Merencana Konstruksi Beton

M E R E N CA N A

K O N ST R U K SI B E T O N

PRELIMINARY DESIGN

BALOK MELINTANG

 × L   = × 500 

h=

= 41,67 cm̴ 45 cm

   = × 45 

b = × h

= 30 cm BALOK MEMANJANG

 × 400   = × 400 

h=

= 33,34 cm̴ 40 cm

   = × 40 

b = × h

= 26,67 cm̴ 30 cm PERENCANAAN KOLOM STRUKTUR t = 425 cm p = 400 cm

   =             =       ×

×

× ×

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   ×× =  × 





33,33 b4 = 96820321,5 cm b =



 96820312,5 96820312,5

= 41,28 cm̴ 45 cm

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PRELIMINARY DESIGN PLAT

PLAT TYPE A BENTANG 500 cm × 400 cm Bentang Bersih

 

B2 30 /40

Sn = 500 - × ( 30 + 30 ) 0 4 / 0 3 2 B

= 500 – 300 = 470 cm

Y L

 

LX

Ln = 400 - × ( 30 + 30 ) = 400 – 300 = 370 cm

   = 0,79 cm = 

β=

Menentukan Harga

∝

Panel Tepi Balok Melintang be =

63 c m

be= bw + ( h – t ) t =

m c 5 4

12 c m

= 30 + ( 45 – 12 ) = 63 cm

= h

  −     −      −     −  

= 30 c m K =b w 1+ 1 × ( ) × (4

= 1+

1 ×(

) × (4

6×

6×

+ 4× + 4×

   −    −  )

   × ( ) ) 

+(

1 ×( ) )

+(

1

= 1,29 × (2,68 + 0,020 )

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= 3,48

Inersia Balok dan Plat

 ×  ×  × ℎ   = × 3,48 × 30 ×45 

Ib =

= 792787,5 cm4

 ×  × t   = × 370 × t  3

Ip =

3

= 53280 cm4

∝

1=

 =   = 14,88 cm   ,

Panel Tengah Balok Melintang be =

96 c m

be= bw + 2 × ( h – t ) t = 12 c m

m c 5 4

= 30 + 2× ( 45 – 12 ) = 96 cm

= h b w = 30 c m

  −     −     −     − 

K = 1+ = 1+

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6×

6×

+ 4× + 4×

   −    −  )

   × ( ) ) 

+(

1 ×( ) )

+(

1

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= 1,58 × (2,68 + 0,041 ) = 4,29


 ×  ×  × ℎ   = × 4,29 × 30 ×45 

Ib =

= 977315,62 cm4

 ×  × t   = × 370 × t  3

Ip =

3

= 53280 cm4

∝

1=

 =   = 18,34 cm   ,

Panel Tepi Balok Memanjang be= bw + ( h – t ) b e = 58 c m t=

m c 0 4

12 c m

= 30 + ( 40 – 12 ) = 58 cm

= h b w = 30 c m

  − 

K = 1+ D EK Y E K A

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1 × ( ) × (4

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6×

+ 4×

   −  )

+(

 

1 ×( ) )

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M E R E N CA N A

  − 

= 1+

1 ×(


 ) × (4 −6 ×  + 4 ×  + ( − 1 × ())     

= 1,28 × (2,56 + 0,025 ) = 3,31


 ×  ×  × ℎ   = × 3,31 × 30 ×40 

Ib =

= 529600 cm4

 ×  × t   = × 470 × 40  3

Ip =

3

= 67680 cm4

∝

1=

 =  = 7,82 cm  

Panel Tengah Balok Memanjang be= bw + 2 × ( h – t ) be =

86 c m

= 30 + 2 × ( 40 – 12 ) t =

m c 0 4

12 c m

= 86 cm

= h b w = 30 c m

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  −     −     −     − 

K = 1+

1 × ( ) × (4

= 1+

1 ×(

) × (4

6×

6×

+ 4× + 4×

   −    −  )

   × ( ) ) 

+(

1 ×( ) )

+(

1

= 1,56 × (2,56 + 0,07 ) = 4,11


 ×  ×  × ℎ   = × 4,11 × 30 ×40 

Ib =

= 657600 cm4

  ×  × t  = × 470 × 40  3

Ip =

3

= 67680 cm4

∝

1=

 =  = 9,71 cm  

Rasio Kekuatan Balok

∝ ∝

∝ ∝

1 = 14,88 2 = 18,34

3 = 7,82 4 = 9,71

Rasio Kekuatan Rata – rata

∝

=

        ,

,

,

,

= 12,68

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∝

Ternyata = 12,68 >2 sehingga untuk menentukan tebal plat minimum menggunakan rumus : ( SNI 11.5.3,3(c))

    h=       h=    ×( ,

)

×( ,

( ,

)

)

= 9,12 cm̴ 10 cm

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PERHITUNGAN PLAT ATAP

Lx = 400 cm 0 0 5

Ly = 500 cm fc = 35 MPa fy = 320 MPa

400

Direncanakan Tebal Plat 10 cm Pembebanan Plat a. Beban Mati Berat sendiri plat Beart plafond + rangka Berat air hujan 5cm Berat lapisan aspal

= 0,1 × 2400 = 240 kg/m2 = 7 +11 = 18 kg/m2 = 50 = 50 kg/m2 = 14 = 14 kg/m2 322 kg/m2

b. Beba hidup = 100 kg/m2 c. Beban ultimate (q u) q u = 1,2 D +1,6 L = 1,2×322 + 1,6 ×100 = 546,4 kg/m Perhitungan Momen Plat Ly/Lx = 500/400 = 1,25̴ 1,3 ≤ 2 digunakan plat 2arah Momen yang terjadi Mlx = 0,001× q × lx2 × clx = 0,001× 546,4× 42 × 31 = 271,01 kg m Mly = 0,001× q × lx2 × cly = 0,001× 546,4× 42 × 19 = 166,11 kg m Mtx = 0,001× q × lx2 × ctx = 0,001× 546,4× 42 × 69 = 603,22 kg m Mty = 0,001× q × lx2 × cty = 0,001× 546,4× 42 ×57 = 498,31 kg m D EK Y E K A

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Momen Perlu (Mu) = Mnominal Factor Reduksi Kekuatan Mnlx = Mlx : 0,8 = 271,01 : 0,8 = 338,76 kg m Mnly = Mly : 0,8 = 166,11 : 0,8 = 207,63 kg m Mntx = Mtx : 0,8 = 603,22 : 0,8 = 754,02 kg m Mnty = Mty : 0,8 = 498,31 : 0,8 = 622,89 kg m

 = 0,0018 0018 ×  = 0,0022       ×  = 0,75 ×        ×  = 0,75 ×  

ρmin = 0,0018 × ,

ρmax

× ×

,

× ,

×

= 0,035

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PERHITUNGAN PENULANGAN PELAT ATAP Penulangan Arah X Lapangan Mnlx = 338,76 kg m = 3387600 N.mm

∅

Direncanakan menggunakan tulangan 8mm dx = ht – s – ½ Ø = 100 – 20 - ½ 8 = 76 mm

 =  = 0,58 N/mm      =  = 10,75 m=           ) 1− ρ = × (1-     × (1-  1 −      ) =   

2

Rn =

×

,

×

×

,

×

× ×

×

,

× ,

,

= 0,0018 ρmax > ρmin > ρperlu

0,035 > 0,0022 > 0,0018 As = ρ × b × d

= 00022 ×1000 × 76 = 167,2 mm2 ̴ dipakai tulangan Ø 8 – 150 = 335,1 mm2

Tumpuan Mntx = 754,22 kg m = 7542200 N.mm

∅

Direncanakan menggunakan tulangan 8mm dx = ht – s – ½ Ø = 100 – 20 - ½ 8

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= 76 mm

 =  = 1,31 N/mm      =  = 10,75 m=           ) 1− ρ = × (1-     × (1-  1 −      ) =   

2

Rn =

×

,

×

×

,

×

× ×

×

,

× ,

,

= 0,004 ρmax > ρperlu > ρmin

0,035 > 0,004 > 0,0022 2

As = ρ × b × d

= 0,004 ×100 × 76 = 304 mm2 Ø8 – 150 = 335,1 mm2 berasal dari tulangan lapangan

  = 167,5  ,

As pasang ≤ As’ maka dipasang tulangan ekstra As sisa = As’ – As pasang = 304 – 167,5 = 136,45 mm2 Dipasang tilangan ekstra Ø8 - 150

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= 335,1 mm2 > 136,45 mm2

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Penulangan Arah Y Lapangan Mnlx = 207,63 kg m = 2076300 N.mm

∅

Direncanakan menggunakan tulangan 8mm dx = ht – s – Ø - ½ Ø = 100 – 20 – 8 - ½ 8 = 68 mm

 =  = 0,44 N/mm      =  = 10,75 m=           ) 1− ρ = × (1-     × (1-  1 −      ) =   

2

Rn =

×

,

×

×

,

×

× ×

×

,

× ,

,


0,035 > 0,0022 > 0,0013 2

As = ρ × b × d

= 00022 ×1000 × 68 = 149,6 mm2 ̴ dipakai tulangan Ø8 – 150 = 335,1 mm2


∅

Direncanakan menggunakan tulangan 8mm dx = ht – s – Ø - ½ Ø = 100 – 20 - 8 - ½ 8 = 68 mm

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 =  = 1,34 N/mm      =  = 10,75 m=           ) 1− ρ = × (1-     × (1-  1 −      ) =   

2

Rn =

×

,

×

×

,

×

× ×

×

,

× ,

,


0,035 > 0,004 > 0,0022 2

As = ρ × b × d

= 0,0042 ×100 × 68 = 285,6 mm2 Ø8 – 150 = 162,1 mm2 berasal dari tulangan lapangan

  = 167,5  ,

As pasang ≤ As’ maka dipasang tulangan ekstra As sisa = As’ – As pasang = 285,6 – 167,5 = 118,05 mm2 Dipasang tilangan ekstra Ø8 - 150

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= 335,1 mm2 > 118,05 mm2

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PERHITUNGAN PELAT LANTAI

Lx = 400 cm 0 0 5

Ly = 500 cm fc = 35 MPa

400

fy = 320 MPa

Direncanakan Tebal Plat 10 cm Pembebanan Plat a. Beban Mati Berat sendiri plat = 0,12 × 2400 Beart plafond + rangka = 7 +11 Berat tegel + keramik = 1 × 24 Berat instalasi listrik = 60 Berat spesi = 3 × 21

= 288 kg/m2 = 18 kg/m2 = 24 kg/m2 = 60 kg/m2 = 63 kg/m2 + = 453 kg/m2

b. Beba hidup = 400 kg/m2 c. Beban ultimate (q u) q u = 1,2 D +1,6 L = 1,2× 453 + 1,6 ×400 = 1183,6 kg/m Perhitungan Momen Plat Ly/Lx = 500/400 = 1,25̴ 1,3 ≤ 2 digunakan plat 2arah Momen yang terjadi Mlx = 0,001× q × lx2 × clx = 0,001× 1183,6 × 42 × 31 = 587,065 kg m Mly = 0,001× q × lx2 × cly = 0,001× 1183,6× 42 × 19 = 359,814 kg m Mtx = 0,001× q × lx2 × ctx = 0,001× 1183,6× 42 × 69 = 1306,69 kg m D EK Y E K A

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Mty = 0,001× q × lx2 × cty = 0,001× 1183,6× 42 ×57 = 1079,44 kg m

Momen Perlu (Mu) = Mnominal Factor Reduksi Kekuatan Mnlx = Mlx : 0,8 = 587,065: 0,8 = 733,83 kg m Mnly = Mly : 0,8 = 359,814: 0,8 = 449,775 kg m Mntx = Mtx : 0,8 = 1306,69: 0,8 =1633,37 kg m Mnty = Mty : 0,8 = 1079,44: 0,8 = 1349,3 kg m

 = 0,0018 0018 ×  = 0,0022       ×  = 0,75 ×        ×  = 0,75 ×  

ρmin = 0,0018 × ,

ρmax

× ×

,

× ,

×

= 0,035

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PERHITUNGAN PENULANGAN PELAT LANTAI Penulangan Arah X Lapangan Mnlx = 733,83 kg m = 7338300 N.mm

∅

Direncanakan menggunakan tulangan 10 mm dx = ht – s – ½ Ø = 120 – 20 - ½ 10 = 95 mm

 =  = 0,677 N/mm      =  = 10,75 m=           ) 1− ρ = × (1-     × (1-  1 −      ) =   

2

Rn =

×

,

×

×

,

×

× ×

×

,

× ,

,


0,035 > 0,023 > 0,0022 As = ρ × b × d

= 0,0023 ×1000 × 95 = 218,5mm2 ̴ dipakai tulangan Ø10 – 150 = 523,6 mm2

Tumpuan Mntx = 1633,37 kg m =16333700 N.mm

∅

Direncanakan menggunakan tulangan 10 mm dx = ht – s – ½ Ø = 120 – 20 - ½ 10 D EK Y E K A

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= 95 mm

 =  = 1,476 N/mm      =  = 10,75 m=           ) 1− ρ = × (1-     × (1-  1 −      ) =   

2

Rn =

×

,

×

×

,

×

× ×

×

,

× ,

,


0,035 > 0,0047 > 0,0022 As = ρ × b × d

= 0,0047 ×1000 × 95 = 446,5mm2 Ø10 – 150 = 523,6 mm2 berasal dari tulangan lapangan

  = 261,8  ,


D EK Y E K A

A RIFIN

/ 10 57 57 2 4 25 2 59

= 523,6 mm2 > 214,25 mm2

Page 18

M E R E N CA N A


Penulangan Arah Y Lapangan Mnly = 449,775 kg m = 4497750 N.mm

∅

Direncanakan menggunakan tulangan 10 mm dx = ht – s – Ø - ½ Ø = 120 – 20 – 10 - ½ 10 = 85 mm

 =  = 0,5184 N/mm      =  = 10,75 m=           ) 1− ρ = × (1-     × (1-  1 −      ) =   

2

Rn =

×

,

×

×

,

×

× ×

×

,

× ,

,


0,035 > 0,0022 > 0,0013 As = ρ × b × d

= 0,0022 ×1000 × 85 = 187 mm2 ̴ dipakai tulangan Ø10 – 150 = 523,6 mm2


∅

Direncanakan menggunakan tulangan 10 mm dx = ht – s – Ø - ½ Ø = 120 – 20 - 10 - ½ 10 = 85 mm

D EK Y E K A

A RIFIN

/ 10 57 57 2 4 25 2 59

Page 19

M E R E N CA N A


 =  = 1,55 N/mm      =  = 10,75 m=           ) 1− ρ = × (1-     × (1-  1 −      ) =   

2

Rn =

×

,

×

×

,

×

× ×

×

,

× ,

,


0,035 > 0,009 > 0,0022 As = ρ × b × d

= 0,0049 ×1000 × 85 = 416,4 mm2 Ø10 – 150 = 523,6 mm2 berasal dari tulangan lapangan

  = 261,8  ,


D EK Y E K A

A RIFIN

/ 10 5724 259

= 523,6 mm2 > 214,25 mm2

Page 20

M E R E N CA N A


PEMBEBANAN PADA TANGGA 25 15

18

3 0 . 8

Optrede

= 18 cm

Aantrede

= 25 cm

t. plat tangga = 12 cm t. plat bordes = 12 cm maka :

√ 



S = 25 + 18 = 30,8 cm Cos ɑ =

   ,

  

ɑ

= arcCos

ɑ

= 35⁰44’19,63”

β

= 90⁰ - 35⁰44’19,63”

,

= 54⁰15’40,37” Sin β =

 

X

= 18 x sin 54⁰15’40,37”

X

= 14,61 cm

Beban Plat Tangga

D EK Y E K A

B.S plat tangga

=

0,12 x 2400

B.S anak tangga

=

  x 2400 

A RIFIN

/ 10 5724 259

,

   = 175,2   = 288

Page 21

M E R E N CA N A

B. spesi ( t = 4 cm )


=

4 x 21

B. keramik (t = 1cm) =

1 x 24 Qdl

Beban hidup ( Qll ) = 300

   = 24    = 571,2   = 84

 

+

Beban Plat Bordes

B.S plat bordes

=

0,12 x 2400

B. spesi ( t = 4 cm)

=

4 x 21

B. keramik (t = 1cm ) =

1 x 24 Qdl

Qll

= 500

 

   = 84    = 24    = 396   = 288

+

Beban Balok Bordes

Dinding diatas balok bordes = 250 x 1,725 Akibat plat bordes

 

= x396 x 1,88 Qdl

Qll

D EK Y E K A

A RIFIN

 

= x500 x 2 = 500

/ 10 5724 259

 

   = 372,24 +   = 539,91   = 431,25

Page 22

M E R E N CA N A


Beban Balok Tangga

   = x571,2 x 2   = 571,2    Akibat beban hidup = xQll plat tangga x L   = x500 x 2   = 500      Akibat plat lantai =( xQd p lantai x Lx x (1-(  ( ) ))       =( x453 x 4 x (1-(  ( ) ))     = 712,72    Akibat qll lantai = xQll plat lantai x Lx   = x 500 x 4   = 666,67   Akibat plat tangga

= xQd plat tangga x L

2

2

Pembebanan pada SAP untuk balok tangga

D EK Y E K A

A RIFIN

/ 10 5724 259

Page 23

M E R E N CA N A


Elevasi + 4,25 Qdl = 571,72 + 539,91 Qll = 500 + 666,67

   = 1166,67  = 1111,63

Elevasi + 8,50 Qdl = 336,288 + 405,552 Qll = 213 + 291,666

D EK Y E K A

A RIFIN

/ 10 5724 259

   = 504,6  = 741,832

Page 24

M E R E N CA N A


Perhitungan Penulangan Plat Tangga Dari hasil SAP :

 Per rmeter  = 0,5 x10053457,1  Per rmeter = 5026728,55 

Mu tumpuan = 10053457,1 Mu lapangan

Fc’ Fy

= 35 mpa = 320 mpa

Digunakan tulangan utama = Ø 12 dan sb = 20 mm dengan tebal plat =120 mm d

= tp – sb – ( 0,5 x Ø TU ) = 120 – 20 – ( 0,5 x 12 ) = 94 mm

d

Menentukan batas rasio penulangan (ρmin dan ρmax ) Untuk fy

= 320 mpa

ρmax = 0,75 xρ b

 x β x     ∗x 0,85 x  = 0,75 x   = 0,75 x

,

1

,

= 0,035

D EK Y E K A

A RIFIN

/ 10 5724 259

Page 25

M E R E N CA N A

` ρmin = 0,0018 ×




 = 0,0018 0018 ×  = 0,0022  

Tulangan utama pada daerahTumpuan Hasil dari SAP : Mu

Mn

Rn

m

ρ

  

= 10053457,1

permeter

 = ∅   = 12566821,38 N mm = = ∗  = 1,42  =  = 10,75 =  ∗  ∗  ∗∗ = x { 1 -  1−  }   x { 1 -  1 −     } =  , = 10053457,1

=

,

12566821,38

,

,

2

,

,

= 0,0045 Cek rasio penulangan

ρmax = 0,038 ≥ 0,0045

≥ ρmin

= 0,0022

ρ pakai = 0,0045 Menentukan As

As

D EK Y E K A

= ρx b x d

A RIFIN

/ 10 5724 259

Page 26

M E R E N CA N A


= 0,0045 x 1000 x 94 = 423 mm2 Berdasarkan tabel luas tulangan maka digunakan = Ø12 – 200 As = 565mm2 Cek spasi tulangan S max

Smax

= 2 x tp = 2 x 120 = 240 mm

S pasang = 200 mm Maka Smax≥ S pasang 240 mm ≥ 200 mm

ok!

Maka tulangan pada tumpuan menggunakan : Ø12 –200



Tulangan utama pada daerah lapangan Hasil dari SAP : Mu

Mn

Rn

m

ρ

D EK Y E K A

= 5026728,55

  

permeter

 = ∅   =6283410,69 N mm = = ∗ ∗ = 0,711  =  = 10,75 =  ∗  ∗  ∗∗ = x { 1 -  1−  }  5026728,55

=

A RIFIN

,

6283410,69

,

,

2

/ 10 5724 259

Page 27

M E R E N CA N A

=

   ,

x{1-


  −     ,

1

,

}


ρmax = 0,038 ≥ 0,0023

≥ ρmin

= 0,0022

ρ pakai = 0,0023

Menentukan As

As

= ρx b x d = 0,0023x 1000 x 94 = 216,2 mm2

Berdasarkan table luas tulangan maka digunakan = Ø12 – 200

As = 565mm2

Cek spasi tulangan S max

Smax

= 2 x tp = 2 x 120 = 240 mm


ok!


Tulangan Susut D EK Y E K A

A RIFIN

/ 10 5724 259

Page 28

M E R E N CA N A


As susut = ρminx b x tp = 0,0023x 1000 x 120 = 276 mm2 Dipasang tulangan Ø10 – 200 ( As = 392,7 mm2 )

D EK Y E K A

A RIFIN

/ 10 5724 259

Page 29

M E R E N CA N A


PERHITUNGAN PLAT BORDES Dari hasil SAP :

 Per rmeter 

Mu tumpuan =8555532

Mu lapangan =0,5 x8555532

 Per rmeter 

= 4277766

Fc’ Fy

= 35 mpa = 320 mpa

Digunakan tulangan utama = Ø 12 dan sb = 20 mm dengan tebala plat =120 mm d

= tp – sb – ( 0,5 x Ø TU ) = 120 – 20 – ( 0,5 x 12 ) = 94 mm

d

Menentukan batas rasio penulangan (ρmin dan ρmax ) Untuk fy

= 320 mpa

ρmax = 0,75 xρ b

 x β x     ∗x 0,85 x  = 0,75 x   = 0,75 x

,

1

,

= 0,035 D EK Y E K A

A RIFIN

/ 10 5724 259

Page 30

M E R E N CA N A

` ρmin = 0,0018 ×




 = 0,0018 0018 ×  = 0,0022  

Tulangan utama pada daerahTumpuan Hasil dari SAP : Mu

Mn

Rn m

ρ

=8555532

  

permeter

 = = 10694415 N mm ∅  = = ∗ ∗= 1,21 =  ∗=  ∗= 10,75  ∗∗ = x { 1 -  1−  }   x { 1 -  1 −     } =  , 8555532

=

,

10694415

,

,

2

,

,


ρmax = 0,038 ≥ 0,0038

≥ ρmin

= 0,0022


As

= ρx b x dx = 0,0038x 1000 x 94

D EK Y E K A

A RIFIN

/ 10 5724 259

Page 31

M E R E N CA N A


= 357,2 mm2 Berdasarkan table luas tulangan maka digunakan = Ø12 – 200 As = 565mm2


Smax

= 2 x tp = 2 x 120 = 240 mm


ok!




Tulangan utama pada daerah lapangan Hasil dari SAP : Mu

Mn

Rn m

ρ

=4277766

  

permeter

 = = 5347207,5 N mm ∅  = = ∗ ∗= 0,605 =  ∗=  ∗= 10,75  ∗∗ 1− = x { 1 -   }   x { 1 -  1 −     } =  , 4277766

=

,

5347207,5

,

,

2

,

D EK Y E K A

A RIFIN

/ 10 5724 259

,

Page 32

M E R E N CA N A



ρmax = 0,038 ≥ 0,0019

≥ ρmin

= 0,0022


As

= ρx b x dx = 0,0022x 1000 x 94 = 206,8 mm2

Berdasarkan table luas tulangan maka digunakan = Ø12 – 200 As = 565mm2


Smax

= 2 x tp = 2 x 120 = 240 mm


ok!

Maka tulangan pada tumpuan menggunakan : Ø12 –200 Cek spasi tulangan S max

Smax

= 2 x tp = 2 x 120 = 240 mm

S pasang = 200 mm

D EK Y E K A

A RIFIN

/ 10 5724 259

Page 33

M E R E N CA N A


Maka Smax≥ S pasang 240 mm ≥ 200 mm

ok!


Tulangan Susut

As susut = ρminx b x tp = 0,0022x 1000 x 120 = 264 mm2 Dipasang tulangan Ø10 – 200 ( As = 392,7 mm2 )

D EK Y E K A

A RIFIN

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M E R E N CA N A


PERHITUNGAN PENULANGAN PLAT BORDES

Direncanakan menggunakan tulangan 4 D 16, As = 804 mm2 Hasil dari SAP = Mu = 52981723 N mm Data data perencanaan :

Fc’

= 35 mpa

sb

= 40 mm

Fy

= 320 mpa

Øsengkang

= 10 mm

b

= 300 mm

β1

= 0,79

h

= 400 mm

cek pemasangan tulangan ( 1 lapis atau 2 lapis )

 ∗ ∗    ≥ 25 mm   ∗ ∗  ≥ 25 mm  (

)–(

(

)–(

Ø

)–(

)–(

Ø

)

)

12 mm ≤ 25 mm ( maka dipasang 2 lapis )

Perhitungan tinggi efektif (d)

Bila penulangan lentur harus digunakan 2 lapis maka harus diisikan jarak bersih antar tulangan pada arah vertikal = 0 mm d

 

= h – sb – Ø sengkang – Ø TU - x jarak bersih vertikal = 400 – 20 – 10 – 16 – ( 0,5 x 0 ) = 354 mm

D EK Y E K A

A RIFIN

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M E R E N CA N A


Kuat nominal penampang

ΣH = 0 sehingga dari distribusi tegangan beton bertulang didapatkan persamaan T = C

T

= As x fy = 804 x 320 = 257280 N

C

= 0,85 x fc’ x b x β1 x X

X

=

X X

∗  ∗∗∗ =  ∗∗∗  ,

257280

,

,

= 33,91 mm

kontrol tulangan leleh / tidak leleh εs

x εc   x 0,003 =   =

( εc = 0,003 )

,

,

= 0,0283 εy

   =  =

= 0,0016

Karena εs ≥ εy maka termasuk tulangan leleh ( fs = fy )

D EK Y E K A

A RIFIN

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M E R E N CA N A


Perhitungan kuat nominal penampang

Mn

 

=Tx(d- )

∗ )   ∗  ) = 804 x 320 x ( 354 –  = As x fy x ( d –

,

,

= 87369264,94 N mm Kontrol kuat nominal penampang

φx Mn ≥ Mu 0,8 x87369264,94 N mm

≥

52981723 N mm

69895411,97 N mm

≥

52981723 N mm

OK

Kontrol rasio penulangan

  =   = 0,0043   ,

,

ρmin

=

ρmax

= 0,75 xρ b

 ∗x β x     ∗x 0,85 x  = 0,75 x   = 0,75 x

,

1

,

= 0,035

ρ

 ∗  = ∗ =

= 0,0075 D EK Y E K A

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M E R E N CA N A

Sehingga


ρmax≥ ρ≥ρmin 0,038 ≥ 0,0075 ≥ 0,0048

OK

Kontrol tulangan minimum

As min = ρminx b x d = 0,0075x 200 x 354 = 531 mm2 Syarat As pasang ≥ As min 804 mm2≥ 531 mm2

OK

Tulangan tekan (As’)

As’ diasumsikan As’

δ=

 = 0,5 

= δ x As

= 0,5 x 804 = 402 mm2 Tulangan terpasang

D EK Y E K A

As

= 4 D - 16 ( As = 804 mm2 )

As’

= 2 D – 16 ( As’ = 402 mm2 )

A RIFIN

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M E R E N CA N A


PERHITUNGAN TULANGAN GESER TANGGA Data perhitungan : Balok

= 300 mm x 400 mm

Fy

= 320 Mpa

F’c

= 35 Mpa

Panjang

= 4000 mm

VU Max

2000

Step 1: Menentukan Vu max (gaya geser ultimate)

 

Vumax= xq uxl Vumax= 61876,79N

diperoleh dari SAP

Step 2: Kontrol dimensi balok terhadap gaya geser ultimate

 ≤ φ∗ √  ∗ ∗ ∗ ∗ √ ∗ ∗

5 f c bw H 6

Vu

0,75

5 6

35 200 400 = 295803,98 N

Sehingga:

 = 61876,79 N < 295803,98 N imensi balok OK

Vu

D EK Y E K A

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M E R E N CA N A


Step 3: Menentukan kuat geser yang disumbangkan oleh beton (Vc)

1 6 1 Vc = 6 Vc =

∗  f′c∗ ∗ ∗√ ∗ ∗

bw d

35 300 400 = 118321,59 N

Step 4: Menentukan persyaratan penulangan geser

φ∗

Vc…………(1)

0,75 x118321,59 = 88741,192 N

1 2

∗φ∗ ∗

Vc…………(2)

0,5 118321,59 = 59160,795 N

Step 5: Cek kondisi penulangan geser pada penampang balok

≥ φ∗ φ∗  ≥ ∗φ∗

Bila Vu Bila

Vc…………(Perlu tulangan geser)

Vc >

1 2

Vc…(Tulangan geser minimum)

Sehingga:

88741,192 > 61876,79 N >

  ∗ > 59160,795 N 

(Perlu tulangan geser

minimum)



perhitungan penulangan geser pada daerah tumpuan Cek kondisi geser pada penampang balok Untuk daerah tumpuan, x = d = 354 mm (4000):2 = 2000 mm



=

D EK Y E K A

−

2000 354 × 59160,795 = 48689,34 N 2000 A RIFIN

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M E R E N CA N A


Direncanakan menggunakan tulangan sengkang diameter Ø10 - 100 mm 1 22 = 2. = 2 × × 10 = 157,14 mm 4 7

 



Vn

= Vc + Vs

Vs

=





∗∗   ∗∗ =  ,

= 178008,192 N

Cek Avmin

∗ ∗∗ ∗ ∗ √ ∗∗ = ∗

Avmin =

= 23,109 mm2 Av = 157,14 mm2 ≥ Avmin = 23,109 mm2

OK

Cek jarak sengkang

Vu

≤

φx ( Vc + Vs )

61876,79 N ≤ 0,75 x (118321,59 + 178008,192) 61876,79 N ≤ 222247,34 N

OK

   = x 354 

Smax = x d

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= 177 mm

Smax = 177 mm ≥ Spasang = 100 mm

OK

Jadi dipasang sengkang d10-100 mm pada daaerah tumpuan



perhitungan penulangan geser pada daerah lapangan Cek kondisi geser pada penampang balok (4000):4 = 1000 mm



=

−

2000 1000 × 59160,795N = 29580,39 N 2000

Direncanakan menggunakan tulangan sengkang diameter Ø10 - 150 mm 1 22 = 2. = 2 × × 10 = 157,14 mm 4 7

 



Vn

= Vc + Vs

Vs

=





∗∗   ∗∗ =  ,

= 111255,12 N

Cek Avmin

∗ ∗∗ ∗ ∗ √ ∗∗ = ∗

Avmin =

= 34,66 mm2 Av = 157,14 mm2 ≥ Avmin = 34,66 mm2 D EK Y E K A

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Cek jarak sengkang

Vu

≤

φx ( Vc + Vs )

59160,795 N ≤ 0,75 x (118321,59 + 111255,12) 59160,795 N ≤ 172182,532 N

OK

   = x 354 

Smax = x d

= 177 mm

Smax = 177 mm ≥ Spasang = 150 mm

OK

Jadi dipasang sengkang d10-150 mm pada daerah lapangan

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PEMBEBANAN BALOK MEMANJANG ATAP

BEBAN MERATA 

Balok memanjang D

LINE 1 

BEBAN MATI

Segitiga

= 1 / 3 * Q * Lx = 1/3 × 322 × 2,5 = 268,34 kg/m2

Trapesium

= 1 / 2 * Q * Lx * (1 - 1 /3 * (lx/ly)2)

 

= × 322 × 1,75 ×(1

−     ×

,

)

= 209,84 kg/m2 Berat sendiri = 0,3 ×04 × 2400 × 2 = 576 kg/m2 Beban total = 268,34 + 209,84 + 576 = 1054,17 kg/m2

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


BEBAN HIDUP

Segitiga

= 1 / 3 * Q * Lx = 1/3 × 100 × 2,5 = 83,34 kg/m2

Trapesium

= 1 / 2 * Q * Lx * (1 - 1 /3 * (lx/ly)2)

 

= × 100 × 1,75 ×(1

−     ×

,

)


LINE 2 

BEBAN MATI

Segitiga

= 1 / 3 * Q * Lx = 1/3 × 322 × 4 = 429,33 kg/m2

Trapesium

= 1 / 2 * Q * Lx * (1 - 1 /3 * (lx/ly)2)

 

= × 322 × 1,75 ×(1

−     ×

,

)

= 263,78 kg/m2 Berat sendiri = 0,3 ×04 × 2400 × 4 = 1152 kg/m2

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Beban total = 429,33 + 263,78 + 1152 = 1845,11 kg/m2



BEBAN HIDUP

Segitiga

= 1 / 3 * Q * Lx = 1/3 × 100 × 4 = 133,34 kg/m2

Trapesium

= 1 / 2 * Q * Lx * (1 - 1 /3 * (lx/ly)2)

     = × 100 × 1,75 ×(1 − ×   )    ,


BEBAN TERPUSAT 

Balok memanjang D

LINE 1 

BEBAN MATI

Segitiga

= 1 / 3 * Q * Lx = 1/3 × 322 × 1,75

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= 187,83 kg/m2

Trapesium

= 1 / 2 * Q * Lx * (1 - 1 /3 * (lx/ly)2)

 

= ×322×2×(1

−    ×

,

)




BEBAN HIDUP

Segitiga

= 1 / 3 * Q * Lx = 1/3 × 100 × 1,5 = 58,34 kg/m2

Trapesium

= 1 / 2 * Q * Lx * (1 - 1 /3 * (lx/ly)2)

 

= ×100×2×(1

−    ×

,

)

= 78,67 kg/m2 Berat sendiri = 0,3 ×04 × 2400 × 2 = 576 kg/m2 Beban total = 78,67 + 58,34 + 576 = 713 kg/m2 LINE 2 

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Segitiga


= 1 / 3 * Q * Lx * 2 = 1/3 × 322 × 1,75 × 2 = 375,67 kg/m2

Trapesium

= 1 / 2 * Q * Lx * (1 - 1 /3 * (lx/ly)2) * 2

 

= ×322×2×(1

−    ×

.

) × 2




BEBAN HIDUP

Segitiga

= 1 / 3 * Q * Lx *2 = 1/3 × 100 × 1.75 × 2 = 116,67 kg/m2

Trapesium

= 1 / 2 * Q * Lx * (1 - 1 /3 * (lx/ly)2)

    = ×100×2×(1 − ×   )    ,

= 157,34 kg/m2 Berat sendiri = 0,3 ×04 × 2400 × 4 = 1152 kg/m2 Beban total = 157,34 + 116,75 + 1152 = 1426 kg/m2

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PERHITUNGAN PENULANGAN BALOK ATAP

Balok atap memanjang D Data penulangan

fc’ = 35 Mpa

Fy = 320 Mpa Selimut beton 40 mm Direncanakan menggunakan tulangan

∅

16

Begel 10 mm h = 400 mm b = 300 mm Data SAP Mlapangan = 31036949,54 N.mm Mtumpuan = 62037899 N.mm d= h – s – b – (1/2 tulangan utama ) = 400 - 40 – 10 – (1/2 × 16 ) = 342 mm Perhitungan ρ beton

 = 10,75     ρmin = 0,0018 × = 0,0022       ×  = 0,035 ρmax = 0,75 ×   m=

,

×

,

× ,

×

Perhitungan pada tumpuan Mtumpuan = 62037899 N.mm D EK Y E K A

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  =  = 1,76        ) 1− ρperlu = × (1 −     × (1 −  1 −      ) =    Rn =

×

×

× ×

×

,

× ,

,

= 0,005



As = × b × d =0,005 × 300 × 342 = 513 mm2 Digunakan tulangan 4 Ø 16 = 804,2 mm2 Direncanakan tulangan tekan As = 20% × 513 = 102,6 mm2 Digunakan 2 Ø 16 = 329 mm2 Perhitungan pada lapangan Mtumpuan = 31036949,54 N.mm

  =   = 0,88        ) ρperlu = × (1 −  1−    × (1 −  1 −      ) =    Rn =

,

×

×

× ×

×

,

× ,

,

= 0,0027



As = × b × d =0,0027 × 300 × 342 = 277.02 mm2

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Digunakan tulangan 2 Ø 16 = 402,1 mm2 Direncanakan tulangan tekan As = 20% × 402,1 = 80,42 mm2 Digunakan 2 Ø 16 = 329 mm2 Penulangan geser Vu = 931,1085 KN Vc = 1/6

  √

× b × d

= 1/6 35 × 300 × 342 = 101,16 KN Vs = Vu – 0.8 Vc = 931,1085 – 0.8 101,16 = 850,16 KN = 850160 mm2 S=

   =    = 20,211 mm   ×

×

×

×

Kontrol jarak sengkang S<½d 20,211 < ½ 342 20,211 < 171 Kontrol terhadap Vs Vs < 1/3

  √

× b × d

850,16 < 1/3 35 × 300 × 342 850,16 KN < 2023,29 KN

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PEMBEBANAN BALOK MEMANJANG LANTAI

BEBAN MEARTA Balok memanjang D LINE 1 BEBAN MATI Segitiga

= 1 / 3 * Q * Lx = 1/3 × 453 × 2 = 302 kg/m2

Trapesium

= 1 / 2 * Q * Lx * (1 - 1 /3 * (lx/ly)2)

 

= × 453 × 1,75 ×(1

−     ×

,

)

= 295,21 kg/m2 Berat sendiri = 0,3 ×04 × 2400 × 2 = 576 kg/m2 Beban total = 302 + 295,21 + 576 = 1173,21 kg/m2 BEBAN HIDUP Segitiga

= 1 / 3 * Q * Lx = 1/3 × 400 × 2 = 266,67 kg/m2

Trapesium

= 1 / 2 * Q * Lx * (1 - 1 /3 * (lx/ly)2)

     = × 400 × 1,75 ×(1 − ×   )    ,

= 260,67 kg/m2 D EK Y E K A

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Berat sendiri = 0,3 ×04 × 2400 × 2 = 576 kg/m2 Bebans total = 266,67 + 260,67 + 576 = 1103,34 kg/m2 LINE 2 BEBAN MATI Segitiga

= 1 / 3 * Q * Lx = 1/3 × 453 × 4 = 604 kg/m2

Trapesium

= 1 / 2 * Q * Lx * (1 - 1 /3 * (lx/ly)2)

     = × 453 × 1,75 ×(1 − ×   )    ,

= 371,08 kg/m2 Berat sendiri = 0,3 ×04 × 2400 × 4 = 1152 kg/m2 Beban total = 604 + 371,08 + 1152 = 2127,08 kg/m2 BEBAN HIDUP Segitiga

= 1 / 3 * Q * Lx = 1/3 × 400 × 4 = 533,34 kg/m2

Trapesium

= 1 / 2 * Q * Lx * (1 - 1 /3 * (lx/ly)2)

 

= × 400 × 1,75 ×(1

−     ×

,

)

= 327,67 kg/m2

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Berat sendiri = 0,3 ×04 × 2400 × 4 = 1152 kg/m2 Beban total = 533,34 + 327,67 + 1152 = 2013,01 kg/m2

BEBAN TERPUSAT Balok memanjang D LINE 1 BEBAN MATI Segitiga

= 1 / 3 * Q * Lx = 1/3 × 453 × 1,75 = 264,25 kg/m2

Trapesium

= 1 / 2 * Q * Lx * (1 - 1 /3 * (lx/ly)2)

    = ×453×2×(1 − ×   )    ,

= 356,36 kg/m2 Berat sendiri = 0,3 ×04 × 2400 × 2 = 576 kg/m2 Beban total = 264,25 + 356,36 + 576 = 1196,61 kg/m2 BEBAN HIDUP Segitiga

= 1 / 3 * Q * Lx = 1/3 × 400 × 1,75 = 233,34 kg/m2

Trapesium D EK Y E K A

= 1 / 2 * Q * Lx * (1 - 1 /3 * (lx/ly)2)

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 

= ×400×2×(1


−    ×

,

)

= 314,67 kg/m2 Berat sendiri = 0,3 ×04 × 2400 × 2 = 576 kg/m2 Beban total = 233,34 + 314,67 + 576 = 1124 kg/m2 LINE 2 BEBAN MATI Segitiga

= 1 / 3 * Q * Lx * 2 = 1/3 × 453 × 1,75 × 2 = 528,5 kg/m2

Trapesium

= 1 / 2 * Q * Lx * (1 - 1 /3 * (lx/ly)2) * 2

 

= ×453×2×(1

−    ×

.

) × 2

= 712,72 kg/m2 Berat sendiri = 0,3 ×04 × 2400 × 4 = 1152 kg/m2 Beban total = 528,5 + 712,72 + 1152 = 2393,23 kg/m2 BEBAN HIDUP Segitiga

= 1 / 3 * Q * Lx *2 = 1/3 × 400 × 1.75 × 2 = 466,67 kg/m2

Trapesium

D EK Y E K A

= 1 / 2 * Q * Lx * (1 - 1 /3 * (lx/ly)2)

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 

= ×400×2×(1


−    ×

,

)

= 629,33 kg/m2 Berat sendiri = 0,3 ×04 × 2400 × 4 = 1152 kg/m2 Beban total = 466,67 + 629,33 + 1152 = 2248 kg/m2

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PERHITUNGAN PENULANGAN BALOK LANTAI Balok atap memanjang D Data penulangan

fc’ = 35 Mpa Fy = 320 Mpa Selimut beton 40 mm Direncanakan menggunakan tulangan

∅

16

Begel 10 mm h = 400 mm b = 300 mm Data SAP Mlapangan = 40004019,72 N.mm Mtumpuan = 80008039 N.mm d= h – s – b – (1/2 tulangan utama ) = 400 - 40 – 10 – (1/2 × 16 ) = 342 mm Perhitungan ρ beton

 = 10,75     ρmin = 0,0018 × = 0,0022       ×  = 0,035 ρmax = 0,75 ×   m=

,

×

,

× ,

×

Perhitungan pada tumpuan Mtumpuan = 80008039 N.mm Rn =

    =   = 2,280 ×

D EK Y E K A

×

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 × (1 −  1 −   )    × (1 −  1 −      ) =    × ×

ρperlu =

×

,

× ,

,

= 0,007



As = × b × d =0,007 × 300 × 342 = 718,2 mm2 Digunakan tulangan 5 Ø 16 = 1005,3 mm2 Direncanakan tulangan tekan As = 20% × 1005,3 = 201,06 mm2 Digunakan 2 Ø 16 = 329 mm2

Perhitungan pada lapangan Mlapangan = 40004019,72 N.mm

  =   = 1,14        ) 1− ρperlu = × (1 −     × (1 −  1 −      ) =    Rn =

,

×

×

× ×

×

,

× ,

,

= 0,0036



As = × b × d =0,0036 × 300 × 342 = 369,36 mm2

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Digunakan tulangan 3 Ø 16 = 603,2 mm2 Direncanakan tulangan tekan As = 20% × 603,2 = 120,64 mm2 Digunakan 2 Ø 16 = 329 mm2 Penulangan geser Vu = 1200,121 KN Vc = 1/6

  √

× b × d

= 1/6 35 × 300 × 342 = 101,16 KN Vs = Vu – 0.8 Vc = 1200,121 – 0.8 101,16 = 1119,193 KN = 1119193 mm2 S=

   =    = 15,352 mm   ×

×

×

×

Kontrol jarak sengkang S<½d 15,352 < ½ 342 15,352 < 171 Kontrol terhadap Vs Vs < 1/3

 

× b × d

√

1119,193 < 1/3 35 × 300 × 342 850,16 KN < 2023,29 KN

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PEMBEBANAN PORTAL MELINTANG

AKIBAT BEBAN MATI (MERATA) Segitiga

= 1 / 3 * Q * Lx * 4 = 1/3 × 453 × 1,5 × 4 = 906 kg/m

Trapesium

= 1 / 2 * Q * Lx * (1 - 1 /3 * (lx/ly)2) * 2

 

= ×453×2×(1

−   ×

) × 2

= 857,68 kg/m Segitiga

= 1 / 3 * Q * Lx * 2 = 1/3 × 453 × 3,5 × 2 = 1057 kg/m

Segitiga

= 1 / 3 * Q * Lx * 2 = 1/3 × 453 × 3 × 2 = 906 kg/m

Trapesium

= 1 / 2 * Q * Lx * (1 - 1 /3 * (lx/ly)2) * 2

 

= ×453×2×(1

−    ×

,

) × 2

= 846,34 kg/m Berat dinding = berat dinding ×tinggi bersih = 250 × (4,25 – 0,45 ) = 950 kg/m

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AKIBAT BEBAN HIDUP (MERATA) Segitiga

= 1 / 3 * Q * Lx * 4 = 1/3 × 400 × 1,5 × 4 = 800 kg/m

Trapesium

= 1 / 2 * Q * Lx * (1 - 1 /3 * (lx/ly)2) * 2

 

= ×400×2×(1

−   ×

) × 2


= 1 / 3 * Q * Lx * 2 = 1/3 × 400 × 3,5 × 2 = 933 kg/m

Segitiga

= 1 / 3 * Q * Lx * 2 = 1/3 × 400 × 3 × 2 = 800 kg/m

Trapesium

= 1 / 2 * Q * Lx * (1 - 1 /3 * (lx/ly)2) * 2

 

= ×453×2×(1

−    ×

,

) × 2

= 747,33 kg/m

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AKIBAT BEBAN MATI (TERPUSAT) 

Akibat P1 Trapesium

= 1 / 2 * Q * Lx * (1 - 1 /3 * (lx/ly)2) * 2 1

= × 453 × 1,5 ×(1 2

−  1 3

×

1,5

2

2

) × 2

= 552,09 kg/m

Segitiga

= 1 / 3 * Q * Lx * 2 = 1/3 × 453 × 2 × 2 = 604 kg/m



Akibat P2 Segitiga

= 1 / 3 * Q * Lx * 2 = 1/3 × 453 × 2 × 2 = 604 kg/m

Trapesium

= 1 / 2 * Q * Lx * (1 - 1 /3 * (lx/ly)2) * 2 1

= × 453 × 1,75 ×(1 2

−   1 3

×

1,75 2 2

) × 2

= 590,43 kg/m 

Akibat P3 Trapesium

= 1 / 2 * Q * Lx * (1 - 1 /3 * (lx/ly)2) * 2 1

= × 453 × 1,75 ×(1 2

−   1 3

×

1,75 2

2

) × 2

= 590,43 kg/m Trapesium

D EK Y E K A

= 1 / 2 * Q * Lx * (1 - 1 /3 * (lx/ly)2) * 2

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1

= × 453 × 1,5 ×(1 2


−  1 3

×

1,5 2 2

) × 2

= 552,09 kg/m 

Akibat P4 Trapesium

= 1 / 2 * Q * Lx * (1 - 1 /3 * (lx/ly)2) * 2 1

= × 453 × 1,5 ×(1 2

−  1 3

×

1,5

2

2

) × 2


= 1 / 3 * Q * Lx * 2 = 1/3 × 453 × 2 × 2 = 604 kg/m



Akibat P5 Segitiga

= 1 / 3 * Q * Lx * 2 = 1/3 × 453 × 2 × 2 = 604 kg/m

Trapesium

= 1 / 2 * Q * Lx * (1 - 1 /3 * (lx/ly)2) * 2 1

= × 453 × 1,5 ×(1 2

−  1 3

×

1,5 2 2

) × 2

= 552,09 kg/m 

Akibat berat Sendir

= 0,3 × 0,45 × 2400 × 500 = 1620 kgm

AKIBAT BEBAN HIDUP (TERPUSAT) 

Akibat P1 Trapesium

= 1 / 2 * Q * Lx * (1 - 1 /3 * (lx/ly)2) * 2 1

= × 400 × 1,5 ×(1 2

−  1 3

×

1,5 2 2

) × 2

= 487,5 kg/m D EK Y E K A

A RIFIN

/ 10 5724 259

Page 64

M E R E N CA N A

Segitiga


= 1 / 3 * Q * Lx * 2 = 1/3 × 400 × 2 × 2 = 533,34 kg/m



Akibat P2 Segitiga

= 1 / 3 * Q * Lx * 2 = 1/3 × 400 × 2 × 2 = 533,34 kg/m

Trapesium

= 1 / 2 * Q * Lx * (1 - 1 /3 * (lx/ly)2) * 2 1

= × 400 × 1,75 ×(1 2

−   1 3

×

1,75 2 2

) × 2

= 521,35 kg/m 

Akibat P3 Trapesium

= 1 / 2 * Q * Lx * (1 - 1 /3 * (lx/ly)2) * 2 1

= × 400 × 1,75 ×(1 2

−   1 3

×

1,75 2 2

) × 2


= 1 / 2 * Q * Lx * (1 - 1 /3 * (lx/ly)2) * 2 1

= × 400 × 1,5 ×(1 2

−  1 3

×

1,5

2

2

) × 2

= 487,5 kg/m 

Akibat P4 Trapesium

= 1 / 2 * Q * Lx * (1 - 1 /3 * (lx/ly)2) * 2 1

= × 400 × 1,5 ×(1 2

−  1 3

×

1,5 2

2

) × 2

= 487,5 kg/m

D EK Y E K A

A RIFIN

/ 10 5724 259

Page 65

M E R E N CA N A

Segitiga


= 1 / 3 * Q * Lx * 2 = 1/3 ×400 × 2 × 2 = 533,34 kg/m



Akibat P5 Segitiga

= 1 / 3 * Q * Lx * 2 = 1/3 × 400 × 2 × 2 = 533,34 kg/m

Trapesium

= 1 / 2 * Q * Lx * (1 - 1 /3 * (lx/ly)2) * 2 1

= × 400 × 1,5 ×(1 2

−  1 3

×

1,5 2 2

) × 2

= 487,5 kg/m 

Akibat berat Sendir

D EK Y E K A

A RIFIN

= 0,3 × 0,45 × 2400 × 500 = 1620 kgm

/ 10 5724 259

Page 66

M E R E N CA N A

D EK Y E K A

A RIFIN

/ 10 5724 259


Page 67

M E R E N CA N A


PEMBEBANAN PORTAL MELINTANG ATAP

AKIBAT BEBAN MATI (MERATA) Segitiga

= 1 / 3 * Q * Lx * 4 = 1/3 × 322 × 1,5 × 4 = 644 kg/m

Trapesium

= 1 / 2 * Q * Lx * (1 - 1 /3 * (lx/ly)2) * 2 1

= × 322 × 2 × ( 1 2

−  1 3

×

2 5

2

) × 2


= 1 / 3 * Q * Lx * 2 = 1/3 × 322 × 3,5 × 2 = 751,34 kg/m

Segitiga

= 1 / 3 * Q * Lx * 2 = 1/3 × 322 × 3 × 2 = 644 kg/m

Trapesium

= 1 / 2 * Q * Lx * (1 - 1 /3 * (lx/ly)2) * 2 1

= × 322 × 2 × ( 1 2

−  1 3

×

2

4,5

2

) × 2

= 601,59 kg/m

D EK Y E K A

A RIFIN

/ 10 5724 259

Page 68

M E R E N CA N A


AKIBAT BEBAN HIDUP (MERATA) Segitiga

= 1 / 3 * Q * Lx * 4 = 1/3 × 100× 1,5 × 4 = 200 kg/m

Trapesium

= 1 / 2 * Q * Lx * (1 - 1 /3 * (lx/ly)2) * 2 1

= × 100 × 2 × ( 1 2

−  1 3

×

2 5

2

) × 2


= 1 / 3 * Q * Lx * 2 = 1/3 × 100 × 3,5 × 2 = 233,34 kg/m

Segitiga

= 1 / 3 * Q * Lx * 2 = 1/3 ×100 × 3 × 2 = 200 kg/m

Trapesium

= 1 / 2 * Q * Lx * (1 - 1 /3 * (lx/ly)2) * 2 1

= × 100 × 2 × ( 1 2

−  1 3

×

2

4,5

2

) × 2

= 186,83 kg/m

AKIBAT BEBAN HIDUP (TERPUSAT) 

Akibat P1 Trapesium

= 1 / 2 * Q * Lx * (1 - 1 /3 * (lx/ly)2) * 2 1

= × 100 × 1,5 ×(1 2

−  1 3

×

1,5 2 2

) × 2

= 121,87 kg/m D EK Y E K A

A RIFIN

/ 10 5724 259

Page 69

M E R E N CA N A

Segitiga


= 1 / 3 * Q * Lx * 2 = 1/3 × 100 × 2 × 2 =133,34 kg/m



Akibat P2 Segitiga

= 1 / 3 * Q * Lx * 2 = 1/3 × 100 × 2 × 2 = 133,34 kg/m

Trapesium

= 1 / 2 * Q * Lx * (1 - 1 /3 * (lx/ly)2) * 2 1

= × 100 × 1,75 ×(1 2

−   1 3

×

1,75 2 2

) × 2

= 130,34 kg/m 

Akibat P3 Trapesium

= 1 / 2 * Q * Lx * (1 - 1 /3 * (lx/ly)2) * 2 1

= × 100 × 1,75 ×(1 2

−   1 3

×

1,75 2

2

) × 2


= 1 / 2 * Q * Lx * (1 - 1 /3 * (lx/ly)2) * 2 1

= × 100 × 1,5 ×(1 2

−  1 3

×

1,5 2 2

) × 2

= 121,87 kg/m 

Akibat P4 Trapesium

= 1 / 2 * Q * Lx * (1 - 1 /3 * (lx/ly)2) * 2 1

= × 100 × 1,5 ×(1 2

D EK Y E K A

A RIFIN

/ 10 5724 259

−  1 3

×

1,5 2 2

) × 2

Page 70

M E R E N CA N A


= 121,87 kg/m

Segitiga

= 1 / 3 * Q * Lx * 2 = 1/3 × 100 × 2 × 2 = 133,34 kg/m



Akibat P5 Segitiga

= 1 / 3 * Q * Lx * 2 = 1/3 × 100 × 2 × 2 = 133,34 kg/m

Trapesium

= 1 / 2 * Q * Lx * (1 - 1 /3 * (lx/ly)2) * 2 1

= × 453 × 1,5 ×(1 2

−  1 3

×

1,5 2 2

) × 2

= 121,87 kg/m 

Akibat berat Sendir

= 0,3 × 0,45 × 2400 × 500 = 1620 kgm

AKIBAT BEBAN MATI (TERPUSAT) 

Akibat P1 Trapesium

= 1 / 2 * Q * Lx * (1 - 1 /3 * (lx/ly)2) * 2 1

= × 322 × 1,5 ×(1 2

−  1 3

×

1,5 2

2

) × 2

= 392,43 kg/m

Segitiga

D EK Y E K A

= 1 / 3 * Q * Lx * 2

A RIFIN

/ 10 5724 259

Page 71

M E R E N CA N A


= 1/3 × 400 × 2 × 2 = 429,34 kg/m 

Akibat P2 Segitiga

= 1 / 3 * Q * Lx * 2 = 1/3 × 322 × 2 × 2 = 429,34 kg/m

Trapesium

= 1 / 2 * Q * Lx * (1 - 1 /3 * (lx/ly)2) * 2 1

= × 322 × 1,75 ×(1 2

−   1 3

×

1,75

2

2

) × 2

= 419,69 kg/m 

Akibat P3 Trapesium

= 1 / 2 * Q * Lx * (1 - 1 /3 * (lx/ly)2) * 2 1

= × 322 × 1,75 ×(1 2

−   1 3

×

1,75 2

2

) × 2


= 1 / 2 * Q * Lx * (1 - 1 /3 * (lx/ly)2) * 2 1

= × 322 × 1,5 ×(1 2

−  1 3

×

1,5 2 2

) × 2

= 392,34 kg/m 

Akibat P4 Trapesium

= 1 / 2 * Q * Lx * (1 - 1 /3 * (lx/ly)2) * 2 1

= × 322 × 1,5 ×(1 2

−  1 3

×

1,5 2

2

) × 2


= 1 / 3 * Q * Lx * 2 = 1/3 ×322 × 2 × 2

D EK Y E K A

A RIFIN

/ 10 5724 259

Page 72

M E R E N CA N A


= 429,34 kg/m 

Akibat P5 Segitiga

= 1 / 3 * Q * Lx * 2 = 1/3 ×322× 2 × 2 = 429,34 kg/m

Trapesium

= 1 / 2 * Q * Lx * (1 - 1 /3 * (lx/ly)2) * 2 1

= × 322 × 1,5 ×(1 2

−  1 3

×

1,5 2

2

) × 2

= 392,34 kg/m 

Akibat berat Sendir

D EK Y E K A

A RIFIN

= 0,3 × 0,45 × 2400 × 500 = 1620 kgm

/ 10 5724 259

Page 73

M E R E N CA N A


PENULANGAN BALOK MELINTANG 

Tulangan Lentur -

tinggi efektif (d) diasumsikan pakai tulangan tarik D22, sengkang ϕ10 mm, dan selimut beton 40 mm.

 

d = h balok – sb – ϕsengkang – . ϕ tulangan = 450

− − −  40

10

.22

= 389 mm -

rasio penulangan

→

f’c = 35 Mpa

β1 = 0,85

−

0,008.( 35

= 0,807



=

min

−

30)

  ,

   ,

=

= 0,004



    .  .       .0,807 .  = 0,75 .  

= 0,75 .

max

,

.

,

′

.

= 0,036 

TUMPUAN Mu

= 171067967

Mn

=

Rn m

 D EK Y E K A

(perhitungan SAP2000)

   = =    = =        = . 1 −  1−  

A RIFIN

        



=

,

.

,

.

′

.

,

.

.

/ 10 5724 259

= 309928385



= 4,71 = 10,75

.

Page 74

M E R E N CA N A

=




 .1 − 1 −         .



<

0,005 <

. ,

,

= 0,016 min

,

0,016

<



<

0,004 , maka dipakai = 0,016 .

max



Menentukan As

-

As = ρ.b.d

= 0,016 .300 .389 = 1867,2 mm2 Dipakai tulangan 6-D22 ( As = 2280,8 mm2 > 2811,6mm2 ). As’ = 20% . As tarik

= 20% .1867,2 = 373,44 mm2 Dipakai tulangan 2-D22 ( As = 760 mm2 > 373,44 mm2 ).



LAPANGAN Mu

= 75874906,18

Mn

=

Rn m



        



(perhitungan SAP2000)

  = 94843632,73     = 2,08 = =    = 10,75 = =        = . 1 −  1−    .1 − 1 −      =    =

,

,

.

,

.

,

.

,

′

.

.

.

.

,

. ,

,



= 0,006 min

<

0,005 <

0,006

maka dipakai D EK Y E K A

A RIFIN

 

<



<

0,004 , maka dipakai = 0,006 .

= 0,0096 .

/ 10 5724 259

max



Page 75

M E R E N CA N A

-


Menentukan As As = ρ.b.d

= 0,006 .300 .389 = 700,2 mm2 Dipakai tulangan 3-D22 ( As = 1140,4 mm2 > 700,2 mm2 )

As’ = 20% . As tarik

= 20% .700,2 = 140,04 mm2 Dipakai tulangan 2-D22 ( As = 760 mm2 > 452,5 mm2 ).



Tulangan Geser Vu

=

126892,99 N

126892,99 126892,99

 .√ f c .bw .d   35 .300 .389 = . √ 

Vc

≤   √  √ ≤  ≤ .

Vumax

0,75.

. f ′ c .bw .d . 35 .300 .389

431504,069 . . . . . .

(dimensi balok OK !!)

′

=

= 115067,75 N ∂Vc

= 0,75 . 115067,75 = 86300,81 N

 ∂Vc =  . 86300,81   = 43150,41 N -

Chek kebutuhan tulangan geser :

≥

bila

Vu

bila

∂Vc >

D EK Y E K A

A RIFIN

∂Vc

Vu

(perlu tulangan geser )

≥

 ∂Vc 

/ 10 5724 259

(tulangan geser minimum) Page 76

M E R E N CA N A

Vu 126892,99

≥ ≥


∂Vc

86300,81 N

maka, perlu dipasang tulangan geser. -

Mencari besar Vs

 –     – 115067,75 =  

Vs

=

,

,

= 54122,91 N

-

untuk tumpuan : direncanakan sengkang ϕ10 – 200 mm As

= 78,57 mm2

Av

= 2. As = 2. 78,57 mm2 = 157,1 mm2

Vs’

        =  .

=

.

, .

.

= 97779,04 N Vs’ 97779,04

≥ ≥

Vs 54122,91 N . . . . . (OK!!)

maka, dipasang sengkang ϕ10 – 200 mm di tumpuan.

D EK Y E K A

A RIFIN

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Page 77

M E R E N CA N A


PERHITUNGAN GAYA GESER GEMPA Data Bangunan o o o

o o o o o

o o

Fungsi bangunan Tinggi Gedung Tinggi tiap lantai

: Ruang Kuliah : 13,20 m : Lt1 = 4,75 m Lt2 = 4,25 m Lt3 = 4,25 m Mutu Beton ( fc‘ ) : 35 mpa Mutu Baja Tulangan : 320 mpa Tebal Pelat atap : 10 cm Tebal Pelat lantai : 12 cm Dimensi balok : 30 / 40 30 / 45 Dimensi Kolom : 45 / 45 Sistem Struktur : SRPMM (Sistem

Rangka

Pemikul

Momen Menengah) o

Wilayah Gempa

:2

o

Jenis Tanah

: Keras

Pembebanan Seluruh Beban Bangunan

WATAP 0 0 6 3

W3 0 0 6 3

W2 0 5 0 4

W1

D EK Y E K A

A RIFIN

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Page 78

M E R E N CA N A


Atap 

Plat Qdl * Luas plat Qll * Luas plat

= 288 * 576 = 100 * 576

= 185472 kg = 57600 kg



Balok Memanjang Volume balok * 2400 * jml  Balok Melintang Volume balok * 2400 * jml  Kolom

= 0,3 * 0,4 * 36 * 2400 * 4

= 41472 kg

= 0,3 * 0,45 * 16 * 2400 * 10 = 51840 kg

Volume kolom * 2400 * jml = 0,45 * 0,45 *2,2 * 2400*40 = 42768 kg + W atap

= 379152 kg

Lantai 3 

Plat Qdl * Luas plat = 453 *(576 - 20) = 251868kg  Balok Memanjang Volume balok * 2400 * jml = 0,3 * 0,4 * 36 * 2400 * 5 = 51840 kg  Balok Melintang Volume balok * 2400 * jml = 0,3 * 0,45 *16 * 2400 * 10 = 51840 kg  Kolom Volume kolom * 2400 * jml = 0,45 * 0,45*4,25* 2400*49 = 101210 kg 

Dinding Q dinding * L * tinggi * jml = 250 * 36 * 4,25 * 3 Q dinding * L * tinggi * jml = 250 *28 * 4,25 * 1 Q dinding * L * tinggi * jml = 250 * 8 * 4,25 * 2

= 114750 kg = 29750 kg = 17000 kg

Q dinding * L * tinggi * jml = 250 * 4 * 4,25 * 1

= 4250 kg

W lantai 3

+

= 621788 kg

Lantai 2 

Plat Qdl * Luas plat = 453 *(576 - 20) = 251868kg  Balok Memanjang Volume balok * 2400 * jml = 0,3 * 0,4 * 36 * 2400 * 5 = 51840 kg  Balok Melintang Volume balok * 2400 * jml = 0,3 * 0,45 *16 * 2400 * 10 = 51840 kg D EK Y E K A

A RIFIN

/ 10 5724 259

Page 79

M E R E N CA N A






Kolom Volume kolom * 2400 * jml = 0,45*0,45*4,25*2400*49

= 101210 kg

Dinding Q dinding * L * tinggi * jml = 250 * 36 * 4,25 * 2 Q dinding * L * tinggi * jml = 250 * 32 * 4,25 * 2

= 76500 kg = 68000 kg

Q dinding * L * tinggi * jml = 250 *8 * 4,25 * 3

= 25500 kg

W lantai 2

+

= 626758 kg

Lantai 1 



Kolom Volume kolom * 2400 * jml = 0,45*0,45*4,25*2400*49 Dinding Q dinding * L * tinggi * jml = 250 * 36 * 4,25 * 2 Q dinding * L * tinggi * jml = 250 * 32 * 4,25 * 2 Q dinding * L * tinggi * jml = 250 *8 * 4,25 * 3



= 101210 kg

Sloof Volume sloof * 2400 * jml Volume sloof * 2400 * jml

= 0,3 * 0,4 * 36 * 2400 * 5 = 51840 kg = 0,3 * 0,45 * 16 * 2400 * 10 = 51840 kg + W lantai 1

Wtot

D EK Y E K A

= 76500 kg = 68000 kg = 25500 kg

=374890 kg

= W atap + Wlt 3 + Wlt 2 + Wlt 1 = 379152 + 621788 + 626758 + 374890 = 2002588 kg

A RIFIN

/ 10 5724 259

Page 80

M E R E N CA N A


Perhitungan periode alami struktur

T

= Ct * h0,75

T

= 0,0731 * 13,20,75

T

= 0,51

Diket : T = periode alami struktur (dtk) Ct = 0,0731 (beton) h = tinggi total gedung

Faktor respons gempa rencana ( C )

Faktor respons gempa ditentukan melalui Tabel 2 SNI 03-1726-2003 Wilayah gempa

:

2

Jenis Tanah

:

keras

T

:

0,51 detik

C = 0,3 ( dari grafik diatas )

Faktor keutamaan gedung ( I )

Faktor keutamaan gedung berdasarkan SNI 03-1726-2003 Kategori gedung

: gedung umum

I

: 1 ( tabel 1 faktor keutamaan gedung )

D EK Y E K A

A RIFIN

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Page 81

M E R E N CA N A


Penentuan parameter daktilitas struktur ( R )

berdasarkan Tabel 2 SNI 03-1726-2003, dengan batasan daktilitas maksimum menurut Tabel 3 SNI 03-1726-2003 > Untuk SRPMM ------ daktail parsial μm = 3,3 (Tabel 3 SNI 03-1726-2003)

Rm = 5,5 (Tabel 3 SNI 03-1726-2003)

Perhitungan gaya geser gempa ( V )

∗∗   ∗∗ = 

V

=

,

,

= 109232,07 kg ( untuk portal melintang ) Portal memanjang (V) = 30 % * V = 30 % * 1091232,07 = 32769,621 kg

Distribusi gaya geser gempa ( F )

Tabel distribusi gaya geser gempa portal memanjang Lantai

Tinggi

Atap

Berat

wi*zi

F=(wi*zi/∑wi*zi)*v

F/(5-1)

12.3

379152

4663569.6

38,491.82

9,622.95

Lt3

9.5

621788

5906986

48,754.63

12,188.66

Lt2

4.25

626758

2663721.5

21,985.62

5,496.41

∑

D EK Y E K A

A RIFIN

13234277.1

/ 10 5724 259

Page 82

M E R E N CA N A


Tabel distribusi gaya geser gempa portal memanjang Lantai

Tinggi

Atap

Berat

wi*zi

F=(wi*zi/∑wi*zi)*v

F/(5-1)

12.3

379152

4663569.6

11,547.54

1,283.06

Lt3

9.5

621788

5906986

14,626.39

1,625.15

Lt2

4.25

626758

2663721.5

6,595.69

732.85

∑

13234277.1

48114.77

48114.77

60943.29

60943.29

27482.03

27482.03

Sketsa pembebanan gempa pada portal melntang

D EK Y E K A

A RIFIN

/ 10 5724 259

Page 83

M E R E N CA N A


12830.6

12830.6

16251.54

16251.54

7328.54

7328.54

Sketsa pembebanan gempa pada portal memanjang

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PERHITUNGAN KOLOM

Kolom Lantai 1

Dimensi Kolom

= 45 × 45

Diameter Tulangan Utama

= Ø 22

Diameter Tulangan Sengkang

= Ø 12

Untuk fc 35 koefisien 0,85 diganti dengan 0,81 φ = 0.65

P = 72768.73 kg = 727687.3 N M = 17118.11 kg m = 171181100 N mm Agr = b × h = 0.45 × 0.45 = 0.2025 cm2 = 202500 mm

Sumbu Ordinat

 =   = 0.195            .

×

× .

×

.

×

× .

×

Sumbu Absis

  = 235.23 mm     =   = 0.523  

e = =

.

.

0,195 × 0.552 = 0.101

 =   = 0.143   ,

r = 0.038

…………………….dari grafik 6.1.d buku Gideon jikid 4 hal 86

ρ=r×β

= 0.0218 × 1.33 D EK Y E K A

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= 0.028

As = ρ × b × h

= 0,028 × 450 × 450 = 5670 mm2 – digunakan tulangan 12 Ø 25 = 5890,5 mm2

PERHITUNGAN TULANGAN GESER KOLOM Vu = 6084,43 Vn =

 =   = 9360,661 kg = 93606,61 N    ,

.

Vc = 1/6 × φ ×

  √

× b × d

= 1/6 × 0.65 × 35 × 450 × 392 = 113056.284 mm Vn > 0.5 Vc 93606,61 > 56528.142

. . . . . . . . perlu tulangan geser

Vs = Vn – Vc = 93606,61 - 113056.284 = 19449,63 N Direncanakan tulangan sengkang Ø 12 Av = 2 × luas tulangan = 2 × 113.1 = 226.2 mm2 Vsa =

      . ×

×

.

= 145.39 mm – digunakan tulangan sengkang Ø12 – 100

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Kolom Lantai 2

Dimensi Kolom

= 35 × 35


= Ø 19


= Ø12


P = 53463,15 kg = 534631,5 N M = 13741,15 kg m = 134754636,3 N mm Agr = b × h = 0.35 × 0.35 = 0.1225 cm2 = 122500 mm

Sumbu Ordinat

 =   = 0,236            ,

×

× .

×

.

×

× .

×

Sumbu Absis

   = 252,051 mm     =   = 0.720   ,

e = =

,

,

0,236 × 0.720 = 0.169

 =   = 0.175   ,

r = 0.016


ρ=r×β

= 0.016 × 1.33 D EK Y E K A

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= 0.0212

As = ρ × b × h

= 0,0212 × 350 × 350 = 2597 mm2 – digunakan tulangan 8 Ø 22 = 3041,1 mm2


 =   = 9199.338 kg = 91993,38 N    ,

.

Vc = 1/6 × φ ×

  √

× b × d

= 1/6 × 0.65 × 35 × 350 × 292 = 65500,863 mm Vn > 0.5 Vc 91993,98 > 32750,43


Vs = Vn – Vc = 91993,98 – 32750,43 = 69243,55 N Direncanakan tulangan sengkang Ø 12 Av = 2 × luas tulangan = 2 × 113.1 = 226.2 mm2 Vsa =

D EK Y E K A

      . ×

×

,

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= 33,385 mm – digunakan tulangan sengkang Ø12 – 100

Kolom Lantai 3

Dimensi Kolom

= 35 × 35


= Ø 19


= Ø12


P = 14181,28 kg = 141812,8 N M = 10358,88 kg m = 101585954,3N mm Agr = b × h = 0.35 × 0.35 = 0.1225 cm2 = 122500 mm

Sumbu Ordinat

 =   = 0,062            ,

×

× .

×

.

×

× .

×

Sumbu Absis

   = 716,33 mm     =   = 2,046   ,

e = =

,

,

2,046 × 0.062 = 0.126

 =   = 0.175   ,

r = 0.0143 D EK Y E K A


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ρ=r×β

= 0.0143 × 1.33 = 0.0191

As = ρ × b × h

= 0,0191 × 350 × 350 = 2339,75 mm2 – digunakan tulangan 6 Ø 22 = 2280,8 mm2


 =   = 11584,015 kg = 115840,15 N    ,

.

Vc = 1/6 × φ ×

  √

× b × d

= 1/6 × 0.65 × 35 × 350 × 292 = 65500,863 mm Vn > 0.5 Vc 115840,15 > 32750.431


Vs = Vn – Vc = 115840,15 – 65500,863 = 50339,287 N Direncanakan tulangan sengkang Ø 12 Av = 2 × luas tulangan = 2 × 113.1 = 226.2 mm2 D EK Y E K A

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Vsa =


      . ×

×

,

= 45,92 mm – digunakan tulangan sengkang Ø12 – 100

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PRELIMINARY DESIGN SLOOF 

Dimensi sloof h= =

1 12 1 12

× L × 500

= 41,7 cm̴ 45 cm 2

b = × h 3

2

= × 45 3

= 30 cm Maka dimensi sloof adalah 30 × 45



Pembebanan sloof Berat sendiri Berat dinding

= 0,3 × 0,45 × 2400 = 4,25 × 250

= 324 = 1062

kg/m kg/m +

q = 1386



kg/m

Penulangan sloof Tulangan lentur Tinggi efektif Diasumsikan memakai tulangan tarik D16 sengkang Ø10mm,dengan selimut beton 60mm. 



1

d = h – sb – Øsenngkang – Øtulangan 2



D EK Y E K A

= 450 – 60 – 10 – 8 = 372 mm Rasio penulangan β1 = 0,85 – 0,008 × (35 – 30) f’c = 35 Mpa = 0,81

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ρmin =

=

1,4

1,4 320

= 0,0043 0,85

ρmax = 0,75 ×

′

×β ×

0,85 35

= 0,75 ×

320

600



600

×0,81 ×

600



600

320

= 0,0368

 LAPANGAN

Mu lap

= 2493,75 kgm = 24937500 Nmm

Mn

=

=

Rn

=

=

m

=

ρ

= × (1-

.d2

24937500 0,8

31171875 300.372

=

0,85.

1

=

1

= 31171875 Nmm

320

= 10,75

0,85.35

 −  − 2.

1

× (1-

10,75

= 0,751

1

.

) 2.10,75.0,751 320

)

= 0,0023

ρmin > ρ 0,0043 > 0,0023 mka dipakai ρ = 0,0043



Menentukan As As = ρ × b × d

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= 0,0043 × 300 × 372 = 479,88 mm2 Dipakai tulangan 3-D16 (As = 603,4 mm2 ) As’ = 20% As tarik = 20 % . 479,88 = 95,976 mm2 Dipakai tulangan 2-D16 (As = 402,1 mm2 ) 

Cek lebar perlu 2sb + 2Øsengkang + n.Øtul + (n-1).Øtul 2.60 + 2.10 + 3.16 + (2-1).16 220 mm

< 300 mm < 300 mm < 300 mm

 TUMPUAN

Mu tum

= 4987,5kgm = 49875000 Nmm

Mn

=

=

Rn

=

=

m

=

ρ

= × (1-

.d2

49875000 0,8

62343750 300.372

=

0,85.

1

=

1

= 62343750 Nmm

320

= 10,75

0,85.35

 −  − 2.

1

× (1-

10,75

= 1,501

1

.

) 2.10,75.1,501 320

)

= 0,0048

ρmin <ρ 0,0043 < 0,0048 mka dipakai ρ = 0,0048



Menentukan As As = ρ × b × d = 0,0048 × 300 × 372

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= 535,68 mm2 Dipakai tulangan 3-D16 (As = 603,4 mm2 )

As’ = 20% As tarik = 20 % . 535,68 = 107,096 mm2 Dipakai tulangan 2-D16 (As = 402,1 mm2 ) 

Cek lebar perlu 2sb + 2Øsengkang + n.Øtul + (n-1).Øtul 2.60 + 2.10 + 3.16 + (2-1).16 220 mm



Vu

Vc

< 300 mm < 300 mm < 300 mm

Tulangan geser = 5985 kg

1

= × 6 1

= × 6

√ √

= 59850 N 5

√

Vumax

≤

φ × ×

59850

≤

0,75 × ×

59850

≤

412646,564 …………… dimensi sloof OK

6

5 6

× bw × d

√

35 ×

300 × 372

× bw × d

35 ×

300 × 372

= 110039,084 N φVc

= 0,75 × 110039,084 = 82529,313 N

1 2

1

φVc = × 82529,313 2

= 41264,65

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


Cek kebutuhan tulangan geser φVc

>

1

Vu

> φVc 2

82529,313 > 59850 > 41264,65 Maka dipasang tulangan geser minimum 



Untuk tumpuan Dipasang sengkang praktis Ø10 – 150 mm

Untuk lapangan Dipasang semgkang praktis Ø10 – 300 mm

2 D16

0 5 , 4

LAP ANG AN

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3 D16

3,00

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3,00 TUMP UAN

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PERHITUNGAN PONDASI TIANG PANCANG

Kondisi I (Dua Tumpuan)

Gambar 4. 39 Kondisi Pengangkatan 1 dan Momen yang Ditimbulkan

  × a    M = ×  × ( − 2) - ×  × a   2

M1 = × 2

2

M1 = M2

 ×  × a =  ×  × (− 2) -  ×  × a    2

2

4 a2 + 4 a L – L2 Dimana: q = Berat tiang pancang



= 0,25 × × 0,30 × 2400 = 169,64 kg/m L=6m D EK Y E K A

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4 a2 + 4 a L – L2 4.a2 + 4.a.6 – 62 = 0 4.a2 + 24.a – 36 = 0 a2 + 6.a – 9 = 0

Didapatkan: a =

       ± (

)

× ×(

)

= 1,243 m

 

M1 = M2 = ×

 



× a2

= × 169,64 × 2,4852 = 131,05 kgm

 

Dmak = × 169,64 × 6 = 508,92 kg

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Kondisi II (Satu Tumpuan)

Gambar 4.40 Kondisi Pengangkatan 2 dan Momen yang Ditimbulkan

  × a    −   D =          D =   2

M1 = ×

)

×(

×

1

×(

×

×

1

(

)

× ×

)

Mx = D1 × x – 0,5 × q × x2

  = 0  ×

x=

 

R 1 – q × x = 0

 =        × ×

(

D EK Y E K A

)

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   −               M =     × ×

M3 = D1

(

× ×

)

[ ×(

)]

× ×

2

[ ×(

)]

M1 = M2

 ×  × a =          × ×

2

[ ×(

)]

2a2 – 4 × a × L + L2 =0 Maka: 2a2 – 4 × a × 6 + 62 =0 2a2 – 24 × a + 36 =0 a2 – 12 × a + 18 =0

Didapatkan: a = = 1,75 m

 

M1 = M2 = ×

 



× a2

= × 169,64 × 1,7512 = 260,05 kgm D1 =

       ×

×

(

D EK Y E K A

× ×

)

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=


           ,

×

× ,

(

,

×

,

×

)

= 299,36 kg

Dari kedua kondisi di atas diambil yang paling menentukan yaitu: M = 299,36 kgm D = 508,92 kg

D = 30

Gambar 4.41 Penampang Tiang Pancang

Data yang digunakan: - Dimensi tiang = ø 30 cm - Berat jenis beton = 2,4 t/m3 - f’c = 35 Mpa - f y = 400 Mpa - h = 300 mm - p = 70 mm - øtulangan = 19 mm - øsengkang = 8 mm - d = h – p – øsengkang – ½ ø tulangan = 300 – 70 – 8 – 9,5 = 212,5 mm

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- d’ = p + øsengkang + ½ øtulangan = 70 + 8 + 9,5 = 87,5 mm Tulangan Memanjang Tiang Pancang

Mu = 299,36 kgm = 2,9936 kNm

 =    = 3,8841 N/m        =  = 13,44 m=           ) 1− ρ = × (1-     × (1-  1 −      ) =    ,

.

2

, . ,

,

×

,

×

× ×

×

,

× ,

,

= 0,011 Dengan rumus abc didapatkan nilai ρ = 0,00027 Pemeriksaan syarat rasio penulangan (ρmin < ρ < ρmax)

  =   = 0,0035     ×    =    ×    = 0,028 =    

ρmin = ρmax

,

,

×

,

×

,

×

,

×

karena ρ < ρmin maka dipakai ρ As = ρ.b.d

= 0,011 . 300 . 212,5 = 701,25 mm2 Digunakan tulangan 4D19 (As = 1134 mm2) Cek Terhadap Tekuk

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Dianggap kedua ujung sendi, diperoleh harga k = 1 r = 0,3 . h = 0,3 . 300 = 90 mm

  =   = 66 ( K > 20 maka kelangsingan diperhitungkan )    lg = ×  ×     = ×  × 300 = 397607820,2 mm  ×

K=

×

Ec = 4700 (fc)0,5 = 27805,57 Mpa

           = 3,41 × 10 =   × × ,

El =

,

×

, × ,

12

,

Pu = 713,617 KN Pcr =

  =          ×

( ×

× ,

)

×

( ×

)

= 9348708,61 N Cm = 1 Cs =



   ,

=

       ,

,

×

×

= 1,133

,

Mn = Cs × Mu = 1,133 × 0,523780 × 107 = 5934427,4 Nmm

 =   = 8,315 mm     - 87,5 = 70,815 mm e = ea + – d’ = 8,315 +     =    = 127,5 mm cb =    =  = 79,95 mm a=         ,

ea =

×

,

×

×

×

,

,

×

×

ab = 0,85 × cb = 0,85 × 127,5 = 108,375 mm

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a < ab , maka dipakai rumus

      As = As’ =            = -94,305 mm =     ×

(

×

)

,

,

(

,

×

×

2

, )

Digunak an As min 1% Ag = 0,01.(1/4.π.(300)2) = 706,858 mm Maka digunakan tulangan 4D19 ( As pasang = 1134 mm2) Penulangan Geser Tiang Pancang

Vu = 508,92 kg = 5089,2 N

  = 8482 N     ×  ×  =  × √ 35 ×300×212,5 = 62858,347 N Vc = ×    Vn =

,

,

Periksa vu > f vc:

 =   = 0,0798 Mpa         vc = = ×      =  × √ 35 = 0,986 Mpa ,

vu =

×

,

fvc = 0,6 × 0,986 = 0,5916

8 D 19 8 - 200

vu < vfc dipakai tulangan praktis Dipakai tulangan sengkang Ø8 - 200



Diketahui

: Psap Msap



= 713,617 KN = 167,87 KNm

PERHITUNGAN TIANG PANCANG Ptotal = Psap + P poer = 72768,73 +4800

D EK Y E K A

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= 77568,73 kg = 77,567Ton Data Tiang Pancang



∅         

Diameter 30 cm

Kedalaman Pemancangan 15 m Atp

= . . =

. .30

= 706,86 cm2 K tp

= 2. .r

= 2. .15

= 94,3 cm f’c



= 35 Mpa

= 350 kg/cm2

Perhitungan Daya Dukung Tiang Tunggal (Single Pile) 

Berdasarkan kekuatan bahan

= 0,33 . f’c = 0,33 . 350 = 115,5 kg/cm2

σ b

Ptiang = Atp . σ b = 706,86 . 115,5 = 81642,33 kg 

Berdasarkan hasil sondir Safety Factor, SF

= 3 dan 5

Nilai hambatan Conus (Qc)

= 70

kg/cm2

      = 

Ptiang =

.

.

,

= 16493,4

kg

Hambatan Cleef rata-rata Diambil per segmen setiap 3m 0–3 D EK Y E K A

→

150

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3–6

6–9

→ →

9 – 12

3–6 6–9

=

  = 0,5 

=

  = 0,833 

=

  = 1,5 

400

 850

→

12 – 15

0–3






1550

→



=

 = 2,33 

2450



=

 = 3 

→      →      →      →      →     , ×

× ,

, ×

× ,

, ×

× ,

, ×

9 – 12

12 – 15

= 2829

= 4713,12 kg = 8487 kg

× ,

, ×

kg

= 13183,14 kg

×

= 16974 kg

+

Q = 46186,26 kg

Daya dukung single pile = 16493,4 + 46186,26 – 1695 = 60966,66 

Menentukan jumlah tiang pancang (n) n



=

  

=

    ,

,

= 1,2 -------- dipasang 2 tiang pancang.

PERHITUNGAN PILE CAP 

Pembebanan Poer

-

Pterpusat

=

= 72768,7

kg

-

Berat sendiri poer

= 2,1 x 0,9 x 0,5 x 2400

= 2268

kg

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M E R E N CA N A


-

Berat sendiri kolom

= 0,45 x 0,45 x 1 x 2400

= 486

kg

-

Urugan Tanah

= vol. tanah x 1673

= 3176,60

kg

Ptot= 81231,3 kg

 Penulangan Pile Cap 

Momen pada titik gulung pondasi Mu

   = . 81231,3. 1,2  = . Wu . l2

= 24369,3 kgm 

Tinggi efektif (d) Direncanakan pakai tulangan ϕ19 mm, dengan selimut beton 75 mm. d

= ht – sb – (0,5 . ϕtul) = 500 – 75 – (0,5 . 19) = 615,5 mm



Rasio penulangan f’c = 35 Mpa

→

β1

= 0,85 = 0,567

−  

0,008.(35

−

30)

ρ min = 0,0018

= 0,0018 = 0,0022

    .  .       . 0,567 .  = 0,75 .  

ρ max = 0,75 .

,

.

,

.

′

= 0,0363 

Tulangan lentur Mu

D EK Y E K A

= 243693000

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Nmm

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Merencana Konstruksi Beton

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