Calculation Sheet Pondasi Tower SUTT 150 kV

CALCULATION DESIGN OF FOUNDATION FOUNDATION 150 kV TRANSMISSION LINE OF KALTENG-1 2X100 MW 4DD6 CLASS 7W DATA TEKNIS PERENCANAAN

Mutu Beton f'c fy g eton g tana* qcrata -rata Koe!"en ge!er b1 $! ec %e&"mut Beton on(a!" %e&"mut Beton Ko&om

= = = = = = = = = = = =

K225 225 400 2.40 1.60 22.30 0.45 0.#5 210000 0.003 )5.00 )5.00

%u(ut ge!er (a&am tana*

=

0.00

DATA BEBAN

= = = =

+aya ,atera&  +aya ekan    +aya ar"k 

kg/cm2 Mpa t/m3 t/m3 kg/cm2

Mpa mm mm 0

33.62 ton 16.46 ton 132.36 ton

DIMENSI PONDASI

,ear &engan pe&at 1 1 ,ear &engan pe&at 2 2 ea& e&at a ,ear !&a * ea& !&a f a&am pon(a!" (f

= = = = = =

6.3 3.15 0.6 3.5 0.) 4

m m m m m m

Kem"r"ngan !tu a) "ngg" *ymey (" ata! tana* *t "ngg" *ymney , ,ear *ymney  ,ear &engan !&a g Back to Back Kengg"an ma ko&om

= = = = = = =

#1.16 0.4 3.13)2) 1 1.25 16.22) 3

m m m m m m

= =

3.13)2) m m3 3 ).5244 to ton

= =

32.3# m3 )).)336 to ton

0

BERAT PONDASI

1 Berat Berat ko&om ko&om 7o&ume Ko&om Berat ko&om 2 Berat Berat pe&at pe&at 7o&ume pe&at Berat e&at BERAT TANAH

Berat tana* !eaga" pena*an gaya 8p &"9 :!u ("mo(e&kan !eaga" entuk ,"ma! %eg"empat terpancung b'

b' " ! b1 b1

anDang !"!" ata! ' = 1 < 2  (f  tan f  7o&ume ,"ma! erpancung 7!u;

=

Vsu=1/3times ital df times left (b1 rSup {size8{2} } +b'rSup { size 8{2}}}} + sqrt {b1 rSup {size 8{2}} timesb'rSup {size8{2} } } right )}{}{

6.30 m

=

15#.)6 m3

= =

123.23 m3 11).0) ton

¿

ota& 7o&ume ana* 7!u = 7!u; - 7tota& Berat ana* ana* untuk 8p &"9 :!u = g tana*  7!u KONTROL PONDASI K#$%&# %(&)**+ ,** UPLIFT .T/

u = :p < :!u u/1.5

= =

202.34 ton 134.# ton



132.36 ton

OK

K#$%&# T(&)**+ G/$, Momen Resistance (MR)

- Mrconc = :p  &engan - Mr!o"& = :!u  &engan ("mana > &engan = 1/2

= 26#.5)#6 ton.m = 36#.))6 ton.m = 3.15 m

Momen Guling (MG)

- M =   (f - Momen ta*anan ak"at gaya tekan pa!"f tana*

= 13 134.4661 ton.m

p = 0.5  (f 2  gtana*  Kp   Mp = p  (f  1/3 - M+ re!u&t = M - Mp - %? = M@/M+  2

= 12.# = 1).0666) = 11).34 = 5. 5.42#51

ton ton.m ton.m  2

=

kg /cm2

OK

K#$%&# T(&)**+ D** D/k/$, T*$*) .B(*&$, C*+*%

qa&& = qcrata-rata /30

ex = ey =

M N

1.11

=

11.15

ton/m2

= .03#0) ton/m2

A

11.15

ton/m2

OK

=

A

11.15

ton/m2

OK

= 0.62)6# m

smak! sm"n

N 6 ex 6 ey σ = ( 1± ± ) A L B

-1.36442 ton/m2

("mana >

s = egangan tana* ak"at gaya-gaya &uar t/m 2  = +aya-Cerka& yang ekerDa pa(a pon(a!" ton H = ,ua! tapak pon(a!" m 2 ,EB = anDang (an &ear pon(a!" eEey= $k!entr"!"ta! ara*  (an y K#$%&# T(&)**+ G** L*%(&*

Fs= δ

x

N >1  ! H

= 2.26#45 

1.5

OK

("mana > d = koe!"en ge!er K#$%&# T(&)**+ G(3(& 1 A&*) 500

d

(! = %e&"mut eton < /2 ( = -(! a = ,/2 - /2 - ( sa = smin+(L-a).( smaks- smin)/L

a

= = = =

#4.50 12 1 215.5 34.5 ).2)3

mm mm mm = k/m2

0.0345

m

d ds L

min a

max

Gaya tekan ke atas dari tanah (Vu) 7u = B.a.F!uneGo < Hq1
h

= 1243)))  = 0.212)4 Mpa

.gtana* a'.geton

=

43200 /m2

=

0.0432 /mm2

=

31200 /m2

=

0.0312 /mm2

d

b

Gaya geser yang dapat ditahan beton ( f.Vc)

f.Vc = f.√f'c.B.d/6 Vu < f.Vc 1E+06 < 2E+06

=

OK

2522460

 =

2522.455 k

K#$%&# T(&)**+ G(3(& 2 A&*) L / 2

L / 2

i!ens i!ensii ko"o! ko"o!## b = $$$ !! b + d = h + d = $$ $$$ + $$ $ $$ d B

b d

d

h

=

-. = hk / bk =

d

2215.5 mm

=

2.2155 m

Gaya tekan ke atas (gaya geser pons) % Vu= &'   * (b+d (b+d) )(h+ (h+d d Qsunetto = 2E+06  = 1, 6 2 k  1

bo= 2  &(bk &(bk + d) d) + (hk (hk + d) d) =  6 2 !! d ds

Gaya geser yang ditahan o"eh beton ( f . Vc):

L

min max

V. = (1+(2/-.)  ( Ö f'c bo  d ) / 6 = 2,,33  = 2,,33 k = 2 , , 3 k  V. = (2 + (as . d) / (bo)).( Ö f'c = 2 6 04 6 11 1  = 2 6 04 6 k 

bo  d ) / 12

V. = 1/4  Ö f'c bo d = 1 3 04 1 6, 0  = 1 3 04 2 k  ipi"ih V. terke.i" f.7c =

130 042 k 0.)5 . 13 = 12))3.)4 k

7u A f.7c

OK

Cek Kelangsingan Struktur KxL r

< 22

23.5#1)# A

r = 0.3   K = 2 kan&eCer

r K

Jp(!t& = 1/12     3

= =

Jp(!t& =

0.5

$c = 4)00  f'c Jp(!t&  $c

$c

22

300 2

mm

#$<10

mm4

.K(*$,3$,*$ .K(*$,3$,*$ b(&+($,*&/)

= 2224.06 Mpa =

2$<15

mm2

=

2$<12

kmm2

=

1

Faktor pembesar momen ( ) δ =

Cm Pu 1− φ × Pc

d

"mana > m = 1 kon!erCaf u =  j = 0.65 c = p2  Jp(!t&  $c / k  , 2

³ 1.00

Ok

m = 1 u = 16.4642 ton 0.65 j = c

= 4.66$<11 ton

Penulangan Pedestal

M1 =   , M2 =   em"n Mt = M1 < M2 M = d  Mt Mnper&u = M/j

M1

=

M2 Mt M

= ).625# ton.m = 11 1 13.0# ton.m = 113.0# ton.m

1 05.464 ton.m 10

Mnper&u = 1)3.#44 ton.m nper&u = 260.)142 ton

nper&u = /j Ek3($%&3%*3 4

e = Mn per&u/nper&u

e em"n

em"n = 15 < 0.03   %yarat > e  e m"n

= 66).33)6 m mm m =

45

66).33# 

45

H%/$, $*

%umu 7erka&

Pu ϕ × Ag × β 1× f ' c

0.13364#

("mana > 1 ton =

III

mm OK

%umu or"onta&

()

Pu e × ϕ × Ag × β 1× f ' c b

0.0#1##

(ar" ("agram "nterak!" ko&omE ("pero&e* > per*akan !umu Cerka& (an *or"onta& maka > 6

r'

7&

b'

= =

0.0150 0.

r

=

0.0135

H!tot

=

13500

n '

= = = 

L/*3 T/*$,*$ 4

H!tot = r  Hg %e*"ngga ("gunakan tu&angan Lum&a* "ameter H! terpa!ang

40

D,/$*k*$ %/*$,*$ /%** 4

mm2

40 21 13#4).4  21

13500

OK

Penulangan er!adap "eser

L/2

Diagram geser

Vud deff Vu

7u =  7u( = penampang kr"! pertama pa(a Darak (e 7u(/7u = ,/2-(e/,/2 (e = &ear efekf (e=  - 0.5("a.u& utama - ("a.%engkang - !e&"mut eton

7u

=

7u( =

33.62

ton

13536 

(ef =

04.5

mm

Kapa!"ta! ge!er

Vc= 1 / 6× √ f ' c ×b ×deff Vn

=

Vud φ

("mana > f = 0.65 %yarat > 7u(  f 7c

Av Av

=

Av =

' c  b 1"""

12"" 12""  fy

("paka"

S=

=

)150)0 ton

7n

=

2146)1 ton

f

=

135 53 36.0 

b  1""" 1""" 3  fy

#! √ f

7c

1/ $  n

  d! 1""" Avu

%yarat > % N 16  % N 4#.("a !engkang % N (/2 % N 600 mm

HC1

= #33.3333 !!2

HC2

= )41.15## !!2

HC HC

= #33.3333 !!2

%

= 1##.456 mm mm

% % % %

N N N N

336 4#0 452.25 600

0.65 464 64) )6

mm mm mm mm

%e*"ngga ("gunakan tu&angan ge!er "ameter Larak antar tu&angan

O %

= =

10 150

mm A 1##.456

D,/$*k*$ %/*$,*$ 3($,k*$, 4

O

10

-

150

P(b(3*$ T*+*k P#$*3

 %econ (

=

1300

mm 

ata - (ata > u&angan 8tama

1

(=a (e= ( - 0.5("a.u& - ("a.u& - !e&"mut eton

H! = 2#3.52#) mm2 ( = 1300 mm (e = 116.5 mm

Hg = 1  1

Hg

=

F=s

!

= .0 .03#0 #0) ) t/m2

=

1

 "g "gunak unakan an u&an u&ang gan +e!er e!er M"n" M"n"mu mum m

mm

22500 mm2

OK

Mu = 0.5  F  1  eP

Mu "

" 8 ×b 1× deff

Mu =

(

= #×fy × 1−"!88 #×

fy f 'c

)

III

.mm Q 400Q 400 -

0.42060Q#2 = 41#1 -

H! = r  1  (e

III

r

=

H!

= 1244.611 mm2

Q2 41#1 0.42060#

0.00104

%e*"ngga ("gunakan tu&angan Larak tu&angan

!

=

150

= 1#0.12

"ameter



=

1

mm2

D,/$*k*$ %/*$,*$ /%** 4

D

18

-

OK

150

Penulangan er!adap "eser apak Pondasi

bc o = 4  c < (e

bc = o =

Vc= 1 / 6× √ f ' c ×b$×deff

1 1#)#6 mm mm

7c 7c

= 1))6#4  = III ton

7 = F  1  e

7 7

= III  = 16.5411 ton

7u = 7/f

7u

= 302.3)0# ton

%yarat > 7u A f7c

302.3)1 A  %econ (

P(b(3*$ T*+*k P#$*3

=

)00

11)#.15

.T*k ((&/k*$ %/*$,*$ ,(3(&

mm 


1

(=a (e= ( - 0.5("a.u& - ("a.u& - !e&"mut eton

H! = 2#3.52#) mm2 ( = 600 mm (e = 46.5 mm

=

1

mm

22500 mm2

Hg = 1  1

Hg

=

F=s

!

= .03#0) t/m2

!

=

III

e

=

1)50

Mu =

III

e = */2 Mu = 0.5  F  1  eP

Mu

(

fy = #×fy × 1−"!88 #× " f 'c " 8 ×b 1× deff

)

/mm 2

.mm Q 400Q 400 -

0.3)65Q52 = 41#1 -

H! = r  1  (e

r

= 0.00033

H!

= 463.3)4 mm2

Q2 41#1 0.3)655


!

=

250

= 1134.115

"ameter



=

1

mm2

D,/$*k*$ %/*$,*$ /%** 4

D

18

-

OK

250


bc o = 4  c < (e

bc = o =

Vc= 1 / 6× √ f ' c ×b$×deff

1 145#6 mm mm

7c 7c

= =

5)25263  III ton

7 = F  1  e

7 7

= =

10)04#5  III ton

7u = 7/f

7u

=

III

%yarat > 7u A f7c

16).#4 A

III

ton

.T*k ((&/k*$ %/*$,*$ ,(3(&

P(&)%/$,*$ P(&)%/$,*$ +(&($*$**$ B*#k *$ S##!

"rencanakan ("men!" !&oof Bean-ean yang ekerDa pa(a !&oof a(a&a* > Berat !en("r" !&oof

=

30 0.3

 0.4

40

cm

2400 =

2## Kg/m

%e*"gga mo(e& pemeanannya !eper pa(a gamar er"kut > q = 5## Kg/m

B

H



= 1/#.q.,2 5## 2.26 = # = 2150.41 Kg.m = 1/12.q.,2

Momen &apangan H R   < 

Momen tumpuan H R   - 

=

5## 2.26 12 = 1433.60) Kg.m = 5/4#.q.,2 2 240 40 2 2.2 .26 6 = 4# = 1)2.00# kg m

Momen &apangan B  < 

D(3*$ %/*$,*$ /%** Tulangan lapangan ( M + )

Tm"n

=

1.4 f y

=

Tmak! = 0.)5 U1 = 0.)5

1.4 = 400 0.#5.f'c

0.0035 600 600 < f y

f y

0.#5

0.#5 22.5 400

600 600 < 400

=

0.01#2 600 a = U1 ( 600 < f y 600 = 0.#5 333.5 600 < 400 = 1)0.0#5 mm amak! = )5V.a = 0.)5 1)0 = 12# mm mm "tung > amak! Mna(a = 0.#5.f'c..amak! ( 2 = 0.#5

22.5

300

12# 333.5 - 63.)#

=

1 1)4 )405 05# #0. 0.)4 )4) ) .mm .mm Mu 21504101.5 Mnper&u = = = 26 26## ##01 0126 26. .1# 1## # .mm .mm S 0.# Karena Mna(a  Mnper&u maka ("gunakan tu&angan tungga& m

=

@n

=

f y 0.#5.f'c Mu 2

S..(

= =

400 0.#5

=

22.5

20.15

21504101.535 0.#

300

111222

= 0.#06 Mpa

"tung > T

=

1 m

1 -

1

1 = 1 20.15 = 0.00206 Kontro& ra!"o tu&angan T A Tm"n A Tmak!

-

-

2.m.@n f y 1

-

33.6#2 400

,/$*k*$ &*3# %/*$,*$ $/

(engan (em"k"an &ua! tu&angan yang ("per&ukan a(a&a* > H! = Tm"n..( = =

0.0035 300 333.5 350 mm2

("gunakan tu&angan

4

S

13

=

531 mm2

Tulangan tumpuan ( M - )

Tm"n

=

1.4 f y

=

Tmak! = 0.)5 U1 = 0.)5

1.4 = 400 0.#5.f'c

0.#5

f y

0.0035 600 600 < f y

0.#5 22.5 400

600 600 < 400

600 333.5 600 < 400 = 1)0.0#5 mm = )5V.a = 0.)5 1)0 = 12# mm mm = 0.#5

amak!

"tung > Mna(a =

0.#5.f'c..amak!

= 0.#5

22.5

(

300

-

amak! 2

12# 333.5 - 63.)#

=

1) 1)405 405#0. #0.) )4) .mm .mm Mu 1433606).) Mnper&u = = = 1) 1)2 200 00#4 #4.6 .612 125 5 .mm .mm S 0.# Karena Mna(a  Mnper&u maka ("gunakan tu&angan tungga& m

=

@n

=

f y 0.#5.f'c Mu S..(2

= =

400 0.#5

=

22.5

20.15

1433606).6 0.#

300

111222

= 0.53) Mpa

"tung > T

=

1 m

1

-

1

-

2.m.@n f y

1 22.4655 1 1 20.15 400 = 0.00136 Kontro& Kontro& ra!"o tu&angan T A Tm"n A Tmak! ,/$*k*$ &*3# %/*$,*$ $/ =

(engan (em"k"an &ua! tu&angan yang ("per&ukan a(a&a* > H! = Tm"n..( =

0.0035

=

350

300 333.5

mm2

("gunakan tu&angan

4

S

13

=

531 mm2

OK999

A$*33 %/*$,*$ Tulangan lapangan ( M + )

Ta(a

=

H!terpa!ang .(

=

530.22 100050

Tm"n A Ta(a A Tmak! a

=

= 0.00531 OK999

H!terpa!ang .f y 0.#5.f'c.

=

531 0.#5

400

22.5

300

=

3)

mm

3)

mm

%e*"ngga momen nom"na& a(a&a* a Mna(a = H!.f y ( 2 = 531 400 333.5 - 1#.51 = 66# 66#55 5516.3 16.32 25) .mm .mm Kontro& Kontro& keamanan pena mpang > Mna(a  Mu OK999 Tulangan tumpuan ( M - )

Ta(a

=

H!terpa!ang .(

=

530.22 100050

Tm"n A Ta(aA Tmak! a

=


= 0.00531 OK999

=

531 0.#5

400

22.5

300

=

%e*"ngga momen nom"na& a(a&a* a Mna(a = H!.f y ( 2 = 531 400 333.5 - 1#.51 = 66# 66#55 5516.3 16.32 25) .mm .mm Kontro& Kontro& keamanan pena mpang > Mna(a  Mu OK999 P((&k3**$ %/*$,*$ ,(3(&

Be!ar gaya ge!er rencana a(a&a* 7u 7n 7c

=

33616.5 kg 7u 33616.51 = = = 5602).5 Kg S 0.6 1 22.5 300 333.5 f'c.X.( = = 6 6

6 )06.5  = )10

Kg

S .7c 6 532 532.2 .24 4 kg b/%/)k*$ %/*$,*$ ,(3(& 7u A S . 7c

7!

6

7u - S7c 33616.51 - 532.24 6 3612.3) = S 0.)5

7c1

6 0.33. fW fW.X.(

=

156611 kg

7c2

6 0 66. fW fW X (

=

313222 kg

kg

Meng*"tung Darak tu&angan !engkang er(a!arkan er(a!arkan per!amaan > "rencan "rencanakan akan menggunak menggunakan an ("ameter ("ameter 10 mm HC = 15) m2 HC.fyt.( = 56).324 %1 6 mm 7! %2 6 (/2 = 166.)5 mm HC.fyt = 5#.052 %3 6 mm 0.35.X %4 6 600 mm ("gunakan % =

166.)5 mm

=

160

mm

OK

P(&)%/$,*$ +(&($*$**$ B*#k P($*$,,* P($*$,,*

 co! Z q =

2##

kg/m co! Z

, "rencanakan menggunakan ("men!"  = 146#.21 kg kg ,' = 5.10 m , = 5.2 %u(ut Z = 30.4# Y  co! Z = 1265.3) kg q co! Z = 24#.211 kg = ., < q., 2/2

Mma

=

30



40

1302).05434 kg k g.m


Tm"n

=

1.4 f y

=

Tmak! = 0.)5 U1 = 0.)5

1.4 = 400 0.#5.f'c

0.0035 600

f y

0.#5

600 < f y

0.#5 22.5 400

600 600 < 400

= a

amak!

0.01#2 600 = U1 ( 600 < f y 600 = 0.#5 332 600 < 400 = 16.32 mm = )5V.a = 0.)5 16 =

12) mm mm

"tung > Mna(a =

0.#5.f'c..amak!

= 0.#5

22.5

(

300

-

amak! 2

12)

332

-

63.5

=

1 156 5634 341 11 1.0 .0## ## .mm .mm Mu 1302)0544 Mnper&u = = = 16 162# 2#3# 3#1) 1). . 3 3 .mm .mm S 0.# Karena M na(a  Mnper&u maka ("gunakan tu&angan tungga& m

=

@n

=

f y 0.#5.f'c Mu S..(

2

= =

400 0.#5

=

22.5

20.15

1302)0543.441) 0.#

300

110224

= 4.24 Mpa

"tung > T

=

1 m

1

-

1

1 = 1 20.15 = 0.01451 Kontro& ra!"o tu&angan Tm"n A T A Tmak!

-

-

2.m.@n f y 1

-

205.1 400

(engan (em"k"an &ua! tu&angan yang ("per&ukan a(a&a* > H! = T..( = =

0.01451

300

332

2

1446 mm

("gunakan tu&angan

#

S

16

=

160# mm2

OK999

A$*33 %/*$,*$ 4 Tulangan tumpuan ( M - )

Ta(a

=

H!terpa!ang .(

=

160#.45 = 0.01615 600


=

,/3

OK999

H!terpa!ang.f y 0.#5.f'c.

=

160# 0.#5

400

22.5

300

=

112

mm ,

%e*"ngga momen nom"na& a(a&a* a Mna(a = H!.f y ( 2 = 160# 400 332 - 56.0) = 1) 1))5 )533 3314 143. 3.0# 0#1 1 .mm .mm Kontro& Kontro& keamanan penampang > Mna(a  Mu OK999 P((&k3**$ %/*$,*$ ,(3(&


=

2)16. kg 7u 2)16.#) = = = 452#.1 Kg S 0.6 1 22.5 300 332 f'c.X.( = = 6 6

6 )#)40.)  = )#)4

Kg

S .7c 6 5 505 05.5 .55 5 kg 7u A S . 7c b/%/)k*$ %/*$,*$ ,(3(& 7!

6

7u - S7c 2)16.#) - 505.55 = 6 31)4#.42 S 0.)5

7c1

fW.X.( 6 0.33. fW

=

43020 kg

7c2

fW.X.( 6 0.66. fW

=

#6040 kg

7! A 7c2

P($*+*$, /k/+

Meng*"tung Darak tu&angan !engkang er(a!arkan per!amaan > "rencana "rencanakan kan menggunak menggunakan an ("ameter ("ameter 10 mm HC = 15) m2 HC.fyt.( = 656.)13 %1 6 mm 7! %2 6 (/2 = 166 mm HC.fyt = 5#.052 %3 6 mm 0.35.X %4 6 600 mm ("gunakan % =

166

mm

=

160

,/3

mm

OK

kg

CALCULATION DESIGN OF FOUNDATION FOUNDATION 150 kV TRANSMISSION LINE OF KALTENG-1 2X100 MW 4##6 CLASS 7W DATA TEKNIS PERENCANAAN


= = = = = = = = = = = =

K225 225 400 2.40 1.60 31.0 0.45 0.#5 210000 0.003 )5.00 )5.00


=

0.00

DATA BEBAN

= = = =



Mpa mm mm 0

22.03 ton 14#.2 ton 122.65 ton

DIMENSI PONDASI


= = = = = =

6.4 3.2 0.65 3 0.6 3.5

m m m m m m

Kem"r"ngan !tu a) "ngg" *ymey (" ata! tana* *t "ngg" *ymney , ,ear *ymney  ,ear &engan !&a g Back to Back Kengg"an ma ko&om

= = = = = = =

#1.16 0.4 2.6#1#6 0.65 1.1)5 .#3 3

= =

1.1330# m m3 3 2.)14 ton

= =

32.024 m3 )6.#5)6 to ton

0

m m m m m m

BERAT PONDASI



b' " ! b1 b1


=

Vsu=1/ Vsu=1/3timesital df times left(b1 rSup {size 8{2}} +b'rSup {size8{2} }+ sqrt {b1 rSup rSup {size 8{2} } timesb'rSup {size 8{2} }} right)} {

6.40 m

=

143.36 m3

= =

110.20 m3 104.6 ton

¿


u = :p < :!u u/1.5

= =

1#4.2) ton 122.#5 ton



122.65 ton

OK



= = =

254.646 to ton.m 335.01) to ton.m 3.2 m


=

)).02# ton.m ))


= = = =

6.3) ).4316) 6.6611 #. #.464)4

=

1.60

Momen Guling (MG)

ton to ton.m to ton.m  2

OK

K#$%&# T(&)**+ D** D/k/$, T*$*) .B(*&$, C*+*%


ex = ey =

M N

smak! sm"n

6 ex 6 ey N σ = ( 1± ± ) A L B

kg/cm2

=

15.5

ton/m2

=

0.46))) m

=

6.#245 ton/m2 A

15.5

ton/m2

OK

=

0.4462 ton/m2 A

15.5

ton/m2

OK

=

3.04244 

OK

("mana >


Fs= δ

x

N >1  ! H

1.5


d


a

= = = =

#4.50 1165.5 .5 6#.043

mm mm mm = k/m2

7u = B.a.F!uneGo < Hq1
=

32)46) 

q1 = .gtana*

=

36000 /m2

=

0.036

=

30000 /m2

0.005

m

d ds L

min a

max

Gaya tekan ke atas dari tanah (Vu) = 0.2314) Mpa

q2 = a'.geton h

d

=

0.03

/mm 2 /mm 2

b


f.Vc = f.√f'c.B.d/6 Vu < f.Vc 4E+0, < 2E+06

=

OK

20)316.4)  =

20)3.164) k

K#$%&# T(&)**+ G(3(& 2 A&*) L / 2

L / 2

i!e i!ens nsii ko"o! o"o!## b = 6, 6,0 0 !! b + d = h + d = 6,0 + $$ $$$ = d B

b d

d

h

-. = hk / bk =

d

1#15.5 mm

=

1.#155 m

Gaya tekan ke atas (gaya geser pons) % Vu= &'   * (b+d (b+d)( )(h+ h+ Qsunetto = 1E+06  = 1 42 0 k  1

bo= 2  &(bk &(bk + d) d) + (hk (hk + d) d) = 3 2 6 2 !! d ds L

min max

Gaya geser yang ditahan o"eh beton ( f . Vc): V. = (1+(2/-.)  ( Ö f'c bo  d ) / 6 = 2E+03  = 200341 k = 20 03  k  V. = (2 + (as . d) / (bo)).( Ö f'c = 2235 223553 5356 56  = 22 0 0 k 

bo  d ) / 12

V. = 1/4  Ö f'c bo d = 144 1442, 2,45 45  = 14 4 4 k  ipi"ih V. terke.i" f.7c = = 7u A f.7c .7c

144 44 4 k 0.)5 . 14 10036. k

OK


31.60## A

< 22

r = 0.3   K = 2 kan&eCer

r K

= =

Jp(!t& = 1/12     3

Jp(!t& =

$c = 4)00  f'c 0.5 Jp(!t&  $c

$c

=

15 2

22

.K(*$,3$,*$ .K(*$,3$,*$ b(&+($,*&/)

mm

1$<10 mm4 2224.1 Mpa

=

3$<14 mm2

=

3$<11 kmm2

Faktor pembesar momen ( ) δ = 1−

Cm Pu φ × Pc

d


=

³ 1.00

1

m = u = j =

1 14#.2 ton 0.65

c

1.1$<11 ton

=

Ok

Penulangan Pedestal

M1 =   , M2 =   e m"n Mt = M1 < M2 M = d  Mt Mnper&u = M/j nper&u = /j

M1

=

5 5.0)1 ton.m

M2 Mt M

= = =

5.13))6 ton.m 64.20) ton.m 64.20) ton.m

Mnper&u = nper&u =

#.)#41 ton.m 22.10# ton

Ek3($%&3%*3 4


e em"n

em"n = 15 < 0.03   %yarat > e  e m"n

=

431.16# mm

=

34.5

431.16# 

34.5

H%/$, $*

%umu 7erka&

Pu ϕ× Ag × β 1× f ' c

0.2))#

("mana > 1 ton =

III

mm OK


()

Pu e × ϕ × Ag × β 1× f ' c b

0.1#43


r'

7&

b'

= =

0.0150 0.

r

=

0.0135

H!tot

=

5)03.)5 mm2

n '

= = = 

20 21 623.)  21

L/*3 T/*$,*$ 4


20

D,/$*k*$ %/*$,*$ /%** 4

5)03.)5

OK


L/2

Diagram geser

Vud deff Vu


7u

=

7u( =

22.03

ton

12664# 

(ef =

554.5

mm

Kapa!"ta! ge!er

Vc= 1 / 6× √ f ' c ×b ×deff Vn

=

Vud φ

("mana > f = 0.65 %yarat > 7u(  f 7c

Av Av

=

Av =

2#441 ton

7n

=

14#43 ton

f

=

1""" ' c  b  1"""

HC1

=

541.66) !!2

HC2

=

4#1.)53 !!2

HC

=

541.66) !!2

%

=

2#.3 mm

% % % %

N N N N

336 4#0 2)).25 600

0.65 1#5212

 "g "gunak unakan an u&an u&ang gan +e!e +e!err M"n" M"n"m mum 

12"" 12""  fy

("paka"

S=

=

1266 664 4#.2 

b  1""" 1""" 3  fy

#! √ f

7c

1/ $  n

  d! 1""" Avu


mm mm mm mm


O %

= =

10 200

mm A 2)).25

D,/$*k*$ %/*$,*$ 3($,k*$, 4

O

10

-

200

P(b(3*$ T*+*k P#$*3

 %econ (

=

1250

mm 


1 =

(=a (e= ( - 0.5("a.u& - (" ("a.u& - !e&"mut eton

H! = ( = (e =

Hg = 1  1

Hg

=

1$ 1 $<0) mm2

F=s

!

=

6 #245 t/m2

1

mm

2#3.52 mm2 1250 mm 1146.5 mm

OK

e = 2-g Mu = 0.5  F  1  eP

e = Mu =

(

Mu

fy = #× fy × 1−"!88 #× " f 'c " 8 ×b 1× deff

3200 III

400Q Q 400

0.325)) = 2 41#1Q -

)

.mm -

r

=

0.000#1

H! = r  1  (e

H!

=

25.15 mm 2


!

=

"ameter



D,/$*k*$ %/*$,*$ /%** 4

D

250

=

2

41#Q1 0.325)6#

= 1134.115

OK

2

1 1 18

III

mm

-

250


bc o = 4  c < (e

bc = o =

1 165#6 m mm m

7c 7c

= =

1.5$<0)  III ton

7 = F  1  e

7 7

= =

III  13.)6# ton

7u = 7/f

7u

=

215.02) ton

%yarat > 7u A f7c

215.02) A

Vc= 1 / 6× √ f ' c ×b$×deff

 %econ (

P(b(3*$ T*+*k P#$*3

=


600

6.)11

mm 

1 = H! = ( = (e =

(=a (e= ( - 0.5("a.u& - ("a.u& - !e&"mut eton

1

mm

2#3.52 mm2 650 mm 546.5 mm

Hg = 1  1

Hg

=

F=s

!

=

!

=

III

e

=

1500

Mu =

III

e = */2 Mu = 0.5  F  1  eP

Mu "

" 8 ×b 1× deff

(

= #× fy × 1−"!88 #×

fy f 'c

1$ 1 $<0)

mm2

6.#245 t/m2 /mm 2

.mm 400Q Q 400

0.15)52 = 2 41#1Q -

)

.T*k ((&/k*$ %/*$,*$ ,(3(&

-

r

=

0.0003

H! = r  1  (e

H!

=

214.32 mm 2


!

=

"ameter



D,/$*k*$ %/*$,*$ /%** 4

D

250

=

= 1134.115

OK

2

1 1 18

2

41#Q1 0.15)51)

mm

-

250


bc o = 4  c < (e

bc = o =

1 14#6 m mm m

7c 7c

= 64)4644  = III ton

7 = F  1  e

7 7

= =

7u = 7/f

7u

=

III

%yarat > 7u A f7c

100.)4 A

III

Vc= 1 / 6× √ f ' c ×b$×deff

642314  III ton ton

.T*k ((&/k*$ %/*$,*$ ,(3(&

P(&)%/$,*$ P(&)%/$,*$ +(&($*$**$ B*#k *$ S##!

"rencanakan ("men!" !&oof Bean-ean yang ekerDa pa(a !&oof a(a&a* >

=

25



35

cm

%e*"gga mo(e& pemeanannya !eper pa(a gamar er"kut > q = 510 Kg/m

B

H



= 1/#.q.,2 510 11.0) = # = )05.2) Kg.m = 1/12.q.,2 = 510 11.0) 12 = 4)0.61# Kg.m

Momen &apangan H R   < 

Momen tumpuan H R   - 

= 5/4#.q.,2 2550 2550 11.0 11.0) ) = 4# = 5##.2)3 kg m

Momen &apangan B  < 

D(3*$ %/*$,*$ /%** Tulangan lapangan ( M + )

Tm"n

=

1.4 f y

=

Tmak! = 0.)5 U1

1.4 = 400 0.#5.f'c

0.0035 600 600 < f y

f y

= 0.)5 0.#5

0.#5 22.5 400

600 600 < 400

=

0.01#2 600 a = U1 ( 600 < f y 600 = 0.#5 2#3.5 600 < 400 = 144.5#5 mm amak! = )5V.a = 0.)5 145 = 10# mm mm "tung > amak! Mna(a = 0.#5.f'c..amak! ( 2 = 0.#5 22.5

250

10# 2#3.5 - 54.22

= 11 11## ##)5 )5)6 )61. 1.53 53  .mm .mm Mu )052)0.4) Mnper&u = = = ##24 ##240# 0##. #.0# 0##5 #54 4 .mm .mm S 0.# Karena Mna(a  Mnper&u maka ("gunakan tu&angan tungga& m

=

@n

=

f y 0.#5.f'c Mu S..(2

= =

400 0.#5 22.5

=

20.15

)052)0.4)0#3334 0.#

250 #03)2.3

= 0.43 Mpa

"tung > T

=

1 m

1 -

1 -

2.m.@n f y

1 1#.3)01 = 1 1 20.15 400 = 0.00111 Kontro& ra!"o tu&angan T A Tm"n A Tmak! ,/$*k*$ &*3# %/*$,*$ $/ (engan (em"k"an &ua! tu&angan yang ("per&ukan a(a&a* > H! = Tm"n..( = 0.0035 250 2#3.5 = 24# mm2 ("gunakan tu&angan

4

S

13

=

531 mm2


Tm"n

=

1.4 f y

=

Tmak! = 0.)5 U1

1.4 = 400 0.#5.f'c

= 0 )5 0 #5

f y

0.0035 600 600 < f y

0.#5 22.5

600



1

600 < f y

600 2#3.5 600 < 400 = 144.5#5 mm = )5V.a = 0.)5 145 = 10# mm mm = 0.#5

amak!

"tung > Mna(a =

0.#5.f'c..amak!

= 0.#5 22.5

( -

250

amak! 2

10# 2#3.5 - 54.22

= 11## 11##)5 )5)6 )61. 1.53 53  .mm .mm Mu 4)061#0.31 Mnper&u = = = 5##2 5##2)2 )25. 5.3 323 236 6 .mm .mm S 0.# Karena Mna(a  Mnper&u maka ("gunakan tu&angan tungga& m

=

@n

=

f y 0.#5.f'c Mu S..(

2

= =

400 0.#5 22.5

=

20.15

4)061#0.313#### 0.#

250

#03)2.3

= 0.23 Mpa

"tung > T

=

1 m

1 -

1 -

2.m.@n f y

1 12.246# 1 1 20.15 400 = 0.000)4 Kontro& Kontro& ra!"o tu&angan T A Tm"n A Tmak! ,/$*k*$ &*3# %/*$,*$ $/ =

(engan (em"k"an &ua! tu&angan yang ("per&ukan a(a&a* > H! = Tm"n..( =

0.0035

=

24#

250 2#3.5

mm2

("gunakan tu&angan

4

S

13

=

531 mm2

OK999

A$*33 %/*$,*$ Tulangan lapangan ( M + )

Ta(a

=

H!terpa!ang .(

=

530.2 = 0.00)4 )0#)5

Tm"n A Ta(a A Tmak! a

=


OK999

=

531

400

0.#5 22.5

250

= 44.42 mm

%e*"ngga momen nom"na& a(a&a* a Mna(a = H!.f y ( 2 = 531 400 2#3.5 - 22.21 = 5540#46.4)4 .mm Kontro& Kontro& keamanan penampang > Mna(a  Mu OK999 Tulangan tumpuan ( M - )

Ta(a

=

H!terpa!ang .(

=

530.2 )0#)5


=


= 0.00)4 OK999

=

531

400

0.#5 22.5

250

= 44.42 mm

%e*"ngga momen nom"na& a(a&a* a Mna(a = H!.f y ( 2 = 531 400 2#3.5 - 22.21 = 5540#46.4)4 .mm Kontro& Kontro& keamanan penampang > Mna(a  Mu OK999 P((&k3**$ %/*$,*$ ,(3(&


=

22026.5 kg 7u 22026.5 = = = 36)10.# Kg S 0.6 1 22.5 250 2#3.5 f'c.X.( = = 6 6

6 56031.6  = 5603

Kg

S .7c 6 4202 4202.3 .3) ) kg b/%/)k*$ %/*$,*$ ,(3(& 7u A S . 7c 7!

7c1

6

7u - S7c 22026.5 - 4202.3) 6 23)65.51 = S 0.)5

6 0.33. fW.X.(

= 11043 kg

kg

7! A 7c2

P($*+*$, /k/+

Meng*"tung Darak tu&angan !engkang er(a!arkan per!amaan > "rencan "rencanakan akan menggunak menggunakan an ("ameter ("ameter 10 mm HC = 15) m2 HC.fyt.( = )4.144 %1 6 mm 7! %2 6 (/2 = 141.)5 mm HC.fyt = )1).)14 %3 6 mm 0.35.X %4 6 600 mm ("gunakan % =

141.)5 mm

=

140

mm

OK

P(&)%/$,*$ +(&($*$**$ B*#k P($*$,,* P($*$,,*

 co! Z q =

210

kg /m co! Z

, "rencanakan menggunakan ("men!"  = 633.4 kg kg ,' = 3.02 m , = 4.25 %u(ut Z = 44.#4 Y  co! Z = 44.1#1 kg q co! Z = 14#.02 kg

25



35

= ., < q., 2/2 = 4042.6050625 kg kg.m

Mma Tulangan tumpuan ( M - )

Tm"n

=

1.4 f y

=

Tmak! = 0.)5 U1

1.4 = 400 0.#5.f'c

0.0035 600 600 < f y

f y

= 0.)5 0.#5

0.#5 22.5 400

= a

0.01#2 600 = U1 600 < f y

600 600 < 400

(

600 2#3.5 600 < 400 = 144.5#5 mm amak! = )5V.a = 0.)5 145 = 10# mm mm "tung > amak! Mna(a = 0.#5.f'c..amak! ( 2 = 0.#5

= 0.#5 22.5

250

10# 2#3.5 - 54.22

= 11 11## ##)5 )5)6 )61. 1.53 53  .mm .mm Mu 4042605 Mnper&u = = = 50 5053 5336 3631 31.3 .32# 2#1 1 .mm .mm S 0.# Karena M na(a  Mnper&u maka ("gunakan tu&angan tungga& m

=

@n

=

f y 0.#5.f'c Mu S..(

2

= =

400 0.#5 22.5

=

20.15

4042605.06251#3 0.#

250

#03)2.3

= 2.515 Mpa

"tung > T

=

1 m

1 -

1 -

1 = 1 20.15 = 0.006)) Kontro& ra!"o tu&angan Tm"n A T A Tmak!

2.m.@n f y 1 -

105.202 400

(engan (em"k"an &ua! tu&angan yang ("per&ukan a(a&a* > H! = T..( =

0.006))

250 2#3.5

A$*33 %/*$,*$ 4 Tulangan tumpuan ( M - )

Ta(a

=

H!terpa!ang .(

=

530.2 = 0.00)4 )0#)5


=

OK999

H!terpa!ang.f y 0.#5.f'c.

=

531

400

0.#5 22.5

250

= 44.42 mm ,/3

%e*"ngga momen nom"na& a(a&a* a Mna(a = H!.f y ( 2 = 531 400 2#3.5 - 22.21 = 5540#46.4)4 .mm Kontro& Kontro& keamanan penampang > Mna(a  Mu OK999 P((&k3**$ %/*$,*$ ,(3(&


=

16064.# kg 7u 16064.# = = = 26))4.6 Kg S 0.6 1 22.5 250 2#3.5 f'c.X.( = = 6 6

6 56031.6  = 5603

Kg

S .7c 6 42 4202 02..3) kg 7u A S . 7c b/%/)k*$ %/*$,*$ ,(3(& 7!

6

7u - S7c 16064.# - 4202.3) = 6 15#16.52 S 0.)5

7c1

6 0.33. fW.X.(

= 350#31 kg

7c2

6 0.66. fW.X.(

= )01663 kg

7! A 7c2

kg

P($*+*$, /k/+

Meng*"tung Darak tu&angan !engkang er(a!arkan per!amaan > "rencana "rencanakan kan menggunak menggunakan an ("ameter ("ameter 10 mm HC = 15) m2 HC.fyt.( = 1125.65 %1 6 mm 7! %2 6 (/2 = 141.)5 mm HC.fyt = )1).)14 %3 6 mm 0.35.X %4 6 600 mm ("gunakan % =

141.)5 mm

=

140

mm

OK

CALCULATION DESIGN OF FOUNDATION 150 kV TRANSMISSION LINE OF KALTENG-1 KALTENG-1 2X100 MW DD$6 CLASS % DATA TEKNIS PERENCANAAN


= = = = = = = = = = = =

K225 225 320 2.40 1.60 100.00 0.45 0.#5 210000 0.003 )5.00 )5.00


=

20.00

DATA BEBAN

= = = =

1#.1) ton 110.24 ton #.)2 ton



Mpa mm mm 0

DIMENSI PONDASI


= = = = = =

5 2.5 0.4 2.2 0.4 3.5

m m m m m m

Kem"r"ngan !tu a) "ngg" *ymey (" ata! tana* *t "ngg" *ymney , ,ear *ymney  ,ear &engan !&a g

= = = = =

#1.16 0.4 3.13)2) 0.)5 0.)25

= =

1.)64)1 m m3 3 4.23531 to ton

= =

11.36 m3 2#.6464 to ton

0

m m m m

BERAT PONDASI



b' " ! b1 b1


=


).55 m

=

13.66 m3

= =

125.6 m3 11.66 ton

¿


u = :p < :!u u/1.5

= =

152.54 ton 101.)0 ton



#.)2

ton to

=

33.33

ton/m2

=

0.40) m

=

OK



= = =

#2.2043 to ton.m 2.152 to ton.m 2.5 m


=

63.52 ton.m 63


= = = =

).35 #.5)5 55.01) 6.315

=

3.33

Momen Guling (MG)

ton ton.m to ton.m  2

OK

K#$%&# T(&)**+ D** D/k/$, T*$*) .B(*&$, C*+*%


ex = ey =

M N

smak! sm"n

6 ex 6 ey N σ = ( 1± ± ) A L B

kg/cm2

.613 ton/m2 A

33.33

ton/m2

OK

= -0.#)204 ton/m2 A

33.33

ton/m2

OK

=

OK

("mana >


Fs= δ

x

N >1  ! H

2.)3035 

1.5


d


a

= = = =

#4.50 )15.5 .5 6.45))

mm mm mm = k/m2

=

303#) 

q1 = .gtana*

=

43200 /m2

=

0.0432 /mm 2

=

1200 /m2

=

0.012 /mm 2

0.005

m

d ds L

min a

max

Gaya tekan ke atas dari tanah (Vu) = 0.201) Mpa

q2 = a'.geton h

d

b


f.Vc = f.√f'c.B.d/6 Vu < f.Vc 4E+0, < 3E+0,

=

OK

33326.4#)  =

33.3264#)2 k

K#$%&# T(&)**+ G(3(& 2 A&*) L / 2

L / 2

i!e i!ens nsii ko"o! o"o!## b = 3, 3,0 0 !! b + d = h + d = 3, 0 + 3 16 = d B

b d

d

h

-. = hk / bk =

d

1465.5 mm

=

1.4655 m

Gaya tekan ke atas (gaya geser pons) % Vu= &'   * (b+d (b+d)( )(h+ h+ Qsunetto = E+0,  = 3124 k 1

bo= 2  &(bk &(bk + d) d) + (hk (hk + d) d) = ,  6 2 !! d ds L

min max

Gaya geser yang ditahan o"eh beton ( f . Vc): V. = (1+(2/-.)  ( Ö f'c bo  d ) / 6 = 553,64  = 553,64 k = 5 5  k  V. = (2 + (as . d) / (bo)).( Ö f'c = 5 4  63 15  = 5 4 3 k 

bo  d ) / 12

V. = 1/4  Ö f'c bo d = 6 6 4 13 05  = 6 64 2 k  ipi"ih V. terke.i" f.7c =

6642 k 0.)5 . = 4)3.)#2 k

7u A f.7c .7c

OK


31.4424 A

< 22

r = 0.3   K = 2 kan&eCer

r K

= =

Jp(!t& = 1/12     3

Jp(!t& =

$c = 4)00  f'c 0.5 Jp(!t&  $c

$c

=

225 2

22

.K(*$,3$,*$ .K(*$,3$,*$ b(&+($,*&/)

mm

3$<10 mm4 2224.1 Mpa

=

6$<14 mm2

=

6$<11 kmm2


Cm Pu φ × Pc

d


=

³ 1.00

1

m = u = j =

1 110.242 ton 0.65

c

1.5$<11 ton

=

Ok

Penulangan Pedestal


M1

=

5) 5).0022 ton.m

M2 Mt M

= = =

4.1340) ton.m 61.1363 ton.m 61.1363 ton.m


4.055# ton.m 16.603 ton

Ek3($%&3%*3 4


e em"n

em"n = 15 < 0.03   %yarat > e  e m"n

=

554.565 mm

=

3).5

554.565 

3).5

H%/$, $*

%umu 7erka&

Pu ϕ× Ag × β 1× f ' c

0.15456

("mana > 1 ton =

III

mm OK


()

Pu e × b ϕ × Ag × β 1× f ' c

0.1142


r'

7&

b'

= =

0.0100 0.

r

=

0.000

H!tot

=

5062.5 mm2

n '

= = = 

20 1 566).)  1

L/*3 T/*$,*$ 4


20

D,/$*k*$ %/*$,*$ /%** 4

5062.5

OK


L/2

Diagram geser

Vud deff Vu


7u

=

7u( =

1#.1)

ton

10364 

(ef =

655.5

mm

Kapa!"ta! ge!er

Vc= 1 / 6× √ f ' c ×b ×deff Vn

=

Vud φ

("mana > f = 0.65 %yarat > 7u(  f 7c

Av Av

=

Av =

3##664 ton

7n

=

1552 ton

f

=

1""" ' c  b  1"""

HC1

=

)#1.25 !!2

HC2

=

64.#36 !!2

HC

=

)#1.25 !!2

%

=

201.062 mm

% % % %

N N N N

304 4#0 32).)5 600

0.65 25 252 2631

 "gun "gunak akan an u&ang &angan an +e!e +e!err M"n" M"n"mu mum m

12"" 12""  fy

("paka"

S=

=

1036 36 3.# 

b  1""" 1""" 3  fy

#! √ f

7c

1/ $  n

  d! 1""" Avu


mm mm mm mm


O %

= =

10 200

mm A 201.061

D,/$*k*$ %/*$,*$ 3($,k*$, 4

O

10

-

200

P(b(3*$ T*+*k P#$*3

 %econ (

=

#00

mm 


1 =

(=a (e= ( - 0.5("a.u& - (" ("a.u& - !e&"mut eton

H! = ( = (e =

Hg = 1  1

Hg

1

mm

2#3.52 mm2 #00 mm 66.5 mm

= 6250000 mm2

OK

e = 2 Mu = 0.5  F  1  eP

Mu

e = Mu =

(

fy = #× fy × 1−"!88 #× " f 'c " 8×b 1× deff

2500 III

0.)650) = 2 26)6Q -

)

.mm 320Q Q 320

-

r

=

0.00234

H! = r  1  (e

H!

=

1633.21 mm 2


!

=

"ameter



D,/$*k*$ %/*$,*$ /%** 4

D

160

=

1 1 18

-

III 2

26)Q6 0.)650)4

= 1))2.055

OK

2

mm 1:0


bc o = 4  c < (e

bc = o =

1 115#6 m mm m

7c 7c

= 63)61#  = III ton

7 = F  1  e

7 7

= =

III  121.142 ton

7u = 7/f

7u

=

1#6.3)3 ton

%yarat > 7u A f7c

1#6.3)3 A

Vc= 1 / 6× √ f ' c ×b$ ×deff

 %econ (

P(b(3*$ T*+*k P#$*3

=

400

422.6

.T*k ((&/k*$ %/*$,*$ ,(3(&

mm 


1 =

(=a (e= ( - 0.5("a.u& - ("a.u& - !e&"mut eton

H! = ( = (e =

1

mm

2#3.52 mm2 400 mm 26.5 mm

Hg = 1  1

Hg

= 6250000 mm2

F=s

!

=

!

=

III

e

=

1100

Mu =

III

e = */ 2 Mu = 0.5  F  1  eP

Mu

(

fy = #× fy × 1−"!88 #× " f 'c " 8×b 1× deff

.613 t/m2

0.40#6) = 2 26)6Q -

)

/mm 2

.mm 320Q Q 320

-

r

=

0.00126

H! = r  1  (e

H!

=

3)4.6# mm 2


!

=

"ameter



D,/$*k*$ %/*$,*$ /%** 4

D

250

=

1 1 18

-

2

26)Q6 0.40#66

= 1134.115

OK

2

mm 250


bc o = 4  c < (e

bc = o =

1 111#6 m mm m

7c 7c

= 2622041  = III ton

7 = F  1  e

7 7

= =

7u = 7/f

7u

=

III

%yarat > 7u A f7c

#2.0041 A

III

Vc= 1 / 6× √ f ' c ×b$ ×deff

5225)5  III ton ton

.T*k ((&/k*$ %/*$,*$ ,(3(&

CALCULATION DESIGN OF FOUNDATION 150 kV TRANSMISSION LINE OF KALTENG-1 KALTENG-1 2X100 MW DD$6 CLASS % DATA TEKNIS PERENCANAAN


= = = = = = = = = = = =

K225 225 320 2.40 1.60 50.00 0.45 0.#5 210000 0.003 )5.00 )5.00


=

20.00

DATA BEBAN

= = = =

1#.1) ton 110.24 ton #.)2 ton



Mpa mm mm 0

DIMENSI PONDASI


= = = = = =

4.5 2.25 0.4 2.2 0.4 3.5

m m m m m m


= = = = =

#1.16 0.4 3.13)2) 0.)5 0.)25

= =

1.)64)1 m m3 3 4.23531 to ton

= =

10.036 m3 24.0#64 to ton

0

m m m m

BERAT PONDASI



b' " ! b1 b1


=


).05 m

=

11#.5# m3

= =

106.)# m3 1)0.#4 ton

¿


u = :p < :!u u/1.5

= =

1.16 ton 132.)) ton



#.)2

ton to

=

16.6)

ton/m2

=

0.40) m

=

12.6#2 ton/m2 A

16.6)

ton/m2

OK

= -1.#0111 ton/m2 A

16.6)

ton/m2

OK

=

OK

OK



= = =

63.)23# to ton.m 3#4.3 ton.m 2.25 m


=

63.52 ton.m 63


= = = =

).35 #.5)5 55.01) #. #.144#

=

1.6)

Momen Guling (MG)


OK

K#$%&# T(&)**+ D** D/k/$, T*$*) .B(*&$, C*+*%


ex = ey =

M N

smak! sm"n

N 6 ex 6 ey σ = ( 1± ± ) A L B

kg/cm2

("mana >


Fs= δ

x

N >1  ! H

2.)3035 

1.5


d

(! = %e&"mut eton < /2 ( = -(! a = ,/2 - /2 - ( sa = smin+(L-a).( smaks-smin)/L

a

= = = =

#4.50 )15.5 .5 126.266

mm mm mm = k/m2

=

303#) 

q1 = .gtana*

=

43200 /m2

=

0.0432 /mm 2

=

1200 /m2

=

0.012 /mm 2

0.005

m

d ds L

min a

max

Gaya tekan ke atas dari tanah (Vu) = 0.201) Mpa

q2 = a'.geton h

d

b


f.Vc = f.√f'c.B.d/6 Vu < f.Vc 4E+0, < 3E+0,

=

OK

33326.4#)  =

33.3264#)2 k

K#$%&# T(&)**+ G(3(& 2 A&*) L / 2

L / 2

i!e i!ens nsii ko"o! o"o!## b = 3, 3,0 0 !! b + d = h + d = 3, 0 + 3 16 = d B

b d

d

h

-. = hk / bk =

d

1465.5 mm

=

1.4655 m

Gaya tekan ke atas (gaya geser pons) % Vu= &'   * (b+d (b+d)( )(h+ h+ Qsunetto = E+0,  = 3124 k 1

bo= 2  &(bk &(bk + d) d) + (hk (hk + d) d) = ,  6 2 !! d ds L

min max

Gaya geser yang ditahan o"eh beton ( f . Vc): V. = (1+(2/-.)  ( Ö f'c bo  d ) / 6 = 553,64  = 553,64 k = 5 5  k  V. = (2 + (as . d) / (bo)).( Ö f'c = 5 4  63 15  = 5 4 3 k 

bo  d ) / 12

V. = 1/4  Ö f'c bo d = 6 6 4 13 05  = 6 64 2 k  ipi"ih V. terke.i" f.7c =

6642 k 0.)5 . = 4)3.)#2 k

7u A f.7c .7c

OK


31.4424 A

< 22

r = 0.3   K = 2 kan&eCer

r K

Jp(!t& = 1/12     3 $c = 4)00  f'c Jp(!t&  $c

= =

225 2

Jp(!t& =

0.5

$c

=

22

.K(*$,3$,*$ .K(*$,3$,*$ b(&+($,*&/)

mm

3$<10 mm4 2224.1 Mpa

=

6$<14 mm2

=

6$<11 kmm2

Faktor pembesar momen ( ) δ =

Cm Pu 1− φ × Pc

d


=

³ 1.00

1

m = u = j =

1 110.242 ton 0.65

c

1.5$<11 ton

=

Ok

Penulangan Pedestal


M1

=

5) 5).0022 ton.m

M2 Mt M

= = =

4.1340) ton.m 61.1363 ton.m 61.1363 ton.m


4.055# ton.m 16.603 ton

Ek3($%&3%*3 4


e em"n

em"n = 15 < 0.03   %yarat > e  e m"n

=

554.565 mm

=

3).5

554.565 

3).5

H%/$, $*

%umu 7erka&

Pu ϕ× Ag × β 1× f ' c

0.15456

("mana > 1

III

mm OK


()

Pu e × ϕ × Ag × β 1× f ' c b

0.1142


r'

7&

b'

= =

0.0100 0.

r

=

0.000

H!tot

=

5062.5 mm2

n '

= = = 

20 1 566).)  1

L/*3 T/*$,*$ 4


20

D,/$*k*$ %/*$,*$ /%** 4

5062.5

OK


L/2

Diagram geser

Vud deff Vu


7u

=

7u( =

1#.1)

ton

10364 

(ef =

655.5

mm

Kapa!"ta! ge!er

Vc= 1 / 6× √ f ' c ×b ×deff Vn

=

Vud φ

("mana > f = 0.65 %yarat > 7u(  f 7c

Av Av

=

Av =

3##664 ton

7n

=

1552 ton

f

=

1""" ' c  b  1"""

HC1

=

)#1.25 !!2

HC2

=

64.#36 !!2

HC

=

)#1.25 !!2

%

=

201.062 mm

% % % %

N N N N

304 4#0 32).)5 600

0.65 25 252 2631

 "gun "gunak akan an u&ang &angan an +e!e +e!err M"n" M"n"mu mum m

12"" 12""  fy

("paka"

S=

=

1036 36 3.# 

b  1""" 1""" 3  fy

#! √ f

7c

1/ $  n

  d! 1""" Avu


mm mm mm mm


O %

= =

10 200

mm A 201.061

D,/$*k*$ %/*$,*$ 3($,k*$, 4

O

10

-

200

P(b(3*$ T*+*k P#$*3

 %econ (

=

#00

mm 


1 =

(=a (e= ( - 0.5("a.u& - (" ("a.u& - !e&"mut eton

H! = ( = (e =

Hg = 1  1

Hg

1

mm

2#3.52 mm2 #00 mm 66.5 mm

= 5062500 mm2

OK

e = 2 Mu = 0.5  F  1  eP

Mu "

" 8 ×b 1× deff

e = Mu =

(

= #× fy × 1−"!88 #×

fy f 'c

2250 III

0.#114 = 2 26)6Q -

)

H! = r  1  (e

.mm 320Q Q 320

r

=

0.0024#

H!

=

1)30.13 mm2

III -

2

26)Q6 0.#11403


!

=

160

= 1))2.055

"ameter



=

1 1

mm2

D,/$*k*$ %/*$,*$ /%** 4

D

18

-

OK

1:0


bc o = 4  c < (e

bc = o =

1 115#6 m mm m

7c 7c

= 63)61#  = III ton

7 = F  1  e

7 7

= =

III  12#.4)# ton

7u = 7/f

7u

=

1).65 ton

%yarat > 7u A f7c

1).65 A

Vc= 1 / 6× √ f ' c ×b$ ×deff

 %econ (

P(b(3*$ T*+*k P#$*3

=

400

422.6

.T*k ((&/k*$ %/*$,*$ ,(3(&

mm 


1 =

(=a (e= ( - 0.5("a.u& - ("a.u& - !e&"mut eton

H! = ( = (e =

1

mm

2#3.52 mm2 400 mm 26.5 mm

Hg = 1  1

Hg

= 5062500 mm2

F=s

!

=

!

=

III

e

=

1100

Mu =

III

e = */ 2 Mu = 0.5  F  1  eP

Mu

(

fy = #× fy × 1−"!88 #× " f 'c " 8 ×b 1× deff

H! = r  1  (e

12.6#2 t/m2

0.5350# = 2 26)6Q -

)

/mm 2

.mm 320Q Q 320

r

=

0.00165

H!

=

4#.041 mm 2

-


!

=

250

= 1134.115

"ameter



=

1 1

mm2

D,/$*k*$ %/*$,*$ /%** 4

D

18

-

2

26)Q6 0.5350#1

OK

250


bc o = 4  c < (e

bc = o =

1 101#6 m mm m

7c 7c

= 23#)63)  = III ton

7 = F  1  e

7 7

= =

7u = 7/f

7u

=

III

%yarat > 7u A f7c

6.6331 A

III

Vc= 1 / 6× √ f ' c ×b$ ×deff

615)  III ton ton

.T*k ((&/k*$ %/*$,*$ ,(3(&

CALCULATION DESIGN OF FOUNDATION 150 kV TRANSMISSION LINE OF KALTENG-1 KALTENG-1 2X100 MW 4CC6 CLASS % DATA TEKNIS PERENCANAAN


= = = = = = = = = = = =

K225 225 320 2.40 1.60 50.00 0.45 0.#5 210000 0.003 )5.00 )5.00


=

20.00

DATA BEBAN

= = = =

2#.#4 ton 15#.1# ton 12#.24 ton



Mpa mm mm 0

DIMENSI PONDASI


= = = = = =

6 3 0.6 2.# 0.5 3.5

m m m m m m


= = = = =

#1.16 0.4 2.#3366 0.)5 1.025

= =

1.533 m m3 3 3.#2544 to ton

= =

25.52 m3 61.24# ton

0

m m m m

BERAT PONDASI



b' " ! b1 b1


=


#.55 m

=

1#).0# m3

= =

15.6 m3 255.4 ton

¿


u = :p < :!u u/1.5

= =

321.01 ton 214.01 ton



12#.24 ton

OK



= = =

15.22 ton.m )6).#21 to ton.m 3 m


=

100.43 ton.m 10


= = = =

).35 #.5)5 2.36#4 10 10.4261

=

1.6)

Momen Guling (MG)


OK

K#$%&# T(&)**+ D** D/k/$, T*$*) .B(*&$, C*+*%


ex = ey =

M N

smak! sm"n

N 6 ex 6 ey σ = ( 1± ± ) A L B

kg/cm2

=

16.6)

ton/m2

=

0.5#33 m

=

.5255 ton/m2 A

16.6)

ton/m2

OK

= -0.)3)5) ton/m2 A

16.6)

ton/m2

OK

=

OK

("mana >


Fs= δ

x

N >1  ! H

2.46#12 

1.5


d


a

= = = =

#4.50 1015.5 .5 4.0))

mm mm mm = k/m2

=

4005#3 

q1 = .gtana*

=

3#400 /m2

=

0.03#4 /mm 2

=

26400 /m2

=

0.0264 /mm 2

0.005

m

d ds L

min a

max

Gaya tekan ke atas dari tanah (Vu) = 0.2665) Mpa

q2 = a'.geton h

d

b


f.Vc = f.√f'c.B.d/6 Vu < f.Vc E+0, < 1E+06

=

OK

16#52#.#1  =

16#5.2##06 k

K#$%&# T(&)**+ G(3(& 2 A&*) L / 2

L / 2

i!e i!ens nsii ko"o! o"o!## b = 3, 3,0 0 !! b + d = h + d = 3,0 + $$ $$$ = d B

b d

d

h

-. = hk / bk =

d

1)65.5 mm

=

1.)655 m

Gaya tekan ke atas (gaya geser pons) % Vu= &'   * (b+d (b+d)( )(h+ h+ Qsunetto = 1E+06  = 1 2 ,5 k  1

bo= 2  &(bk &(bk + d) d) + (hk (hk + d) d) = 3 0 62 !! d ds L

min max

Gaya geser yang ditahan o"eh beton ( f . Vc): V. = (1+(2/-.)  ( Ö f'c bo  d ) / 6 = 1300614  = 130061 k = 13 00 5 k  V. = (2 + (as . d) / (bo)).( Ö f'c = 135 135, ,4 4  = 13 5 5 k 

bo  d ) / 12

V. = 1/4  Ö f'c bo d = 1144 114450 503, 3,  = 11 44 5 k  ipi"ih V. terke.i" f.7c =

114 445 k 0.)5 . 11 = #504.30) k

7u A f.7c

OK


2#.)436 A

< 22

r = 0.3   K = 2 kan&eCer

r K

= =

Jp(!t& = 1/12     3

Jp(!t& =

$c = 4)00  f'c 0.5 Jp(!t&  $c

$c

=

225 2

22

.K(*$,3$,*$ b(&+($,*&/)

mm

3$<10 mm4 2224.1 Mpa

=

6$<14 mm2

=

6$<11 kmm2


Cm Pu φ × Pc


d

=

³ 1.00

1

m = u = j =

1 15#.1#5 ton 0.65

c

1.#$<11 ton

=

Ok

Penulangan Pedestal


M1

=

#1 #1.)255 ton.m

M2 Mt M

= = =

5.312 ton.m #).65)4 ton.m #).65)4 ton.m


134.#5# ton.m 243.361 ton

Ek3($%&3%*3 4

e = Mn per&u/nper&u em"n = 15 < 0.03   %yarat > e  e m"n

e em"n

=

554.14) mm

=

3).5

554.14) 

3).5

H%/$, $*

%umu 7erka&

Pu ϕ × Ag × β 1× f ' c ("mana >

0.221)#

mm OK


()

Pu e × ϕ × Ag × β 1× f ' c b

0.163#)


r'

7&

b'

= =

0.0140 0.

r

=

0.0126

H!tot

=

)0#).5 mm2

L/*3 T/*$,*$ 4


n '

= 24 = 21 = #30#.44   21

24

D,/$*k*$ %/*$,*$ /%** 4

)0#).5

OK


L/2

Diagram geser

Vud deff Vu


7u

=

7u( =

2#.#4

ton

15213) 

(ef =

654.5

mm

Kapa!"ta! ge!er

Vc= 1 / 6× √ f ' c ×b ×deff Vn

=

Vud φ

("mana > f = 0.65 %yarat > 7u(  f 7c

Av Av

=

Av =

b  1""" 1""" 3  fy

#! √ f

1""" ' c  b  1""" 12"" 12""  fy

1/ $  n

=

3##0)1 ton

7n

=

23405) ton

f

=

1521 213 3).1 

("paka"

S=

7c

  d! 1""" Avu


HC1

=

)#1.25 !!2

HC2

=

64.#36 !!2

HC

=

)#1.25 !!2

%

=

201.062 mm

% % % %

N N N N

336 4#0 32).25 600

0.65 252246

mm mm mm mm


O %

= =

10 200

mm A 201.061

D,/$*k*$ %/*$,*$ 3($,k*$, 4

O

10

-

200

P(b(3*$ T*+*k P#$*3

 %econ (

=

1100

mm 


1 =

(=a (e= ( - 0.5("a.u& - (" ("a.u& - !e&"mut eton

H! = ( = (e =

Hg = 1  1

Hg

1

 "g "gunak unakan an u&an u&ang gan +e!e +e!err M"n" M"n"m mum 

mm

2#3.52 mm2 1100 mm 6.5 mm

= 000000 mm2 t/m2

OK

e = 2 Mu = 0.5  F  1  eP

Mu

e = Mu =

(

fy = #× fy × 1−"!88 #× " f 'c " 8 ×b 1× deff

3000 III

0.5201 = 2 26)6Q -

)

.mm 320Q Q 320

-

r

=

0.00163

H! = r  1  (e

H!

=

1625.1 mm 2


!

=

"ameter



D,/$*k*$ %/*$,*$ /%** 4

D

160

=

1 1 18

-

III 2

26)Q6 0.52005

= 1))2.055

OK

2

mm 1:0


bc o = 4  c < (e

bc = o =

1 151#6 m mm m

7c 7c

= =

1.2$<0)  III ton

7 = F  1  e

7 7

= =

III  1)1.461 ton

7u = 7/f

7u

=

263.)#6 ton

%yarat > 7u A f7c

263.)#6 A

Vc= 1 / 6× √ f ' c ×b$ ×deff

 %econ (

P(b(3*$ T*+*k P#$*3

=

500

)3.1#5

.T*k ((&/k*$ %/*$,*$ ,(3(&

mm 


1 =

(=a (e= ( - 0.5("a.u& - ("a.u& - !e&"mut eton

H! = ( = (e =

1

mm

2#3.52 mm2 600 mm 46.5 mm

Hg = 1  1

Hg

= 000000 mm2

F=s

!

=

!

=

III

e

=

1400

Mu =

III

e = */ 2 Mu = 0.5  F  1  eP

Mu

(

fy = #× fy × 1−"!88 #× " f 'c " 8 ×b 1× deff

.5255 t/m2

0.23204 = 2 26)6Q -

)

/mm2

.mm 320Q Q 320

-

r

=

0.000)2

H! = r  1  (e

H!

=

35).#65 mm 2


!

=

"ameter



D,/$*k*$ %/*$,*$ /%** 4

D

250

=

1 1 18

-

2

26)Q6 0.23203#

= 1134.115

OK

2

mm 250


bc o = 4  c < (e

bc = o =

1 13#6 m mm m

7c 7c

= 54#)53  = III ton

7 = F  1  e

7 7

= =

7u = 7/f

7u

=

III

%yarat > 7u A f7c

123.1

A

III

Vc= 1 / 6× √ f ' c ×b$ ×deff

)#4461  III ton ton

.T*k ((&/k*$ %/*$,*$ ,(3(&

$KH H@JK +$%$@ @6<0 10#0.3) #).21 1)#.06 k 46<0 k 46<3 1640.52 13 22 33. k k 46< 1662.444 12#.43 32.))# k 46<12 1551.) 125#.0) 2#2.3 k 4BB6<6 1;:0<81 120=<18 21:<0> k 4BB6<3 1453.332 1201.4# 21.53 k

4.4 Perhitungan Perencanaan Sloof dan Kolom Perletakan 4.4.1 4.4.1 Perhit Perhitunga ungan n per perenc encanaa anaan n Sloo Sloof f

Direncanakan dimensi sloof =

20

x

30

cm

Beban-beban yang bekerja pada sloof adalah : Berat sendiri sloof Berat dinding !#2 bata

0.2

0.3

200 =

! "g#m

3

2$0

=

%$0 "g#m

=

&' "g#m

(ehigga model pembebanannya seperti pada gambar berik)t : *

=

&' "g#m

L

+omen lapangan ,  

= !#2.*.2 &' 20.2$ = 2 =

+omen t)mp)an , - 

%$.3!3 "g.m

= !#!2.*.2 =

&' 20.2$ !2

= Desain tulangan utama

Tulangan lapangan ( M + )

!$0&./3 "g.m

!.

ρmin

=

!. =

=

f y

0.00

3$0

ρmaks = 0.%$ β1

0.&$.fc

/00

f y

/00  f y

0.&$ = 0.%$

2$

/00

0.&$ 3$0

=

/00 

3$0

0.02$ /00

a b

= β1

d /00  f y /00

= 0.&$

2/ /00 

= amaks

3$0

!32.0/3 mm

= %$4.a b

=

0.%$

!32 =

''

mm

1it)ng : +nada =

0.&$.fc.b.amaks

amaks d

2

= 0.&$

= +nperl) =

2$

200

''

2/

- '.$2

&2%0/'/3./$3 .mm +)

%$3!2$ =

Φ

=

'2&'0/.2$

.mm

0.&

"arena +nada 5 +nperl) maka dig)nakan t)langan t)nggal

f y m

3$0

=

= 0.&$.fc

6 n

= !/.%0/ 0.&$

+)

2$ %$3!2$

=

=

Φ.b.d

= 0.%%' +pa

2

0.&

200

/0$!/

1it)ng : 2.m.6 n

!

ρ

=

!

-

!

f y

m !

2$.//2/

=

!

-

!

-

!/.%0/ =

3$0

0.0022%

"ontrol rasio t)langan

ρ 7 ρmin 7 ρmaks

dig)nakan rasio t)langan minim)m

dengan demikian l)as t)langan yang diperl)kan adalah : 8s

= ρmin.b.d =

0.00

200

2/

= !'/.& mm2 dig)nakan t)langan

Φ

2

!2

=

Tulangan tumpuan ( M - ) !.

ρmin

=

!. =

f y

ρmaks = 0.%$ β1

=

0.00

3$0 0.&$.fc

/00

f y

/00  f y

22/ mm2

0.&$ = 0.%$

2$

/00

0.&$ 3$0

=

/00 

3$0

0.02$ /00

a b

= β1

d /00  f y /00

= 0.&$

2/ /00 

= amaks

3$0

!32.0/3 mm

= %$4.a b

=

0.%$

!32 =

''

mm

1it)ng : +nada =

0.&$.fc.b.amaks

amaks d

2

= 0.&$

= +nperl) =

2$

200

''

2/

- '.$2

&2%0/'/3./$3 .mm +)

!$0&/2$0 =

=

Φ

!&&$%&!2.$

.mm

0.&

"arena +nada 5 +nperl) maka dig)nakan t)langan t)nggal f y m

3$0

=

= 0.&$.fc

6 n

= !/.%0/ 0.&$

+)

2$ !$0&/2$0

=

=

Φ.b.d

2

= !.$$& +pa 0.&

200

/0$!/

1it)ng : 2.m.6 n

!

ρ

=

!

-

!

f y

m !

$!.32$!

=

!

-

!

-

!/.%0/ =

3$0

0.00/3

"ontrol rasio t)langan

ρmin 7 ρ 7 ρmaks dengan demikian l)as t)langan yang diperl)kan adalah : 8s

= ρ.b.d =

0.00/3

=

22&

200

2/

mm2

dig)nakan t)langan

Φ

3

!2

=

33' mm2

Analisis tulangan utama

Tulangan lapangan ( M + ) 8sterpasang

ρada

=

22/.!'$ =

=

b.d

'200

ρmin 7 ρada 7 ρmaks

OK!!!

8sterpasang.f y a

=

0.00/

22/

3$0

= 0.&$.fc.b

= !&./3 mm 0.&$

(ehingga momen nominal adalah a =

2$

200

.

nada

s

-

y

2 =

22/

3$0

2/ -

'.3!

= !&%3%''%.22%% .mm "ontrol keamanan penampang : +nada 5 +)

OK!!!

Tulangan tumpuan ( M - ) 8sterpasang

ρada

=

33'.2'2 =

=

b.d

'200

ρmin 7 ρada7 ρmaks

OK!!!

8sterpasang.f y a

0.00/'

33'

=

3$0

=

= 2%.' mm

0.&$.fc.b

0.&$

2$

200

(ehingga momen nominal adalah a +nada = 8s.f y d

2

=

33'

3$0

2/ -

!

= 2%$$3'%2.&%&% .mm "ontrol keamanan penampang : +nada 5 +)

OK!!!

Pemeriksaan tulangan geser

Besar gaya geser rencana adalah *. 9)

=

&'

.$

= 2

= 2

20!!.$

"g

Vu=1/2.q.L Vud

246 mm

Vu=1/2.q.L

2250 mm

Berdasarkan gambar di atas< maka besar gaya geser , 9 )d  adalah 9)d

22$0 -

2/

= 20!!.$

22$0 200

9)d

20!!.$

=

= !%'!.$&

"g

22$0 1it)ng  9n

9) =

!%'!.$& =

= 2'&/

;

0./

! 9c

2$

200

2/

fc.b.d =

= /

=

"g

/

!000

 = !00

"g

check : 2. fc.b.d

2

$

200

2/

= 3

3 =

9) =

!/000  =

!/00 "g

-

-

c

; =

-!!!.0

"g

>adi : !/00

5

-!!!.0

(Penampang ukup

Aheck nilai Φ.9c : 0./

!00

=

2/0

"g

5

!%'!.$& "g

dig)nakan t)langan sengkang praktis : diambil jarak sengkang s 7 d#2 s = 8@ =

!2

cm

b.s

=

!20

200

!20

3

8s = !#2.8@

!23 cm

mm

= 22.&/ mm2

= 3.f y

=

3$0

= !!.3 mm2

g)nakan sengkang ; &

-

!20

4.4. 4.4." " Kolo Kolom m Perle Perleta taka kan n

Direncanakan mengg)nakan dimensi kolom 0 cm x 0 cm< dengan panjang kolom perletakan = 2 m , 200 cm . dari hasil analisa (8? 2000 @ersi ' diperoleh : ?)

=

!!!!%.! "g

+) =

!02.!'

"g . m

9)

$&&./%

"g

=

Desain kolom perletakan

")at tekan rencana maksim)m , ;?n,maks  komponen str)kt)r tekan adalah :

Φ?n,maks

= 0.&0.Φ., 0.&$.fc.,8g - 8st  f y.8st 

8st

= )as )as t)la t)lang ngan an tota totall =

rt.8g

= 0.02

Φ?n,maks

=

32

=

0.&

=

0

0

cm2 0./$ 2!2.$ !/00 -

23!$0

32



!!2000

"g

"ontrol kek)atan penampang : ?)

≤ Φ?n,maks

!!!!%.! ≤

23!$0

OK!!!

6adi)s girasi dari penampang adalah :  r

2!3333.333

=

=

=

8

!!.$%

!/00

ata) r

= 0.3.h =

0.3

=

!2

0

Cnt)k komponen str)kt)r tekan dari rangka tak bergoyang< faktor panjang efektif , "  har)s diambil sama dengan !.0< ".)

! =

200 =

. r

!2

( Kelangsingan dia#aikan 

ksentrisitas dari penampang adalah : et =

+)

!02!' =

?)

= 0.%2$%%

cm

!!!!%.!

Cnt)k menghit)ng rasio t)langan dengan mengg)nakan grafik < maka besaran tak berimensi yang dib)t)hkan adalah : -

Cnt)k s) s)mb) @e @ertikal ?) "! =

Φ.8gr.0.&$.fc =

!!!!%.! 0.&$ !/00 0.&$ 2!2.$

= 0.$%/ -

Cnt) Cnt)k k s)m s)mb) b) hori horiso sont ntal al et "2 =

"!

0.%2$%% = 0.$%/

h

0

dari grafik diperoleh : r =

β = !

0.0!2

r = r.β =

0.0!2

=

0.0!2

(ehingga :

1

= 0.0!02

8st =

r 8g

=

0.0!2

=

!'.2

!/00 cm2

=

3 Φ

dig)nakan t)langan

!'20 !/

mm2

pada d)a sisi

Analisis Kolom Perletakan

)as t)langan terpasang adalah : 8s =

/ !#2.p.D2

=

/ 02.!2

= 2!2.% mm2 5

!'20

OK!!!

1arga r dari t)langan terpasang adalah : 8st

r =

2.!2% =

= 0.0!$0&

8g

!/00

6adi)s girasi t)langan terpasang adalah :

r r =

0.0!$0& =

= 0.0!$0&

E

!

Besarnya besaran tak berdimensi pada s)mb) @ertikal adalah : ?) "! =

Φ.8gr .0.&$.fc =

!!!!%.! 0.&$ !/00 0.&$ 2!2.$

= 0.$%/ dengan harga "! dan r di atas < maka dari grafik diperoleh

nilai "2 =

0.0!3

< sehingga : +r

"2

= ;.8gr .0.&$.fc.h

+r

= "2., ;.8gr .0.&$.fc.h  =

0.0!3

0.&$ !/00 0.&$ 2!2.$

0

= !2%%3& "g.cm "ontrol keamanan penampang dan pen)langannya : +r

≥

+)

!2%%3& ≥ !02!'

OK!!!

Pemeriksaan tulangan geser $

9) 9n

= 9)d =

$&&./% "g

9) =

$&&./% =

= '&!.!!%

; 9c

0./ )

fc

!.8g

/

b.d

= ! 

!!!!%!.

$

= ! 

00 !

!/0000

/

= !./2''' 0.%/& 00

3&%

= !'3&$&  =

!'3&$.& "g

check : 2. fc.b.d

2 !.$&

0

= 3

"g

3

3&.%

3&%

= 9) -

9c

!$03.&$$' "g

= '&!.!!% - !'3&$.&

; = -!&0.//& "g >adi : !$03.&$$' 5

-!&0.//&

(Penampang ukup

Aheck nilai Φ.9c : 0./

!'3&$.& = !!/3!.$ "g

5

$&&./% "g

dig)nakan t)langan sengkang praktis : diambil jarak sengkang s 7 d#2 s = 8@ =

!&

cm

b.s

=

!&0

00

!&0

8s = !#2.8@

mm

= /&.$% mm2

= 3.f y

= !'.3$ cm

3

3$0

= 3.2' mm2

g)nakan sengkang ; !0

-

!&0

2.2$

Calculation Sheet Pondasi Tower SUTT 150 kV

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