Contoh Soal · 7 Diketahui: f'c

= 25 Mpa

e-'cu

= 0,0032

Es

= 2 . 105 Mpa = 0,002

t-y

PL =25 /KN

qo = 10 KN/m

~I ~

q,= IS KN/m

fHyAAi~

-~- 4000 ~m :

8000 mm

T

Pertanyaan: a. Sketsa rencana penulanga n lengkap dengan kontrol regangan dan kekuatan balok. b. Gambar diagram regangan pegangan.

Solusl A Sketsa Rencana Penulangan Step l.Analisa Pembebanan

= 36

pu

= I ,2 ( 10) + I ,6( 15) = I ,6 (25)

Mu

= 1/R . 36 . 64 + 1/.1 . 40 . 8

= 368 KN/m'

qu

KN/rn'

= 40KN

Mu 368 Mn perlu = - = - - = 460 KN!m' = 460. 106 Nmm'


Bab l _ Balok

0,80

29

Step 2. Mendesain Dimensi b,d

ECU

=0,0032 d

Nt =As.

:EH = O

Hk.Hook Es = fy

Ne = Tfy

=Es. cy =2. 105 . 0,00175

r 0,85 rc . a . b = p . b . d . fy

0,85 c . a . b = As . f y= 350 Mpa

p balance

= 0,85f'c· a = ~ 1 fy ·d

p balance

=~ 1

0,85f'c -~ fy d

0,85f'c ·( ,&'cu ) fy E cu+&y

=0,85

0,85 * 25 ( 0,0032 ) 400 0,0032 + 0,002

=0,0278 pmax .

pmm

30

=0,75 Pa =

1,4

400

= 0,02084 = 0,0035

Coba ambil PR = 75% } pmax = 0,01563

Seton Bertulang

m

fy = _____::...:...___ 0,85f'c

= -400 -0,85*25

= 18,8235 Rn

= p. fy (1 - Yz p . m) =0,01563 * 400 (1- Yz. 0,01563. 18,8235) =5,3323 N/mm2

Rn

= Mnper1u bd 2

ambil b d

=280 mm'

=

Mn perlu = b.Rn

460.10 6 280 * 5,3323

=555 mm' kontro] dengan ratio penampang r =b/d

=0,2- 0,6

untuk perencanaan r = 0,5.

R

= b/d =

280 555 =0,5045 memenuhi syarat.

Step 3.Menghitung As perlu untuk b =280 mm', d Rn

=555 mm'

_ Mn perlu _ 460.106 bd 2 - 280 * 555 2

= 5,3335 N/mm 2

p perlu= =

~(1- ~1- lm:n J 1 (1 18,8235

= 0,0156 Bab l _Balok

1- 2 * 18,8235 * 5,3335) 400

konvergen dengan p rencana= 0,01563 "OK" 31

Aspcrlu = pperl u . bd

= 0,01563 * 280 * 555 = 2429 mm2 5D25

40

40~

200 H 602

d =555 5 D25

14'7. . · · · · . . . ..... .. L-----------------~ 33.3

33.4

33,1

~~~

I~

280

·I

Ukuran balok dikontrol dengan dimensi. Mcnumt SK. SNI. T. 15 - 199 1 03 , Diman a sengkang D.lO. H = 555 + 25/2 + 10 + 25 = 602 mm. Jadi beton dicor dengan split 2/3.

32

Seton Bertu lang

Solusi B Kontrol Regangan dan kekuatan balok dilengkapi dengan diagram.

H 602

280

1:H

=0

Ne

=T

0 ,85 f'c 0a 0b =As 0fy

a

= AS(5D25)ofy 0,85f ' coh

a

=~ 1

o

X

X

2454,37 * 400 = 164 999 mm' "" 165 mm' 0,85.25 * 280 '

165 =-j3a = -0,85 = 194' 12 1

z

=d -

c

=T

a/2

=555 - -165 = 472 '5 mm' 2

= As (5D25)0 fy = 2454,37

* 400

=981747 ,7N

= 982 KN

Bab l _ Balok

33

Mo l
=T .Z

= As. fy . Z

= 982* 472,5. 10 - 3 KNm' Mo t Mn perlu 464 KNm' > 460 KNm'; memenuhi syarat, kelebihan momen 1,3%

1. 2 Perencanaan Penampang Persegi terhadap Momen Lentur dengan Tulangan Rangkap (REF, PB'89 ; SNI T14-199 1 -03) Pada prinsipnya pemasangan tulangan tekan (tulangan rangkap) untuk mengurangi lendutan/tidak menambah kekuatan. Perhitungan tulangan rangkap dapat diuraikan sebagai berikut: Menentukan apakah tulangan tekan itu diperlukan atau tidak, dengan cara membandingkan kekuatan momen perlu dengan kekuatan momen dari penampang tersebut tapi memakai tulangan tarik maksimum saja. Atau d il akukan kontro l lendutan dengan menentukan As dan As'. Memeriksa As yang diberikan tidak menghasilkan harga X (garis netral) lebih besar dari 75% Xb, dan tulangan tekan memang sudah leleh. Bila tulangan tekan belum leleh, maka dihitung lagi dengan persamaan kesetimbangan dan memasukkan nilai sebenarnya.

rs

Disarankan diambil X = 0,375 Xb.

1.2. 1 Kriteria Kondisi Tulangan Tekan (As 1) Leleh Step-step perhitungan:

34

( 1).

As 1 As p = bxd ; p = bxd

2 ( ).

p

1

b = /}1x0,85X;tc ( 600 fy 600 + fy

J Beton Bertulang

(3).

p max = 0,75 pb

h

d As

{p - '( 1 _0,85f' c)~ os 5 p (f'cxd'J( 600 J p fy ' fyxd 600+ fy

(4)

1

J ika Es ~ Ey~ untuk cek tulangan tekan leleh.

(5)

't· = 60j f

1_ 0,85fi,xf'cxd

. " vl

J

1

(10- p') .

fyxd

1.2.2 Prosedur/Mekanisme Perhitungan Diketahui: b, d, d 1, As, As 1, f 1c, fy ( 6).

As 1 p = bxd ; p

(7).

. 1,4 pmm = fy

As 1 = bxd

Kontrol p > p min. Jik
{p - p'}~ 0,85~tcxd'xP1 ( fyxd

Bab l_Balok

600 600+ fy

J

3S

1

= fy

Jika step 5 ya, tegangan leleh baja tekan f s Jika step 5 tidak, maka: 1

f

.

(8).

1

_

S -

( _ 0,85/31 x/cxd 600 1 . I) fyxd - p

(f)

j31 x0,8S~tc( 600 P.Jl = fy 600+ fy

)

t·

< Y

J d .tarnb.l I

p = O,Spb

(9).

p

~ 0,75ph + p'f's p .

fy

Jika step 9 tidak memcnuhi, penampang diperbesar Jika step 9 ya, maka:

( 10) . a

= AsJ..jy -

1

1

A sxf s

0,85xf 1cxb

c!>Mn = <1>l(A.qfy- As 1xf' s )(d- ~)+ A 1s~t s(d - d 1 )j dimana~ 1 =f* (f c) 1

Catatan: Tulangan tekan berperan utama untuk mengurangi rangkak beton clan lendutan yang tetjadi . Sedikit perannya untuk meningkatkan kekuatan

Contoh Soal - 8 Diketahui:

36

b

= 355 mm

d dl

=660mm

A 1s

= 1100 mm2

As

= 3300 mm 2

= 75mm

Beton Bertulang

1

= 35 MPa = 400 Mpa

f c

fy

Pertanyaan: hitung Momen Nominal (Mn) penampang. Solusi: =

p

~ = hxd

3300 = 0 014085 355x660 ' ··

PI = AI s =

bxd

1100 = 0,004695 355x660

1 p min = ,4 =~=00035 f y 400 , Kontrol:

p > p min; OK! (

c)

1_ 0,85xt p \ fy 1{

_

p

1

2': 0 ,85xl cxd x/31 (

fyxd

600 ) 600+ fy

0,008692 2': 0,004057 (OK!)

berarti £s 2': £y atau f 1s = fy b=

p

1

/31x0,85xf c ( fy

600 ) 600+.fy

= 0,8x0,85x35 ( 600 ) 400 600 +400

= 0,0357 l t •l

p ~ 0,75xpb+ P ·. s .fy

0 0004695 400 x 0,014085 ~ 0,75x0,0375 + • 400 0,014085 ~ 0,031470 a= As~fy - A sxf s 0,85xf 1cxb 1

Bab l_Balok

1

37

_ (3300x400)- (1100x400) 0,85x35x355

= 83,32 mm a=~~* X

83 32 • = 10415 mm 08 ' ' 104 15 • x100%0 = 15 78%0 660. ' X =

15,78 % < (3/8)x d

(bagus)

ct>Mn = l(AsxfY- A 1sxf 1 y )(d - a/2)+ (A I s~t s xd -

d1 )J

Mn = 0,8[(3300x400)- (1100x400)(660- 41,66)+ (1100x400)(660-75)] cf>Mn = 641231360 Mn

= 641231360 = 80 1539200N- mm 0,8

= 801539,2N-M = 801,5392 KN-M = 80,15392 T- M (Momen Nominal) Kapasitas Momen $Mn

=Mu

0,8 * 801,539 =Mu Mu

= 641,23 KN- M

(Momen perencanaan/momen terfaktor)

1.2.3 Perencanaan Tulangan Perencanaan Tulangan Rangkap adalah sebagai berikut:

Langkah-langkah: Dalam Tulangan tunggal dihitung momen maksimum yang bisa dipikul oleh penampang, Mn Cek apakah Mn yang didapat Iebih besar dari Mu yang bekerja (Mn ;:::: Mu) Jika Ya, maka dite.ntukan L\ Mn = Mn- Mn~. kemudian L\ Mn sebagai penyebab tulangan tekan

38

Beton Bertulang

Jika Tidak, maka Mn =Mu dan dengan tulangan As~ 0,75 * Xb Den gan asumsi tulangan tekan dalam poin a diasumsikan meleleh. Kemudi an hitung tulangan tekan A 1s Jika tulangan tidak meleleh, maka perhitungan harus dimulai lagi dengan f 1s adalah f 1s sesuai dengan perhitungan yang didapat. 1 Sebaiknya dicek Jendutan yang terjadi dengan As dan A s yang ada.

Contoh Soal - 9 Diketahui penampang balok dengan ukuran seperti gambar dibawah ini: f 1c

= 35 MPa

fy

=400 MPa

Es

=200000 Mpa *10mm

=70mm

Beban-beban yang bekerja: Akibat beban mati (DL)

= 280 KN- m

Akibat beban hidup (LL)

= 500 KN- m

Ditanya: Rencanakan/hitunglah tulangan tarik As d~n tulangan tekan A s perlu, untuk dapat menahan beban/momen diatas. 1

Solusi: Step 1. Mu = ( 1,2 * DL) + ( 1,6 * LL)= (1,2 * 280) + (1,6 * 500) = 1136 KN - m Bab l_Balok

39

Step 2. <1>

Mn =Mu =Mu=1136 = 1420KN - m
Mn

Step 3. Garis Netral dalam kcadaan " balanced" atau berimbang

ecu

=0,003

1~1

lx

- -·- - -·

)I esb = ey

Xb

=

0,003 d 0,003 + 0£sb

400 Esb = fy = = 0 002 Es 200000 ' Xh =

X ada

0 003 + 630 = 378 mm ' 0,003 + 0,002

= 0,75 * Xb = 0,75 * 378 = 283,5 mm

40

Beton Bertulang

Garnbar diagram tegangan

a 1

fc

=P1* X , =35 maka p, =0,8 = 0,8

* 283,5

=226,8 * f 1c * b * p, * X = 0,85 * 35 * 350 * 0,8 * 283,5 = 2361555 N

C m ax = 0,85

=2361 ,555 KN LH As 1 * fy

=C max

=

Cmax

fy 2361 ,555 * 1o'

= -----

5903,898

=5903,898 mm2 LM

=0

Mn t

= C * ( d - 0,5 a)

=2361 ,555 * (630 -

113,40) 10 -3

= 1219,98 KN-m Cek:

Mn 1= 1219,98 < Mn balanced =1420 KN · M akajika pcrlu tulangan tekan Bab l_Balok

41

Step 4. Momen yang dipikul tulangan tekan adalah ~ Mn. Maka:

~

Mn = 1420- 1219,98

=200,02 KN -

m

!::J.Mn sehingga: Cs = - d-d1

=

200,02 * 10 3 630 - 70

=357, 18 KN

Cek Tulangan Tekan £

1

c=

630-70 630

0,003 = 0,00267 > cY

Maka tulangan tekan sudah leleh.

42

Seton Bertulang

1

As

d

Dari diagram tegangan di atas didapat persamaan:

T2 1

A s

=Cs

= As2 T, = -fy

835,54 * IOJ

= - --

400

-

= 2088,86 mm 2

= 2088, 86 mm 2 As

I = As+ Asz = 5903,888 + 2088,85

=7992,738 mm 2

Bab l _Balok

43

700 As = 7992,735 xtx70

I(

350 mm

>I

Contoh Soal- 10 Diketahui: momen, beban mati =300KN - M Momen beban hidup =500KN - M

=0,8

b d

= 350mm = 660mm

cCU

= 0,0030 =80 mm

d' Es f'c fy

=2 * 105 MPa = 35 Mpa =400 Mpa

Pertanyaan: Hitung As dan As' ? Solusi: Momen berfaktor : Mn = 1,2 * 300 + 1,6 * 500 = 1160 KN - M Mn 1160 Mn perlu = - = - - = 1450 Kn - M
44

Seton Bertulang

Garis netral:

=

0 003 · x660 = 396 mm 400 0 003+ ' 200000

X maks

=0,75 * 396 = 297 mm

a

=0,8 * 297 =237,6 mm

~I

=0 85 '

0 05 [ '

35 28 ] 69 )

= 0,799 z 0,8 C

max

=2474 KN

T=C

As

max ,

As

max

Mn ,

T tulangan tunggal = .fy

2474 J = --xlO =6185 mm2 400

(max)

Mn1

tulangan tunggal ; Mnz

= 2474(660-

237 6 · ) 2

* w- 3

= 1040 KN -M Mn 1 < Mn pcrlu Jadi perlu tulm:gan tekan As'=? Mnz

= Mn-Mn1

Mn 2

= 1450- 1040 = 410 KN- M

Cs perlu =

Mn 2

- =

d-d

410 x10 3 = 707 KN 660-80

Mengecek tulangan tekan, apakah leleh?

Bab 1_ Balok

45

1.3 Penampang T/Balok T 1.3.1 Umum Struktur balok T dilapangan seperti terdapat pada konsol, fondasi clan balok menerus pacla sistem lantai sebagai berikut: a. Konsol

4

a

b. Fondasi

~-JT

I

d I

r+ b

I

I

T ~M .. . .l~ . ~-

I Pcnampang b dan c

46

Pcnampang d

Beton Bertulang

Penampang d bukanlah balok T Penampang T

! ! t t! t trtt t !IT:u l

~

!I

·

!I

if

!g

i

.

~-- .

I . .I .

· I I·

~e

I

I . .I I

Penampang e, penampang T

,...t....L-L-J )

M

Tx

- · - -- -- -- -T- ·-·-

Penampang f dan g, hukan penampang T

} ·--)M -·-·-·-·-· Bab l _Balok

47

Kenapa terjadi penampang T ?

A

-r)~. xl

h

I(

bm Gmnbar 1

· r~-M

lt bm

71

Gambar2

lo '

Dari gambar 1 diatas Lerlihat penampang perseg1, dimana volume bclOn besar tetapi bagian A tidak diperhitungkan , atau mernikul keku atan. Scdangkan dari gambar 2 jika b < bm, sedemikian rupa seh ingga garisne t.ral seperti X maka terjadi balok; pe nampang T, maka volume beton berkurang.

1.3.2 lebar Efektif Penampang Seton yang Menerus (bE), menurut SK SNI T- 15- 1991 - 03

I<

>I

>I

I<

I<

>I

IL hw/2

In

I~

4 8

Beton 8ertulang

Lebar Efektif untuk lantai menerus Balok Tengah I.

L

bE~bw+ -

12

I

2. bE ~ bw + - Ln 2

Balok Tepi I.

bE~

1 - L

bE~

bw + 8 t

4

bE::;:; Ln

NB : Dari kctiga ketidaksamaan diatas diambil nilai terkecil. Lcbar Efektif untuk lantai tidak menerus (balok berdiri sendiri) bE::=:; 4 bw t :=;; 0,5 * bw t = tebal pelat

1. 3. 3 Pemeriksaan Penampang Balok T terhadap Lentur 1. Jika Letak Garis Netral (X) Berada dalam Flens

&c

=0,003

I
.,

.1- ·-·-·-·-·-·-·-lf.lih·-·-·-I~ . .I ' I

I

I h

d As

t:

d-a/2

T

0

I<

Bab l _ Balok

c

& <& s

)

>I

49

0 85 * /c*b Jika a ~ t, maka As< ' £ fy

Untuk X = -

t

/31

Berlaku kescrimbangan dengan asumsi tulangan tarik sudah leleh C-T=O C =T

Dimana:

c = 0,85 * f 1c * be * a T = As* fy

Sehingga: 1 0,85 * f c * bE * a- As * fy As * fy a= ·0 ,85*f1c*bc: L.M=O Mn - As* fy

* ( d - 0,5 * d )

2. Jika Letak Garis Netral ( X ) > tebal Flens

1~-- > 1

0,0031 1 ~ ;:.

Q,85* f 1c 1~1 a/2

1 .-C

d - a/2

As £ _, < £ 1

1~ ~I

so

Seton Bcrtulang

Ct

=0,85 * f 1e *a* bw

C2

= 0,85

1:H

=0

Tt

=Ct

T2

= c2

T

=TI +T2

* f 1e *(bE- bw) * t

=C 1+C2. sehi ngga: 1

1

0 ,85 * f e * a* b w + 0,85 * f e * (bE - hw)

*t

= As* fy 1 c*(b £ - b )*t a= A\'' * .~~~. - 085f > 0 ,85 * j I C *h... IV

a=

T - C, 1 0,85 * f c *h.,.

Contoh soal - 11 Diketahui penampang T seperti di bawah ini:

0,85 * f c 1

_ 1' 200..

f.-- 2500 mm

~~

-+j

_ - ·- - ·- ·- - ·-

\!1

a/2

~ llO...-·-·-·-·-~ ~-1--4-1-...L--.+- C 600mm

(d -0,5*d)

As

-

Lso

V 1....---~ T

1<:30(~1

Bab l_Balok

51

Jika f 1c

fy

= 28 MPa = 320 MPa

As

= 6000 mm2

Hitung Kuat Nominal, Mn penampang.

Solusi: Cek tintuk X = 130 mm, berapakah As perlu? C

=0,85 * f 1c *a* bE

Asumsi c.s > c.y

T

=As* fy

C-T

=0

0,85 * f 1c *a * bE- As* fy = 0 0,85

* 28 * 130 * 2500- As * 320 =0

320 *As

= 7735000

As

=2447 1,875 mm2

Jika tinggi (a) dari balok persegi Jebih kecil dari (t) untuk As ada = 4051,6 mm 2 , j adi T perlu.

* As = 320 * 6000 = 1920000 Newton

T

= fy

C

= 0,85 * f c *a* bE

a

=

L.M

=0

Mn

= T*(d -~a)= 1920000*(600 -

1

C 0,85 * f' c *b£

=

1920000 = 32 27 mm 0,85*28*2500 '

~ 32,27)

= 1121020800 N - mm = 1121,02 KN - m

Con"toh Soal - 12 Ditentukan penampang T seperti gambar dibawah ini. Jika garis netral (X ) > t, t aclalah tebal flens.. 52

Seton Bertulang

0,003

t/2

X

L...-----+ T

Ji ka: f 1c

=22 MPa,

maka, ~

= 0 ,85

a

= 0 ,85 X

fy

= 320 MPa

As

= 8000 mm

2

Pertanyaan: Tentukan Momen Nominal (Mn) dan Penampang T di atas Solusi: (Asumsikan £S > £y sehi ngga fs > fy) Menurut SK SNI T- 15 - 1991 - 03 bE::;; 4 bw Cck Lcbar Efck tif, bE = 800 ::; ( 4 * 350 mm) = 800::;; 1400 mm ( OK ! )

Cek Tebal Flens, t < bw/ 2 150 mm < 350 I 2 150 mm< 175 m m (OK !)

Jadi ukuran- ukuran di atas memenuhi syarat. Tahap perhitungan se lanj utnya:

Bab l _ Balok

53

1

Cs

= t * CbE- bw ) * 0,85 * f c = 150 * ( 800- 350 ) * 0,85 * 22 = 980950 Newton

Cc

= 0,85 * f c * bw * a = 0,85 * 22 * 350 * a =6545 *a

T

=As* fy = 8000 * 320 = 2560000 Newton

1

LH =0 Cs- Cc- T = 0 980950 + ( 6545 * a) - 2560000 = 0

a a X

=391 mm

=0,85 * X 391 = -=48875mm 0,85 ,

Sehingga: Cc = 6545 * 391 = 2559095 Newton

LM=O 1 a Mn - Cs *(725- -) -Cc *(725--) =0

2

.

2

150

Mn - 980950 « (725- -

2

391 )- 2559095 * (725 - - ) = 0

2

9

Mn = 63617500N - mm*l,355 *10 Mn = 1992658303N - mm Mn = 1992.66KN- m

54

Seton Bertulang

Contoh Soal-13 Tulangan Tarik ( As ) maksimum penampang T p maks = 0,75 pb ( balanced) Garis netral balanced, Xb

Xb =

.

0 ,003

%

0,003+ · E•s

data lainnya: f 1c

=24MPa

= 360 MPa =2 ,1 * 105 MPa

fy Es

Penampang T seperti gambar di bawah ini.

lI

~50

I X

I

ab

700 ·- ·- ·- - ·- ·- - ·-·-· As

·

= _..fy_ = E,

·>I

I~

l~>l

c: \"

>.ey

£s

350m

360 s = 0,00171 2,L * 10·

Dari diagram regangan : 0 003 • Xb = 700 = 445,86mm 0,003 + 0017 j clari tepi atas .

/3, = 0,85 Bab l _Balok

o,os( I

I

c- 28,6 ) 7.15 55

p, p, ab

=0,85 untuk f 1c > 28,6 MPa =0,85 jika f 1c < 28,6 MPa dan nilai minimum p, =0,65 =P1 * Xb = 0,85

* 445,86 "" 379 mm

ab> t, maka penampang adalah T murni Cc

=CRm = 0,85

* f 1c * bw * ab

=0,85 * 24 * 350 * 379 = 2706060 Newton

Csl

= Cflens

=0,85 * f1 c * ( bE - bw ) * t =0,85 * 24 * ( 750-350 ) * 150 =1224000 Newton L H=O T -Cs 1 - Cc =0 As * fy - 2706060 - 1224000 =0 As * 360 = 3930060 Tulangan B alanced ; Asb

=10916,8 mm2

Tulangan maksimum

= 0,75 Asb

As maks

=8187,625 mm 2

1. 3.4 Perencanaan Balok Penampang Balok T Yang dimaksud dengan perencan aan disini adalah merencanakan luas tulangan yang perlu dengan dimensi penampang sudah diketahui .

Contoh Soal- 14 Diketahui penampang seperti gambar berikut ini:

56

Beton Bertulang

Jika Momen terfaktor, Mu 1 Mutu beban, f ·~ Mutu Baja, fy

= 1500 KN -m =22MPa = 360Mpa

Pertanyaan: I. Apakah a> t? 2. Rencanakan tulangan As. Solusi: Akibat momen lentur yang terjadi = 0,80 I. Cek a> t Jika a

=t

C

= 0,85 * f

1 C

* bE * t

= 0,85 * 22 * 2000 * 200 = 7480000 Newton =7480 KN Mn

Bab l_Balok

=

c(

d-

~) = 7480(0,65- O,~ 2 ) = 4413,20KN

s 7

* Mn

= 0,85 * 4413,2

Mn

=3751,2 KN - m< M terfaktor = 1500 KN - m = 1500 = 1875 0.8

2. Formula Momen Perlawanan a)

Rn = Mu(momenterfaktor) rp*b*d 2

atau

*

Rn = p

n{

dengan m =

C

1-

~ p *m). fy

0,85 * j 1c

=0,85 * f 1c *bE* a = 0,85 * 22 * 2000 * a = 37400 * a .. .......( 1)

I.H = 0 C-T = O sehingga C

T

=T

= As* fy =As* 360

= 360 * As Dari persamaan T = C,

58

Seton Bertulang

M aka:

360

* As = 37400 * a a=

360 As =0,0096256 * As 37400

Dicoba dengan As= 6000 mm 2 , maka a

= 57,754 mm < t, maka

c

=37400 * 57,574 =2160000N = 51620 KN

Mn

=

c(

d- ;) = 2 160(0,65-

·~ )

0 58

Mu 1500 = 1341 ,38 KN - m < - < - - < 1875 KN - m 4' 0,8 Harga Mn ini sebagai harga kontrol. Dicoba lagi dengan harga As = 7000 mm2

a C Mn

= 67.379 mm < t =2520 KN 0 06 379

= 252o(0,65 - • ~ = 1553 KN -

)

m < 1875 KN - m

Dicoba lagi dengan harga As = 7500 mm2

a C

= 72, 192 mm < t = 2699,98 KN

Mn

= 2699 98( 0 65 '

'

0 072 1 • ) 2

= 1657,6 KN - m < 1875 KN- m

Bab l_Balok

59

Dicoba lagi dengan harga As = 8600 mm a = 82,78 mm < t C

= 3096 KN

Mn

= 3096( 0,65-

2

0 0 278

•~ )

= 1884 KN- m> 1875 KN- m

Contoh Soal-15 Diketahui balok penampung segi tiga

Bahan beton, mutufc = 28 Mpa Bahan baja tulangan, mutu =400 Mpa Koefisien keamanan: SK-SNI-TlS-1993

Pertanyaan Untuk keadaan ''Under Reinforced" hitunglah besar As maksimum, dan besarnya momen terfaktor Mu.

60

Seton Bertulang

As untuk keadaan "Under Reinforced" berarti tulangan maksimum peraturan ( SK-SNI-TlS -1993), yaitu 75% tulangan keadaan "balanced". Menghitung tulangan "balanced" tersebut, dari gambar (a), didapat

Es (baja) = 2,1 x 10 Mpa Es =Ey

=fy I Es =

Yu,b

400Mpa = 1,9 X 103 Mpa 2,1 x105 Mpa h

_e_n_: = ecc +esy ~

Bab l _ Balok

Y u.b =

0,003 70 0,003+0,0019 X

61

b. =

37 14 • .40 =18 57cm 80 ,

Yu,b = 0,612 x 70 cm = 42, 86 cm

Cc,b = 0,85 f'c (luas trapesium)

Cc,h = 0,85 x 28 x (

T s,b = A .fJ,·fy

A s,b

400+ 185,7 2

ab

=0,8 . 42,86 = 34, 29 cm

) . 428 ,6 =2987269 Newton

= 2987269

= 2987269 = 7468 mm 2 400

As, maksimum

=0 ,75 x Asb = 0,75 x 7468 mm2 =5601 mm2

b. Underreinforced

T:f =A .f.fy =560lx400 =2240400Newton

62

Seton Bertulang

Cc

= O85 /c[400x800

Cc

=0 ,85 x 28 ( 160000-0,25 ( 800 =3808000- 5,95 ( 800- a )2

'

2

400.(800-af] 2x800 a)

2

)

l:H=0 -7 T·' = Cc 2240400 =3808000-5,95 ( 800- a )2 2240400 = 3808000- 5,95 ( 640000- 1600 a+ a2 ) 2

1567600- 3808000 + 9520 a- 5,95 a = 0 - 5,95 a2 + 9520 a- 2240400 = 0 a2 - 1600 a+ 376537,8 =0 a,~

_

1600± ~(1600)2 -4x tx 376537,8 2

= 800 ±_!_.J2560000-1506151,2 2

= 8oo ±

LJt o53848,8 2

= 800±_!_.1026 =800±51 3,3 2

a1

=286,7 mm

400IMI

-

pq

=b"

a* = titik berat penampaqgitrapesium pqrs ,..,.

3 = 513 800• .400 =256 ,65

Bab l_Balok

63

titik berat, a* adalah :

a*

= 256,65x286,7xl43,35+ ~xl43,35x286,7 x~x286,7 = 256,65x286,7 + Xx143,35x286,7

=

1248522,62 93944,425

= 132,9 mm lengan m omen, z

=( 700 -

maka momen terfaktor, Mu

132,9 ) =567, 1 mm

=Ts.z = 2240400 N x 567, 1 mm

=12705308,840 N -

mm

;-· 1270,5 KN - m ( Momen terfaktor )

64

Beton Bertulang

Bab2

J

TULANGAN GESER

d

------ -H----------... If

V

Bab 2_Tulangan Geser

65

Gaya-gaya yang bekerja pada keretakan miring - a dan tulangan geser.

I:H Vn

=0 =Vcz + Vd +Vay+ Vs, atau

Vn

=Vc +Vs

Dim ana: Vc =Vcz+Vd+Vay Vc = Gaya geser beton tanpa tulangan V = Gaya geser Vn = Gaya geser nominal Vs = gaya tul angan geser

Catatan: Untuk lebih rinci mengenai tulangan geser dapat dipelajari dalam buku "Design of Reinforced Concrete" by; Wang at all. Vu = VcR +Vs VeR = gaya geser, terjadinya keretakan diagonal pertama sekali tanpa tulangan geser. Vu = tegangan geser terfaktor Vs = gaya geser yang dipikul oleh sengkang Dimana gaya geser dengan miring, a. terhadap horisontal adalah :

Vs=

A v* fy* d

s

(Sina+ Cos a)

dimana: d = tinggi effektif balok S = jarak sengkang/tulangan geser Av = luas total tulang sengkang Jika a.= 90°, maka

Av * fy*d Vs=----=;..:;__S

Langkah - Langkah Perencanaan: Hitung besarnya gaya geser berfaktor, Vu Hitung kuat geser beton, Vc

66

Seton Bertulang

a) Cek apakah ; Vn = kuat geser nominal V c = kuat geser beton ; <1> = factor reduksi

Vn ~ {Vc+ ~ ~J'c }w·d, bila kondi si tak memenuhi maka penampang diperbesar.

b) Pakai tulangan geser minimum, A v bila <1> Vc > Vn > ~ <1> Vc

. .

bw.S

Tulangan m1mmum, A v = - 3fy

c) Bila Vn > <1> Vc, tulangan geser harus diberikan sehingga Vn ~ ~(Vc+ Vs) A v.fy.d . d V: A v.fy.d di mana : Vs = , untuk sengkang vert1ca1 an s = .

s

(Sin

s

a d +Cos a ), untuk sengkang miring sebesar d, terhadap horizontal.

. . bw.S A m1mmum= - -

3jy

d) Jarak sengkang, S S maksimum = ~ 2

~ 600 mm

1 Jika Vs > .!. lrl:f c *b IV *d, 3 "1/JC

maka:

Untuk menghitun g Kuat geser tanpa tulangan ( Vc ): PB 89 mempunyai rumus sederhana sebagai berikut: Bab 2_Tulangan Geser

67

1

(1)

Vc = (~ ~f

c ) *b,. * d , satuan Newton

(2)

1 Vc = (_!_ lrl::"f 6 vJ c

c)(J+ 14Nu*

Ag

) *b

w

* d, satuan Newton

dimana:

= gaya normal tekan terfaktor, N = lebar web, mm = luas penampang balok, mm 2 = tinggi efektif penampang, mm

Nu bw Ag d

Contoh Soal- 1 Ba lok di atas dua tumpuan sederhana, diketahui bentang bersih 6 m, harus memikul beban mati merata sebesar 20 KN/m 1 (termasuk berat sendirinya), 1 dan beban hidup merata sebesar 35 KN/m Muru beton, f =28 MPa Mutu baja, fy = 300 MPa Ukuran penampang 300 mm Penutup beton = 50 mm 1

* 500 mm

Pertanyaan: tentukan tulangan sengkangnya? Solusi:

500 mm

Berat Sendiri (B.S) Beban Hidup (L.L)

6 8

=20 KN/m 1 = 35 KN/m 1

Seton Bertulang

Beban terfaktor: Wu = 1,2 * B.s + 1,6 * LL = I ,2 * 20 + I ,6 * 35 = 80 KN/m 1

=R2 = .!_ * Wu *lt 2 = .!_*80 *3 2 = 360K.N(=V•) 2 2

Vu

=.!_ * Wu * (lt - d? 2 =

~ *80*(3 -

0,5Y

=250KN Vc

= .!_ ~!Ic * b w * d = _!_6 * 6~1 c

.J28 * 300 * 450 = 119058'79 KN

= 119,059 KN

Cek kekuatan penampang terhadap geser:

(~ffc + ~ ffc)*b.,. *d = iffc *b. *d =

~.J28 * 300 * 450

6 = 595,294 KN

Bab 2_Tulangan Geser

69

~ 250 Vn = - 1 =-=416,67KN < Vc = 595,29 KN fjJ 0,6

Berarti tidak perlu tulangan geser Pakai tulangan geser untuk sengkang Coba dengan diameter (D) 12 mm, sehingga: As= 113,04 mm 2 Av = 2 As= 226,08 mm2 Jarak sengkang, S S _ Av * .fy * d _ 226,08 * 300*450*10-3

-(~ -Vcr

S=

(~~~ -119,059]

30520 8 • = I02 55 mm 297,61 ,

Maka: S maksimum =

~ 4

Kontro1:

.!3-.JJC ~f'c *b

IV

*d =!m *300*45o 3 ,

= 238,118 Newton Vu 250 (Vn- Vc) = - = --119,059 = 297,61 Newton fjJ 0,6

70

Seton Bertulang

(Vn- Vc) <

~ ffc * b.., *d

450 maka S maksimum = E:_ = = 225 mm 2 2

Geser Vu = 1.4 V DL + V Lll Vu kritis pada j arak ' d' dari muka tumpuan Prinsip perencanaan Vu ~ <1> (Vc +Vs) ~ aVmax = (V6+Vs) Kekuatan geser yang disumbang beton

•

Geser + Ientur V c = _!_ ffi bw.d 6

•

geser + axial

•

geser + torsi

•

semua harus memenuhi Vc ~ 0.3ffi bw.d

V= _!_6 .f]2 bw.d (1+ .!!!!:._) 14Ag c

,; 0.3./fr' bw.d. ~I + 0.3Nu Ag

Kasus: Vu ~ 0.5 <1> V c 0.5 <1> Vc< Vu <1>

tak perlu tulangan geser ~ <1> Ye diperhitungkan ada gaya geser minimal

Vs min =

<1>

bw.

S = Av. fy.d v smin

}

. 3Av.fy Smm= ----=-.:.... bN

s ~d/2 S

~

12 inci

Bab 2_Tulangan Geser

71

Vu> Ye

perlu tulangan tekan

Vs s

=Yu - Yc A v. fy.d = Vs

s ~ d/2 s

~

12 inci

Satuan yang digunakan fc' Mpa fyMpa bw,dmm Lebar efektif Nilai terkecil dari bw + 16t

L

•-

4

L

= bentang

space center to center (Jarak pusat ke pusat)

2.1

Perencanaan Tulangan Geser

Berdasarkan PB ' 89 (SNI-T- 15- 1991- 03)'

72

Seton Bertulang

Contoh Soal - 2

Mutu bahan: Beton: Fe'= 24 Mpa POT a-a d = 500 - 40 - 10

=450 mm

Baja: Fy = 280 Mpa lebar perletakan = 400 mm ybeton: 24 KN/m 3 (ditumpukan A dan B) beban mati lainnya = W 0 ' = 2 t/m'

Pertanyaan: Hitunglah gaya geser rencana maksimum dengan prinsip "strength method" Hitung juga jarak tulangan geser (sengkang) yang diperlukan seluruh balok

Solusi 1 - Gaya Geser Rencana Makimum dengan Prinsip Strength Method Strenght methode: ACI code 318 - 83 atau PB 89

Bab 2_Tulangan Geser

73

S

. ) = fjJAv.fy.d (cosa+ sma fjJVs

4> Vs= Vu- q>Vc

Menurut PB' 89: Vn ~ 4> Vu dimana Vn 4> Vu

= gaya geser nominal = gaya terfaktor rencana (terfaktor)

dead load

Wo

= 300 * 500 * 24 KN/m3 * 10 - + 6 1 = 23.6 KN/m

live load

W1

= 4 t/m

6

1

* 10Kn/m 1

=40 KN/m 1

geser maksimum sejauh d = 450 mm dan tepi per1etakan dengan panjang balok yang diperhitungkan terhadap hidup = 4000 - 200 - 450 = 3350 mm (efektif)(penampang kritis), gaya geser pada penampang kritis: Vn

= 1.2 * 23.6 (2 - 0.2- 0.45) 2 = 38.23 KN + 1.4 * 40 (4- 0.2- 0.45) I 8 =78.56KN

total Vn = 116.79 KN gaya geser ditengah bentang (Ln/2) (untuk beban hidup Y2 L ). Vucv2) = 1/8 * 1.4 * 4 0 * (4.0- 0.2) = 26,6 KN (kondisi kritis beban hidup)

Solusi 2 - Menghitung Jarak Sengkang S =? PB ' 89 a. Vc=0.166..[ft' =0.166 £4=0,81 Mpa (tegangan geser beton izin) Vc = 4> Cc. b. d =0.80 * 0.81 * 300 * 450 * 10 = 87.48 KN

74

Beton Bertulang

Maksimum geser pada penampang -kriti·s: <)>Vs =Vu-<)>Vc Vc = 116.79-87.48 = 29.31 KN b. batasan tulangan geser: <)>Vc =2<)>Vc <)> Vc = 2 * 87.48 = 174.96 KN <)>Vs

= 174.96 maks <)>Vs = 87.48 KN

artinya jarak tulangan geser S ~ d/2 = 450 I 2 = 225 mm

S, per1u daerah kritis S = (jJAv.fy.d = 0.8*157 *280*450 *10 3 = 15825.6 (jJVs

(jJVs

(jJVs

. . = 15825.6 S perIu pem. kntts

540 mm

29.31

d/2 = 225 Maka dipakai S = 225 mm atau D 1Omm C 225 mm

S, perlu daerah minimum tulangan geser PB '89 min <)>Vs=<)> (0.345). b. d (KN) Min <)>Vs=<)> (0.345)

* 300 * 450 * 10- 3

= 37.26 KN Maks S

=

15825.6 37.26

= 425 mm > d/2 = 225 m Jadi pilih D 10 mm C 225 mm

Bab 2_Tulangan Geser

(sama dengan daerah kritis)

75

2.2 Aneka Ragam Soal diambil dari Buku Reinforced Concrete Design, CH U - KIA W ANG dan CHARLES G SALMON dan Solutions Manual to Accompany Reinforced Concrete Design, Third Edition, 1979. Contoh Soal- 3 Consider a beam with a 12 in, width and an effective depth of 19,5 in , 1 which has 3 - # 7 bars in the tension face. The concrete has f c = 3000 psi, and the steel has a minimum specified yield strength of 40000 psi·. Determine the allowable bending moment. Is the beam under reinforced, ideally reinforced (balanced), or over reinforced according to the working stress (cara elastis) method? Locate neutral axis using transformed section: 12 *X* (

~) = 16,2 * (19,5- X)

X2 + (2,7 *X)

= 52,7 +(1 ,35) 2 + 1,83

76

x

= Js4,53 -1,35

X

=7,39 - 1,35 =6,04 in.

Seton Bertulang

If concrete controls, fc = 1350 psi fs

n

6 04 • ) 6,04

= 1350*( 19·5 -

6 =1350*( 136,04 ,4 )=3010 psi> 2222 NG

If steel controls, fs

n

= 20000 =2222 psi; 9

fc

6 04 = 2222 * • = 996 psi < 1350 OK ! .13,46

c

= .!.2 * 0 '996 * 12 * 6' 04 = 36 '0 k

T

= I ,80 * 20 = 36,0 k

M

= C (arm)= 360 = 36 ,0

*

(19,5-

6

·~)* 1~

* 17 ,49 * ._I 12

= 52,5 ft. kips

since steel governs, beam is under reinforced.

stress distributioq

2222

Bab 2_ Tulangan Geser

77

Contoh Soal - 4 For the beam of "Contoh Soal- 1" use 3 - # 10 bars instead of 3 - # 7. Determine the allowable bending moment. Is the beam underreinforced, ideally reinforced, or over reinforced according to the working stress method ?

Allowables fs

= 1350 Psi = 20000 Psi

n

=9

fc

Locate neutral axis: 2 12 *X = 34,3*(19,5 - X) .2

2

X + c5,72

* x) = 111 ,5

2

+ (2,86) + 8, 1'7 X= .Ju9,67 - 2,86 = 10,95 -2,86 = 8,09 in If concrete controls, fc = 1350 Psi

78

Beton Bertul ang

11 1 19 5 8 09 • - · ) = 1350* ( ,4 ) = 1905 < 2222 n 8,09 8,09 Psi ......... OK fs = 1350(

C

= ..!_2 * 1'35 * 12 * 8' 09 =65 '6K

T

=3,81 * 1905 * 9 =65,4k

M

=C (arm) = 65,5

1 8 09 ( 19,5- • ) * - = 91,7 ft. kips 3 12

Since concrete govens, beam is Over- Reinforced

Contoh Soal - 5 What are the theoretically required effective depth and area of tension 1 reinforcement for a rectangul ar beam 16 in. Use f c = 4000 Psi and fy = 60000 Psi· Allowables: fc = 0,45 * f 1c = 1800 Psi fs = 24000 Psi n

=8 1800 k' d (1 - k)*d

For ideally reinforced (balanced)

Bab 2_Tulangan Geser

79

k

1800 = - --1800+ 3000

=0,375 J

=1-%'

R

= 0,5

* fc * k * j

= 0,5

* 1800 * 0,375 * 0,875

= 295 Psi M,,.

R

= b *c/ 2

b * d2

= M.,.

= 400 *12000

R

295

= 16250 d

=

~16250 16

=31,9 in As

12 = _ M,. _..:.:;_ =- -400* - ---

f<;* j*d

24*0,875*3,19

=7.16 sq in Contoh Soal - 6 What are the required widht and the required area of tension sreel for a rectangular beam with a 28 in, effective depth which is to resist a service load moment of 400ft - kips. Use f 1c =4000 Psi and fy =60000 Psi Allowabies: fc = 1800 Psi fs = 24000 Psi n

Ideal Constants (See "Contoh Soal 3") k =0 j =0,875 R =295 Psi

=8 b

=

M

R*d 2

400*12000 295*28

= - - - 2-

= 20,7 in 80

Seton Bertulang

As

M =--= - -- - fs* j*d 24*0,875*28 400* 12

=8,15 sq in Contoh Soal - 7 A beam having 12 in, width is to carry service load moments of 70 ft-kips live load and 40 ft-kips dead load. Using f 1c = 4000 Psi and Grade 50 steel, select the overall cross section dimension and select bars. Show sketch of the final design, and check stress. Allowables: f 1c

fy

= 4000 Psi, =50000 Psi,

n

=8

fc fs

= 1800 Psi = 20000 Psi

Ideal Constants:

k

=

1800 20000/ = 0,4! 9 1800+ /8

J

=

1-%

= 0,860

R

1

= 2

* fc* k* j

=324 Psi

Reqd ........ b

* d2

- (70 + 40)* 12000 324

= 4074 in 3

Bab 2_ Tulangan Geser

Bl

b 12

d 18,4 + 2,5 = 20,9

Try 12 * 21 d = 18,5 in

M

110*12 =----fs * j * d 20 * 0,860 * 18,5 = 4,15 sq in

Reqd ........ As=

Try 2 - # 8 and 2 - # 10. As= 4,12 sq in

Min Width = 11,8 in (2* 1,5)+(2* 0,375) +(4* 1,27)+(3* 1)

Check: 12 2

2

*X =412*8 *( 185-X ) 2 , ,

X + 5,49

* x + (2,27) 2 = 101,6 + 7,5

X= .jiOJ,9- 2,74= 7,70 in

fs = 180( n '

18 5 7 70 ' - ' ) = 2 525 > 2 5 7,70 ' '

Sligtly Over reinforced but close to ideally rein forced

82

Beton Bertulang

2,525

T = 20

* 4'

Mw

= 82 ' 4 * 15' 93 * -12

12 = 82 4k ' 7 70 Arm "" 18.5 - • = 15,93 in ' 3

1

= 109,4 1k""

Jl0 1k ( OK )

12in

I~

USE 12

I

* 21

With 2- # 8 and 2 - # 1Q

Bab 2_ Tulangan Geser

83

Contoh Soal - 8 Design a reinforced concrete beam to carry a live load moment of 225 ft kips on a simply supported span of 20 ft. Use f 1c = 4000 Psi and fy = 60000 Psi. Use a beam width of 14 in. Show of the cross section with selected bars, Check Stress. Allowabies: fc = 1800 Psi, fs = 24000 Psi, n = 8 Ideal Design constants: k

= 0,375,

J R

= 0,875, = 295 Psi

Loads Moment ( Mu) LL

= 225 1k

DL

= 0,4 * 202 =2Q1 k

Mu

8 =LL+DL=245 1k

Mu corected = 225 + 21 ,9 = 246,9

cl

=

245*12000 295 *14

= 26,68 in Add ::::: 2,5 11 for one layer Add ::::: 3,5 11 for one layer Try 1 layer: h

= 26,68 + 2,5 = 29,18 in

Say 30 in , cl = 27,5 Wt

84

=

14 30 * 0 15 = 0 437 144 ' '

Beton Bertulang

d

=

As

=

246,9*12000 295 *14

=26,78 in

246,9 * 12 . = 5 12 .sqm 24 * 0,875 * 27 ,5 '

2 - # 11 and 2 - # 9

d = 30 - 2 in (cover+ stirrup) 0,63 ( Arg Barrad) = 27 ,73 Accept S light Over Stress USE .. . 14 * 30 With 2 - # 11 and 2 - #_.2

I· Check S tress : 14 *

x( ~ } =

Bab 2_Tulangan Geser

5, 12

* 8 * c27,37-

x)

as

X + 5,85 * x = 160 + (2,92) 2 + 8,55 2

X

= .j168,55- 2,92 = 10,06 in

arm

= 27,37 - 3,35 =24,02 in C=T

M arm = 123,2k

2 123 2 * • =I 755 Psi 14 * lO 06 ' ' 123 2 • = 24,05 Psi ....... . .. .... ( OK ) fs = 5,12 fc =

Contoh Soal - 9 In order to reuse forms and eliminate multiple fraiming details, a reinforced concrete cross section 15 in. wide by 30 in. Deep overall is to be used in a location where the maximum live load moment is 100 ft-kips. The simply supported span is 24 ft. Use f 1c = 3000 Psi and Grade 40 steel. Assume cover is 2,5 in, from the center of the bars to the bottom of the beam. Determine steel requirement using "exact" method Determine steel requirement using "approximate" method. Select bars, show placement in cross section, and check stresses. Allowables: fc

= 1350 Psi

fs

= 20000 Psi,

n =9

Ideal Design Constants: K

= 0,378

J

=0,874 =223 Psi

R

86

Seton Bertulang

3- # 8 2 -# 6

,, .1

.

•••• •

Ideal Moment Capacity of This Setion M

=R * b * d 2 = 223 * 15*275

2

12000

Assumse d = 30-2,5 = 27,5 in Actual Moment ( LL + DL) LL

= 100 1k

DL

= 15 *30 015*242 =33,7 144 ' 8

LL+DL

= lOO+ 33,7 = 133,ik < 2 10,5 1k'

fc

fs = 2222 n

Bab 2_ Tulangan Geser

87

Under Reinforced

M

=C *j*d 133,7

fc

J

* 12

= _!_ fc '~ X * 15 * j 2

* 27,5

=2,222( 27,5X-

J

= _ 1

X

X = 82,5-X 3 * 27,5 82,5

X 133 7 * 12 = _!_ * 2 222 * ( ' 2 ' 27,5- X

288,5

=(

J* X * (82'5 - X )

X ]*X *(82,5- X ) 27,5 -;- X

7940- c28,5 * x ) = 82,5 * X2 - x3 x3 - 82.5 * X 2 - 28.8.5 * x + 7940 Use Homers Method 1 - 82,5 - 288,5 ± 8 - 596 - 7070 - 74,5 - 884,5 ± 870 ± 8 -532 - 66,5 - 14 16,5 ±8 - 58,5

-58,5 -34,7 - 57,9 - 1451 ,2 ± 0,6

X j *d

BB

o

1 + 794ot

- 1415,5 -870

8

± 870

0

= 8,6 in = 27,5 - 2,87 = 24,63 in Beton Bertulang

(a) " EXACT" As

=

133,7*12 --325 . . . sq m 20*24,63

(b) "APPROXIMATE" As

=

133,7*12 20 * 0,874 * 27,5

CHOICES: 2-#9and 1-#10 3 - # 8 and 2 - # 6 4 - # 7 and 2 - # 6

. =3.34 sa m Actual d For 2 in "Cover" Any of these

As= 3,27 As= 3,27 As= 3,27

27,41 in 27,54 in 27,58 in

are

Probably equally acccptabley

Contoh Soal - 10 Suppose the cross section and materials of prob "Contoh Soal - 9" is to be used where the live load moment is 200 ft-kips and the simply supported span is 26 ft. Select bars and show placement in the cross section for reinforcement in the tensionface only. Check the stress on the chosen section. Allowables: fc = 1350 Psi , fs = 20000 Psi, n =9

3011

•••• • • • • 15 11

Bab 2_Tulangan Geser

89

Ideal Design Constants : k

= 0,378,

J

= 0,874,

R

= 223 Psi

Moment Capacity of ideay reinforced section : = 21 0,5 1k (Prob "Contoh Soal - 7")

Actual Moments (LL + DL) : LL=200

DL =

262 15 30 * *015 * =396 144 , 8 ) = 200 + 39,6

LL+DL

= 239,6 ft-kips Since 239,6 > 210,5 , beam is to be Over Reinforced M=C*j*d ~Estimate

239 )6

two layer

* 12 =_.!_2 * 1350 *X* 1. * 26 ,5

J

239 ,6 * 12

= I_

= 79,5- X 3 * 26,5 7 9,5

X

5 = _!_1 * 1350 * X *( 7979 · -XJ*265 * 15 5 ' '

90

Beton Bertu lang

851

= 79,5

X 2 -79,5

* x - X2

* x + (39,7) 2

= - 851 + (39,7) 2 X

= .J-851 + 1582 + 39,75

X

=- 27,0 + 39,75 = 12,75 in, arms = 22,25 in

fs = 1,350 * 26,5 -12,75 n 12,75

=1,4SS,

= I ,458 * 9

fs

= 13,1Psi M 239,6 * 12 . = 985 sq tn As = - -- = fs *arms 13,1* 22,25 ' Use I0 - # 9. As = I 0

Check Stress :

15* X 2 - = I0 2

* 9 * ( 26,5 -

X)

X2 + 12 * X= 318,5 X = 12,82 in ann = 22,23 in

c fc

T fs

= 96,2 * fc = 1342 Psi = 10 * fs = 12,9 Psi

C = T = !!!_ = 239,6* 12 arm 22,23

= 129,0k Bab 2_ Tulangan Geser

91

Contoh Soal - 11

Design the cross section for a beam to carry a live load of 5 kif on a span of 28 ft. Use f1 c = 3000 Psi and Grade 60 steel. (See Sec. 3.8 practical proportions) Show sketch of chosen cross section with placement, and check stresses.

~ 1111111111111111111111111111111111111111111 ~ w 281 - ott

Ideally Reinforced Constants: k

=

J

= 0,888

R

= J_ 2 * 1350 * 0 ,396 * 0 ,888

1350 1350 + 2667

=0 336 ,

fc fc n

= 1350 = 24000 = 2667 9

= 201,5 Psi

LOADS:

= .!.. *5*28

2

8

=491 MoL

= .!..*o 8 ,9*28

2

= 88 = 579 1k

92

Beton Bertulang

For Ideally Reinforced Section :

b

b * d2 =M= 579*12000 = 34,500 R 201,5

18 20 22

d

43,8 41,5 39,6

Total Depth= 39,6 + 1,5 + 0,5 + 0,5 = 42,1 in Bar Radius - - • Stirrup Est Cover

TRY 22 * 44 Wt

22 42 * *0,15 144 = 0,875

=

d Reqo = 39,5 in M = 577,7 1k (OK) USE 22 * 42 REINFORCEMENT: d = 39,5 in As

577,7 * 12 M =---=----fr;*j*d 24 * 0,888 * 39,5 = 8,22 sq in

4-#lland 2-#9 4-#lOand 4-#8 6-#9and 2-#10

MinWeb=l8,5in MinWeb=2l,Oin MinWeb=21,0in

As= 8,24 As= 8,24 As= 8,27

Any of the above reinforcement choises OK, For the fewest bars USE 4 - # 11 and 2 - # 9

Bab 2_Tulangan Geser

93

Contoh Soal - 1 Design a reinforced concrete simply supported one- way floor slab to carry a uniformly distributed Jive load of 175 Psi on a span of 18 ft center to center of supports. ·Use f 1c = 3000 Psi and grade 40 steel . Make design and check stresses. ( Note ACI - 7.12)

Design Constants :

k

= 0,378

J R

= 0,874 = 223 Ps i

Loads: W .(Live load) = 175 Psf W (Dead load)= J 25 Psf (est 10 in Slab)

9 4

Seton Bertulang

Slab Thickness: Mw

= i*(0,175+0,125)*(8Y

= 12 15 ik/ '

I ft

~ R~ b =

12,15*12000 223 * 12

= 7 ,38 in

Reqd d

=

Thickness

= 7,38 + 0,75 (Cover)+ 0,50 (Wear Surf) + 0,25 (est. bar rad) = 8,88 in 9*150 = 112,5 Psf) 12

Try 9 in.( Wt = M = 12,15 *

:~~5

2

= 11 ,641ft, d = 7 ,23, Thickness

d = 7,5 in

USE 9 in SLAB As=

M f';* j*d

=

= 8, 73 in

11,64*12 = 107 sq*in/ 20*0,874*7,5 ' /ft

TRY : #8@ 8

As= 1,18

#7@ 6

As= 1,20

# .6 @ 5

As

= 1,06

USE

TEMP + SHRINKAGE As

=p*b*h =0,002

* 12 * 9

=0,22 USE#4@ 10

Bab 2_Tulangan Geser

95

Check Streeses: 12

2

* X =9,54 * ( 7,5 2

X)

2

X + I ,59 * X = 11 ,92 =2,75in X arm = 6,58 in

fs=

M

As* arm

=

11,64* 12 1,06 * 6,58

= 20 Psi; fc

M

11,64*12

_!_*X *b*arm 2

_!_*2 75*12*6 58 2 , '

=- - -- - = ---=----- = 1,29 Psi .... .. ... .. . OK

Contoh Soal - 12

Determine the areas of tension and compression reinforcement required in abeam 20 in. Wide an effective depth of 29 in, to the centroid of the tension steel, which is to carry a bending moment of 475 ft-kips. Assume that d 1 = 1 2,5 in. Use f c = 4000 Psi and grade 60 steel.

96

Beton Bertulang

Allowables Stresses: fc = 1800 Psi; fs

=24000 Psi ;

n

=8

Ideal Design Constants:

=0,375; =0,875;

K

J R

= 295 Psi

RM of ideally Reinforced Section

=R * b * d 2 = 295 *20 * 29

M

2

12000

I

=413,5 1k Mz= M- Mt = 475-413,5 = 61,5 1k Tension Reinforcement 1 _ = =··--- .. M __ 413,5 * 12 j~· '''-.T"' d . 24 * 0,875 * 29

= 8,15 sq in

M?

= j~· * (cJ ~ d I)= As

61,5 * 12 24 * (29- 2,5)

= 1,16 sq in = As t + As2= 9,31 sq in

Bab 2_Tulangan Geser

97

USE 2 Layers: 4 - # 10 bottom 2 - # 10,2- # 9 Top ( As= 9,62 sq in) Compression Reinforcement: k

* d =0,375 * 29 = I0,88 in 8 38 fc = 1800*( • ) I 10,88

= 1386 Psi =2

* 8 * 1386

= 22200 Psi < 24000 Psi As 1

( OK )

61 ,5 * 12 = ~------:-;-----:(22,2 - 1,386)(29 - 2,5)

= I ,34 sq in USE 2 - # 8 ( As = I ,58 sq in )

le= 1800

10,8811

Contoh Soal - 13 Redesign the beam of Prob "Contoh Soal - 8" using compression reinforcement which may be assumed to be centered at 2,5 in , from the comoression face of the beam .

98

Seto n Bertul ang

Allowable Stresses: Fe fs

= 1350 Psi;

11

=9

= 20000 Psi;

Ideal Design Constants: k

=0,378;

J R

= 0,874; = 223 Psi

Loads: = 200

= 15 * 30 * 0 15 26 144 , 8

2

= 39,6 M

= MLL+MoL = 200 + 39,6 = 239.6 1k MLL = 2001k

x,,, , , , ,,,,,,,,,, li ,,, ,,J)\4' ~

1...

26' -

1..

Qll

·I

' 1511

Capacity of Ideally Reinforced Beam M = I

223 * 15 * 26 5 ' 12000

Bab 2_Tulangan Geser

2

= 196 1k

99

M 2 = 239,6 - 196

=43,6 1k Tension Reinforcement

=

As 1

M1

.fy * .i * d

=- -196* - -12- -

Revisi : for one layer d =27,5 11 M1 =210,5 1k M2 = 29,1 1k As 1 = 5,46 ; As 2 =0,73 As =5,46 + ,73 =6,19 sq in USE 4 - # I I ; As = 6,24

20 * 0,874 * 26,5

= 5,08 sq in

A M2 43,6*12 2 1 s = fs*(d - d )= 20*(26,5-2,5)

= 1,09 sq in = As + As =5,08 + 1,09

As

1

2

= 6,17 sq in Compression Reinforcement

k*d

= 0,375

* 27,5 = 10,4 11

fc = 1350 * 10,4-2,5 I 10 4 ' = 1025 Psi; 2

* n * fc 1 = 2 * 9 * 1025 = 18450 Psi < 20000 ( OK )

29,1 * 12 = --;----____:.,.---.,.-----:(18,45 -1,03)* (27,5- 2,5)

= 0,80 sq in USE 2- # 6; As = 0,88 sq in

100

Seton Bertulang

Contoh Soal - 14

For the beam of the accompany figure: Determine the allowable moment capacity Mw Compute the transformed cracked section moment of inertia wich would be needed for a deflection calculation. Use concrete with fie= 4000 Psi and Grade 60 steel Alloweable Stresses fc

= 1800 Psi;

fs

= 24000 Psi;

n

=8

Locate neutral axis Assume use of 2 n for comp. Steel

16

*x

2

+70,2*(X - 2,5)

2

X 2 +(8,77*X-21,9) 2

X + 15 02 '

*x

X= .J248,8 -7,51

Bab 2_Tulangan Geser

= 50,0 * ( 27,3 - X ) = 170,5 - ( 6,25

*X )

= 192,4 + (7,5Il + 96,5

= 15,78-7,51 =8,27 in

10 1

Determine applicable stress distribution

If : fc = 1800: =

fs

n

4150

> 24000

8

= 3000 Psi Thus , steel controls.

fc, = JC *( 8,2;,~72,5 )

= 1303*

5 77 • 8,27

=910 Psi 2*n

* fc 1 = 16 * 91 0

= 14560 Psi < 24000 Psi ......... OK Internal Forces

C = .!_*fie* X * b = .!_ * 1303 * 8,27 *16

2

I

2

=86 ' lk C2

= A 1s(f 1s- fc 1) = 3 * I ,56 * ( 14,56- 0,91 ) = 63,9 k

T = 4 * 1,56 = 149,8 k

* 24

Moment capacity

Mw

=861 *(273 ' ,

8 27 ' 3

)*-

1 1 +639 *(273 - 25)12 ' ' , 12

= 175,5 + 132,0 = 307,5 1k Moment of Inettia

h

102

2

= _!_ * 16 *8,27 3 +70,2(8,27 - 2,5) +50,0 *(27,3 - 8,27) 3 4 = 3015 + 2335 + 18100 = 23450 in

2

Beton Bertulang

Contoh Soal - 15 For the given beam cmpute the transformed cracked section moment of inertis Icr that would be needed for a deflection calculation; i.e. , use modular ratio n (instead of 2n) when transforming the compression steel. Use f 1c = 4000 Psi and fy = 60000 Psi

/

f

•••••• 6-#7

2, 5 4

d =36,1 11

10-# ll

lB~; ,,, At= 10*1 ,56~8

~I

=12,0

At =6*0,6*(8-1) == 25,2 sq in

Allowables Stresses fc

= 1800 Psi;

fs

= 24000 Ps i; =8

n

Locate neutral axis 18* X

-

2

-

2

+ 25,2*(X -2,5)

X2 + t6,69 X

= 125,0 * ( 36,1- X)

* x + c8,34) 2 = 508,4 + 69,6

= J578 - 8,34 =24,04 1

8,34 = 15,70 in

f er

= 3 * 18 * (15,70) + 3,6 * 7 * (15.70 - 2,5) + 125,0 * (36,1 - 15,7Y

f er

=23220 + 4390 + 52020 = 79630 in4

Bab 2_Tulangan Geser

3

2

103

Contoh Soal - 16

The simply supported beam of 16ft span is to carry a uniform ldead load of I ,5 Ki ps/ft (inci. Beam weight) and a uniform live load of 2,4 kips/ft. Determine the adequacy of the# 3 U - stirrups which are spaced at 8 in. Use the simplified method with constant Vc. Draw the curve of shear capacity
Wu

= I ,4 * I ,5 + I ,7 = 2, I+ 4,08

* 2,4

=6, 18 k/ft At

=
104

= 6, 18 * 8 = 49,5k

Seton Bertulang

At d from face; Vu

16 2 04 =21 *98-204)+408* ( - ' ) , , ' 2*16

2

= 12,52 + 24,72

= 37,2 k At A ft from
Vu

(16 - 4) = 2,1 * 4 + 4,08 --'-----'2*16 = 8,4 + 18,35 = 26,8 k

At midspan; Vu

4,08 * 16 =--8 = 8,2 k

* Vc = *Ne* bw *d = o 85 ,

*

=26,0

k

= * (2 *ffc)* bw * d

1 (2 * .J35oo)* 14* 18 5 * -' 1000

*Vs = * AN * fy * d = 0,85 * 0,22 * 40 * 18,5

s

s

138 ,4 k. S . =- 1ps, atuan S d aIam m.

s

s

* Vs

* Vn =* Vc + * 43,3 k > 37,2k at face (OK) 41,4 k > 37,2k at face (OK) ---.

8 9 d

Max2

=9,25

Bab 2_ Tulangan Geser

Max * Vs for Vs = 4 *

ffc = 52,0

k

105

Design is in adequate over the distance "A", where stirrups are required for min Vs= 50 P si with S s;; d ,

2

Min <1>* Vs for Vs= 50 Psi : Min * Vs= 11

Ok < 15,4k; . -d controIfor mm. 2 USE One additional space @ 9 11 : l.e : USE 4 @ 9

11

Contoh Soal - 17 T he beam of the accompanying fi gure carries a uniform live load of 3,0 kif in additional to its own weight. Assume a support width of 12 in., and use f 1c = 3000 Psi and fy = 40000 Psi, Draw the maximum factored shear Vu envelope D raw the cur ve of required stirrup spacing using the s implifed procedure permitted by ACI - 11. 3 .1 .1. show also the spacing provided. Determine if the spac ings used in the figure satisfy the ACI cod!e requirements according to the simplified method. Repeat (b) using the more detailed procedure of ACI - 11.3.2. 1; s how also the spacings provided. Detennine if the spacing used in the figure are satisfactory according to the analysis of (d)

106

Seton Bertulang

5-#9

-

• • • • • 11

'· 12 · f,lc

~I

=3000 psi

fy =40,000 psi

Requires one more stirrups, I I

12 =1q,s 11

plus a sti-rrups at midspan

I I I I

5 6@ to ll

Wu

=3,0* 1,7+0,313

* 14

=5, 1 + 0,44 = 5,54 k/ft Wt

=

12 * 25 * 0 15 144 '

= 0,313 k/ft At
Vu

= 5,54 * 20 =55 ,4k 2

Bab 2_ Tulangan Geser

107

At d from


Vu

= 55,4-5,10*2,3(20-1,15) -(2 ,3 * 0 ,44 ) 20

=43 ,3 k At S 1 from support; Vu

(S*0,44)

= 55,4-5,10*(20-1,15) 20

= 30,9k At mid span; Vu

5,1 * 20

= - -8 = 12,7 k

Simplified Method; Vc

=Ne* bw = 2*

*d

1 ffc * 12 * 21 '6 * -1000 - = 28 4 ,

* Vc = 0,85

k

* 28,4 = 24,1k

0,85 * 0,22 *40 * 21,6 = 161,6 k' (: #3 *Vs= * AN*"·*d n = - 1ps; 10r _

s

s

s

stinups Max Vu - * Vc Max

* Vc for Ns

= 43,3-24,1 = 19,2k = 4*

ffc ; * V c

= 48,2 k Min * Vs for Ns

=50 Psi;

*Vs

=ll,Ok

Max S (Control Section)

lOB

= 161,6 =8 411 19 2 '

'

Seton Bertulang

Max S (Min. Percentage) = 161,6 =14,611 >:!:._=10,811 2 11,0

t

Min *Vs

p*v*d

More Details Method using ..:..___ _ M

p

=

50 • 12*21,6

0,0193

at d from face of support ;

M

= 43,3

* 2,3 +

0 44 2 32 • * ' 2

99,5 + 1, 2

V * d = 43,3 * 1,8 = 0 77 < 1 0 ( OK) M 100,7 ' '

= 1 9 * ffc

Ne

'

2500*p*V + _ _....:...__*d _ M

= 1,9* ..}3000 + (2500 * 0,0193 * 0,77)

=141Psi <1>

* Vc

= * 0,141 * bw * d =0,85 * 0,141 * 12 * 21,6 <1>

= 31 ,I k

1

At S from M


=30' 9 * 5 + 0,442*52

= 154,5 + 5,5 = 160,0 lk

V * d = 30,9 * 1,8 = 0 35 < 1 0 ( OK ) M 160 0 ' ' ' Bab 2_Tulangan Geser

109

Ne

= 104 + (2500 * 0,0193 * 0,35 = 104 + 17) = 121 Psi

* V c = * 0 , 121 * bw * d = 0 ,85 * 0 ,121 * 12 * 2 1,6= 26, 7k

t2.7>0,5*cp*Yc= lt,o• Pactored Shear

22.0

v.

cp• Ye

(kips)

Since stirrups required until Vu:::; O,S**Vc, entire span required stirrups 11

I

M•>
lO

9 8 7 Need Stirrups here

5

Max S (at critical section) = (161,6 I (43,3 - 31 ,3)) contro ls ; m ax S = I 0 ,8in

=

13,2 in; O,S* d

Contoh Soal - 18 For the port ion of the conti nuous beam show n, w ith the given maxi mum shear e nvelope Vu, determine the spacing to be used fo r # 3 U - stirrups.

1 10

Belon Bertulang

Dimension and show the stirrups on the given portion of beam. Use the 1 sim plified method of constant Vc, with f c = 3500 Psi and fy = 60000 psi

•••• I< 12 >I 11

Vu= 1,4 • Vd t

,7 ' VL

. 18k I

Vc

1 = 2 *ffc *bw*d = 2 *.J3500 * 12 *21 *- . 1000

=29,8k

* Vc = 0,85 * 29,8

=25 ,3k <1>

*Vs=

s 11

10,4 Max. S =d/2 11 = 10,5

4> *A *fy * cl N

S

=

0 85 * 0 22 * 60 * 21 ' '

S

= -235-'-6 S


22,7

Min
= 12,6k

Bab 2_Tulangan Geser

1 1 1

Max cJ> *Vs

= 59,6k for Ns =

Min * Vs

4ffc

= 12' 6 k

for Ns

=50 Psi

At Critical section:

* Vs

=Vu - *Vc

= 48-25,3 =22 ,7k Since Vu

18 at mid span>

* Vc 2

= 12,65 k

Stirrups needed over entire span USES - 1011

Contoh Soal-19 The given 20 ft simply supported beam must carry adead load of 6 kips/ft (including beam weigth) and a live load of 10 kips/ft. Use f 1c = 4000 Psi and fy = 40000 Psi.

1 12

Seton Bertulang

Using the simplified method of ACI- ll .e for the U - stirrups. Explicitly show the length over which stirrups are requred. Repeat (a) hut use lhe more detailed procedure of ACI - 11.3.2.1.

I I -@ 4.5~ ... ~@

6~l~ 2@

1811 ...

I

# 4 stirrups Wd = 6 * I ,4' = 8,4 k/ft Wd =6 *lA·'= 8,4 k/ft Support';

= 8,4+17,0 * 20 =254k

Straight

2

line

.

1000

1

Stirrups Required over Entire Span. 35 Total For 20 Span

* Vs

= cp* AN* fy*d = 0,85*0.22*40*36,3 = 271,5 for# 3

s

s

d

For max S

= 2'

Max * Vs

= 141 k.

based on Ns

=4*ffc

Bab 2_Tulangan Geser

s

'

1 1 3

k

At cl from face,

S

----

11

M ax. S (Critical Section)= 4,7 d . Max S = - = 18 · ' 2 ' min. Percenlage required
5 6

*Vs

10

98,7k 82,3 54,9 49,4

12 14

41 ,1 35 ,3

18 Max

27,4 (27,8 min)

9

(b) M ore Detai Is Procedure Using Straightline Approximation for Vu At d from face ; (use maximum moment) M

= 2; 4 * 3,77 * 16,23 =7771k;

V* d M Ne

=

174 * 3,02 777

= 1,9*

= 63,5 + 7,80 = 0 02 16 p

18*36,3

'

= 0,68 < 1 ( OK )

ffc +(2500*p*;*d)

= 120 + (2500*0,0216*0,68) = 120 + 37 = 157 Psi ;

* Vc = * Ne * be* d = 87,2k

At S

1

-

0 from
M

1 14

= 25,4 * 5 * 15 = 9521k 2

Seton Bertulang

V*d M

= 148,6 * 3,02 = O 47 < 1 ( OK ) 952 '

Ne

= 120 + 25= 145 Psi; * Vc = 80,5k

At mid span:

M=

25 4 ' *10*10= 12701k 2

V *d

= 42,5 * 3,02

M

Ne

= O, I

1270

= 120 + 5

= 125 Psi;

* Ye= 69,4k



* Vs * S

= 493 ,7k

( # 4)

At d ; * Vs = 174 - 87,2 = 86,8

#4 Stirrups

s = 5 11 86.8k

Vs 6 lps)

Bab 2_ Tulangan Geser

1 15

493,7 M ax S = - 86,8

= 5,7 in AtS

1

-

0;

= 67,5 Max S

= 493,7 67,5

= 7,3 in At midspan;


Vu= 42,5 > - - = 35,1

2

Stirmps needed for entire span (Vu- * V c) 1

31 Total Stirrups For 20 Span

Contoh Soal - 20 For a simply supported span of 32 ft, with uniform dead load of 2,3 kips/ft (incl. Beam weight) and live load of 3,7 kips/ft, determine and detail the# 4 vertical U - stirrups spacing using 1 in. multiples. Use f 1c =3750 Psi and fy = 50000 Psi, and the simplfied method with constant Vc. The support width is 12 in, and the beam section is shown in the accompanying figure.

Wu

= (2,3

* 1,4) + (3,7 * 1,7)

= 9,51 k/ft At Cl.Support; Vu

1 16

=9,51

32

=152k

2

Beton Bertulang

At midspan;

= 6,29

Vu

32 = 25k 8

At Critical Seclion;

Vu

= 6 •29 * (32 -

* Ye

=

2

2 54 • ) + (3,22 * (16 - 2,54)) 2*32

= 128,6k

* Ne * bw * d =

s 4,4411 5 6

8 10

12

Bab 2_ Tulangan Geser

=


s

=

0,85*0,4*50 *24,5

s

=

416,5

s

* Vs 132-38,3 83,3 69,4 52,1 41,7 34,7

1 17

Vu

s=

Kips)

11

12

#4

d

ForS =- max·

2

Max * Vs

'

=76,6k * (Ns = 4 * ~ f

1

c)

.·. Where76,6k,upperlimitonS = ForNs

=50 Psi;

* Vs

= 15,6kmin

~

=6,1 11

USE 55-# 4 Stirrups for Entire Beam Csengkang seluruh balok)

Contoh soal - 21 The beam in accompanying figure is to carry dead load of 1,4 kips/ft (incl. Beam weight) and live load of 1,6 kips/ft. Use f 1c = 3500 Psi, fy = 40000 1 1a

Seton Bertulang

Psi, and neglect any compression steel effect. Using the simplifi ed procedure, calcul ate and detail the # 3 single U - stirrups spacing for the beam. Repeat (a) using the more detailed procedure involving (p*V *d I M ). How many stinups could be eliminate as compared to the simplified method?

9 1 - 0 11

2 - #9

2-#8---2-#8

= ( I ,6 * I ,7 ) + ( 1,4 * I ,4 )

U

=2,72 + 1.96 =4.68 k/ft

43,7

Same stirrups for conservative and more exact methods

33.7~ rr-r-r-7'...

Vu

Kips

5''

More detail method

J

More Detail

~~~~:~~- --- ~*V~~~~~~~-~

~~ .

Since Vu= O,S • ~vc, min. stirrups rcqcl on

cantilever

7@to•.l 5 1 - I 0 11

Bab 2_Tulangan Geser

1 19

Based on Simplified Method:

* Vc=
* bw * d

1 1 =085 -1000 , . *(2"-J~f JCc)* 12*215* '

=25,9k For Ns

=50 Psi;

Min * Vs

= 10,9k

Cantilever;

Vu

= 4,68

*8

= 37,4 Ri ght End - Main Span :

= 4 68 * lO + 4.68 *(8Y *-I , 2 20

Vu

= 54,3k At d From Right End;

= 1,791'•

d 1

At S

1

-

d+6

= 2,291

Vu

18 85 =54 ,3- (2,72 *2' 29* 20 • ) -(196 , *2' 29) = 43 ' 9k

0 From Right End;

Vu

17 5 = 543 = 32 ,6k ' - ( 272*5 , ,0* 20• )-(196*5) ,

At Mid Span:

Vu

1 20

= 54,3 - (2,72*1 0 *

~~) - (1,96*10) = 14,3k

Seton Bertulang

Location

Vn

Vu

Mu

V*d

--;;;:/

V*d p *M

More Detailed Ne

*Vc (Psi)

Kips)

43f

199

-

Left end d from S 1 - 0

33,3

152

R1 ,4 1k

0,7

0,0113

141

30,9k

from left Midspan

22,0

100

134,5

0,29

0,0045

124

27,2

3,7

17

135,0

0,05

0 ,0007

11 5

25,2

54,3

248

4 3,9

200

-44,2

143

31,4

32,6

149

1,0*

0,0122

From right

+37,7

128

28,1

14,3

65

1,0

0,0061

Midspan

Support d from

37,4

170

-

Suppott

26,8

122

76,4

0,63

0,0038

122

26,8

Right end d from Right end S

1

-

0

-

Cantilever at

* Computed Value exceeds 1,0; Use Left end;

p=

Right end; p

=

4

*

12 2 15 ,

I ,0

=0,0155

4*074 ' 12*21,5

= 0,0 122

*

=112,5

1,9

.J3500

s, from right end, p = 0,0061

Simplified Ne: 2*

ffc = 118,4 Psi

Cantilever Support;

Bab 2_Tulangan Geser

p = 0,0 122

12 1

Stirrups;

* Vs

* AN* fy * d

=

s

0,85 * 0,22 * 40 * 21 ,5

s

=

160,8

s

At Left End; Max (Vu - * Vc) = 33,3-25,9 = 7,4k < Min * Vs= 10,9k

At Critical Section : 160 8 Max S = • 10,9

Same Either Method

= 14,7 11 > :!_2

d Max S = - Contro ls = ·10,7 2

At Right End;

=43,9-25,9 = 18,0k; = 160,8 =8 911 < !!:._

M ax (Vu - * Vc) M ax S

18,0 Using Ne

max S

=

2

'

2[jl;

= 43 ,9 -

31,4 = 12,5 k;

= 160,8

= 12 9 > d

12,5

'

2

Using more detail ed Ne; Max S

d

=2

No Difference in Number of Stirrups Reauired by the Two Methods.

122

Beton Bertulang

Contoh Soal - 22 The beam of the accompanying figure is to carry 1,6 kips/ft live load and 0,9 kpis/ft dead load (incl. Beam weight). Using f 1c = 3000 Psi and fy = 40000 Psi, investigate the beam for sti rrups adequacy according to the simplified method using constant Vc.

Wo

=0,9 * 1,4 = 1,26 k/ft ;

WL

= 1,6 * 1,7 = 2,72 k/ft

Max Vu at support

3,98*22

=---

2 = 43 gk '

At d from face; Vu

= 43,8 -

2 2 72 21 * • * -(1,26*2) =36,1k 22 t "

~

At midspan;

Vu

2,72 *22

= - --

8

= 7 5k )

Bab 2_ Tulangan Geser

1 23

Reqd extent of stirrups ups Vu

K p

s



* Vs =

* An * fy * d S

=

0,85 * 0,22 * 40 * 18

S

134,6

=- S

ForS = 6 11 ; * Vs= 22,4k

At d from face; Max ( Vu- * Vc ) = 36, 1 -20,2 =15,9k < * Vs = 22,4k forS = 611

(

OK )

Answer: Stirrup Capacity is adequate, but more stirrup are requi:ed

Contoh Snal - 23 Completely design and detail the stirrups for the beam of prob ( Contoh Soal- 20 )( ignore the spacing given ), using the more detaild ( pVd I M ) method of ACI- I 1.3.2. 1. Using f 1c = 3000 Psi and fy = 40000 Psi 3c

124

Beton Bertulang

Vu Kips

At d from face; Mu

= 36,1 + 2,0 + ( 0,5 = 74,ik

* ( 2,52 ) * 2,0

At S1 - 0 from et Support; Mu

= 1;26 * 5 * 17 + 2,72 * 17 *8,5 * 5 = 143 1 lk 2

22

'

At midspan; Mu

= 1,26*222 +2,72*11*5*11 8

=

22

155 ,4 1k

Bab 2_Tulangan Geser

125

At d from face;

V*d = 36,1*1,5 = . 0725 M 747 ' ' , 2500 * p

*

V

*d

= 40 Psi

M

At S1 - 0 from
V*d 26,7*1,5 = - -M 143,1

= 0,28;

2500 * p * V *d M

= 15,5 Psi

At midspan; V *d

M

2500

=

7,5* 1,5 158,4

=0,07;

V*d M

= 4 Psi

* p * -~ 1

~

1,9*...;f c = 104 PsiNc = 1,9*...;1 c + 2500 * Ye

*p*

V*d M ;

= * Ne * bw * d 1 =0,85*Nc* 12* 18* - - =0,1836Nc#3 Stimtps 1000



* Vs * S

= 134,6 (from prob" Contoh Soal- 21 ")

Max ( Vu - * Vc )

126

<1> * Vs

S

9,7

13,9 > 0,5

11

* d = 911 max

Seton Bertulang

Contoh Soal - 24 The deep beam of the accompanying figure has been designed by using fy= 50000 Psi and f 1c = 3750 Psi. Determine if web reinforcement is required If so, detrmine the required for # 3 vertical U - stirrups.

55,3

Vu

= L.46 nf - --

Take GFiijcal Section q,t d from face of support

L=~=274

d

17,5

,

Factored Loads; WL = 3

* 1,7 = 5,1

WD

= 2 * 1,4

=2,8

PL

= 12 * 1,7 = 8 * 1,4

= 20,4

Pn

Bab 2_Tulangan Geser

= 11,2

127

· Shear at support; Vu

= 7,9 * ( 4,0- I ,0 ) + 31,6 = 55,3k

Mu

= 7,9(3,0)2 +(31,6*3,5) 2 1 = 146 k

~ = Mu 146 * 12 = I 81 > 1 ' d Vu*d 55,3*I7,5

Member Sho1ud probably be Considered a reguler beam Ve

=Ne* bw * d

et>* Ve

= 0,85 * Ne * 14 * I7 ,5 = 208,3 Ne = 10 4k.

Min * Vs based on Ns

' ' =50 Psi

(a) Stirrups using simp1ified Conservatif method Ne

= 2*ffc;

* Ve= 25 5k · '

'

et>* Ve= Vu- * Ve = 46,6-25,5 = 21,1 k > 10,4 k Max. Spacing for strength at d from face;

s

=

0,85 * 0,22 * 50*17.5 21,1

= 7,8 11 For Strength d 11 Max. S=- = 8,75 2

128

Seton Bertulang

(b) Stirrups using more exact procedure 2 Mu at d = 7 •9 *(l,54 ) +(31,6*2,04) 2

p

=

=76,41k

4*0,79 14*17,5

= 0,0129; Vu* d _ 55,3 * 1,46 _ U --- 1,06 > 1 , se 1,0 Mu

Ne

et>* Vc et>* Vs

76,4

= 1,9 * .J3750 + (2500 * 0,0129 * 1,0)

= 149 Psi; = 31,0k =Vu - ct>*Vc = 46,6-31,0 = 15,6k

S = 0,85*0,22*50*17,5 = 10511> d = 875 11 15 6 ' 2 ' '

For Strength

Contoh Soal - 25 The beam of the accompanying figure carries a live load of 2,7 kips/ft in addition to the weight of the slab and beam. Calculate and detail the vertical U- stirrups spacing for# 3 Stirrups. Use f 1c = 3000 Psi and fy = 40000 Psi. Use simplified procedure of ACI - 11.3.1.1. Use more detailed procedure of ACI - 11.3.2.1.

Bab 2_Tu langan Geser

129

_ _ _ Symmetrical about
d = 2211

H

~--- bE=6911 _ _~

~~

~

9 11 ~~----

I

I Il

I I I I I I I l

Vu Kip~

I I I

13,2k

5,5 11 7@ IJ II

30 Stirrups/beam

130

Seton Bertulang

62,5k

_9_11_ IJII

*Vs =25,6k

Vu Kips

5,5 11

26 Stirrups I beam

Weight of beam and slab=

WL= 2,7

45* 57 ' 144

+ 12*255] · ' * 0,15 = 0.586 klft 144

= 4,6 klft

* 1,7

Wo = 0,59

[

* 1,4

Vu ( Support ) =

= 0,83 k/ft 5 43 * 23 ' 2

=62,5k

4,6 * (20,67) Vu = ---'----~ 2 * 23,0 2

At d from face;

0,83

Bab 2_Tulangan Geser

=42,6

)

50,2k

* 9,17 =7,6

13 1

L At-·

4'

} 34,5'

Vu= 4,6*(17,25f =29,7 2*23,0 0,83 * 5,75

=4,8

At mid Span;

Vu

- 4,6 * (11,5) 2 *23,0

2

= 13,2 k (a) Simplified Method ACI- 11.3.1

~ * V c = * (2 * ffc)* bw * d =0'85 * (2 *.J3000 )* 12 * 22 * -10001

Max *Vs

=Vu max- * Vc =50,2-24,6 = 25,6k

*Vs

=49,2k based on Ns = 4*ffc

S max. for strength =

At Ns = 50 Psi: *Vs= 11,2k min

132

A N

* fy * d * ~ = 0'22 * 40 * 22 * 0'85 = *~

*~

ct>*Vs 25,6 20,6 18,3 16,5 15,0

164 6

•

$*~

s 6,4 8 9 10 11

max. 0,5

*d

Beton Bertulang

(b) More Detailed Method, ACI- 11.3.2.1 p(8#9)=

80 • =0,0303; 12*22

p ( 6 # 9 )=

60 = 0,0227 • 12 *22

At d from face: Mu

= ( 50,2 * 2,23 ) + 0,83 * (2,23 )2 2 =] l4,Jk

V * d 50,2 * 1,83 --= M 114,1 = 0,805; Ne

= 104 + ( 2500 * 0,0227 * 0,805)

=150 Psi et>* Vc = 0,85 * 0,150 * 12 * 22 = 33,7k

At 31 -911 from
Vu ( max )

Corresp Mu

4 6 19 25 = • * ( · ) + (0 83 * 7 75) 2*23 ' ' = 37,0 + 6,4 = 43,4k 0 83 = (370*375)+ •2 *375*1925 ' ' , ) = 139,0 + 30,0 = 169,01k

Bab 2_Tulangan Geser

133

4 1 83 V* d = 43 • * • = 0 47· Ne= 104 + 27 = 13 1 Psi ( p = 0,0227) M 169,0 ' ' Ne



= 104 + 36 = 140 Psi ( p = 0,0303 )

* Ve =29,4k


At 51 - 911 from
= ( 29 ' 7 * 5 ,75) +

0 83 1 •2 * 5 '75 * 17 ,25 = 2 J?- k

= 34,5 * 1,83 =O 3 . 212

' '

= 104 + 23 = * Vc = 28,5k

Ne

127 Psi

At 81 - 011 from
= 4,6*(15)2 +083*35 2 * 23 ' '

=22,5 + 2,9 = 25 4k '

Mu

= ( 22.5 * 8) +

0 83 • *8*1 5 2

= 180 +50 = 2301k V *d

M

Ne

25,4 * J,83 230 =0,20;

=

= 104 + 15 = 119 Psi

* Vc =26.7 k

134

Beton Bertulang

Bab3

J

KOLOM

3. 1

Kolom Pendek

Kekuatan Nominal (nominal stren gth), Pn: Pn

=Kc . rc . Ac + fy . Ast -+l····K~··:···F~y···:··A~pl ..... ... ............ (K- t ) l ...............................................,,;

Kete.rangan : Kc

= 0 ,85, Koefisien = f (uji silinder beton)

f'c

=kuat tekan bcton si linder umur 28 hari

Ac

= Juas penampang beton netto = Juas kasar penampang dengan kolom pengikat

Ast

Bab 3_Kolom

= luas inti (core) kolom dengan spiral. = luas tul angan pokok 1 3

s

Fy

= tegangan le1eh tu1angan pokok

Ks

= konstansta : 1,5 - 2,5, rata - rata= 1,95

fsy

= tegangan leleh tulangan spiral

Asp

= volume dari tulangan spiral persatuan panjang kolom

[.'.'.'.'.'.'.'.'!

= bila tidak ada spiral maka suku tersebut hilang dalam persamaan di atas.

Formula (K-1) tanpa tulangan spiral menj adi sebagai berikut: Sebut kekuatan nominal maksimum , Pn = Po Untuk kolom pendek berlaku: P0

=0,85f'cAn+fy.Ast ................ . ................ (K-la)

= 0,85 f'c (Ag - Ast) + fy . Ast keterangan: P0

= kapasitas penampang batas dengan eksensitas, c = 0

=[KN] An

= luas penampang beton netto ?

= Ag- Ast, [mm-] Ag

= Iuas penampang brute [mm2] = b . h 2

- Ast = luas penampang tulangan vertical [mm

]

f'c

= kuat tekan beton sekunder [MPa]

fy

= tegangan leleh tulangan (baja)[MPa]

- Ast

= luas penampang tulangan memanjang total

At

=At+A2 = luas penampang tulangan tarik

A2

= luas penampang tulangan tekan(=As')

Po dapat juga diformulakan sebagai berikut: Po

= Ag { 0,85f'c (1-p) + fy. pg}

= Ag [0,85 f'c + p(fy- 0,85fc)]

136

.................. . .... (K-2)

Beton Bertulang

kerja . an c2.

menurut PB '89 , Ps 10.3,5 kuat tekan rencana, <1> Pn maksimum adalah: komponen struktur: non prategang <1> Pn = 0,85 <1> [0,85 f'c (Ag- Ast)- fy. Ast] ......... (K-3a) tulangan sengkang <1> = 0,65 untuk kolom dengan pengikat (tulangan geser), PP' 89 . Ps . 9. 3. 2. 2

Contoh Soal - 1 Kolom pendek harus memikul beban - beban sebagai berikut: P beban mati total = I 000 KN P beban hidup = 800 KN Momen Lentur = 35 KN- m (akibat beban hidup) Mu tu bahan : f' c

fy

= 28 Mpa (=N/mm2) 2 = 400 Mpa (=N/mm )

Pertanyaan: Rencanakan kolom pendek di atas, dengan pengikat (sengkang).

Bab 3_Kolom

137

Solusi: p =P1 + P2

•

dicoba p1

=3% (Trial & Error)

Pu

= 1,2 DL+ 1,6LL= 1,2x 1000+ 1,6 x800 =2480KN

Mu

= 1,6 * 35 =56 KN -

m

Mu 56 = --= 0,02258 m Pu 2480

e=-

Eksentrisitas

= 0,85 f' c (Ag - Ast) + fy Ast

Po

= Ag [0,85 f'c ( 1-Pg) + fy. Pg] = Ag [0,85 * 28 (1-0,03) + 400 * 0,03] = untuk kolom dengan pengikat : Pn (maks) = 0,80 Po

= Pu

Pn

=

Pu fjJ

jadi Pu = 0,80 Ag [0,85

* 25 (1 -

0,03) + 400 * 0,03]

fjJ


= 0,65

Pu

= 0,65

* 0,80 Ag [0,85 * 25 (1 -

0,03) + 400 * 0,03]

2,480 = 18,2447 Ag, Ag

= [m2]

Ag

= 0 , 1359 m 2 '·

Ag

=b xh

Kolom bujur sangkar, b = h h2 ~ 0,1359 pilih h = 0,370 m (Ag1 = 0 ,1369 m2)

Pn (m aks)

Po

138

=

OK

= 0,8 Pu

Beton Bertulang

=

2480 0,65

III

=3815,4KN

I<

Po

>I

3700 mm

3815,4

= _ ____:___ 0,80 =4769,2 KN

P0

= 0,85 f' c (Ag - Ast) + fy . Ast = 0,85 f'c Ag + fy. Ast- 0,85 f'c. Ast

4,7692 = 0,85

* 28 * 0,1369 + (400- 0,85 * 28) Ast

4,7692 = 3,2582 + 376,2 Ast Ast

= 4016 mm

p ""'2,9 % OK""'

2

I 3% :::; p :::; 8%

Pakai tulangan D 25 mm , n1

=

4016 2

± n(25)

= 8 bh Pakai 8 D 25 mm Eksentrisitas e = 22,58 e/h

=

22,58 370

= 0,06 < 0, I (kriteria kolom konsentris PB'89) OK

Bab 3_Kolom

139

(Pn) Po

0,8 Po=Pn

..__.._-~•

maks

Kutva

~ " INTERACTION DIAGRAM"

Mo

0

Contoh Soal - 2 Penampang kolom seperti gambar di bawah in i:

300mm

I< I<

413 550mm

>I >I

Ref I Standard: 1. Code SK: SNI T-15-1991-03

2. SNI 1727 - 1984 F 140

Seton Bertulang

spec material: beton f' c

=2 1 Mpa.

Baja tulan gan,

fy = 400 Mpa Standar I codes: P B '89 I ACI Code 318 - 83 (dapat anda pilih sendiri )

Pertanyaan: Hitung bcsar beban tekan Ph dan hitung juga ekstensitas "balance" eb untuk suatu kondi si regangan " balance" llitung kuat tekan minimal Pn kolom diatas jika diberikan ckstensitas e = 1 meter.

Catatan: Perhitungan harus memakai metode statis - analitis dengan prinsip metode kekuatan (strenght method). Reganga batas beton , Ecu = 0,003 Modulus elastisita baja tul angan Es = 210000 M pa a = 0 ,80X, d imana X j arak gari s nctraJ dari serat tepi bcton yang tertekan dan a ad alah panj ang " stress block" a;;u msi dari dalam pcrhitungan dapat ditentukan sendiri.

olusi 1 Mcnghitung Besa r Beban Tekan Pb Step 1: Mcnentukan lctak sumbu netral, Xb As

= As'= 4

* 7t/4 (25) 2

= 1963 mm 2 Es'

Xb 0,003 = Xb+68,5

ES

= tSb = fy = 40000 Es 2,1XIOM

=0,000 19 Bab 3_Kolom

141

=

0,003 0,003 + 0.00019

Xb

=

0 003 * 48 1,5 = 452,8 mm • 0,0031 9

Ab

= 0,8

Xh

48 1,5

* 452,8

= 362,3 mm 68.5

68,5

300 mm

I<

:

71

550

' ~

;

~

i

~~ ~<~c,_·---'-- 71 _.. Pb =Pn

i _,. ah

i"

V~

~ ..........

1'--/

V

.,___ T

Cc

/ ......_

10.85 re

'I' Cs

ab= 0,80 Xb

1 42

Seton Bertulang

Step 2: Menghitung gaya Cc , Cs dan T Cc = 0,85 * 2100 * 0,30 * 0,3623 = 194 KN T

= I 963

* 10 - 6 * 40000

= 78,52 KN

t-s'

=

452 8 * 0,003 = 0,00026 > £y ' 452,8 + 68,5

= 0,00019 (OK) Cs

= 1963

* I0- 6 * 40000 -0,25 * 2100 * 1963 = 75 KN

Step 3: Menghitung Ph dan eh LH Pb

= 0 ; Pb + T - Cc - Cs = 0 = Cc +Cs -T= 194+75 - 78,52 = 190,48 KN

LM

= 0 di Sb penarnpang kolorn

Pb. eb =

Pb. eh = 78,5

* (275- 68,5) +

194 + (275 - 117,9) + 75(275 - 68,5)

eb

= 16210,25 + 30477,4 + 15487,5 = 62175 KN- MM

eb

= 326 mm , Mb

= 62 KN -

M

Solusi 2 Menghitung Pn jika e = 1000 mm Berarti penarnpang berada dalam kondisi "kontrol " tarik. (dalam keruntuhan larik). es > ey menghitung gaya - gaya dalam Cc, Cs dan T

T

= As. Fy = 1963

* 40000 * 10 - 6

= 78,5 KN

Bab 3_Kolom

1 43

Cs

=As' (fy - 0,85 f'c) =75KN

* 2100 * 0,30 *X= 535,5 ?'; X?

Cc

= 0,85 f'c. b. a= 0,85

:EH

= 0 (pada penampang kolom)

Pn + T - Cc - Cs = 0 *)Pn =Cc+Cs-T = 535,5X + 75 - 78,52 = 535X -3,52 (pengaruh tulangan pacta beton yang tertekan diperhitungkan) **):EM= 0; di sumbu tulangan tarik T Pn(l + 0,2065)- 535,5X(0,4815 - 0,4X) -75(0,4815- 0,0685)= 0 I ,2065Pn - 257 ,84X + 214,2 X2 - 30,975 = 0

1,2065(535,5X- 3,52)- 257,84X + 214,2X2 - 30,975 = 0 645,48X - 4,25 - 257,84X + 214,2X2 - 30,475 = 0 214,2X2 + 387,64 - 35,22 = 0

x,.2

1 = - -(- 387,64 ± ~(387,64 2 )+4* 214,2 *35,22) 428,4 I = - - (- 387,64±424,78) 428,4

= 0,087 meter Jadi Cc = 535,5 * 0,087 =46,54 KN Kuat rencana adalah:

144

Pn

= 535,5 * 0,087- 3,52 =43KN

Mn

-:: 1,2065 * 43 =51 ,88 KN - meter,

e

= 1000 mm Seton Bertulang

Contoh Soal - 3 .l ika e Pn

= 0;

= ? KN

Step 1 - Kolom struktur non pratekan dengan pengi kat (faktor tekan diabaikan) Po =0,8l(0,30 * 0,55-2 * 1963 * 10- 6 ) * 0,85 * 2100 * 10- 6 + 40000 Po

Pn

* 1963 * 2 *

I0 - 6]

= 355,6 KN =0,65 * 355,6

=23 1 KN (SK SNI T lS- J 991 - 63 , pasal 3.3)

Step 2 - Interaction diagram Pn = 0; Mo = ... (penampang dalam kondisi under reinfo rced) X

=

0,003 X dengan E s = :;:. 0 ,003 (dan. d"1agram ) 0,003+£'s X +68,5 I

Asumsikan c's > ey Cs

= f's . As'

Es

= 2, I * 10 - G kg/cm2 = 2,1

Cs

= T = f's. As'= f's. As= 78,52 KN (tulangan simetri)

* 10 8 KN/m2

Karcna tulangan simetri Cs= T sehingga yang memberikan momen pada pen am pang adalah komponen/gaya. :EM

= 0 (terhadap sumbu T) Cc

* (0,48 15 -

0,4X)

= Ml

Cc

=0,85 f'c. b. 0,8X 0,85

Bab 3_Kolom

* 2100 * 0,30 * 0,8X (0,4815 - 0,4X) = M 1 1 45

MI

=206,27X -

:EH

=0

171 ,36X2

.................. (1)

Cc - T= 0 (mulanya dianggap As'= 0) jadi As1 ? 0,85 * 2100

* 0,30 * 0,8X- fs, * 1963 * 10-6 =0

.. ....... .. ..... . (2)

= Es. Es 1 = 0,003 d - 0,0030 X

ES1

= 0,003 0,4

815

........ ...... ... (3)

- 0,0030

X

substitusi persamaan (3) ke persamaan (2) diperolch : 0,85 * 2100 * 0,30 * 0,8X - 2,1

X

I0 8 * [0,003 0,4

815

- 0,003] * 1963 = 0

X

428,4X - [

303345

630000]

* 1963 * 10 - 6 = 0

X

428,4 x

2

-

454,47 + 1236,693X = o

2

X + 2,89X - 1,39

=o

x,,2

= Y2 ( - 2,89 ± ~(-2,89) +4*1 ,39)

x1.2

= Y2 (- 2,89 ± 3,73)

X1

=0,42 m

2

Persamaan ( 1) :

MI

=206,27 (0,420 -

M1

=56,4 KN - M

171 ,36 (0,42)

2

Persamaan (3): Es l

=0,0030 *

815 0,4 - 0 0030 0,42 ,

=0,000440 >Ey (OK) Asumsi benar 146

Seto n Bertulang

(Ey

40000

= 2,1X iOR =0,0019)

Pn (KN) " balance line"

231

tarik 190,48 c= I m

43

62

Contoh Soal - 4 Di ketahui: kolom harus memikul beban mati dan beban hidup, climana bcban hiclup sama dcngan 2 kali beban mati yang bckerja. Mutu beban:

fy

= 400 Mpa

f"c

= 25 Mpa = 6D25 mm

Ast Bab 3_Kolom

14 7

Ko mpatibilitas Ecu = 0,003

Pertanyaan: Jika eksentisitas Pn = 500 mm , Hitunglah beban ke1ja aktual ko lom ini, dengan mengabaikan kehilangan beton akibat pcmasangan tulangan Catatan: se lesaikan dcn gan prinsip" statika" CL

(1)

(2)

I

38

~

----71

X

~0

I

I

(3)

0,003 (
es' es

~~~( 200, Pn

I (4)

T

I

I

I cc[ tcs a

148

11 t o.85f'c

)

Beton Bertulang

Solusi: c

= 50 mm

= 0,1 h

Ao

= 1,2 : AL

= 1,6 ;

<1>

= 0,65; ~~

=0,85

Statika dasar: Asumsikan baja tulangan tcrtekan

* 3 = 1472,6 mm2

As'

= As = lA 1t (25)2

Cs

= As' (fy - o,85f 'c) = 1472,6 (400 - 0,85

Cs

= 588,78 KN

Cc

= 0,85 f 'c . a. b = 0,85 * 21 *a* 500

* 2 1)

= 8,925 a KN = 7,586 K

Dari gambar (3) di atas,

cS

=

X -

400

.0,0030

X

T

=As. fs =As. Es. Es = 1472,6. x -

400

.0,0030 * 10s

X

T

=883,56

X -

400

(KN)

X

Menghitung Jctak garis netral " X" :EM a12 a

= 0 di garis ke1ja Pn : diperoleh: = 0,425 X ; = 0,85X : X= garis netral (terletak diluar pen)

Cc (200 - 0,425X) + Cs (200 - 50) - T (200 + 50) 7,586 X (200- 0,425X) + 588,78 (150) - 883,56

X-

400

* 250= 0

X

1517,2 X 2 - 3,224 x3 + 88317 x- 220890 x + 97 191600 = o 3,224 x3 - 1517,2 x2 +1 32573X + 97191600 = o Bab 3_ Kolom

149

(dihitung dengan bantuan komputer didapat): X

z

507 mm

£y =

400 2,1* 105

= 0,002

Cc

440 * 0,0030 = 0,0004 < 0,002 (= £y) 507 = 7,586 * 507 = 3846 KN

T

= 883 56 ,

Pn

=Cc + Cs + T = 3846 + 588,78 + 116,8

Pn

= 4552 KN

Jadi cs =

507

*

507 440 = 116 8 KN 507 ,

Menghitung beban kerj a aktual (service load): Pn

=Pu = AI) . PI) + AL . PL

0,65

* 4552 =

1,2 Po + 1,6 . PL

ditentukan (Po :PL = 1 : 2 Po jad i 0,65

= 2 PL)

* 4552 = l ,2 * 2PL+ 1,6 PL= 4PL

PL

= 739,7 KN (maksimal beban hidup aktual nom·m al load)

P0

= Y2 Pt. = Y2 * 739,7 = 369,85 (maksimal bcban mati aktual noramal load)

Mcmcriksa/analis is pcnampang dengan tulangan rangkap

f'c

= 35 M pa

fy

= 400 Mpa

d'

= 65 mm

d

= 550 mm

b

= 350 mm

As

= 3280 mm2 = 775 m 2

As'

15 0

Beton Bertulang

3280 p =As/bd= 350* 550

p'

00170 , .

=As' I bd 775

= 350*550 = 0,0040 As - As' = 3280 - 775 = 2505 mm 2

p - p' = 0,0170-0,0040 = 0,013 meli hat tulangan leleh atau tidak

p-p' ;:::: 0,85/3 , I' c.d' ( 600 ) = 0,85 *0,8 * 35* 65 ( 600 ) fy.d 600 - fy 400 * 500 600-400 =0,021

p-p' = 0,013 < 0,021

f's < fy tulangan tidak leleh. 's

= 600 (I - 0,85/3 J ' cd')

(p- p')fyd

= 600 1--085*08*35*65] ·-·- -'- - ( 0,013 * 400 * 550 = 600

* 0,4591

= 275,45 K- N/mm 2

a

= Asfy - As' fs' _ 3280 * 400-775 * 375,45 0,85 f' cb 0,85 * 35 * 350 = 105,5 mm

Bab 3_Kolom

1

s

1

105 5 · 08

=

X

'

= 131' 87 mm

Dari perbandingan gambar. 0,003

0,85 f'c

a B.X

\

M) Es'

0,003

=

x - d'

=

X

131,87 -65 131,87

13 187 65 • 131,87

ts'

=

f's

= cs'

2/3

* 0,003 = 0,00152

* Es

=0,00152 * 200000 = 304 N/mm 2

pb

= fJ

0,85 * f' c * I

pb

fy

600 600+ fy

= 0,8 *0,85 * 35 * 600 = 0 0357 , 400 600 + 400

maka p maks = 0,75 p ~ 0 ,75

* 0,0357 = 0,0268

f'i' pb+p' _. fy

00 ,0268 + 0,0040

1 52

275 5 0 02955 - • ,4 400

=0' 02955 > p = 0'0170 Seton Bertulang

a

=

M.fy- As' fs 0,85f'cb

=

3280 * 400-775 * 304 0,85 * 35 * 350

= 103,375 mm Mn

=As. Fy - As'. fs' (d- a/2) +As' . fs' (d- d')

Mn

=(3280 * 400 - 775 * 304) (550 (775 * 304) (550- 65)

*

103,375 I 2) *

1 1000*1000

l + 1000* 1000

=650.659515 KNrn '

= 650,6 KN m Mn

=Mn

= 0,8 * 650,6 = 520,5 KN-m Contoh Soal - 5 Perencanaan Penampang Tul angan Rangkap Akibat Lentur

Mo . ML 660 mm

OKN - m OKN - m

f'c d'

fy d

=400MPa = 66'0 mm

As

300 rnm

Solusi: Mn

= I ,2 * 270 + 1,6 * 530 = 1172 KNm

Mu 1172 Mnperlu = - = - - = 1465 KNm f/J . 0,8

Bab 3_Kolom

1 53

Xb

£ cu

* d=

= - -c: cu

Xmax = 0,75

+£ y

0,003 *660 400 0003+ ' 200000

= 396 mm

* 396

= 297mm Cmax = 0,85 f'c b ~ 1 X

=o,ss * 34 * 350 * o,s * 297. w- 3 =2403,32 KN Asfy As

Mn

=C 2403,320 = --400 =6008,3 mm2 = Asfy(d - Y2 a)

=2403,32 (660- Y2 * o,8 * 297) * w- 3

= 1300,6R KNm b.M

= 1465 -

< Mn vol = 1465 KNm

1300,68

= 164,32 KNm Csperlu

= -6.M- = d - d'

164,32 * 1000 1,60 - 80

= 283,3 KN

Cek tulangan tekan pada X= 297, apakah sudah leleh?

Es'

= 297 -

80

297

*0 003 '

= 0,002192 > £y = 0,002 OK leleh

1 54

Cs

=As' (fy - 0,85 f'c)

As'

=

Es .fy - 0,85f'c

=

283300 = 763 4 mm 2 400 - 0,85*34 '

Beton Bertu lang

Tmax

=Ccmax + Cs =2403 ,32 + 243,3 = 2686,62 KN

Asperlu

= -T = -2686,62 - fy

400

=67 16,65 mm cs' ;::: cy T

= p. b. d. fy

Cc

= 0,85 f'c ~~

Cs

= (fy- 0,85 f'c) p' b d

T

=Cc+C

X

= o.s~~f'c[p-

cs'

= &cu(x -d')

*d

X

atau

X ;:::

;::: ey 600(x-d') ;::: fy

X

600d' 600- fy

600d' ;::: fyd 600 - .fy 0,85/3 1f' C

p

p'(I-o,s;·c )]

[p-p'(

1_0,85f' cJJ

fy

_,(t -

0,85f'cJ ;::: fyd ·[ _ '(l- 0,85f'cJJ fy 0,85/3 J'c p P fy

p

Contoh Soal - 6 Data- data: f'c

= 220 kg I cm2 (beton)

fy

=2800 kg I cm2 (baja)

Bab 3_Kolom

1

55

kondisi balance: £sb cCU

2800 =0,00133(ba·a) 2 1* 106 ~ ' =0,0030 (batang)

=

tulangan simetris 10 # 32 mm 4> 32 mm Aist

.

= -1l4 (32) "- =8,042 cm?

Pertanyaan: Tentukan gaya normal tekan balanced,

156

Seton Bertulang

Pn

= Pu

dengan Pb (eksensitas balanced)

f/J

Terjadi momen lentur di sumbu kolom dengan cara kesetimbangan (statika dasar) Dengan memperhitungkan pengaruh beton yang ditempuh tulangan tekan.

Cs,

= 4 (8,042) (2800- 0,85

* 220)

= 84055 kg

Cs2

= 2 (8,042) (1556 - 0,85

* 220)

= 22019 kg Cc

= 0,85 (220) (0,30) (37 ,41) (50) = 279827 kg

T

= 4 (8,042) (2800) = 90070

'LV

=0

Xb

=

0,0030 *54 0,0030 + 0,00133

= 37,4 1 cm ab

= 0,8 *37,4 1

= 29,93 cm f's2

= £S2. Es * 106

= 0,00074 1 * 2, l = 1556 kg/cm

2

= (37 ,4!- 30 ) * 0 003 =0,000741 30

Pn

'

=Pb = 295,83 ton

Bab 3_Kolom

1 57

Eksentrisitas balanced , eb, dilakukan :EM Eb

=

84,015(24) + 22,014(0) + 279,827(15,035) + 90,07 (24) 295,83

= 28,35 cm Sehi ngga Mb

=295,83 * (28,35 * 10 - 2 ) = 83,87 ton meter

3.2

Perencanaan Kolom

Pastikan kolom brace frame I unbrace frame LE/ - . -kolom ' L - • H 1tung fj/ = -L=:E1 - -balok L

Cara menghitung El fc' (MPa) Kolom:

Ec = 4700 ES

ff

=200.000

Mpa

Mpa

El = 0.2 Ec . Jg + Es . Is } Ambit yang besar

El = 0.4 Ec . Ig Bal,ok:

Icr Blk = 'h Ig Ig = 1/12 bh3

El blk = Ec . Icr From salmon ha! 69 Faktor panjang efektif

KS 1 (brace frame) K 1 (unbrace frame)

•

lu cara nilai K r

r = 0.3 h kolom sengkang

r =0.25 h kolom spiral

158

Seton Bertulang

termasuk kolom pendek bila: _ M 18 untuk brace frame K lu :::; 34 12 r M 28 untuk unbarce frame

IM 2b I > IMtb I emin

M 18

=Et>

kurva tunggal 0 kurva)?anda

M2B

= ( 15 + 0.03h) mm

perhitungan kolom pendek Pn

emin :::;

Pu

Mu

=-

eperlu

f/J

e :::; 0.1 h

=-

Pu

beban konsentris

Pn = 0.8 [0.85 fc' (Ag- Ast) + Ast. Fy] e

> emin

K= E H

=

f/JMn fc'.Ag.h Dengan y, fc', fy dari Mn diagram didapat p.

K

=

f/JPn fc'.Ag

Ast= p. Ag bila p > 8% ukuran kolom diperbesar.

KONTROL RET AK (BEAM~ SLAB) Dibatasi dengan nilai Z

Z

= fsVdc.A

Fs

=60% fy (ACI-10.6.4)

de

= jarak serat tarik beton ke pusat tulangan tarik paling bawah

A

=----

Bab 3_Kolom

Ae m

2ds.b m

159

Os m Z

= jarak serat tarik beton ke pusat kumpulan tulangan tarik = jumlah tulangan tarik

batas ~

Zhalas ~

30 MN I m (interior)

25 MNim (exterior)

Nb: Zbatas ZbaUlS

(interior) diperhitungkan untuk lebar retak 0.016 inci (041 mm) (exterior) 0.013 inci (0.3 1 mm)

CATATAN Concrete:

E

=2.1

v

= 0.2 (pbs '71)

* 105 kglcm 2

beton normal y = 2200 - 2500 kglm3

= 1 .10·5 I 0c beton ri ngan y~ '1900 kglm 3

=10° c between part

y

=2.4 Tlm 3 = 15° c environment

V

= 2.1 * 106 kg/cm2 =0.3 (pbbi)

y

= 7.85 Tlm 3

Steel: E

160

~T

Seton Bertulang

a

= 12 .10·6 I 0c

~T

= 15° c between part= 30° c environment

LEKATAN BETON DENGAN STEEL Oup

= 4<1> .Jabk'

<1> = Reduction factor= 0.65

= (ksi)

o bk

= compresive strength (ksi)

3. 3 Aneka

Ragam Soal diambil dari Buku Reinforced Concrete Design , CHU - KIA WANG dan CHARLES G SALMON

Contoh Soal - 1 Determine the compressive service load that the column may be ommitted to carry at an eccentricity cy = 0,1 * h = I ,8 in., according to strength design 1 procedures. Asume that load and live load are aqua!. Use f c = 3000 Psi and fy = 40000 Psi. Use basic principles of statics to abtain solution, considering the effect 'Jf compression concrete displaced by steel ; compare the result with that obtained by the Whitney formula, Eq. ( 13.14.8 ), and compare with the maximum Pn given by ACI - 10.3.5.

(a) Use Basic Statics. Assume Comp. Steel Yields Cs

1

=As * (fy - 0,85*f1c) = 3,0 * ( 40- 0,85 * 3 ) = 3,0 * ( 40 - 2,55 ) = 112,35k

Cc

* a = 0,85 * 3 * 18 * a = 45,9 * a OK ......... 0,85 * 45,9 * X 1

= 0,85 * ( f c ) * b

= 39,0 *X T

= As*0,003 *(X - 15)(29000)= 26 I *(X - 15)

X

Bab 3_Kolom

X

1 6 1

·:. l ·:. T t___. 1 : ' 6 - t9

.___•_-T-i'__

0,003

LM

A!30UT

1,811

Pn = 0

15 26 1 *(X ) * 7 8 -- 0 112 ,35 * 4 '2 + (( 39 '0 * X ) * ( 7 '2 - 0 ' 425 * X )) X ' 3 2 c47 1* x ) + c280,8 * X ) - c 16,575 * X ) - c2035,8 * x ) + c30,537 )= o X3 - 16.95

* X2 + 94.3 * X - 1843 - 0 -Neglects Displaced conrete At T, which is actually compression.

1- 16,75

+ 17,56 +0,61 X Cs

+ 94,3 + 10,7 + 105,0

= 17,5611 = 112,35 k

2 56 c:s = 0.003 * - · - < c:y 17,56

Cc

162

-1843 + 1843

~

0

Considering Dis~1 aced Concrete at T; X3 - c 16,94 * x- ) + c90,72 * x ) 1841,65 = 0 X =17 696 11 Pn = 834,86 k ( Computer Solution )

.

= 39,0 * 17,56 =685 k

Seton Bertulang

261 * 2,56 = -- 17,56

T

= 38,0 k = 112,35 + 685 + 38,0 = 835,4 k

Pn

0 ,7

* 835,4 = 1,4 * !_ + 1,7 * p 2

2

= 1,55 * p Pw

= 377 k

(b) Whitney Formula

Pn

=

b*h* f 1c

As 1 * .fy

e 1+ 0,5 d- d

+

3*h2*e+ 1,18 d

=

3*40

324*3 + 3* 8* 8 1,8+0,5 I I, +1,18 12 15 2

= 184,8 + 60,2 =787k (c) ACI - 318-77 Maximum ACI- 10.3.5.1. Pn max = 0,85*(0,85* f 1c*(Ag - Ast)+(jy*Ast))

=0,85 * Po = 0,85 * 1050 = 893 k

Bab 3_ Kolom

1 63

Ultimate Strength Interaction

__ ~ A~I - 318 - 77 ; Pn max

=893k

)11!5-r__;:_=~::!!.-:=.2..::::!......!~-- eb =8,2i P = 232,2 k ( X

1

=6, 1511 ) c = 9,0

11

10

•• 10 -# 11 •

0 00237 > ty

Cs2

164

Cc

Cs,

Beton Bertulang

Contoh Soal - 2 Determine the balanced condition Pb = U I in addition to eb with respect to bending about the strong axis for the column of the accompanying figure. Use the basic statics method with f c = 3000 Psi and fy =40000 Psi. a. Consider the effect of the compression concrete displaced by steel. b. Neglect the effect of the compression concrete displaced by steel. 1

Cs1

= ( 4* 1,56 ) * ( 40 ( 0,85 * 3 )) = 6 24 * ( 40 - 2 55 ) = 233 5k '

'

Cc

= 0,85 * 3 * 20 * 0,85 = 623 k

CsL

=2

T

= 6,24 * 40

'

* 14,38

* 1,56 * ( 14,5 -2,55) = 37,3 k

= 249,3 k 233,5 623,0 37,3 893,8 - 249,3

Pn

= Pb = 644.5 k -considering Effectof Displaced Concrete;

= 668 '7 1k eb = 623 * (12 - 6,11) + (233,5 * 9 )+ (249,3 * 9) = Mb

644,5

Bab 3_Kolom

3670 + 4350 = ,4511 12 644,5

165

Neglecting Effect of Displaced Concrete ; Cs, Cs2

=T; = 45,2 ph

= 623 + 45,2

=668~ =

3670 + 4480 668,2

= 12.20

11

Contoh Soal - 3 Using the basic statics method and taking account of the compression concrete displaced by steel, calculate and plot the Pn versus Mn interaction capacity diagram for the column of "Contoh Soal- 2". Take bending with respect to the strong axis (h = 24 in). Use f 1c = 3000 Psi and fy = 40000 Psi. To obtain points for the diagram, compute Pn for the following cases in addition to eh ( see "Contoh Soal- 2" ) e=O(Po)

(d)

e=0,7*h

=0,1 * h e =0,3 * h

(e)

e=h

e

(a) Find Po

166

(f)

e=0

Cs,

= 6,24 * ( 40 - 2,55) = 233,5

Cs2

= 3,12 * ( 40 - 2,55) = 116,75

C S3

= 6,24 * ( 40- 2,55 ) = 233,5

cc

= 0,85 * 3 * 24 * 20 - 1124,0 PO

- 1807.75k

e

-o

Beton Bertulang

(b) Find Pn At e

=0,1 * h;

e =2,4

11

Cs1 = 233,5k

C , = 3 12 * 0,003 *(X - 12,0) * 29000 s_ ' X =

c

2

~·5 (x -1 2,o)

= 6 24 * o,003*(X -2 t,o)*29000 = 543,o(x - 21 o) S3

Cc

:E M

Bab 3_Kolom

X

'

= 2,55 * 20 * 0,85 = 43,4 *X ABOUT

X

'

*X

Pn =0

167

233,5 *(9,0 -2,4) - (43,4 * X * (0,425* X -9,6)) -

2

~·5 *(X -

12,0) * 2,4

543 0 X • (x - 21,o) * 11,4 = o (1538 * X) - (18,43 *X3)+ (417 * X~)- (65 1 *X) + (7815) - (61 95 * X) + ( 130000) = 0 X3 - (22 ,65 * X 2) + (288 * X) - (7470) = 0;

Sehingga: X= 23,8

~ = 20,2'

11

;

1

1: M check

* 6,6 =

= 233,5k

+ 1540

=3, 12*( 40 -2,55)=116,75 k*2,4 = = 0,003 *

11 8 • 23,8

= 0,00149 > Ey

Cs3

= 6,24 * 10,23 Psi

Cc

= 43,4

* 23,8 Pn

Mu

= 63 8 k ' = 1033 k - 1447 k

*1 1,4 = + 728 * 0,5 = - 517 + 15 zO

=289,4 1k

(c) FindPn at e = 0,3

16 B

280

* h:

Cs ,

= 233,5

Cc

= 43,4 * X

e - 7,2

11

Seton Bertulang

Cs2

= 3,12 * 0·~03 * (x -12)* 29000 = ~

T

= 6,24 *

27

44

(X -12)

·~ * (21- X)* 29000 = 54~88 * (21 -

0 03

X)

23~5*(9-7,2) + 54288 *(21-X)*(9+ 7,2)-(4~4* X *(0,425* X - 4,8))X

27

~

44

*(X - 12)*7,2

=0

(420,3 * X) - ( 8794,66 * X) + (184687,78) - (18445 * X3) + (208,32 * X2 ) - (1954,37 *X)+ (23452,42) =0 (X3 )

Bab 3_ Kolom

- (

11,29 * X2) + (559,97 *X)- (11284,37) = 0 - 11,29

+ 559,97

+11284,37

+ 17,10 + 5,81

+ 99,35

+ 11274,37

+ 659,32

-10,0:::::0

169

Sehingga:

X

= 17, 111 ;

a

= 14,54 11 141 ' = 0 ,00247 > cy; = 0 ,003 * -171 ,

Cs1

= 233,5k

Cc

=43,4 * 17,1 =742,1 k =

27 144*~ , 17,1

= 80 96 k . , ,

T

39 = 542 88 * • = 123,8 k , 17,1

Pn

= 233,5 + ( 80,96 - 7,96) + 742,1 - 123,8 = 924,8 k;

Mn

= 554,9

Check L. M = 0

( 233,5

ABOUT·

lk

Cl

* 9) + ( 123,8 * 9) + ( 742,1 * ( 12 -

7,27 ))

=( 3215+3510)

= 6725 1k Pn

=

6725 7,2

= 934k::::: 925; Displaced Concrete at Cs2 was not Accounted for in X eqn

170

Beton Bertulang

11

(d) Find Pn at e =0,7 * h ; e = 16,8 > eb Cs =233,5k ( assumed) Cc

= 43,4 *X

T,

=249,3 k

T2

= 3,12 * 0·~03 *(12- X)*29000 = 27_i

LM

ABOUT·

44

* (12-

X)

=0

Pn

23~5*7,8+{43,4X *(0,425* X +4.8))- 27 l 44(12- X)* 16,8-(24Q3*2~8) = 0 X

< s2o * x ) + < 18,4 * x ) + c208 ( 54600 ) - ( 6435 * X ) = 0 3

cX 3 )

+ ( 11,3 ,~ X 2 )

-

* x2 ) + <4555 * x ) -

c 3,26 * x ) - 2967 = o

+ 11,3 + 11 ,48

3,3

- 2967

+ 261,5

+ 2965

+ 22,78

+ 258,2

sehingga:

X A

11

= 11 ' 48 ,• 11 = 9,77

es

8 8 = 0,003 * ,4 = 0,0022 > ey 11,48

Cs

= 2,335

Cc

= 43,4

T,

= 249,3

T2

= 27 1 44 *

OK

* 11,48 = 498,0 k

,

0 52 • 11,48

= 12,3 k

Bab 3_ Kolom

1 71

= 498,0 + 233,5 = 657 ' 9 1k

Pn

Mn

Check L M = 0

ARO UT

249,3 - 123 = 470k ;


( 233,5 + 249,3) * 9 = 4350 lk 498,0

Pn

* ( 12-4,89)

= 7895 = 470 16,8 =

tk

1

= 3545 k =7895 1k OK

=2,411 > eb

(e) Find Pn ate= h; e Cs

= 233,5k

Cc

=43,4 *X

(Assumed)

= 249,3k = 271,44 *(12 _ X) X I. M

ABOUT

Pn

=0

(23~5*1 5)+(43,4* X*(0,425+12))- 2 7l4 *(12-X)*24-(24Q3*33) =0 X

(3502,5 *X)+ (18,4 * X3) + (520,8 * X 2) + (65 14,6 * X)- (78,438)(8226,9 * X)

=0

x3 + c28,3 * X 2 ) + c97,3 * x ) - 4255 = o 1

1 72

+28 ,3

+ 97,3

- 4255

+ 9,42

+ 355

+4260

+ 37,72

+ 452,3

""'0

19,42 maka : X= 9,42

11

;

a= 8,00

11

Beton Bertulang

6 2 ,4 = 0,0020 > c_y 9,42

1

s

= 0,003 -

Cs

= 233,5 k

Cc

T,

= 43,4 * 9,42 = 408,8 = 249,3 k

T2

2 58 = 27144* • = 743k ' 9,42 ,

Pn

= 233,5 + 408,8 - 74,3

£

Check :EM= 0 9

k

* ( 233,5 + 249,3)

=4345

* ( 12 -4,00 )

= 3270

408,8

= 7615

=~; Mn

=637 lk

Pn

=

7618 24 =317,4 k

(f) Find Mn for Pn = 0

Cs Cc

= 233,5k = 43,4 ~' X

(Assumed)

=249,3k

= 271,44*(l 2 -X) X

:EF=O

( 233,5 ) + ( 43,4 * X) - ( 249,3)-

271 44 • * (12- X) == 0 X

( 233,5 *X)+ ( 43,4 * X2 ) - ( 249,3 *X)- ( 3257,28 ) + ( 27 1,44 * X ) = 0

Bab 3_ Kolom

173

+ ( 5,89 *X) + 6,2 + 12,09

-75,05

=0

~

:::.0;

Maka: X = 6,i1

+ 74,96 - 0,9

32 c 1s= 0,003*-'- =0,0155>cy 6,2 k

Revise T2= 3,12 * 40 = 124,8

:EF=O ( 542,8 *X)--'( 1628,64)- ( 15,91 *X)+ ( 43,4

a

* X 2 ) - ( 249,3 * X ) -

= 3,91

(124,8

* X) = 0

+ 3,52 *X

-37,53

+ 4,6

+ 37,35

+ 8,12

-0,18

=0 zO;

11

16 = 0003 * • ' 6,2 = 0,0104 > cy

OK

Cs1

= 6,24 * ( 30,26-2,55) = 172,9 k

Cc

= 43,4

Mn

=9

* 4,6 = 199,64 k

* ( 172,9 + 249,3) + ( 199,64 * ( 12 -

1,95 ))

= 3799,8 + 2006,4 = 5806,2 = 484lk 12

174

Beton Bertulang

Check L: F = 0

172,9 + 199,6 - 124,8 - 249,3

= 372,5-374,1

1--r

•

•

10 . #11

• •

I

11 • 20 •

•

,...

..,

_l

h= 3 11 f1c =3000 Psi fy =40000 Psi

100

200

300

400

500

600

700

Mn·:.

Bab 3_Kolom

17 5

P. 13.7 And P 13.8 Results Only

250

P = 686 k; e = 16,54°

200

400

600

1000

800

722 r~

=Mo

Mn ft- kips

Contoh Soal For the section of the accompanying figure, using ba. ic statics determine plot the interaction diagram of Pn - Mn for bending about the X - axis. Compute the balanced condition in addition to tho e point indicated for "Contoh soal- 2". Use f 1c = 3000 Psi and fy = 40000 Psi.

1 76

Beton Bertulang

1,5 0,375 0,635 2,51

(a) e

=0 PO

= ( 0,85

* f 1c * b *h)+

(As* ( fy- 0,85

* f 1c))

= ( 0,85*3 * 14 * 18 ) +(5 ,08*(40 -2,55)) = 642,6 + 190,2 (b) e =0 I

Cs

* h = L8 in = 2,54 * ( 40 -

=833 k

2,55 )

= 95, 1 k Cc

=0,85 * 3 * 14 * ( 0,85 *X) = 546, 1 k

0,003

Bab 3_Kolom

177

=

T

2,54*[ 0·~03 *(X - 15,49)*29000 - (0,85*3)] =24,3k

X 3 - 16,94 * X 2 + 103,3

(c) e

X

= 18,0 in

Pn

=665 k

* x - 2200 = o

=0.3 * h =5.4 in

Cc

Cs

~ 0,003

•I

I·X : \2,68 !

Cs

= 2,54

* ( 40 -

=95,1

k

2,55 )

= 30,34* X = 384,6 k

Cc

= 22l*(X -

T

X

L

15.49)

=_ 491 k '

Same as at 0, I *h but tension so there is no subtracted term X X

3

-

c84,7 * X 2 ) + ( 195,7 * x)- c3156) = o = 12,675 in

Pn=431 178

k

Seton Bertulang

(d) e =0,7

* h = 12,6 in

Cs

= 2,54 * ( 40 = 95,1

2,55 )

k

cc

=( 30,34* X) =208,4 k

T

= 2,54 * 40 = 101 ,6 k

x3 + c 84,7 * x)- 1o5,3 =o X

= 6,87 in

Pn

= 202 k

Bab 3_Kolom

1 7 9

(e)e=h=l8in

X= 4,43 11

Cs

= 2,54*[29000' 0,003* X

~2,51 -(0,85*3)]

= 89,2 k C

= 30,34* X= 134,3 k

T

= 2,54 * 40 = 101 ,6 k

X3 + (21,18

(f)

* X2 ) -

(

1,49 * X)-495,0=0

X Pn

= 4,43 in = 122 k

Cs

2 51 = 254*[87*(X - • ) , X

Mo

Cc T

255] =164k ' '

= 30,34 *X= 35,7 *a = 85,0 k

=2,54 * 40 = l01,6k

X2 + ( 3,72 * X) - 18,28 = 0 ; maka : X= 2,80 11 Mn

1 80

;

a= 2,38

11

1 = [85,0*(15,49 -1,19)+(16,4*12,98)]*- = 119 1k 12

Seton Bertulang

Po:::: 833 ~

Nomin,al Streng~_h

P;..

Mo = 119 1k

Nominal Strength, Mn

Contoh Soal- 5 For the section of "Contoh Seal - 4" detennine the nominal ultimate capacity Pn for an eccentricity of 5 in, with respect to the X axis. Use f 1c = 5000 Psi and fy = 60000 Psi.

Bab 3_Kolom

18 1

Cc

t:>..

Cs

~0,003

b~

Balanced Condition

xh

11

pb

= 9,17 = 426 k

Cb

= 9,95

11

Actual Condition T

= 29000 * 0,003 * lS,49 - X * 2,54 X

= 44 0 k '

182

Cs

= 2,54 * ( 60 = 141 ,6 k

Cc

= 14 * 0,85 = 614,8 k

( 0,85

* 5 ))

* S * 0,8 * X = 47,6 * X

Beton Bertulang

x3 - c w,o * X2 ) + c 122,3 * x ) - 2065,7 = o X

= 12,92

Pn

=~ =~

Mn

11

Contoh Soal - 6 Using basic principles, determine the nominal, ultimate capacity Pn for an 1 eccentricity of 30 in, on the column of the accompanying figure. Use f c = 3500 Psi and fy =40000 Psi.

Bab 3_Kolom

183

Assume Tension Controls Cc

= 0,85

Cs

=2

* 3,5 * 16 * * 0,85 * X ) = 40,5 * X

* 0,6 * ( 40 -

2,97 ) = 44,4

L =2

k

Assumes Yield

* 0,6 * 40 = 48,0 k

L

Assumes Yield

= 2 * 1,56 * 40 = 124 ' 8 k

L. M ABOUT Pn ((40,5 * X ) * ( 15 + 0,425 42) =0

* X )) + ( 44,4 * 17,3 1 ) -

( 48 * 30) -( 12,48

*

c607,5 * x) + c 11,2 * X2 ) +770 -1 440-5290= 0 2 2 X + c35,3 * x ) + c 17,65 ) = 346 + 311 ,5 X

= .J657,5 -1 7,65 = 8,01 in = 0,003 *5 70 8,0 1 , = 0,00214 > ey ......... OK

0,003 es At T2 -- - * 6 , 99 > ey ......... OK 8,01

Pn

1 B4

= ( 405 * 8,01) + 44,4-48- 124,8 = 196 k

Seton Bertulang

Contoh Soal - 7 Design a square tied column with symmetrical reinforcement of about 3,5% to carry a dead load of P = 200 kips and M= 75 ft-kips and live load of P = 1 175 kips and M = 65 ft-kips. Use 1- in, increments with f c 4000 Psi and fy =50000 Psi.

=

- (1,4*200)+(1,7*175) 0,70

Reqd: Pn

= 825 k Reqd: Mn

- (1,4*75)+(1 ,7*65) 0,70

=308 k e= (308*12) = 448 11 825 ,

(a) Determine size for balanced section

xb

= ( 3 + 31,73 ) * d =o,635 * d

ab

=0,85 * Xb =0,85 * 0,635 * d

Pb

= 0,540 * d =Cc+Cs -

T

Since Cs ""'T assuming Cs yields and neglecting displaced concrete Pb

=0,85 * f 1c * b * ~

=0,85 * 4 * b * ( 0,504 * d ) = 1,835 * b * d =825; b*d

825 1,835 = 450 sg.in

Bab 3_Kolom

185

Assuming d :::: 0,9 * h ; b *h =

450 = 500 sg.in balanced 0,9

Try column w ith Ag < 500 ; :. Compression controls

(b)Est "LO" Use Eq ( 13.14,19) to get trial section

4 48 !!... :::: • = 0224· Say 0,2 h 20 , d- d 1 =- h

15 = - Say 0,75 20 d = - = 0 875 h

Pn

= Ag *

'

f 1c 3

e

-J:' * -+ 118 '='- h ,

186

825

= Ag*

Ag

. = -825 = 259 sq.tn 3,18

P * !Y

K(e) + 1 - *-

+2

r

h

4 + i035* 50 =Ag*(2,04+1 ,1 4 ) '2 *0,2+ 1,18 -*0,2+ 1 0,75 ( 0,875) 3

Beton Bertulang

rl

I

1

Cs2

:4 I

c;::;:==='"" I I

. •.. .

~.

..___ X= 13,14

Try 17

11

11

0,003

0,00243

--j

col.:

4 8 e = ,4 = 0 264 · h 17 ' '

12

- = 071 'Y =17 ) ·,

~=

14 5 • = 0,854 17

825

ReqdAg = - -1,77 + 1,12

= 285 sq.in. Bab 3_Kolom

187

Could use 17 11 ; 18 is better dimension.

11

Try 18 11 column ;

As

= 0,0035 * 259 = 9,08

4 8 !!._ = ,4 = 0 249· h 18 ' '

'Y

= 13,02 = 0 725 . 18 , '

): ...,

= 15,51 = 0 863 18 '

Pn

= 324 *

[~+ 0,028*50]= 865 k 2,18

1,687

If only 6 bars are used in formu la; Pn

= 324 * ( 1,835 + 0,65 ) = 806 k < 825 k ; Reqd

Probably OK Check by Statics Cs 1 = 165,0k

= 3,54 * ( 50 - 3,4)

·~

0 03

* (x

- 9) * 29000 - 3,4

Cs2

= 48,1 k = 2,0* [

T

=- 55,5 = - 3,54 * [

c

=0,85*4*0,85*13, 14 *18

·~

0 03

J

* (15,51- X)* 29000]

= 683,7 k = 0,85 * f 1c * 0,85 * X

188

Beton Bertulang

X 3 - (10,64 X= l 3,14 11 Pn

* X 2 )+ (171 ,83 * x)- 2691 ,8 = o

= 683 ,7 + 165,0 + 48, 1-55,5 = 841,3 k > 825

k .......... OK

Use 18* 18sectionwith4-#10and4 As Shown above.

#9

Contoh soal - 8 Redesign the column of "Contoh Soal- 7" using a rectangular column not over 15 in wide instead of asquare column. Required : 3,5 % reinforcement; dead load P = 200 kips and M = 75 ft-kips; live load P = 175 kips and M = 65 ft-k ips; I- in, increments; f 1c = 4000 Psi; fy = 50000 Psi. - (1,4*200)+(1,7* 175) 0,70

Reqd:Pn

= 825

k

- (1,4*75)+(1,7* 65) 0,70

Reqd: Mn

= 308 k - (308*12) 825

e

= 4,475

11

Balanced Condition =

0,003 *d 0,003 + 0,001724

=0,635 * d Pb "'"Cc assuming Cs = T pb

:= 0,85 * f 1c * b * ( 0,85 * xb )= 0,85 * 4 * 0,85 * 0,635 * b * d =

1,835 Bab 3_Kolom

*b *d

189

For Pb

b

=825 k

* d ( balanced )

825 = -1,835

=450 sq.in = 450 = 3011.

d ( balanced )

15

h=

'

For section smaller than about 15

30

09 '

= 33 '4 11

* 33,4, compression will likely control.

Use Eq ( 13.14.9) to examine what size is needed for pg"" 3,5%

Compression Controls:

Pn

=Aa * "

e Est. h

=0,2; =

=

Pn

d-d 1 h

cl h

=0,8 ;

=0,854

= 825 =Ao * 0

4

3

-( )2 *0,2+ 1,18

0035*50

+ ;

0,9

825

19 0

*[

=A g

- *0,2 + 1 0,8

4,0 + 1,75] = 3 249 *A 1,921 1,50 ' g

Seton Bertu lang

825 11 = 254 • 3,249 '

Reqd: A = g

h = 254 = 16 911 15 '

For 15

* 18

4 475 = 0 249· Est. !!._ = • h 18 ' '

825 =A

"{ =

13 -=0 72. 18 ' '

~ = 15,5 = 0 86 ]8

'

1 75

40

• + • ]= 2,863 *A * [ 1,92 g 1 1,69 g

Reqd : 825

= 2,863 = 288 11 . '

h

= 19 '2 11

· As

= 0,035

* 288

=10,1 sq.in Since formula is conservative when well up in the compression controls region , Try 15

* 18 wi th

10-#9;

d1

11 =2437 , '•

d

= 15,563

11

Capacity by statics; Pn = 809 k < 825 k ...... ... reqd within 2% ........ 0K Use h = 18 11 :

b = 15 11 ; with 10 - # 9 in two faces

It is preferable to use some bars in opposite faces as Bab 3_Kolom

19 1

Acceptable Choices Section 1 b * h; 15 * 18 0

Bars 10- # 9

2,437 15,56 809k

Mn 301 6 '

lk

X= 14,23 in

x3 - c10,65 * X2 ) + ( 234,2 * x ) - 4056 = o 2. 15

* 19

"10- # 9

2,437 16,56 830,5k 309,7 lk

( 4 bars each face ; 2 bars mid - depth )

Contoh Soal - 9

Redesign the column of "Contoh Soal - 7" as a spirally reinforced circular column. Required : 3,5% reinforcement; dead load P = 200 kips and M= 75 ft-kips; live load P = 175 kips and M= 65 ft-kips; l- in, increments ; f 1c = 4000 Psi ; fy = 50000 Psi. Reqd: Pn

=

(1,4*200)+(1,7*175) 578 = -0,75 0,75

=772 k

Reqd: Mn

= (1,4*75)+(1,7*65) = 215,5 0,75 0,75 = 308 k

e

= 287 * 12 = 4 47 11 772 '

0,8. h

192

Beton Bertulang

Balanced Condition; Using Equiv. rectanguler Xb=

0,003 *d 0,003 + 0,001724

= 0,635

* d "" ( 0,635 * ( 0,45 * h)) = 0,29 * h

Pb "" Cc assuming Cs "" T

Pb = 0 85 '

* f 1c * (

A,~~ J* 0'85 * Xb

08* h '

L.. b = 0,8*h Ag = 0,85 * 4 *

A g

* 0,85 * ( 0,29 * h )

0,8 * h = 1,05

* Ag

11

Ds"" h- 4 ./ .4 1,5 ll Cover+ bar dia d "" (2/3) * ( h - 4 ) + 0, 1 * h X = 0,57 - ( 8/3 )

Say: 0,45

*h

2

tr* h If: Pn = Pb , then 735 11 = - 4

h = 30,6 11 for balanced

Desire Smaller Column; :.Compression Control ! Use Eq. ( 13.15.1)

For Preliminary size.

A *fl.

Pn

c

1:

=Aae *

9,6* el

--=---...,..-------'--/..:...: hh_...,..,..- + 118

(o,8 + (o,67 *

pg

ny,JY '

+ p 11 *p g *fy 3* el I h +1

D:ltz

= 0,035 ( Desire as per statement of prob)

!!_ = 4 ,4? =z025· Ds z08 h h ' ' h '

Lsay 20 11 or so

Bab 3_Kolom

193

*

Pn

-A g

Pn

= Ao * e

4,0

0,035 *50

9,6 * 0,25 + 1,18 (o,8Y *(t,67Y

3 * 0,25 + 1 o,8

------~-----+-------

1 75 0 fI 24 •52 + · ] = Ag * ( 1,585 + 0 ,90) J 937 ' ' '-

= 2,485 * Ag = 772 = 31011. 2,48 ' h

= 19,9

11

11 Try h = 20 ;As= 0,035

* 3 14 =

110

11 - # 9 9 - # 10 ( 11 ,42) 7 - # 1 ] ( ] 0,9)

11

With 7- # 11

p8

= 0,0347

Recheck By Eq (13 . 15.1)

!:. = h

4 47 • = 0 2235· 20 ' '

__ Ds _ 20 -3-0,75 - 1,41 =0,74

h

40 1 735 ] =314 = 314*[ • + • 2,45 1,905

Pn

20

* ( 1,63 +0,91)

= (314 * 2,54) = 799k > 772 k Use h = 20 11 Diameter with 7- # 11

Statics Using equivalent rectangle: 11

20 dia; 7- # 11, equiv. rect. h = 16 b

1 94

= 19' 6

11

11

;

•

'

Seton Bertulang

=3, 13 11,• = 12,87

11

x 3 - c8,31 * x 2 ) + c 180,4 * x) -

2376,2 = o;

11

X

= 11,18

Pn

= 816,3 k > 772 k ... .. .... .OK

20 11 dia; 8 - # I 0 Pn

= 797,9 k > 772 k ......... OK

Contoh Soal - 10 Design a square tied column with symmetrical reinforcement of about 4% to carry a dead load of P = 225 kips and M= 220 ft-kips and a live load of P 1 = 200 kips and M= 190 ft-kips . Use 1- in , increments with f c = 4000 Psi and fy = 40000 Psi.

=

Reqd: Pn

(1,4 * 225) + (1,7 * 200) 655 = -0,70 0,70

= 935

k

Reqd: Mn

=2 =902 kf

e

=---

902 * 12 935 = 11 ,57 u

For Balance:

= Cs

0,003 *d 0,003 + 0,00 138

=A's

* ( fy-(0,85 * f 1c))

= A Is

* ( 40 -

3,4 )

= 36,6 *A's Bab 3_ Kolom

19

s

* fy = 40,0 * As

T

= As

Cc

= 0,85 * f 1c * b

* ( 0,85) * 0,685 *d )

*b*d

= 1,977

=Cc+ Cs - T = (1 ,977 * b *d) - ( 3,4 * ( 0,02 * b * h) If : d b and 935

= 0,9 * h; =h; = 935

k

= ( I ,977 = 1,848

* 0,9 * h2 ) -

(

2

3,4 * (0,02) * h

* h2

= 935 = 506· 1,848 ' h

= 22,5

11

Try smaller section so that compression conytol.

Use Eg. ( 13, 14.9) to betler estimate size;

Pn

=A • g

Use 21

11

r

I

+ p' • fy

Ic

l;, ·~+ l,l8 ~·(~) + !

as basis for ratios;

= d - d I = .!i = 0 7 62 ' h 21 ' '

e h

196

=

1

~~

7

= 0,55;

~ = !._ = 18'5 = 0 88

..,

h

21

'

( May be high enough for tension controls)

Beton Bertulang

Pn

4

-A g*

3 2 (

0,88)

0,044*40 + 2 *0,55 + 1,18 --*0,55 + 1 0,762

=A8 * ( 1,205 + 0,657) =1,862 * Ag 935 =-1,862

Reqd.: Ag

= 502 sq.in;

=22 11 +

h

Appears that section with be nearly balanced when pg = 0,04 compression controls formula is not conservative in this range near the balanced condition. Try 23

11

* 23 11 ;

As = 0,04

* 502 = 20 ll+ sq.in

Try 14 - # 11 ; As =21 ,84 Statics Check; Cst

=5

* 1,56 * 36,6 = 285,5 k

= 3,12*87*(X -7,71) = (271,5* x) - 2090

X

X

= 116,8 k

= 3,12*87*(X-15,30)= (271,5*X) -4150 =- 354 k X

X

'

= 5 * I ,56 * 40 =312 k Cc

Bab 3_Kolom

=0,85 * 4 * 23 * ( 0,85 * X ) =66,4 * X =898 k

197

I 11 11

T,

Ill Ill Il lI

Tz

Cs1

Cc

:E M

ABOUT

Pn

=0

(285,5 *X *2,78)+(271,5 * X- 2090)*7,78+(27L5 *X -4150)* 15,37 - 3 12 * (20,37)* X + (66,4 * X 2 * (0,425 *X+ 0,07 )) = 0 3

X + co, 165 X= 13,52

* X2 ) + c2,57 * x ) - 2850 = o ;

11

Pn = 285,5 + 116,8 - 35,4 - 3 12 + 898 = 953 k > 935 k ... ..... reqd ... .. ..OK Use 23

* 23 with 14- #

lO

24 * 24 with 16 - # 10 ( 6 in outer faces, d1 = 2,51 11 2 at 6 from outer layers

198

11

);

Pn = l016,8k

Seton Bertulang

1

24 * 24 with 14- # 10 ( 5 in outer faces, d = 2,51

11

);

Pn = 937,5

k

2 at 6 11 from outer layers Good Alternative; 24*24with14-#10

Contoh Soal - 11 Redesign the column of "Contoh Soal - 10" as a spirally reinforced circular column. Required : 4% reinforcement ; dead load P = 225 kips and M = 220 ft-kips; live load P = 200 kips and M= 190 ft-kips; l- in, increments; f 1c = 4000 Psi ;

fy = 40000 Psi.

= (1,4*225)+(1,7 *200)= 655 0,75 0,75

Reqd: Pn

= 874 k Reqd: Mn

631 = (1,4*220)+(1,7*190) =- 0,75 0,75

=842 k e

=

842 *12 874

= 11 ,57 11

For Balance;

=

3 *d 3+ 1,38

= 0,685

Bab 3_Kolom

*d

199

Assume : Pb z Cc -- 0 , 85 * f 1c *A seg.

Reqd: A seg

874 0,85*4

= - --

= 257 = 0,582 * d

ao

Say: d

= 0,8 * h

ab

=0,582 * ( 0,8 ) * hb

Fig. 13.13.4;

= 0,465;

11 Find Coeff A

seg

= 0,36

= 0,36"' hb 2 ;

Req d · : hb-"

257

= 0 36 =7 15 ·' '

200

h

= 26 71! '

Seton Bertulang

Try 26 11 - Compression Probably Controls : Use Eq. ( 13.15. I ) 1

Ax * f c 96*h*e

Ast * fy +--....::....::...3*e ' ? + 1,18 - + 1,00 (0,8*h+0,67*Ds)Ds

Pn

=

Pn

f'c Pg * fy= A * ______;;_-:----+ ----"--,-

9 6* el

11

'

lh

(o,8 + 0,67 * D~y

3* el _ /_hh +1 00

+118

'

D~

'

For Design Use

Estimate:

e = 11,57 = 0 445 . 26

h

Ds

= 26 -

h

874

874

'

'

3 - 0,75-1,0 = 21,25 26 26

=Ag

=A

*

h

Bab 3_Kolom

=

'

40

0 04*40

'

+--'---

9,6*0,445 + 1,18 2 (0,8 + 0,67 * (0,818 ))

3*0,445 +1,00 0,818

6 *[~+_!_.· ]= 1'743 * Aa ~ 3 531 2 63 0

'

=

=0818

874 1,743

'

= 50011

~50~*4 = 25,311 201

As = 0,04 * 500

= 20,0 11

9 - # 14

As =20,2

13 - # 11

As = 20,3

16 - # 10

As = 20,35 11

M ax. Bars in 26 Dia.Col -

20- # 9

As = 20,2

Try 26 11 Dia.Co l W ith 16 - # 10 Statics on Equi v. Rectangle

Pn = 866,8 k < 874 k Reqd but within 1% . . .. OK Coumputer Solid X= 12 '06 11 ·, 11 Very Slight! · n Cqptrols Xb = 12,03

Use 26 11 Di a. Col. Wi th 16 - # 10 B ars

Contoh Soal - 12 Des ign a square tied column with about 4 % reinforcement to carry a dead load of P = 250 kips and M = 90 fl-kips, and a live load of P = 185 kips and M= 70 ft-kips. Use 1-in, increments with f 1c = 4000 Psi and fy = 60000 Psi.

Reqd:Pn

= (1 ,4 * 250)+(1,7* 185) = 0,70

Reqd : Mn

20 2

= (1,4* 90)+(1,7* 70) = 0,70

665 = 950 0,70

k

245 =350k 0,70

Seton Berlulang

e

350 * 12 = - -950 =442 11 '

Balance Condition: 0,003 *d 0,003 + 0,001 38

=

= 0,592 * d Assuming Cs

z

T;

Pn =Cc =Pb at balance pb

= 950 = 0,85

* f 1c * b * ( 0,85 * x

b)

* 4 * 0,85 * 0,592 * b * d =I ,705 * b * d = 0,85

b*d

=

950 = 557 sq.in· b 1,705 '

*h

=

557 = 620 0,9

11

Since Smaller Section is Desired, Compression Control ;

Estimate Size Using Eg ( 13.14.9) U se 20

11

f

. . . or estnnatmg ratios;

4 2 !!._ = ,4 = 0 22· h 20 ' ,

Pn

Bab 3_Kolom

= Ag *

):. =

"

!!... =17 '5 = 0 87 5. h

20

'

'

075 r =d -hd ' = _!2.= 20 '

'c _ ____.:I:___ + PK * !Y l_ * ~ + 1,18 ~2 h

~ * (!!._) + 1

r

h

203

4

-A g * - -- -- - -+

950

3

* 0,22 + 1,18

(o,875Y

0 04*60 ' 2 - - *0,22+ 1 o,7s

= A * ( 4,0 + 2,4) = 3,47 * A g 2,04 1,59 g 950 . 11 Reqd. Ag = - - = 274 sq.tn; h = 16,5 square 3,47 Try h = 1711 4 2 !!._ = ,4 h 17

=0 26· '

'

=Ag *(~ + 2•4 ) =316*A · 2,248 1,74 , g.

950

12 r = - = 0,705 17

= !!:._ = 14,5 = 0 855· s h 17 ' ' t

Reqd : A 2 =300 sq. in -

= 17,3 11

h

Either h = 17 11 or 18 11 may work As

= 0,04 * 300 = 12,0 sq.in

10-#10; As=l2,70

Acceptable Choices (pilihan yang dapat diterima) Size 17

* 17

Bars 10- # 10

2,51

d

Layers

14,49

2

Pn

Mn

357 lk

OK

360,5 lk

OK

11

X= l337 X3 - ( 9,6 * X2 ) + ( 248,5 * x ) - 3997,5 = o 18*18

10 -#9 X = 14,19 X3 -

204

(

2

978,7 k

* X2 ) + ( 188,6 * x) -

3361,6 = o

2,44

15,56

11

10,78

Beton Bertulang

17*17

8-#11 X= 13,13

X 18 * 18

3

2,705

14,295

2

950,1 k

350,8 lk

OK

360,5 lk

OK

11

c9,58 * X2 ) + c242,8 * x)- 3800,3 = o

-

10-#9

2,44

15,56

930,3 k

3

( 4 in faces, 2 at mid- depth )2, 17% Low X = 13,54 11 X 3 - c 10,78 * x 2 ) + c 184,3 * x)- 3002,3 = o

Use 18 * 18 with 10 - # 9 ( # 4 above )

Contoh Soal - 13

\

Design a square tied column with symmetrical reinforcement of about 2% to carry a dead load of P = 25 kips and M= 125ft- kips and live load of P = 10 kips and M = 50 ft-kips . Use 1-in, increment f 1c = 4000 Psi and fy = 60000 Psi. Reqd: Pn

= (1,4*25) +(1,7*10) = 52 -0,70

=74,4 Reqd: Mn

0,70

k

·= (1,4*125)+(1,7*50)= 260 0,70 0,70

=372 k e

372 * 12 = --74,4

= 60 11 Certainly for this large eccentricity, tension will control. Use Eq. ( 13.16.7 ) to estimate size.

Bab 3_Kolom

205

Estimate: p = 0,5 * pg= 0,01 e1

= distance from tension steel I

-e =

d _ hi +e

12

d

d

d1 - "" 0 15 · d ' ' 74 4 · = b*d 3,4

1

e ::::::45 : : : -+2 d '

m=

fy

0,85 * j

IC

=

60 0,85 * 4

= 17,65

*{0,99 - 4,5+~(- 3,5) 2 + 0,02*[(16,65*0,85)+4,5]}

21,85 = b * cl* {- 3,51 + .JI2,25 + 0,373 } 21,85 = b Reqd: b

* d * ( - 3,51 + 3,5529) = 0,0429 * b * d

*d =

21,85 0,0429

= 510 sq.i n Say 24

Try 21

* 24 ;

b

* d = 21,5 * 24 = 515

* 21 p

t__=

= 0,01; £=18,5 - 10,5 +60 =367· d 18,5 ' ' d

2,5 18,5

=0,135; m

206

= 17,65

Seton Bertulang

21,85 =b* d * [

~- 2,68+~(- 2,67)

2

+0,02* [(16,65*0,865)+3,67] ]

* d * ( - 2,68 + 2,74) = b * d * 0,06

=b

Reqd. b * d =

21,85 0,06

= 365; 21 * 21; b*d=389 .. .. .. .. .0K

Determine Reinforcement: 1 Pn::::: Cc= 0,85 * f c *a* b 74,4

= 0,85 * 4 * a* 21;

a

=1,04

11

74 4 '

* ( 60 '0- lO' 5 +

1 04 • ) 2

= As

* 60,0 * ( 18,5 -

2,5 )

Pn

* (e-~ +~)

= As* fy

* ( d- d 1 );

Approx . Eqn L. M about Cc

Try: 10- # 8;

As

= 7,90

8 - # 9;

As

= 8,00

6- # 10;

As

=7,62

Bab 3_Kolom

207

Statics Check: I. 21

* 21 section, 10 - # 8 ( 3layers); d 1 = 2,375

x 3 + c 116,5 * x 2 ) - c 190,0 * x ) -

2. 21

= 4,18

Pn

= 77,2 k > 74,4 k

section,

;

d = 18,625

11

;

d = 18,561

11

1313,2 = o

11

X

* 21

11

. ....... OK

8- # 9 ( 3layers);

1

d = 2,439

11

x 3 + c 116,5 * X2 ) - c249,2 * x)- 1276,9 = o

3. 21

X

=4,44 11

Pn

= 77,4 k > 74,4 k

* 21

........ OK

section, 6- # 10 ( 2layers); d 1 = 2,51

11

;

d = 18,49 11

x 3 + c 116,5 * x 2 ) - c39,7 * x)- 1677,6 = o X

=3,58 11

Pn

= 74,4 k > 74,4 k Reqd......... OK

Preferred Choice (pilihan utama):

Use 21 in square with 8- # 9 208

Beton Bertulang

Contoh Soal- 14

Design a square tied column with about 4% reinforcement uniformy distributed around its sides. The loads are: dead load: P 225 kips, Mx 236 ft-kips, and Mv = 108 ft-kips, live load: P = 200 kips, Mx = 204 ft-kips, and Mv = 64 ft-kips ; Select 1 size in whole inches that are multiples of two, using f c = 4000 Psi and fy = 4000 Psi.

=

=

(a) Loads Reqd: Pn

= (1,4 * 225)+ (1,7 * 200) = 936 k 0,70

Reqd: Mnx

= (1,4*236)+(1,7*204) =967k 0,70

Reqd: Mnv

= (1,4*108)+(1,7*64) = 371 k 0,70

Reqd: ey

=

Reqd :ex

=

M/IX

P,, MIIY

P,,

= =

967 *12 936 371*12 936

= 12,41 in = 4,76 in

(b) Equivalent Mox

Equiv. Mox

~M,x+ M,, •(::: )*('1)

Estimate ~ ::::: 0,65 ; for square column with symmetrical reinforcement , Mox = Mov

967 + { 371 * I *

(I ~,~;s)}

Equiv. Mox

z

Equiv. ey

= M ox = 1167 * 12 = 15 0 in P,, 936 '

Bab 3_Kolom

= 1167 "

209

(c) Balanced Condition

xb

= d

*( ececey ) = +

0,003*d = 0,685 0,003 + 0,001 38

*

d

* f1c * b * ~~ * Xb * 4 * b * 0,85 * ( 0,685 * d ) 1,986 * b * d

Pb == Cc = 0,85 = 0,85

= For

Pb = 936 kips ;

. Ba Ianc ed : b* d == -936 = 472 sq.m 1,98 For

b*h = Ag=22x 22; d::::: 19,5;

Balanced: Ag =

472 0,89

=530 sq.in .

d = 0 89

h

'

23 x 23 Section

!!... = _!2. for balanced section= 0,65

h 23 This eccentricity ratio is likely section to be in the tension controls zone.

Assume tension controls ! Est. 26 x 26 for ratio Use Eq ( 13.16.7 )

Pn ~o.85*f'c*b*d*{-p+1-: +(1-:)' +2i(m-{1-~)+ :]} Estimate p = 0,02 (one - half gross percentage) Equiv. e

+ _e' == 23,5 -13+15 = 1085 d

~::::: d

21 0

23,5

25 · = 0 11· 23 ,5 ' '

'

m=

fy = 11 76 ' 0,85* f'c

Beton Bertulang

Pn=0,85*4

*b*d~0,02+ 1-1,085+~(-0,08~2 + O,o4(1Q76*0,89)+ 1,08~}

= 3,4 * b * d * {- 0,105 + ~0,007 + 0,426} = 1,88 * b

*d

For : !!:_ = 0,905 h Reqd. Ag =

936 = 550 sq.in; 1,88 * 0,905

23,5

X

23,5

Try 24 x 24 in tension controls formula! 1

Est.

_e = 21,4 - 12 + 15 = 14 1 21,4 ' d 1

_d = 1,5+0,5+0,65 = 0 12 214 ' d '

Pn

=

3,4*b*d*{-0,16+~0,444}

= I ,725 * b * d Reqd. Ag

936 1,725 *0,88

= -- - = 616 sq.in;

24,8 x 24,8

For Ag to equal 576 ( 24 x 24 ), p = 0,045

(d) Investigate 26 x 26 section- 1/P; Method

Reqd. As

=0,04 * 61 6

=24,6 sq.in Try 16- # 11;

As

= 24,96 sq.in

d1

=1,5 + 0,5+0,71

=2,7 1 in Bab 3_Kolom

2 1 1

Use interaction relationship for uniaxial bending , Pn - Mn . Computer Program CE x Comp For:

1

P; 1000

ey

= 12,41 in

~

Px = 1168 kips

ex

= 4,76 in

~

Py = 2109 kips

e

=0

]

~

P0 = 3212 kips

1

=-+--Px Pr P0 1000 1168

1000 2109

1000 3212

--=--+-----

P;

= 0,856 + 0,474-0.311

= 1,019 Provided: Pn ""'Pi

1000 =- =950 k >936 k ...... OK 1,019

Use 26 x 26 column shown above.

212

Beton Bertulang

(e) Investigate 26 x 26 section- Parmesi B Method.

q

=

p 11

* fy = 24,96 *(40) = 0 369 j 1c (26Y 4 '

Pn = 936 :;: O 29 P0 3212 ' find

From Fig. 13.23.11 d ;

~::::

0,62

From Fig. 13.23.12 967 A=062· Mnx = =075 · ... , , M ox 1288 , ,

Find M"y =0,47 M OY

LFrom Below answer on next sheet Using SP - 7; Chart 44

Pu

= j I C * b *t =

K

* P,1 0,7 * 936 = j I C * b * h 4 * 26 * 26

= 0,243

Pt *m = pg *m = ( 0,0369

* 11,76)

= 0,434 Find : K *!!... = t

Mu

j 1c*b*h 2

= 0,154 M OX

= 0,154*4*26*(26)2 = 0,154*4*26*(26)2 = 1288 lk cl>*12

0,7*12

At

Moy = 1288 lk Because of simmetry;

Bab 3_Kolom

.

Pn=936k

213

1168 936

e=l2,41 11

1288 lk

. Mn

Contoh Soal- 15 For the column of the accompanying figure, determine the nominal strength Pn at an eccentricity of21 in. Use f 1c =4000 Psi and fy = 50000 Psi. Apply the basic principles of statics. (a) Use the Whi tney - Hognestad equ ivalent system for the reinforcement along with the circular concrete section. (b) Use more exact treatment of the reinforcement in its correct position in the cross section.

2 14

Seton Bertulang

2

1

9

3

4

5

6

7 8

(a) Use Whitney - Hognestad Equivalent Rectangle

= 28- 3,0- ( 2 * 0,625 ) -

Ds

=22,057 d- d

1

= 3. * Ds =3. * 22 057 3

= 14 ,7 1 Bab 3_ Kolom

1,693

11

3

'

11

215

= ~ * 1l' * (28 4

= 615

r

11

1

e s Assumed> ey

50

- -- - " - - - ey= 29000 =0,00172

es= 0,003 *(18,55-X)

X

!Pn

llll~llr~ Cc

216

e- 11 ,2 u =9,8 11

Cs

* 4 * 27,5 * ( 0,85 *X)= 79,5 *X

Cc

= 0,85

Cs

= 18,0 * ( 50 -

T

= 180 *

87

X

( 0,85

* 4 )) =838,8 k

* (18 55 - X)= '

29000

X

- 1566

Beton Bertulang

L.M about load= 0 79,5* X(9,8+0,425* X)+(838,8*(9,8+3,84))-

~ *(9,8+18,55)

29 00

+ (1566 * (9,8 + 18,55)) = 0 X 3 + ( 23,06

* X 2) + ( 1653- X)- 24382 = 0 ;

maka:X= 11.81

11

c1s

=

0 003 *(1 1,81-3,84) = 0,00203 > cy ........... OK • 11,81

cy

=

0 003 · * (18,55 - 11,81) = 0,0017 < 0,00172 ........ OK 11,81

Cc

= 79,5

Cs

= 838,8

T

= 29000 -1566 11,81

* 11,81 = 938,5 k

= 894,1 < ( As * fy ) =900

Pn

= 938,5- 55,3 = 883,2 k

(b) Use more exact Treatment of the reinforcement in its correct position in the cross - section Assume X is between bars 5 and 6, say between l 0 11 - 14 11 T I= 2,25

* 50

T2= 4,5 *50 87

= 112,5 k = 225,0 k

T3. = 4,5 * X * (21 ' 8- X)

Bab 3_ Kolom

;

X = 14,04 11

112,5 225,0 =

391 5 ' *(21,80- X) X

216,4

217

T4 = 4,5 * ~ * (18,22- X)

=

4,5*~ *(14,00-X)

=

T5 =

C6 = 4,5* ~*(X - 9,78) C1 =

=

39 5 ;· *(18,22 - X)

116,6

3

~'5 *(14,00 -X )

-1,1

3

~'5 *(X -9,78)

118,8

4,5* [~ *(X -6,2)-3,4] = 39;'5 *(X - 6,2X)-15,3

Cs C9

=4,5*(50 -3,4) =210k =2,25*(50 -3,4) =105k

Cc

= 0,85 * 4 * A seg

203,3

= 3,4 * A seg

:E M about Pn = 0 Momof = 112,5 * ( 25,03 + 7) =- 3603,4 = 225 * ( 24,2 + 7) = - 7020,0 =

218

3

~'5 *(21,8 + 7)*(21,8 - X) = - 1 1~75 ' 2 *(21,8 - X)

Beton Bertulang

8221 5 391 5 • * (14,00+ 7)* (14,00 - X)= • * (14,00- X)

=

Ts

~·5 *(18,22+7)*(18,22-X)= - 9 ~ 3 ·6 *(18,22- X)

3

=

X

3

=

X

~· 5 *(9,78+7)*(X -9,78)= - 6~9 ,4 *(X -

9,78)

- 15,3 * 16,78 =- 256,7 3

=

~· 5 *(6,20+7)*(X -6,20) = - 5 ~ •8 *(X -6,20) 7

- 15,3 * 13,20 = - 202,0

= 210 * ( 3,80 + 7 )

C8

= + 2268,0 C9

= 105 * ( 2,97 + 7) = + 1046,9

Cc

=3,4 * A seg * ( 21 -

33340" ,.) -

637086 5 *(21o - x)) =0 X • + (34*A ' .
Trial a= 0,85 *X

Mo

X)

A

seg

A~·..:

1 I ,9

081*h 3 032*h2 ' ' 1l =0,425

I 1,95

0 815 * h3 0 325* h2 7 02

Bab 3_ Kolom

I.

708

11

'

I. I .

11

Der a

Der X

X=Mo 3 4* A *(? 1 0- -X ) , seg - '

11873,6

14,09 11 11 ,98

12111 ,1

14,0211 11 ,9 1

I

I

219

11

= 0,427

Cc

= 3,4

* ( 0,322* h2 ) = 858,3

Say X- 14 04 11 =- 112,5-225,0-216,4-116,6 + 1,1

Pn

+ 118,8 + 203,3 +

210 + 105 +Cc =826 k Considerably Lower Than Obtained By Equiv. Rect. Section ( P = 898 k)

Contoh Soal 16 L

Kami level

h = 6m

d'"' IOcm p

Kolom

d ., 70 cm h, =80 ~m

A s = A 's = 32 c m 2

PBf71

~

Beton : K300 B c~a

tulangan : V 32

Be rat sendiri balok clan kolom diabaikan Pe nampang sama seluruh balok dan kolom Load factor = 1,5 (PBI:7 1)

220

Beton Bertulang

Pertanyaan 1. Buatlah Interaction diagram penampang 2. Berapa Vmax yang bias dipikul oleh kolom 3. Akibat P max, cek gescr di potongan A-A dan tulan gi

Solusi:

Ea;:

+

os

3.5%<>

"'

'l

r---- - - -----~:-----!------r-

10

32

l'.w =

- ~:.

. lto.>Uy

a,.

'

,....

------ - If

-;~---------:_::::::: ----------C,.•

60

-- '

32" 10

I~

Ts

(a)

Balanced

I. a. Keadaan seimbang 8 ,-"

=

2780 2.lx 10 6

= 1,3 %o

3,5

v,h = 1,3+3,5 x70cm =51 ,04 cm

. .

Cc.b = 0,8 x 51.04 x 0,5 x 300 x 40 kg

= 244,992 ton = 245 ton

Bab 3_Kolom

221

Tulangan simetris, -7 sama-sama Jeleh -7

I:H = 0-7 N ,,,b = Cc,b = 244,99 ton

=C(,.( 802 -

O,SxSl,04 )+32x2780x60 2

= 101,36 ton = 101 ,4 ton

b. Hanya normal tekan saja -7

Nu,o

= (40x80-64) x 0,5x300+64x2780

=648,32 ton

=648 c. Lentur Murai -7

----------;r -- --

-- ----

- ---- ---.........----'""--'

32

(a}

£se =

Yu -JO

(b)

x3.5 o/on

Yu

2 22

Set on Bertulang

Cs = Yu -lO x 0,0035x2,1x106x32 Yu = 235200 Yu - lO Yu Ts = 32x2780 = 88960 kg Cc = 0,8 yu x 0,5 x 300 x 40 = 4800 yu

r.f

:I:H = 0 ~ = cc+CS Subtitusi dan penyelesaiannya memberikan yu = 11,64 cm

c.f

1 64 = • x235200=33149kg 11,64

Cc

= 4800 x 11 ,64 = 55872 kg

Mu

= 33 149 x 60 + 55872 ( 70 -

~ 1•64 ) kg cm

O,Sx

= 37,07 ton cm

2. Dari konstmksi yang ada

M1-1.u = Pu.L

~

& coba dengan beberapa harga sekarang dan

plot pada diagram

Pu =50 ton~

Mu= 100 ton m

~

titik(a)

~

Mu= 50 ton m

~

titik (b)

Pu = 25 ton

Bab 3_Kolom

223

Pu = 10 ton

~Mu=

Titik a, b, dan c

~

20 ton m~ titik c

dihubungkan (karena linear)

Didapat Mu yang dicari Mu= 46 ton m

~

~

:. Nu = Pu = 23 ton

23 :. P = - ton= 15, 33 ton. 1,5 3. Mencari dan menulangi Shear (geser) di AA Seperti bias a. 224

~

Seton Bertulang

Bab4

j

BRAKET /KONSOL PENDEK (CORBEL)

Bab 4_ 8rakeVKonsol Pendek (Corbe~

225

d Syarat: h

~

1

Nu~Vu

Nu ~0.2 Vu d

~2dt

d 1 = tin ggi manfaat tepat pacta ujung luar perletakan Vn max = Vumax t/J

~ 0.2f' cbw.d~ (800psi) bw.d

5.52 bw. D

lcntur * Penulangan Vu = 1.4 VoL + 1.7 VLL Mu

= 1115 Vu = Vu * a + Nu (h- d)

m

=

Nu

fy

0.85.fc'

Mf,p Rn

=

p

=~

hd 2

(I_JI_2~n

)

pm in = 0.04 fc'

j)r

reinforcement: Due to bending •

Af = p . b. d

Due to HRZ force An=

Nu

- -- f/J = 0.9

As=

f/Jfy

Take the

2/3 A vf + An Large } value

•

226

Due to verl force

Avf =

Vu t/J .fy.Jl

1-1= 1.

A =As + A n

Beton Bertulang

*

Penulangan horizontal Ah perlu;;::: 1/3 Avf Ah

=n * 11<1 1td2 * 2

n =jurnlah lapis tulangan Tulangan horisontal harus berada di 2/3 d.

* Tegangan lekat antara baja dan beton For <1> ~ D 36

~ = 16.753 ffr' db

Mpd;

fc' (Mpd);

db (mm) Tegangan lekat batas dari tulangan tekan dalam semua ukuran

~ = 1.08

ffi

~5.6Mpa

~56 kg/cm 2

H = 0.4 (2400)

*

* '0i 1t . d2

Geser friksi Steel bar (fy) Betonl

Assumsi } Bidang retak Beton U

Vn

Di cor pada umur berbeda

=A vf * fy * Jl

Bab 4_BrakeVKonsol Pendek {Corbe~

227

Dimana:

1.1

= 1A

beton yang dikasarkan

= 0.6 A beton yang tak dikasarkan

*

=1

beton normal

A

=0.85

beton ringan berpasir

A

-= 0.75 beton ringan total

Panjang penyaluran dasar untuk tulangan tarik Untuk tulangan ~ # 1 I (# 35 M)

4Jb

=0.019 A~= 0.058 db. fy -vfc'

Untuk tulangan # 14 (# 45 M) Ldb

=

26.Jy

ffi

Untuk tulangan # 18 (# 45 M)

4lb

34 ..fy

= ffi

U ntuk kawat berprofi l (tapi bukan kawat anyaman)

L db

=

0.36.db.Jy

ffi

Panjang penyaluran tulangan tarik

41

= ~b * faktor -

faktor ;: -: 300 mm.

Panjang penyaluran dasar untuk tulangan tekan = 0.24 fy.db

ffi

Ambil yang terbesar

= 0.044 fy. db Panjang penyaluran tulangan tekan

22 8

Ld

= Ldb * faktor- faktor

;: -: 200 mm

Seton Bertulang

*

T ingkat Daktilitas 2 (D aktili tas terbatas) •

K mm= 2 D esain balok fortal Terhadap lentur Mn . h = 1.2 Mo . b + 1.6 Mo . b = 1.05 (Mo. b + Mo) Te rhadap geser V n . b 1.05 (Vo . b+ Vo . b ± 4/k Vo . b)

=

•

•

D is ain kolom portal T e rhadap lentur T e rhadap aksial T erhadap geser

Mo .k= 1.05 (Mo . k +M L.k ±codNE. k ) No . k = 1.05 (No .k + NL. k ± ffidNE. k) Vn . k

= 1.05 (Vd. k +V L. k ±

ffid NE . k)

S yarat - syarat ya ng harus dipenuhi a. Pada balo k I . No . b :5: Ag. fc ' * 1/1 0 2. Ln 4d, kecuali untuk blh perangkat shear wall 3. b/h '2: 0.25 4. b '2: 200 m m b .be= ¥1 h

b d 5 . p "'): pI '2: 1.4 fy w. . pada sembarang tempa t

6. sengka ng pertam a S :5: 50 mm diukur dari muka kotom, 3 ;spas inya S :5: d/4; 10 db ; 24 ds :5: 300 m m :5: fy * As 1 bw (harus I kaki sengkang) 7. untuk daerah yang tak meme rlukan sengkang S max :5: d/2 pacta seluru h batang 8. sengkang terhitun g boleh di b. P ada kolom 1. bk a tau l1k ~ 250 mm 2. bk/ hk ~ 0.4 3. hk I bk S 25 4 . xp :5: 6 % clan p :5: cls untuk daerah sambun gan 5. sengkang harus d ipasang pada seluruh tin ggi kolom .

Bab 4_ BrakeVKonso l Pendek

(Corbe~

22 9

4. 1 Perencanaan Konsol Pendek Syarat konsol pendek

alht ~ l htmax 2y bila arla gaya Hrz hams dikontrol tegangan. Tarik pada tulangan pokok.

=

Langkah - langkah penulangan: cari M= Pu. a •

cari Cu=

h

-.==== 1.5M

2K oO"I>I:'b cr

< qmax = (I - 3)(7350 +a- bn) 0" bk =q . 2 Ko - . b . h (tulangan pokok)

• A

•

2205

=0

o- av

0" bk Amin = 0.03 b . ht . - -

o- av

Tulangan ge er 0" 1,

= 0.5 cr'bk

<1>

b. tetap

230

Seton Bertulang

er"

= 0.6 cr'bk
b. sementara

4.2 Contoh Soal Data yang didapatkan dari perencanaan suatu bangunan yang meliputi: 1. Denah dan sistem struktur bangunan 2. Gambar 1 : Denah struktur bangunan 3. Gambar 2: Denah struktur potongan A - A 4. Gambar 3 : Denah potongan B - B 5. Gambar 4 : Gambar Struktur Corbel 6. Dimensi untuk seluruh kolom 50 I 50 7. Dimensi seluruh balok 30 I 45 8. Tebal pelat Jantai 12 cm

B

B

Gambar 4.1 Oenah Struktur Bangunan

Bab 4_ BrakeVKonsol Pendek (Corbe~

2 31

Gambar 4.2 Potongan A - A

9m

12m

Gambar 4.3 Potongan B - B

232

Beton Bertulang

~~

Um _.1 Gambar 4.4 Struktur Corbel

Wilayah gempa bangunan berada Daerah tcmpat perencanaan Corbel yang akan dibangun masuk ke Zone 3, dengan jenis tanah lu nak. Data Pembebanan Karcna yang dianalisis aclalah Corbel maka yang akan dihitung beban gravitas i adalah beban yang akan bcrpengaruh terhadap perencanaan corbel saja.

1. Beban mati Berat scndiri Corbel ( 0,8

* 0,5 * 2400 )

Berat scncliri rei 0 Jumlah beban Mati

= 960 Kg = 145 Kg = 1105 Kg

2. Beban Hidup Bcban hiclup untuk Cm·hel adalah 200 Kg/m Bab 4_BrakeV Konsol Pendek ( Corbe~

233

Koefisien Reduksi adalah 1 ( untuk pabrik ) Maka beban hidupnya adalah 1 * 200 = 200 Kg/m Beban hidup keran P

~ V'*rL+G);{l - u) +

Dimana: L G

G ; }:

2 ... ....... ... ( 3. 1 )

= beban yang diangkat 50 ton = berat keran angkat 51 ton = bentang atau lebar bangunan 12 m

=berat keran seluruhnya 83 ton u

= jarak as rel ketengah- tcngah kerekan 2 m

p

= 12*{(50 + 51)*(12 -

P'

= I 2 * {(50 + 51 )* 2 + 83} : 2 =35 ton

H

= 0,1 * P =0, 1 * 75400 = 7540 Kg

Hw

= -*7540 7

H'

=0, I * P = 0, I * 35000 =3500 Kg

Hw'

I = -*3500 7

,

'

I

12

12

2) + 83} : 2 =75,4ton 2

2

= 1077 Kg = 500 Kg

Karcna P dan Hw besar maka kedua beban terscbut digunakan untuk beban hidup keran. sedangkan umuk beban hidup total adalah 76677 Kg.

3. Perhitungan gaya gcser dasar horisontal akibat gcmpa dan distribusi ke epanjang tinggi gedung

2 34

Seton Bcrtulang

A) Berat Bangunan Beban Mati Atap

* 12 * 0,12 * 2400 = 138240 Kg Balok ( 2 * 40 +6 * 12) * 0,3 * 0,33 * 2400 = 36115 Kg = 24000 Kg Kolom 40 * l * 0,5 * 0,5 * 2400 = 24000 Kg Plafond 12 * 40 * 50 P1at 40

Aspal40

* 12 * 14

Jumlah Beban Mati Beban Hidup Beban hidup untuk pabrik adalah

=

6720Kg

=

229075 Kg

= 400 Kg/m2

Koefisien reduksi untuk pabrik adalah

= 0,9

Jadi beban hidup atap 400 * 12 * 40 * 0,9

=

172800 Kg

Beban Mati Titik 2 ( Daerah Corbel ) Berat sendiri Corbel I * 6 * 0,8 * 0,5 Berat rellintasan 145 Berat kolom 40

* 2400

5760 Kg

* 2 * 40

11600 Kg

* 0,5 * 0,5 * 2400

= 72000 Kg

89360 Kg

J urn lab beban mati dititik ke 2 adalah Beban hidup dititik 2 ( daerah Corbel ) Beban hidup

=

200 Kg/m2

Koefisien reduksi

=

0,9

Maka beban hidup 200 * 0,9 * 0,5 * 0,5 * 6

= 135 Kg

Beban hidup karena keran

p

= 12*~(50+51)*(12-2)+83} '

L

12

2

P'

= 1 2*~ (50 +51)*2 + 83}

H

= 0,1 * p =

'

L

Bab 4_Brake1;/Konsol Pendek

12 0,1 * 150,8

(Corbe~

2

= 75,4 ton

=

35 ton

=

15,08 ton

235

1 -* 15,08

Hw

=

H'

= 0,1 * P = 0,1

= 2,154 ton

7

* 70 = 7 ton

Hw'= _!_ *3500 7

= 500 Kg

Beban hidup total=t35 +150800+70000+2154+1000 = 224089 Kg Berat total daerah Corbel 89360 + 224089 = 313449 Kg Berat beban atap + beban titik kedua ( daerah Corbel) W

Total

= 40875 + 313449 = 354324 Kg

B) Waktu Getar Bangunan

T X = Tz = 0,06 * ( H ) Tx = Tz = 0,06

213

* ( 12 )213

Tx = Tz = 0,387 detik

C) Koefisien Gempa dasar T x = T z = 0,387 detik, masuk ke zone 3 dan jenis tanah lunak maka diperoleh nilai C =0,07

D) Gaya geser horisontal Vx

=Vy

= C * T* K * Wt = 0,07 * 1 * 1 354324 = 24802,68 Kg

E) Distribusi gaya geser horisontal akibat beban gempa kesepanjang tinggi

gedung a) Arah X

H 12 - = -=1<3 A 12

236

F. = rX

w.' *h.'

L,W, *h.,

*V

x

Seton Bertulang

b) Arah Z F. ,r

=

W; *h; *V I:W. *h. r I

I

Tabel 3. 1 Distribusi gaya geser dasar hori sontal total akibat gempa kesepanjangtinggi gedun g dalam arah x dan y dalam tiap portal.

Titik

w1 (ton)

hi (ton)

h1 * W1 (ton)

F1 total (ton)

Untuk tiap portal

1

-* F IX 6

1 2

12 9

401,875 3 13,449

4822,5 2821,041 7643,54 1

15,649 9, 154 24,8

*t

_!_* F *t 2 /Z 7,842 4,577 12,401

2,608 1,526 4,134

7,824 t - --+ Il > .----..,...--- - - . -----.------.-----.

12m

I

8m

I

8m

Gambar 4.5 Gambar Distribusi

Bab 4_BrakeVKonsol Pendek (Corbe~

Bm

Arah Sumbu Z

237

> I Gambar 4.6 Gambar Distnbusi Gempa Arah Sumbu X

Perhitungan 1. Data Setelah dilak:ukan perencanaan awal dan dilak:ukan perhitungan- perhitungan pembebanan yang dipikul oleh Corbel, dan data pembebanan itu adalah: : 1105 Kg/m Beban mati yang bekerja pada Corbel Beban hidup yang bekerja pada Corbel : 76677 Kg/m Beban gempa lantai atap arah X : 2608 Kg : 1526 Kg Beban gempa terhadap Corbel arah X Beban gempa lantai atap arah Z : 7824 Kg : 4,577 Kg Beban gempa terhadap Corbel arah Z

2. Perhitungan Struktur Corbel Menurut SK SNI T - 15 - 1991 - 03 Dalam perencanaan penulangan Corbel ini struktur yang digunakan beton yang dicor secara mono! it dengan nilai J..l : 1,4, kemudian untuk mu tu beton yang digunakan adalah f 1c: 25 MPa dan mu tu baja yang digunakan fy: 400 MPa dengan faktor reduksi kekuatan : 0,65 a) Beban berfaktor

U

=Vu

Vu

* D ) + ( 1,6 * L ) =( 1,2 * 1105 ) + ( 1,6 * 76677 )

Vu

= 124009,2 Kg

Vu

= 1240092 N

= ( 1,2

238

Beton Bertulang

Sedangkan untuk nilai N

uc

adalah

N

uc

= 1526 Kg ( Gaya gempa yang ada di titik Corbel)

N

uc

= 0,2

N

uc

= 248018,4 N

* 1240092 ( syarat minimum)

Maka yang diambil gaya-gaya tarik berfaktor adalah yang terbesar yaitu: 248018,4 N

b) Tul angan geser A Vu ~

yang dibutuhkan untuk menahan gaya geser

* Vn

Vn ;;::: Vu

A vf

vf

= 1240092 0,65

= 1907834 N

VII

= ! _,.*j.J.

Karena hubungan Corbel dengan kolom monolit J..l = 1,4 A vf

= 1907834 400* 1,4

= 3407 mm2

c) Tulangan A r yang dibutuhkan menahan momen Mu

= ( Vu

* a) + ( N uc * ( h -

d)

Untuk a digunakan 0,25 m, sedangkan untuk ( h - d ) digunakan ( 80-75 ) = 5 cm= 0,05 m Mu

= ( 0,25 * 1240092) + ( 248018,4

* ( 0,8 -

0,75 )

= 322423,92 Nm Dari tabel SK SNI T-15- 1991-03 nilai- nilai p min teoritis, karena menggun akan f 1c = 25 MPa dan fy = 400 MPa maka p yang digunakan adalah 0,001 9. Ar

=p * b * d

Ar

= 0,0019 * 500 * 750 =7125,5 mm2

Bab 4_BrakeVKonsol Pendek (Corbe~

~.

239

d) Tulangan yang dibutuhkan untuk menahan gaya tarik N

uc

=* An* fy

N uc

An

= *fy

An

= 0,65*400

248018

954 mm 2

e) Tulangan tarik utama As

= Af+ An = 7 12,5 + 954 = 1~66,5 mm

2

As

= 2 * A,:r

+ An

3

. 1 . R as10 penu angan p mm .

= 2 *3407 +954 =3225 mm2 3

= 0,04fy*fie

pmm =

0,04 * 25 0 0025 · 400 = '

Untuk mencari luas tulangan yang digunakan adalah luas tulangan yang terbesar yaitu 3225 mm2 , maka tulangan As menggunakan tulangan 9D22 = 3421,2 mm 2.

f) Tulangan sengkang tertutup

Ah

= All,. = 3407 = 1135 67 mm 2

Ah

= 0,5

3

3

'

* ( As - An ) = 0,5 * ( 3225-925 )

= 11 35,5 mm2

240

Seton Bertulang

Maka yang digunakan adalah luas tulangan yang terbesar yaitu 1135,67 mm 2 , oleh karena itu digunakan tulangan 9013 = 119,6 mm 2 dipasang sepanjang ( 2/3 ) * d = ( 2/3 ) * 750 = 500 mm, dengan spasi 56 mm.

g) Menghitung luas plat landasan Tebal plat digunakan 2 cm Luas plat

=

12400092 Vu =-----<1> * 0,65 * 500 0,70 * 0,65 * 500

= 5450,95 mm

2

Ukuran plat landasan adalah 73,83 mm* 73,83 mm= 74 mm * 74 mm

3. Perhitungan Struktur Corbel Menurut ACI (American Concrete Institue) Dalam perencanaan penulangan struktur Corbel, beton yang digunakan secara monolit dengan J.L = I ,4 , kemudian mutu beton yang digunakan f 1c = 25 MPa dan mutu baja fy = 400 MPa dengan faktor reduksi 0,85. a) Beban berfaktor U

=Yu=(l,4*D)+(l,7*L)

Vu

= ( 1,4 * 1I 05 ) + ( 1,6 * 76677 )

Vu

= 131897,9 Kg

Vu

= 1318979 N

Sedangkan untuk nilai N

uc

adalah

N

uc

= 1526 Kg (Gaya gempa yang ada di titik Corbel arah X)

N

uc

= 0,2

N

liC

= 26379,58 N

* 13 I 897,9 ( syarat minimum )

Maka yang diambil gaya - gaya tarik berfaktor adalah yang terbesar yaitu: 26379,58 N

Bab 4_BrakeVKon~ol Pendek (Corbe~

241

b) tinggi Corbel untuk geser 2

d min

=

Vu .................. ( untuk satuan lb/inch