perhitungan plat balancing tank

8/16/2019 Perhitungan Plat Balancing Tank

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Perhitungan Pelat Tebal Plat minimum ditentukan harus memenuhi syarat ‐ syarat

h min < h < h maxß = ln

Sn

direncanakan :panjang plat arah x terpanjang = 3000 mmpanjang plat arah y terpanjang = 000 mm

ln panjang bersih arah x= 3000 ‐0!" # " $ " % = &'" mm

Sn panjang bersih arah y= 000 ‐0!" # " $ " % = (&'" mm

ß = ln = (!"Sn

h max= 3000 x # ( $ 3&0) ("00 % = ("'!" mm*

h min = 3000 x # ( $ 3&0) ("00 % = (00+, mm* $ # & x ß %

dari persyaratan diatas maka h untuk plat diambil = ("0 mm

Plat Type -Penentuan syarat batas.x = 3000 .y =(!".y = 000 .x

Pembebanan Plat/eban ultimate SS12 3+++(! hal (3 = (! 4. $ (!5 ..

/eban 6ati # 4. %Berat sendiri plat & Struktur = 0!*, x *00 = ((" kg

Berat air = (0! , x (000 = (0,00 kg

/eban 7idup # .. %Beban Orang/benda bergerak = ("0kg

/eban ultimate SS12 3+++(! hal (3 = (! 4. $ (!5 ..

= (! x ((&" $(!5 x ("0= (*",!* kg)m = (*",* 1)m


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.x = 3000 .x = (!" plat dua arah.y = 000 .y

6lx = 6tx= 0!00( x 8u x .x9 x = 0!00( x (*", x & x "3= 5&&!, 1m

6ly = 0!00( x 8u x .x9 x = 0!00( x (*", x & x 3,= *&5,!" 1m

6ty = 0!00( x 8u x .x9 x = 0!00( x (*", x & x 3,= *&5,!" 1m

Perhitungan tulangan

Tebal plat (t) = ("0 mmPenutup Beton (p) = " mmMutu Beton (fc') = 0!'" MpaMutu Baja (f) = *0 Mpa!ebar plat ang ditinjau (b) = (000 mm"iameter tulangan utama = (5 mm # dalam dua arah x dan y %

Tinggi e;ekti; untuk arah xdx = h ‐ p ‐ 0+" dia = ("0 ‐ " ‐ ,

= ((' mm Tinggi e;ekti; untuk arah ydy = h ‐ p ‐ 0+" dia ‐ dia

= ("0 ‐ " ‐ , ‐ (5= (0( mm

d# d ("0 mm

ati tulangan dalam keadaan berimbang SS12 3+(+*+3! hal ,pb = 0!," x ;c> x ß # 500 %

;y 500 x ;y= 0!," x 0!'" x (!" # 500 %

*0 500 x *0

= 0!000*",ati tulangan minimum SS12 3+3+"+(! hal *p min = (!* = (!* = 0!00",3

;y *0ati tulangan maximum

p max= 0!'" x pb SS12 3+3+3+3! hal 3= 0!0003*3

Perbandingan teganganm = ;y = (3!50'*

0!," x ;c>


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Penulangan .apanganPenulangan lapangan arah x

6lx = 5&&!, 1m =5&&,** 1mme?sien reduksi = 0!, SS12 Pasal 3++3+

e?sien la@an untuk perencanaan kekuatann = 6lx

0!, x b x

dx9= 0!53

Prsentase tulangan yang dibutuhkanp perlu = ( # (‐ A

m= ( # ( ‐ A

(3!50'*

p perlu = 0!005, p min= 0!00",3

=

#

#

5&&,**0!, x (000 x (35,&

( ‐ x m x n%

;y( ‐ x (3!50'* x 0!53

%*0

p max= 0!03*3

Bika p perlu < p max dipakai tulangan tunggal

p perlu < p min dipakai p min = 0!00",3p min < p perlu < p max jadi dipakai 0!00",3

-s perlu = p x b x d = 0!00",3x (000 x ("0=,'*!" mm

dipakai dengan tulangan diameter 4 = (3‐ ("0# -s = ,," mmCke DDD

ntrl jarak tulangan menurut Pasal 3+5+* butir # SS12! hal 5' %S max< h =("0 < *0 Cke DDD

Penulangan lapangan arah y

6ly = *&5,!"1m = *&5," 1mme?sien reduksi = 0!, S$S% Pasal

e?sien la@an untuk perencanaan kekuatann = 6lx = *&5,"

0!, x b x dx9 0!, x (000 x (35,&= 0+*"

Prsentase tulangan yang dibutuhkanp perlu = ( # ( ‐

Am

= ( # ( ‐ A

(3!50'*

p perlu = 0!00(,& p min= 0!00",3

(#

(#

p max

‐ x m x


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perhitungan plat balancing tank

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