jiunkpe-is-s1-1980-273058-25172-kristal-chapter4
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IV . PERENCANAAN DAN PERHITUNGAN :I V . 1 . Perh itungan Bangunan Atas :IV. 1 * 1 . Pembebanan ;
Bentang t e o r i t i s ;jembatan = 37 ,0 m„ Pembebanan a k ib a t d a r i muatan hidup :
* Susunan muatan bergerak dan t e r b a g i r a ta digunakan P .M .U .J .J .R . No. 12 /1 9 7 0 .
* K o e f i s i e n k e j u t :- . 20 , . 20
5 o“ T T = 1 += 1 ,2 3 .
t 1.0 5,SO m\y____2 ^ir>y\'/>cJLox,iH'tnijy iti rrrrrrrrri,.i|jn 111!"I muct&uij imuuiiiui
nuiiumm I |J.l t H I ‘ 1111IIH111 |l|||| |[| |
1,0.1 b
1t/JcjLtH33MSJ i g a i i s i I tox/jq
mmmnnmm nnnnnn
g b . 4 . 1 . pembebanan a k ib a t m. h idu p .* Muatan hidup t e r d i r i d a r i :
a . Muatan t e r b a g i r a t a ;bentang 37 ,0 m -------> 30 m ( 1 < 60 m
P " 2 ’ 2 - EOT ( 1 - 30 >- 2 ,2 - ( 37 - 30 )■ 2 ,072 t / m ’ / J a l u r .
Mu; i tan g a r i s : bentang 3 7 ,0 m Muatan t r o t o i r :muatan hidup = 60% X 500 kg/m2 = 300 kg/m
I V .1 * 2 , RENCANA GELAGAR INDUK TENGAB
t------- ----- /jyotv—----------->
g b . 4 . 2 . g e la g a r induk tengah.
Luas g e la g a r induk (B) :B » 1 3 5 .1 7 ,5 + 2 . - . 1 5 . 1 0 + 1 4 2 ,5 .2 0 + 5 0 . 1 5
+ 2 . - . 1 5 . 4 0 « 6712,50 cm2 .
l e t a k g a r i e n e t r a l d a r i s i s i bawah :y . 6 7 1 2 ,5 - 1 5 . 5 0 . 7 , 5 + 4 0 .1 5 . ( 1 5 + 3 . 40 )
+ 1 4 2 ,5 .2 0 . ( 1 5 + 7 1 ,2 5 ) + 1 5 .1 0 . ( 1 5 7 ,5 - ^ . 1 0 ) + 1 3 5 .1 7 ,5 .1 6 6 ,2 5
y « 101,948 cm. j a d i V * 101,948 cm.
V'= 175 - 101,948 - 9 3 ,0 5 ? cm.Momen i n e r s i a :
I = 1 / 1 2 . 5 0 . 153 + 5 0 .1 5 . ( 1 0 1 ,9 4 8 - 7 , 5 ) 2+ 2 . 1 / 3 6 . 1 5 .4 0 3+ 1 5 .4 0 . ( 1 0 1 ,9 4 8 - 15 -
1 / 3 . 4 0 ) 2+ 1 / 1 2 . 2 0 . 1 4 2 , 53+ 2 0 . 1 4 2 , 5 ^(101 ,948 -1 5-71 , 2 5 ) 2 + 2 . 1/ 36 .3 ,5 .103
+ 1 5 .1 0 . (7 3 ,0 5 2 - 17 ,5 - 1 / 3 . 1 0 ) *+ 1 / 1 2 . 1 3 5 . 1 7 , 53 + 1 3 5 .1 7 ,5 . ( 7 3 , 0 5 2 -
8 , 7 5 ) 2= 25772722 cm2 .
IV.
S t a t i s momen terhadap g a r i s n e t r a l :S = 1 3 5 .1 7 ,5 .6 4 ,3 0 2 + 5 5 , 5 5 2 . 2 0 . 1 / 2 . 5 5 , 5 5 2
+ 2 . 1 / 2 . 1 0 . 1 5 . ( 7 3 , 0 5 2 - 17 ,5 - 1 / 3 . 1 0 )= 190606,51 cm3 .
W = | , = ~73^q'3'|~ = 352799,67 cm3 .
K = |p « 52 ,559 cm.
W 3 7 * “ T0T7578 K» = | - 37,661 cm.
Direncanakan l e t a k c . g * s . = 13 cm.
25772722 „ 252802>62 cm3 #
ek . 101 ,948 - 13 - 52 ,559 = 36 ,389 cm.
e '= 101 ,948 - 13 + 37,661 « 126,609 cm. k
1 . 3 . RENCANA GELAGAR INDUK DIPERLETAKAN ;US c*n-—
/47,»
Kaj &i&u„’ *“}v ““ •J.SJiJO
*•C
2-
V— 4'Otvn,—>gb . 4 . 3 . g e la g a r induk d i p e r l e t a k a n .
B = 1 3 5 .1 7 ,5 + 1 5 7 ,5 .5 0 « 10237,5 cm2 . Letak g a r i s n e t r a l d a r i s i s i bawah : y . 10237,5 = 50.1 5 7 , 5 . 1 / 2 . 1 5 7 , 5 + 1 7 ,5 .1 3 5 .
(175 - 1 /2 . 1 7 5 ) y = 9 8 ,94 cm. j a d i V = 98 ,94 cm.
V ’ = 175 -9 8 ,9 4 = 76 ,06 cm.Momen i n e r s i a :
9
• I = 1 / 1 2 . 1 3 5 . 1 7 , 53 + 1 3 5 . 1 7 , 5 . ( 7 6 , 0 6 - 8 , 7 5 ) 2+ 1 / 1 2 . 5 0 , 1 5 7 , 53 + 1 5 7 , 5 . 5 0 . ( 1 / 2 . 1 5 7 , 5 - 9 8 , 9 4 ) :
= 30253155 cm2 .S t a t i s momen terhadap g a r i s n e t r a l :
S » 1 7 ,5 .1 3 5 .6 7 ,3 1 + 5 0 . 5 8 , 5 6 . 1 / 2 . 5 8 , 5 6 = 244751,71 cm3 .
W' = 7 ' “ = 397753,81 cm3 .
K = |p = 38 ,853 cm.
W - I - = 305772,74 cm3 .W
98,94
I V .1 . 4 . RENCANA GELAGAR MELINTMG: <*----------------------- bm---------------- >
aocfw.
S" C»v : b =m 6 - ho + bo» 6 . 1 7 , 5 + 20« 125 cm.
g h . 4 .4 g e la g a r m e l in ta n g .
10
I V . 1 . 5 . PERHITUNGAN MOMEN iIV . 1 . 5 . A. PEMBEBANAN DAN MOMEN 3? A DA GELAGAR TEPI AKI
BAT MUATAN HIDUP.* Akibat beban hidup rnerata :
- ben tang t e o r i t i s 37 /n.- l e b a r la w ta i kendaraan 7 ,5 0 m ( t id a k terma
suk t r o t o i r ) .- t r o t o i r ?x a 1 ,20 m.- p = 2 ,072 t / m ' .
-tj a «P~) s'? -1,0? it/p" /yJU\ .
I—gb. 4 . 5 . beban hidup merata pada g e l . t e p i .
s ’ = l e b a r pengaruh beban hidup = 1,65 - 1 ,20 = 0 ,4 5 m.
p ' = 2 ,072 = 0 ,3 39 t / m ' .
Mp1= 1 / 8 . 0 , 3 3 9 . 372 . 1*23 = 71 ,365 tm.
* Akibat beban hidup g a r i s :____ ,P?Ut/ja.lu^m u ; TjUT
1 Ug b . 4 . 6 . beban hidup g a r i s pada g e l . t e p i .
s ' =« 0 ,4 5 m.P ' = Jt 75x ° » 4 5 - 1 ,964 t .
Mp2 1 / 4 . 1 , 9 6 4 . 3 7 . 1 , 2 3 » 22,345 tm.
* A k iba t beban hidup pada t r o t o i r :
'~I7T6,ho,s'i7r*
p = 60%.500 = 300 kg /n T . = 0 , 3 t /m2 .
g b . 4 . 7 . beban hidup pada t r o t o i r .
M 3 = 1 / 8 . 0 , 3 . 1 , 2 . 3 7 2 . 1 ,23 = 75 ,774 tm.M t o t a l g e l . t e p i =71,365 + 22 ,345 + 75 ,774= 169,35
p tm.
11
I V . 1 . 5 . B . PEMBEBANAN DAN MOMEN PADA GELAGAR TENGAHAKIBAT MUATAN HIDUP.
*Akibat beban hidup merata :,,0>t _ </v^
urI'vhs t '►VU'
SWjbVYnigb . 4 . 8 . beban hidup morata pada g e l . tengah,
e » 1 ,6 5 m.Dari t a b e l , f a k t o r d i s t r i b u s i = 1 , 6 5 . ( l i h a t p e n je la s a n & keten iuan tambahan P .M .U .J .J .R . No. 1 2 /1 9 7 0 ) .
Mp1 “ 1/ 8 . § ^ 2 . ~ * f | . 3 7 2 .1 ,23 =* 159 tm.
*Akibat beban hidup g a r i s jIji-CA*,, / ,_______ .gv.ro.Uv
tfe.
a*Y
g b . 4 . 9 . beban hidup g a r i s pada g e l . tengah.
MP2 ,/ 4 . ^ 5 . j ^ | | . 3 7 . 1 ,23 « 49 ,65 tm.
M t o t a l - 159 + 49 ,65 ■ 208,65 tm.Karena M t o t a l pada g e l . t e n g a h ^ M t o t a l g e l .
P pt e p i , maka dimensi g e l . t e p i disamakan dengan d imensi d a r i g e l . tengah.
12
I V . 1 *5 .C . PERHITUNGAN MOMEN AKIBAT BEBAN MATI :
- Akibat b . s . g e la g a r utama:g = 0 ,6 7 1 2 5 . 2 , 5 = 1 ,678 t / m ' .
Mg1= 1 /8 .1 , 6 7 8 . 372 = 287,169 t a .
- Akibat g e l . l i n t a n g yang d i c o r pada g e l . induk:P - 2 . 0 , 2 . 2 , 5 . ( 1 / 2 . 1 , 0 2 5 . 0 , 5 7 5 - 1 / 2 . 0 , 1 . 0 , 1 5 )
» 0 ,2 8 7 t .Mg2= 0 , 2 8 7 . 1 2 , 3 3 - 3,541 tm.
- Akibat p e r te b a la n pada g e l . utama:g - 1 / 2 . ( 0 , 9 2 5 + 1 , 4 2 5 ) . 0 , 1 5 . 2 , 5 . 2 = 0,881 t / m ' .
M 1 / 2 . 0 , 8 8 1 . 1 , 7 5 2 » 1 ,349 tm.
- A kibat g e l . l i n t a n g yang d i c o r kemudian:P = 1 / 2 . ( 0 , 3 + 1 , 4 5 ) . 1 , 0 2 5 . 0 , 2 . 2 , 5 - 0 ,4 4 8 t .
Mg4= 0 ,4 4 8 . 1 2 , 3 3 = 5 ,529 tm.
- Akibat p l a t d iafragma ( t e b a l 17 ,5 cm, l e b a r 30 cm)g * 0 , 3 . 0 , 1 7 5 . 2 , 5 «* 0,131 t /m * .
Mg 5 = 1 / 8 . 0 , 1 3 1 , 3 7 2 = 22,461 tm.3
- Akibat b*s» a sp a l s e t e b a l 6cm. dengan 2 t/m .g n 0 , 0 6 . 7 , 5 . 2 - 0 ,9 t / m ' .
Mgg« 1 / 8 . 0 , 9 . 372 , ^ - 25 ,668 tm.- Akibat pen inggian t r o t o i r (sebanyak 2 buah, t i n g -
g i 20 cm, l e b a r 120 cm, s e r t a = 2 ,4 t /m 3 ) .g » 2.1 , 2 . 0 , 2 . 2 , 4 = 1 ,1 5 t / m ( .
Mg ? » 1 /8 .1 , 1 5 . 372 .£• * 32 ,586 tm.
- Akibat sandaran, lampu d l l . :g d ia m b i l 0 ,1 5 t /m 1.
M g - 1 / 8 . 0 , 1 5 . 3 7 2 . ^ - 4 ,276 tm.
I V . 1 . 5 . D . PERHITUNGAN MOMEN TOTAL YANG BEKERJA :Momen t o t a l yang b e k e r ja :M = M « 287,169 + 3,541 + 1 ,349 + 5 ,529 + 22,461ulclX * ^
+ 25 ,66 8 + 32,856 + 4 ,2 78 + 2 0 8 , 6 5 = 591 ,501 tm.
Mmin = Mg = 287»169 + 5 » 541 + 1 »349 = 292,059 tm.
Gaya pratekane.n ? w = = 467,187 t .
S y a r a t : j p . B . CPb ^ Fw 4*fc "fc
^ ^ . 6 7 1 2 , 5 . 1 2 0 <Fw < l ^ 2 l § . 6712 f 5.120
336,247 < F w< 469,1 52' t .( O . K . ) .
1 . 5 . E . KONTROL TEGANGAN YANG TERJADI :4 6 7 ,1 8 7 .1 0 0 0 .^ 6 ,3 8 9
552799, t l- ^ i b a t Fw! 0 -^ =
■ 48 ,187 kg/cm .rr 1'. e , . , 4 6 7 . 1 8 7 . 1 0 0 0 . 1 2 6 . 6 0 9 Ubb= ~ k ----------~ 2'5~285'^752
- -2 3 3 ,9 7 7 kg/cin2 .M 5
- A k i ta t °~ba“ = - S 2 ' 783 k g / cmM 2 9 2 ,0 5 9 .1 0G .b= ^g= “ ^ 55^02 ,62 “ 115 ,528 kg/cm
M.- Akibat Mmax#: C7ba= r^= 591 ,501 .10 ' “ 167 ,66 kg/cm ‘
^ b= = 2 3 5 *977 k s /cm :M
DIAGRAM TEGANGAN:- Akibat V dan M. w nun.
+-
-3y,> fei/c,
14
- A kibat F dan w
NOTE:
Menurut T .Y .L in
M : max.
b i l a : ^ < 2 0 % - 30%.q
( p r o f i l s y m e t r i s ) .
« £ > 20% - 30%. q
( p r o f i l a s y m e t r i e ) .
Dari perh itungan d ia ta a :
S * = H r t w e10C)% " 4 9 ,3 7 5 % > 3° ^
Jadi anggapan k i t a bahwa p r o f i l a sym etr is benar*
IV. 1 . 6 . PERHITUNGAN MOMEN PAM TIAP PENJLMPANG :
- Akibat b . s . b a l o k :' = a k ib a t b . s . p r o f i l = 1 ,6 7 8 t /m * .
g 2 , , = a k ib a t p e r te b a la n = 0,881 t /m * .
rr-L i i QTT-frrj'H i i ,i i i,i i i i i i it t t : i i i i m i i i i i i i i h
........................ .......... L -i-L L-
g b . 4 . 1 0 . momen a k ib a t b . s . p r o f i l .
Mx = 1 / 2 . g l ' . 1 . x - 1 / 2 . S 1 ' . x 2 = 1 / 2 . g 1 ' . x . ( l - x )
(.,7 f v ■ 0 , - x . d - X )
tgir*---------------- ----------------- ----- ----- ---------------------------------------- mtm> a A *, S i -f 8 jr------- • Vtv-
W i t rg b . 4 .1 1 . momen a k ib a t p e r te b a la n .
M1=M2=M3*M4«M5=M6«:M7»M8=M9 = 1 ,542.1 ,75
= 1 ,350 tm.
- Akibat g e l . m e l in ta n g :P « 0 ,7 35 t .
J, *!* 'Vf*
f Itgb . 4 . 1 2 . momen a k ib a t b . s . g e l . m e l i n t a n g .
Mx = P .x -------------------- (x $: 1., ) .Mx » P .x - P (x - 1 1 ) - * ( x > 1 1 ) .
= P .x - P .x + P . l^= p . l 1
- A k ibat beban mati sekunder:( a s p a l + t r o t o i r + sandaran + p l a t p e n g i s i ) .
g - 1 / 6 . ( 1 , 1 5 + 0 ,1 5 + 0 , 9 ) + 0,13': = 0 ,497 t/m*p £ c */>"'
JTTTTI I I I I I I I M I M l l I I i l‘l l l I I II I I ) M I ~EXl.t-U_LJJ*«. %---------- : » ------- . , • , -------U-‘ J - I— :--------
gb . 4*13. momen a k ib a t bebar s e k ^ id e r .
Mx = 1 / 2 . g . 1 . x - 1 / 2 . g . x 2 = 1 / 2 : . g . x . ( l - x )= C g . x . (1 - x ) .
- Akibat beban g a r i sP a ^ 75 .1 *23 « 5 ,3 6 ? t .
P
. ~ -------_ . H — ........... .. ...... ...............f---------------------------------- ------£ *i};0 K,----------------- ----------------------!-
gb* 4 »14 . momen a k ib a t beban g a r i s .x.
Mx - “ - j —“ »P*x - | * x . ( l - x )
» c ^ . x ( l - x ) .
- Akibat beban hidup t e r b a g i r a t a ;
p« ^ 7 2 . 1 ^ 2 3 = 6 ,927 t /m * .f * Oj yzj -fyW
p r m n i n i o : .»’.i i j J i - U T i m j i i m m m i i r q-------.. _ . .........— - - r = - f
X a g b . 4^15^ momen a k ib a t beban t e r b a g i rata*
Mx *» 1 / 2 . p . 1 . x - 1 / 2 . p . x 2 = 1/ 2 . p . x . ( l - x )- C<j#X> ( l "* x ) *
I V . I . 7 . RENCANA LETAK TEWDON DAN KABEL ;
Digunakan angker hidup type 12 S c , 2' : 5 x 21 5 mm'sedangkan angker raati type 12 P
ir------------ ;------- MS City " ------ ' 200 x 200 mm2
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y....— —g b . 4*16 . rencana tendon d i u jung b a lo k ,
i--------------------
19
PERSAMAAN LENGKUNG TENDON :
* Arah v e r t i c a l :1 = 37 ,0 m.Kabel A:f * 30 - 8 ,8 5 = 21 ,15 cm = 0 ,2115 m.„ 4 . f . z . ( l - z ) 4 . 0 , 2 1 1 5 . z . (37 - z ) y a - r ------ = --- e— n-----s------- u
= 0 ,0229 z - 0 ,000618 z 2
= 0 ,0229 - 0 ,001234 z .
Kabel B :f - 55 - 8 ,8 5 = 46 ,15 cm = 0 ,4615 m.yB= 1 ^ 1 615 t 2 «,{37 - 2 ) = o ,0499 z - 0 ,00135 z 2
37dyB^ = 0 ,0 49 9 - 0 ,00270 z .
Kabel C:f = 80 - 8 ,8 5 = 71 ,1.5 cm = 0 ,7115 m.y c = 4 - * ? 1 1 = 0 ,0769 Z - 0 ,002 08 z 2
37d y Pg " 1 = 0 ,0769 - 0 ,00416 z .
Kabel D :f = 105 - 8 ,8 5 = 71 ,15 cm = 0 ,7115 m. y = 1 . 0 ^961 5,.Z_ J 37_.- , z ) = 0>1039 z _ 0 ,0 0 281 z 2
37<3yD
= 0 ,1039 - 0 ,00562 z .
Kabel E:f = 130 - 2 1 ,3 = 108,7 cm = 1,087 m.
37. s 'Wy E= i - l ’0 8 1 = 0 , 1 1 7 5 z - 0 , 0 0 3 1 8 z:
dy™= 0 ,1 175 - 0 ,00635 Z .
Kabel F :f - 155 - 2 1 ,3 = 133,7 cm = 1 ,337 m.
4^ ,,337^ : i , y 7 . - _z ) = 0 > U 4 5 z _ Qj37
dyp^ = 0 ,1445 - 0 ,0 07 8 z .
Arah horizontal :Kabel A = Kabel B :f = 16 ,15 cm = 0 ,1615 m.„ „ 4 . f . z . ( l - z ) 4 . 0 , 1 6 1 5 . z . (37x a=xb ----------- j r ------------ --- — — p r —
» 0 ,0 1 7 5 z - 0 ,000472 2 Z dx . dxwd i ^ = d T 1 = ° » 0175 - 0 ,00094 z .
Kabel 0 = Kabel D = Kabel E = Kabel Ff = 1 / 2 . 7 , 7 = 3 ,8 5 cm = 0 ,0385 m.„ 4 . 0 , 0 3 8 5 . z . (37 - z )c d e f ~ — ~ r ------37
= 4 , 1 6 2 . 1 0 " 3 . z - 1 ,1 2 5 ,1 0 H dxp dxT) dx-rj, dx™ ,
4,162.10-* - 2,25.10*
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22
I V . 1 . 8 . PERHITUNGAN KEHILANGAN GAYA PRATEKANAN :
I V . 1 . 8 . A. Akibat susut (krimp) pada b e t o n :Ditentukan b erd a sa r R . V . B . ' 6 7 a r t* 8 . 2 . 1 . t ' k c o = 3 . 1 0_ 7 . p . (100 - R)dimana p = banyaknya pasta semen dalam v o l . %
R = kelembaban r e l a t i f p » 0 , 1 0 . ( w c . g c + &c / f Q)
w = water cement r a t i o = 0 ,475C 'Zg c - kadar semen = 400kg/m .2fc = b e r a t j e n i s semen = 3,15
p = 0 , 1 0 . ( 0 , 4 7 5 . 4 0 0 + 4 0 0 /3 ,1 5 )= 31,698%.
R = kelembaban r e l a t i f , diarnbil 75% (k o n s t ru k s i d i udara t e rb u k a ) .
i* = 3 . 1 0 "7 . 3 1 ,6 9 8 . (1 0 0 - 75)« 2 3 ,7 7 .1 0
Untuk k o n s t r u k s i beton pratekan dengan metode p o s t - t e n s i o n i n g , penyusutan yang t e r j a d i dapat d i r e d u k s i , dengan menggunakan perumusan menu- r u t R . V . B . *67 a r t 8 . 2 . 5 .i \ t ■= 0 . 0 0 5 . 0 . 4 ^
Dimana t diarnbil 28 h a r i , yakni jumlah h a r i s a -a t pengecoran sampai penegangan.£ «k t = 0 ,0 0 5 .2 8 .2 3 ,7 7 .1 0 " -^
= 3 , 3 2 8 . 1 0 “ 5< 0 , 4 . 2 3 , 7 ? . 1 0 -5< 9 , 5 1 . 1 0 “ 5
(O .K . )O _ <■ » - S Ik ~ £ ko, kt
= (2 3 ,77 - 3 , 3 2 8 ) . 1 0 ‘ 5 = 2 0 ,4 4 2 .1 0 ~ 5 (nrnrupakan perpendekan s p e s i f i k yang d ip e r h i tu n g k a n )
Ea = 1 ,9 8 .1 0^ kg /cm 2 = 1 ,9 8 .1 0 ^ kg/mm2 . ( l i h . b rosu rV . S . L . ) .
. Jadi kehilarigan t e g . a k ib a t susut beton == 2 0 ,4 4 2 .1 0 ~ 5 . 1 , 9 8 . 1 0 4 = 4 ,047 kg/rnm .
23
I V . 1 . 8 . B . Akibat c re e p b e t o n :D itentukan b e rd a sa r R .V .B .*67 a r t . 8 . 3 . 1 .
£ ' k r c ? = 100 ' ( 4 b - 0,36R5<r, w2Q ^ ’ k^>" ^ 'k t ^ *dimana CP bag = t e g . tekan i n i t i a l r a t a - r a t a d i
suatu k on s tru k s i*R = kelembaban r e l a t i f , d iam bil 75%.
CT‘ w28 = kokoh tekan kubus umur 28 h a r i * 400 kg/cm2 ’
£ 'k = £ \ c ? ~ £ ' k t * nieerupakan p e r p e n - j dekan s p e s i f i k be ton a k ib a t susut .
^ b a g = dimana : F& = 1,15*4.67,187 = 537 ,3 t*B = 6712,5 cm2 *
(T'bag = = 8 0 ,0 kg/cm2 .£»k = 2 0 ,4 4 2 * 1 0 "5 .J a d i , d i p e r o l e h :
^ ' k r o? = 100*"("40°- 0 , 3 . 7 5 ) .4 0 6 * 2 0 »442 *10 5 = 2 3 ,3 6 2 .1 0 “ 5 .
En = 1 ,9 8 * 1 0 4 kg/mm2 *clJad i keh i lan ga n tegangan a k ib a t c re e p b e t o n :
= 2 3 , 3 6 2 . 1 0 " 5.1 * 9 8 .104 = 4 .626 kg/mm2 .
I V . 1 * 8 . C. Akibat r e l a x a s i b a j a :D ih itung berd a sar b r o s u r V .S .L . a r t* 4 « 2 .2 * c *_. 2Tegangan b a ja p u tu s , Oar = 189,46 kg/mm ( l i h .digram 0~- £ pada la m p ir a n ) .Tegangan b a ja i n i t i a l yang d i i j i n k a n = 0 , 7 . ^ a r
( f a i = 0 , 7 . 1 8 9 , 4 6 = 13 2 ,68 kg/mm2 ,Besarnya susut dan c re e p beton yang t e r j a d i s e t e l a h pratekanan ad a la h :
£*k + £ ' k r ^ = C2 0 »442 + 2 3 , 3 6 2 ) . 1 0 ' 5 = 4 3 ,8 0 4 .1 0 “ 5 .
Kehilangan t e g . a k ib a t susut dan c re e p b e ton *= 4 3 , 8 0 4 . 1 0 " 5 .1 , 9 8 . 1 0 4
___________________=8.673 kg/mm2 .________________________________
24
% keh i langan t e g . a k ib a t susut dan c re e p == Tf 100% = 6,521%*
Jadi % keh i langan t e g . a k ib a t r e l a x a s i == 14(1 - = 13,087%.
Jadi keh ilangan tegangan a k ib a t r e l a x a s i == 3 2 ,6 8 = 17 ,364 kg/mm2 .
I V . 1 . 8 . D , Penin.jauan kembali .jmlh. keh ilangan t e g . s e r t a k o n t r o l t e g , a k ib a t Jumlah keh i la ngan tegangan: p
- a k ib a t susut = 4 ,047 kg/mm'p- a k ib a t c re e p = 4 ,626 kg/mm2
- a k ib a t r e l a x a s i = 17 ,364 kg/mm------------- --------------- _+
T o ta l d a r i = 26 ,037 ■kg/mm .Penentuan juinlah k a b e l :
Direncanakan 6 buah tendon type 12 Sc dengan jumlah kawat unta ian 8 buah p e r - t e n d o n .Luas k a b e l t o t a l = 6 .7 90 = 4740 ram2 .Fw = 467 ,187 t . _0;w= 9 8 - 562 kS/ ™ 2 4 CT w= 0 , 55 .<Tar
= 104,203 kg/mm ,T ota l d a r i a CT= 26 ,037 kg/mm2 .Jadi tegangan t o t a l yang dibutuhkan =
= 36 ,037 + 98 ,562 = 124,599 kg/mm2 , « r a i = 132 ,680 kg/mm2 ) .
I n i t i a l F orce -ya ng d ib u tu h k a n (F ^ ) :Fa = 124 ,5 99 .4740 = 590,599 to n .
Kekuatan t a r i k max. 6 tendon = 6 .1 05 = 630 t .( > F ± = 590,599 t o n ) .
K ontro l t e g . beton a k ib a t( d i t i n j a u penampang tengah)
Akibat F± dan *• p.(T Fi * Gk x M * 3 9 0 5 9 9 . 3 6 , 3 8 9 . 2 9 2 . 0 5 9 . . 1oZ
ba = " T 1 vr = 3527997570 3 5 2 ^ 9 , ™= - 2 1 ,B67 kg /cm2 .
25
rr- _ Fi > ek< \ -590599>1 26 .609 . 292, 0 5 9 . 103 bb “ W + W = £ 5 ^ 0 2 , 6 2 + 252802,62
= -1 8 0 ,2 5 6 kg /cm2 ><fbi = 168 kg/cm2Akibat P. dan M,------------ ^ “ ek------ftla x * 5^ i q 59 0599 .36 ,389 5 9 1 ,0 5 1 .105
b a = “i p ~ ~ + w = — y s m r fr i— y & i w & r ~17 evt m = -1 0 6 ,6 1 5 kg /cm 2^.Crb/ = 120 kg /cm2 ,
rr _ i . q - 5 9 0 5 9 9 .1 2 6 ,6 0 9 . 591 ,051 . 10 bb V? + W = "2 $2 8 o£ ,6 2 + 252862 ,62“
= -6 1 ,9 8 5 kg/cm2 .Karena Obb (a k ib a t I\ & ^ m i n ^ maka d i lakukanpengejackan secara b e r ta h a p .Pengejackan 100% di lakukan s e t e l a h p l a t diafragma s e l e s a i d i c o r .Jad i Mg = 292,059 + 5 ,529 + 22,461 + 32,856
= 352,905 tm.K on tro l t e g , yang t e r j a d i :
F j . e k ’ -5 9 0 5 9 9 .1 2 6 ,6 0 9 . 3 5 2 , 9 0 5 . 105 u bb 83 T ~ + w = ...25-28^73'i------ +
= -1 5 6 ,1 9 kg /cm 2 <C<7bi = 168 kg /cm2(O.K. )
I V . 1 . 8 . E . Kehilangan pra-tekanan a k ib a t wobble e f f e c t dan gesekan antara tendon dengan c o n d u i t :D ih itung berdasar b r o s u r V .S .L . a r t 4 . 2 . 2 . e , dengan formula : Pv . + "1 + ? l x >A Xdimana Fx = gaya pra - tekanan pada jarak l x d a r i u j .
k i r i .= goya pra -tekanan yang d ik e r ja k a n .
/*> ~ k o e f i s i e n gesekan.= perubahan sudut dalam rad . da r i ja ck ke
t i t i k Z yang d i t i n j a u (dim. arah v e r t . ) .= k o e f i s i e n gesekan p a r a s i t e r .
(rj = perubahan sudut da.lain ra d . d a r i j a c k ke t i t i k Z yang d i t i n j a u (dim. arah h o r i z . ) .
Dari b r o s u r V .S .L . t a b e l 4 . 2 . 2 . d i p e r o l e h :
= 0 , 2 0 .^ = 0 , 0 1 0 .
Pengejackan dimulai d a r i kabel C,D,B,E,A Kabel D,B dan P d i Jack d a r i ujung k i r i . Kabel C,E dan A d i Jack d a r i u jung kanan
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I V . 1 .8 .1 ’ . Perhitungan s l i p anpker :D ih itung b erda sar b j jospr V .S .L . a r t . 4 . 2 . 2 . g , dengan formula.: S = m.x
17T Ta cDari b r o s u r t- horga s l i p ( £ ) , d iam bil = 6 mm.
g b . 4 . 1 8 . diagram tegangan.* Kabel A :
B . 1 0 2 9 6 3 ,^ 5 ^ - . 9408 8 ,09 . 2>399kg/cnl.
'i RT eT TT W 0 , 6 . 1 .9 8 .1 0 S . 7 , 9x =V ma~ a _ * V ' ---------
= 1978 cm "^1 /2 . ben tang = 1850 cm.(d ip e r lu k a n tambahan gaya p r a - tekanan, untuk meng- a t a s i a k ib a t s l i p d itengah b e n ta n g ) ,AF = 2mx = 2 . 2 ,3 9 9 .1 9 7 8 = 9490,44 kg .AF s l i p
J,9,7 8 ^ ^ 1 8 50 . 9490^44 = 6 1 4 , 1 4 k g .
- rj?<? Cfy -
18 So
gb . 4 . 1 9 . s l i p pada kabel A.c
* KABEL B
m = 10: 5 3 7 ^ - , 9 3 5 9 1 ,7 2 = 2>68? kg/cm.
:6~7X =y 0 *6 •1 = 1868,91 cm > 1850 cm.
(d ip e r lu k a n tambahan gaya pra*»tekanan).
29
F * 2.2,687•1868,91 = 10039,8 kg.
Fs l i p * 1868f e ] 85° ^ 0° 3 9 . 8 * 101 *60 k8-
g b . 4 .2 0 . s l i p pada kabel B,*KABE1 C :
m
x =\/ = 1820,75 cm < 1850 cm.
AF = 2 .2 ,8 3 1 .1 8 2 0 ,7 5 = 1 0 3 0 9 ,1 kg .
tSio^sa^rg b . 4 . 2 1 . s l i p pada kabe l C.
*KABE1 D :
m
x
- l M ^ ,,r .9 ? 8 2 6 ^ ,= 5#12 kg/cm<
J\f 0 j 6 . l ^ l g T T o f . 7 19 = 1734140 Cm < 1850 cm.
AF = 2 . 3 , 1 2 . 1 7 3 4 , 4 0 = 10822,65 kg .
30
| ig&iz, Cs kg
'S-a-U
4------ ----------y ---------^gb . 4 . 2 2 . s l i p pada k a be l D.
» KABEL E : m = 1104669j 4 1 D- , 9258 ^ 6 4, 3>2g kg/cBU
x AJ y1Q6 • 7-S-2 = 1696 ,73 < 1850 cm.
AF = 2 . 3 , 2 6 . 1 6 9 6 , 7 3 = 11062,68 kg .//063,6a j
liSOCtn
g b . 4 . 2 3 . s l i p pada kabe l E.*KABEL F :
m . 105231 j 0 3 n- , 9 ? q m £ . 3 , 555 kg /cm .
x ^9 | ^ ° •7j 9 = 1624,81 < 1850 cm.
AF = 2 .3 ,5 5 5 .1 6 2 4 ,8 1 = 11552,40 kg .
31
lb l^S I u*v-gb. 4 .2 4 . s l i p pada k a be l F.
I V . 1 . 8 . G . KEHILANGAN GAYA PRaTEKANAN AKIBAT PERPENDEKANELASTIS BETON.Ea = 1 ,9 8 .1 0 ^ kg/cm2 .e" =v 3 3 3 .1 0 3 kg/cm2 ,
a 1 , 9 8 . 106 c = 5$3b06 = 5 ,9 4 .
* Kabel G d i j a c k du lu , jurnlah gaya yang menyebabkanhi lan gnya pra-tekanan kabel G :£F- = 593149,23 - 93336,06 = 499813,17 kg.
* n .F - 5 ,9 4 .4 9 9 8 1 3 .1 7 . . . ™ ,__/__2 A i *TT~ ..Z7\ ? , $— — = 442 *293 kg/cm .
* S e te la h i t u kabel I) d i ja c k .t F = 499813,17 - ,104371 ,89 = 395441 ,28 kg.
- 4 = - 5 4 9 ,9 y k g /= m 2 .
* Kemudian kabel B d i jn ck .= 395441,28 - 103537,5? = 291903,76 kg.
5 . .94 .291903.76 „ m , . . / . _ 2 = —*—giyyg-- c,---- *---- = 258,31 kg/cm .
* Menyusul kabel E d i Jack.■$F. = 291903,76 - 92582 ,64 = 199321,12 kg.
5, ^4.199321,12 2= "17HT~,"5— — = ’ '®»382 kg/cm .
* Baru kemudian kabel A d i ja ck .<F. = 199321 ,1? - 94088 ,09 = 105231,
5,94.105231 ,03 n7 ^n,../..2 = " " 6?'ii,'5— — ~ 93,1 22kg/cm .* Terakhir kabe l F d i ja c k ,
03 kg.
Atf i = o .
32
Jadi pa d a .masing-masing kabe l mengalami keh i langa n pra tekanan s e b e s a r :- Kabel C = 4 4 2 ,2 9 3 .7 ,9 = 3494,11 kg.- Kabel D = 3 4 9 , 9 3 . 7 , 9 = 2764,45 kg.- Kabel B = 2 5 8 , 3 1 . 7 , 9 = 2040,65 kg.- Kabel E = 1 7 6 ,3 8 2 .7 ,9 = 1393,42 kg.- Kabel A = 9 3 , 1 2 2 . 7 , 9 = 735 ,67 kg.- Kabel P = 0 kg.
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34
RINGKASAN :
KABEL GAYA JACKING (KG) TEG. JACK (KG/MM2 )
A 103577,66+735,67= 104313,33 104313,33 /790= 132 ,04
B 103639,12+2040,65= 105679,77 105679 ,77 /790= 133,77
C 103810,52+3494,11= 107304,63 107304 ,63 /790= 135,83
D 104371,89+2764,45= 107136,34 107136 ,34 /790= 135,61
E 104660,41+1393,42= 106053,83 106053 ,83 /790= 134 ,24
F 105231,03 + 0 = 105231,03 105231,03 /790= 133 ,20
Menurur b ro s u r V .S .L . a r t . 4 . 2 . 3 maka t e g . a k ib a t o v e r dapat d iam bi l 75% d a r i 0"ar.J a d i s y a r a t t e g . jack d i a t a s ^ o v e r t e n s i o n i n g = 0,75(>ar
< 6 ,7 5 .1 8 9 ,4 6 < 1 42 ,0 95 kg/mm2 .
Dari t a b e l d i a t a s tern yata teg* ja ck t id a k ada yangmelampaui t e g . yang d iperbo leh kan , j a d i perencanaank i t a memenuhi s y a r a t .
35
I V . 1 . 9 . Penentuan Daerah L im it Kabel :I V . 1 . 9 . A. Perhitungan l e t a k e . g . s . t i a p - t i a p potongan :* U.iung p e r le ta k a n k i r i :
J.L
— — c <\.c .- -/V-
& OK.J4. » '
g b . 4 . 2 5 . l e t a k k a be l d i ujung k i r i ,Kabel B D EF, 94702,23 93599,32 93336,06 93549 ,23 92582,64 93678,63
w 75041,55 71964 ,08 73832 ,50 71739 ,65 73236,50 71689,53Jarak d r a la s
30 55 80 105 130 155
a b s i s x .(cm)
Letak e . g . s . F ( d a r i s i s i bawah):y ± . 561408,12 = 9 4 70 2 ,23 .30 + 9 3 5 99 ,32 .5 5+ 9 3 3 36 ,0 6 .8 0+
9 3549 ,23 .10 5 + 9 2 58 2 ,6 4 .13 0 + 9 3 67 8 ,6 3 .1 5 5 y A= 9 2 ,3 3 cm.
Letak c . f t . s . F ( d a r i s i s i b^wah):y w. 437503,81 = 75041 ,5 5 .3 0 + 7 1 9 6 4 ,0 8 .55+7383:2 , 50.80+
7 1739 ,65 .1 05 + 73 23 6 ,5 .1 3 0 + 71 6 89 ,5 3 .1 55 y w= 92 ,07 cm.
35 .
* Pbtongan 1 :*
Kabel A Bg j ( k g ) 95134 ,68 94042,44 93792,82 94067,20 93116,411 94259 ,95^ ( k g ) 75383,64 72511 ,05 74193 ,80 72377,54 73658,73 72409 ,35Jarak dr a l a s 26 ,18 46,68 67,18 87,68 110,4 130,90
a b s i s xI cm ) •2,91 2,91 - 0,70 0 ,7 0 - 0,70 0 ,7 0
Letak e . g . s . F-j ( d a r i . s i s i bawah) :
y ± . 564413,5 = 9 5 1 3 4 ,6 8 .2 6 ,1 8 + 9 4 0 4 2 ,4 4 .4 6 ,6 8 + 9 3 7 9 2 ,8 2 .6 7 , IE +94067 ,2 .87 ,68+93116 ,41 .11 0 ,4 +94 259 ,95 .1 3Q9
y A = 78 ,04 cm.
Letak e . g . s . F^ (d a r i s i s i bawah) : y w. 440534,11= 75383 ,64 .26 ,18+72511 , 0 5 . 4 6 , 6 8 + 7 4 1 9 3 , 8 . 6 7 , 1£
+ 7 23 7 7 ,5 4 .8 7 ,6 8 + 7 3 6 5 8 ,7 3 .1 1 0 ,4 + 7 2 4 0 9 ,3 5 . 130 ,9
y w = 77 ,86 crnl
37
* Potongan 2 :
Kabel B DFjQcg) 9.5587,52 94580 ,43 94374,17 94678,32 93765,30 94259 ,95
|*w(kg) 75741,97 73172 ,34 74563 ,67 73134 ,78 74172,04 73286,46Jarak dr a l a s 2 2 ,10 37,79 5 3 ,48 69 ,16 8 9 ,4 8 105,15
a b s i s x (cm) -6 +6 - 1 , 4 3 +1 ,43 - 1 , 4 3 +1 ,43
Letak e . g . s . F^ ( d a r i s i s i bawah);567960,08 = 9 5 5 8 7 ,5 7 .2 2 ,1 0 + 9 4 5 8 0 ,4 3 .3 7 ,7 9 + 9 4 3 7 4 ,1 7 .5 3 ,4 8
+ 9 4 6 7 8 ,3 2 .6 9 ,1 6 + 9 3 7 6 5 ,3 .8 9 ,4 8 + 9 4 2 5 9 ,9 5 .1 0 5 ,1 5y A. = 62 ,65 cm.
Letak e . g . s . F ( d a r i s i s i bawah):7 5 7 4 1 ,9 7 .2 2 ,1 0 + 7 3 1 7 2 ,3 4 .3 7 ,7 9 + 7 4 5 6 3 ,6 7 .5 3 ,4 8
+ 7 3 1 3 4 ,8 .6 9 ,1 6 + 7 4 1 7 2 ,0 4 .8 9 ,4 8 + 7 3 2 8 6 ,4 6 .1 0 5 ,1 5yw. 444071,26
y w = 62 ,65 cm.
* Potongan 3
Kabel B D EJ,i (k g ) 96081,94 95139,09 94945,12 95320,75 94414,20 95687 ,25Fw(kg)(76132,92 73849 ,98 75105 ,32 73916 ,78 74685,34 74155 ,25
Jarak Ir a l a s 18,56 30,10 41 ,60 53 ,13 71 ,35 82 ,80
i b s i s x (cm) - 8 ,7 4 8 ,7 4 - 2 , 0 8 2 ,0 8 - 2 , 0 8 2 ,0 8
Letak c . g . s . F ( d a r i s i s i bawah):y ± .5 7 1 588,35 » 96081 ,9 4 .1 8 ,5 6 + 9 5 1 3 9 ,0 9 .3 0 ,1 0 + 9 4 9 4 5 ,1 ? .4 1 ,6 0
+95320-,7 5 . 5 3 , 13+944M ,2 .71 ,3 5 + 9 5 6 8 7 ,2 5 .8 2 ,8 0 y L - 49 ,55 cm.
Letak c . g . s . F ^ ( d a r i s i s i bawah):y w. 447845,59 « 7 6 1 3 2 ,9 2 .1 8 ,5 6 + 7 3 8 4 9 ,9 8 .3 0 ,1 0 + 7 5 1 0 5 ,3 2 .4 1 ,60
+ 7 3 9 1 6 ,7 8 .5 3 ,1 3 + 7 4 6 8 5 ,3 4 .7 1 ,3 5 + 7 4 1 5 5 ,2 5 .8 2 ,8 0
w 49 ,47 cm.
38
* Potongan 4 :
Kabel A B G D £ F
F jO tg ) 96476,17 95677,06 95516 ,08 95963,17 95084,02 96392,65*wOtg) 76523 ,88 74294 ,26 75556,97 74698 ,78 75215,20 75032 ,36Jarak d r a l a s 15 ,57 23 ,58 3 1 ,5 8 3 9 ,57 56 64
a b s i s x (cm) -11 ,03 11 ,03 - 2 , 6 2 2 ,6 2 - 2 , 6 2 2 ,6 2Letak e . g . s . l\ ( d a r i s i s i bawah :y j . 575209,15= 9 6 5 7 6 ,1 7 .1 5 ,5 7 + 9 5 6 7 7 ,0 6 ,2 3 ,5 8 + 9 5 5 1 6 ,0 8 .3 1 ,5 8
+ 9 5 9 6 3 ,1 7 .3 9 ,5 7 + 9 5 0 8 4 ,0 2 .5 6 ,0 + 9 6 3 9 2 ,6 5 .6 4 ,0 y j = 38 ,36 cm.
Letak e . g . s . ? w ( d a r i s i s i bawah):y w. 451321,45= 7 6 5 2 3 ,8 8 .1 5 ,5 7 + 7 4 2 9 4 ,2 6 .2 3 ,5 8 + 7 5 5 5 6 ,9 7 .3 1 ,5 8
+ 7 4 6 9 8 ,7 8 .3 9 ,5 7 + 7 5 2 1 5 ,2 0 .5 6 ,0 + 7 5 0 3 2 ,3 6 .6 4 ,0 y w = 38,33cm.
* Potongan 5:
Kabel A B C D E . PPjCkg) 97060,10 96215,01 96097 ,42 96574,30 95764,32 97117 ,55
76906 ,68 75172 ,55 76016 ,84 75456,01 75753,34 75917 ,79Jarak i r a l a s 13 ,13 18 ,26 2 3 ,3 8 28,49 43 ,50 . 48 ,50
a b s is x (cm) -1 2 ,9 0 1 2 ,90 - 3 , 0 6 3,06 - 3 , 0 6 3 ,0 6
Letak e . g . s . I’ , ( d a r i s i s i bawah ) :y i . 578828,70= 9 7 0 6 0 ,1 0 .1 3 ,1 3 + 9 6 2 1 5 ,0 1 .1 8 ,2 6 + 9 6 0 9 7 ,4 2 .2 3 ,3 8
+ 9 6 5 7 4 ,3 0 .2 8 ,4 9 + 9 5 7 6 4 ,3 2 .4 3 ,5 0 + 9 7 1 1 7 ,5 5 .4 8 ,5 0 y^ = 2 9 ,20 cm.
Letak e . g . s . F.. ( d a r i s i s i bawah ) :y w-455 223,21= 7 6 9 0 6 ,6 8 .1 3 ,1 3 + 7 5 1 7 2 ,5 5 .1 8 ,2 6 + 7 6 0 1 6 ,8 4 .2 3 ,3 8
yw = 2 9 ,1 8 cm.
39
■* Potongan 6 :
Kabel A B C I) E Fp i ( k g ) 97554 ,33 96773,71 96668 ,40 97227 ,15 96423 ,70 97831 ,9;i:fw(k g ) 77297 ,64 75850,20 76468,50 76246,27 76274,92 76794 ,90Jarak d r a las 11 ,22 14 ,13 1 7 ,0 19 ,89 3 3 ,8 36 ,5Lbsis x (cm) - 1 4 ,3 5 14,35 - 3 , 4 3 ,4 - 3 , 4 3 ,4
Letak e . g . s . F_ ( d a r i s i s i bawah):y i . 582479,22*: 97554 ,33 .1 1 ,22+96773,71 . 1 4 ,1 3 + 9 6 6 6 8 ,4 0 .1 7 ,0
+ 9 7 2 2 7 ,1 5 .1 9 ,8 9 + 9 6 4 2 3 ,7 0 .3 3 ,8 0 + 9 7 8 3 1 ,9 3 .3 6 ,5 y i = 22 ,09 cm.
Letak e . g . s . P._, ( d a r i s i s i bawah):yw. 458932,43= 77297,64 .1
+76246,27.1 y w = 22 ,02 cm.
1 ,22+75850 , -2 0 .1 4 ,1 3 + 7 6 4 6 8 ,5 0 .1 7 ,0 9 ,8 9 + 7 6 2 7 4 ,9 2 .3 3 ,8 0 + 7 6 7 9 4 ,9 0 .3 6 ,5
* .Potongan 7 :
]iabel A B G D E F^ ( k g ) 98048 ,55 97322,03 97270 ,50 97859,15 97083,04 9 8 5 77 ,8«'^ ( k g ) 77688 ,59 76519,67 76944,77 77020,02 76796,50 7 7 6 9 6 ,8<Jarak
'ir a las 9 ,8 7 11 ,20 12 ,48 13 ,77 26 ,87 28 ,0
altisis x (cm) - 1 5 , 4 15 ,40 - 3 , 6 5 3 ,65 - 3 , 6 5 3 ,65 .Letak c . g . s . F . (d ar i s i s i bawah) •
y± . 586161,18 = 9 8 0 4 8 ,5 5 .9 ,8 7 + 9 7 3 2 2 ,0 3 .1 1 ,2 + 9 7 2 7 0 ,5 0 .1 2 ,4 8 + 9 7 8 5 9 ,1 5 .1 3 ,7 7 + 9 7 0 8 3 ,0 4 .2 6 ,8 7 + 9 8 5 7 7 ,8 9 .2 8 ,0
y A = 17,04 cm.Letak e . g . s . F . ( d a r i s i s i bawah):jrw.46 26 66 ,5 4 = 77 688 ,59 .
+77020,02 . y w = 17 ,04 cm.
9 ,8 7 + 7 6 5 1 9 ,6 7 .1 1 ,2 0 + 7 6 9 4 4 ,7 7 .1 2 ,4 8 1 3 ,7 7 + 7 6 7 9 6 ,5 .2 6 ,8 7 + 7 7 6 9 6 ,9 9 .2 8 ,0
40
* Potongan 8;
Kabel B E^ ( k g ) 98542 ,78 97880,66 97862,21 98511 ,98 97784 ,30 99169,80
78079 ,54 77197,33 77412 ,85 77810,28 77351 ,20 78447,20Jarak dr a l a s 9 ,0 5 9 ,45 9 ,8 0 10 ,13 22 ,77 22,80
a b s i s x (cm)
-1 6 ,01 16,01 - 3 , 8 0 3 ,80 - 3 , 8 0 3 ,8 0
Letak e . g . s . ( d a r i s i s i bawah):y ± . 589751,73= 9 8 5 4 2 ,7 8 .9 ,0 5 + 9 7 8 8 0 ,6 6 .9 ,4 5 + 9 7 8 6 2 ,2 1 .9 ,8 0
+98511 ,9 8 .1 0 ,1 3 + 9 7 7 8 4 ,3 0 .2 2 ,7 7 + 9 9 1 6 9 ,8 .2 2 ,8 0 3^ = 14 ,0 cm.
Letak e . g . s . F ( d a r i s i s i bawah):y w.46 62 98 ,40= 7 8 0 7 9 ,5 4 .9 ,0 5 + 7 7 1 9 7 ,3 3 .9 ,4 5 + 7 7 4 1 2 ,8 5 .9 ,8 0
+ 7 7 8 1 0 ,2 8 .1 0 ,1 3 + 7 7 3 5 1 ,2 0 .2 2 ,7 7 + 7 8 4 4 7 ,2 0 .2 2 ,8 0 = 14,01 cm.
* PotonKan 9 :
£abel^ ( k g )
(k g )98433,17
B98433,17 98433,17
D98433,17 98433,17
E F98433,17
w 77864 ,50 77864,50 77864 ,50 77864 ,50 77864 ,50 77864 ,50Jarak Ir a la s 8 ,85 8 ,8 5 8 ,8 5 8 ,8 5 21,3 2 1 , 5i b s i s x (cm) -16,1,5 16 ,15 - 3 ,8 5 3 ,85 -3 ', 85 3,85
Letak c . r . s . F ( d a r i s i s i bawah):590599,00 = 4 . 9 8 4 3 3 , 1 7 . 8 , 8 5 + 2 .9 8 4 3 3 ,1 7 .2 1 ,3
= 13 cm.Leltak c . / t . s . F (d a r i s i s i bawah):
4.7786.13 cm.
y w. 467187,00 = 4 . 7 7 8 6 4 ,5 0 .8 ,8 5 + 2 .7 7 8 6 4 ,5 0 .2 1 ,3yw
41
* Po'toflg®*1 8 1 «
Kabel A B^ ( k g )
D 197943,81 97996,37 98019,41 97806,93 98765,7277340,77 77540,211 77329,27 77369 ,12 78057 ,08 77306,7 ' B
Jarak i r a la s 9 ,0 5 9 ,4 5 9 ,8 0 10,13 22 ,77 22 ,80
labsis x (cm) -1 6 ,0 1 16,01 - 3 , 8 0 3,80 - 3 , 8 0 3 ,8 0
Letak e . g . s . .b\ ( d a r i s i s i bawah):y A. 588260,36= 9 7 9 4 3 ,8 1 .9 ,0 5 + 9 7 9 9 6 ,3 7 .9 ,4 5 + 9 8 0 1 9 ,4 1 .9 ,8 0
+ 9 7 8 9 6 ,9 3 .1 0 ,1 3 + 9 8 7 6 5 ,7 2 .2 2 ,7 7 + 9 7 7 2 8 ,1 2 .2 2 ,8 y i = 14,01 cm.
l e t a k c . / y . s . g.. ( d a r i s i s i bawah.):y w. 46 494 3 ,2 3* 7 7 3 4 0 ,7 7 .9 ,0 5 + 7 7 5 4 0 ,2 1 .9 ,4 5 + 7 7 3 2 9 ,2 7 .9 ,8 0
+77369 ,12 .10 ,1 3+78057 *08.22 ,-77+77306 ,78 .22 ,8 -y w = 14,01 cm.
* Potongan 7 ’ :
Kabel A B D ' P(k g ) 97430 ,45 97426,91 97422,1
f'w(k g )91749,39 98097 ,73 96970,45
76801,26 77089 ,74 76608 ,3 76848 ,98Jarak 3r a l a s
77246,57 76707,439 ,8 7 11 ,20 12 ,48 13,77' 26 ,87 28 ,0
i b s i s x (cm) - 1 3 , 4 15 ,4 - 3 , 6 5 3 ,65 - 3 , 6 5 3,65
Letak e . g . s . ( d a r i s i s i bawah):y A. 584497,03= 974 3 0 ,4 5 .9 ,8 7 + 9 7 4 2 6 ,9 1 .1 1 ,2 0 + 9 7 4 2 2 ,1 0 .1 2 ,4 8
+ 9 7 1 4 9 ,3 9 .1 3 ,7 7 + 9 8 0 9 7 ,7 3 .2 6 ,8 7 + 9 6 9 7 0 ,4 5 .2 8y i = 17 ,04 cm.
Letak e . g . s . ( d a r i s i s i bawah):yw. 461298,29= 7 6 8 0 1 ,2 6 .9 ,8 7 + 7 7 0 8 9 ,7 4 .1 1 ,2 0 + 7 6 6 0 8 ,3 1 .12 ,48
+ 7 6 8 4 8 ,9 8 .1 3 ,7 7 + 7 7 2 4 2 ,5 7 .2 6 ,8 7 + 7 6 7 0 7 ,4 3 .2 8 y w = 17 ,04 cm.
42
* Potongan 6 1:
Kabel B 33 E* j ( k g ) 36937,69 96878,16 96845,56 96502,30 97419,29 96233 ,83V * g ) 76054,89 76655 ,66 75903 ,79 76337,10 76419 ,79 76124,74Jarak dr alsis 11 ,22 14 ,13 17,0 19,89 3 3 ,8 3 6 ,5
a b s i s x (cm) - 1 4 , 3 5 14 ,35 - 3 , 4 3 ,4 - 3 , 4 3 ,4
Letak e . g . s . F ( d a r i s i s i bawah);y i . 580816,83= 9 6 9 3 7 ,6 9 .1 1 ,2 2 + 9 6 8 7 8 ,1 6 .1 4 ,1 3 + 9 6 8 4 5 ,5 6 .1 7
+ 9 6 5 0 2 ,3 .1 9 ,8 9 + 9 7 4 1 9 ,2 9 .3 3 ,8 + 9 6 2 3 3 ,8 3 .3 6 ,5 y.* = 2 2 ,08 cm.
Letak e . g . s . F ( d a r i s i s i bawah);yw. 454795,97= 7 6 0 5 4 ,8 9 .1 1 ,2 2 + 7 6 6 5 5 ,6 6 .1 4 ,1 3 + 7 5 9 0 3 ,7 9 .1 7
+ 7 6 3 3 7 ,1 0 .1 9 ,8 9 + 7 6 4 1 9 ,7 9 .3 3 ,8 + 7 6 1 2 4 ,7 4 .3 6 ,5 y w = 22 ,09 cm*
Kabel
* Potongan 5 * :
A B D E(k g ) 96444 ,92 96319 ,06 96258,64 95865,61 96761 ,82 95518,26
Fw(kg) 75454 ,55 76213 ,39 75191 ,05 75833,46 75613 ,56 75558 ,70Jarak dr a l a s 13 ,13 18,26 2 3 ,38 28 ,49 43 ,5 4 8 ,5
a b s i s x (cm) - 1 2 , 9 12 ,9 - 3 , 0 6 3,06 - 3 , 0 6 3 ,06
Letak e . g . s . F ( d a r i s i s i bawah);y r 577168,31= 9 6 4 4 4 ,9 2 .1 3 ,1 3 + 9 6 3 1 9 ,0 6 .1 8 ,2 6 + 9 6 2 5 8 ,6 4 .2 3 ,3 8
+ 9 5 8 6 5 ,6 1 .2 8 ,4 9 + 9 6 7 6 1 ,8 2 .4 3 ,5 + 9 5 5 1 8 ,2 6 .4 8 ,5 y i - 29 ,2 cm.
Letak e . g . s . F ( d a r i s i s i baw^h);y w. 453864,71= 7 5 4 5 4 ,5 5 .1 3 ,1 3 + 7 6 2 1 3 ,3 9 .1 8 ,2 6 + 7 5 1 9 1 ,0 5 .2 3 ,3 8
+ 7 5 8 3 3 ,4 6 .2 8 ,4 9 + 7 5 6 1 3 ,5 6 .4 3 ,5 + 7 5 5 5 8 ,7 .4 8 ,5 yw = 29 ,2 cm.
tI.
43
* Potongan 4 1:
Kabel A B C D E 3?^ ( k g ) r95993 ,30 95780 ,62 95682 ,07 95228,91 96072,82 94813 ,16Fw(k g ) 74870,49 75787 ,49 744 86 ,53 75329,83 74744,22 75000 ,98Jarak dr a l a s 15 ,57 2 3 ,5 8 31‘ , 58 39,57 56 64a b s i s x
(cm) - 1 1 , 0 3 11 ,03 - 2 , 6 2 2,62 - 2 , 6 2 2 ,62
Letak e . g . s . J?\, ( d a r i s i s i bawah'i:y.j_. 573570,88= 95993 ,3 .1
+95228,91 . Y i = 38 ,34 cm.
Letak e . g . s . F ( d a r i s
5 ,5 7+ 9 57 8 0 ,6 2 .2 3 ,5 8 + 95 68 2 ,0 7 .3 1 ,58 39 ,57 +96 0 72 ,82 .5 6+ 94 8 1 3 ,1 6 .6 4
i s i bnwah):1 5 ,5 7 + 7 5 7 8 7 ,4 9 .2 3 ,5 8 + 7 4 4 8 6 ,5 3 .3 1 ,5 8 39 ,57 +74 7 44 ,22 .5 6+ 75 0 0 0 ,9 8 .6 4
’ y w. 450219,54= 74 870 ,49 .+75329,83.
y w = 38 ,36 cm.
* Poton^an 3 ‘ :Kabel A B C D E F
^ ( k g ) 95459,35 95221 ,52 95105 ,54 94613 ,1? 95488 ,58 94076,54
Fw(kg ) 74253,86 75345,22 73782,01 74801,44 73992 ,83 74418 ,28
Jaiak dr a la s 18 ,56 1 30 ,10 41 ,60 53,13 71 ,35 82 ,80
a b s i s y. ( cm) - 8 , 7 4 8 ,74 - 2 , 0 8 2 ,0 8 - 2 , 0 8 2 ,0 8
Letak e . g . s . F . . (d ' -r i sjLsi bawah ) :?1 ,5 2 .3 0 ,1 +951*05,54.41 ,6 3 8 ,5 8 .7 1 ,3 5 + 9 4 0 7 6 ,5 4 .8 2 ,8
4 5 ,2 2 .3 0 ,1 + 7 3 7 8 2 ,0 1 .4 1 ,6 3 2 ,8 3 .7 1 ,3 5 + 7 4 4 1 8 ,2 8 .8 2 ,8
y i . 569964,65= 95459 ,35 .18 ,56+95?^ +9461 3 ,1 2 .5 3 ,1 3+954*
y i = 49 ,52 cm.
y .446593 ,64= 742 5 3 ,8 6 .1 8 ,5 6+ 7 53+74801 ,4 4 .53 ,13+739
V = 49 ,55 cm.J w
i1......■............................
44
* Potongan 2 ' :
£abel A B C H E FPfCkg) 95007,77 94714 ,18 94539,37 93934,70 94799 ,67 93392,54 .FwU g ) 75653,53 74699,75 73077,47 74330,82 73186 ,62 73860,56Jarak Ir a l a s 22., 1 37 ,79 53 ,48 69,16 8 9 ,48 105,15
i b s i s x (cm) - 6 6 - 1 , 4 3 1 ,43 -1 ,413 1 ,4 3
l e t a k e . g . s . F.. ( d a r i s i s i bawah] :y ^ . 566388 ,23* 9 5 0 0 7 ,7 7 .2 2 ,1 + 9 4 7 1 4 ,1 8 .3 7 ,7 9 + 9 4 5 3 9 ,3 7 .5 3 ,4 8
+ 9 3 9 3 4 ,7 .6 9 ,* 6 + 9 4 7 9 9 ,6 7 .8 9 ,4 8 + 9 3 3 9 2 ,5 4 .1 0 5 ,1 5 y i = 62 ,74 cm.
Letak e . g . s . I' ( d a r i s i s i bawah):y w. 442808,75= 7 3 6 5 3 ,5 3 .2 2 ,1 + 7 4 6 9 9 ,7 5 .3 7 ,7 9 + 7 3 0 7 7 ,4 7 .5 3 ,4 8
+ 7 4 3 3 0 ,8 2 .6 9 ,1 6 + 7 3 1 8 6 ,6 2 .8 9 ,4 8 + 7 3 8 6 0 ,5 6 .1 0 5 ,1 5 y w = 62,81 cm.
* Potongan 1 ’ :
Kabel A B G D E F^ ( k g ) 94515 ,02 94144 ,76 93973,25 93339,86 94131,72 92666 ,50Bw(kg ) 73036 ,90 74290,24 7238:9,37 73835,45 72372,11 73302 ,83Jarak 3r a la s 2 6 ,1 8 4 6 ,6 8 67 ,18 87 ,68 110 ,4 130 ,93.bsis
(cm) -2 ,9 1 2,91 - 0 , 7 0 ,7 - 0 , 7 0 ,7
l e t a k e . g . s . F . ( d a r i s i s i bawah):
y i . 562771 ,1 1 = 9 4 5 1 5 ,0 2 .2 6 ,1 8 + 9 4 1 4 4 ,7 6 .4 6 ,6 8 + 9 3 9 7 3 ,2 5 .6 7 ,1 8 + 9 3 3 3 9 ,8 6 .8 7 ,6 8 + 9 4 1 3 1 ,7 2 .1 1 0 ,4 + 9 2 6 6 6 ,5 .1 3 0 ,9
y A - 7 7 ,9 8 crn.Letak e . g . s . F (d a r i s i s i bawah):y w. 439226,90= 7 3 0 3 6 ,9 .2 6 ,1 8 + 7 4 2 9 0 ,2 4 .4 6 ,6 8 + 7 2 3 8 9 ,3 7 .6 7 ,1 8
+ 7 3 8 3 5 ,4 5 .8 7 ,6 8 + 7 2 3 7 2 ,1 1 .1 1 0 ,4 + 7 3 3 0 2 ,8 3 .1 3 0 ,9 y w = 78 ,09 cm.
45
(a b e l A B G D E FPi (k g ) 94087,22 93696,32 93501,42 92826 ,82 93597 ,73 92077,72
72836,69Pw(kg ) 72571,91 74108 ,50 71809,11 73397 ,88 71727 ,85Jarak di a l a s 30 55 80 105 130 155
a b s is x (cm) 0 0 0 0 0 0
* Ujung p e r le ta k a n Ice nan :
Letak e . g . s . F ( d a r i s i s i bawah) :9 4 08 7 ,2 2 .30 + 9 36 9 6 ,32 .5 5 + 9 3 0 5 1 ,4 2 .8 0 + 9 2 8 2 6 ,8 2 .1 0 5 + 9 3 5 9 7 ,7 3 . 130+92077 ,72 .155 = 5 5 9 7 8 7 , 2 3 ^
y * 9 2 ,2 5 cm.Letak e . g . s . Fw ( d a r i s i s i bawah) :7 2 57 1 ,9 1 .30 + 7 41 0 8 ,50 .5 5 + 7 1 8 0 9 ,1 1 .8 0 + 7 3 3 9 7 ,8 8 .1 0 5 + 7 1 7 2 7 ,8 5 . 130+72836 ,69 .155 = 436451 ,9 4 . y,w
w 9 2 ,3 8 cm.
IV. 1 . 9 . Mencari Kn & Kn , t i a p - t i a p p o to n g a n ;* P r o f i l pada ujjung p e r l e t a k a n :
W * 305772,74 c i AK' = 29 ,868 cm.K = 38 ,853 cm.
* p r o f i l tengah ( potongan 2 - 2 ' ) :W = 252802,62 cm3 .K 37,661 cm.K = 52^559 cm. __
Untuk F. :K_. - K' , dimana Obi" = 1 6 8 kg/cm^.
Untiflk F :K„ --------------w nw
xOT>l.W K* , dimana CTb1 = 120 kg/cm .
' w
46
POT. * i ( k g ) Fw(k g ) W(cm^) OM ' ,W(cm) K'(CM] Kn i (CK0 Kn w ^...... IT" ..." .fewUj kr 561408,12 437503,81 305772,74 91 ,50 83 ,868 29 ,868 61 ,632 54 ,0
1 564413,50 440534,11 305772,74 91,01 83,291 2 9 ,868 61,142 53 ,4232 567960,08 444071,26 252802,62 74 ,778 68 ,314 37,661 37,117 30 ,6533 571588,35 447845,59 252802,62 74 ,303 67 ,738 37,661 36 ,642 31 ,0964 575209,15 451321 ,45 252802,62 73,835 67 ,217 37,661 36,174 31 ,0435 578828,70 455223,21 252802,62 73 ,374 66 ,640 37,661 35 ,173 28 ,9796 582479,22 458932,43 252802,62 72 ,914 66 ,102 37,661 35,253 28,4417 586161,18 462666,54 252802,62 72 ,456 65 ,56 8 37,661 34 ,795 27 ,9078 589751,73 466298,40 252802,62 72 ,015 65 ,058 37,661 34 ,534 27 ,3979 590599 ,0 467187,0 252802,62 71,911 64 ,934 37,661 34,250 27 ,2738 1 588260,36 464943,23 252802,62 72,197 65,247 37,661 34,536 27 ,5867* 584497,03 461298,23 252802,62 72,662 65 ,763 37,661 35,001 28 ,1026 ' 580816,83 457495,97 252802,62 73 ,123 66,309 37,661 35,462 28 ,6485 ' 577168,31 453864,71 252802,62 73 ,585 66 ,840 37,661 35,924 29 ,1794 ’ 573570,88 450219,54 252802,62 74,046 67,381 37,661 36,385 29,7203 ' 569964,65 446593,64 252802,62 74,515 67 ,928 37,661 36,854 30,2672 ' 566388,23 442808,75 252802,62 74 ,985 68,509 37,661 37,324 30 ,8481 ' 562771,11 439226,90 305772,74 9 1 ,2 8 0 83,539 29 ,868 61 ,412 53,671
Uj kn 559787,23 436451,94 305772,74 91,767 84,070 29 ,868 61,899 54,202
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48
Tabel harga En , dimana kabe l F belum d i jack .
K- n i ' =t f b i ’ .W - K1.^ 7 7
POT. V' ■ — i ........ .
W(CM5 )Obi.W
Fi . k * (cm) Kn iUj kr 467729,49 305772,74 109,828 29 ,868 79 ,96
1 470153,55 305772,74 109,262 29 ,868 7 9 ,3942 472985 ,73 252802,62 89 ,793 37,661 52,11323 475901,1 252802,62 89 ,243 37,661 51,5824 478816,5 252802,62 88 ,699 37,661 51 ,0385 481711,15 252802,62 88 ,167 37,661 50,5066 484647,29 252802,62 87,632 37,661 49,9717 487583,29 252802,62 87 ,105 37,661 49 ,4448 490581 ,93 252802,62 86 ,572 37,661 48,9119 492165,85 252802,62 ' 86 ,293 37.66T 48 ,6328 ' 490532,24 252802,62 86,581 37,661 48 ,9207* 487526 ,58 252802,62 87,11 5 37,661 49 ,4546 ' 484583,0 252802,62 87,644 37,661 49 ,9835 ' 481650,05 25*2802,62 88,177 37,661 50,5164 ' 478757,72 252802,62 88,710 37,661 51,0493 ' 475888,111 252802,62 89 ,245 37,661 51,5842 ' 472995,69 252802,62 89,791 37,661 52 ,1301 ' 470104,61 305772,74 109,273 29 ,868 79 ,40 5
Uj kn 467709,51 305772,74 109,833 29 ,868 79 ,965
Untuk potongan U j. k i r i , 1 , 1 ' , & U j . kanan• . > K l i m i t kabe l diteritukan o l e h K & K1. m • '
Untuk potongan 2 s / d 2 'Kn i < K —o l i m i t kabe l d i ten tukan o l e h Kn i & K*
49
Tabel untuk rnencari daerah l i m i t k a b e l , dimana kab e l F belurr d.i j a c k .
POT.Jarak d r u.i. MgiCkgcm) (kgcm) Fi ’ y 1 i ,=y kn i
U j kr 0 • 0 0 467729,49 0 1 65 , 0871 1 ,7 5 53t0500 10725400 470153,55 11,295 53, 7922 3 ,85 1084300C 22063700 472985,73 22,924 25, 8923 5 ,95 15635000 31904600 475901 ,1 33 ,855 117, 5134 8 ,0 5 19688000 40251100 478816,5 41 ,118 9 , 7925 10 ,15 23000000 46800100 481711,15 47,746 3, 6966 12 ,25 25573000 52459600 484647,29 52,766 “ 0 , 7897 14 ,35 27405000 56167900 487583,29 56,206 - 3 , 7028 16 ,45 28497500 58375000 490581,93 58,089 - 5 , 0229 18 ,50 28850000 59157200 492165,85 58,618 - 5 , 3028* 20 ,55 28497500 58375000 490532,24 58,095 - 5 , 0677 ’ 22 ,65 27405000 56167900 487526,58 56,212 - 3 , 7186* 24 ,75 25573000 52459600 484583,0 52,773 - 0 , 8085* 26 ,85 23000000 46800100 481650,05 47,752 3 , 684 • 28 ,95 19688000 40251100 478757,72 41,123 9 , 7763* 31 ,05 15635000 31904600 475888,11 32,854 17, 512* 33 ,15 10843000 22063700 472995,69 22,924 26 , 8941 • 35 ,25 5310500 10725400 470104,61 11,296 53, 791
U3 kn 37 ,0 0 0 467709,51 0 65 , 087
50
Tabel untuk mencari daerah l i m i t k a b e l , dimana kabe l F t e l a h di, j a c k , j a d i t o t a l , dimana pengangkeran di lakukan s e t e l a h p l a t d iafragma dan g e l . m elin tangt e l a h s e l e s a i d i c o r .
POT.%- _ . .... ,
Jarak d r u j . M^2 (kgcm) * i t o t a l k u - 1!
0y 1 i =3rk n i “
o
Uj kr 0 0 561408,12 0 65 ,08 71 1 ,7 5 6105200 564413,50 110,82 54 ,2672 3 ,8 5 12504400 567960,08 22,02 42,8113 5 ,9 5 18067900 571588,35 31,61 33,6964 8 ,0 5 22796700 575209,1 5 39,63 26,1445> 10 ,15 26689400 578828,70 46,11 20,6656 12 ,25 29747400 582479,22 51,07 15,6257 14 ,35 31821500 586161 ,18 54,29 12,8639 16 ,45 33054600 589751,73 56 ,05 11,3649 18 ,50 33452700 590599,00 56,64 V O vfl 03
8 * 20 ,55 33054600 588260,36 56,20 11 ,2127 ' 22 ,65 31821500 584497,03 54,44 t,2,5076 * 24 ,75 29747400 580816,83 51,22 15 ,2665* 26 ,85 26689400 577168,31 46 ,24 19 ,7844* 28 ,95 22796700 573570,88 39,74 25 ,8233* 31 ,05 18067900 569964,65 31 ,70 33 ,3942 ’ 3 3 ,15 12504400 566388,23 22 ,07 42,5541 • 35 ,25 6105200 562771,11 10,85 54,237
Uj kn 3 7 ,00 0 559787,23 0 65 ,087
51
Daerah l i m i t k abe l
52
I V . 1 . 1 0 . K ontro l terhadap momen h a n cu r :Anggapan: - b e ton dianggap hancur , b i l a 3 , 5 ° / o c
- u luran max. b a ja = u luran pada t e g . k e r js + 5° / o o 4 l 1 0 ° / o o .
- t e g . hancur b e t o n := 0 ,6 7 .4 0 0=■ 268 kg/cm^.
r ton ] ^ Fw 467187 rto , . _ /_ _ 2 ,2 J °Iw“ T= “7770" = 98 ,562 kg/cm .
F = 467,187 t o n ] “ tA = 4740 ram'
Uluran yang t e r j a d i pada t e g . k e r j a :4 w = T * * f ; ^ f 5 4 - 1 0 0 0 ° / o ° = 4 , 9 7 8 ° / o o .
&Uluran max. b a ja = 4 ,9 7 8 + 5 >? 9 , 9 7 8 ° / o o < 1 0 ° / o o .
( O . K . ) .
* Keadaan b a l a n c e :' — .... /(tlTcwv "1 1 \ £\a4,~ %0 ,
*5 JS.O i*r.
I-2 Ct v 7 S %0gb . 4 . 2 7 . Diagram t e g . & u luran dalam ke
adaan b a la n c e .*1 n
^7 5IT 5 =Tf, — x i - 6 6 ,7 0 cm.b * 4f x ( l - x ) = 4 U ^ b u » .4 9 ;2 . (1 3 3 .4 - 4 9 .2 )
1 * P 1 3 3 ,4 5« 0 , 9 3 l ° b u , #
a. = 0 , 0 6 9 ° b u ' .c . ( 1 33 t 4,,,-_4,5y87J, „ o ,9O 20bu ' .' 1 3 3 , 42
53
Nb , = 2 / 3 . 2 0 . 6 6 , 7 . (Thu' + (T65 -2 0 ) . i7 ,5 .0 ,9 3 1 .G tm » + 2 /3 . ( 1 6 5 - 2 0 ) . 1 7 , 5 . 0 , 0 6 9 . °bu' + 2 . 1 / 2 . 1 5 .1 0 . 0 , 9 0 2 . (Jt)ul
= 3503.77.6bu*.» 939010,57 kg .
£ au * 9 ,9 7 S ° / oo.Dari s t r e s s - s t r a i n diagram ( l i h a t la m p ir a n ) , d i d a -
. pa tkan : Fu = 38900 l b s = 17660 ,6 kg , untuk A « 0 ,1 537 i n 2 » 0 ,9 9 1 6 cm2 .
O-au - T 1 ■ * ^ 8 1 0 , 2 0 5 kg/cm2.
N - 1 7 8 1 0 ,2 0 5 .4 7 ,4 0 . 844203,71 kg < Nh , .au u(under-reinforced)* D icoba untuk x ^= 60 cm.
g b . 4 . 2 8 . Diagram t e g . & uluran untuk Xg d iam bil t* 60 cm*
<^b2 ' 60 _ <5 Q>iiOyrtri- j r “ (1 75 -13 7 -6 T )— > ^ b 2 ' 2,941 / 0 °* b «. 4.gBu>.(2.xr,t~42,5) B 4.(Tbu».42.5.(2.7 1.4-42.5 )
2 " (2 x 2 * )2 142,82■ 0 ,8366bu'.
a « 4 . Cibu ' . 6 0 . ( 2 . 7 1 , 4 - 6 0 2 _ o > 8 360bu' « 0 ,138(Jbu ' .— 2__________ .142-,az2_____________________________________________
54
0 , = 1 7 - ( 2 : 7 1 , 4 - 3 9 , 1 7 ) . 0 , 7 96SbU. .142,18
Nb ,= 2 / 3 . 2 0 . 6 0 . ^ b u ' + (165 - 2 0 ) . 1 7 , 5 . 0 , 8 3 6 . 0 b u ' + 2 / 3 . ( 1 6 5 - 2 O ) .1 7 ,5 .O ,1 3 8 0 b u ' + 2 . 1 / 2 . 1 5 . 1 0 . 0,796.<Jbu'
= 3274,19(Tbu‘ .= 877485,57 k g ) N &u = 844203*71 kg.
* D icoba x „ = 5 7 .0 cm.— y - ------------------
H P ” = (T 7 B -Iy )-B 7 -------► ^b3 » = 2 > 7 U ° / o o .b . 4 > ^ u » . 3 9 , 5 . ( 2 x 3> 3 9 t 5) a 4 . 0 b u « . 3 9 , 5 . ( 2 . 7 3 . 5 - 3 9 .
(2 x 3 * )2 147= 0,786(Jbul .
a 3 = l ^ ' * ^ ( 2 . 7 3 , 5 - 5 7 ) . 0 , 786(rb u , . 0 ,164(Tbu«.5 147
c , - 3 , 6 . 1 7 . ( 2 . 7 3 ^ - 3 6 , 1 7 ) „ 0>742(rbut#p 147
Nb ,» 2 /3 .2O .6O .0bu* + (165 - 2 0 ) . 1 7 , 5 . 0 , 7 8 6 . Gbu' +2 / 3 . ( 1 6 5 - 2 0 ) .1 7 ,5 .0 ,1 6 4 . C f t u * + 2 . 1 / 2 . 1 5 . 1 0 .0,742.CTbu'.
■ 3183,21(TBu'.» 853099,82 k g ^ N au . 844203,71 kg.
* Dico'ba x ^= 5 6 , 0 cm.
■b4' 56 . £ v „ , a 2 .6415 ° / o o .- 5— =TT75"-'T3"5-56-------* b4b , . l ^ ., 3 « ^ . ( 2 r7 4 >2 - 3 8 , 5 ? = 0 f 7 6 8 a 5 u i .
4 _ 1 4 8 , 4 ^ a 4 = 4 -g^u.1» % .± 2 . 7 4_i 2 - 56) _ 0 f 76B(Pbu , = 0 ,172<Jbu' .
4 1 48 ,4e . 1 ^ - 3 5 , 1 7 . ( 2 . 7 4 , 2 - 3 5 , 1 7 ) = Of7230Bu. .
4 1 4 8 , 4Nb ,= 2 / 3 . 2 0 . 6 0 . ( f b u 1 + (165 - ? 0 ) . 1 7 , 5 . 0 , ?68.(Jbu» +
2 / 3 . (1 6 5 - 2 0 ) . 1 7 , 5 . 0 , 1 72.(Tbu' + 2 . 1 / 2 . 1 5 . 1 0 . 0 , 7 2 3 . f r b u ' .
= 3 1 4 9 ,2 1 Obu1.
:>)
55
= 843988 ,28 kg<-Nau
y = [ 2 / 3 . 2 0 . 5 6 . Gbu».3 / 8 . 5 6 + ( 1 6 5 - 2 0 ) . 1 7 , 5 . 0 , 7 6 8 . Cbu.1 / 2 . 1 7 , 5 + 2 / 3 . ( 1 6 5 - 2 0 ) . 1 7 , 5 . 0 , 1 7 2 . J b u » . 3 / 8 . 1 7 , 5 +2 . 1 / 2 . 1 5 . 1 0 . 0 , 7 2 3 . ^ b u ' . ( 1 7 5 + 1 0 / 3 ) ] / { 2 / 3 . 2 0 . 5 6 .+ ( l 6 5 - 2 0 ) . 1 7 , 5 . 0 , 7 6 8 . C b u ‘ + 2 / 3 . ( 1 6 5 - 2 0 ) . 1 7 , 5 .0 ,172.crbu» + 2 . 1 / 2 . 1 5 . 1 0 . 0 , 7 2 3 . ^bu*|
3 6 9 0 0 ,48<Jbu • Jy = ~3B9?t8S’Cbu»
= 11 ,923 cm.a - 175 - 13 - 11 ,923 = 150,077 cm.
= 8 4 4 2 0 3 ,7 1 .1 5 0 ,0 7 7 = 1 ,2 6 7 .1 0 9 kgcmU
» 1266,95 tm.
Syarat : Mu > 1 , 7 5 Mg + 2 ,2 5 Mp
> 1 , 7 5 . 3 8 3 , 2 8 + 2 ,2 5 .2 0 8 ,2 9 2 > 1 1 3 9 ,3 9 7 tm.( O . K . ) .
I V . T . 1 1 . K ontro l terhadnp Moroen r e ta k :Menurut R . V . B . '67 a r t 7 . 3 . 6 . k o n s tru k s i mulai
r e ta k pada be ton dengan t e g . t a r i k l e n t u r s eb esar 0 ,33<7V.Dimana 0 ,3 3 0=b ' = 0 ,3 3 .1 2 0 = 39 ,6 kg/cm2 .Dari diagram t e g . be to n a k i b s t beban Fw, d i d ap at ;
° b b ' = -2 3 3 ,9 7 7 kg /cm2 ( l i h a t I V . 1 . 5 . E h a l .132Jadi beda t e g . t a r ik beton s e b e s a r 39 ,6 kg/cm dengan
pt e g . tekan b e to n s e b e s a r -2 3 3 ,9 7 7 kg/cm ad a la h :
- 39 ,6 + 233,977 » 273,577 kg/cm2 .W * 252R02,6Z cm3.
Jadi besarnya rnornen r e ta k = 2 7 3 ,5 7 7 .2528 02 ,62= 69160982 kgcm = 691,609 tm.
Menurut R .V .B . '6 7 a r t 1 0 .1 , b e sa r momen d i b a t a s i :- untuk k o n s tr u k s i d i l u a r :
Mr > 1 .1 M g + 1,25Mp^ 1 , 1 . 3 8 3 , 2 8 + 1 ,2 5 .2 0 8 ,2 9 2
__________________£6 .81 ,973 tm. (O^K.) ,_______________________________
56
- s e l a i n i t u harus pula menjenuhi: dimana M I V . 1 .1 0 )
Mr < 0 , 8 K , dimana M = 1266,95 tm. ( l i h a t
Mr < 0 , 8 . 1 2 6 6 , 9 5 <1013 ,56 tin.
Jadi perencanaan p r o f i l memenuhi s y a r a t .
I V . 1 . 1 2 . Perhitungan gaya l i n t a n g d i t i a p potongan:r—~r r~H\s
I "I II i' i H I I nit -f &
Dx = 1 / 2 . q . l - q . x = q ( 1 / 2 . l - x ) = c ^ ( 1 / 2 . l - x ) dimana
Dx = P / l . ( l - x ) = c 2 ( l - * x ) dimana Cg- P /1 ,
a . a k ib a t b e r a t s e n d i r i p r o f i l : r% -'M st/K,!I TI m 11111 |I I I 1 l 111 1 111'I H Ij^n gg a 1 ,67 8t /m V
I * 5 t f 2 1'j }\'tny % * 'V*s A
g b . 4 . 2 9 . Muatan a k ib a t b . s . p r o f i l .b . a k ib a t p e r te b a la n p r o f i l :
j r a ----------- ------------------------------------------------_n®j
g b . 4 . 3 0 . Kuatan a k ib a t per teb a la n p r o f i l .( D « r ) ud = 1 »542 t .(Dg^')^ = O, demikian untuk t i t i k - t i t i k l a i n
c* a k ib a t beban znati sekunder : ( a s p a l + t r o t o a r + san-daran + p l a t p e n g i s i )
|UliIi LiiJ.u,;! i : iJ [iTOLnil mill T iTI'l iTITT TI i 111 I^C-, =gg= 0 , 4 9 7 t / m ' .
gb . 4 . 3 1 . Muatan a k ib a t beban sekunder.
d . a k ib a t b . s . g e la g a r m el in tang :| V : oi 9 is t | P °v? ■
£? '/ i "3 'i V * 7 S v5 A-j? Hr T'
gb . 4 . 3 2 . Fiuatan* a k ib a t b . s . g e l . m el intang .
( Dg 3 ) u j =(:Dg3 ) l “ (Dg3 ) 2SS!(3)g3 ) 3=(3)63 ) 43(I)s3 ) 5,s(I)g3 )6 “
= 0 ,7 35 to n .( ( Dg^) q= ( Dg ^ ) 0 .
e. Akibat beban hidup t e r b a g i r a t a : p , . • 0 ,9 2 7 t/m 1,
0 ,9 2 7 t / m ' !f . A k ibat beban g a r i s :
P2 " 1 »2 3 *f 4 5 * 1 »65 = 8 ,856 t o n *c 2=P2/ l = 8 ,8 5 6 /3 7 = 0 ,2 39 t / m ' .
g . Akibat tekanan ke a ta s d a r i tendon:
Kabel f (m ) V k g )8Fw. f / l 2
( t / m ' )A 0 ,2115 77864 ,50 0 ,0962B 0 ,4615 77864 ,50 0,2099C 0,7115 77864 ,50 0 ,3237D 0,9615 77864 ,50 0,4375£ 1 ,087 77864 ,50 0,4946P 1 ,337 77864 ,50 0,6084
q 2 ,1 7 t / m «.J a d i c r = q «* 2 ,1 7 t / m ' .
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POT.S
FT”
r esh ) g 1 +d/ | t
-r- DS
(kg/Em2 ) Fw(kg) B(cm2 )
Prr w Ubn=Er
(kg/cm )
Uj kr 0 ,000162 -1 ,365 4 ,455 *37503,81 10237,5 42 ,735 0 ,459
11 i t -1 ,463 3,886 440534,11 i i 43,031 0 ,3 4 8
2 0,000370 -2 ,921 8 ,003 444071,26 67 12 ,5 66 ,156 0 ,954
3 i i - 2 , 5 0 3 7 ,1 29 447845,59 11 66 ,718 0 ,7 53
4 i t - 2 ,0 8 4 6,257 451321,45 i i 67 ,236 0 ,577
5 11 -1 ,665 5,382 455223,21 t i 67 ,817 0 ,425
6 i i -1 ,2 4 6 4 ,509 458932,43 i i 68 ,370 0 ,296
7 i i - 0 ,8 2 7 3,364 462666,54 • t 68 ,926 0 ,164
8 t i - 0 ,4 0 9 2,490 466298,40 i i 69 ,467 0 ,089 7
9 i t 0 1 ,637 467187,0 11 69 ,599 0 ,039
8 ' • i - 0 ,4 0 9 2 ,490 464943,23 i i 69 ,265 0 ,0899
7 ' i i - 0 ,8 2 7 2,364 461298,29 i i 68 ,722 0 ,164
6 ' i i -1 ,2 4 6 4 ,509 457495,97 i i 68 ,156 0 ,297
5 ’ i i -1 ,665 5,382 453864,71 i t 67 ,615 0 ,429
4 ’ 11 - 2 ,0 8 4 6 ,257 450219,54 t » 67 ,072 0 ,579
3 ’ t i - 2 ,5 0 3 7 ,1 29 446593,64 11 66 ,532 0 ,755
2* 11 -2 ,921 8 ,0 03 442808,75 i i 65 ,968 0 ,957
1 ' 0 ,000162 - 1 , 4 6 3 3,886 439226,90 10237 ,5 42 ,904 0 ,349
Uj kn ♦ i - 1 , 3 6 5 4 ,455 4 36451,94 *« 42,633 0,461
Pb = 0 ,0 6 t ^ , « 0 , 0 6 . 1 2 0 = 7 , 2 kg/cm2 ,Dari t a b e l d i a ta s t e rn y ata max. = 0 ,9 5 7 kg/cm2 *
Jadi e e c a r a t e o r i t i s hanya dibutuhkan b e u g e l p r a k t i s . Beugel min. yang dibutuhkan (menurut K . B . I . I r . Sutami a r t * 5 . 4 . 1 * )
2min.= 0^002 .b Q cm /cm . lintuk 1 m dibutuhkan:
min.= 0 , 0 0 2 . b Q.100 cm2 .
60
Dimana b0 ■ 10 +V 0,3.b.h.- untuk profil ujung:
b0 « 10 ♦ 0,3V50,175 - 38,06 cm.A^min. • 0,002.38,06.100 « 7,612 cm2.Dipakai beugel 0 10 mm, A « 1,57 cm2.Jumlah beugel yang dibutuhkan « * 5 biji/m.Jadi digunakan beugel 010 - 20.
- untuk profil tengah:10 ♦ 0 ,3 1 /2 0 .17 ,5 » 27 ,75 cm.
A^min. 0 ,0 0 2 .2 7 ,7 5 .1 0 0 - 5 ,55 cm .Dipakai beugel 0 10 mm, Ag « 1 ,57 cm .Jumlah beugel yang dibutuhkan * * 4 b i j i /m .Digunakan beugel 010 - 20.
IV .1 .1 3 . Perhitungan pembesian meman.jang:Dipergunakan perumusan I r . Sutami (K .B .I . a r t 5*4 . 2) dimana pembeeian memanjang min. 0,25% dari b0h.Amin“ ° » 2 5%.bo .h .
- 0 ,2 5 /1 0 0 .2 0 .1 7 5 «= 8 ,7 5 cm .Satu s i s i : Afflin • 1 /2 .8 ,7 5 - 4 ,3 75 cm2 . (4 0 1 2 ) .
I V .1*14 . Kontrol Lendutan:*Lendutan akibat p restressin g (P^ ) :
7i. - 2.3C-•f ,. 1
F±m 561408,12 kg.L * 37 m.Dipergunakan cara euperposiei:
Pi.el.l2 1" 5EI—
5 .F i .e 2 .L ‘ — JBEI------
dimana : e1 « 9 8 ,94 - 92 ,33 * 6 ,61 cm.e2 - 98 ,94 - 6 ,61 - 13 - 7 9 ,33 cm.
« 0 ,7 4 ♦ 7 ,4 * 8 ,1 4 cm. tt)
61
- Lendutan akibat b .s . p r o f i l : ^,-hys-^- OjSSl t/sin '
^ 2 * I s f e l T . + ^ ' L4 8 e ! ^ " 2a ) 2
* l i l ! i ; l6 ^ ? 5 5 7 7 i S 7 5 2 , 8 , 4 S . ' i ? w i ! I ^ i ? J 2 ’ 2 , 1 ? ?
» 4 ,766 + 0 ,142 » 4 ,9 0 8 cnu(4J *^ t o t a l - 8 ,1 4 - 4 ,9 08 « 3 ,232 cm < ^ . 3 7 0 0 = 4 ,625 cm.
♦Lendutan jangka panjang:(memperhitungkan pengaruh kehilangan pratekanan&creep).Cc - || ( l ih a t T .Y .Lin h a l. 4 0 ) .dimana Cc = k oefisien creep.
St m uluran seketika + uluran akibat creep
B^e + ^krc-j Si ■ uluran seketika ■ £ *
J a d i : “ I ? E i r g ? l l ‘? i t f 3 y 6 g g -' " 19 » 5 5 .1 0 ~ 5£k rw - 2 3 ,3 6 2 .1 0 "5 ( l ih a t h a l. 23)
H krc* (1 9 .5 5 «■ 23 .362 )1 0~ 5 «GC “ — ------------ --- " ----------- 2 »195*©
- karena b .e . p r o f i l :A * Cc .^ 2 - 2 ,1 9 5 *4 ,9 1 « 11 ,084 cm.
- karena akibat beban sekunder:(aepal+trotoar+sandaran+pipa+plaat pengiai ) •g ■ 0 ,4976 t/m *. 4
» 193 • 3 *10 CBU ^• karena akibat b . s . gelagar m elintang:
P ■ 0 ,7 35 ton._<2,32t . '^3yvi...... ^ j j
—_ — . |_ - ----- ------------- >
* - H s - l f - ' c
A * C T inl500D 7L 7T 777-2 . 195 - 0.337 o>. UJ>
-M
62
- karena akibat beban hidup terbagi rata:' p » 0 ,9 2 7 t / m ' . • *
5 0 ,9 2 7 .1 0 .3 7 0 0 0 ^ , , ,5^ 000;£577372? - 2,64 cm. UJ
- karena akibat beban garis:
P * 5 ,367 t o n .a PL3 5 3 6 7 .37003 ^ _ , , vA = TSET = 4 S .3 3 3 6 0 i0 5 7 7 '? 7 S ? = ° » 66 CIR* ^
- karena akibat gaya pra-tekanan Fw:F = 437503,81 kg. ?
w F . e ^ . L . L 5F . e9 .L A / W 1 , W d , \ „W = SeI---- + 55E1 ;*Cc
dimana e^= 98 ,94 - 92 ,07 = 6 ,8 7 cm.e2= 9 8 ,94 - 13 - 6 ,8 7 = 7 9 ,07 cm.
a / 4 3 7 5 0 3 .8 1 .6 .8 7 . 3 7 0 0 2w s l%735T6$Q-f5777722------ +
5 .4 3 7 5 0 3 ,8 1 .7 9 . 0 7 . 3 7 0 0 2 \ 0W .J 3 W m r 2 f7 T 2 T 2 7 ^ ------
= 13 ,92 cm. i f )
^ t o t a l a k h ir * 1 1 ,0 8 4 + 3 ,1 0 + 0 ,3 3 7 + 2 ,6 4 + 0 ,6 6 -1 3»923 ,902 cm < £ ^ . 3 7 0 0 • 4 ,6 25 cm.
(O .K . )I V . 1 . 1 5 . Perhitungan Plaat:
f ■- 10 to*'
gb . 4 . 2 9 . muatan t i t i k pada p l a t .p
Mutu b a ja yang d ip ak a i » CTjU* = 2080 kg/cm .Hutu be ton B^qq , 6 ^ , = 0^m, - k . s = 0 , 9 5 . 4 0 0 - 2 , 0 5 . 4 6
63
( J ^ , * 286 kg /cm 2
Beban-beban yang b e k e r j a pada p l a t r- b . s . p l a a t = 0 ,1 7 5 * 2 ,5 .1 = 0 ,4375 t /m»*- b . s . a s p a l = 0 , 0 6 . 2 . 1 = 0 ,1 2 t /m**- a k ib a t muatan T, b e k e r ja beban t i t i k s e b e s a r 10 t
gb. 4.30. beban muatan titik. a a lebar kerja plat = 20 + 17,5 * 37,5 cm. sa= 3/4.a + 3/4.r.lx ( lihat P.B.I.'71 a r t 13.4).
= 3/4.0,375 + 3/4.1/2.1,65=r 0 ,9 0 m.
Momen yang b e k e r . ja :( d i dimensi terhadap M ap )
M- ak . b . s . « r y . (0 ,4375 -»• 0 ,1 2 )1 ,5 5 ^ 0 ,112 tn P 0 .7 5 .1 0 .1 .5 5 , non
Mt o t “ 5 >341 tnM « 3341 tm.
W °> 0533q » i/o.
d a r i nomogram; ^ » 0 ,9 35 a x .
* 17 ,18 cm (0 1 9 -1 5 )
A *s 019 - 15.T «l- pembagi= 209617,18 * 3 , 44cm2 (010 -2 0 )
64
Petiy e l.i^ i Van t p'rVi;} rlffn n n n s !
-r ru i , ?• i uuuu4>Pu= rTTa+FJ^ETJHT = ? I T 7 7 T ^ 7 T ^ 7 T 7 3
= 4 ,0 8 kg/cm2< r * pin>u = 1 ,02* 286 =17,25 kg/cm2 .U IIi p u.( p r a k t i s tak d ip er lu k a n t u l . g e s e r ) .
bpm,u
P la a t Tfrpi;Beban-beban yang b e k e r j a :
- b . s . p l a t = 0 , 1 7 5 . 2 , 5 . 1 « 0 ,4 375 t /m * .- b . s . t r o t o i r = 0 , 2 . 2 , 4 . 1 * 0 ,4 8 t / m ’ .- m. h idup t r o t o i r = 1 . 0 , 3 ** 0 ,3 0 t /m * .
Akibat sandaran = 0 , 1 5 * 1 * 0 , 1 5 t o n .Akibat gay& h o r i s o n t a l pada saiidaran; M « 2 . 0 , 1 ; 1 , 1
(sandaran b e r ja r a k 2 m, beban sandaran 0 ,1 t / m ' , be k e r j a pada jarak 0 , 9 m d a r i a ta s l a n t a i kendaraan)
= 0 ,7 5 8 tm.M y g . t e r j a d i pada p l a t t e p i < M pada p l a t tengah . Jadi penulangan p l a t t e p i disamakan s a ja dengan p e - nulangan p l a t tengah.
IV . 1 .1 6 . Kontro l terhadap kekakuafi k o n s t r u k s i :
Munurut p . B . I . ' 7 1 p a sa l 10 .5 ayat 1:
q = 1 ,217$ t/m*
- 0 ,2 2 tm
Mmax. * 1 / 2 . 1 , 2 1 7 5 . 0 , 8252 + 0 , 1 5 . 0 , 8 2 5 + 0 ,2 2
^ 0 , 6 2 cm.f yang t e r j a d i = f g . fiJiEl & .-t..P.*.
_ _ _
dimana: t g = " f i g j )h- 10' 6 (cm)
f = lendutan e l a s t i s a k ib a t beban k e r j a .tsOZ = t e g . t a r i k b a ja yg b e k e r j a d i ternpatCl
lendutan a k ib a t beban k e r j a .1Q = bentang p l a t .a * k o e f . bergantung pada j e n i s b a j a , untuk
penampang b u la t p o l o s = 2 0 ,8 .^ * k o e f » jarak g a r i s n e t r a l d i tempat len du ta n .
pDari perh itungan h a l . 63 , A = A' = 1 7 , 1 8 e r a .
j = 2 ,5 . -B3 . 0 au?(rbk'' f 0 c 1 7 .1 8 2080
f = * T 5 5 .1k = 0 ,1 3 4 3 . ~
* 2080.1 55 6 „Jadl e = 20 ,g ' : (1 " -07 l3^3" ) -10 = 0 ,185 cnu
MfT. k i . k 0 + M_f » f .Sr _____ £e I * ; + M g p
Mg = momen max, a k ib a t beban mati «* 0 ,1 1 2 tm.M = momen max. a k ib a t beban hidup= 3 ,229 tm. k = k o e f . yang bergantung pada bentuk penam
pang , b a lo k I =2.k = ------V - * 0 , 5 .
1k1 .k = 2 . 0 , 5 = 1 ^ 0 , 6 ( O . K ) .
f - ° . 185- g ' T l l ' * ° l 1 ^ . K . ) f ' ° ’ 62 cm-
I V . 1 .17 Kontrol turhadap r e ta k :Menurut P . B . I . ' 7 1 pasa l 10 .7 ^yat 1b, l e b a r re ta k max. w = 0,1 mm.Menurut P . B . 1 , ' 7 1 pasa l 10 .7 ay at 3 , q
l e b a r re tak w = o<(CU.c + C,~4“ ) ( ^ 7 “ — )1 0 “ ^(cm)5 4 CeJ-p a (zJp
dimana: c = t e b a l penutup b e ton = 1 ,5 cm. d = d iam eter batang = 1 ,9 cm.
_____________ ^a = t e g , t a r i k b a ja vang b e k e r ja d i tempat___
65
662
re ta k (kg/cm ) .oU k o e f . yang bergantung pada j e n i s b a ja t u la n g -
an, untuk batang p o l o s = 1 , 2 .Dari t a b e l 1 0 .7 .1 P . B . 1 , ' 7 1 d i p e r o l e h :C3= 1 , 5 .C4 = 0 , 0 4 .Cc- 7 , 5 .<»l_ A 1 7 ,18 nn-t- ic
p T5HT = = *
Jadi w * 1 , 2 . ( 1 , 5 . 1 , 5 + 0 , 0 4 . ft.^ f 3 ) ( 20 80-y - ^ g )1 1= 1 , 2 . 8 , 858. 1427, 8 3 . 10"6 s 0 ,01 cm = 0,1 mm.
( O . K . ) .
I V . 1 . 1 8 . Perhitungan Sandaran:Ukuran sandaran 10 x 15 cm dengan jarak antara 2 m. Gaya h o r i s o n t a l yg b e k e r ja pada t ia n g sandaran s e -
■ b e s a r H « 100 k g / m ' , pada k e t in g g ia n 90 cm d a r i la n t a i kendaraan ( F .K .U .J .J .R . No: 1 2 /1 9 7 0 ) .
*■—. ...3 TC a. v a. ir. Kutu beton K225
M yg b e k e r j a = ( 2 . 1 0 0 ) . ( 0 , 9 + 0 , 2 ) = 220 kgm
Mutu ba ja
Kutu beton ^-^25
Check g e s e r : D = 200 kg .
(cukup b e u g e l p r a k t i s 06- 2Ocm)
67
IV. 1 , 1 91. Perhitungan Gaya B e la h :I V . 1 , 1 9 . A. Perhitungan gaya b e la h ke arah v e r t i k a l :
* Jack ing I y Kabel G.Gaya ja ck k a be l C = = 107304,63 Ig*
= 3Komponen arah v e r t i k a l d ia b a ik a n , maka 106988*7
liSCM/-----1 o r tu /c^ - ,
\W*uir v O w ' A®
i m ,•—* • —« •-*
I V38,1 6**,
jL
c. 8ff-io -►
g b .4 .3Cg b . 4 . 3 0 . j a c k in g pada kabe l C.B =10237,5 cm2 .W'=397753,81 cm3 .W =305772,74 cm3 .^ F .F .e 1 0 6 9 8 8 ,7 4 .1 0 6 9 8 8 ,7 4 ,1 8 ,9 4°ba' = ~ T 0 ’?3T75"h— 5977537ST—
= - 5 ,356 kg /cm 2.CTbb. . . | . % e a .1Q6988X74M 106988 ^ 4 . 1 8 ,94
T>-\T 1 6 2 3 7 , 5 36£7y2,74= - 17 ,077 kg /cm2 .
Mq = 1 / 2 . 5 8 5 , 9 4 . 802 + 1 / 2 . 2 6 7 , 9 1 . 2 / 3 . 802 » 2446549,3 kgcm.
“ b u r s t i n g 1” 3* ■ % * 2446549,5 kgcm. 1 /2Tegangan i j i n untuk b e u g e l ( f ) = (■ ''■ ! ■ c— —)
8 6 2dimana Es = mod* e l a s t i s i t a e b a j a = 2 , 1 . 1 0 kg/cm .f c '= kekuatan tekan s i l i n d e r beton
= 0 ,8 3 .4 0 0 = 332 kg /cm 2 *w = l e b a r r e ta k , d ia m b i l w = w =0,01 cm.A_ = lu a s beu g e l = 1/4.77*. 1 , 2 2= 1 ,1 3 cm2 .D
J a d i ’ f = ( 4 . 2 . 1 . 1 0 6 .Vr33~2,0J0 l ) l / 2 a 1163 ,8 2 kg /cm2M 1,13
Ft =■ » dimana: F ^ gaya t a r i k t o t a l ,M = momen l o n g i t u d i n a l max.
_________________ __ z = j arak antara u.iung b a lo k k e -
68
t i t i k b e r a t t u l penampung gaya belahl / 2 . h .
h= t i n g g i b a lo k .Dari perh itungan d i a t a s , d ida p a t :
Ft = 2446^49, 3 = 18122,59 k g .(harga z d i t a k s i r = 40 cm) .
Luas b e s i yg dibutuhkan = = 15 ,57 cm2 .Dipasang beu g e l 4 s n e d ig .
1 5 57Banyak j a j a r a n = = 3 , 4 5 ^ 4 ,
$2-1**.- JrJack ing I I , Kabel D.
« 2764 ,5 + 928!- 95079 ,45 kg .
i fraAeJt 'Saw. feugjl S.<f u*\ J
FjD “ 27 64 ,5 + 92826 ,82 » 95591 ,32 kg .
HDPHC « 106988,74
= 106988,74 - 66 4 ,7 * 106324,04 kg .\ H=>of. ‘Qj/uJ-
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gb* ‘4 .3 1 . j a c k in g pada kabe l D.F ^reeu ltan = 95079 ,45 + 106324,04 = 201403,5 kg... 9 5 07 9 .4 5 .10 5 + 10 632 4 .04 .80 __ y R - — 20140575---------- ----------- 3 91 ,80 cm*
201403.5 . 2 0 1 4 0 3 .5 .7 ,1 4 nfi , . r , rm2 ° b a ‘ ~ TU ZTT't' + “ 5 9 7 7 * 3 7 ^ ---------1 6 ,0 6 kg/cm .
201403.5 2 0 1 4 0 3 .5 .7 .1 4 i._/_^.2^b b '* “ m > v r , i ------------3 ' ' 0 5 7 ^ 7 7 l " * ~2 4 *376 k^ cm -Mc = 1 / 2 . 1 0 2 8 , 6 8 . 802 + 1/ 2 . 8 0 2 . 2 / 3 . 1 9 0 , 1 2
« 3697365,3 kgcm.Mc+5= 1 / 2 . 1 0 1 6 , 8 . 852 + 1 / 2 . 852 . 2 /3 .2 0 2 - 10 632 4 ,0 4 .5
» 3628053,1 kgcm.Mc+10= 1 / 2 . 1 0 0 4 , 9 . 902 +1/ 2 . 902 . 2 / 3 . 2 1 3 ,8 8 6 - 1 0 6 3 2 4 ,0 4 .1 0
69
MC+1CT 3584096,8 kgcm.\ = 1/2.91 ,8 2 . 1000,64+1/2.91,8 2 . 2 /3 .218,16 ■
106324.04.11.8 = 3574521,9 kgcm.KR+5 = 1/ 2 . 9 6 ,82 . 988 ,75+1 /2 .96 ,82. 2 /3 .230 ,05 -
106324.04.16.8 = 3564709,8 kgcm.Mr+10= 1 /2 .1 0 1 ,8 2 . 976 ,87+1 /2 .101 ,8 .2 /3 .241 ,93 .
106324.04.21.8 = 3579631,3 kgcm.
MD « 1 / 2 . 1052. 969 ,27+1 /2 .1052. 2 /3 .249 ,53 -106324,04.25 = 3602022,5 kgcm.
Jadi rynax. = 3697365,319 kgcm.
' * JACKING I I I , Kabel B.F^b * 2040,65 + 93696 ,32 = 95736,97 kg.
a Fb « 8 4 , 5 2 . 7 , 9 = 667 ,74 kg.PHD " 95079 ,45 - 667 ,74 = 94411,71 kg . FH0 * 106324,04 - 667 ,74 » 105656,3 kg.
gb . 4 .3 2 ja c k in g pada kabel B.
FHR = 295584,53 kg. y R = 79,91 cm.
Banyak jajax-an ^
( jaAaJt AjScfui &eujpQ
95516 ,523 kg n.F-n* IJC ■'■D =
70(T . 295584,53 . 295584,53 .19 .03 u 7 , Vc/rm2ba'----TO7 W ? r + ----* kg/cm *
^ . 295584,53 295584,53.19,03 , 7 „ 7 it(r/ cm2M)b1 ------------- W H T T T rf------- “ 47,27 kg/cm .Mg = 1 / 2 . 552 . 1852,16 + 1/ 2 . 5 5 2 . 2 / 3 . 5 1 1,34 =3316993,1 kgc Mg+5 « 1 / 2 . 602 . 1 8 0 5 ,7+ 1 /2 .602. 2 /3 .5 57 ,8 3 -9 5516 ,52 3 .5
= 3442073,4 kgcm.Wb+10- 1/ 2 . 6 5 2 . 1759,2+1/ 2 . 6 5 2 . 2 /3 .604 ,3 1 -9 551 6 ,5 23 .10
= 3612214,6 kgcm.MB+15" 1/ 2 . 7 0 2 . 1712,7+1/ 2 . 7 0 2 . 2 /3 .6 5 0 ,8 -9 5 516 ,5 23 .1 5
a 3826340,5 kgcm.MB+2oa3 l / 2 . 7 5 2 .1 6 6 6 ,2 1 + 1 /2 .7 5 2 .2 /3 .6 9 7 ,3 -9 5 5 1 6 ,5 2 3 .2 0
= 4085322,7 kgcm.= 1 / 2 . 7 9 , 912 . 1 6 2 0 ,56 + 1 /2 .7 9 ,912. 2 /3 .7 4 2 ,9 3 -
9 5 51 6 ,5 2 3 .2 4 ,9 1 = 4376167,1 kgcm.Mc » 1 / 2 . 802 . 1 6 1 9 , 7 3 + 1 /2 .802 . 2 /3 .7 4 3 ,7 7 - 9 5 5 1 6 ,5 2 3 .2 5
=> 4381932,3 kgcm.Mc+5 = 1 / 2 . 852 . 1 5 7 3 , 2 4 + 1 /2 .852 . 2 / 3 . 7 9 0 ,2 6 - 9 5 5 1 6 ,5 2 3 .3 0 -
1 0 5 65 6 ,3 .5 = 4192761,8 kgcm.Mc +10= 1 / 2 . 902 . 1 5 2 6 , 7 6 + 1 /2 .902 . 2 / 3 . 8 3 6 ,7 4 -9 5 5 1 6 ,5 2 3 .3 5 -
1 0 56 5 6 ,3 .10 = 4042934,7 kgcm.MC+15“ 1/ 2 . 9 5 2 . 1 4 8 0 , 3 + 1 / 2 . 952 . 2 / 3 .8 8 3 ,2 - 9 5 5 1 6 ,5 2 3 .4 0 -
10 56 5 6 ,3 .1 5 = 3931308,2 kgcm.Mc +20= 1/ 2 . 1 0 0 2 . 1 4 3 3 , 7 8 + 1 /2 .1002 . 2 / 3 . 9 2 9 , 7 1 - 9 5 5 1 6 , 5 2 3 . 4 5 -
1 0 56 5 6 ,3 .20 = 3856564 kgcm.Md = 1 / 2 . 1052 . 1 3 8 7 , 3 + 1 / 2 . 1052 . 2 / 3 . 9 7 6 ,2 - 9 5 5 1 6 ,5 2 3 .5 0 -
10 565 6 ,3 .25 = 3817792,5 kgcm.Jadi MLmax. =4381932,3 kgcm.
Ft = 32458,76 kg.
Banyak ja j a r a n = g y g g = 6,7<^> 7*Ma- * 'H *
i
71
Jack ing IV, Kabel E.Pjj ; = 106053,83 kg .F = 105329,22 kg.
™ 5 ,9 4 .1 0 5 3 2 9 ,2 2 _ _ , 2 *°E “ - * — g'7 f g 95" ----- * 93,21 kg/cm .*TE » 9 3 , 2 1 . 7 , 9 = 736 ,34 kg.FHB “ 95516 ,523 - 73 6 ,34 = 94780 ,18 kg.
= 94411,71 - 736 ,34 * 93675 ,37 kg .Fu„ = 105656 ,3 - 7 36 ,3 4 = 104919,96 kg.
-----------------HC
*• I3S
V
0 « _< f> ,T '/-to
IlaSf, kj
34»t
&
/u*xg b . 4 .3 3 ja c k in g pada kabe l E.
FHR “ 598704,73 kg . y R * 9 3 ,1 4 cm.J 1 .. 398704.73 . 3 9 8 7 0 4 .7 3 .5 .8 ,__.2t fba '3 T 0 2 3 7 t5 " + " 3~9 7 7 5 j , 8 1 ' ' c " 33» 13 k£ ' cin .
. 398704.73 . 3 9 8 7 0 4 ,7 3 .5 .8 _ ^ , _ 2° b b ‘ “ iT5'237l5----------53577ft ;7 4 ----------46,51 kg' c “ ‘
Mb » 1 / 2 . 552 . 2 1 1 5 , 2 4 + 1 / 2 . 552 . 2 / 3 . 2 1 0 , 2 6 - 3411312,6 kgcm Mb +5 » 1 / 2 . 602 . 2 0 9 6 , 1 3 + 1 /2 . 6 0 .2 /3 .2 2 9 ,4 - 9 4 7 8 0 ,1 8 * 5
« 3574413,1 kgcm.1/ 2 . 7 0 2 . 2057 ,9+1 / 2 . 7 0 2 . 2 / 5 . 2 6 7 , 6 - 9 4 7 8 0 , 1 8 . 1 04531133,1 kgcm.l / 2 . 7 5 2 . 2 0 3 8 , 8 + l / 2 . 7 5 2 . 2 / 3 . 2 8 6 , 7 1 - 9 4 7 8 0 , 1 8 . 1 5
M.B+10"
^B+1
M,4850003,5 kgcm.1 / 2 . 802 , 2019,7+1/ 2 . 8 0 2 . 2 /3 .3 0 5 , 8 3 - 9 4 7 8 0 , 1 8 . 2 0
= 5219873,7 kgcm.MC+5 = 1 / 2 . 852 . 2 0 0 0 , 5 6 + 1 /2 .852 . 2 /3 .3 2 4 ,9 4 - 9 4 7 8 0 , 1 8 . 2 5 -
1 0 4 9 1 9 ,9 6 .5 = 5115482,5 kgcm.M
Mr
C+10= 1/ 2 . 9 0 2 . 1981 , 4 4 + 1 / 2 . 902' . 2 / 3 . 3 4 4 ,0 6 -9 4 7 8 0 , t 8 .3 0 -1 0 4 91 9 ,9 6 .10 a 5061189 kgcm.
= 1 / 2 . 9 3 , 142 . 1 9 6 9 , 4 4 + 1 / 2 . 9 3 , 142 . 2 /3 . 3 5 6 , 0 6 -
72
Mgn
M100
9 4 7 8 0 , 1 8 . 3 3 , H - 1 0 4 9 1 9 , 9 6 .13 ,14 = 5052455,1 kgcm.
MR+5 = 1 / 2 . 9 8 , 142195 0 , 3 2 + 1 / 2 . 9 8 , 142 . 2 / 3 . 3 7 5 , 2 -94780,1 8 .3 8 ,1 4 - 1 0 4 9 1 9 , 9 6 .1 8 ,1 4 = 5078624 kgcm.
= 1 / 2 . 9 8 , 942 . 1 9 4 7 , 3 + 1 / 2 . 9 8 , 942 . 2 / 3 . 3 7 8 , 2 -9 4 7 8 0 ,1 8 .3 8 , 9 4 - 1 0 4 9 1 9 ,9 6 . 1 8 ,9 4 = 5087338 kgcm.
■ 1 / 2 . 1 0 0 2 . 1 9 4 3 ,2 1 + 1 /2 .1 0 0 2 . 2 / 3 . 3 8 2 , 3 - 9 4 7 8 0 , 1 8 . 4 0 -10 491 9 ,9 6 .2 0 = 5100776,8 kgcm.
x 1 / 2 . 1052 . 1 9 2 4 , 1 + 1 / 2 . 1052 . 2 / 3 . 4 0 1 , 4 - 9 4 7 8 0 , 1 8 . 4 5 -1 049 1 9 ,9 6 .2 5 = 5193639 kgcm.
- 1 / 2 . 1102 . 1 9 0 4 ,9 8 + 1 /2 .1 1 0 2 . 2 / 3 . 4 2 0 , 5 2 - 9 4 78 0 ,1 8 .5 0 -1 0 4 9 1 9 ,9 6 .3 0 -9 3 6 7 5 ,3 7 .5 =* 4866241 ,4 kgcm.1/ 2 . 1152 . 1 8 8 6 , 4 3 + 1 /2 . 1 1 52 . 2 / 3 . 4 4 0 , 0 7 - 9 4 7 8 0 , 1 8 . 5 5 -1 0 4 9 1 9 ,9 6 .3 5 -9 3 6 7 5 ,3 7 .1 0 =4592131,7 kgcm.
5=1/2 * 1 2 0 2 . 1867 ,3+1 / 2 . 1 2 0 2 . 2 / 3 . 4 5 9 , 2 - 9 4 7 8 0 , 1 8 . 6 0 -
MD+5
MD+1 O'
M.L + 2 0 ‘
1 0 4 9 1 9 ,9 6 .4 0 -9 3 6 7 5 ,3 7 .1 5 » 4359981,1 kgcm.r
4174513,6 kgcm.1/ 2 . 1 2 5 2 . 1 8 4 8 , 1 7 + 1 / 2 . 1252 . 2 / 3 . 4 7 8 , 3 3 - 9 4 7 8 0 , 1 8 . 6 5 -
ME
Jadi
1 0 4 9 1 9 ,9 6 .4 5 -9 3 6 7 5 ,3 7 .2 01/ 2 . 1 3 0 2 . 1 8 2 9 , 0 3 + 1 /2 .1302 . 2 / 3 . 4 9 7 , 5 - 9 4 7 8 0 , 1 8 . 7 0 - 1 0 4 1 9 ,9 6 .5 0 - 9 3 6 7 5 ,3 7 .2 5 = 4035391,81 kgcm.
5219873,7 kgcm.
Ft = m 38665»73 kE‘Banyak jajaran * 4 ^ 1 B2 * 7 ,3 5 ^ 8 .
i A>C*v 9 fS t o
hi-MjyJk. tc C**,
♦JACKING V, Kabel A. F3A = 104313,33 kg . FHA “ 1°3672,01 k g .AC3A = 5>67 12°567^ " “ 91 »74 kg /cm2 .*F a » 7 , 9 . 9 l ! 7 4 « 724 ,25 kg .Fhe = 105329,22 - 72 4 ,25 = 104604,47 k g .
I X
■HB?HD?HC
= 94 780 ,18 - 724 ,75 = 94055 ,43 k g .= 93675 ,37 - 724 ,75 = 92950 ,62 k g .= 104919,46 - 724 ,75 = 10419,21 kg .
wc
22
gb. 4.34 jacking padia kabel A.Fhr = 499477,74 kg.yR «= 80,04 cm.
^ 499477,74 . 499477.74.18,9 _Cba*3 * n r m T t r + "" * A i i h f ; z r ------------25 ,05 kg/cm_ 499477.74 499477.74.18,9 _ ™ ^ i . _ / _ 2Qbb'= “ ’~TG73Tp"5-----------3U57f?rJT~ -79,66 kg/cm .
» A - 1 / 2 . 302. 3514,9+1/2 .302. 2/3.468,1 « 1722135 kgcm.MA+5 “ 1 / 2 . 352 . 3436,89+1/ 2 . 3 5 2. 2 /3 .546,1-103672,01.5
« 1809725,9 kgcm.MA+10“ 1 / 2 . 402 . 3358,87+1/2 .402 . 2/3.624,13-103672,01*10
« 1983245,2 kgcm.MA+15“ 1/ 2 .4 5 2. 3280,86+1/ 2 . 4 5 2. 2/3 .702,14-103672,01.15
o 2240735,1 kgcm.MA+20“ 1 / 2 . 502. 3202,84+1/2 .502. 2/3 .780,16-103672,01.20
- 2580243,1 kgcm.Mb = 1 / 2 . 552. 3124,82+1/2 .552 . 2/3.858,17-103672,01 *25
* 2999811,3 kgcm.MB+5 " 1 / 2 . 602 . 3046,81+1/2.602. 2 /3 .936 ,19-103672,01 .30-
94055,43.5 = 3027248,6 kgcm.Hg+1o= 1 / 2 . 652 . 2968,8+1/2.652. 2/3.1014,2-103672,01 .35-
94055,43.10 = 3130847 kgcm.Mb+15- 1 /2 .7 0 2 . 2890,78+1/2 .702 . 2/3.1092,22-103672,01.40-
94055,43.15 » 3308658,4 kgcm.Mb +20= 1 / 2 . 752. 2812,76+1/2.752 . 2/3.1170,24-103672,01.45-
74
M,9 4 0 5 5 ,4 3 .2 0 = 3558738,4 kgcm.1/ 2 . 8 0 2 . 2734,74+1/ 2 . 8 0 2 . 2 /3 .1 2 4 8 ,2 6 -1 0 3 6 7 2 ,0 1 . 5 0 -9 4 05 5 ,4 3 .2 5 = 3879136,3 kgcm.
Mc+5 = 1 / 2 . 852 . 2 6 5 6 , 7 3 + 1 /2 .852 . 2 /3 .1 3 2 6 ,2 7 -1 0 3 6 7 2 ,0 1 . 5 5 - 9 4 0 5 5 ,4 3 .3 0 - 1 0 4 1 9 5 ,2 1 .5 = 3746938,1 kgcm.
MC+10" 1 / 2 . 902 . 2 5 7 8 , 7 1 + 1 /2 .902 . 2 /3 .1 4 0 4 , 2 9 - 1 0 3 6 7 2 , 0 1 . 6 0 - 9 4 0 5 5 ,4 3 .3 5 - 1 0 4 1 9 5 ,2 1 .1 9 = 3681145,9 kgcm.
Mc+15« 1 / 2 . 952 . 2 5 0 0 , 7 + 1 / 2 . 952 . 2 / 3 . 1 4 8 2 , 2 3 - 1 0 3 6 7 2 , 0 1 . 6 5 - 9 4 0 5 5 ,4 3 .4 0 - 1 0 4 1 9 5 ,2 1 .1 5 « 3679834,7 kgcm.
MC+20'
MD
1/ 2 . 1 0 0 2 r2 4 2 2 ,7+1/ 2 . 1 0 0 2 . 2 /3 .1 5 6 0 ,3 -1 0 3 6 7 2 ,0 1 . 7 0 - 9 4 0 5 5 ,4 3 .4 5 - 1 0 4 1 9 5 ,2 1 .2 0 = 3741060,8 kgcm.
= 1 / 2 . 1052 . 2 3 4 4 , 7 + 1 / 2 . 1052 . 2 /3 .1 6 3 8 , 3 - 1 0 3 6 7 2 , 0 1 . 7 5 -3862858,8 kgcm.9 4 0 5 5 ,4 3 .5 0 - 1 0 4 1 9 5 ,2 1 .2 5
Md+5 = 1 / 2 . 1 102 . 2 2 6 6 , 6 5 + 1 / 2 . 1102 . 2 /3 .1 7 1 6 ,3 5 - 1 0 3 6 7 2 ,0 1 . 8 09 4 0 5 5 ,4 3 .5 5 - 1 0 4 1 9 5 ,2 1 .3 0 - 9 2 9 5 0 ,6 2 .5 = 3578425,6 kg^m
Md+i o = 1/ 2 . 1152 . 2188,63+1/ 2 . 1 1 5 2 . 2 /3 .1 7 9 4 ,3 7 - 1 0 3 6 7 2 ,0 1 . 8 5 9 4 0 5 5 ,4 3 .6 0 - 1 0 4 1 9 5 ,2 1 * 3 5 - 9 2 9 5 0 ,62.10»335071 2 ,5 kgciji
Md+15= 1/ 2 . 1202 . 2110,62+1/ 2 . 1 ? 0 2 . 2 / 3 . 1 8 7 2 , 4 - 1 0 3 6 7 2 , 0 1 . 9 0 - 9 4 0 5 5 ,4 3 .65>-104195,21 .40 -9 2 9 5 0 ,6 2 .1 5 = 3 1 7 7 8 3 4 ,3 kgcdi
MD+20B
ME
1/ 2 . 1 2 5 2 . 2032 ,6+1 / 2 . 1 2 5 2 . 2 / 3 . 1 9 5 0 , 4 - 1 0 3 6 7 2 , 0 1 . 9 5 - 9 4 0 5 5 ,4 3 .70 -10 4 19 5 ,21 .45 -9 2 9 5 0 ,6 2 .2 0 = 3 0 5 7 5 0 3 ,2 kgcrji 1/ 2 . 1 3 0 2 . 1954 ,5+1 / 2 . 1 3 0 2 . 2 /3 .2 0 2 8 ,4 -1 0 3 6 7 2 ,0 1 . 1 0 0 - 94055m43.7 5 -1 0 4 1 9 5 ,2 1 .5 0 -9 2 9 5 0 ,6 2 .2 5 = 2 9 8 8 1 3 9 ,5 kgcn
Jadi Mmmax.= 3879136,3 kgcm.T7r ” '4'&?- = 28734,342 kg .
,ifC
75
Jack ing V I , Kabel F. F_.p = 92077 ,72 kg .J’ = 91131 ,216 kg ,
5 .9 4 .9 1 1 3 1 ,2 1 6 2
Fha « 103672,011 - 6 3 7 ,0 8 = 103034,93 kg, Fhe = 104604,47 - 6 3 7 ,0 8 = 103967,40 k g . Fjjg = 94055 ,43 - 6 3 7 ,0 8 = 93418,35 kg . Fhd = 92950 ,62 - 6 3 7 ,0 8 = 92313 ,54 k g . Fhc = 104195,21 - 63 7 ,0 8 = 103558,13 kg .
Ma = 1/ 2 . 302 . 3 3 5 5 , 7 + 1 / 2 , 302 . 2 /3 .2 1 1 ,81 = 1 57 3 60 8 kgcm. MA+5 = 1 / 2 . 3 5 ?3 3 2 0 ,4 + 1 / 2 . 352 . 2 /3 * 2 4 7 ,1 - 1 0 3 0 3 4 ,9 3 .5
= 1619469,5 kgcm.Ma+i 0 = ' / 2 . 4 ° 2 . 3 2 8 5 ,1 + 1 /2 . 4 0 2 . 2 / 3 .2 8 2 ,4 - 1 0 3 0 3 4 ,9 3 * 1 °
= 1748344 kgcm.
MA+15= 1/ 2 . 4 5 2 . 3249 ,8+1 / 2 . 4 5 2 . 2 /3 .3 1 7 ,7 - 1 0 3 0 3 4 ,9 3 .1 5 = 1 9 59 3 4 6 ,1kgcm.
Ma+20= 1 / 2 . 5 0 2 . 3 2 1 4 , 5 + l / 2 . 5 0 2 . 2 / 3 . 3 5 3 - 1 0 3 0 3 4 , 9 3 . 2 0 = 22 51593 kgcm.
Mg = 1 / 2 . 552 . 3 1 7 9 , 2 + 1 / 2 . 552 . 2 / 3 . 3 8 8 ,3 - 1 0 3 0 3 4 ,9 3 .2 5 = 2624202,6 kgcm.
Mb+5 = 1 / 2 . 602 . 3 1 4 3 , 9 + 1 / ? . 602 . 2 / 3 . 4 2 3 , 6 - 1 0 3 0 3 4 , 9 3 . 3 0 -
g b . 4 .3 5 j a c k in g pada k a b e l F.Fhr = 587423,56 kg .
i s
9 3 4 1 8 , 3 5 . 5 = 2 6 0 9 2 0 0 ,4 kgcm,
Mb+1q= 1 / 2 . 6 5 2 . 3 1 0 8 , 6 + 1 / 2 . 6 5 ? . 2 / 3 . 4 5 8 , 8 - 1 0 3 0 3 4 , 9 3 . 3 5 -9 3 4 1 8 , 3 5 . 1 0 = 2 6 7 2 7 9 5 ,6 kgcm.
Mb + i 5= 1 / 2 . 7 0 2 r3 0 7 3 , 3 + 1 / 2 . 7 0 2 . 2 / 3 . 4 9 4 , 2 - 1 0 3 0 4 3 , 9 3 . 4 0 -
9 3 4 1 8 , 3 5 . 1 5 = 2 8 1 4 1 0 5 ,9 kgcm.
MB+20= 1/ 2 . 7 5 2 . 3 0 3 7 , 9 7 + 1 / 2 . 7 5 2 . 2 / 3 . 5 2 9 , 5 3 - 1 0 3 0 4 3 , 9 3 . 4 5 -9 3 4 1 8 , 3 5 . 2 0 = 3 0 3 2 2 2 0 ,5 kgcm.
Mq = 1 / 2 . 8 0 2 . 3 0 0 2 , 6 7 + 1 / 2 . 8 0 2 . 2 / 3 . 5 6 4 , 8 3 - 1 0 3 0 3 4 , 9 3 . 5 0 -
9 3 4 1 8 , 3 5 . 2 5 = 3 3 2 6 3 0 9 ,3 kgcm.
M0+5 = 1 / 2 . 8 5 2 . 2 9 6 7 , 4 + 1 / 2 . 8 5 2 . 2 / 3 . 6 0 0 , 1 - 1 0 3 0 3 4 , 9 3 . 5 5 -9 3 4 1 8 , 3 5 . 3 0 - 1 0 3 5 5 8 , 1 3 . 5 = 3 1 7 7 7 0 9 ,9 kgcm.
Mc +10= 1/ 2 . 9 0 ? . 2 9 3 2 , 1 + 1 / ? . 9 0 2 . 2 / 3 . 6 3 5 , 4 - 1 0 3 0 3 4 , 9 3 . 6 0 -9 3 4 1 8 , 3 5 . 3 5 - 1 0 3 5 5 B , 1 3 .1 0 = 3 1 0 3 2 6 6 ,5 kgcm.
M0+ 1 5 = 1 / 2 . 9 5 2 . 2 8 9 6 , 7 7 + 1 / 2 . 9 5 2 . 2 / 3 . 6 7 0 , 7 3 - 1 0 3 0 3 4 , 9 3 . 6 5 - 9 3 4 1 8 , 3 5 . 4 0 - 1 0 3 5 5 8 , 1 3 . 1 5 = 3102077 ,1 kgcm.
Mc +2o= 1/ 2 . 1 0 0 2 . 2 8 6 2 , 5 + 1 / 2 . 1002 . 2 / 3 . 7 0 6 - 1 0 3 0 3 4 , 9 3 . 7 0 - 9 3 4 1 8 , 3 5 . 4 5 - 1 0 3 5 5 8 , 1 3 . 2 0 = 3 1 7 3 3 9 9 ,7 kgcm.
Md = 1 / 2 . 1 0 5 2 . 2 8 2 6 , 2 + 1 / 2 . 1052 . 2 / 3 . 7 4 1 , 3 - 1 0 3 0 3 4 , 9 3 . 7 5 - 9 3 4 1 8 , 3 5 . 5 0 - 1 0 3 5 5 8 , 1 3 . 2 5 = 3 3 1 6 2 1 4 ,8 kgcm.
MD+5 = 1 / 2 . 1102 . 2 7 9 0 , 8 6 + 1 / 2 . 1102 . 2 / 3 . 7 7 6 , 6 4 - 1 0 3 0 3 4 , 9 3 . 8 0 -
9 3 4 1 8 , 3 5 . 5 5 - 1 0 3 5 5 8 , 1 3 . 3 0 - 9 2 3 1 3 , 5 4 . 5 = 3 0 6 8 0 3 5 ,4 kgcmMD+10= 1/ 2 * 1 1 5 2 “ 27!55 , 5 + 1 / 2 . 1 1 5 2 . 2 / 3 . 8 1 1 , 9 - 1 0 3 0 3 4 , 9 3 . 8 5 -
+ 9 3 4 1 8 , 3 5 . 6 0 - 1 0 3 5 5 8 , 1 3 . 3 5 - 9 2 3 1 3 , 5 4 . 1 0 = 2 8 8 9 1 2 8 , 1 kgcm.
MD+15= 1/ 2 * 1 2 0 2 . 2 7 2 0 , 3 + 1 / 2 . 1202 . 2 / 3 . 8 4 7 , 2 - 1 0 3 0 3 4 , 9 3 . 9 0 -9 3 4 1 8 , 3 5 . 6 5 - 1 0 3 5 5 8 , 1 3 . 4 0 - 9 2 3 1 3 , 5 4 . 1 5 = 2 7 8 0 3 5 6 , 7 kgcm.
Md +2 0 = 1 / 2 . 1 2 5 2 . 2 6 8 4 , 9 + 1 / 2 . 1252 . 2 / 3 ^ 8 8 2 , 6 - 1 0 3 0 3 4 , 9 3 . 9 5 -9 3 4 1 8 , 3 5 . 7 0 - 1 0 3 5 5 8 , 1 3 . 4 5 - 9 2 3 1 3 , 5 4 . 2 0 = 2 7 3 8 6 6 6 , 4 kgcm.
Me = 1 / 2 . 1302 . 2 6 4 9 , 6 + 1 / 2 . 1302 . 2 / 3 . 9 1 7 , 9 - 1 0 3 0 3 4 , 9 3 . 1 0 0 -
9 3 4 1 8 , 3 5 . 7 5 - 1 0 3 5 5 8 , 1 3 . 5 0 - 9 2 3 1 3 , 5 4 . 2 5 = 2 7 6 4 3 4 2 , 5 kgcm.ME+5 = 1 / 2 . 1 3 5 2 . 2 6 1 4 , 3 + 1 / 2 . 1352 . 2 / 3 . 9 5 3 , 2 - 1 0 3 0 3 4 , 9 3 . 1 0 5 -
9 3 4 1 8 , 3 5 . 8 0 - 1 0 3 5 5 8 , 1 3 . 5 5 - 9 2 3 1 3 , 5 4 . 3 0 - 1 0 3 9 6 7 , 4 , 5« 2 3 3 6 4 2 2 ,8 kgcm.
Me +10= 1/ ? *1402 . 2579 + 1/ 2 . 1 4 0 2 . 2/ 3 . 9 8 8 , 5 - 1 03034 , 9 3 . n o - 934 1 8 , 3 5 . 8 5 - 1 0 3 5 5 8 , 1 3 . 6 0 - 9 2 3 1 3 , 5 4 . 3 5 - 1 0 3 9 6 7 , 4 . 10
= 19738 6 3 ,1 kgcm.
ME+15= 1/ 2 . 1 4 52 . 2 5 4 3 , 7 + 1 / 2 . 1452 . 2 / 3 . 1 0 2 3 , 8 - 1 0 3 0 3 4 , 9 3 . 1 1 5 - 9 3 4 1 8 , 3 5 . 9 0 - 1 0 3 5 5 8 , 1 3 . 6 5 - 9 2 3 1 3 , 5 4 . 4 0 - 1 0 3 9 6 7 , 4 . 1 5
= 1 6 7 5 7 7 9 ,4 kgcm. ME+20= 1/ 2 . 1 5 0 2 . 2 5 0 8 , 4 + 1 / 2 . 1502 . 2 / 3 . 1 0 5 9 , 2 - 1 0 3 0 3 4 , 9 3 . 1 2 0 -
9 3 4 1 8 , 3 5 . 9 5 - 1 0 3 5 5 8 , 1 3 . 7 0 - 9 2 3 1 3 , 5 4 . 4 5 - 1 0 3 6 9 7 , 4 . 2 0 = 1 4 4 1 2 8 9 ,7 kgcm. = 1 / 2 . 1552 * 2 4 7 3 , 1 5 + 1 / 2 . 1552 . 2 / 3 . 1 0 9 4 , 3 5 - 1 0 3 0 3 4 , 9 3 . 1 2 ^
9 3 4 1 8 , 3 5 . 1 0 0 - 1 0 3 5 5 8 , 1 3 . 7 5 - 9 2 3 1 3 , 5 4 . 5 0 - 1 0 3 6 9 7 , 4 . 2 5S mr= 1269711 kgcm.
Jadi M max. = 3 3 2 6 3 0 9 ,3 kgcm. m -
________________________________________________________________________ 77_
“ = 24639 ,33 kg .24639,33Banyak j a j a r a n = 4 J , 1 3 . 1 jS TTffg * 4 , 6 8 c * 5
KESIMPULAN:Mmmax. t e r j a d i pada ja c k in g ke IV, dimana “banyak j a j a r a n yang d ip e r lu k a n = 8.Jad i a k ib a t gaya b e la h ke arah v r t i k a l i n i d ipasang t u l . b eu g e l 4 sn e d ig b e r ja r a k 10 cm Berta d ipasang sampai s e - jauh 175 cm d a r i t e p i .
I V . 1 . 1 9 . B . Perhitungan gaya be lah pada p l a t :Tegangan max. a k ib a t gaya b e lah yang t e r j a d i pada p l a t , t imbul pada penge jackan ke VI, y a i t u pada k a b e l F.( l i h a t h a l . 7 5 ) .Tegangan p l a t r a t a - r a t a = 4§x54.+ 4 9 ,H - 1 7 , 5 - 837 ,8125 kgcijiK max. yang t e r j a d i = 1 / 2 . 8 3 7 , 8 1 2 5 . 8 2 , 52= 2851180,6 kgcm.Harga h = 165 cm, harga z d i t a k s i r 60 cm. p 2851180,6 0 7 1 c . 1 ,Ft = T5 T '- 5 o~ 271 54,1 k g ‘Digunakan tulangan 0 19 mm.A = 1 /4 . 77*. 1 , 9 2 = 2 , 83 5 c m L , f . ( 4 . 2 , 1 . 1 0 " . 332 .0 .01 % , ._ /__2
** "27^55 2 715 4. 1 2lu a s tu langan yang dibutuhkan = = 3 6 ,96 kg/cm .a & f Pt- , 10
•) = 734 ,76 kg/cm'
Banyak ja j a r a n = = 13.cm-. * • „ i x t 60 -4 ja rak t u l . = — g" = 9 cm.
(d ipasang sampai se jauh 165 cm)r
Jadi t i a p m‘ , butuh A a 3 1 ,5 cm'3 c i t .
f t* j-
78
Resume t u l . p l a t :A kibat muatan T butuh A = A' = 17 ,18 cm2 . ( l i h a t h a l 6 3 ) . A kibat gaya b e lah butuh t u l . = 3 1 ,5 cm . (sam p ai se jauh 165' cm).Jad i sampai 165 cm d a r i ujung butuh t u l . t o t a l *
a 3 1 ,5 + 2 . 1 7 , 1 8 s 65 ,86 cm2 .Dipasang rangkap 3 , j a d i t u l . yang dibutuhkan =
= 1 / 3 . 6 5 , 8 6 = 21 ,96 cm2 .( j* 19 - 12 ,5 cm).
I V . 1 . 1 9 . G . Perh itungan gaya b e la h searah l e b a r b a l o k . ( h o r i s o n t a l )
D i t i n j a u ja c k in g Y I , dimana F ^ ** 587423,56 kg.Gaya yg t e la h didukung p l a t * 837 ,8 125 .165 « 138239,06 kf;Gaya F yang harus didukung t u l . searah l e b a r b a lo k o
•= 587423,56 - T?8239,06 = 449184,5 kg.Dipergunakan g r a f i k lyen gara ( l i h a t la m p lr a n ) .a = 215 mm. ?1 cd * 50 cm. J 1 = 501 - “ 0 ,4 3Dari g r a f i k : Z = 0 ,1 2 6 .4 4 9 1 8 4 ,5 = 56597,25 kg.Digunakan t u l 0 12 mm.
= 1,131 cm2 , f = 1163,82 kg/cm2 , s sDipasang b e u g e l 9 s n e d ig .Banyak ja j a r a n a '1 1'jT ** 4 »8 07 5
(d ip a sa n g se jauh 50 cm).
( r la C4„ j
79
I V , 2 , Perhitungan Landasan:landasan direncanakan menggunakan " F r e y s i n e t Rubber B ear ing P ad" ,Data-dat3. yang hairus d i k e t a h u i :- beban v e r t i k a l max, t i a p p l a t .................................... Pmax(t)- beban v e r t i k a l min, t i a p p l a t , . . ......................, ,=? Pm in(t )- pergesekan h o r i s o n t a l m a x , / / s b . jem batan , , . * a. (cm)- pergesekan h o r i s o n t a l max. -L s b , j e m b a t a n . . ,= A.j(cra)- r o t a s i h o r i s o n t a l max. / / s b . jembatan............ .. 04 ( r a d )- r o t a s i h o r i s o n t a l max. J_ sb . jembatan............ * ( r a d )- gaya h o r i s o n t a l max. / / s b . jembatan................. = H ( t )- gaya h o r i s o n t a l max. _L sb . jembatan.................= ( t )- ruangan yg ada untuk landasan / / s b . jemb, , .= : ^ o (cm)- ruangan yg ada untuk landasan J_ s b . j e m b . . . . = (cm)- ukuran p l a t landasan / / s b , jembatan................. .. X ( c m )- ukuran p l a t landasan sb . jembatan.................=- sudut d i s t o r s i a k ib a t H atau A ....................T f (rad )- t e b a l t o t a l da r i k a r e t .................................................. d (cm)
I V . 2 . 1 . P e rh i tu ngan beban v e r t i k a l :Beban v e r t i k a l max. 1 g e la g a r (D ^ ) = 6 8 r 5t5 t o n ,Beban v e r t i k a l min. 1 g e la g a r (D g ) = 42 ,514 to n .( l i h a t h a l . 5 8 ) .Akibat gaya gempa v e r t i k a l = k^.D (b e rd a s a r "P eren ca -
D \naan tahan gempa jembatan ja la n raya " a r t , 5 . 3 ) , dimana k = k o e f . gempa rencana v e r t i k a l = 0 , 1 0 .Jadi gaya gernp/n v e r t i k a l = 0,1 .4 2 ,51 4 = 4 ,2 5 ton .Jadi P = 68 ,515 + 4 ,2 5 = 72 ,765 ton .
I V . 2 . 2 . Perhitungan beban h o r i s o n t a l :- Akibat gaya rem:
Gaya rem d iperh itungkan 5% d a r i muatan D tanpa k o e f . k e j u t (b e rd a s a r P . M . U . J . J . R , ’ 7 0 ) .Muatan D = ( ?|Q7||. .1 , 6 5 .1 , £5 + ^ 1 ^ . 1 f 6 5 ) , 53 ,198
ton »Jadi gaya rem Hr = 5%.1/2 * 5 3 * 198 = 1 ,3 3 ton*
80
- Akibat gaya gempa:D ih itung b e r d a s a r "Perencanaan tahan gempa jembatan ja la n ra ya " a r t . 3 . 3 .K o e f i s i e n gempa rencana h o r i s o n t a l d i ten tu kan deng- an rumus: kh * W v 3 - k o ( l i h a t a r t . 4 . 2 . 1 ) . dimana: v 1 = f a k t o r daerah gempa, d iam bil daerah I I
= 1 ,0 .= f a k t o r kepent ingan ( l i h a t a r t . 4 . 3 . 2 . )= 1 ,0 .
kQ = k o e f . gempa rencana h o r i s o n t a l standard = 0 , 1 4 .
k, = 0 , 5 . 1 . 1 . 0 , 1 4 = 0 , 0 7 > k , m i n . = 0 , 0 3 .( O . K . ) 1?
Gaya gempa h o r i s o n t a l , adalah t e r k e c i l d a r i :
Dimana f = k o e f . geseran s t a t i s antara landasan dan k o n s t r u k s i .
f = 0 ,3 5 (untuk landasan type A = c h e m ic a l ly p r o t e c t e t e x t e r n a l l y ) .
VGaya gempa h o r i s o n t a l = k, . P .h m m
= 0 ,0 7 . 4 2 , 5 1 4 = 2 ,976 ton .- Akibat gaya a n g in :
2Pengaruh tekanan angin = 100 kg/cm . ( P . M . U . J . J . R . ’ 7 0 ) .Gaya angin t o t a l = ( 1 , 5 . 1 , 7 5 . 3 7 + 2 . 3 7 )0 ,1 = 17 ,112 t .
17 112Tiap p e r le ta k a n rnenerima gaya angin = —4f-g—= 1,426 t o n .
Kombinasi gaya h o r i s o n t a l :- a k ib a t gaya gempa / / sb . jembatan
H = 2 ,976 ton .= 0 ton .
- a k ib a t gaya gempa ± sb . jernbatan
= 0 , 5 .Vg = f a k t o r keadaan tanah ( l i h a t a r t . 4 . 3 . )
H = 0 ton & H1 = 2 ,976 to n .
81
- a k ib a t gaya rem dan a n g in :H = 1 ,33 t o n .H.| = 1 ,426 t o n .
I V . 2 .3 * Perhitungan perpindahan h o r i s o n t a l :- Akibat su su t dan c r e e p :
£ = * k ‘ + ^ k r ’£= 2 0 ,4 4 2 .1 0 “ 5+ 23 ,36 2*10“ 5» 4 3 ,8 0 4 .1 0 ~ 5 *
( l i h a t h a l . 2 3 ) .Bentang t e o r i t i s jembatan = 37 m* c t . j : a _ 4 3 ,8 0 4 .1 0 .37
t i a p p e r le ta k a n ~ 5= 8 , 1 . 1 0 “ 3m * 8,1 mm.
- Akibat perbedaan suhu:K o e f . muai 1 0 .1 0 “ ^
At • 10°G*A = X . l . A t
1 0 . 10“ 6 . 37*10 = 3 ,7 * 1 0 ’*3m.
J ad i A t i a p pe r ie ta k a n “ 1/ 2 *5 *7 a 1 »85 mm* Kombinasi perpindahan h o r i s o n t a l :
* * 8,1 + 1 , 8 5 = 9 ,9 5 mm.0
I V . 2*4* Perh itungan perputaran s u d u t :Beban-beban yang b e k e r j a :
- a k ib a t b . s . g e l a g a r ........................................ . . . . = 1 ,678- a k ib a t beban mati sekunder ...................... .. 0 ,4976
( a s p a l , t r o t o a r , sandaran, p l a t p e n g i s i )- a k ib a t beban hidup t e r b a g i r a t a ......................» 0 ,9 3- a k ib a t t e k . lengkung tendon ................. . * . . . * « - 2 , 1 7
1 ' q l 5 1 0 , 9 3 6 . 373 .1 07x = 27“ *kI 7 , - 2? *5$300o.257727'22
= 2 , 3 . 1 0 ” rad .- a k ib a t p e r t e b a la n :
0 . 9 3 6a \
3} ^ -----------------------------------------------
82
* - 15X-2a), 0 , 8 8 1 . 1 , 7 5 ( 3 . 3 7 - 2 . 1 , 7 5 ) 1 0 0 n, 10- 3 r . fl
T5:3330'6b725772722' — ----------0 , 0 3 .10 rad.- a k ib a t beban9g a r i s P = 8 ,856 ton d itengah bentang.
, 1 PI* =TZ ‘ET „ 7
1 8 ,8 5 6 .3 7 .10 _ Q„ ^ - 3 „ ^ ^ ’^ * 5 3 5 0 0 0 7 2 5 7 7 2 7 2 ? ~ r a d »
- a k ib a t g e la g a r l i n t a n g , P = 0 ,735 t o n .
« . * * § y i ± b l^ 0 ,7 35 .1 ? , 3 3 . 2 4 . 6 7 { (37+12 .33)+ (3 7 + 2 4 .6 7 )}lOTo(= -------- 6 .$3300b7S57727£2:$7 .................a = 0,1 3 . 1 0“ 3rad .
Jadi t o t a l = ( 2 ,3 + 0 ,0 3 + 0 ,8 8 3 + 0 ,1 3 )10~ 3 = 3 , 3 4 3 . 1 0 " 3ra d .
I V . 2 . 5 . Dimensi Landasan:A. Kombinasi beban mati + hidup + gempa:
Keadaan I : arah gempa i . sumbu jembatan.
W - 7 2 >765 t o n -p» i n = 4 2 .51 4 t o n .H = 0 to n .H1 = 2 ,976 t o n .^ = 0 ,9 9 5 cm.^ = 0 cm. oC = 3 , 3 4 3 . 1 0 “ 3ra d .
Penentuan t e b a l landa san :
d >/c7£s 0 .9 95
^ 1 ,658 cm.d > -------——qrr------
0 7(1 )u » ' u Prnax'
^ 1 , 4 1 2 cm.
D i p a k a i : d = 2 x 10 mm = 2 cm.Dari nomo-gram ( l i h a t la m p ir a n ) :
2Didapat A .x^ = 15 x 40 cm .Syarat ^ ^ 5d
> 5 . 2> 1 0 cm. (OK. ).
Penentuan r e a k s i h o r i s o n t a l : _______ _______________
Rh15.^402 1 ^ + 0 )^ + (0+2,976 ) 2
= 4 ,773 ton .S y a r a t :
r c 8 ,75% “ ^fooo
/ 1 5 .4 0 .8 ,7 5 43 ~ 1 'T m< 5 , 2 5 to n . ( O . K . ) .
Keamanan terhadap g e s e k a n ( s l i p ) :
Rh >p—■— < f , dimana f = 0 ,35 min angka keamanan v » 1 , 4 .
5 ,2 5 / Or35 42,514 ^ T?T~0 ,1 2 3 < 0 , 2 5
( O . K . ) .
Keadaan I I : arah gempa//sumbu jembatan.
Pmax 88 7 2 »765 t o n *Pmin = 4 2 , 5 1 4 ton .H = 2,976 ton.A. = 0 ,995 cm. oC = 3 ,3 4 3 .1 0 _ 3ra d .
Penentuan t e b a l landasan: r i > ° l 995 d
^ 1 ,658 cm. 0 ,995 a ) -------- 1.
0 , 7 . ( 1- ^ -7^5 )^ 2 , 2 4 9 cm.
.Dipakai.d. = .? x 1.5,5 mm. _____________.
84
B'
Dari nomo-gram ( l i h a t la m p ir a n ) :Didapat 7-x /<• = 20 x 40 cm2 .Syarat X^5d
^15 ,5 cm.Penentuan r e a k s i h o r i s o n t a l :
« A h o , 9 9 5 . 2 0 . 4 0 . 1 2 , 5 “ 7 7 ^ 2h = V c TooOT5tr-------- + 2 »9 7 6 '= 6 ,186 ton .
S y a r a t :D , 2 0 . 4 0 . 8 , 7 5 Rh ~ — Tooo
4 7 t o n . ( O . K . ) .Keamanan terhadap gesekan ( s l i p ) :
^ — < 0 , 2 5 rnin
6 1184 / oc 4 ^ T 7 ^ 0 ,2 5 0 ,145 < 0 ,2 5
( O . K . ) .
Korobinasi beban mati + hidup + angin + rem:
Pmax “ 7 2 ’ 765 ton .Pmin = ^ , 5 1 4 t o n .H = 1 ,3 3 to n .H1 = 1 ,426 ton .& - 0 ,9 9 5 cm.<*. = 3 , 3 4 3 . 1 0 “ 3rad .
Penentuan t e b a l landa san : n O ,995
a > ^ 7 5 -^1 ,658 cm.
d ' ° » " 5f/ 5 1 ’V5. *0 ,7 (1
>,1,7 cm.Dipakai d = 2 x 15 ,5 cm.Dari nomo-gram ( l i h a t la m p ir a n ) :
Didapat ?*x = 20 x 40 cm2 .Syarat ^ 5 . 3 , 1
________________ Zl3* 5__cHL ______ _____________________
Penentuan r e a k s i h o r i s o n t a l :Rh 4 1 . 3 3 5 ) ^ + (0 +1 , 4 2 6 ) 2
= 4 ,776 to n .Syarat :
n , 2 0 .4 0 * 8 ,7 5 h ~ ------TSUT) —
7 to n . (O .K ) .Keamanan tErhadap g e s e k a n ^ s l ip ) ;
R>,r r -~ < 0 ,2 5
min
4 , 7 7 6 , 0 9t- 4£75T4 < 0 ’ 250 , 11 2 <0 , 2 5
( O . K . ) .
Jadi Landasan k a re t yang d ip ak a i adalah landasan depngan t e b a l (2 x 15,5)mm, s e r t a ukuran 20 x 40 cm .
; V ; " : r ; .v r ;,»v, .v
f _fio CiK^/ *(S”
g b . 4 . 3 6 . landasan k a r e t .
86
I V . 3. Perhitungan k o n s t r u k s i bag ian ba>/ah r' embatan: Digunakan mutu beton &225*
mutu b a ja ^24*Direncanakan s b b . : z
_|3o
j I f l f C K
V L Bo UK, « O C * ^ It > f / I f OCUu, ' 1,
■ A-KX T ■ {
'pcXrncjxX* - ftYdUntXnjCXnJ
) 8c**v
g b . 4 . 3 6 . rencana k o n s tr u k s i bawah jembatan.
Reno ana p o e r ::
090CM,
•330 C*ks
LhO )— i 1 1 1— 1u,*J□ UkJ r- 1< 1 1 •—(-J
!
M r- -1L.J
(A,*J r —1LJ
C\iJC—t » ; *- * -»
<3,/J r—1< 1 L-_tr—1 L.J
C&A) 1—iUJ
«5j (ffk. b&n*t t.f>.y• IOSXa . ('t#....-
c or**iL.J L.J i.A
m1 ! iflC-tivX 1---Jvr-'T 1 t
ecur— 1 1 <
16,0 r -*1 t 1(t,yr"i1 Il» —• -m L ‘--H
;c
4 « v
<roc*»v/
-1-- /o.<rc**. , yr-jf. -gb . 4.3,7. rencana poer,
88
Beban-beban yang beker. ja pada abutm ent/kepa la .jembatan:A. Muatan pr im er :
- a k i b a t bangunan a t a s :beban mati = 6 x 42 ,514 = 255,84 t o n .beban hid up = -g.1 , 2 3 ( f f f i - - . 5 , 5 . 3 7 + g--y j ~ . 2 . 3 7 +
v>"2 * 5 ,5 + + ■ £ .2 ,0 ,5 .3 7= 147,39 t o n .
- a k ib a t b e r a t s e n d i r i abutment:I : 0 ,2 5 .1 , 7 8 7 . 2 , 5 . 9 , 9 = 11 ,057. t o n .I I { 1 , 2 . 0 , 7 5 . 2 , 5 . 9 , 9 = 22 ,275 t o n .I l l : 1 , 2 5 . 0 , 9 . 2 , 5 . 9 , 9 - 27 ,844 t o n .IV : 1 , 2 . 0 , 7 5 . 2 , 5 . 9 , 9 - 22 ,275 t o n .
83,451 t o n .- a k ib a t b e r a t tanah:
2 , 2 8 7 . 1 » 2 . 1 , 8 . 9 , 9 - 48 ,905 ton .- a k ib a t tekanan tanah(tekanan tanah p a s i f t id a k d i p e r -
h i tu ngk a n )tkarena data tanah t id a k lengkap , maka tanah d i b e - lakang abutment d ia m b i l :
3 0 ° .45°+ 1 / 2 . = 6 0 ° .1 ,8 t /m 3 .t g 2 ( 4 5 ° - 1/2,£>) . t g 230°= 0 ,3 3 3 .
H s* gaya h o r i s o n t a l a k ib a t tekanan tanah.Ht = p .h .A a « 1 , 0 8 . 3 , 0 3 7 . 0 , 3 3 3 . 9 , 9 » 10 ,813 t o n .H2« l / 2 . / . h 2 .Aa = 1 /2 .1 , 8 , 3 , 0372 . 0 , 3 3 3 . 9 , 9 = 27 ,366 ■;
B. Muatan sekunder:SRT = gaya a k ib a t s u s u t , rnngkak dan temperatur,R as gaya rcm dan t r a k a i .
Diperhitungkan 5% d a r i muatan D tanpa k o e f , k e j u t , y^ng bek er ja 1 ,2 m d i a t a s permuka&n l a n t a i kendaraan.
89
oA = muatan a n g in , diarnMl = 100 kg/m .
H angin a k .b a n g .a ta s = ?*1 >5-1 , 7 5 . 3 7 . 0 , 1 = 4,85f>H 1angin ak.muatan hidup = •^•.2.37.0,1 = 3 ,7 t o n .
Muatan khusus:P = tekanan g e s e r d a r i tumpuan yang b e r g e r a k .A^= a l i r a n arus dan hanyutan, dimana untuk abutment
dianggap = 0 .Gb= gaya a k ib a t gempa bumi.
Gb = kh*beban niati = 0 , 0 7 . beban mati ( l i h . h a l 8C Hg bang, a ta s = 0 , 0 7 .2 5 5 ,8 2 = 17,91 ton .HGb bang, bawah = 0 ,0 7 .8 3 ,4 5 1 = 5 ,842 t o n .
Kombinasi pembebananiI J K+H+K+Ta+Ah— -------------- -^100%.I I : M+Ta+Ah +F+A+SRT------------- >125%*I I l l Komb. I+R+J'+A+SRT------------ 1 40%*IV i M+Ta+Ah+Gb------------------------->150%.
I V . 3 *1 . Perhitungan t ia n g pancan^io
Direncanakan berukuran 30x30 cm1' .<^,= 50 kg/cm2 .Tegangan konus pada kedalaman 17 ,50 m adalah seb esar p190 kg/cm . ( l i h a t la m p ira n ) ,^>pada kedalaman + 17 ,5 m d ia m bi l 19°»Panjang t ia n g pancang direncanakan «
* 17 ,5 - 3 , 0 + 0 , 75 + 0 ,5“ 15 ,25 + + 0 ,5= 15,25 + 4 ' / t ? ( 4 5 ° + | ^ 0 , 3 + 0 ,5 = 17,4 m w 17 ,5 m,
Bay a dukunp; 1 t ia n g paneana:- berdaBarkan kokuatan bahan, dimana A d i t a k B i r 1,5%B.
~‘ l t i a n g " ( 3 0 - T‘° + n ; A>f f=-- (900 + 15 . j i g . 900)50 a 55125 kg
- berctasarkan daya ciukung ta n a h (p o in t b e a r in g p i l e ) :T> - 30. 30.1 90 _ ^7000 Xce1 1 t i a n g " 3 _ “ ' ^ U0J Kg*
Jadi untuk 1 t i a n g , 1 = 55 ,125 ton .
90
Daya dukung berd a sa r kelompok t i a n g :P = p
kelompok i n d i v i d u i l<X = e f f i c i e n c y o f p i l e group (<*(.1).
Digunakan "S iile r -K e e n e y " Formulae untuk menentukan o(0 , 3m+n* - 1 1 • ° ’ 4 7 9 ( 7 ^ ? ) ( ^
dimana: m = jumlah b a r i s .n = jumlah kolom.S = ja rak t ia n g r a t a - r a t a dalam m.
01 - 1 1 - ° - 4 7 9 (7 - ¥ t 7 7 : ) ( I ^ )J +1 , 4 - 0 , 0 9 3 o U 0 ,686 + 0 ,0 3 3 = 0 ,7 1 9 .
,3r3
Jadi daya dukung 1 t ia n g pancang dalam kelompok t ia n g P = 0 , 7 1 9 . 5 5 , 1?5 = 39 ,64 ton .
Penulangan t ia n g pancang berd a sa r pengangkatan:Saurin menganjurkan pemilangan t ia n g pancang b e r d a - sarkan pengangkatan sbb :
Mmin = “ TSqL‘ maxq = 0 , 3 . 0 , 3 . 2 5 0 0 = 225 kg /m « .
Mmax = V 22 5*1 7 » 5? = 3828,125 kgm Pembebanan t e t a p : k = 0 , 5 . 0 = 0 , 5 .
;u = T 3 7 W H 7 W = 2 ,710 ] ? mil2 1 0 , 5 7 > > um£
s = 1 .
Dari nomo-gram d ip e r o le h ; <fu = 0,91<j>umax
q = 0 » H 8 ) q min
mm = ° ’ ° 492 max = 0 ,949
= 0 .0533^mmq = ex?Hmax
Mu 1 ,5 .3 8 2 8 ,1 2 5 . . . 2 , , ^o- u , h = 5teo.o,5i .55 * 12
au^u r w • ■
T u l . yang d ip ak a i t o t a l = 12020. 9 _________________________________ (3 7 .7 cm2 ) . . i
3oc*v
91
Penulangan t ia n g pancang berdasarkan beban yang harus d i p i k u l :Kombinasi I : M+H+R+Ta+A^.• . .100%.
4V = 25 5 ,84+83 ,451+48 ,905+ 147 ,39 = 535,568 t o n . O j 1H = 10 ,813+27 ,366 = 38,179 t o » . ( - * )
Letak t i t i k b e r a t rangkai^n t ia n g pancang(Z^ ) :Women s t a t i s terhadap t ia n g pada k o lo m ( 1 ) ,6 . E . 1 ,35 + 6 F .2 ,4 = 18F.X x = 1 ,2 5 m d a r i t ia n g kolom ( 1 ) *
terhadap t i t i k Z1 : a k ib a t b . s . abutment: 1 1 ,0 5 7 .0 ,3 7 5 + 2 2 ,2 7 5 .1 ,1 + 2 7 ,8 4 4 .
0 ,0 5 - 2 2 ,2 7 5 .1 = 7 ,7 66 tm. (A) a k ib a t b . s , tanah: 4 8 ,9 0 5 .1 ,1 = 53,796 tm. ( A ) a k ib a t tekanan tanah: 1 0 ,8 1 3 .3 ,0 3 7 .^ + 2 7 ,3 6 6 .3 ,0 3 7 .^ -
* 44 ,123 t m . ( r V ) a k ib a t R = ( 2 5 5 ,8 4 + 1 4 7 ,3 9 ) 0 ,2 = 80 ,642 tm. ( . n o
*MZ1 = - 7 , 7 6 6 -5 3 »7 9 6 + 4 4 ,123+80,642 = 63 ,203 tm. C<"*j Beban yang harus d i p i k u l o le h masing-masing t ia n g t e *- gak a k ib a t V & Ms
P1V » - ~ | 4 - ( t i a n g - t i a n g pada k o l o m ( l ) ) .
dimana £x^= 6 ( 1 ,2 5 ^ + 0 ,1 2 + 1 ,0 5 2 ) = 16 .05 m2 .P1V = —Th— - ~ 1'g ;Q5'1 z-"~ = 24 ,832 t < P =39,64 t .
T ia n g - t ia n g pada kolom ( 2 ) :
P2y = + = 30 » 147 t < P = 3 9 »64 t *T ia n g - t ia n g pada kolom ( 3 ) :
P3V * + ' ," l 6 ; Q 5'1|t~~ = 34 »285 11 < P s 3 9 *64Akibat £ H = 38 .179 t o n :
Karena pada kombinasi I i n i beban yang b e k e r j a adalah beban t e t a p , maka gaya .j. dianggap d i te r im a o le h t ia n g pancang m ir in g .Dipergunakan methode a n a l i t i s , dengan t ia n g pamcang- t ia n g pancang pada kolom (2 ) dan kolom ( 3 ) d ib u a t seba ga i t ia n g pancang mir ing dengan kemiringan 1 : 10 . Jadi tg tC = 0,1 dan cos°< = 0 ,9 9 5 .
P2 - - 50" 298 ton < 5
p 3 " - 34 ,455 ton < 5P2H~ >2V *^°^ = 3®*147.0 ,: f - 3*0147 t
I?,, fyjA** P3H= P3v . t g * = 3 4 ,4 5 5 .0 ,1 = 3 ,4455 t Jadi t i a n g pancang m ir ing t o t a l dapat menerima gaya Hs e b e s a r = (3 ,0 1 4 7 + 3 ,4 4 5 5 )6 = 38,761 t ^ H yang timbul
= 38 ,179 t o n .K on tro l tegangan yang t e r j a d i :Pmax yf,r>s b e k e r j a = 34 ,455 t o n .Beban yang b e k e r j a adalah beban s e n t r i s , d i s e l e s a i k a ndengan cara S c h r i e r .l c =l^.= 1 / 3 . 1 7 , 5 = 5 ,85 m (d a r i Dunham).
A= l c / l - t = 585 /30 = 19,5") d a r i t a b e l S c h r i e r d ip e r o le h d/h^= 4 /3 0 = 0 ,1 33 J k o e f . tekuk » = 1 , 7 .B = 30 x 30 = 900 cm2 . pTulangan yang d ip a k a i 12 20 = 37 ,7 cm . nT , B.tfb + At.uau*
u = y1 ,5 .3 4 4 5 5 = 2QQ. ? ^ + 37,7.♦ 2080
* 9 •d^yang t e r j a d i =21 ,203 kg/cm2 <G^ = 50 kg /cm 2.Jadi penulangan 12 20 dapat d ip a k a i .
Kombinasi I I : K+Ta+A^+F+A+SRT............ 1?5%.Karena pada kombinasi I I i n i b e k e r j a gaya ang in yang arahnya J. sumbu jembatan, maka dalam pen in jauan gaya H dibedakan dalam 2 h a l , y a i t u :
jr * •» “* jm Ph x , searah sb . memanjang jembatan.
y » arah sumbu memanjang jembatan.Untuk beban F+A-fSRT besarnya gaya R t o t a l « 6 .7 ton
« 42 ton .( l i h a t la nda san , halaman 84 )*Dimana tg f> = |j£
= 0 9 5 l4 o 7 T O T T ? T 5 = ° * 4 4 4 . ( l i h a t p e r h i t .10OO.3,1 landasan ad« B h a l 84 /
Jadi s i n f » = 0 ,406 & c o s = 0 ,9 1 4 .
93
Besarnya gaya-gaya yang b e k e r ja pada kombinasi I I : £ Hy = i ^ s i n ^ s 4 2 .0 ,4 0 6 = 17 ,052 t o n .£Hx = R ^ c o s + a k ib a t Ta = 42 .0 ,9 14 + 10 ,8 13 + 27 ,3 66
« 76 ,567 to n .£V = 255 ,84+83 ,451+48 ,905 = 388,196 t o n . (Jr)
^Momen terhadap t i t i k 2 ^ :\ = 1 7 ,0 5 2 .1 ,2 8 7 = 21 ,946 ton®.
v
= 83 ,184 tonrn. ( A )Beban yang d i p i k u l o l e h masing-roasing t i a n g a k ib a t V dan K:
Beban max. d i p i k u l o l e h t ia n g P ( 6 , 3 ) .V IViv 1 m V ,r r A R
K o n tro l a k ib a t gaya h o r i s o n t a l :Karena pada kombinasi I I i n i beban yang b e k e r j a adalah beban sementara, maka gaya H dianggap seba g ian d i t e r i - ma o l e h tekanan perlawanan tanah dan baru siBanya d i p i k u l o le h t ia n g pancang m ir in g .
D i s i n i dianggap tanah s o f t c l a y , menurut Bowles harga Es = 250 - 600 p s i .
L = ja rak d a r i permukaan sampai dimana eudjit d e f l e k s i n o l .
Mq = r e s t r a n i n g moment pada p i l e - t o p .E & I , d a r i t ia n g pancang s e n d i r i .
Mx = -7 ,7 6 6 -5 3 ,7 9 6 + 4 4 ,1 2 3 + 2 5 5 ,8 4 .0 ,2 + 4 2 .0 ,9 1 4 .1 , 2 8 7
» 28 ,104 ton
Dipergunakan rumus Chang:
Dimana Es = mod. e l a s t i s i t a s tanah
94
Beban sementara : E = 9600 V"bk"f£=9600^22!) = 144000 k g / cm2 .
Es d ia m b i l = 400 p s i = 28 ,12 kg /cm2 .I - 1 / 1 2 . 3 0 . 3 0 3= 67500 cm4 .Jadi
1 |»28^12
"144000.67500' * 5 , 1 8 , 1 0 ” 3cm”*1 = 0 ,5 1 8 m~1 .
L = 075T8M = 76 ,567 ox — 1------
= 6 ,0 6 5 m.= 4 ,106 tm.
M.oyAkibat Hx :
t im bul A x =
1 8 . 2 . 0 , 5 1 8
= T5t5TT)75T8 = 0 .914 tm.
765671 8 . 4 .1 4 4 0 0 0 . 6 7 5 0 0 .5 ,1 8 % 1 0 ^9
= 0 ,7 87 cm.Akibat V
t imbul -J ^9~ 1 8 .4 .1 4 4 0 0 0 .6 7 5 0 0 .5 ,1 8 .10
» 0 ,2 1 7 c m.A - to ta l p e r - t i a n g =Y^x + A-y2*" = / 0 , 7872 +0 , 2172
= 0 ,8 2 cm.Munurut Wayne C. Teng , * yang d i i j i n k a n p e r - t i a n g
( S - 1 , 27cm)A . t o t a l p e r - t i a n g c S
( O . K . ) .
D i t i n j a u t ia n g ( 6 , 3 ) . i * karena b e k e r ja momen 2 arah , M &y ,r * rt/j ' OX
2 4;Ol f»*v/
I
, *>
oymaka d i s e l e s a i k a n dengan b i a x i a l - b e n d i n g .Beton K225 -> pCfJ= 75 kg/cm2 .Baja U 24 pg c = 1400 kg/cm2.y a e x s e n t r i e i t a s d a r i beban
Mox 4 ,106 ^" ~ * 287245 = 0 ,1 4 5 m-
x " t F - " 0 ,0 3 2 m*d? = be ton dekking a 4 cm.
d2 _ 4 n 3- a y j = 0 , 1 3 3 .P 28245 n , 1P
p ~ 7 5 7 d * T57W T50 ~ ° ’ 4 1 8*CCs = d /b - 30 /30 a 1.
95
e = transformed radial eccentricity of load= y2+(sx)2 0 , 1 4 5 2+ 0 ,0 322 = 0 , 1 4 8 .
“ p~TE7d* I ’ ■ ° ' 206*Cv c cdimana Mg = transform ed r a d i a l f a i l u r e moment*Dari c h a r t 3 , dengan ( l i h . lam piran ) ]
i ^ 2 = 0 ,206c c
pDd* c c
d i d a p a t : N = 0 , 0 7 )r . p sc - 0,25*
2 - 1 . 3 .0 , 0 7 5 . 0 .
K3
F = 1 + K1K3 = 1 + 0 ,4 * 0 ,0 7 5 - 1 ,0 3 * / « b (1 + I-,K2K4 ) = 1 *Dari c h a r t 11, dengan N * 0 ,0 75 ( l i h * lam piran ) tg e = ^*~ 1
Ja d i i
= 1*0 ,032 =0.22 U7T75
P a^G
d i d a p a t : G - 1 ,03 .
c co FPyG o _ -p
Tbd2 ~ p *bd " Pc c »hd* d= 1 ,03*0 ,41 8*°* f§ 5 » 1 ,>03 , 0 , 2 1 4 .
Dari c h a r t 3 dengan:
ccP
7 ^ 2 - 0 ,214 ► d id ap at r ,,^ sc 0 ,2 8 .
c c 3 d - ° ' 418-» 75 _
cc
Jadi r = 0 , 2 8 . ^ ^ = 0,01 5 - 1 ,596.
r a ^ s c * ^ s tbd
Aoa+A,,+= rbd » 1,5%*30*30 = 13 ,5 cm *SC oA t o t a l = 13 ,5 cm^< 12 0 20*
Jadi penulangan 12 0 20 a k ib a t pengangkatan l e b i h me- nentukan.N0TE:A__,A„J. = lu a s tulangan tekan dan t a r i k .1 SC S X
F = c o n s ta n t r e l a t i n g the f a i l u r e s u r f a c e s o f two columns i d e n t i c a l e x c e p t f o r c o v e r r a t i o s .
* f a c t o r s d e f in in g ; F and gf.
96
Mg = transformed r a d i a l f a i l u r e moment.N = f a c t o r d e v i a t i e .p = t e g . tekan .yang d i i j i n k a n pada b e t o n .L Vp« = t e g . b a ja yang d i i j i n k a n .
o Cr = s t e e l r a t i o . i> ~ column c o n s t a n t .e - a n g le between Y and the transformed f a i l u r e
p la n e .
Kombinasi I I I :Kombinasi I +R+F+A+SRT.. . . . . 1 4 0 % .Pad* kombinasi I I I i n i b e k e r ja gaya rem s e b e s a r R =10,47$ ton yang b e k e r ja 1 ,2 m a i a t a s permukaan l a n t a i kenaaraan Gaya R i n i mcnyebabkah timbulnya momen s e b e s a r =
M = 1 0 ,4 7 8 ( 1 ,2 + 1 ,7 5 ) = 30,91 tm.Momen yang t imbul i n i menyebabkan timbul gaya v e r t i k a l ,
M = Vr .37
VR = 0 ,8 35 t o n .Jadi a k ib a t gaya R i n i t imbul tambahan gaya V s 0 ,835 t .
Rh = 42 t o n .
tg ^ =0 . 9 (j 5 > 2 o ! 4 0 . i 2 . 5 " T T h = ° * 51 3*1000 .3 ,1 ' 007
( l i h a t perh itungan landasan ad.B h a l 8 5 ) .s in i> = 0 ,2 9 9 .cos i> = 0 ,9 5 4 .
Besarnya gaya-gaya yang b e k e r j a pacia kombinasi I I I :£V = 535 ,568 + 0 ,835 =536,403 t .£Hy = 42 s in i = 12 ,558 t o n .IVL = 38 ,179 + 4 ? c o s i = 78 ,247 t o n . (,-v)A
Momen terhaiiap t i t i k 2M = 1 2 ,55 8 .1 ,287 =» 16 ,162 t o n .M = 4 2 .0 ,9 5 4 .1 ,2 8 7 + 6 3 ,2 0 3 + 0 ,8 3 5 .0 ,2 = 114,97 tm.
Beban max. d i p i k u l o l e h t ia n g P ( 6 , 3 ) . J536 ,4 03 . 1 1 4 .9 7 .1 ,1 5 . 1 6 .1 6 2 .4 .5
( 6 , 3 ) v ” ~ r $ — * 11 ,0 5 " 4 — f n r r r 1-= 41 ,976 ton .
F( 6 , 3 ) = = 4 2 ’ 187 t o n * ** * s 1» 4 .3 9 ,6 4 t .
97
Kontrol akibat gaya horisontal:- akibat H = 7 8 , 2 4 7 ton,
+ - > i a — 78247 , qximDui 1 8 . 4 . -| 4 4 0 0 0 .6 7 5 0 0 .5 ,1 8 .10= 0 , 8 0 4 cm.
- akibat H = 12,558 ton,y i 2558
timbul Ay= 1 8 . 4 . 1 4 4 0 0 0 . 6 7 5 0 0 . 5 , 1 8 3 . 1 0 “ 9
= 0 , 1 2 9 cm._total per-tiang = V o , 8 0 4 +0,129
Kox Per“ tiang * Hx
= 0,814 cm < Z = 1,27 cm. (O . K . ) .
78,247 = 4 j9 6 tm.2 ? = 1 8 .2 : 0 , ^ 8
M per-tiang = Hy = 12 , 558 « 0,673 tm.
Bitinjau tiang (6 , 3 ) :
L-.i
Ho
**v
XT777X
* 1C '
Diselesaikan dengan biaxial-bending:
y - I 22 = h . r r l ? - ° * ° " m-
x = 5 r^ = f i ; r l 7 ' ° . ° 16 ” •
d2 0 .0 4 n 1^7 cf~= 0 5 ” = * • *s = § 4 = 1 .
e = Vy^ + t s x )^ ^0,099^+ 0.0162" = 0,1 m.])ari chart 3, didapat:42187 roc
7^"5T)".3o=
Mr n P e 0 coc 0,1' ° > 625- s t ?c c
= 0 ,2 0 8 .
N = 0, 0 00 .
r *Psc = 0,Pcc
^ = 0 ,5 .
itCvJ ,3 .K, = 0 ,075
-t* '
u 0 •F = 1 +K1K3 = 1 + 0 , 5 . 0 , 0 7 5 = 1 ,0 3 7 5 .i = a(1 + ^ K g ^ ) = 1.Dari ch a r t 11 dengan K = 0 , 0 8 . j d id a p a t k a n :
tg e = *£ = 1 . § ^ | =0,162[ G = 1 ,02 5 .
* F-pc e7Bd*d^ = 1 ,0375 .0 ,625 .° ^ 94 ; '3'i— =0. 219
pDari ch a r t 3 d e n g a n : -— ■ =0,625 d id a p a tk a n :
98
...1cc
h e ? = n . ? iq l - r T ^ = °>35.r ‘ Psc
P c c ^ 2 = ° ’ 219J Pcc
J a d i : r = 0 , 3 5 . ^ ^ = 0 ,01875 = 1 ,875%.
r - Atot r " b .dAt o t = 1 *875%.30 .30 = 16 ,875 cm2 < 12020.
Jadi penulangan 12^20 a k ib a t pengangkatan l e b i h menen- tukan.
KQgLkinaM...lv : K+Ta+Ah -:Gb ......................150%.Pada kombinasi IV i n i t erdapat gaya gempa bumi , makapada perhitungan i n i d i t i n j a u dalam 2 keadaan.Kc-adaanI: Gaya gempa. burri yang t e r j a d i searah dengan
sumbu jembatan.H^bangunan a ta s = = 42 t on .Hnvbanpunan bawah= 0 ,0 7 .8 3 ,4 5 1 - -5*842 to n .
h . h . bH ^ tanah lo n g s o ra n = 0 ,0 7 .1 / 2. — e0 , 0 7 . 0 , 5 . 1 , 8 . 3 . 0372 . 9 ,9
= 3,321 ton .Jadi besarnya gaya yang b e k e r ja pada kombinasi IV i n i a d a la h : £H = 42 + 5 ,842+3,321 +10,81 3+27 ,366 = 89 ,342 t .
£V = 3P8.196 ton . f— »^Komen terhaciap t i t i k 2^:£?■} - - 7 ,7 6 6 -5 3 ,7 9 6 4 4 4 ,1 2 3 + 2 5 5 ,8 4 .0 ,2 + 4 2 .1 ,287
+ 3 , 3 ? 1 . | . 3 , 0 3 7 + ( 0 ,0 7 ) (1 1 ,057 .1 *0185 +22 ,27 5 .6 ,3 7 5 +27,84 4 . 0 , 6 2 5+ 2 2 ,2 7 5 .0 ,3 7 5)
= 11 3 ,9 0 tm. cr>)Beban max. d i p i k u l o le h t ian g pancang -t iang pancang pada koloni ( 3 ) »
388,196 1 1 3 . 9 . 1 . 1 5 nn 3V " ~ Ta + 1 S', C5 “ 29 ,73 ton .
V * - 2 ° , P8 ton < P = 1 , 5 . 3 9 , 6 4 t o n ,Kontrol alcjb^t gaya h o r i e o n t a l :
= H' / M ? tor*
99
y-mg t e r j p d i = _______ 8 9 3 4 2 ___________1" H.4.1 44000 .6 7 5 0 0 .5 ,1 8 ^ .1 0 “
= 0 ,9 1 8 cm < a = 1 ,2 7 cm.(O.K.).
D i t i n j a u t ia n g pan ca n g -t ia n g pancang pada kolom ( 3 ) : +f ^ . K. =
89^342
L-
4,791 tm.
'o2
___ =
= 2^ 1 ^ = ° » 16 m*= 1 / 3 0 . h .= 0,01m............ min = 0 ,0 2 m
e = 0 ,1 6 + 0 ,0 2 = 0,18m.oe o 0 , 1 8 „ ^F7 = U730 = ° ’ 6
> _ c c (_____ I k ) y, _ 1 5 94 _..g,jtO.^ *o 31 “ 1 2 ' i 0 0 . i i + nt “ 1*b » J4 '1 0 0 . 0 , 3
= 0,08R m. x
c 1 = 1 .
= 0 ,1 5 h. ' , i 5 . 0 , 3 = 0 ,0 45 m.e t o t - 0 ,1 P^O,085+0 ,045 =-- 0,31 m.Beban sementara .......... Nu = 1 , 0 5 . 2 9 , 8 8 = 31 ,374 t o n ,
ko = 0 , 6 . j 6 - 0 ,6 *21 rt 0 ,0 3 4 8 t /cm^ -- 3 ,486 Vg/cm^,
(To
Nub .h . V.). ?0
* ,486? n r o 7 j J 7 ’ 2 X D T 5 “ ° - 0129-
e t o t 0,31 4TT^- ” Ct55 * 1 *0 3 3*Dari g r a f i k I r . Wiratman d ip E ro le h , Q + _+a-i » 0 , 2 2 ,
- t o t . b . h t . 2 . k o . ( 7 b k ' x o ia xat o t a l 5au*= S i.? ? t | § ^ P .a,? a ° i i .r.?25 « 25,71 cm2 12^20
J a d i penulangan 12^20 a k ib a t pengangkatan l e b i h menen- tukan*
Keadaan I I : Gaya gempa buini yang t e r j a d i sumbu jemb, r v> = A? to®.AT------------- mT n ,< Hy 2 ,9 7 6 . 1000.2
I tg 0 = ^ - o,3 9 5 . T 5 . t ( r ' i T r ir^ ---- i = 0 ,797 ( l i h a t landasanad*
* Vi T Q^ ^A keadaan I
««• h a l > 83),s in 0 = 0 ,6 2 3 . c o s 0 = 0 ,7 8 2 .
Besarnya gsya -gaya yang b e k e r j a pada kombinasi IV k e - adaan 2 i n i a d a la h :
£V = 388,196 t o n ,*Hy = 42 s in 0 + 5 ,842 + 3,321 = 35 ,329 ton .^Hx = 42 cos 0 + 10 ,813 +27,366 = 71 ,023 ton*
£ Momen terhadap t i t i k :M = 42s in 0 . 1 , 2 8 7 + 0 ,0 7 ( 1 1 , 0 5 7 .1 ,0 1 8 + 2 2 ,2 7 5 .0 ,3 7 5
+ 2 7 , 8 4 4 . 0 , 6 2 5 + 2 2 , 2 7 5 . 0 , 3 7 5 )+ 3 ,3 2 1 . 2 / 3 . 3 , 0 3 7 = 4-3,609 tm.
t'x = 4 2cos 0.1 , 2 8 7 - 7 ,7 6 6 - 5 3 ,7 9 6 + 4 4 ,1 2 3 + 2 5 5 ,8 4 .0 ,2 = 7 6 ,0 tm.
Beban maximum d i p i k u l o l e h t ia n g P ( 6 , 3 ) .
'2k ] b T - + '4- t7 8 ? ^ - 2 8 *165 tOT 1 ( 6 , 3 ) = = 28,31 ton < P s 1 ,5 .39 ,64- t o n .
K ontro l a k ib a t gaya h o r i s o n t a l :Akibat £Hx = 71 ,023 ton ,
71 0 2 3timbul a x 8 T57471 4 4 0 0 0 .6 7 5 0 0 .5 , t S^TT0 “ 9
» 0 ,7 3 cm.Akibat £Hy = 35 ,329 ton ,
*5 S t imbul Ay = 1440007^7500 .5 ,1857TcT9
- 0 ,3 63 c m . ___________<d t o t a l p e r~ t ia n g 0,73*"+ 0 ,363 = 0 ,816 cm ( 2L =1 , 2*r
cm. ( O . K . ) .
I)i t in ,jau t ian g 1 ' ( 6 ,3 ) J.. . . 71 ,023
■H MoM vp e r - t i a n g = j A A a1'e rg *= 3 ,809 tm.OX I C f t * U | v I O
Moyrer-tiang = TB72tE^TTi “ 1,895 tm‘D.ltif-1 es;aiknn dengan bi a x i a l -b e n d i n g i
U.Roq ri , ,c* • = 0 ,1 3 5 m.
101
~ - 4 , - = 0*419p cc d ' 7 5 .3 0 .3 0 »
Mg ■ o _ P 1 - n A - i o ° » 1 ,Pc c . * . d ‘ p _ . b . d . * d ~ ° » 4 1 9 . - J ^ y -c c* 0 ,211.
d a r i ch a r t 3 , d i p e r - !>oleh:
N = 0 , 0 8 . r *Psc = 0 , 2 8 .
P * 1 + K1K3 « 1 + 0 ,4 4 .0 ,0 7 5 - 1 ,0 3 3 .0 = s (1+K1K2K4 ) = 1 (1 + 0 )= 1 .Dari ch a r t 11 dengan N = 0 ,0 8
CCK1 = 0 , 4 4 . K2 = 1 , 3 .K3 = 0 , 0 7 5 . E4 = 0 .
1 d i p e r o l e h ,
Pc c . t . d2 - F. r y .G . i ” “ K d ' d -cc
= 0 ,2 2 9 .Iteri ch a r t 3 d e n g an : -— —r » 0 , 4 1 9 .
" Pcc;- L - aMe
1 , 0 3 3 . 0 , 4 1 9 . ^ * ^ 1 ^ - ^
Jadi r = 0 , 3 2 . ^ - ^ = 0 ,0 1125 - 1 ,125%. A t o t .
pc c - * - d0 ,2 2 9 .
d id a p a t :rp sc = 0 ,?
c c
r = b . d
‘t o t a l = 1 ,1 25% .30 .30 » 10 ,125 cnT < 12020Jadi penulangan 12020 a k ib a t pengangkptan l e b i h menen- tukan.
I V . 3 . 2 . Perh itungan Poer ;Anggapan-anggapan:~ a k ib a t tekanan tanah p a s i f t id a k
t
d ip e rh itu n g k a n .- l e b a r p o e r yang d i t i n j a u adalah
satuan l e b a r p oer yang t e r p i n g g i r , y a i t u = 90+45 * 135 cm.
/APttou
H ,3
Beban-beban yang b e k e r j a :* a k ib a t b . s . poer = 0 ,7 5 .1 , 3 5 . 2 , 5
« 2,531 t /m * .
102
Kombinasibeban
P( t o n )
Hx(t©n)
Hy( t o n )
Mox(tm)
Moy(tm)
1 ( 100%) 34 ,283 3 ,445 - - —
11(125%) 28 ,104 4 ,254 0 ,9 4 7 4,106 0 ,9 1 4
HI (14096) 41 ,976 4 ,347 0 ,6 9 8 4,196 0 ,6 7 2
I V (150%) 29 ,730 4 ,963 - 4,791 -
28 ,165 3,946 1 ,963 3,809 1 ,895
J a d i yang menentukrn untuk beban t e t a p .............kombinasi Iuntuk beban sem en ta ra . . kombinasi I I I
* Beban te ta p (k o m b in a s i I ) :L ih a t p o t . a - a :
P = 34 ,283 t o n .H = 3 ,445 t o n .Ka - a “ 3 4 , 2 8 3 . 0 , 7 5 - 1 / 2 . 2 , 5 3 t . 1 ,2
= 23 ,89 tonm.Beban t e t a p ; n . » 21,
(W_d 3' 7-JJ 1 + S 1
~ 7
I
74* CUVt
-3- = -Ll-oj'jtrHj
b xs 135 cm.h^= 75 cm
Kueoi = hh eo 2 “ 1 / 3 0 . 0 , 7 5 = 0 ,2 5
1,. 5 ( ? 3 , 8 9 - 0 , 3 7 5 . 3 , 4 4 5 ) 1 , 5 . 3 , 4 4 5
h ** 67 cm*
6 ,5 6 m»
m,
e o
= 6 ,5 6 +0 ,2 5 = 6,81 in.
9 ,0 8 1 . . . .6,81 = <57ft
e 11 s
1 . 7 , ( ^ ^ -§ 77 5 ) » 0 . 75 » 0 ,0054 m.
7 .U
eg *= 0 , 1 5 . 0 , 7 5 = 0 ,112 5 m* e t o t = 6 ,5 6 +0 ,25+0 ,0054+0,1125 n 6 ,9 2 8 m. eau = 6 ,9 2 8 + 0 ,3 7 5 - 0 ,0 8 = 7 ,223 m.
u = ££L\ in i3 2 n n ~ 2 Z l1 /2 .0 ,5 .225 .1 ,35
= 0 ,2
6 » ? 2 5 *1 q = 0 ,0 5 3 3 . J^u*3 ^*^48
103
1i = 1
iA
1 : r r = i - = 1 *095#JUq 7 ,2 2$
q ^ k o f^ f tb k ' .b .h 0 .0 6 5 2 .2 .0 ^ 5 .2 2 5 .1 3 5 * 6 7crau*
= 63 ,79 3 cm .A = 58,529 cm2 .A '= 0 , 2 . 6 3 , 7 9 3 = 12,759 cm2 ,
p e r - s a tu a n l e b e r 1 m:A * 5 8.i| 59 = 43>155 Cm2 .
A«> = 9 *451cin2*Check ragam keruntuhan:
S = 0 , 2 .A Cfau* 58 ,259 2080 „ „ „ „
“> - ^ -S k o .W * 1 « :6 7 -S .0 ,5.555 ‘ ° ’ 7 •Hub= ^7550+4d'8'u«' - V ) b . h . 2 k o . < T 6 k '
-< T O § y g g H0- - 0 . 0 7 ) 1 3 5 . 6 7 . 2 . 0 , 5 . 2 2 5
= 492030,86 k g > Hu = 1 ,5 .3 4 4 5 = 5167 ,5 k g .keruntuhan tarfik ( O . K . ) .
Check gearer :V = 34 ,283 - 1 / 2 . 2 , 5 3 1 . 1 , 2 « 32 ,764 ton . max
'Cbu ■ § i - T iot o- gr l s - 8 - 03 W c m 2 < r bu* - 9 , 5 m^cukup tulangan p r a k t i e e a j a ) . cm
* Bebnn sementara(kombinasi I I I ) t L ihat potongan a - a :
3 1 1 •(£)>—is ./t~.*r5W
p = 41 ,976 ton .Hx *= 4 ,3 4 7 ton*Mox= 4 ,196 tonm.'jaya searah sumbu $r d ia ba ika n , sebab kekakuan kearah y b e s a r .M « /p , 9 7 6 . 0 , 7 5 - 1 / 2 . 2 , 5 3 1 .1 ,2cl^ci
-4 ,1 9 625,464 tonm.
j Beban sementarai n = 14.sko “ 0 , 6 . z'ss *5 1,05*
~o2 - i -/ - - . 2 5 m.eQ = 5 ,4 83+ 0 ,2 5 = 5 ,733 m.
1 ° - 5 ,733 _ 7 <aa 1 r - 7h. " ' 0,75 “ 7 *644 1 ..........°2 " 7 *G1 = 1 «
e 1 = 1 ‘ 7 ( T 6 D ^ 7 Y 5 ) ? e 0 ' 75 = ° » ° ° 5 4 m. e 2 = 0 , 1 5 . 0 , 7 5 = 0 ,1125 m. e t ( j t = 5 ,733+0,0054+0,11 25 = 5,851 m. e , „ = 5 ,8 5 1 + 0 ,3 7 5 - 6 ,0 8 = 6 ,146 m.<3. u
°u *]>i'.0^4ji4?.s,l4fl = 7,751 * ■ftSTFrp'ST. O S "
£ = 0 , 2i = ---------- --------- <7777- = 1 , 1 1 5 .
1 ** ° ^ 4 8 * ^ T 7 6 1A = 0 , 0 5 ^ 6 ^ ^ 6 7 s ^8>705 cm2^
A = * 52 ,65 cm2 .A' = 0 , 2 . 5 8 , 7 0 5 = 11,741 cm2 .Per-sa.tuan l e b a r 1 m :A e 5 ^ a -,q
q = 0 , 0 5 3 3
j& v T ° ’ 948
. _ U 4 4 1 _ o . 1,35 8 ,7 cm'
Check rapam keruntuhan:S = 0 , 2 .
5 2 ,6 5 2080 ^q = = ° ’ 045-
= ( 7 ^156+2^^0 ~ 0 , 0 4 5 ) 1 3 5 . 6 7 . 2 . 0 , 6 . 2 2 5
- 651^°C ,7° ^ > H u = 1 ,0 5 .4 3 4 7 = 4564,35 kg. keruntuhan t a r i k ( O . K . ) .
Check pc 'cer :Dm«v = 4 1 , 9 7 6 - 1 / 2 . 2 , 5 3 1 . 1 ,2 = 4 0 ,2 7 7 ton.T ~ -- f —■■ t -?■■■* £X7 - 6 c)2 kp/crr2 Z * = 11 kg/cm'Lbu 1 0 0 . 6 , ^ . 6 8 »- bus 11 KS / cm
(cukup t u l p r a k t i s s a j a ) .
KESIM-ULAN: TUlangan Poer : A = 43 ,155 cm2 ( 0 2 5 - 1 1 ) .A '= 9 ,5 cm2 (0 1 2 - 1 1 ) .
Tul. pembagi = 0 , 2 . 4 3 * 1 0 5 = 8 ,6 3 cm2 . ( A' ( 0 1 2 - 1 3 ) .
2Tul. pembagi .= 0 , 2 . 9 , 5 = 1 , 9 cm .( A ' ) ' ( 0 8 - 2 0 ) .
K on tro l terhadap tegangan p o n s :Pit!?]i jau terh adap beban t e t a p :
Pu = 1 ,5 .3 4 , 2 8 3 = 51 ,4245 t o n . ' A i L ^ 0 -r"v Pu 51424.5; _~y Lbp » u =2 ( a : ; r a t ) F t = 2 ( 36+3 0 ^ 7 7 5 7 7 51 L_l? = 1»^32 kg/cm = 15 kg
( O . K . ) . cm
105
D i t i n j a u terhadap beban sementara:P = 1 ,05 .41 ,976 = 44 ,0748 tot).
T b p , u =- T n T O ^ 7 7 5 ) = 1 » 399 kg /cm24rbpm,us = 16 M( O . K . ) . Cm
D i t i n j a u P o t . b - b ( l i h a t gb. 4 . 36 h a l . 86)* Kombinasi pembebanan I :N+H+Ta+A^.. . . . . . . . . 1 0 0 % .
M yang b e k e r ja = 1/ 2.1 , 8 .1 , 7 8 7 . 1 , 7 8 7 . 0 . 3 3 3 . 9 , 9 . ^ ^
+ 1 . 0 8 . 1 , 7 8 7 . 0 , 3 3 3 . 9 , 9 . 1 / 2 . 1 , 7 8 7 = 11, 34 tm.
* Kombinasi pembebanan I V ; Ivi+Ta+A^+Gb...............150%.M yang b e k e r j a = 1 1 , 3 4 + 0 . 0 7 ( 0 , 2 5 . 1 , 7 8 7 . 9 , 9 . 2 , 5 +
1 / 2 . 1 , 8 . 1 ^ 7 8 7 . 9 , 9 . 2 / 3 . 1 . 7 8 7 ^tg
= 12 ,80? tmYang menentukan adalah kombinasi I (b eb a n t e t a p ) . Pembebanan t e t a p : M=t 11, 34 tm.
rs t = 1 *5 *0,, ^iiA -r rTT’Z'An ~~~ = 7 ,599;u si /r 7 r r r m n
r?.T) 7 r ^ 7 r t-9£ = 0 ,2
> q min = ° ’ ° ” 3
A = 0 , 0 5 3 3 . 9 9 0 . 2 1 . 2 , 0 , 5 . 2 2 5 / 2 0 8 0 = 119,87cm2/ 9 , 9 m. A1= 23 ,97 cm2/ 9 , 9 m.A t i a p m = 12,1 cm2? (025 - 10) •’U p _ = G ,2 .12-1
..A1 t i a p m = 2 , 42 cm (08 - 2 0 ) . J _2 4 cm*(08-2O)