Perhitungan Penulangan Balok.docx

Perhitungan penulangan balok Perhitungan penulangan balok portal melintang 6-6 Data-data perhitungan : Dimensi balok

: 20/40 cm

Fc’

: 25 Mpa

Fy

: 240 Mpa

Dicoba : Untuk lantai 1-4 Ø sengkang

= 10 mm

Ø tulangan pokok

= 16 mm

Selimut beton

= 40 mm 1

d = h - p - Ø sengkang - 2 Ø tulangan pokok d = 400 - 40 – 10 -

1 2

. 16

d = 342 mm 1

d’ = p + 2 Ø tulangan pokok + Ø sengkang 1

d’ = 40 + 2 . 16 + 10 d’ = 58 mm  Penulangan tumpuan melintang H-H 1.

Momen tumpuan rencana Untuk momen tumpuan rencana ada dua jenis tulangan tumpuan, yaitu tulangan tumpuan positif dan tulangan tumpuan negatif. Penulangan tersebut berdasarkan nilai momen positif dan terbesar tiap lantai dalam satu portal. Adapun nilai momen maksimum positif dan negatif tersebut diambil dari kombinasi :



Momen kombinasi akibat beban mati dan beban hidup

Mk1 = 1,2 MD + 1,6 ML 

Momen kombinasi akibat beban mati dan beban gempa

Mk2 = 0,9 MD + ME 

Momen kombinasi akibat beban mati, beban hidup dan beban gempa

Mk2 = 1,05 ( MD + 0,9 ML + ME )

2.

penulangan tumpuan 

perhitungan balok lantai 1

M tumpuan

= 63,092 KNm

Tulangan tumpuan negatif Mu = 63,092 KNm b

= 200 mm

h

= 400 mm

d

= 342 mm

Mu

fy

= ρ.0,8.fy. (1-0,5882 ρ . fC′ )

𝑏𝑑2

63,092 .106 Nmm 200 .3422

= ρ.0,8.240. (1-0,5882 ρ .

240 25

2,697 = 192 ρ – 892,170 ρ2 892,170 ρ2 - 192 ρ + 2,697 = 0 𝜌1,2 = 𝜌1 = 𝜌2 =

−𝑏 ± √𝑏2 − 4 𝑎𝑐 2𝑎

192 + √1922 − 4 ( 892,170) (2,697 ) 2 ( 892,170) 192 −√1922 − 4 ( 892,170) (2,697) 2 ( 892,170)

= 0,200 = 0,015

ρ = 0,015 As = ρ.b.d = 0,015.200.342 = 1026 mm2 As’ = 0,5.As = 0,5 . 1026 mm2 = 513 mm2

)

Dalam hal ini tulangan terpasang diambil berdasarkan nilai maksimum dari 2 jenis tulangan tumpuan Asatas

= 1026 mm2

Asbawah

= 513 mm2



perhitungan balok lantai 2

M tumpuan

= 58,832 KNm

Tulangan tumpuan negatif Mu = 58,832 KNm b

= 200 mm

h

= 400 mm

d

= 342 mm

Mu

fy

= ρ.0,8.fy. (1-0,5882 ρ . fC′ )

𝑏𝑑2

58,832 .106 Nmm 200 .3422

= ρ.0,8.240. (1-0,5882 ρ .

240 25

)

2,514 = 192 ρ – 892,170 ρ2 892,170 ρ2 - 192 ρ + 2,514 = 0 𝜌1,2 = 𝜌1 = 𝜌2 =

−𝑏 ± √𝑏2 − 4 𝑎𝑐 2𝑎

192 + √1922 − 4 ( 892,170) (2,514 ) 2 ( 892,170) 192 −√1922 − 4 ( 892,170) (2,514 ) 2 ( 892,170)

= 0,201 = 0,014

ρ = 0,014 As = ρ.b.d = 0,014.200.342 = 957,6 mm2 As’ = 0,5.As = 0,5 . 957,6mm2 = 478,8 mm2 Dalam hal ini tulangan terpasang diambil berdasarkan nilai maksimum dari 2 jenis tulangan tumpuan Asatas

= 957,6 mm2

Asbawah

= 478,8 mm2



perhitungan balok lantai 3

M tumpuan

= 57,398KNm

Tulangan tumpuan negatif Mu = 57,398KNm b

= 200 mm

h

= 400 mm

d

= 342 mm

Mu

fy

= ρ.0,8.fy. (1-0,5882 ρ . fC′ )

𝑏𝑑2

57,398 .106 Nmm 200 .3422

= ρ.0,8.240. (1-0,5882 ρ .

240 25

)

2,453 = 192 ρ – 892,170 ρ2 892,170 ρ2 - 192 ρ + 2,453 = 0 𝜌1,2 = 𝜌1 = 𝜌2 =

−𝑏 ± √𝑏2 − 4 𝑎𝑐 2𝑎

192 + √1922 − 4 ( 892,170) (2,453 ) 2 ( 892,170) 192 −√1922 − 4 ( 892,170) (2,453 ) 2 ( 892,170)

= 0,205 = 0,0136

ρ = 0,014 As = ρ.b.d = 0,0136.200.342 = 930,24 mm2 As’ = 0,5.As = 0,5 . 930,24mm2 = 465,12 mm2 Dalam hal ini tulangan terpasang diambil berdasarkan nilai maksimum dari 2 jenis tulangan tumpuan Asatas

= 930,24 mm2

Asbawah

= 465,12 mm2



perhitungan balok lantai 4

M tumpuan

= 63,828 KNm

Tulangan tumpuan negatif Mu = 63,828 KNm

b

= 200 mm

h

= 400 mm

d

= 342 mm

Mu 𝑏𝑑2

fy

= ρ.0,8.fy. (1-0,5882 ρ . fC′ )

63,828 .106 Nmm 200 .3422

= ρ.0,8.240. (1-0,5882 ρ .

240 25

)

2,728 = 192 ρ – 892,170 ρ2 892,170 ρ2 - 192 ρ + 2,728 = 0 𝜌1,2 = 𝜌1 = 𝜌2 =

−𝑏 ± √𝑏2 − 4 𝑎𝑐 2𝑎

192 + √1922 − 4 ( 892,170) (2,728 ) 2 ( 892,170) 192 −√1922 − 4 ( 892,170) (2,728 ) 2 ( 892,170)

= 0,199 = 0,015

ρ = 0,014 As = ρ.b.d = 0,015.200.342 = 1026 mm2 As’ = 0,5.As = 0,5 . 1026 mm2 = 513 mm2 Dalam hal ini tulangan terpasang diambil berdasarkan nilai maksimum dari 2 jenis tulangan tumpuan

3.

Asatas

= 1026 mm2

Asbawah

= 513 mm2

penulangan geser  perhitungan balok lantai 1 momen nominal aktual ( asumsi balok T ) Tulangan atas

= 5 D 16 = 1026 mm2

Tulangan bawah

= 3 D 16 = 513 mm2 1

d2 = h-p- Ø sengkang - 2 Ø tulangan pokok 1

d2 = 400 – 40 - 10 - 2 . 16

d2 = 342 mm 1

1

d1 = d2 - 2 Ø tulangan pokok – 25 - 2 Ø tulangan pokok 1

1

= 342- 2. 16 – 25 - 2 .16 = 301 mm 1

As1 = 3 x 4 . π . 162 = 603,185 mm2 1

As2 = 3 x

. π . 202 = 603,185 mm2 +

4

= 1026,37 mm2

As d=

As1 .d1 + As2 .d2 As

=

603,185 .301 + 603,185 .342 1026,37

= 321,5 mm

1

d’ = p + 2 Ø tulangan pokok + Ø sengkang 1

d’ = 40 + 2 . 16 + 10 d’ = 58 mm 1

As’ = 3 x 4 . π . 162 = 603,185 mm2 a=

(𝐴𝑠−𝐴𝑠′ ).𝑓𝑦 0,85. 𝑓𝑐 ′ 𝑏

=

(1026,37 – 603,185 ). 240 0,85.25.200

= 34,062 mm

β1 =0,85 𝑎

c = 𝛽1 =

34,062 𝑚𝑚 0,85

= 40,073 mm

cek kelelehan : 𝑓𝑦

240

εy = 𝐸𝑠 = 200000 = 0,0012 εs = εs’ =

𝑑−𝑐 𝑐

. 𝜀𝑐 ′ =

𝑐−𝑑′ 𝑐

321,5−40,073

. 𝜀𝑐 ′ =

40,073 40,073 − 58 40,073

. 0,003 = 0,021 > 𝜀𝑦 . 0,003 = − 0,0013 < 𝜀𝑦

kondisi II ( 0,85 . fc’. b . 𝛽1 ) c2 + ( 600 As’ – As.fy ) c – 600 As’ d’ = 0

( 0,85 . 25 . 200 . 0,85 ) c2 + ( 600 . 603,185 –1026,37 . 240) c – 600 . 603,185. 58= 0 3612,5 c2 + 116302,2 c – 20990838 = 0 𝜌1,2 = 𝜌1 = 𝜌2 =

−𝑏 ± √𝑏2 − 4 𝑎𝑐 2𝑎

− 116302,2 + √116302,2 2 − 4 ( 3612,5) (−20990838) 2 ( 3612,5) − 116302,2 − √116302,2 2 − 4 ( 3612,5) (−20990838) 2 ( 3612,5)

= 61,811 mm = - 94,005 mm

𝑐 = 61,811 mm 𝑎 = 𝛽1 𝑥 𝑐 = 0,85 𝑥 61,811 = 52,539 mm < hf = 120 mm ( asumsi benar ) Bagian 2 As2 = =

600 𝐴𝑠′ 𝑥

𝑐−𝑑′ 𝑐

𝑓𝑦 600 . 603,185 𝑥

61,811 − 58 61,811

240

= 92,974 mm2

Mn2 = Ts2 ( d-d’ ) = As2 . fy ( d-d’ ) = 92,974. 240 ( 321,5 -58 ) = 5,879 KNm

Bagian 1 As1 = As – As2 = 1206,37 mm2 – 603,185 mm2 = 603,185 mm2 Mn1 = Ts1 ( d- a/2) = As1 . fy ( d- a/2) = 603,185. 240. ( 321,5 – 52,539 /2 ) = 42,738 KNm Mnak‘= Mn = Mn1 + Mn2 = 42,738 KNm + 5,879 KNm

= 48,617 KNm

Momen nominal aktual ( asumsi balok persegi ) Dimisalkan a < hf a=

(𝐴𝑠−𝐴𝑠′ ).𝑓𝑦 0,85. 𝑓𝑐 ′ 𝑏

=

(1026,37 – 603,185 ). 240 0,85.25.200

= 34,062 mm

β1 =0,85 𝑎

c = 𝛽1 =

34,062 𝑚𝑚 0,85

= 40,073 mm

cek kelelehan : 𝑓𝑦

240

εy = 𝐸𝑠 = 200000 = 0,0012 εs = εs’ =

𝑑−𝑐 𝑐

. 𝜀𝑐 ′ =

𝑐−𝑑′ 𝑐

321,5−40,073

. 𝜀𝑐 ′ =

40,073 40,073 − 58 40,073

. 0,003 = 0,021 > 𝜀𝑦 . 0,003 = − 0,0013 < 𝜀𝑦

kondisi II ( 0,85 . fc’. b . 𝛽1 ) c2 + ( 600 As’ – As.fy ) c – 600 As’ d’ = 0 ( 0,85 . 25 . 200 . 0,85 ) c2 + ( 600 . 603,185 –1026,37 . 240) c – 600 . 603,185. 58= 0 3612,5 c2 + 116302,2 c – 20990838 = 0

𝜌1,2 = 𝜌1 = 𝜌2 =

−𝑏 ± √𝑏2 − 4 𝑎𝑐 2𝑎

− 116302,2 + √116302,2 2 − 4 ( 3612,5) (−20990838) 2 ( 3612,5) − 116302,2 − √116302,2 2 − 4 ( 3612,5) (−20990838) 2 ( 3612,5)

= 61,811 mm = - 94,005 mm

𝑐 = 61,811 mm Cs = 600 x As’ x

𝑐−𝑑′ 𝑐

= 600 x 603,185 x

61,811 −58 61,811

= 22313,873 N

Mn2 = Cs ( d-d’ ) = 22313,873 N ( 321,5 -58 ) mm = 5,879 KNm Cc = 0,85 fc’ b . β1 . c = 0,85.25.200.0,85. 61,811 = 267950,685 N Mn1 = Cc ( d- a/2) = 267950,685 N ( 321,5 –56,277/2 ) = 78,606 KNm Mnak‘= Mn = Mn1 + Mn2 = 78,606 KNm + 5,879 KNm = 84,485 KNm

 perhitungan balok lantai 2 momen nominal aktual

( asumsi balok T ) Tulangan atas

= 5 D 16 = 957,6 mm2

Tulangan bawah

= 2 D 16 = 478,8 mm2 1

d2 = h-p- Ø sengkang - 2 Ø tulangan pokok 1

d2 = 400 – 40 - 10 - 2 . 16 d2 = 342 mm 1

1

d1 = d2 - 2 Ø tulangan pokok – 25 - 2 Ø tulangan pokok 1

1

= 342- 2. 16 – 25 - 2 .16 = 301 mm 1

As1 = 3 x 4 . π . 162 = 603,185 mm2 1

As2 = 2 x 4 . π . 162 = 402,123 mm2 + = 1005,308 mm2

As d=

As1 .d1 + As2 .d2 As

=

603,185 .301 + 402,123 .342 1005,308

= 317,39 mm

1

d’ = p + 2 Ø tulangan pokok + Ø sengkang 1

d’ = 40 + 2 . 16 + 10 d’ = 58 mm As’ = 2 x a=

1 4

. π . 162 = 402,123 mm2

(𝐴𝑠−𝐴𝑠′ ).𝑓𝑦 0,85.

𝑓𝑐 ′

𝑏

=

(1005,308 – 402,123 ). 240 0,85.25.200

β1 =0,85 𝑎

c = 𝛽1 =

34,062 𝑚𝑚 0,85

cek kelelehan :

= 40,073 mm

= 34,062 mm

𝑓𝑦

240

εy = 𝐸𝑠 = 200000 = 0,0012 𝑑−𝑐

εs =

𝑐

εs’ =

. 𝜀𝑐 ′ =

𝑐−𝑑′ 𝑐

317,39−40,073 40,073

. 𝜀𝑐 ′ =

40,073 − 58 40,073

. 0,003 = 0,020 > 𝜀𝑦

. 0,003 = − 0,0013 < 𝜀𝑦

kondisi II ( 0,85 . fc’. b . 𝛽1 ) c2 + ( 600 As’ – As.fy ) c – 600 As’ d’ = 0 ( 0,85 . 25 . 200 . 0,85 ) c2 + ( 600 . 402,123 –1005,308. 240) c – 600 . 402,123. 58= 0 3612,5 c2 + 0 c – 13993880 = 0 𝜌1,2 = 𝜌1 = 𝜌2 =

−𝑏 ± √𝑏2 − 4 𝑎𝑐 2𝑎

− 0 + √0 2 − 4 ( 3612,5) (−13993880) 2 ( 3612,5) − 0 − √0 2 − 4 ( 3612,5) (−13993880) 2 ( 3612,5)

= 62,239 mm = -62,239 mm

𝑐 = 62,239 mm 𝑎 = 𝛽1 𝑥 𝑐 = 0,85 𝑥 62,239 = 52,903 mm < hf = 120 mm ( asumsi benar ) Bagian 2 As2 = =

600 𝐴𝑠′ 𝑥

𝑐−𝑑′ 𝑐

𝑓𝑦 600 . 402,123 𝑥

62,239 − 58 62,239

240

= 68,469 mm2

Mn2 = Ts2 ( d-d’ ) = As2 . fy ( d-d’ ) = 68,469. 240 ( 317,39 -58 ) = 4,262 KNm

Bagian 1 As1 = As – As2 = 1005,308 mm2 – 402,123mm2

= 603,185 mm2 Mn1 = Ts1 ( d- a/2) = As1 . fy ( d- a/2) = 603,185. 240. ( 317,39 – 52,903 /2 ) = 42,117 KNm Mnak‘= Mn = Mn1 + Mn2 = 42,117 KNm + 4,262 KNm = 46,379 KNm

Momen nominal aktual ( asumsi balok persegi ) Dimisalkan a < hf a=

(𝐴𝑠−𝐴𝑠′ ).𝑓𝑦 0,85.

𝑓𝑐 ′

𝑏

=

(1005,308 – 402,123 ). 240 0,85.25.200

= 34,062 mm

β1 =0,85 𝑎

c = 𝛽1 =

34,062 𝑚𝑚 0,85

= 40,073 mm

cek kelelehan : 𝑓𝑦

240

εy = 𝐸𝑠 = 200000 = 0,0012 εs =

𝑑−𝑐 𝑐

. 𝜀𝑐 ′ =

321,5−40,073 40,073

. 0,003 = 0,021 > 𝜀𝑦

εs’ =

𝑐−𝑑′ 𝑐

. 𝜀𝑐 ′ =

40,073 − 58 40,073

. 0,003 = − 0,0013 < 𝜀𝑦

kondisi II ( 0,85 . fc’. b . 𝛽1 ) c2 + ( 600 As’ – As.fy ) c – 600 As’ d’ = 0 ( 0,85 . 25 . 200 . 0,85 ) c2 + ( 600 . 402,123 –1005,308. 240)c – 600 . 402,123. 58= 0 3612,5 c2 + 0 c – 13993880 = 0 𝜌1,2 = 𝜌1 = 𝜌2 =

−𝑏 ± √𝑏2 − 4 𝑎𝑐 2𝑎

− 0 + √0 2 − 4 ( 3612,5) (−13993880) 2 ( 3612,5) − 0 − √0 2 − 4 ( 3612,5) (−13993880) 2 ( 3612,5)

= 62,239 mm = -62,239 mm

𝑐 = 62,239 mm Cs = 600 x As’ x

𝑐−𝑑′ 𝑐

= 600 x 402,123 x

62,239 −58 62,239

= 16432,777 N

Mn2 = Cs ( d-d’ ) = 16432,777 N ( 317,39 -58 ) mm = 4,262 KNm Cc = 0,85 fc’ b . β1 . c = 0,85.25.200.0,85.62,239 = 224838,387 N Mn1 = Cc ( d- a/2) = 224838,387 N ( 317,5 –56,277/2 ) = 65,034 KNm Mnak‘= Mn = Mn1 + Mn2 = 65,034 KNm + 4,262 KNm = 69,296 KNm

 perhitungan balok lantai 3 momen nominal aktual ( asumsi balok T ) Tulangan atas

= 5 D 16 = 930,24 mm2

Tulangan bawah

= 3 D 16 = 465,12 mm2 1

d2 = h-p- Ø sengkang - 2 Ø tulangan pokok 1

d2 = 400 – 40 - 10 - 2 . 16 d2 = 342 mm 1

1

d1 = d2 - 2 Ø tulangan pokok – 25 - 2 Ø tulangan pokok 1

1

= 342- 2. 16 – 25 - 2 .16 = 301 mm 1

As1 = 3 x 4 . π . 162 = 603,185 mm2 1

As2 = 2 x 4 . π . 162 = 402,123 mm2 + = 1005,30 mm2

As d=

As1 .d1 + As2 .d2 As

=

603,185 .301 + 402,123 .342 1005,30

= 317,40 mm

1

d’ = p + 2 Ø tulangan pokok + Ø sengkang 1

d’ = 40 + 2 . 16 + 10 d’ = 58 mm 1

As’ = 3 x 4 . π . 162 = 402,123 mm2 a=

(𝐴𝑠−𝐴𝑠′ ).𝑓𝑦 0,85. 𝑓𝑐 ′ 𝑏

=

(1005,30 – 402,123 ). 240 0,85.25.200

= 34,062 mm

β1 =0,85 𝑎

c = 𝛽1 =

34,062 𝑚𝑚 0,85

= 40,073 mm

cek kelelehan : 𝑓𝑦

240

εy = 𝐸𝑠 = 200000 = 0,0012 𝑑−𝑐

εs =

𝑐

εs’ =

. 𝜀𝑐 ′ =

𝑐−𝑑′ 𝑐

317,4−40,073

. 𝜀𝑐 ′ =

40,073 40,073 − 58 40,073

. 0,003 = 0,020 > 𝜀𝑦 . 0,003 = − 0,0013 < 𝜀𝑦

kondisi II ( 0,85 . fc’. b . 𝛽1 ) c2 + ( 600 As’ – As.fy ) c – 600 As’ d’ = 0 ( 0,85 . 25 . 200 . 0,85 ) c2 + ( 600 . 402,123–1005,30. 240) c – 600 . 402,123. 58= 0 3612,5 c2 + 0 c – 13993880 = 0 𝜌1,2 = 𝜌1 = 𝜌2 =

−𝑏 ± √𝑏2 − 4 𝑎𝑐 2𝑎

− 0 + √0 2 − 4 ( 3612,5) (−13993880) 2 ( 3612,5) − 0 − √0 2 − 4 ( 3612,5) (−13993880) 2 ( 3612,5)

= 62,239 mm = - 62,239 mm

𝑐 = 62,239 mm 𝑎 = 𝛽1 𝑥 𝑐 = 0,85 𝑥 62,239 = 52,903 mm < hf = 120 mm ( asumsi benar ) Bagian 2

As2 = =

600 𝐴𝑠′ 𝑥

𝑐−𝑑′ 𝑐

𝑓𝑦 600 . 402,123 𝑥

62,239 − 58 62,239

240

= 68,469 mm2

Mn2 = Ts2 ( d-d’ ) = As2 . fy ( d-d’ ) = 68,469. 240 ( 317,4 -58 ) = 4,262 KNm

Bagian 1 As1 = As – As2 = 1005,30mm2 – 402,123mm2 = 603,177 mm2 Mn1 = Ts1 ( d- a/2) = As1 . fy ( d- a/2) = 603,177. 240. ( 317,4 – 52,903 /2 ) = 42,118 KNm Mnak‘= Mn = Mn1 + Mn2 = 42,118 KNm + 4,262 KNm = 46,380 KNm

Momen nominal aktual

( asumsi balok persegi ) Dimisalkan a < hf a=

(𝐴𝑠−𝐴𝑠′ ).𝑓𝑦 0,85. 𝑓𝑐 ′ 𝑏

=

(1005,30 – 402,123 ). 240 0,85.25.200

= 34,062 mm

β1 =0,85 𝑎

c = 𝛽1 =

34,062 𝑚𝑚 0,85

= 40,073 mm

cek kelelehan : εy =

𝑓𝑦

εs =

𝑑−𝑐

=

𝐸𝑠

𝑐

εs’ =

240 200000

. 𝜀𝑐 ′ =

𝑐−𝑑′ 𝑐

= 0,0012

317,4−40,073

. 𝜀𝑐 ′ =

40,073 40,073 − 58 40,073

. 0,003 = 0,020 > 𝜀𝑦 . 0,003 = − 0,0013 < 𝜀𝑦

kondisi II ( 0,85 . fc’. b . 𝛽1 ) c2 + ( 600 As’ – As.fy ) c – 600 As’ d’ = 0 ( 0,85 . 25 . 200 . 0,85 ) c2 + ( 600 . 402,123–1005,30. 240) c – 600 . 402,123. 58= 0 3612,5 c2 + 0 c – 13993880 = 0 𝜌1,2 = 𝜌1 = 𝜌2 =

−𝑏 ± √𝑏2 − 4 𝑎𝑐 2𝑎

− 0 + √0 2 − 4 ( 3612,5) (−13993880) 2 ( 3612,5) − 0 − √0 2 − 4 ( 3612,5) (−13993880) 2 ( 3612,5)

= 62,239 mm = - 62,239 mm

𝑐 = 62,239 mm Cs = 600 x As’ x

𝑐−𝑑′ 𝑐

= 600 x 402,123x

62,239 −58 62,239

Mn2 = Cs ( d-d’ ) = 16432,777 N ( 317,4 -58 ) mm = 4,262 KNm Cc = 0,85 fc’ b . β1 . c = 0,85.25.200.0,85. 62,239 = 224838, 387 N Mn1 = Cc ( d- a/2)

= 16432,777 N

= 224838, 387 N ( 317,4 –56,277/2 ) = 65,037 KNm Mnak‘= Mn = Mn1 + Mn2 = 65,037 KNm + 4,262 KNm = 69,299 KNm

 perhitungan balok lantai 4 momen nominal aktual ( asumsi balok T ) Tulangan atas

= 5 D 16 = 1026 mm2

Tulangan bawah

= 3 D 16 = 513 mm2 1

d2 = h-p- Ø sengkang - 2 Ø tulangan pokok 1

d2 = 400 – 40 - 10 - 2 . 16 d2 = 342 mm 1

1

d1 = d2 - 2 Ø tulangan pokok – 25 - 2 Ø tulangan pokok 1

1

= 342- 2. 16 – 25 - 2 .16 = 301 mm 1

As1 = 3 x 4 . π . 162 = 603,185 mm2

1

As2 = 3 x 4 . π . 202 = 603,185 mm2 + = 1026,37 mm2

As d=

As1 .d1 + As2 .d2

=

As

603,185 .301 + 603,185 .342 1026,37

= 321,5 mm

1

d’ = p + 2 Ø tulangan pokok + Ø sengkang 1

d’ = 40 + 2 . 16 + 10 d’ = 58 mm 1

As’ = 3 x 4 . π . 162 = 603,185 mm2 a=

(𝐴𝑠−𝐴𝑠′ ).𝑓𝑦 0,85. 𝑓𝑐 ′ 𝑏

=

(1026,37 – 603,185 ). 240 0,85.25.200

= 34,062 mm

β1 =0,85 𝑎

c = 𝛽1 =

34,062 𝑚𝑚 0,85

= 40,073 mm

cek kelelehan : 𝑓𝑦

240

εy = 𝐸𝑠 = 200000 = 0,0012 εs = εs’ =

𝑑−𝑐 𝑐

. 𝜀𝑐 ′ =

𝑐−𝑑′ 𝑐

321,5−40,073

. 𝜀𝑐 ′ =

40,073 40,073 − 58 40,073

. 0,003 = 0,021 > 𝜀𝑦 . 0,003 = − 0,0013 < 𝜀𝑦

kondisi II ( 0,85 . fc’. b . 𝛽1 ) c2 + ( 600 As’ – As.fy ) c – 600 As’ d’ = 0 ( 0,85 . 25 . 200 . 0,85 ) c2 + ( 600 . 603,185 –1026,37 . 240) c – 600 . 603,185. 58= 0 3612,5 c2 + 116302,2 c – 20990838 = 0 𝜌1,2 = 𝜌1 = 𝜌2 =

−𝑏 ± √𝑏2 − 4 𝑎𝑐 2𝑎

− 116302,2 + √116302,2 2 − 4 ( 3612,5) (−20990838) 2 ( 3612,5) − 116302,2 − √116302,2 2 − 4 ( 3612,5) (−20990838)

𝑐 = 61,811 mm

2 ( 3612,5)

= 61,811 mm = - 94,005 mm

𝑎 = 𝛽1 𝑥 𝑐 = 0,85 𝑥 61,811 = 52,539 mm < hf = 120 mm ( asumsi benar ) Bagian 2 As2 = =

600 𝐴𝑠′ 𝑥

𝑐−𝑑′ 𝑐

𝑓𝑦 600 . 603,185 𝑥

61,811 − 58 61,811

240

= 92,974 mm2

Mn2 = Ts2 ( d-d’ ) = As2 . fy ( d-d’ ) = 92,974. 240 ( 321,5 -58 ) = 5,879 KNm

Bagian 1 As1 = As – As2 = 1206,37 mm2 – 603,185 mm2 = 603,185 mm2 Mn1 = Ts1 ( d- a/2) = As1 . fy ( d- a/2) = 603,185. 240. ( 321,5 – 52,539 /2 ) = 42,738 KNm Mnak‘= Mn = Mn1 + Mn2 = 42,738 KNm + 5,879 KNm = 48,617 KNm

Momen nominal aktual ( asumsi balok persegi ) Dimisalkan a < hf a=

(𝐴𝑠−𝐴𝑠′ ).𝑓𝑦 0,85. 𝑓𝑐 ′ 𝑏

=

(1026,37 – 603,185 ). 240 0,85.25.200

= 34,062 mm

β1 =0,85 𝑎

c = 𝛽1 =

34,062 𝑚𝑚 0,85

= 40,073 mm

cek kelelehan : 𝑓𝑦

240

εy = 𝐸𝑠 = 200000 = 0,0012 εs = εs’ =

𝑑−𝑐 𝑐

. 𝜀𝑐 ′ =

𝑐−𝑑′ 𝑐

321,5−40,073

. 𝜀𝑐 ′ =

40,073 40,073 − 58 40,073

. 0,003 = 0,021 > 𝜀𝑦 . 0,003 = − 0,0013 < 𝜀𝑦

kondisi II ( 0,85 . fc’. b . 𝛽1 ) c2 + ( 600 As’ – As.fy ) c – 600 As’ d’ = 0 ( 0,85 . 25 . 200 . 0,85 ) c2 + ( 600 . 603,185 –1026,37 . 240) c – 600 . 603,185. 58= 0 3612,5 c2 + 116302,2 c – 20990838 = 0 𝜌1,2 = 𝜌1 =

−𝑏 ± √𝑏2 − 4 𝑎𝑐 2𝑎

− 116302,2 + √116302,2 2 − 4 ( 3612,5) (−20990838) 2 ( 3612,5)

= 61,811 mm

𝜌2 =

− 116302,2 − √116302,2 2 − 4 ( 3612,5) (−20990838) 2 ( 3612,5)

= - 94,005 mm

𝑐 = 61,811 mm Cs = 600 x As’ x

𝑐−𝑑′ 𝑐

= 600 x 603,185 x

61,811 −58 61,811

= 22313,873 N

Mn2 = Cs ( d-d’ ) = 22313,873 N ( 321,5 -58 ) mm = 5,879 KNm Cc = 0,85 fc’ b . β1 . c = 0,85.25.200.0,85. 61,811 = 267950,685 N Mn1 = Cc ( d- a/2) = 267950,685 N ( 321,5 –56,277/2 ) = 78,606 KNm Mnak‘= Mn = Mn1 + Mn2 = 78,606 KNm + 5,879 KNm = 84,485 KNm

𝑀tumpuan

1. Vub = 0,7 Ø(

𝑙𝑛

) + 1,05Vg

2. Vubterpakai = [1,05Vg − 0,7 (

𝑀tumpuan 𝑙𝑛

Gaya gesser Maksimum Balok (Vub)

)] +

𝑙𝑛−𝑑 𝑙𝑛

𝑀tumpuan

[Vub − [1,05Vg − 0,7 Ø (

𝑙𝑛

)]]

Lantai 1 Vub = 98,52 KN(lihat table gaya geser rencana) Lantai 2 Vub = 92,265 KN(lihat table gaya geser rencana) Lantai 3 Vub = 91,112 KN(lihat table gaya geser rencana) Lantai 4 Vub = 101,79 KN(lihat table gaya geser rencana)

Gaya gesser Maksimum Balok (Vub) Lantai 1 Vubterpakai = 96,281 KN(lihat table gaya geser rencana) Lantai 2 Vubterpakai = 90,428 KN(lihat table gaya geser rencana) Lantai 3 Vubterpakai = 89,275 KN(lihat table gaya geser rencana) Lantai 4 Vubterpakai = 99,547 KN(lihat table gaya geser rencana)

Penulangan Geser a. Penulangan Geser Balok Lantai 1 Skema Gaya Geser rencana :

Vujung = Vub-(

𝑉𝑢𝑏−𝑉𝑢𝑏𝑡𝑒𝑟𝑝𝑎𝑘𝑎𝑖 𝑑 98,52−96,281

Vujung = 98,52-(

0,342

𝑥 𝑙𝑛)

𝑥4,25)

Vujung = 70,696 KN Penulangan Geser Dalam Daerah Sendi Plastis Vs =

𝑉𝑢𝑏𝑡𝑒𝑟𝑝𝑎𝑘𝑎𝑖 Ø

=

96,281 0,75

= 128,374 𝐾𝑁 = 128374,66 𝑁

Dipakai Sengkang ∅ 10 𝑚𝑚

S=

𝐴𝑣.𝑓𝑦.𝑑

=

𝑉𝑠

1 4

2 𝜋102 .240.342 128374,66

S= 100,433 mm >

𝑑

342

=

4

= 100,433

4

= 85,5

Dipakai Jarak 85,5 mm Jadi, Dipakai Sengkang∅10 − 85,5 mm

Penulangan Geser Diluar Daerah Sendi Plastis Vub’ = Vujung +

𝑙𝑛−2ℎ

Vub’ = 70,696 +

4,250−2.0,4

𝑙𝑛

. (𝑉𝑢𝑏 𝑚𝑎𝑘𝑠𝑖𝑚𝑢𝑚 − 𝑉𝑢𝑗𝑢𝑛𝑔)

4,250

. (98,52 − 70,696)

Vub’ = 93,282 KN √𝑓𝑐′

Vc = ( Vs = S=

𝑉𝑢𝑏′ ∅

) . 𝑏𝑤. 𝑑 = (

6

− 𝑉𝑐 =

𝐴𝑣.𝑓𝑦.𝑑 𝑉𝑠

=

√25 ) . 200.342 6

93,828 0,75

− 57 = 68,104 𝐾𝑁 = 68104 𝑁

1 4

2 𝜋102 .240.342 68104 𝑑

S= 189,314 mm >

4

= 57000 𝑁 = 57 𝐾𝑁

=

= 189,314

342 2

= 171

Dipakai Jarak 171 mm Jadi, Dipakai Sengkang∅10 − 171 mm

b. Penulangan Geser Balok Lantai 2 Skema Gaya Geser rencana :

Vujung = Vub-(

𝑉𝑢𝑏−𝑉𝑢𝑏𝑡𝑒𝑟𝑝𝑎𝑘𝑎𝑖 𝑑 92,265−90,428

Vujung = 92,265-(

0,342

𝑥 𝑙𝑛)

𝑥4,25)

Vujung = 69,436 KN Penulangan Geser Dalam Daerah Sendi Plastis Vs =

𝑉𝑢𝑏𝑡𝑒𝑟𝑝𝑎𝑘𝑎𝑖 Ø

90,428

=

0,75

= 120,570 𝐾𝑁 = 120570 𝑁

Dipakai Sengkang ∅ 10 𝑚𝑚 S=

𝐴𝑣.𝑓𝑦.𝑑

=

𝑉𝑠

1 4

2 𝜋102 .240.342 120570

S= 106,934 mm >

𝑑 4

= 106,934

342

=

4

= 85,5

Dipakai Jarak 85,5mm Jadi, Dipakai Sengkang∅10 − 85,5 mm Penulangan Geser Diluar Daerah Sendi Plastis 𝑙𝑛−2ℎ

Vub’ = Vujung + Vub’ = 69,436 +

𝑙𝑛

. (𝑉𝑢𝑏 𝑚𝑎𝑘𝑠𝑖𝑚𝑢𝑚 − 𝑉𝑢𝑗𝑢𝑛𝑔)

4,250−2.0,4 4,250

. (92,265 − 69,436 )

Vub’ = 87,967 KN √𝑓𝑐′

Vc = ( Vs = S=

𝑉𝑢𝑏′ ∅

) . 𝑏𝑤. 𝑑 = (

6

− 𝑉𝑐 =

𝐴𝑣.𝑓𝑦.𝑑 𝑉𝑠

=

√25 ) . 200.342 6

87,967 0,75

1 4

− 57 = 60,289 𝐾𝑁 = 60289 𝑁

2 𝜋102 .240.342 60289

S= 213,854 mm >

𝑑 2

=

= 57000 𝑁 = 57 𝐾𝑁

= 213,854

342 2

= 171

Dipakai Jarak 171 mm Jadi, Dipakai Sengkang∅10 − 171 mm

c. Penulangan Geser Balok Lantai 3 Skema Gaya Geser rencana :

Vujung = Vub-(

𝑉𝑢𝑏−𝑉𝑢𝑏𝑡𝑒𝑟𝑝𝑎𝑘𝑎𝑖 𝑑 91,112−89,275

Vujung = 91,112-(

0,342

𝑥 𝑙𝑛)

𝑥4,25)

Vujung = 68,283 KN Penulangan Geser Dalam Daerah Sendi Plastis Vs =

𝑉𝑢𝑏𝑡𝑒𝑟𝑝𝑎𝑘𝑎𝑖 Ø

89,275

=

0,75

= 119,033 𝐾𝑁 = 119033 𝑁

Dipakai Sengkang ∅ 10 𝑚𝑚 S=

𝐴𝑣.𝑓𝑦.𝑑

=

𝑉𝑠

1 4

2 𝜋102 .240.342 119033

S= 108,315mm >

𝑑 4

= 108,315

342

=

4

= 85,5

Dipakai Jarak 85,5mm Jadi, Dipakai Sengkang∅10 − 85,5 mm

Penulangan Geser Diluar Daerah Sendi Plastis 𝑙𝑛−2ℎ

Vub’ = Vujung + Vub’ = 68,283 +

𝑙𝑛

. (𝑉𝑢𝑏 𝑚𝑎𝑘𝑠𝑖𝑚𝑢𝑚 − 𝑉𝑢𝑗𝑢𝑛𝑔)

4,250−2.0,4 4,250

. (91,112 − 68,283 )

Vub’ = 86,814 KN √𝑓𝑐′

Vc = ( Vs = S=

𝑉𝑢𝑏′ ∅

) . 𝑏𝑤. 𝑑 = (

6

− 𝑉𝑐 =

𝐴𝑣.𝑓𝑦.𝑑 𝑉𝑠

=

√25 ) . 200.342 6

86,814 0,75

− 57 = 58,752 𝐾𝑁 = 58752 𝑁

1 4

2 𝜋102 .240.342 58752

S= 219,449mm >

𝑑 2

=

= 57000 𝑁 = 57 𝐾𝑁

= 213,854

342 2

= 171

Dipakai Jarak 171 mm Jadi, Dipakai Sengkang∅10 − 171 mm

d. Penulangan Geser Balok Lantai 4 Skema Gaya Geser rencana :

Vujung = Vub-(

𝑉𝑢𝑏−𝑉𝑢𝑏𝑡𝑒𝑟𝑝𝑎𝑘𝑎𝑖 𝑑 101,79 − 99,547

Vujung = 101,79 -(

0,342

𝑥 𝑙𝑛) 𝑥4,25)

Vujung = 73,916 KN Penulangan Geser Dalam Daerah Sendi Plastis Vs =

𝑉𝑢𝑏𝑡𝑒𝑟𝑝𝑎𝑘𝑎𝑖 Ø

=

73,916

= 98,554 𝐾𝑁 = 98554 𝑁

0,75

Dipakai Sengkang ∅ 10 𝑚𝑚 S=

𝐴𝑣.𝑓𝑦.𝑑

=

𝑉𝑠

1 4

2 𝜋102 .240.342 98554

S= 130,822 mm >

𝑑 4

= 130,822

342

=

4

= 85,5

Dipakai Jarak 85,5mm Jadi, Dipakai Sengkang∅10 − 85,5 mm

Penulangan Geser Diluar Daerah Sendi Plastis Vub’ = Vujung + Vub’ = 73,916 +

𝑙𝑛−2ℎ 𝑙𝑛

. (𝑉𝑢𝑏 𝑚𝑎𝑘𝑠𝑖𝑚𝑢𝑚 − 𝑉𝑢𝑗𝑢𝑛𝑔)

4,250−2.0,4 4,250

. (101,79 − 73,916 )

Vub’ = 96,543 KN √𝑓𝑐′

Vc = (

6

) . 𝑏𝑤. 𝑑 = (

√25 ) . 200.342 6

= 57000 𝑁 = 57 𝐾𝑁

Vs = S=

𝑉𝑢𝑏′ ∅

− 𝑉𝑐 =

𝐴𝑣.𝑓𝑦.𝑑 𝑉𝑠

=

96,543 0,75

1 4

− 57 = 71,724𝐾𝑁 = 71724 𝑁

2 𝜋102 .240.342 71724

S= 179,759 mm >

𝑑 2

=

= 179,759

342 2

= 171

Dipakai Jarak 171 mm Jadi, Dipakai Sengkang∅10 − 171 mm

Penulangan lapangan melintang 6-6 1). Momen lapanan Rencana Nilai nilai dari momen lapangan rencana dapat dilihat dari table 2). Penulangan lapangan Data-data perhitungan : L

= 4250 mm

Ln

= 4250 mm

Lebar efektif untuk lantai 1,2,3 dan 4 1

1

4

4

Beff ≤ L= 4250 = 1062,5 mm Beff ≤ 16 hf + bw = 16. 120 + 200 = 2120 mm Beff ≤ bw + ln = 200 + 4250 = 4450 mm a. Perhitungan Balok Lantai 1 Mu = 35,60 KNm Asumsi letak garis netral di dalam sayap a ≤ hf dan c ≤ hf Batasan tulangan 𝜌 min = 𝜌 min =

√𝑓 ′ 𝑐 4.𝑓𝑦

=

√25 4.240

= 0,0052

1,4 1,4 = = 0,0058 𝑓𝑦 240

As min = ρmin . bw . d = 0,0052 . 200 . 342 = 355,68 mm2 𝐴𝑚𝑎𝑘𝑠 = 0,75 . ( 𝐴𝑚𝑎𝑘𝑠 = 0,75 . (

0,85. 𝛽. 𝑓𝑐 ′ 600 . ) 𝑏𝑤. 𝑑 𝑓𝑦 600 + 𝑓𝑦

0,85.0,85.25 600 . ) 200.342 240 600 + 240

𝐴𝑚𝑎𝑘𝑠 = 2757,756 𝑚𝑚2 𝑀𝑢 𝑎 = 0,85. 𝑓𝑐 ′ . 𝑎. 𝑏 (𝑑 − ) ∅ 2 35,60. 106 𝑎 = 0,85.25. 𝑎. 1062,5 (342 − ) 0,8 2 𝑎 43,825.106 = 22567,5 𝑎 (342 − ) 2 𝑎2 − 684 𝑎 + 3883,903 = 0

𝑅𝑢𝑚𝑢𝑠 𝑎𝑏𝑐 =

𝑎12 =

−𝑏 ± √𝑏 2 − 4𝑎𝑐 2𝑎

684 ± √(−684)2 − 4. (1). (3883,903) 2. (1)

𝑎1 = 678,273 𝑚𝑚 𝑎2 = 5,726 𝑚𝑚 Ambil a yang terkecil a = 5,726 mm 𝑐=

𝑎 5,726 𝑚𝑚 = = 6,736 < ℎ𝑓 = 120 𝑚𝑚 (𝑎𝑠𝑢𝑚𝑠𝑖 𝑏𝑒𝑛𝑎𝑟) 𝛽1 0.85

Cc = Ts 𝐴𝑠 =

0,85. 𝑓𝑐 ′ . 𝑐. 𝑏𝑒𝑓𝑓 0,85.25.6,736.1062,5 = = 633,692 𝑚𝑚2 𝑓𝑦 240

Dipakai 3 D 16 (As = 603,2 mm2)

b. Perhitungan Balok Lantai 2 Mu = 33,380 KNm Asumsi letak garis netral di dalam sayap a ≤ hf dan c ≤ hf Batasan tulangan 𝜌 min = 𝜌 min =

√𝑓 ′ 𝑐 4.𝑓𝑦

√25

= 4.240 = 0,0052

1,4 1,4 = = 0,0058 𝑓𝑦 240

As min = ρmin . bw . d = 0,0052 . 200 . 342 = 355,68 mm2 𝐴𝑚𝑎𝑘𝑠 = 0,75 . ( 𝐴𝑚𝑎𝑘𝑠 = 0,75 . (

0,85. 𝛽. 𝑓𝑐 ′ 600 . ) 𝑏𝑤. 𝑑 𝑓𝑦 600 + 𝑓𝑦

0,85.0,85.25 600 . ) 200.342 240 600 + 240

𝐴𝑚𝑎𝑘𝑠 = 2757,756 𝑚𝑚2 𝑀𝑢 𝑎 = 0,85. 𝑓𝑐 ′ . 𝑎. 𝑏 (𝑑 − ) ∅ 2 33,380 . 106 𝑎 = 0,85.25. 𝑎. 1062,5 (342 − ) 0,8 2

𝑎 41,725.106 = 22567,5 𝑎 (342 − ) 2 𝑎2 − 684 𝑎 + 3697,795 = 0

𝑅𝑢𝑚𝑢𝑠 𝑎𝑏𝑐 =

𝑎12

−𝑏 ± √𝑏 2 − 4𝑎𝑐 2𝑎

684 ± √(−684)2 − 4. (1). (3697,795) = 2. (1)

𝑎1 = 678,550 𝑚𝑚 𝑎2 = 5,449 𝑚𝑚 Ambil a yang terkecil a = 5,449 mm 𝑐=

𝑎 5,449 𝑚𝑚 = = 6,410 < ℎ𝑓 = 120 𝑚𝑚 (𝑎𝑠𝑢𝑚𝑠𝑖 𝑏𝑒𝑛𝑎𝑟) 𝛽1 0.85

Cc = Ts 𝐴𝑠 =

0,85. 𝑓𝑐 ′ . 𝑐. 𝑏𝑒𝑓𝑓 0,85.25.6,410.1062,5 = = 603,024 𝑚𝑚2 𝑓𝑦 240

Dipakai 3 D 16 (As = 603,2 mm2)

c. Perhitungan Balokk Lantai 3 Mu = 32,324KNm Asumsi letak garis netral di dalam sayap a ≤ hf dan c ≤ hf Batasan tulangan 𝜌 min = 𝜌 min =

√𝑓 ′ 𝑐 4.𝑓𝑦

√25

= 4.240 = 0,0052

1,4 1,4 = = 0,0058 𝑓𝑦 240

As min = ρmin . bw . d = 0,0052 . 200 . 342 = 355,68 mm2 𝐴𝑚𝑎𝑘𝑠 = 0,75 . ( 𝐴𝑚𝑎𝑘𝑠 = 0,75 . (

0,85. 𝛽. 𝑓𝑐 ′ 600 . ) 𝑏𝑤. 𝑑 𝑓𝑦 600 + 𝑓𝑦

0,85.0,85.25 600 . ) 200.342 240 600 + 240

𝐴𝑚𝑎𝑘𝑠 = 2757,756 𝑚𝑚2

𝑀𝑢 𝑎 = 0,85. 𝑓𝑐 ′ . 𝑎. 𝑏 (𝑑 − ) ∅ 2 32,324 . 106 𝑎 = 0,85.25. 𝑎. 1062,5 (342 − ) 0,8 2 𝑎 40,405.106 = 22567,5 𝑎 (342 − ) 2 𝑎2 − 684 𝑎 + 3580,813 = 0

𝑅𝑢𝑚𝑢𝑠 𝑎𝑏𝑐 =

𝑎12 =

−𝑏 ± √𝑏 2 − 4𝑎𝑐 2𝑎

684 ± √(−684)2 − 4. (1). (3580,813) 2. (1)

𝑎1 = 678,724 𝑚𝑚 𝑎2 = 5,275 𝑚𝑚 Ambil a yang terkecil a = 5,275 mm 𝑐=

𝑎 5,275 𝑚𝑚 = = 6,205 < ℎ𝑓 = 120 𝑚𝑚 (𝑎𝑠𝑢𝑚𝑠𝑖 𝑏𝑒𝑛𝑎𝑟) 𝛽1 0.85

Cc = Ts 𝐴𝑠 =

0,85. 𝑓𝑐 ′ . 𝑐. 𝑏𝑒𝑓𝑓 0,85.25.6,205.1062,5 = = 583,738 𝑚𝑚2 𝑓𝑦 240

Dipakai 3 D 16 (As = 603,2 mm2)

d. Perhitungan Balokk Lantai 4 Mu = 40,504 KNm Asumsi letak garis netral di dalam sayap a ≤ hf dan c ≤ hf Batasan tulangan 𝜌 min = 𝜌 min =

√𝑓 ′ 𝑐 4.𝑓𝑦

√25

= 4.240 = 0,0052

1,4 1,4 = = 0,0058 𝑓𝑦 240

As min = ρmin . bw . d = 0,0052 . 200 . 342 = 355,68 mm2 𝐴𝑚𝑎𝑘𝑠 = 0,75 . (

0,85. 𝛽. 𝑓𝑐 ′ 600 . ) 𝑏𝑤. 𝑑 𝑓𝑦 600 + 𝑓𝑦

𝐴𝑚𝑎𝑘𝑠 = 0,75 . (

0,85.0,85.25 600 . ) 200.342 240 600 + 240

𝐴𝑚𝑎𝑘𝑠 = 2757,756 𝑚𝑚2 𝑀𝑢 𝑎 = 0,85. 𝑓𝑐 ′ . 𝑎. 𝑏 (𝑑 − ) ∅ 2 40,504 . 106 𝑎 = 0,85.25. 𝑎. 1062,5 (342 − ) 0,8 2 𝑎 50,63.106 = 22567,5 𝑎 (342 − ) 2 𝑎2 − 684 𝑎 + 4486,983 = 0

𝑅𝑢𝑚𝑢𝑠 𝑎𝑏𝑐 =

𝑎12 =

−𝑏 ± √𝑏 2 − 4𝑎𝑐 2𝑎

684 ± √(−684)2 − 4. (1). (4486,983) 2. (1)

𝑎1 = 677,375 𝑚𝑚 𝑎2 = 6,624 𝑚𝑚 Ambil a yang terkecil a = 6,624 mm 𝑐=

𝑎 6,624 𝑚𝑚 = = 7,792 < ℎ𝑓 = 120 𝑚𝑚 (𝑎𝑠𝑢𝑚𝑠𝑖 𝑏𝑒𝑛𝑎𝑟) 𝛽1 0.85

Cc = Ts 𝐴𝑠 =

0,85. 𝑓𝑐 ′ . 𝑐. 𝑏𝑒𝑓𝑓 0,85.25.7,792.1062,5 = = 733,036 𝑚𝑚2 𝑓𝑦 240

Dipakai 3 D 16 (As = 603,2 mm2)

Perhitungan penulangan balok Perhitungan penulangan balok portal melintang 1-1 Data-data perhitungan : Dimensi balok

: 20/40 cm

Fc’

: 25 Mpa

Fy

: 240 Mpa

Dicoba : Untuk lantai 1-4 Ø sengkang

= 10 mm

Ø tulangan pokok

= 16 mm

Selimut beton

= 40 mm 1

d = h - p - Ø sengkang - 2 Ø tulangan pokok d = 400 - 40 – 10 -

1 2

. 16

d = 342 mm 1

d’ = p + 2 Ø tulangan pokok + Ø sengkang 1

d’ = 40 + 2 . 16 + 10 d’ = 58 mm  Penulangan tumpuan melintang 1-1 1. Momen tumpuan rencana Untuk momen tumpuan rencana ada dua jenis tulangan tumpuan, yaitu tulangan tumpuan positif dan tulangan tumpuan negatif. Penulangan tersebut berdasarkan nilai momen positif dan terbesar tiap lantai dalam satu portal. Adapun nilai momen maksimum positif dan negatif tersebut diambil dari kombinasi :



Momen kombinasi akibat beban mati dan beban hidup

Mk1 = 1,2 MD + 1,6 ML

2.

penulangan tumpuan 

perhitungan balok lantai 1

M tumpuan

= 45,238 KNm

Tulangan tumpuan Mu = 45,238 KNm b

= 200 mm

h

= 400 mm

d

= 342 mm

Mu

fy

= ρ.0,8.fy. (1-0,5882 ρ . fC′ )

𝑏𝑑2

45,238.106 Nmm 200 .3422

= ρ.0,8.240. (1-0,5882 ρ .

240 25

)

1,933= 192 ρ – 892,170 ρ2 892,170 ρ2 - 192 ρ + 1,933 = 0 𝜌1,2 = 𝜌1 = 𝜌2 =

−𝑏 ± √𝑏2 − 4 𝑎𝑐 2𝑎

192 + √1922 − 4 ( 892,170) (1,933 ) 2 ( 892,170) 192 −√1922 − 4 ( 892,170) (1,933 ) 2 ( 892,170)

= 0,204 = 0,010

ρ = 0,010 As = ρ.b.d = 0,010.200.342 = 684 mm2 As’ = 0,5.As = 0,5 . 684 mm2 = 342 mm2 Dalam hal ini tulangan terpasang diambil berdasarkan nilai maksimum dari 2 jenis tulangan tumpuan Asatas

= 684 mm2

Asbawah

= 342 mm2



perhitungan balok lantai 2

M tumpuan

= 41,725 KNm

Tulangan tumpuan Mu = 41,725 KNm b

= 200 mm

h

= 400 mm

d

= 342 mm

Mu 𝑏𝑑2

fy

= ρ.0,8.fy. (1-0,5882 ρ . fC′ )

41,725 .106 Nmm 200 .3422

= ρ.0,8.240. (1-0,5882 ρ .

240 25

)

1,783 = 192 ρ – 892,170 ρ2 892,170 ρ2 - 192 ρ + 1,783 = 0 𝜌1,2 = 𝜌1 = 𝜌2 =

−𝑏 ± √𝑏2 − 4 𝑎𝑐 2𝑎

192 + √1922 − 4 ( 892,170) (1,783 ) 2 ( 892,170) 192 −√1922 − 4 ( 892,170) (1,783 ) 2 ( 892,170)

= 0,205 = 0,009

ρ = 0,009 As = ρ.b.d = 0,009.200.342 = 615,6 mm2 As’ = 0,5.As = 0,5 . 615,6 mm2 = 307,8 mm2 Dalam hal ini tulangan terpasang diambil berdasarkan nilai maksimum dari 2 jenis tulangan tumpuan Asatas

= 615,6 mm2

Asbawah

= 307,8 mm2



perhitungan balok lantai 3

M tumpuan

= 40,719 KNm

Tulangan tumpuan negatif

Mu = 40,719 KNm b

= 200 mm

h

= 400 mm

d

= 342 mm

Mu

fy

= ρ.0,8.fy. (1-0,5882 ρ . fC′ )

𝑏𝑑2

40,719 .106 Nmm 200 .3422

= ρ.0,8.240. (1-0,5882 ρ .

240 25

)

1,740 = 192 ρ – 892,170 ρ2 892,170 ρ2 - 192 ρ + 1,740 = 0 𝜌1,2 = 𝜌1 = 𝜌2 =

−𝑏 ± √𝑏2 − 4 𝑎𝑐 2𝑎

192 + √1922 − 4 ( 892,170) (1,740) 2 ( 892,170) 192 −√1922 − 4 ( 892,170) (1,740) 2 ( 892,170)

= 0,205 = 0,009

ρ = 0,009 As = ρ.b.d = 0,009.200.342 = 615,6 mm2 As’ = 0,5.As = 0,5 . 615,6 mm2 = 307,8 mm2 Dalam hal ini tulangan terpasang diambil berdasarkan nilai maksimum dari 2 jenis tulangan tumpuan Asatas

= 615,6 mm2

Asbawah

= 307,8 mm2



perhitungan balok lantai 4

M tumpuan

= 46,067 KNm

Tulangan tumpuan negatif Mu = 46,067 KNm b

= 200 mm

h

= 400 mm

d

= 342 mm

Mu

fy

= ρ.0,8.fy. (1-0,5882 ρ . fC′ )

𝑏𝑑2

46,067 .106 Nmm 200 .3422

= ρ.0,8.240. (1-0,5882 ρ .

240 25

)

1,969 = 192 ρ – 892,170 ρ2 892,170 ρ2 - 192 ρ + 1,969 = 0 𝜌1,2 = 𝜌1 = 𝜌2 =

−𝑏 ± √𝑏2 − 4 𝑎𝑐 2𝑎

192 + √1922 − 4 ( 892,170) (1,969 ) 2 ( 892,170) 192 −√1922 − 4 ( 892,170) (1,969 ) 2 ( 892,170)

= 0,204 = 0,010

ρ = 0,010 As = ρ.b.d = 0,010.200.342 = 684 mm2 As’ = 0,5.As = 0,5 . 684 mm2 = 342 mm2 Dalam hal ini tulangan terpasang diambil berdasarkan nilai maksimum dari 2 jenis tulangan tumpuan Asatas

= 684 mm2

Asbawah

= 342 mm2

3. penulangan geser  perhitungan balok lantai 1 momen nominal aktual ( asumsi balok T ) Tulangan atas

= 3 D 16 = 684 mm2

Tulangan bawah

= 2 D 16 = 342 mm2 1

d2 = h-p- Ø sengkang - 2 Ø tulangan pokok 1

d2 = 400 – 40 - 10 - 2 . 16 d2 = 342 mm 1

1

d1 = d2 - 2 Ø tulangan pokok – 25 - 2 Ø tulangan pokok

1

1

= 342- 2. 16 – 25 - 2 .16 = 301 mm 1

As1 = 3 x 4 . π . 162 = 603,185 mm2 1

As2 = 2 x 4 . π . 162 = 402,123 mm2 + = 1005,308 mm2

As d=

As1 .d1 + As2 .d2 As

=

603,185.301 + 402,123.342 1005,308

= 317,4 mm

1

d’ = p + 2 Ø tulangan pokok + Ø sengkang 1

d’ = 40 + 2 . 16 + 10 d’ = 58 mm 1

As’ = 2 x 4 . π . 162 = 402,123 mm2 a=

(𝐴𝑠−𝐴𝑠′ ).𝑓𝑦 0,85. 𝑓𝑐 ′ 𝑏

=

(1005,308 – 402,123 ). 240 0,85.25.200

= 34,062 mm

β1 =0,85 𝑎

c = 𝛽1 =

34,062 𝑚𝑚 0,85

= 40,073 mm

cek kelelehan : 𝑓𝑦

240

εy = 𝐸𝑠 = 200000 = 0,0012 εs = εs’ =

𝑑−𝑐 𝑐

. 𝜀𝑐 ′ =

𝑐−𝑑′ 𝑐

317,4−40,073

. 𝜀𝑐 ′ =

40,073 40,073 − 58 40,073

. 0,003 = 0,020 > 𝜀𝑦 . 0,003 = − 0,0013 < 𝜀𝑦

kondisi II ( 0,85 . fc’. b . 𝛽1 ) c2 + ( 600 As’ – As.fy ) c – 600 As’ d’ = 0 ( 0,85 . 25 . 200 . 0,85 ) c2 + ( 600 . 402,123–1005,30. 240) c – 600 . 402,123. 58= 0 3612,5 c2 + 0 c – 13993880 = 0 𝜌1,2 =

−𝑏 ± √𝑏2 − 4 𝑎𝑐 2𝑎

𝜌1 = 𝜌2 =

− 0 + √0 2 − 4 ( 3612,5) (−13993880) 2 ( 3612,5) − 0 − √0 2 − 4 ( 3612,5) (−13993880) 2 ( 3612,5)

= 62,239 mm = - 62,239 mm

𝑐 = 62,239 mm 𝑎 = 𝛽1 𝑥 𝑐 = 0,85 𝑥 62,239 = 52,903 mm < hf = 120 mm ( asumsi benar ) Bagian 2 As2 = =

600 𝐴𝑠′ 𝑥

𝑐−𝑑′ 𝑐

𝑓𝑦 600 . 402,123 𝑥

62,239 − 58 62,239

240

= 68,469 mm2

Mn2 = Ts2 ( d-d’ ) = As2 . fy ( d-d’ ) = 68,469. 240 ( 317,4 -58 ) = 4,262 KNm

Bagian 1 As1 = As – As2 = 1005,30mm2 – 402,123mm2 = 603,177 mm2 Mn1 = Ts1 ( d- a/2) = As1 . fy ( d- a/2) = 603,177. 240. ( 317,4 – 52,903 /2 ) = 42,118 KNm Mnak‘= Mn = Mn1 + Mn2 = 42,118 KNm + 4,262 KNm = 46,380 KNm

Momen nominal aktual ( asumsi balok persegi ) Dimisalkan a < hf a=

(𝐴𝑠−𝐴𝑠′ ).𝑓𝑦 0,85. 𝑓𝑐 ′ 𝑏

=

(1005,30 – 402,123 ). 240 0,85.25.200

= 34,062 mm

β1 =0,85 𝑎

c = 𝛽1 =

34,062 𝑚𝑚 0,85

= 40,073 mm

cek kelelehan : 𝑓𝑦

240

εy = 𝐸𝑠 = 200000 = 0,0012 εs = εs’ =

𝑑−𝑐 𝑐

. 𝜀𝑐 ′ =

𝑐−𝑑′ 𝑐

317,4−40,073

. 𝜀𝑐 ′ =

40,073 40,073 − 58 40,073

. 0,003 = 0,020 > 𝜀𝑦 . 0,003 = − 0,0013 < 𝜀𝑦

kondisi II ( 0,85 . fc’. b . 𝛽1 ) c2 + ( 600 As’ – As.fy ) c – 600 As’ d’ = 0 ( 0,85 . 25 . 200 . 0,85 ) c2 + ( 600 . 402,123–1005,30. 240) c – 600 . 402,123. 58= 0 3612,5 c2 + 0 c – 13993880 = 0 𝜌1,2 = 𝜌1 =

−𝑏 ± √𝑏2 − 4 𝑎𝑐 2𝑎

− 0 + √0 2 − 4 ( 3612,5) (−13993880) 2 ( 3612,5)

= 62,239 mm

𝜌2 =

− 0 − √0 2 − 4 ( 3612,5) (−13993880) 2 ( 3612,5)

= - 62,239 mm

𝑐 = 62,239 mm Cs = 600 x As’ x

𝑐−𝑑′ 𝑐

= 600 x 402,123x

62,239 −58 62,239

= 16432,777 N

Mn2 = Cs ( d-d’ ) = 16432,777 N ( 317,4 -58 ) mm = 4,262 KNm Cc = 0,85 fc’ b . β1 . c = 0,85.25.200.0,85. 62,239 = 224838, 387 N Mn1 = Cc ( d- a/2) = 224838, 387 N ( 317,4 –56,277/2 ) = 65,037 KNm Mnak‘= Mn = Mn1 + Mn2 = 65,037 KNm + 4,262 KNm = 69,299 KNm

 perhitungan balok lantai 2 momen nominal aktual ( asumsi balok T ) Tulangan atas

= 3 D 16 = 615,6 mm2

Tulangan bawah

= 2 D 16 = 307,8 mm2

1

d2 = h-p- Ø sengkang - 2 Ø tulangan pokok 1

d2 = 400 – 40 - 10 - 2 . 16 d2 = 342 mm 1

1

d1 = d2 - 2 Ø tulangan pokok – 25 - 2 Ø tulangan pokok 1

1

= 342- 2. 16 – 25 - 2 .16 = 301 mm 1

As1 = 3 x 4 . π . 162 = 603,185 mm2 1

As2 = 2 x 4 . π . 162 = 402,123 mm2 + = 1005,308 mm2

As d=

As1 .d1 + As2 .d2 As

=

603,185.301 + 402,123.342 1005,308

= 317,4 mm

1

d’ = p + Ø tulangan pokok + Ø sengkang 2

1

d’ = 40 + 2 . 16 + 10 d’ = 58 mm 1

As’ = 2 x 4 . π . 162 = 402,123 mm2 a=

(𝐴𝑠−𝐴𝑠′ ).𝑓𝑦 0,85.

𝑓𝑐 ′

𝑏

=

(1005,308 – 402,123 ). 240 0,85.25.200

= 34,062 mm

β1 =0,85 𝑎

c = 𝛽1 =

34,062 𝑚𝑚 0,85

= 40,073 mm

cek kelelehan : 𝑓𝑦

240

εy = 𝐸𝑠 = 200000 = 0,0012 εs =

𝑑−𝑐 𝑐

. 𝜀𝑐 ′ =

317,4−40,073 40,073

. 0,003 = 0,020 > 𝜀𝑦

εs’ =

𝑐−𝑑′ 𝑐

. 𝜀𝑐 ′ =

40,073 − 58 40,073

. 0,003 = − 0,0013 < 𝜀𝑦

kondisi II ( 0,85 . fc’. b . 𝛽1 ) c2 + ( 600 As’ – As.fy ) c – 600 As’ d’ = 0 ( 0,85 . 25 . 200 . 0,85 ) c2 + ( 600 . 402,123–1005,30. 240) c – 600 . 402,123. 58= 0 3612,5 c2 + 0 c – 13993880 = 0 𝜌1,2 = 𝜌1 = 𝜌2 =

−𝑏 ± √𝑏2 − 4 𝑎𝑐 2𝑎

− 0 + √0 2 − 4 ( 3612,5) (−13993880) 2 ( 3612,5) − 0 − √0 2 − 4 ( 3612,5) (−13993880) 2 ( 3612,5)

= 62,239 mm = - 62,239 mm

𝑐 = 62,239 mm 𝑎 = 𝛽1 𝑥 𝑐 = 0,85 𝑥 62,239 = 52,903 mm < hf = 120 mm ( asumsi benar ) Bagian 2 As2 = =

600 𝐴𝑠′ 𝑥

𝑐−𝑑′ 𝑐

𝑓𝑦 600 . 402,123 𝑥

62,239 − 58 62,239

240

= 68,469 mm2

Mn2 = Ts2 ( d-d’ ) = As2 . fy ( d-d’ ) = 68,469. 240 ( 317,4 -58 ) = 4,262 KNm

Bagian 1 As1 = As – As2 = 1005,30mm2 – 402,123mm2 = 603,177 mm2 Mn1 = Ts1 ( d- a/2) = As1 . fy ( d- a/2)

= 603,177. 240. ( 317,4 – 52,903 /2 ) = 42,118 KNm Mnak‘= Mn = Mn1 + Mn2 = 42,118 KNm + 4,262 KNm = 46,380 KNm

Momen nominal aktual ( asumsi balok persegi ) Dimisalkan a < hf a=

(𝐴𝑠−𝐴𝑠′ ).𝑓𝑦 0,85. 𝑓𝑐 ′ 𝑏

=

(1005,30 – 402,123 ). 240 0,85.25.200

= 34,062 mm

β1 =0,85 𝑎

c = 𝛽1 =

34,062 𝑚𝑚 0,85

= 40,073 mm

cek kelelehan : 𝑓𝑦

240

εy = 𝐸𝑠 = 200000 = 0,0012 εs = εs’ =

𝑑−𝑐 𝑐

. 𝜀𝑐 ′ =

𝑐−𝑑′ 𝑐

317,4−40,073

. 𝜀𝑐 ′ =

40,073 40,073 − 58 40,073

. 0,003 = 0,020 > 𝜀𝑦 . 0,003 = − 0,0013 < 𝜀𝑦

kondisi II ( 0,85 . fc’. b . 𝛽1 ) c2 + ( 600 As’ – As.fy ) c – 600 As’ d’ = 0 ( 0,85 . 25 . 200 . 0,85 ) c2 + ( 600 . 402,123–1005,30. 240) c – 600 . 402,123. 58= 0 3612,5 c2 + 0 c – 13993880 = 0

𝜌1,2 = 𝜌1 = 𝜌2 =

−𝑏 ± √𝑏2 − 4 𝑎𝑐 2𝑎

− 0 + √0 2 − 4 ( 3612,5) (−13993880) 2 ( 3612,5) − 0 − √0 2 − 4 ( 3612,5) (−13993880) 2 ( 3612,5)

= 62,239 mm = - 62,239 mm

𝑐 = 62,239 mm Cs = 600 x As’ x

𝑐−𝑑′ 𝑐

= 600 x 402,123x

62,239 −58 62,239

= 16432,777 N

Mn2 = Cs ( d-d’ ) = 16432,777 N ( 317,4 -58 ) mm = 4,262 KNm Cc = 0,85 fc’ b . β1 . c = 0,85.25.200.0,85. 62,239 = 224838, 387 N Mn1 = Cc ( d- a/2) = 224838, 387 N ( 317,4 –56,277/2 ) = 65,037 KNm Mnak‘= Mn = Mn1 + Mn2 = 65,037 KNm + 4,262 KNm = 69,299 KNm

 perhitungan balok lantai 3 momen nominal aktual

( asumsi balok T ) Tulangan atas

= 3 D 16 = 615,6 mm2

Tulangan bawah

= 2 D 16 = 307,8 mm2 1

d2 = h-p- Ø sengkang - 2 Ø tulangan pokok 1

d2 = 400 – 40 - 10 - 2 . 16 d2 = 342 mm 1

1

d1 = d2 - 2 Ø tulangan pokok – 25 - 2 Ø tulangan pokok 1

1

= 342- 2. 16 – 25 - 2 .16 = 301 mm 1

As1 = 3 x 4 . π . 162 = 603,185 mm2 1

As2 = 2 x 4 . π . 162 = 402,123 mm2 + = 1005,308 mm2

As d=

As1 .d1 + As2 .d2 As

=

603,185.301 + 402,123.342 1005,308

= 317,4 mm

1

d’ = p + 2 Ø tulangan pokok + Ø sengkang 1

d’ = 40 + 2 . 16 + 10 d’ = 58 mm 1

As’ = 2 x 4 . π . 162 = 402,123 mm2 a=

(𝐴𝑠−𝐴𝑠′ ).𝑓𝑦 0,85. 𝑓𝑐 ′ 𝑏

=

(1005,308 – 402,123 ). 240 0,85.25.200

β1 =0,85 𝑎

c = 𝛽1 =

34,062 𝑚𝑚 0,85

cek kelelehan :

= 40,073 mm

= 34,062 mm

𝑓𝑦

240

εy = 𝐸𝑠 = 200000 = 0,0012 𝑑−𝑐

εs =

𝑐

εs’ =

. 𝜀𝑐 ′ =

𝑐−𝑑′ 𝑐

317,4−40,073 40,073

. 𝜀𝑐 ′ =

40,073 − 58 40,073

. 0,003 = 0,020 > 𝜀𝑦 . 0,003 = − 0,0013 < 𝜀𝑦

kondisi II ( 0,85 . fc’. b . 𝛽1 ) c2 + ( 600 As’ – As.fy ) c – 600 As’ d’ = 0 ( 0,85 . 25 . 200 . 0,85 ) c2 + ( 600 . 402,123–1005,30. 240) c – 600 . 402,123. 58= 0 3612,5 c2 + 0 c – 13993880 = 0 𝜌1,2 = 𝜌1 = 𝜌2 =

−𝑏 ± √𝑏2 − 4 𝑎𝑐 2𝑎

− 0 + √0 2 − 4 ( 3612,5) (−13993880) 2 ( 3612,5) − 0 − √0 2 − 4 ( 3612,5) (−13993880) 2 ( 3612,5)

= 62,239 mm = - 62,239 mm

𝑐 = 62,239 mm 𝑎 = 𝛽1 𝑥 𝑐 = 0,85 𝑥 62,239 = 52,903 mm < hf = 120 mm ( asumsi benar ) Bagian 2 As2 = =

600 𝐴𝑠′ 𝑥

𝑐−𝑑′ 𝑐

𝑓𝑦 600 . 402,123 𝑥

62,239 − 58 62,239

240

= 68,469 mm2

Mn2 = Ts2 ( d-d’ ) = As2 . fy ( d-d’ ) = 68,469. 240 ( 317,4 -58 ) = 4,262 KNm

Bagian 1 As1 = As – As2 = 1005,30mm2 – 402,123mm2 = 603,177 mm2

Mn1 = Ts1 ( d- a/2) = As1 . fy ( d- a/2) = 603,177. 240. ( 317,4 – 52,903 /2 ) = 42,118 KNm Mnak‘= Mn = Mn1 + Mn2 = 42,118 KNm + 4,262 KNm = 46,380 KNm

Momen nominal aktual ( asumsi balok persegi ) Dimisalkan a < hf a=

(𝐴𝑠−𝐴𝑠′ ).𝑓𝑦 0,85. 𝑓𝑐 ′ 𝑏

=

(1005,30 – 402,123 ). 240 0,85.25.200

= 34,062 mm

β1 =0,85 𝑎

c = 𝛽1 =

34,062 𝑚𝑚 0,85

= 40,073 mm

cek kelelehan : 𝑓𝑦

240

εy = 𝐸𝑠 = 200000 = 0,0012 εs = εs’ =

𝑑−𝑐 𝑐

. 𝜀𝑐 ′ =

𝑐−𝑑′ 𝑐

317,4−40,073

. 𝜀𝑐 ′ =

40,073 40,073 − 58 40,073

. 0,003 = 0,020 > 𝜀𝑦 . 0,003 = − 0,0013 < 𝜀𝑦

kondisi II ( 0,85 . fc’. b . 𝛽1 ) c2 + ( 600 As’ – As.fy ) c – 600 As’ d’ = 0

( 0,85 . 25 . 200 . 0,85 ) c2 + ( 600 . 402,123–1005,30. 240) c – 600 . 402,123. 58= 0 3612,5 c2 + 0 c – 13993880 = 0 𝜌1,2 = 𝜌1 = 𝜌2 =

−𝑏 ± √𝑏2 − 4 𝑎𝑐 2𝑎

− 0 + √0 2 − 4 ( 3612,5) (−13993880) 2 ( 3612,5) − 0 − √0 2 − 4 ( 3612,5) (−13993880) 2 ( 3612,5)

= 62,239 mm = - 62,239 mm

𝑐 = 62,239 mm Cs = 600 x As’ x

𝑐−𝑑′ 𝑐

= 600 x 402,123x

62,239 −58 62,239

= 16432,777 N

Mn2 = Cs ( d-d’ ) = 16432,777 N ( 317,4 -58 ) mm = 4,262 KNm Cc = 0,85 fc’ b . β1 . c = 0,85.25.200.0,85. 62,239 = 224838, 387 N Mn1 = Cc ( d- a/2) = 224838, 387 N ( 317,4 –56,277/2 ) = 65,037 KNm Mnak‘= Mn = Mn1 + Mn2 = 65,037 KNm + 4,262 KNm = 69,299 KNm

 perhitungan balok lantai 4 momen nominal aktual ( asumsi balok T ) Tulangan atas

= 3 D 16 = 684 mm2

Tulangan bawah

= 2 D 16 = 342 mm2 1

d2 = h-p- Ø sengkang - 2 Ø tulangan pokok 1

d2 = 400 – 40 - 10 - 2 . 16 d2 = 342 mm 1

1

d1 = d2 - 2 Ø tulangan pokok – 25 - 2 Ø tulangan pokok 1

1

= 342- 2. 16 – 25 - 2 .16 = 301 mm 1

As1 = 3 x 4 . π . 162 = 603,185 mm2 1

As2 = 2 x 4 . π . 162 = 402,123 mm2 + = 1005,308 mm2

As d=

As1 .d1 + As2 .d2 As

=

603,185.301 + 402,123.342 1005,308

= 317,4 mm

1

d’ = p + 2 Ø tulangan pokok + Ø sengkang 1

d’ = 40 + 2 . 16 + 10 d’ = 58 mm 1

As’ = 2 x 4 . π . 162 = 402,123 mm2 a=

(𝐴𝑠−𝐴𝑠′ ).𝑓𝑦 0,85. 𝑓𝑐 ′ 𝑏

β1 =0,85

=

(1005,308 – 402,123 ). 240 0,85.25.200

= 34,062 mm

𝑎

c = 𝛽1 =

34,062 𝑚𝑚 0,85

= 40,073 mm

cek kelelehan : 𝑓𝑦

240

εy = 𝐸𝑠 = 200000 = 0,0012 𝑑−𝑐

εs =

𝑐

εs’ =

. 𝜀𝑐 ′ =

𝑐−𝑑′ 𝑐

317,4−40,073 40,073

. 𝜀𝑐 ′ =

40,073 − 58 40,073

. 0,003 = 0,020 > 𝜀𝑦 . 0,003 = − 0,0013 < 𝜀𝑦

kondisi II ( 0,85 . fc’. b . 𝛽1 ) c2 + ( 600 As’ – As.fy ) c – 600 As’ d’ = 0 ( 0,85 . 25 . 200 . 0,85 ) c2 + ( 600 . 402,123–1005,30. 240) c – 600 . 402,123. 58= 0 3612,5 c2 + 0 c – 13993880 = 0 𝜌1,2 = 𝜌1 = 𝜌2 =

−𝑏 ± √𝑏2 − 4 𝑎𝑐 2𝑎

− 0 + √0 2 − 4 ( 3612,5) (−13993880) 2 ( 3612,5) − 0 − √0 2 − 4 ( 3612,5) (−13993880) 2 ( 3612,5)

= 62,239 mm = - 62,239 mm

𝑐 = 62,239 mm 𝑎 = 𝛽1 𝑥 𝑐 = 0,85 𝑥 62,239 = 52,903 mm < hf = 120 mm ( asumsi benar ) Bagian 2 As2 = =

600 𝐴𝑠′ 𝑥

𝑐−𝑑′ 𝑐

𝑓𝑦 600 . 402,123 𝑥

62,239 − 58 62,239

240

= 68,469 mm2

Mn2 = Ts2 ( d-d’ ) = As2 . fy ( d-d’ ) = 68,469. 240 ( 317,4 -58 ) = 4,262 KNm

Bagian 1 As1 = As – As2 = 1005,30mm2 – 402,123mm2 = 603,177 mm2 Mn1 = Ts1 ( d- a/2) = As1 . fy ( d- a/2) = 603,177. 240. ( 317,4 – 52,903 /2 ) = 42,118 KNm Mnak‘= Mn = Mn1 + Mn2 = 42,118 KNm + 4,262 KNm = 46,380 KNm

Momen nominal aktual ( asumsi balok persegi ) Dimisalkan a < hf a=

(𝐴𝑠−𝐴𝑠′ ).𝑓𝑦 0,85.

𝑓𝑐 ′

𝑏

=

(1005,30 – 402,123 ). 240 0,85.25.200

= 34,062 mm

β1 =0,85 𝑎

c = 𝛽1 =

34,062 𝑚𝑚 0,85

= 40,073 mm

cek kelelehan : 𝑓𝑦

240

εy = 𝐸𝑠 = 200000 = 0,0012 εs =

𝑑−𝑐 𝑐

. 𝜀𝑐 ′ =

317,4−40,073 40,073

. 0,003 = 0,020 > 𝜀𝑦

εs’ =

𝑐−𝑑′ 𝑐

. 𝜀𝑐 ′ =

40,073 − 58 40,073

. 0,003 = − 0,0013 < 𝜀𝑦

kondisi II ( 0,85 . fc’. b . 𝛽1 ) c2 + ( 600 As’ – As.fy ) c – 600 As’ d’ = 0 ( 0,85 . 25 . 200 . 0,85 ) c2 + ( 600 . 402,123–1005,30. 240) c – 600 . 402,123. 58= 0 3612,5 c2 + 0 c – 13993880 = 0 𝜌1,2 = 𝜌1 = 𝜌2 =

−𝑏 ± √𝑏2 − 4 𝑎𝑐 2𝑎

− 0 + √0 2 − 4 ( 3612,5) (−13993880) 2 ( 3612,5) − 0 − √0 2 − 4 ( 3612,5) (−13993880) 2 ( 3612,5)

= 62,239 mm = - 62,239 mm

𝑐 = 62,239 mm Cs = 600 x As’ x

𝑐−𝑑′ 𝑐

= 600 x 402,123x

62,239 −58 62,239

= 16432,777 N

Mn2 = Cs ( d-d’ ) = 16432,777 N ( 317,4 -58 ) mm = 4,262 KNm Cc = 0,85 fc’ b . β1 . c = 0,85.25.200.0,85. 62,239 = 224838, 387 N Mn1 = Cc ( d- a/2) = 224838, 387 N ( 317,4 –56,277/2 ) = 65,037 KNm Mnak‘= Mn = Mn1 + Mn2 = 65,037 KNm + 4,262 KNm = 69,299 KNm

𝑀tumpuan

1. Vub = 0,7 Ø(

𝑙𝑛

) + 1,05Vg

2. Vubterpakai = [1,05Vg − 0,7 (

𝑀tumpuan 𝑙𝑛

)] +

𝑙𝑛−𝑑 𝑙𝑛

𝑀tumpuan

[Vub − [1,05Vg − 0,7 Ø (

Gaya gesser Maksimum Balok (Vub) Lantai 1 Vub = 77,146 KN(lihat table gaya geser rencana) Lantai 2 Vub = 73,137 KN(lihat table gaya geser rencana) Lantai 3 Vub = 72,015 KN(lihat table gaya geser rencana) Lantai 4 Vub = 80,306 KN(lihat table gaya geser rencana)

Gaya gesser Maksimum Balok (Vub) Lantai 1 Vubterpakai = 75,947 KN(lihat table gaya geser rencana) Lantai 2 Vubterpakai = 72,031 KN(lihat table gaya geser rencana) Lantai 3 Vubterpakai = 70,936 KN(lihat table gaya geser rencana) Lantai 4 Vubterpakai = 79,085 KN(lihat table gaya geser rencana)

𝑙𝑛

)]]

Penulangan Geser a. Penulangan Geser Balok Lantai 1 Skema Gaya Geser rencana :

Vujung = Vub-(

𝑉𝑢𝑏−𝑉𝑢𝑏𝑡𝑒𝑟𝑝𝑎𝑘𝑎𝑖 𝑑 77,146−75,947

Vujung = 77,146 -(

0,342

𝑥 𝑙𝑛) 𝑥4,25)

Vujung = 62,246 KN Penulangan Geser Dalam Daerah Sendi Plastis Vs =

𝑉𝑢𝑏𝑡𝑒𝑟𝑝𝑎𝑘𝑎𝑖 Ø

=

75,947 0,75

= 101,262 𝐾𝑁 = 101262 𝑁

Dipakai Sengkang ∅ 10 𝑚𝑚 S=

𝐴𝑣.𝑓𝑦.𝑑

=

𝑉𝑠

1 4

2 𝜋102 .240.342 101262

S= 127,324 mm >

𝑑 4

= 127,324

342

=

4

= 85,5

Dipakai Jarak 85,5 mm Jadi, Dipakai Sengkang∅10 − 85,5 mm

Penulangan Geser Diluar Daerah Sendi Plastis 𝑙𝑛−2ℎ

Vub’ = Vujung + Vub’ = 62,246 +

𝑙𝑛

. (𝑉𝑢𝑏 𝑚𝑎𝑘𝑠𝑖𝑚𝑢𝑚 − 𝑉𝑢𝑗𝑢𝑛𝑔)

4,250−2.0,4 4,250

. (77,146 − 62,246 )

Vub’ = 62,760 KN √𝑓𝑐′

Vc = ( Vs =

6

𝑉𝑢𝑏′ ∅

) . 𝑏𝑤. 𝑑 = (

− 𝑉𝑐 =

√25 ) . 200.342 6

62,760 0,75

= 57000 𝑁 = 57 𝐾𝑁

− 57 = 26,68 𝐾𝑁 = 26680 𝑁

S=

𝐴𝑣.𝑓𝑦.𝑑 𝑉𝑠

=

1 4

2 𝜋102 .240.342 26680 𝑑

S= 483,249 mm >

342

=

4

= 483,249

2

= 171

Dipakai Jarak 171 mm Jadi, Dipakai Sengkang∅10 − 171 mm

b. Penulangan Geser Balok Lantai 2 Skema Gaya Geser rencana :

Vujung = Vub-(

𝑉𝑢𝑏−𝑉𝑢𝑏𝑡𝑒𝑟𝑝𝑎𝑘𝑎𝑖 𝑑 73,137−72,013

Vujung = 73,137 -(

0,342

𝑥 𝑙𝑛) 𝑥4,25)

Vujung = 59,169 KN Penulangan Geser Dalam Daerah Sendi Plastis Vs =

𝑉𝑢𝑏𝑡𝑒𝑟𝑝𝑎𝑘𝑎𝑖 Ø

72,013

=

0,75

= 96,017 𝐾𝑁 = 96017 𝑁

Dipakai Sengkang ∅ 10 𝑚𝑚 S=

𝐴𝑣.𝑓𝑦.𝑑 𝑉𝑠

=

1 4

2 𝜋102 .240.342 96017

S= 134,279mm >

𝑑 4

= 134,279

342

=

4

= 85,5

Dipakai Jarak 85,5 mm Jadi, Dipakai Sengkang∅10 − 85,5 mm

Penulangan Geser Diluar Daerah Sendi Plastis Vub’ = Vujung + Vub’ = 62,246 +

𝑙𝑛−2ℎ 𝑙𝑛

. (𝑉𝑢𝑏 𝑚𝑎𝑘𝑠𝑖𝑚𝑢𝑚 − 𝑉𝑢𝑗𝑢𝑛𝑔)

4,250−2.0,4 4,250

. (73,137 − 62,246 )

Vub’ = 71,086 KN √𝑓𝑐′

Vc = ( Vs = S=

𝑉𝑢𝑏′ ∅

) . 𝑏𝑤. 𝑑 = (

6

− 𝑉𝑐 =

𝐴𝑣.𝑓𝑦.𝑑 𝑉𝑠

=

√25 ) . 200.342 6

71,086 0,75

− 57 = 37,782 𝐾𝑁 = 37782𝑁

1 4

2 𝜋102 .240.342 37782

S= 341,249 mm >

𝑑 4

=

= 57000 𝑁 = 57 𝐾𝑁

= 341,249

342 2

= 171

Dipakai Jarak 171 mm Jadi, Dipakai Sengkang∅10 − 171 mm

c. Penulangan Geser Balok Lantai 3 Skema Gaya Geser rencana :

Vujung = Vub-(

𝑉𝑢𝑏−𝑉𝑢𝑏𝑡𝑒𝑟𝑝𝑎𝑘𝑎𝑖 𝑑 72,015−70,936

Vujung = 72,015 -(

0,342

𝑥 𝑙𝑛) 𝑥4,25)

Vujung = 58,606 KN Penulangan Geser Dalam Daerah Sendi Plastis Vs =

𝑉𝑢𝑏𝑡𝑒𝑟𝑝𝑎𝑘𝑎𝑖 Ø

=

70,936 0,75

= 94,581 𝐾𝑁 = 94581 𝑁

Dipakai Sengkang ∅ 10 𝑚𝑚 S=

𝐴𝑣.𝑓𝑦.𝑑 𝑉𝑠

=

1 4

2 𝜋102 .240.342 94581

S= 136,318 mm >

𝑑 4

=

= 136,318

342 4

= 85,5

Dipakai Jarak 85,5 mm Jadi, Dipakai Sengkang∅10 − 85,5 mm

Penulangan Geser Diluar Daerah Sendi Plastis 𝑙𝑛−2ℎ

Vub’ = Vujung + Vub’ = 58,606 +

𝑙𝑛

. (𝑉𝑢𝑏 𝑚𝑎𝑘𝑠𝑖𝑚𝑢𝑚 − 𝑉𝑢𝑗𝑢𝑛𝑔)

4,250−2.0,4 4,250

. (72,015 − 58,606 )

Vub’ = 69,490 KN √𝑓𝑐′

Vc = ( Vs = S=

𝑉𝑢𝑏′ ∅

) . 𝑏𝑤. 𝑑 = (

6

− 𝑉𝑐 =

𝐴𝑣.𝑓𝑦.𝑑 𝑉𝑠

=

√25 ) . 200.342 6

69,490 0,75

− 57 = 35,653 𝐾𝑁 = 35653 𝑁

1 4

2 𝜋102 .240.342 35653

S= 361,627 mm >

𝑑 4

=

= 57000 𝑁 = 57 𝐾𝑁

= 139,154

342 2

= 171

Dipakai Jarak 171 mm Jadi, Dipakai Sengkang∅10 − 171 mm

d. Penulangan Geser Balok Lantai 4 Skema Gaya Geser rencana :

Vujung = Vub-(

𝑉𝑢𝑏−𝑉𝑢𝑏𝑡𝑒𝑟𝑝𝑎𝑘𝑎𝑖 𝑑 80,306−79,085

Vujung = 80,306 -(

0,342

𝑥 𝑙𝑛) 𝑥4,25)

Vujung = 65,132 KN Penulangan Geser Dalam Daerah Sendi Plastis Vs =

𝑉𝑢𝑏𝑡𝑒𝑟𝑝𝑎𝑘𝑎𝑖 Ø

=

79,085 0,75

= 105,446 𝐾𝑁 = 105446 𝑁

Dipakai Sengkang ∅ 10 𝑚𝑚

S=

𝐴𝑣.𝑓𝑦.𝑑

=

𝑉𝑠

1 4

2 𝜋102 .240.342 105446 𝑑

S= 122,272 mm >

4

= 122,272

342

=

4

= 85,5

Dipakai Jarak 85,5 mm Jadi, Dipakai Sengkang∅10 − 85,5 mm

Penulangan Geser Diluar Daerah Sendi Plastis 𝑙𝑛−2ℎ

Vub’ = Vujung + Vub’ = 65,132 +

𝑙𝑛

. (𝑉𝑢𝑏 𝑚𝑎𝑘𝑠𝑖𝑚𝑢𝑚 − 𝑉𝑢𝑗𝑢𝑛𝑔)

4,250−2.0,4 4,250

. (80,306 − 65,132 )

Vub’ = 77,449 KN √𝑓𝑐′

Vc = ( Vs = S=

) . 𝑏𝑤. 𝑑 = (

6

𝑉𝑢𝑏′ ∅

− 𝑉𝑐 =

𝐴𝑣.𝑓𝑦.𝑑 𝑉𝑠

=

√25 ) . 200.342 6

77,449 0,75

− 57 = 46,265 𝐾𝑁 = 46265 𝑁

1 4

2 𝜋102 .240.342 46265

S= 278,679 mm >

𝑑 4

=

= 57000 𝑁 = 57 𝐾𝑁

= 278,679

342 2

= 171

Dipakai Jarak 171 mm Jadi, Dipakai Sengkang∅10 − 171 mm Penulangan lapangan melintang 1-1 1). Momen lapanan Rencana Nilai nilai dari momen lapangan rencana dapat dilihat dari table 2). Penulangan lapangan Data-data perhitungan : L

= 4250 mm

Ln

= 4250 mm

Lebar efektif untuk lantai 1,2,3 dan 4 1

1

Beff ≤ 4L= 4 4250 = 1062,5 mm Beff ≤ 16 hf + bw = 16. 120 + 200 = 2120 mm Beff ≤ bw + ln = 200 + 4250 = 4450 mm Diambil Beff = 1062,5

a. Perhitungan Balok Lantai 1 Mu = 25,492 KNm Asumsi letak garis netral di dalam sayap a ≤ hf dan c ≤ hf Batasan tulangan 𝜌 min = 𝜌 min =

√𝑓 ′ 𝑐 4.𝑓𝑦

√25

= 4.240 = 0,0052

1,4 1,4 = = 0,0058 𝑓𝑦 240

As min = ρmin . bw . d = 0,0052 . 200 . 342 = 355,68 mm2 𝐴𝑚𝑎𝑘𝑠 = 0,75 . ( 𝐴𝑚𝑎𝑘𝑠 = 0,75 . (

0,85. 𝛽. 𝑓𝑐 ′ 600 . ) 𝑏𝑤. 𝑑 𝑓𝑦 600 + 𝑓𝑦

0,85.0,85.25 600 . ) 200.342 240 600 + 240

𝐴𝑚𝑎𝑘𝑠 = 2757,756 𝑚𝑚2 𝑀𝑢 𝑎 = 0,85. 𝑓𝑐 ′ . 𝑎. 𝑏 (𝑑 − ) ∅ 2 25,492 . 106 𝑎 = 0,85.25. 𝑎. 1062,5 (342 − ) 0,8 2 𝑎 31,865.106 = 22567,5 𝑎 (342 − ) 2 𝑎2 − 684 𝑎 + 2823,972 = 0

𝑅𝑢𝑚𝑢𝑠 𝑎𝑏𝑐 =

𝑎12 =

−𝑏 ± √𝑏 2 − 4𝑎𝑐 2𝑎

684 ± √(−684)2 − 4. (1). (2823,972) 2. (1)

𝑎1 = 679,846 𝑚𝑚 𝑎2 = 4,153 𝑚𝑚 Ambil a yang terkecil a = 4,153 mm 𝑐=

𝑎 4,153 𝑚𝑚 = = 4,885 < ℎ𝑓 = 120 𝑚𝑚 (𝑎𝑠𝑢𝑚𝑠𝑖 𝑏𝑒𝑛𝑎𝑟) 𝛽1 0.85

Cc = Ts 𝐴𝑠 =

0,85. 𝑓𝑐 ′ . 𝑐. 𝑏𝑒𝑓𝑓 0,85.25.4,885.1062,5 = = 459,558 𝑚𝑚2 𝑓𝑦 240

Dipakai 2 D 16 (As = 402,2 mm2)

b. Perhitungan Balok Lantai 2 Mu = 23,688 KNm Asumsi letak garis netral di dalam sayap a ≤ hf dan c ≤ hf Batasan tulangan 𝜌 min = 𝜌 min =

√𝑓 ′ 𝑐 4.𝑓𝑦

√25

= 4.240 = 0,0052

1,4 1,4 = = 0,0058 𝑓𝑦 240

As min = ρmin . bw . d = 0,0052 . 200 . 342 = 355,68 mm2 0,85. 𝛽. 𝑓𝑐 ′ 600 𝐴𝑚𝑎𝑘𝑠 = 0,75 . ( . ) 𝑏𝑤. 𝑑 𝑓𝑦 600 + 𝑓𝑦 𝐴𝑚𝑎𝑘𝑠 = 0,75 . (

0,85.0,85.25 600 . ) 200.342 240 600 + 240

𝐴𝑚𝑎𝑘𝑠 = 2757,756 𝑚𝑚2 𝑀𝑢 𝑎 = 0,85. 𝑓𝑐 ′ . 𝑎. 𝑏 (𝑑 − ) ∅ 2 23,688 . 106 𝑎 = 0,85.25. 𝑎. 1062,5 (342 − ) 0,8 2 𝑎 29,61.106 = 22567,5 𝑎 (342 − ) 2 𝑎2 − 684 𝑎 + 2624,127 = 0

𝑅𝑢𝑚𝑢𝑠 𝑎𝑏𝑐 =

𝑎12 =

−𝑏 ± √𝑏 2 − 4𝑎𝑐 2𝑎

684 ± √(−684)2 − 4. (1). (2624,127) 2. (1)

𝑎1 = 680,141 𝑚𝑚 𝑎2 = 3,858 𝑚𝑚 Ambil a yang terkecil a = 3,858 mm 𝑐=

𝑎 3,858𝑚𝑚 = = 4,538 < ℎ𝑓 = 120 𝑚𝑚 (𝑎𝑠𝑢𝑚𝑠𝑖 𝑏𝑒𝑛𝑎𝑟) 𝛽1 0.85

Cc = Ts 𝐴𝑠 =

0,85. 𝑓𝑐 ′ . 𝑐. 𝑏𝑒𝑓𝑓 0,85.25.4,538.1062,5 = = 429,914 𝑚𝑚2 𝑓𝑦 240

Dipakai 2 D 16 (As = 402,2 mm2)

c. Perhitungan Balokk Lantai 3 Mu = 22,916 KNm Asumsi letak garis netral di dalam sayap a ≤ hf dan c ≤ hf Batasan tulangan 𝜌 min = 𝜌 min =

√𝑓 ′ 𝑐 4.𝑓𝑦

√25

= 4.240 = 0,0052

1,4 1,4 = = 0,0058 𝑓𝑦 240

As min = ρmin . bw . d = 0,0052 . 200 . 342 = 355,68 mm2 𝐴𝑚𝑎𝑘𝑠 = 0,75 . ( 𝐴𝑚𝑎𝑘𝑠 = 0,75 . (

0,85. 𝛽. 𝑓𝑐 ′ 600 . ) 𝑏𝑤. 𝑑 𝑓𝑦 600 + 𝑓𝑦

0,85.0,85.25 600 . ) 200.342 240 600 + 240

𝐴𝑚𝑎𝑘𝑠 = 2757,756 𝑚𝑚2 𝑀𝑢 𝑎 = 0,85. 𝑓𝑐 ′ . 𝑎. 𝑏 (𝑑 − ) ∅ 2 22,916 . 106 𝑎 = 0,85.25. 𝑎. 1062,5 (342 − ) 0,8 2 𝑎 28,64.106 = 22567,5 𝑎 (342 − ) 2 𝑎2 − 684 𝑎 + 2538,163 = 0

𝑅𝑢𝑚𝑢𝑠 𝑎𝑏𝑐 =

𝑎12 =

−𝑏 ± √𝑏 2 − 4𝑎𝑐 2𝑎

684 ± √(−684)2 − 4. (1). (2538,163) 2. (1)

𝑎1 = 680,268 𝑚𝑚 𝑎2 = 3,731 𝑚𝑚

Ambil a yang terkecil a = 3,731 mm 𝑐=

𝑎 3,731 𝑚𝑚 = = 4,389 < ℎ𝑓 = 120 𝑚𝑚 (𝑎𝑠𝑢𝑚𝑠𝑖 𝑏𝑒𝑛𝑎𝑟) 𝛽1 0.85

Cc = Ts 𝐴𝑠 =

0,85. 𝑓𝑐 ′ . 𝑐. 𝑏𝑒𝑓𝑓 0,85.25.4,389.1062,5 = = 412,897 𝑚𝑚2 𝑓𝑦 240

Dipakai 2 D 16 (As = 402,2 mm2)

d. Perhitungan Balokk Lantai 4 Mu = 29,244 KNm Asumsi letak garis netral di dalam sayap a ≤ hf dan c ≤ hf Batasan tulangan 𝜌 min = 𝜌 min =

√𝑓 ′ 𝑐 4.𝑓𝑦

√25

= 4.240 = 0,0052

1,4 1,4 = = 0,0058 𝑓𝑦 240

As min = ρmin . bw . d = 0,0052 . 200 . 342 = 355,68 mm2 𝐴𝑚𝑎𝑘𝑠 = 0,75 . ( 𝐴𝑚𝑎𝑘𝑠 = 0,75 . (

0,85. 𝛽. 𝑓𝑐 ′ 600 . ) 𝑏𝑤. 𝑑 𝑓𝑦 600 + 𝑓𝑦

0,85.0,85.25 600 . ) 200.342 240 600 + 240

𝐴𝑚𝑎𝑘𝑠 = 2757,756 𝑚𝑚2 𝑀𝑢 𝑎 = 0,85. 𝑓𝑐 ′ . 𝑎. 𝑏 (𝑑 − ) ∅ 2 29,244. 106 𝑎 = 0,85.25. 𝑎. 1062,5 (342 − ) 0,8 2 𝑎 36,55.106 = 22567,5 𝑎 (342 − ) 2 𝑎2 − 684 𝑎 + 3239,171 = 0

𝑅𝑢𝑚𝑢𝑠 𝑎𝑏𝑐 =

𝑎12 =

−𝑏 ± √𝑏 2 − 4𝑎𝑐 2𝑎

684 ± √(−684)2 − 4. (1). (3239,171) 2. (1)

𝑎1 = 679,231 𝑚𝑚 𝑎2 = 4,768 𝑚𝑚 Ambil a yang terkecil a = 4,768 mm 𝑐=

𝑎 4,768 𝑚𝑚 = = 5,609 < ℎ𝑓 = 120 𝑚𝑚 (𝑎𝑠𝑢𝑚𝑠𝑖 𝑏𝑒𝑛𝑎𝑟) 𝛽1 0.85

Cc = Ts 𝐴𝑠 =

0,85. 𝑓𝑐 ′ . 𝑐. 𝑏𝑒𝑓𝑓 0,85.25.5,609.1062,5 = = 527,669 𝑚𝑚2 𝑓𝑦 240

Dipakai 3 D 16 (As = 603,2 mm2)

Perhitungan penulangan balok Perhitungan penulangan balok portal memanjang C-C Data-data perhitungan : Dimensi balok

: 20/40 cm

Fc’

: 25 Mpa

Fy

: 240 Mpa

Dicoba : Untuk lantai 1-4 Ø sengkang

= 10 mm

Ø tulangan pokok

= 16 mm

Selimut beton

= 40 mm 1

d = h - p - Ø sengkang - 2 Ø tulangan pokok d = 400 - 40 – 10 -

1 2

. 16

d = 342 mm 1

d’ = p + 2 Ø tulangan pokok + Ø sengkang 1

d’ = 40 + 2 . 16 + 10 d’ = 58 mm  Penulangan tumpuan memanjang C-C 1. Momen tumpuan rencana Untuk momen tumpuan rencana ada dua jenis tulangan tumpuan, yaitu tulangan tumpuan positif dan tulangan tumpuan negatif. Penulangan tersebut berdasarkan nilai momen positif dan terbesar tiap lantai dalam satu portal. Adapun nilai momen maksimum positif dan negatif tersebut diambil dari kombinasi :



Momen kombinasi akibat beban mati dan beban hidup

Mk1 = 1,2 MD + 1,6 ML

2.

penulangan tumpuan 

perhitungan balok lantai 1

M tumpuan

= 62,370 KNm

Tulangan tumpuan Mu = 62,370 KNm b

= 200 mm

h

= 400 mm

d

= 342 mm

Mu

fy

= ρ.0,8.fy. (1-0,5882 ρ . fC′ )

𝑏𝑑2

62,370.106 Nmm 200 .3422

= ρ.0,8.240. (1-0,5882 ρ .

240 25

)

2,666 = 192 ρ – 892,170 ρ2 892,170 ρ2 - 192 ρ + 2,666 = 0 𝜌1,2 = 𝜌1 = 𝜌2 =

−𝑏 ± √𝑏2 − 4 𝑎𝑐 2𝑎

192 + √1922 − 4 ( 892,170) (2,666) 2 ( 892,170) 192 −√1922 − 4 ( 892,170) (2,666) 2 ( 892,170)

= 0,200 = 0,014

ρ = 0,014 As = ρ.b.d = 0,014.200.342 = 1026 mm2 As’ = 0,5.As = 0,5 . 1026 mm2 = 513 mm2 Dalam hal ini tulangan terpasang diambil berdasarkan nilai maksimum dari 2 jenis tulangan tumpuan Asatas

= 1026 mm2

Asbawah

= 513 mm2



perhitungan balok lantai 2

M tumpuan

= 57,882 KNm

Tulangan tumpuan Mu = 57,882 KNm b

= 200 mm

h

= 400 mm

d

= 342 mm

Mu

fy

= ρ.0,8.fy. (1-0,5882 ρ . fC′ )

𝑏𝑑2

57,882 .106 Nmm 200 .3422

= ρ.0,8.240. (1-0,5882 ρ .

240 25

)

2,474 = 192 ρ – 892,170 ρ2 892,170 ρ2 - 192 ρ + 2,474 = 0 𝜌1,2 = 𝜌1 = 𝜌2 =

−𝑏 ± √𝑏2 − 4 𝑎𝑐 2𝑎

192 + √1922 − 4 ( 892,170) (2,474 ) 2 ( 892,170) 192 −√1922 − 4 ( 892,170) (2,474 ) 2 ( 892,170)

= 0,201 = 0,013

ρ = 0,013 As = ρ.b.d = 0,013.200.342 = 889,2 mm2 As’ = 0,5.As = 0,5 . 889,2 mm2 = 444,6 mm2 Dalam hal ini tulangan terpasang diambil berdasarkan nilai maksimum dari 2 jenis tulangan tumpuan Asatas

= 889,2 mm2

Asbawah

= 444,6 mm2



perhitungan balok lantai 3

M tumpuan

= 57,122 KNm

Tulangan tumpuan negatif Mu = 57,122 KNm

b

= 200 mm

h

= 400 mm

d

= 342 mm

Mu

fy

= ρ.0,8.fy. (1-0,5882 ρ . fC′ )

𝑏𝑑2

57,122 .106 Nmm 200 .3422

= ρ.0,8.240. (1-0,5882 ρ .

240 25

)

2,441 = 192 ρ – 892,170 ρ2 892,170 ρ2 - 192 ρ + 2,441 = 0 𝜌1,2 = 𝜌1 = 𝜌2 =

−𝑏 ± √𝑏2 − 4 𝑎𝑐 2𝑎

192 + √1922 − 4 ( 892,170) (2,441 ) 2 ( 892,170) 192 −√1922 − 4 ( 892,170) (2,441 ) 2 ( 892,170)

= 0,201 = 0,013

ρ = 0,013 As = ρ.b.d = 0,013.200.342 = 889,2 mm2 As’ = 0,5.As = 0,5 . 889,2 mm2 = 444,6 mm2 Dalam hal ini tulangan terpasang diambil berdasarkan nilai maksimum dari 2 jenis tulangan tumpuan Asatas

= 889,2 mm2

Asbawah

= 444,6 mm2



perhitungan balok lantai 4

M tumpuan

= 63,630 KNm

Tulangan tumpuan negatif Mu = 63,630 KNm b

= 200 mm

h

= 400 mm

d

= 342 mm

Mu 𝑏𝑑2

fy

= ρ.0,8.fy. (1-0,5882 ρ . fC′ )

63,630 .106 Nmm 200 .3422

= ρ.0,8.240. (1-0,5882 ρ .

240 25

)

2,720 = 192 ρ – 892,170 ρ2 892,170 ρ2 - 192 ρ + 2,720 = 0 𝜌1,2 = 𝜌1 = 𝜌2 =

−𝑏 ± √𝑏2 − 4 𝑎𝑐 2𝑎

192 + √1922 − 4 ( 892,170) (2,720 ) 2 ( 892,170) 192 −√1922 − 4 ( 892,170) (2,720 ) 2 ( 892,170)

= 0,199 = 0,010

ρ = 0,015 As = ρ.b.d = 0,015.200.342 = 1026 mm2 As’ = 0,5.As = 0,5 . 1026 mm2 = 513 mm2 Dalam hal ini tulangan terpasang diambil berdasarkan nilai maksimum dari 2 jenis tulangan tumpuan Asatas

= 1026 mm2

Asbawah

= 513 mm2

3. penulangan geser  perhitungan balok lantai 1 momen nominal aktual ( asumsi balok T ) Tulangan atas

= 5 D 16 = 1026 mm2

Tulangan bawah

= 3 D 16 = 513 mm2 1

d2 = h-p- Ø sengkang - 2 Ø tulangan pokok 1

d2 = 400 – 40 - 10 - 2 . 16 d2 = 342 mm 1

1

d1 = d2 - 2 Ø tulangan pokok – 25 - 2 Ø tulangan pokok 1

1

= 342- 2. 16 – 25 - 2 .16 = 301 mm

1

As1 = 3 x 4 . π . 162 = 603,185 mm2 1

As2 = 3 x 4 . π . 202 = 603,185 mm2 + = 1026,37 mm2

As d=

As1 .d1 + As2 .d2 As

=

603,185 .301 + 603,185 .342 1026,37

= 321,5 mm

1

d’ = p + 2 Ø tulangan pokok + Ø sengkang 1

d’ = 40 + 2 . 16 + 10 d’ = 58 mm 1

As’ = 3 x 4 . π . 162 = 603,185 mm2 a=

(𝐴𝑠−𝐴𝑠′ ).𝑓𝑦 0,85. 𝑓𝑐 ′ 𝑏

=

(1026,37 – 603,185 ). 240 0,85.25.200

= 34,062 mm

β1 =0,85 𝑎

c = 𝛽1 =

34,062 𝑚𝑚 0,85

= 40,073 mm

cek kelelehan : 𝑓𝑦

240

εy = 𝐸𝑠 = 200000 = 0,0012 εs = εs’ =

𝑑−𝑐 𝑐

. 𝜀𝑐 ′ =

𝑐−𝑑′ 𝑐

321,5−40,073

. 𝜀𝑐 ′ =

40,073 40,073 − 58 40,073

. 0,003 = 0,021 > 𝜀𝑦 . 0,003 = − 0,0013 < 𝜀𝑦

kondisi II ( 0,85 . fc’. b . 𝛽1 ) c2 + ( 600 As’ – As.fy ) c – 600 As’ d’ = 0 ( 0,85 . 25 . 200 . 0,85 ) c2 + ( 600 . 603,185 –1026,37 . 240) c – 600 . 603,185. 58= 0 3612,5 c2 + 116302,2 c – 20990838 = 0 𝜌1,2 = 𝜌1 =

−𝑏 ± √𝑏2 − 4 𝑎𝑐 2𝑎

− 116302,2 + √116302,2 2 − 4 ( 3612,5) (−20990838) 2 ( 3612,5)

= 61,811 mm

𝜌2 =

− 116302,2 − √116302,2 2 − 4 ( 3612,5) (−20990838) 2 ( 3612,5)

= - 94,005 mm

𝑐 = 61,811 mm 𝑎 = 𝛽1 𝑥 𝑐 = 0,85 𝑥 61,811 = 52,539 mm < hf = 120 mm ( asumsi benar ) Bagian 2 As2 = =

600 𝐴𝑠′ 𝑥

𝑐−𝑑′ 𝑐

𝑓𝑦 600 . 603,185 𝑥

61,811 − 58 61,811

240

= 92,974 mm2

Mn2 = Ts2 ( d-d’ ) = As2 . fy ( d-d’ ) = 92,974. 240 ( 321,5 -58 ) = 5,879 KNm

Bagian 1 As1 = As – As2 = 1206,37 mm2 – 603,185 mm2 = 603,185 mm2 Mn1 = Ts1 ( d- a/2) = As1 . fy ( d- a/2) = 603,185. 240. ( 321,5 – 52,539 /2 ) = 42,738 KNm Mnak‘= Mn = Mn1 + Mn2 = 42,738 KNm + 5,879 KNm = 48,617 KNm

Momen nominal aktual ( asumsi balok persegi ) Dimisalkan a < hf a=

(𝐴𝑠−𝐴𝑠′ ).𝑓𝑦 0,85. 𝑓𝑐 ′ 𝑏

=

(1026,37 – 603,185 ). 240 0,85.25.200

= 34,062 mm

β1 =0,85 𝑎

c = 𝛽1 =

34,062 𝑚𝑚 0,85

= 40,073 mm

cek kelelehan : 𝑓𝑦

240

εy = 𝐸𝑠 = 200000 = 0,0012 εs = εs’ =

𝑑−𝑐 𝑐

. 𝜀𝑐 ′ =

𝑐−𝑑′ 𝑐

321,5−40,073

. 𝜀𝑐 ′ =

40,073 40,073 − 58 40,073

. 0,003 = 0,021 > 𝜀𝑦 . 0,003 = − 0,0013 < 𝜀𝑦

kondisi II ( 0,85 . fc’. b . 𝛽1 ) c2 + ( 600 As’ – As.fy ) c – 600 As’ d’ = 0 ( 0,85 . 25 . 200 . 0,85 ) c2 + ( 600 . 603,185 –1026,37 . 240) c – 600 . 603,185. 58= 0 3612,5 c2 + 116302,2 c – 20990838 = 0 𝜌1,2 = 𝜌1 =

−𝑏 ± √𝑏2 − 4 𝑎𝑐 2𝑎

− 116302,2 + √116302,2 2 − 4 ( 3612,5) (−20990838) 2 ( 3612,5)

= 61,811 mm

𝜌2 =

− 116302,2 − √116302,2 2 − 4 ( 3612,5) (−20990838) 2 ( 3612,5)

= - 94,005 mm

𝑐 = 61,811 mm Cs = 600 x As’ x

𝑐−𝑑′ 𝑐

= 600 x 603,185 x

61,811 −58 61,811

= 22313,873 N

Mn2 = Cs ( d-d’ ) = 22313,873 N ( 321,5 -58 ) mm = 5,879 KNm Cc = 0,85 fc’ b . β1 . c = 0,85.25.200.0,85. 61,811 = 267950,685 N Mn1 = Cc ( d- a/2) = 267950,685 N ( 321,5 –56,277/2 ) = 78,606 KNm Mnak‘= Mn = Mn1 + Mn2 = 78,606 KNm + 5,879 KNm = 84,485 KNm

 perhitungan balok lantai 2 momen nominal aktual ( asumsi balok T ) Tulangan atas

= 5 D 16 = 889,2 mm2

Tulangan bawah

= 2 D 16 = 444,6 mm2

1

d2 = h-p- Ø sengkang - 2 Ø tulangan pokok 1

d2 = 400 – 40 - 10 - 2 . 16 d2 = 342 mm 1

1

d1 = d2 - 2 Ø tulangan pokok – 25 - 2 Ø tulangan pokok 1

1

= 342- 2. 16 – 25 - 2 .16 = 301 mm 1

As1 = 3 x 4 . π . 162 = 603,185 mm2 1

As2 = 2 x 4 . π . 162 = 402,123 mm2 + = 1005,308 mm2

As d=

As1 .d1 + As2 .d2 As

=

603,185.301 + 402,123.342 1005,308

= 317,4 mm

1

d’ = p + Ø tulangan pokok + Ø sengkang 2

1

d’ = 40 + 2 . 16 + 10 d’ = 58 mm 1

As’ = 2 x 4 . π . 162 = 402,123 mm2 a=

(𝐴𝑠−𝐴𝑠′ ).𝑓𝑦 0,85.

𝑓𝑐 ′

𝑏

=

(1005,308 – 402,123 ). 240 0,85.25.200

= 34,062 mm

β1 =0,85 𝑎

c = 𝛽1 =

34,062 𝑚𝑚 0,85

= 40,073 mm

cek kelelehan : 𝑓𝑦

240

εy = 𝐸𝑠 = 200000 = 0,0012 εs =

𝑑−𝑐 𝑐

. 𝜀𝑐 ′ =

317,4−40,073 40,073

. 0,003 = 0,020 > 𝜀𝑦

εs’ =

𝑐−𝑑′ 𝑐

. 𝜀𝑐 ′ =

40,073 − 58 40,073

. 0,003 = − 0,0013 < 𝜀𝑦

kondisi II ( 0,85 . fc’. b . 𝛽1 ) c2 + ( 600 As’ – As.fy ) c – 600 As’ d’ = 0 ( 0,85 . 25 . 200 . 0,85 ) c2 + ( 600 . 402,123–1005,30. 240) c – 600 . 402,123. 58= 0 3612,5 c2 + 0 c – 13993880 = 0 𝜌1,2 = 𝜌1 = 𝜌2 =

−𝑏 ± √𝑏2 − 4 𝑎𝑐 2𝑎

− 0 + √0 2 − 4 ( 3612,5) (−13993880) 2 ( 3612,5) − 0 − √0 2 − 4 ( 3612,5) (−13993880) 2 ( 3612,5)

= 62,239 mm = - 62,239 mm

𝑐 = 62,239 mm 𝑎 = 𝛽1 𝑥 𝑐 = 0,85 𝑥 62,239 = 52,903 mm < hf = 120 mm ( asumsi benar ) Bagian 2 As2 = =

600 𝐴𝑠′ 𝑥

𝑐−𝑑′ 𝑐

𝑓𝑦 600 . 402,123 𝑥

62,239 − 58 62,239

240

= 68,469 mm2

Mn2 = Ts2 ( d-d’ ) = As2 . fy ( d-d’ ) = 68,469. 240 ( 317,4 -58 ) = 4,262 KNm

Bagian 1 As1 = As – As2 = 1005,30mm2 – 402,123mm2 = 603,177 mm2 Mn1 = Ts1 ( d- a/2) = As1 . fy ( d- a/2)

= 603,177. 240. ( 317,4 – 52,903 /2 ) = 42,118 KNm Mnak‘= Mn = Mn1 + Mn2 = 42,118 KNm + 4,262 KNm = 46,380 KNm

Momen nominal aktual ( asumsi balok persegi ) Dimisalkan a < hf a=

(𝐴𝑠−𝐴𝑠′ ).𝑓𝑦 0,85. 𝑓𝑐 ′ 𝑏

=

(1005,30 – 402,123 ). 240 0,85.25.200

= 34,062 mm

β1 =0,85 𝑎

c = 𝛽1 =

34,062 𝑚𝑚 0,85

= 40,073 mm

cek kelelehan : εy =

𝑓𝑦

εs =

𝑑−𝑐

εs’ =

𝐸𝑠

𝑐

=

240 200000

. 𝜀𝑐 ′ =

𝑐−𝑑′ 𝑐

= 0,0012

317,4−40,073

. 𝜀𝑐 ′ =

40,073 40,073 − 58 40,073

. 0,003 = 0,020 > 𝜀𝑦 . 0,003 = − 0,0013 < 𝜀𝑦

kondisi II ( 0,85 . fc’. b . 𝛽1 ) c2 + ( 600 As’ – As.fy ) c – 600 As’ d’ = 0

( 0,85 . 25 . 200 . 0,85 ) c2 + ( 600 . 402,123–1005,30. 240) c – 600 . 402,123. 58= 0 3612,5 c2 + 0 c – 13993880 = 0 𝜌1,2 = 𝜌1 = 𝜌2 =

−𝑏 ± √𝑏2 − 4 𝑎𝑐 2𝑎

− 0 + √0 2 − 4 ( 3612,5) (−13993880) 2 ( 3612,5) − 0 − √0 2 − 4 ( 3612,5) (−13993880) 2 ( 3612,5)

= 62,239 mm = - 62,239 mm

𝑐 = 62,239 mm Cs = 600 x As’ x

𝑐−𝑑′ 𝑐

= 600 x 402,123x

62,239 −58 62,239

= 16432,777 N

Mn2 = Cs ( d-d’ ) = 16432,777 N ( 317,4 -58 ) mm = 4,262 KNm Cc = 0,85 fc’ b . β1 . c = 0,85.25.200.0,85. 62,239 = 224838, 387 N Mn1 = Cc ( d- a/2) = 224838, 387 N ( 317,4 –56,277/2 ) = 65,037 KNm Mnak‘= Mn = Mn1 + Mn2 = 65,037 KNm + 4,262 KNm = 69,299 KNm

 perhitungan balok lantai 3 momen nominal aktual ( asumsi balok T ) Tulangan atas

= 5 D 16 = 889,2 mm2

Tulangan bawah

= 2 D 16 = 444,6 mm2 1

d2 = h-p- Ø sengkang - 2 Ø tulangan pokok 1

d2 = 400 – 40 - 10 - 2 . 16 d2 = 342 mm 1

1

d1 = d2 - 2 Ø tulangan pokok – 25 - 2 Ø tulangan pokok 1

1

= 342- 2. 16 – 25 - 2 .16 = 301 mm 1

As1 = 3 x 4 . π . 162 = 603,185 mm2 1

As2 = 2 x 4 . π . 162 = 402,123 mm2 + = 1005,308 mm2

As d=

As1 .d1 + As2 .d2 As

=

603,185.301 + 402,123.342 1005,308

= 317,4 mm

1

d’ = p + 2 Ø tulangan pokok + Ø sengkang 1

d’ = 40 + 2 . 16 + 10 d’ = 58 mm 1

As’ = 2 x 4 . π . 162 = 402,123 mm2 a=

(𝐴𝑠−𝐴𝑠′ ).𝑓𝑦 0,85. 𝑓𝑐 ′ 𝑏

β1 =0,85

=

(1005,308 – 402,123 ). 240 0,85.25.200

= 34,062 mm

𝑎

c = 𝛽1 =

34,062 𝑚𝑚 0,85

= 40,073 mm

cek kelelehan : 𝑓𝑦

240

εy = 𝐸𝑠 = 200000 = 0,0012 𝑑−𝑐

εs =

𝑐

εs’ =

. 𝜀𝑐 ′ =

𝑐−𝑑′ 𝑐

317,4−40,073 40,073

. 𝜀𝑐 ′ =

40,073 − 58 40,073

. 0,003 = 0,020 > 𝜀𝑦 . 0,003 = − 0,0013 < 𝜀𝑦

kondisi II ( 0,85 . fc’. b . 𝛽1 ) c2 + ( 600 As’ – As.fy ) c – 600 As’ d’ = 0 ( 0,85 . 25 . 200 . 0,85 ) c2 + ( 600 . 402,123–1005,30. 240) c – 600 . 402,123. 58= 0 3612,5 c2 + 0 c – 13993880 = 0 𝜌1,2 = 𝜌1 = 𝜌2 =

−𝑏 ± √𝑏2 − 4 𝑎𝑐 2𝑎

− 0 + √0 2 − 4 ( 3612,5) (−13993880) 2 ( 3612,5) − 0 − √0 2 − 4 ( 3612,5) (−13993880) 2 ( 3612,5)

= 62,239 mm = - 62,239 mm

𝑐 = 62,239 mm 𝑎 = 𝛽1 𝑥 𝑐 = 0,85 𝑥 62,239 = 52,903 mm < hf = 120 mm ( asumsi benar ) Bagian 2 As2 = =

600 𝐴𝑠′ 𝑥

𝑐−𝑑′ 𝑐

𝑓𝑦 600 . 402,123 𝑥

62,239 − 58 62,239

240

= 68,469 mm2

Mn2 = Ts2 ( d-d’ ) = As2 . fy ( d-d’ ) = 68,469. 240 ( 317,4 -58 ) = 4,262 KNm

Bagian 1 As1 = As – As2 = 1005,30mm2 – 402,123mm2 = 603,177 mm2 Mn1 = Ts1 ( d- a/2) = As1 . fy ( d- a/2) = 603,177. 240. ( 317,4 – 52,903 /2 ) = 42,118 KNm Mnak‘= Mn = Mn1 + Mn2 = 42,118 KNm + 4,262 KNm = 46,380 KNm

Momen nominal aktual ( asumsi balok persegi ) Dimisalkan a < hf a=

(𝐴𝑠−𝐴𝑠′ ).𝑓𝑦 0,85. 𝑓𝑐 ′ 𝑏

=

(1005,30 – 402,123 ). 240 0,85.25.200

β1 =0,85 𝑎

c = 𝛽1 =

34,062 𝑚𝑚 0,85

= 40,073 mm

cek kelelehan : 𝑓𝑦

240

εy = 𝐸𝑠 = 200000 = 0,0012

= 34,062 mm

𝑑−𝑐

εs =

𝑐

εs’ =

. 𝜀𝑐 ′ =

𝑐−𝑑′ 𝑐

317,4−40,073

. 𝜀𝑐 ′ =

40,073 40,073 − 58 40,073

. 0,003 = 0,020 > 𝜀𝑦 . 0,003 = − 0,0013 < 𝜀𝑦

kondisi II ( 0,85 . fc’. b . 𝛽1 ) c2 + ( 600 As’ – As.fy ) c – 600 As’ d’ = 0 ( 0,85 . 25 . 200 . 0,85 ) c2 + ( 600 . 402,123–1005,30. 240) c – 600 . 402,123. 58= 0 3612,5 c2 + 0 c – 13993880 = 0 𝜌1,2 = 𝜌1 = 𝜌2 =

−𝑏 ± √𝑏2 − 4 𝑎𝑐 2𝑎

− 0 + √0 2 − 4 ( 3612,5) (−13993880) 2 ( 3612,5) − 0 − √0 2 − 4 ( 3612,5) (−13993880) 2 ( 3612,5)

= 62,239 mm = - 62,239 mm

𝑐 = 62,239 mm Cs = 600 x As’ x

𝑐−𝑑′ 𝑐

= 600 x 402,123x

62,239 −58 62,239

= 16432,777 N

Mn2 = Cs ( d-d’ ) = 16432,777 N ( 317,4 -58 ) mm = 4,262 KNm Cc = 0,85 fc’ b . β1 . c = 0,85.25.200.0,85. 62,239 = 224838, 387 N Mn1 = Cc ( d- a/2) = 224838, 387 N ( 317,4 –56,277/2 ) = 65,037 KNm Mnak‘= Mn = Mn1 + Mn2 = 65,037 KNm + 4,262 KNm = 69,299 KNm

 perhitungan balok lantai 4 momen nominal aktual ( asumsi balok T ) Tulangan atas

= 5 D 16 = 1026 mm2

Tulangan bawah

= 3 D 16 = 513 mm2 1

d2 = h-p- Ø sengkang - 2 Ø tulangan pokok 1

d2 = 400 – 40 - 10 - 2 . 16 d2 = 342 mm 1

1

d1 = d2 - 2 Ø tulangan pokok – 25 - 2 Ø tulangan pokok 1

1

= 342- 2. 16 – 25 - 2 .16 = 301 mm 1

As1 = 3 x 4 . π . 162 = 603,185 mm2 1

As2 = 3 x 4 . π . 202 = 603,185 mm2 + = 1026,37 mm2

As d=

As1 .d1 + As2 .d2 As

=

603,185 .301 + 603,185 .342 1026,37

= 321,5 mm

1

d’ = p + 2 Ø tulangan pokok + Ø sengkang 1

d’ = 40 + 2 . 16 + 10 d’ = 58 mm 1

As’ = 3 x 4 . π . 162 = 603,185 mm2 a=

(𝐴𝑠−𝐴𝑠′ ).𝑓𝑦 0,85. 𝑓𝑐 ′ 𝑏

=

(1026,37 – 603,185 ). 240 0,85.25.200

= 34,062 mm

β1 =0,85 𝑎

c = 𝛽1 =

34,062 𝑚𝑚 0,85

= 40,073 mm

cek kelelehan : 𝑓𝑦

240

εy = 𝐸𝑠 = 200000 = 0,0012 𝑑−𝑐

εs =

𝑐

εs’ =

. 𝜀𝑐 ′ =

𝑐−𝑑′ 𝑐

321,5−40,073

. 𝜀𝑐 ′ =

40,073 40,073 − 58 40,073

. 0,003 = 0,021 > 𝜀𝑦 . 0,003 = − 0,0013 < 𝜀𝑦

kondisi II ( 0,85 . fc’. b . 𝛽1 ) c2 + ( 600 As’ – As.fy ) c – 600 As’ d’ = 0 ( 0,85 . 25 . 200 . 0,85 ) c2 + ( 600 . 603,185 –1026,37 . 240) c – 600 . 603,185. 58= 0 3612,5 c2 + 116302,2 c – 20990838 = 0 𝜌1,2 = 𝜌1 = 𝜌2 =

−𝑏 ± √𝑏2 − 4 𝑎𝑐 2𝑎

− 116302,2 + √116302,2 2 − 4 ( 3612,5) (−20990838) 2 ( 3612,5) − 116302,2 − √116302,2 2 − 4 ( 3612,5) (−20990838) 2 ( 3612,5)

𝑐 = 61,811 mm 𝑎 = 𝛽1 𝑥 𝑐 = 0,85 𝑥 61,811 = 52,539 mm < hf = 120 mm ( asumsi benar ) Bagian 2

= 61,811 mm = - 94,005 mm

As2 = =

600 𝐴𝑠′ 𝑥

𝑐−𝑑′ 𝑐

𝑓𝑦 600 . 603,185 𝑥

61,811 − 58 61,811

240

= 92,974 mm2

Mn2 = Ts2 ( d-d’ ) = As2 . fy ( d-d’ ) = 92,974. 240 ( 321,5 -58 ) = 5,879 KNm

Bagian 1 As1 = As – As2 = 1206,37 mm2 – 603,185 mm2 = 603,185 mm2 Mn1 = Ts1 ( d- a/2) = As1 . fy ( d- a/2) = 603,185. 240. ( 321,5 – 52,539 /2 ) = 42,738 KNm Mnak‘= Mn = Mn1 + Mn2 = 42,738 KNm + 5,879 KNm = 48,617 KNm

Momen nominal aktual ( asumsi balok persegi ) Dimisalkan a < hf a=

(𝐴𝑠−𝐴𝑠′ ).𝑓𝑦 0,85. 𝑓𝑐 ′ 𝑏

=

(1026,37 – 603,185 ). 240 0,85.25.200

= 34,062 mm

β1 =0,85 𝑎

c = 𝛽1 =

34,062 𝑚𝑚 0,85

= 40,073 mm

cek kelelehan : 𝑓𝑦

240

εy = 𝐸𝑠 = 200000 = 0,0012 𝑑−𝑐

εs =

𝑐

εs’ =

. 𝜀𝑐 ′ =

𝑐−𝑑′ 𝑐

321,5−40,073

. 𝜀𝑐 ′ =

40,073 40,073 − 58 40,073

. 0,003 = 0,021 > 𝜀𝑦 . 0,003 = − 0,0013 < 𝜀𝑦

kondisi II ( 0,85 . fc’. b . 𝛽1 ) c2 + ( 600 As’ – As.fy ) c – 600 As’ d’ = 0 ( 0,85 . 25 . 200 . 0,85 ) c2 + ( 600 . 603,185 –1026,37 . 240) c – 600 . 603,185. 58= 0 3612,5 c2 + 116302,2 c – 20990838 = 0 𝜌1,2 = 𝜌1 = 𝜌2 =

−𝑏 ± √𝑏2 − 4 𝑎𝑐 2𝑎

− 116302,2 + √116302,2 2 − 4 ( 3612,5) (−20990838) 2 ( 3612,5) − 116302,2 − √116302,2 2 − 4 ( 3612,5) (−20990838) 2 ( 3612,5)

= 61,811 mm = - 94,005 mm

𝑐 = 61,811 mm Cs = 600 x As’ x

𝑐−𝑑′ 𝑐

= 600 x 603,185 x

61,811 −58 61,811

Mn2 = Cs ( d-d’ ) = 22313,873 N ( 321,5 -58 ) mm = 5,879 KNm Cc = 0,85 fc’ b . β1 . c = 0,85.25.200.0,85. 61,811 = 267950,685 N

= 22313,873 N

Mn1 = Cc ( d- a/2) = 267950,685 N ( 321,5 –56,277/2 ) = 78,606 KNm Mnak‘= Mn = Mn1 + Mn2 = 78,606 KNm + 5,879 KNm = 84,485 KNm

𝑀tumpuan

1. Vub = 0,7 Ø(

𝑙𝑛

) + 1,05Vg

2. Vubterpakai = [1,05Vg − 0,7 (

𝑀tumpuan 𝑙𝑛

)] +

𝑙𝑛−𝑑 𝑙𝑛

𝑀tumpuan

[Vub − [1,05Vg − 0,7 Ø (

Gaya gesser Maksimum Balok (Vub) Lantai 1 Vub = 94,02 KN(lihat table gaya geser rencana) Lantai 2 Vub = 89,502 KN(lihat table gaya geser rencana) Lantai 3 Vub = 88,327 KN(lihat table gaya geser rencana) Lantai 4 Vub = 97,777 KN(lihat table gaya geser rencana)

Gaya gesser Maksimum Balok (Vub) Lantai 1 Vubterpakai = 92,367 KN(lihat table gaya geser rencana) Lantai 2 Vubterpakai = 87,968 KN(lihat table gaya geser rencana)

𝑙𝑛

)]]

Lantai 3 Vubterpakai = 86,813 KN(lihat table gaya geser rencana) Lantai 4 Vubterpakai = 96,091 KN(lihat table gaya geser rencana)

Penulangan Geser a. Penulangan Geser Balok Lantai 1 Skema Gaya Geser rencana :

Vujung = Vub-(

𝑉𝑢𝑏−𝑉𝑢𝑏𝑡𝑒𝑟𝑝𝑎𝑘𝑎𝑖 𝑑 94,02−92,367

Vujung = 94,02 -(

0,342

𝑥 𝑙𝑛)

𝑥4,25)

Vujung = 73,478 KN Penulangan Geser Dalam Daerah Sendi Plastis Vs =

𝑉𝑢𝑏𝑡𝑒𝑟𝑝𝑎𝑘𝑎𝑖 Ø

=

92,367

= 123,156𝐾𝑁 = 123156 𝑁

0,75

Dipakai Sengkang ∅ 10 𝑚𝑚 S=

𝐴𝑣.𝑓𝑦.𝑑 𝑉𝑠

=

1 4

2 𝜋102 .240.342 123156

S= 104,689 mm >

𝑑 4

=

= 104,689

342 4

= 85,5

Dipakai Jarak 85,5 mm Jadi, Dipakai Sengkang∅10 − 85,5 mm

Penulangan Geser Diluar Daerah Sendi Plastis Vub’ = Vujung + Vub’ = 73,478 +

𝑙𝑛−2ℎ 𝑙𝑛

. (𝑉𝑢𝑏 𝑚𝑎𝑘𝑠𝑖𝑚𝑢𝑚 − 𝑉𝑢𝑗𝑢𝑛𝑔)

4,250−2.0,4

Vub’ = 90,153 KN

4,250

. (94,02 − 73,478 )

√𝑓𝑐′

Vc = ( Vs = S=

𝑉𝑢𝑏′ ∅

) . 𝑏𝑤. 𝑑 = (

6

− 𝑉𝑐 =

𝐴𝑣.𝑓𝑦.𝑑 𝑉𝑠

=

√25 ) . 200.342 6

90,153 0,75

− 57 = 63,204𝐾𝑁 = 63204 𝑁

1 4

2 𝜋102 .240.342 63204

S= 203,991 mm >

𝑑 4

=

= 57000 𝑁 = 57 𝐾𝑁

= 203,991

342 2

= 171

Dipakai Jarak 171 mm Jadi, Dipakai Sengkang∅10 − 171 mm

b. Penulangan Geser Balok Lantai 2 Skema Gaya Geser rencana :

Vujung = Vub-(

𝑉𝑢𝑏−𝑉𝑢𝑏𝑡𝑒𝑟𝑝𝑎𝑘𝑎𝑖 𝑑 89,502−87,968

Vujung = 89,502 -(

0,342

𝑥 𝑙𝑛) 𝑥4,25)

Vujung = 70,439 KN Penulangan Geser Dalam Daerah Sendi Plastis Vs =

𝑉𝑢𝑏𝑡𝑒𝑟𝑝𝑎𝑘𝑎𝑖 Ø

=

87,968

= 117,290 𝐾𝑁 = 117290 𝑁

0,75

Dipakai Sengkang ∅ 10 𝑚𝑚 S=

𝐴𝑣.𝑓𝑦.𝑑 𝑉𝑠

=

1 4

2 𝜋102 .240.342 117290

S= 109,924 mm >

𝑑 4

=

= 109,924

342 4

= 85,5

Dipakai Jarak 85,5 mm Jadi, Dipakai Sengkang∅10 − 85,5 mm

Penulangan Geser Diluar Daerah Sendi Plastis 𝑙𝑛−2ℎ

Vub’ = Vujung + Vub’ = 70,439+

𝑙𝑛

. (𝑉𝑢𝑏 𝑚𝑎𝑘𝑠𝑖𝑚𝑢𝑚 − 𝑉𝑢𝑗𝑢𝑛𝑔)

4,250−2.0,4

. (89,502 − 70,439)

4,250

Vub’ = 85,913 KN √𝑓𝑐′

Vc = ( Vs = S=

𝑉𝑢𝑏′ ∅

) . 𝑏𝑤. 𝑑 = (

6

− 𝑉𝑐 =

𝐴𝑣.𝑓𝑦.𝑑 𝑉𝑠

=

√25 ) . 200.342 6

85,913

− 57 = 57,551𝐾𝑁 = 57551 𝑁

0,75

1 4

2 𝜋102 .240.342 57551

S= 224,029 mm >

𝑑 4

=

= 57000 𝑁 = 57 𝐾𝑁

= 224,029

342 2

= 171

Dipakai Jarak 171 mm Jadi, Dipakai Sengkang∅10 − 171 mm

c. Penulangan Geser Balok Lantai 3 Skema Gaya Geser rencana :

Vujung = Vub-(

𝑉𝑢𝑏−𝑉𝑢𝑏𝑡𝑒𝑟𝑝𝑎𝑘𝑎𝑖 𝑑 88,327−86,813

Vujung = 88,327 -(

0,342

𝑥 𝑙𝑛) 𝑥4,25)

Vujung = 69,512 KN Penulangan Geser Dalam Daerah Sendi Plastis Vs =

𝑉𝑢𝑏𝑡𝑒𝑟𝑝𝑎𝑘𝑎𝑖 Ø

=

86,813 0,75

= 115,750 𝐾𝑁 = 115750 𝑁

Dipakai Sengkang ∅ 10 𝑚𝑚 S=

𝐴𝑣.𝑓𝑦.𝑑 𝑉𝑠

=

1 4

2 𝜋102 .240.342 115750

= 111,387

S= 111,387 mm >

𝑑 4

342

=

4

= 85,5

Dipakai Jarak 85,5 mm Jadi, Dipakai Sengkang∅10 − 85,5 mm

Penulangan Geser Diluar Daerah Sendi Plastis 𝑙𝑛−2ℎ

Vub’ = Vujung + Vub’ = 69,512 +

𝑙𝑛

. (𝑉𝑢𝑏 𝑚𝑎𝑘𝑠𝑖𝑚𝑢𝑚 − 𝑉𝑢𝑗𝑢𝑛𝑔)

4,250−2.0,4 4,250

. (88,327 − 69,512 )

Vub’ = 84,785 KN √𝑓𝑐′

Vc = ( Vs = S=

𝑉𝑢𝑏′ ∅

) . 𝑏𝑤. 𝑑 = (

6

− 𝑉𝑐 =

𝐴𝑣.𝑓𝑦.𝑑 𝑉𝑠

=

√25 ) . 200.342 6

84,785 0,75

1 4

− 57 = 56,046𝐾𝑁 = 56046 𝑁

2 𝜋102 .240.342 56046

S= 230,044 mm >

𝑑 4

=

= 57000 𝑁 = 57 𝐾𝑁

= 230,044

342 2

= 171

Dipakai Jarak 171 mm Jadi, Dipakai Sengkang∅10 − 171 mm

d. Penulangan Geser Balok Lantai 4 Skema Gaya Geser rencana :

Vujung = Vub-(

𝑉𝑢𝑏−𝑉𝑢𝑏𝑡𝑒𝑟𝑝𝑎𝑘𝑎𝑖 𝑑 97,777−96,091

Vujung = 97,777 -(

Vujung = 96,825 KN

0,342

𝑥 𝑙𝑛) 𝑥4,25)

Penulangan Geser Dalam Daerah Sendi Plastis Vs =

𝑉𝑢𝑏𝑡𝑒𝑟𝑝𝑎𝑘𝑎𝑖 Ø

=

96,091

= 128,121 𝐾𝑁 = 128121 𝑁

0,75

Dipakai Sengkang ∅ 10 𝑚𝑚 S=

𝐴𝑣.𝑓𝑦.𝑑

=

𝑉𝑠

1 4

2 𝜋102 .240.342 128121 𝑑

S= 100,632 mm >

4

= 100,632

342

=

4

= 85,5

Dipakai Jarak 85,5 mm Jadi, Dipakai Sengkang∅10 − 85,5 mm

Penulangan Geser Diluar Daerah Sendi Plastis 𝑙𝑛−2ℎ

Vub’ = Vujung + Vub’ = 96,825 +

𝑙𝑛

. (𝑉𝑢𝑏 𝑚𝑎𝑘𝑠𝑖𝑚𝑢𝑚 − 𝑉𝑢𝑗𝑢𝑛𝑔)

4,250−2.0,4 4,250

. (97,777 − 96,825 )

Vub’ = 97,597 KN √𝑓𝑐′

Vc = ( Vs = S=

) . 𝑏𝑤. 𝑑 = (

6

𝑉𝑢𝑏′ ∅

− 𝑉𝑐 =

𝐴𝑣.𝑓𝑦.𝑑 𝑉𝑠

=

√25 ) . 200.342 6

97,597

− 57 = 73,130𝐾𝑁 = 73130 𝑁

0,75

1 4

2 𝜋102 .240.342 73130

S= 176,303 mm >

𝑑 4

=

= 57000 𝑁 = 57 𝐾𝑁

= 176,303

342 2

= 171

Dipakai Jarak 171 mm Jadi, Dipakai Sengkang∅10 − 171 mm

Penulangan lapangan melintang C-C 1). Momen lapanan Rencana Nilai nilai dari momen lapangan rencana dapat dilihat dari table 2). Penulangan lapangan Data-data perhitungan : L

= 4250 mm

Ln

= 4250 mm

Lebar efektif untuk lantai 1,2,3 dan 4

1

1

Beff ≤ 4L= 4 4250 = 1062,5 mm Beff ≤ 16 hf + bw = 16. 120 + 200 = 2120 mm Beff ≤ bw + ln = 200 + 4250 = 4450 mm Diambil Beff = 1062,5

a. Perhitungan Balok Lantai 1 Mu = 34,920 KNm Asumsi letak garis netral di dalam sayap a ≤ hf dan c ≤ hf Batasan tulangan 𝜌 min = 𝜌 min =

√𝑓 ′ 𝑐 4.𝑓𝑦

√25

= 4.240 = 0,0052

1,4 1,4 = = 0,0058 𝑓𝑦 240

As min = ρmin . bw . d = 0,0052 . 200 . 342 = 355,68 mm2 𝐴𝑚𝑎𝑘𝑠 = 0,75 . ( 𝐴𝑚𝑎𝑘𝑠 = 0,75 . (

0,85. 𝛽. 𝑓𝑐 ′ 600 . ) 𝑏𝑤. 𝑑 𝑓𝑦 600 + 𝑓𝑦

0,85.0,85.25 600 . ) 200.342 240 600 + 240

𝐴𝑚𝑎𝑘𝑠 = 2757,756 𝑚𝑚2 𝑀𝑢 𝑎 = 0,85. 𝑓𝑐 ′ . 𝑎. 𝑏 (𝑑 − ) ∅ 2 34,920 . 106 𝑎 = 0,85.25. 𝑎. 1062,5 (342 − ) 0,8 2 𝑎 43,65.106 = 22567,5 𝑎 (342 − ) 2 𝑎2 − 684 𝑎 + 3868,394 = 0 −𝑏 ± √𝑏 2 − 4𝑎𝑐 𝑅𝑢𝑚𝑢𝑠 𝑎𝑏𝑐 = 2𝑎 𝑎12 =

684 ± √(−684)2 − 4. (1). (3868,394) 2. (1)

𝑎1 = 678,296𝑚𝑚 𝑎2 = 5,703 𝑚𝑚

Ambil a yang terkecil a = 5,703 mm 𝑐=

𝑎 5,703𝑚𝑚 = = 6,709 < ℎ𝑓 = 120 𝑚𝑚 (𝑎𝑠𝑢𝑚𝑠𝑖 𝑏𝑒𝑛𝑎𝑟) 𝛽1 0.85

Cc = Ts 𝐴𝑠 =

0,85. 𝑓𝑐 ′ . 𝑐. 𝑏𝑒𝑓𝑓 0,85.25.6,709.1062,5 = = 631,152 𝑚𝑚2 𝑓𝑦 240

Dipakai 3 D 16 (As = 603,2 mm2)

b. Perhitungan Balok Lantai 2 Mu = 32,864 KNm Asumsi letak garis netral di dalam sayap a ≤ hf dan c ≤ hf Batasan tulangan 𝜌 min = 𝜌 min =

√𝑓 ′ 𝑐 4.𝑓𝑦

√25

= 4.240 = 0,0052

1,4 1,4 = = 0,0058 𝑓𝑦 240

As min = ρmin . bw . d = 0,0052 . 200 . 342 = 355,68 mm2 𝐴𝑚𝑎𝑘𝑠 = 0,75 . ( 𝐴𝑚𝑎𝑘𝑠 = 0,75 . (

0,85. 𝛽. 𝑓𝑐 ′ 600 . ) 𝑏𝑤. 𝑑 𝑓𝑦 600 + 𝑓𝑦

0,85.0,85.25 600 . ) 200.342 240 600 + 240

𝐴𝑚𝑎𝑘𝑠 = 2757,756 𝑚𝑚2 𝑀𝑢 𝑎 = 0,85. 𝑓𝑐 ′ . 𝑎. 𝑏 (𝑑 − ) ∅ 2 32,864 . 106 𝑎 = 0,85.25. 𝑎. 1062,5 (342 − ) 0,8 2 𝑎 41,08.106 = 22567,5 𝑎 (342 − ) 2 𝑎2 − 684 𝑎 + 3640,633 = 0

𝑅𝑢𝑚𝑢𝑠 𝑎𝑏𝑐 =

𝑎12 =

−𝑏 ± √𝑏 2 − 4𝑎𝑐 2𝑎

684 ± √(−684)2 − 4. (1). (3640,633) 2. (1)

𝑎1 = 678,635 𝑚𝑚 𝑎2 = 5,364 𝑚𝑚 Ambil a yang terkecil a = 5,364 mm 𝑐=

𝑎 5,364 𝑚𝑚 = = 6,311 < ℎ𝑓 = 120 𝑚𝑚 (𝑎𝑠𝑢𝑚𝑠𝑖 𝑏𝑒𝑛𝑎𝑟) 𝛽1 0.85

Cc = Ts 𝐴𝑠 =

0,85. 𝑓𝑐 ′ . 𝑐. 𝑏𝑒𝑓𝑓 0,85.25.6,311.1062,5 = = 593,710 𝑚𝑚2 𝑓𝑦 240

Dipakai 3 D 16 (As = 603,2 mm2)

c. Perhitungan Balokk Lantai 3 Mu = 31,940 KNm Asumsi letak garis netral di dalam sayap a ≤ hf dan c ≤ hf Batasan tulangan 𝜌 min = 𝜌 min =

√𝑓 ′ 𝑐 4.𝑓𝑦

=

√25 4.240

= 0,0052

1,4 1,4 = = 0,0058 𝑓𝑦 240

As min = ρmin . bw . d = 0,0052 . 200 . 342 = 355,68 mm2 𝐴𝑚𝑎𝑘𝑠 = 0,75 . ( 𝐴𝑚𝑎𝑘𝑠 = 0,75 . (

0,85. 𝛽. 𝑓𝑐 ′ 600 . ) 𝑏𝑤. 𝑑 𝑓𝑦 600 + 𝑓𝑦

0,85.0,85.25 600 . ) 200.342 240 600 + 240

𝐴𝑚𝑎𝑘𝑠 = 2757,756 𝑚𝑚2 𝑀𝑢 𝑎 = 0,85. 𝑓𝑐 ′ . 𝑎. 𝑏 (𝑑 − ) ∅ 2 31,940 . 106 𝑎 = 0,85.25. 𝑎. 1062,5 (342 − ) 0,8 2 𝑎 39,92.106 = 22567,5 𝑎 (342 − ) 2 𝑎2 − 684 𝑎 + 3537,830 = 0

𝑅𝑢𝑚𝑢𝑠 𝑎𝑏𝑐 =

−𝑏 ± √𝑏 2 − 4𝑎𝑐 2𝑎

𝑎12 =

684 ± √(−684)2 − 4. (1). (3537,830) 2. (1)

𝑎1 = 678,788 𝑚𝑚 𝑎2 = 5,211 𝑚𝑚 Ambil a yang terkecil a = 5,211 mm 𝑐=

𝑎 5,211 𝑚𝑚 = = 6,130 < ℎ𝑓 = 120 𝑚𝑚 (𝑎𝑠𝑢𝑚𝑠𝑖 𝑏𝑒𝑛𝑎𝑟) 𝛽1 0.85

Cc = Ts 0,85. 𝑓𝑐 ′ . 𝑐. 𝑏𝑒𝑓𝑓 0,85.25.6,130.1062,5 𝐴𝑠 = = = 576,682 𝑚𝑚2 𝑓𝑦 240 Dipakai 3 D 16 (As = 603,2 mm2)

d. Perhitungan Balokk Lantai 4 Mu = 39,960 KNm Asumsi letak garis netral di dalam sayap a ≤ hf dan c ≤ hf Batasan tulangan 𝜌 min = 𝜌 min =

√𝑓 ′ 𝑐 4.𝑓𝑦

√25

= 4.240 = 0,0052

1,4 1,4 = = 0,0058 𝑓𝑦 240

As min = ρmin . bw . d = 0,0052 . 200 . 342 = 355,68 mm2 𝐴𝑚𝑎𝑘𝑠 = 0,75 . ( 𝐴𝑚𝑎𝑘𝑠 = 0,75 . (

0,85. 𝛽. 𝑓𝑐 ′ 600 . ) 𝑏𝑤. 𝑑 𝑓𝑦 600 + 𝑓𝑦

0,85.0,85.25 600 . ) 200.342 240 600 + 240

𝐴𝑚𝑎𝑘𝑠 = 2757,756 𝑚𝑚2 𝑀𝑢 𝑎 = 0,85. 𝑓𝑐 ′ . 𝑎. 𝑏 (𝑑 − ) ∅ 2 39,960 . 106 𝑎 = 0,85.25. 𝑎. 1062,5 (342 − ) 0,8 2 𝑎 49,95.106 = 22567,5 𝑎 (342 − ) 2

𝑎2 − 684 𝑎 + 4426,719 = 0

𝑅𝑢𝑚𝑢𝑠 𝑎𝑏𝑐 =

𝑎12 =

−𝑏 ± √𝑏 2 − 4𝑎𝑐 2𝑎

684 ± √(−684)2 − 4. (1). (4426,719) 2. (1)

𝑎1 = 677,465 𝑚𝑚 𝑎2 = 6,534 𝑚𝑚 Ambil a yang terkecil a = 6,534 mm 𝑐=

𝑎 6,534𝑚𝑚 = = 7,687 < ℎ𝑓 = 120 𝑚𝑚 (𝑎𝑠𝑢𝑚𝑠𝑖 𝑏𝑒𝑛𝑎𝑟) 𝛽1 0.85

Cc = Ts 𝐴𝑠 =

0,85. 𝑓𝑐 ′ . 𝑐. 𝑏𝑒𝑓𝑓 0,85.25.7,687.1062,5 = = 723,158 𝑚𝑚2 𝑓𝑦 240

Dipakai 3 D 16 (As = 603,2 mm2)

Perhitungan penulangan balok Perhitungan penulangan balok portal memanjang B-B Data-data perhitungan : Dimensi balok

: 20/40 cm

Fc’

: 25 Mpa

Fy

: 240 Mpa

Dicoba : Untuk lantai 1-4 Ø sengkang

= 10 mm

Ø tulangan pokok

= 16 mm

Selimut beton

= 40 mm 1

d = h - p - Ø sengkang - 2 Ø tulangan pokok d = 400 - 40 – 10 -

1 2

. 16

d = 342 mm 1

d’ = p + 2 Ø tulangan pokok + Ø sengkang 1

d’ = 40 + 2 . 16 + 10 d’ = 58 mm  Penulangan tumpuan memanjang B-B 1. Momen tumpuan rencana Untuk momen tumpuan rencana ada dua jenis tulangan tumpuan, yaitu tulangan tumpuan positif dan tulangan tumpuan negatif. Penulangan tersebut berdasarkan nilai momen positif dan terbesar tiap lantai dalam satu portal. Adapun nilai momen maksimum positif dan negatif tersebut diambil dari kombinasi :



Momen kombinasi akibat beban mati dan beban hidup

Mk1 = 1,2 MD + 1,6 ML

2.

penulangan tumpuan 

perhitungan balok lantai 1

M tumpuan

= 45,660 KNm

Tulangan tumpuan Mu = 45,660 KNm b

= 200 mm

h

= 400 mm

d

= 342 mm

Mu

fy

= ρ.0,8.fy. (1-0,5882 ρ . fC′ )

𝑏𝑑2

45,660.106 Nmm 200 .3422

= ρ.0,8.240. (1-0,5882 ρ .

240 25

)

1,951 = 192 ρ – 892,170 ρ2 892,170 ρ2 - 192 ρ + 1,951 = 0 𝜌1,2 = 𝜌1 = 𝜌2 =

−𝑏 ± √𝑏2 − 4 𝑎𝑐 2𝑎

192 + √1922 − 4 ( 892,170) (1,951 ) 2 ( 892,170) 192 −√1922 − 4 ( 892,170) (1,951 ) 2 ( 892,170)

= 0,204 = 0,010

ρ = 0,010 As = ρ.b.d = 0,010.200.342 = 684 mm2 As’ = 0,5.As = 0,5 . 684 mm2 = 342 mm2 Dalam hal ini tulangan terpasang diambil berdasarkan nilai maksimum dari 2 jenis tulangan tumpuan Asatas

= 684 mm2

Asbawah

= 342 mm2



perhitungan balok lantai 2

M tumpuan

= 42,012 KNm

Tulangan tumpuan Mu = 42,012 KNm b

= 200 mm

h

= 400 mm

d

= 342 mm

Mu

fy

= ρ.0,8.fy. (1-0,5882 ρ . fC′ )

𝑏𝑑2

42,012 .106 Nmm 200 .3422

= ρ.0,8.240. (1-0,5882 ρ .

240 25

)

1,795 = 192 ρ – 892,170 ρ2 892,170 ρ2 - 192 ρ + 1,795 = 0 𝜌1,2 = 𝜌1 = 𝜌2 =

−𝑏 ± √𝑏2 − 4 𝑎𝑐 2𝑎

192 + √1922 − 4 ( 892,170) (1,795 ) 2 ( 892,170) 192 −√1922 − 4 ( 892,170) (1,795 ) 2 ( 892,170)

= 0,205 = 0,009

ρ = 0,009 As = ρ.b.d = 0,009.200.342 = 615,6 mm2 As’ = 0,5.As = 0,5 . 615,6 mm2 = 307,8 mm2 Dalam hal ini tulangan terpasang diambil berdasarkan nilai maksimum dari 2 jenis tulangan tumpuan Asatas

= 615,6 mm2

Asbawah

= 307,8 mm2



perhitungan balok lantai 3

M tumpuan Tulangan tumpuan

= 42,044 KNm

Mu = 42,044 KNm b

= 200 mm

h

= 400 mm

d

= 342 mm

Mu

fy

= ρ.0,8.fy. (1-0,5882 ρ . fC′ )

𝑏𝑑2

42,044 .106 Nmm 200 .3422

= ρ.0,8.240. (1-0,5882 ρ .

240 25

)

1,797 = 192 ρ – 892,170 ρ2 892,170 ρ2 - 192 ρ + 1,797 = 0 𝜌1,2 = 𝜌1 = 𝜌2 =

−𝑏 ± √𝑏2 − 4 𝑎𝑐 2𝑎

192 + √1922 − 4 ( 892,170) (1,797 ) 2 ( 892,170) 192 −√1922 − 4 ( 892,170) (1,797 ) 2 ( 892,170)

= 0,205 = 0,009

ρ = 0,009 As = ρ.b.d = 0,009.200.342 = 615,6 mm2 As’ = 0,5.As = 0,5 . 615,6 mm2 = 307,8 mm2 Dalam hal ini tulangan terpasang diambil berdasarkan nilai maksimum dari 2 jenis tulangan tumpuan Asatas

= 615,6 mm2

Asbawah

= 307,8 mm2



perhitungan balok lantai 4

M tumpuan

= 46,615 KNm

Tulangan tumpuan Mu = 46,615 KNm b

= 200 mm

h

= 400 mm

d

= 342 mm

Mu

fy

= ρ.0,8.fy. (1-0,5882 ρ . fC′ )

𝑏𝑑2

46,615 .106 Nmm 200 .3422

= ρ.0,8.240. (1-0,5882 ρ .

240 25

)

1,992 = 192 ρ – 892,170 ρ2 892,170 ρ2 - 192 ρ + 1,992 = 0 𝜌1,2 = 𝜌1 = 𝜌2 =

−𝑏 ± √𝑏2 − 4 𝑎𝑐 2𝑎

192 + √1922 − 4 ( 892,170) (1,992) 2 ( 892,170) 192 −√1922 − 4 ( 892,170) (1,992) 2 ( 892,170)

= 0,204 = 0,010

ρ = 0,010 As = ρ.b.d = 0,010.200.342 = 684 mm2 As’ = 0,5.As = 0,5 . 684 mm2 = 342 mm2 Dalam hal ini tulangan terpasang diambil berdasarkan nilai maksimum dari 2 jenis tulangan tumpuan Asatas

= 684 mm2

Asbawah

= 342 mm2

3. penulangan geser  perhitungan balok lantai 1 momen nominal aktual ( asumsi balok T ) Tulangan atas

= 3 D 16 = 684 mm2

Tulangan bawah

= 2 D 16 = 342 mm2 1

d2 = h-p- Ø sengkang - 2 Ø tulangan pokok 1

d2 = 400 – 40 - 10 - 2 . 16 d2 = 342 mm

1

1

d1 = d2 - 2 Ø tulangan pokok – 25 - 2 Ø tulangan pokok 1

1

= 342- 2. 16 – 25 - 2 .16 = 301 mm 1

As1 = 3 x 4 . π . 162 = 603,185 mm2 1

As2 = 2 x 4 . π . 162 = 402,123 mm2 + = 1005,308 mm2

As d=

As1 .d1 + As2 .d2 As

=

603,185.301 + 402,123.342 1005,308

= 317,4 mm

1

d’ = p + 2 Ø tulangan pokok + Ø sengkang 1

d’ = 40 + 2 . 16 + 10 d’ = 58 mm As’ = 2 x a=

1 4

. π . 162 = 402,123 mm2

(𝐴𝑠−𝐴𝑠′ ).𝑓𝑦 0,85. 𝑓𝑐 ′ 𝑏

=

(1005,308 – 402,123 ). 240 0,85.25.200

= 34,062 mm

β1 =0,85 𝑎

c = 𝛽1 =

34,062 𝑚𝑚 0,85

= 40,073 mm

cek kelelehan : 𝑓𝑦

240

εy = 𝐸𝑠 = 200000 = 0,0012 εs = εs’ =

𝑑−𝑐 𝑐

. 𝜀𝑐 ′ =

𝑐−𝑑′ 𝑐

317,4−40,073

. 𝜀𝑐 ′ =

40,073 40,073 − 58 40,073

. 0,003 = 0,020 > 𝜀𝑦 . 0,003 = − 0,0013 < 𝜀𝑦

kondisi II ( 0,85 . fc’. b . 𝛽1 ) c2 + ( 600 As’ – As.fy ) c – 600 As’ d’ = 0 ( 0,85 . 25 . 200 . 0,85 ) c2 + ( 600 . 402,123–1005,30. 240) c – 600 . 402,123. 58= 0 3612,5 c2 + 0 c – 13993880 = 0

𝜌1,2 = 𝜌1 = 𝜌2 =

−𝑏 ± √𝑏2 − 4 𝑎𝑐 2𝑎

− 0 + √0 2 − 4 ( 3612,5) (−13993880) 2 ( 3612,5) − 0 − √0 2 − 4 ( 3612,5) (−13993880) 2 ( 3612,5)

= 62,239 mm = - 62,239 mm

𝑐 = 62,239 mm 𝑎 = 𝛽1 𝑥 𝑐 = 0,85 𝑥 62,239 = 52,903 mm < hf = 120 mm ( asumsi benar ) Bagian 2 As2 = =

600 𝐴𝑠′ 𝑥

𝑐−𝑑′ 𝑐

𝑓𝑦 600 . 402,123 𝑥

62,239 − 58 62,239

240

= 68,469 mm2

Mn2 = Ts2 ( d-d’ ) = As2 . fy ( d-d’ ) = 68,469. 240 ( 317,4 -58 ) = 4,262 KNm

Bagian 1 As1 = As – As2 = 1005,30mm2 – 402,123mm2 = 603,177 mm2 Mn1 = Ts1 ( d- a/2) = As1 . fy ( d- a/2) = 603,177. 240. ( 317,4 – 52,903 /2 ) = 42,118 KNm Mnak‘= Mn = Mn1 + Mn2 = 42,118 KNm + 4,262 KNm = 46,380 KNm

Momen nominal aktual ( asumsi balok persegi ) Dimisalkan a < hf a=

(𝐴𝑠−𝐴𝑠′ ).𝑓𝑦 0,85. 𝑓𝑐 ′ 𝑏

=

(1005,30 – 402,123 ). 240 0,85.25.200

= 34,062 mm

β1 =0,85 𝑎

c = 𝛽1 =

34,062 𝑚𝑚 0,85

= 40,073 mm

cek kelelehan : 𝑓𝑦

240

εy = 𝐸𝑠 = 200000 = 0,0012 εs = εs’ =

𝑑−𝑐 𝑐

. 𝜀𝑐 ′ =

𝑐−𝑑′ 𝑐

317,4−40,073

. 𝜀𝑐 ′ =

40,073 40,073 − 58 40,073

. 0,003 = 0,020 > 𝜀𝑦 . 0,003 = − 0,0013 < 𝜀𝑦

kondisi II ( 0,85 . fc’. b . 𝛽1 ) c2 + ( 600 As’ – As.fy ) c – 600 As’ d’ = 0 ( 0,85 . 25 . 200 . 0,85 ) c2 + ( 600 . 402,123–1005,30. 240) c – 600 . 402,123. 58= 0 3612,5 c2 + 0 c – 13993880 = 0 𝜌1,2 = 𝜌1 =

−𝑏 ± √𝑏2 − 4 𝑎𝑐 2𝑎

− 0 + √0 2 − 4 ( 3612,5) (−13993880) 2 ( 3612,5)

= 62,239 mm

𝜌2 =

− 0 − √0 2 − 4 ( 3612,5) (−13993880) 2 ( 3612,5)

= - 62,239 mm

𝑐 = 62,239 mm Cs = 600 x As’ x

𝑐−𝑑′ 𝑐

= 600 x 402,123x

62,239 −58 62,239

= 16432,777 N

Mn2 = Cs ( d-d’ ) = 16432,777 N ( 317,4 -58 ) mm = 4,262 KNm Cc = 0,85 fc’ b . β1 . c = 0,85.25.200.0,85. 62,239 = 224838, 387 N Mn1 = Cc ( d- a/2) = 224838, 387 N ( 317,4 –56,277/2 ) = 65,037 KNm Mnak‘= Mn = Mn1 + Mn2 = 65,037 KNm + 4,262 KNm = 69,299 KNm

 perhitungan balok lantai 2 momen nominal aktual ( asumsi balok T ) Tulangan atas

= 3 D 16 = 615,6 mm2

Tulangan bawah

= 2 D 16 = 307,8 mm2

1

d2 = h-p- Ø sengkang - 2 Ø tulangan pokok 1

d2 = 400 – 40 - 10 - 2 . 16 d2 = 342 mm 1

1

d1 = d2 - 2 Ø tulangan pokok – 25 - 2 Ø tulangan pokok 1

1

= 342- 2. 16 – 25 - 2 .16 = 301 mm 1

As1 = 3 x 4 . π . 162 = 603,185 mm2 1

As2 = 2 x 4 . π . 162 = 402,123 mm2 + = 1005,308 mm2

As d=

As1 .d1 + As2 .d2 As

=

603,185.301 + 402,123.342 1005,308

= 317,4 mm

1

d’ = p + Ø tulangan pokok + Ø sengkang 2

1

d’ = 40 + 2 . 16 + 10 d’ = 58 mm 1

As’ = 2 x 4 . π . 162 = 402,123 mm2 a=

(𝐴𝑠−𝐴𝑠′ ).𝑓𝑦 0,85.

𝑓𝑐 ′

𝑏

=

(1005,308 – 402,123 ). 240 0,85.25.200

= 34,062 mm

β1 =0,85 𝑎

c = 𝛽1 =

34,062 𝑚𝑚 0,85

= 40,073 mm

cek kelelehan : 𝑓𝑦

240

εy = 𝐸𝑠 = 200000 = 0,0012 εs =

𝑑−𝑐 𝑐

. 𝜀𝑐 ′ =

317,4−40,073 40,073

. 0,003 = 0,020 > 𝜀𝑦

εs’ =

𝑐−𝑑′ 𝑐

. 𝜀𝑐 ′ =

40,073 − 58 40,073

. 0,003 = − 0,0013 < 𝜀𝑦

kondisi II ( 0,85 . fc’. b . 𝛽1 ) c2 + ( 600 As’ – As.fy ) c – 600 As’ d’ = 0 ( 0,85 . 25 . 200 . 0,85 ) c2 + ( 600 . 402,123–1005,30. 240) c – 600 . 402,123. 58= 0 3612,5 c2 + 0 c – 13993880 = 0 𝜌1,2 = 𝜌1 = 𝜌2 =

−𝑏 ± √𝑏2 − 4 𝑎𝑐 2𝑎

− 0 + √0 2 − 4 ( 3612,5) (−13993880) 2 ( 3612,5) − 0 − √0 2 − 4 ( 3612,5) (−13993880) 2 ( 3612,5)

= 62,239 mm = - 62,239 mm

𝑐 = 62,239 mm 𝑎 = 𝛽1 𝑥 𝑐 = 0,85 𝑥 62,239 = 52,903 mm < hf = 120 mm ( asumsi benar ) Bagian 2 As2 = =

600 𝐴𝑠′ 𝑥

𝑐−𝑑′ 𝑐

𝑓𝑦 600 . 402,123 𝑥

62,239 − 58 62,239

240

= 68,469 mm2

Mn2 = Ts2 ( d-d’ ) = As2 . fy ( d-d’ ) = 68,469. 240 ( 317,4 -58 ) = 4,262 KNm

Bagian 1 As1 = As – As2 = 1005,30mm2 – 402,123mm2 = 603,177 mm2 Mn1 = Ts1 ( d- a/2) = As1 . fy ( d- a/2)

= 603,177. 240. ( 317,4 – 52,903 /2 ) = 42,118 KNm Mnak‘= Mn = Mn1 + Mn2 = 42,118 KNm + 4,262 KNm = 46,380 KNm

Momen nominal aktual ( asumsi balok persegi ) Dimisalkan a < hf a=

(𝐴𝑠−𝐴𝑠′ ).𝑓𝑦 0,85. 𝑓𝑐 ′ 𝑏

=

(1005,30 – 402,123 ). 240 0,85.25.200

= 34,062 mm

β1 =0,85 𝑎

c = 𝛽1 =

34,062 𝑚𝑚 0,85

= 40,073 mm

cek kelelehan : 𝑓𝑦

240

εy = 𝐸𝑠 = 200000 = 0,0012 εs = εs’ =

𝑑−𝑐 𝑐

. 𝜀𝑐 ′ =

𝑐−𝑑′ 𝑐

317,4−40,073

. 𝜀𝑐 ′ =

40,073 40,073 − 58 40,073

. 0,003 = 0,020 > 𝜀𝑦 . 0,003 = − 0,0013 < 𝜀𝑦

kondisi II ( 0,85 . fc’. b . 𝛽1 ) c2 + ( 600 As’ – As.fy ) c – 600 As’ d’ = 0 ( 0,85 . 25 . 200 . 0,85 ) c2 + ( 600 . 402,123–1005,30. 240) c – 600 . 402,123. 58= 0 3612,5 c2 + 0 c – 13993880 = 0

𝜌1,2 = 𝜌1 = 𝜌2 =

−𝑏 ± √𝑏2 − 4 𝑎𝑐 2𝑎

− 0 + √0 2 − 4 ( 3612,5) (−13993880) 2 ( 3612,5) − 0 − √0 2 − 4 ( 3612,5) (−13993880) 2 ( 3612,5)

= 62,239 mm = - 62,239 mm

𝑐 = 62,239 mm Cs = 600 x As’ x

𝑐−𝑑′ 𝑐

= 600 x 402,123x

62,239 −58 62,239

= 16432,777 N

Mn2 = Cs ( d-d’ ) = 16432,777 N ( 317,4 -58 ) mm = 4,262 KNm Cc = 0,85 fc’ b . β1 . c = 0,85.25.200.0,85. 62,239 = 224838, 387 N Mn1 = Cc ( d- a/2) = 224838, 387 N ( 317,4 –56,277/2 ) = 65,037 KNm Mnak‘= Mn = Mn1 + Mn2 = 65,037 KNm + 4,262 KNm = 69,299 KNm

 perhitungan balok lantai 3 momen nominal aktual

( asumsi balok T ) Tulangan atas

= 3 D 16 = 615,6 mm2

Tulangan bawah

= 2 D 16 = 307,8 mm2 1

d2 = h-p- Ø sengkang - 2 Ø tulangan pokok 1

d2 = 400 – 40 - 10 - 2 . 16 d2 = 342 mm 1

1

d1 = d2 - 2 Ø tulangan pokok – 25 - 2 Ø tulangan pokok 1

1

= 342- 2. 16 – 25 - 2 .16 = 301 mm 1

As1 = 3 x 4 . π . 162 = 603,185 mm2 1

As2 = 2 x 4 . π . 162 = 402,123 mm2 + = 1005,308 mm2

As d=

As1 .d1 + As2 .d2 As

=

603,185.301 + 402,123.342 1005,308

= 317,4 mm

1

d’ = p + 2 Ø tulangan pokok + Ø sengkang 1

d’ = 40 + 2 . 16 + 10 d’ = 58 mm 1

As’ = 2 x 4 . π . 162 = 402,123 mm2 a=

(𝐴𝑠−𝐴𝑠′ ).𝑓𝑦 0,85. 𝑓𝑐 ′ 𝑏

=

(1005,308 – 402,123 ). 240 0,85.25.200

β1 =0,85 𝑎

c = 𝛽1 =

34,062 𝑚𝑚 0,85

cek kelelehan :

= 40,073 mm

= 34,062 mm

𝑓𝑦

240

εy = 𝐸𝑠 = 200000 = 0,0012 𝑑−𝑐

εs =

𝑐

εs’ =

. 𝜀𝑐 ′ =

𝑐−𝑑′ 𝑐

317,4−40,073 40,073

. 𝜀𝑐 ′ =

40,073 − 58 40,073

. 0,003 = 0,020 > 𝜀𝑦 . 0,003 = − 0,0013 < 𝜀𝑦

kondisi II ( 0,85 . fc’. b . 𝛽1 ) c2 + ( 600 As’ – As.fy ) c – 600 As’ d’ = 0 ( 0,85 . 25 . 200 . 0,85 ) c2 + ( 600 . 402,123–1005,30. 240) c – 600 . 402,123. 58= 0 3612,5 c2 + 0 c – 13993880 = 0 𝜌1,2 = 𝜌1 = 𝜌2 =

−𝑏 ± √𝑏2 − 4 𝑎𝑐 2𝑎

− 0 + √0 2 − 4 ( 3612,5) (−13993880) 2 ( 3612,5) − 0 − √0 2 − 4 ( 3612,5) (−13993880) 2 ( 3612,5)

= 62,239 mm = - 62,239 mm

𝑐 = 62,239 mm 𝑎 = 𝛽1 𝑥 𝑐 = 0,85 𝑥 62,239 = 52,903 mm < hf = 120 mm ( asumsi benar ) Bagian 2 As2 = =

600 𝐴𝑠′ 𝑥

𝑐−𝑑′ 𝑐

𝑓𝑦 600 . 402,123 𝑥

62,239 − 58 62,239

240

= 68,469 mm2

Mn2 = Ts2 ( d-d’ ) = As2 . fy ( d-d’ ) = 68,469. 240 ( 317,4 -58 ) = 4,262 KNm

Bagian 1 As1 = As – As2 = 1005,30mm2 – 402,123mm2 = 603,177 mm2

Mn1 = Ts1 ( d- a/2) = As1 . fy ( d- a/2) = 603,177. 240. ( 317,4 – 52,903 /2 ) = 42,118 KNm Mnak‘= Mn = Mn1 + Mn2 = 42,118 KNm + 4,262 KNm = 46,380 KNm

Momen nominal aktual ( asumsi balok persegi ) Dimisalkan a < hf a=

(𝐴𝑠−𝐴𝑠′ ).𝑓𝑦 0,85. 𝑓𝑐 ′ 𝑏

=

(1005,30 – 402,123 ). 240 0,85.25.200

= 34,062 mm

β1 =0,85 𝑎

c = 𝛽1 =

34,062 𝑚𝑚 0,85

= 40,073 mm

cek kelelehan : 𝑓𝑦

240

εy = 𝐸𝑠 = 200000 = 0,0012 εs = εs’ =

𝑑−𝑐 𝑐

. 𝜀𝑐 ′ =

𝑐−𝑑′ 𝑐

317,4−40,073

. 𝜀𝑐 ′ =

40,073 40,073 − 58 40,073

. 0,003 = 0,020 > 𝜀𝑦 . 0,003 = − 0,0013 < 𝜀𝑦

kondisi II ( 0,85 . fc’. b . 𝛽1 ) c2 + ( 600 As’ – As.fy ) c – 600 As’ d’ = 0

( 0,85 . 25 . 200 . 0,85 ) c2 + ( 600 . 402,123–1005,30. 240) c – 600 . 402,123. 58= 0 3612,5 c2 + 0 c – 13993880 = 0 𝜌1,2 = 𝜌1 = 𝜌2 =

−𝑏 ± √𝑏2 − 4 𝑎𝑐 2𝑎

− 0 + √0 2 − 4 ( 3612,5) (−13993880) 2 ( 3612,5) − 0 − √0 2 − 4 ( 3612,5) (−13993880) 2 ( 3612,5)

= 62,239 mm = - 62,239 mm

𝑐 = 62,239 mm Cs = 600 x As’ x

𝑐−𝑑′ 𝑐

= 600 x 402,123x

62,239 −58 62,239

= 16432,777 N

Mn2 = Cs ( d-d’ ) = 16432,777 N ( 317,4 -58 ) mm = 4,262 KNm Cc = 0,85 fc’ b . β1 . c = 0,85.25.200.0,85. 62,239 = 224838, 387 N Mn1 = Cc ( d- a/2) = 224838, 387 N ( 317,4 –56,277/2 ) = 65,037 KNm Mnak‘= Mn = Mn1 + Mn2 = 65,037 KNm + 4,262 KNm = 69,299 KNm

 perhitungan balok lantai 4 momen nominal aktual ( asumsi balok T ) Tulangan atas

= 3 D 16 = 684 mm2

Tulangan bawah

= 2 D 16 = 342 mm2 1

d2 = h-p- Ø sengkang - 2 Ø tulangan pokok 1

d2 = 400 – 40 - 10 - 2 . 16 d2 = 342 mm 1

1

d1 = d2 - 2 Ø tulangan pokok – 25 - 2 Ø tulangan pokok 1

1

= 342- 2. 16 – 25 - 2 .16 = 301 mm 1

As1 = 3 x 4 . π . 162 = 603,185 mm2 1

As2 = 2 x 4 . π . 162 = 402,123 mm2 + = 1005,308 mm2

As d=

As1 .d1 + As2 .d2 As

=

603,185.301 + 402,123.342 1005,308

= 317,4 mm

1

d’ = p + 2 Ø tulangan pokok + Ø sengkang 1

d’ = 40 + 2 . 16 + 10 d’ = 58 mm 1

As’ = 2 x 4 . π . 162 = 402,123 mm2 a=

(𝐴𝑠−𝐴𝑠′ ).𝑓𝑦 0,85. 𝑓𝑐 ′ 𝑏

=

(1005,308 – 402,123 ). 240 0,85.25.200

β1 =0,85 𝑎

c = 𝛽1 =

34,062 𝑚𝑚 0,85

= 40,073 mm

= 34,062 mm

cek kelelehan : 𝑓𝑦

240

εy = 𝐸𝑠 = 200000 = 0,0012 𝑑−𝑐

εs =

𝑐

εs’ =

. 𝜀𝑐 ′ =

𝑐−𝑑′ 𝑐

317,4−40,073 40,073

. 𝜀𝑐 ′ =

40,073 − 58 40,073

. 0,003 = 0,020 > 𝜀𝑦 . 0,003 = − 0,0013 < 𝜀𝑦

kondisi II ( 0,85 . fc’. b . 𝛽1 ) c2 + ( 600 As’ – As.fy ) c – 600 As’ d’ = 0 ( 0,85 . 25 . 200 . 0,85 ) c2 + ( 600 . 402,123–1005,30. 240) c – 600 . 402,123. 58= 0 3612,5 c2 + 0 c – 13993880 = 0 𝜌1,2 = 𝜌1 = 𝜌2 =

−𝑏 ± √𝑏2 − 4 𝑎𝑐 2𝑎

− 0 + √0 2 − 4 ( 3612,5) (−13993880) 2 ( 3612,5) − 0 − √0 2 − 4 ( 3612,5) (−13993880) 2 ( 3612,5)

= 62,239 mm = - 62,239 mm

𝑐 = 62,239 mm 𝑎 = 𝛽1 𝑥 𝑐 = 0,85 𝑥 62,239 = 52,903 mm < hf = 120 mm ( asumsi benar ) Bagian 2 As2 = =

600 𝐴𝑠′ 𝑥

𝑐−𝑑′ 𝑐

𝑓𝑦 600 . 402,123 𝑥

62,239 − 58 62,239

240

= 68,469 mm2

Mn2 = Ts2 ( d-d’ ) = As2 . fy ( d-d’ ) = 68,469. 240 ( 317,4 -58 ) = 4,262 KNm

Bagian 1 As1 = As – As2 = 1005,30mm2 – 402,123mm2

= 603,177 mm2 Mn1 = Ts1 ( d- a/2) = As1 . fy ( d- a/2) = 603,177. 240. ( 317,4 – 52,903 /2 ) = 42,118 KNm Mnak‘= Mn = Mn1 + Mn2 = 42,118 KNm + 4,262 KNm = 46,380 KNm

Momen nominal aktual ( asumsi balok persegi ) Dimisalkan a < hf a=

(𝐴𝑠−𝐴𝑠′ ).𝑓𝑦 0,85.

𝑓𝑐 ′

𝑏

=

(1005,30 – 402,123 ). 240 0,85.25.200

= 34,062 mm

β1 =0,85 𝑎

c = 𝛽1 =

34,062 𝑚𝑚 0,85

= 40,073 mm

cek kelelehan : 𝑓𝑦

240

εy = 𝐸𝑠 = 200000 = 0,0012 εs = εs’ =

𝑑−𝑐 𝑐

. 𝜀𝑐 ′ =

𝑐−𝑑′ 𝑐

kondisi II

317,4−40,073

. 𝜀𝑐 ′ =

40,073 40,073 − 58 40,073

. 0,003 = 0,020 > 𝜀𝑦 . 0,003 = − 0,0013 < 𝜀𝑦

( 0,85 . fc’. b . 𝛽1 ) c2 + ( 600 As’ – As.fy ) c – 600 As’ d’ = 0 ( 0,85 . 25 . 200 . 0,85 ) c2 + ( 600 . 402,123–1005,30. 240) c – 600 . 402,123. 58= 0 3612,5 c2 + 0 c – 13993880 = 0 𝜌1,2 = 𝜌1 = 𝜌2 =

−𝑏 ± √𝑏2 − 4 𝑎𝑐 2𝑎

− 0 + √0 2 − 4 ( 3612,5) (−13993880) 2 ( 3612,5) − 0 − √0 2 − 4 ( 3612,5) (−13993880) 2 ( 3612,5)

= 62,239 mm = - 62,239 mm

𝑐 = 62,239 mm Cs = 600 x As’ x

𝑐−𝑑′ 𝑐

= 600 x 402,123x

62,239 −58 62,239

= 16432,777 N

Mn2 = Cs ( d-d’ ) = 16432,777 N ( 317,4 -58 ) mm = 4,262 KNm Cc = 0,85 fc’ b . β1 . c = 0,85.25.200.0,85. 62,239 = 224838, 387 N Mn1 = Cc ( d- a/2) = 224838, 387 N ( 317,4 –56,277/2 ) = 65,037 KNm Mnak‘= Mn = Mn1 + Mn2 = 65,037 KNm + 4,262 KNm = 69,299 KNm

𝑀tumpuan

1. Vub = 0,7 Ø(

𝑙𝑛

) + 1,05Vg

2. Vubterpakai = [1,05Vg − 0,7 (

𝑀tumpuan 𝑙𝑛

)] +

𝑙𝑛−𝑑 𝑙𝑛

𝑀tumpuan

[Vub − [1,05Vg − 0,7 Ø (

Gaya gesser Maksimum Balok (Vub) Lantai 1 Vub = 70,565 KN(lihat table gaya geser rencana) Lantai 2 Vub = 66,645 KN(lihat table gaya geser rencana) Lantai 3 Vub = 65,854 KN(lihat table gaya geser rencana) Lantai 4 Vub 73,857 KN(lihat table gaya geser rencana)

Gaya gesser Maksimum Balok (Vub) Lantai 1 Vubterpakai = 69,355 KN(lihat table gaya geser rencana) Lantai 2 Vubterpakai = 65,532 KN(lihat table gaya geser rencana) Lantai 3 Vubterpakai = 64,740 KN(lihat table gaya geser rencana) Lantai 4 Vubterpakai = 72,621 KN(lihat table gaya geser rencana)

𝑙𝑛

)]]

Penulangan Geser a. Penulangan Geser Balok Lantai 1 Skema Gaya Geser rencana :

Vujung = Vub-(

𝑉𝑢𝑏−𝑉𝑢𝑏𝑡𝑒𝑟𝑝𝑎𝑘𝑎𝑖 𝑑 70,565 −69,355

Vujung = 70,565 -(

0,342

𝑥 𝑙𝑛) 𝑥4,25)

Vujung = 55,528 KN Penulangan Geser Dalam Daerah Sendi Plastis Vs =

𝑉𝑢𝑏𝑡𝑒𝑟𝑝𝑎𝑘𝑎𝑖 Ø

=

69,355

= 92,473 𝐾𝑁 = 92473 𝑁

0,75

Dipakai Sengkang ∅ 10 𝑚𝑚 S=

𝐴𝑣.𝑓𝑦.𝑑

=

𝑉𝑠

1 4

2 𝜋102 .240.342 92473

S= 139,425 mm >

𝑑 4

= 139,425

342

=

4

= 85,5

Dipakai Jarak 85,5 mm Jadi, Dipakai Sengkang∅10 − 85,5 mm

Penulangan Geser Diluar Daerah Sendi Plastis Vub’ = Vujung +

𝑙𝑛−2ℎ

Vub’ = 55,528 +

4,250−2.0,4

𝑙𝑛

. (𝑉𝑢𝑏 𝑚𝑎𝑘𝑠𝑖𝑚𝑢𝑚 − 𝑉𝑢𝑗𝑢𝑛𝑔)

4,250

. (70,565 − 55,528 )

Vub’ = 67,734 KN √𝑓𝑐′

Vc = ( Vs =

6

𝑉𝑢𝑏′ ∅

) . 𝑏𝑤. 𝑑 = (

− 𝑉𝑐 =

√25 ) . 200.342 6

67,734 0,75

= 57000 𝑁 = 57 𝐾𝑁

− 57 = 33,312 𝐾𝑁 = 33312 𝑁

S=

𝐴𝑣.𝑓𝑦.𝑑 𝑉𝑠

=

1 4

2 𝜋102 .240.342 33312

S= 387,040 mm >

𝑑 4

=

= 387,040

342 2

= 171

Dipakai Jarak 171 mm Jadi, Dipakai Sengkang∅10 − 171 mm

b. Penulangan Geser Balok Lantai 2 Skema Gaya Geser rencana :

Vujung = Vub-(

𝑉𝑢𝑏−𝑉𝑢𝑏𝑡𝑒𝑟𝑝𝑎𝑘𝑎𝑖 𝑑

𝑥 𝑙𝑛)

66,645 −65,532

Vujung = 66,645 -(

0,342

𝑥4,25)

Vujung = 52,813 KN Penulangan Geser Dalam Daerah Sendi Plastis Vs =

𝑉𝑢𝑏𝑡𝑒𝑟𝑝𝑎𝑘𝑎𝑖 Ø

=

65,532

= 87,736 𝐾𝑁 = 87736 𝑁

0,75

Dipakai Sengkang ∅ 10 𝑚𝑚 S=

𝐴𝑣.𝑓𝑦.𝑑 𝑉𝑠

=

1 4

2 𝜋102 .240.342 87736

S= 146,953 mm >

𝑑 4

=

= 146,953

342 4

= 85,5

Dipakai Jarak 85,5 mm Jadi, Dipakai Sengkang∅10 − 85,5 mm

Penulangan Geser Diluar Daerah Sendi Plastis Vub’ = Vujung + Vub’ = 52,813 +

𝑙𝑛−2ℎ 𝑙𝑛

. (𝑉𝑢𝑏 𝑚𝑎𝑘𝑠𝑖𝑚𝑢𝑚 − 𝑉𝑢𝑗𝑢𝑛𝑔)

4,250−2.0,4 4,250

. (66,645 − 52,813 )

Vub’ = 64,041 KN √𝑓𝑐′

Vc = ( Vs = S=

𝑉𝑢𝑏′ ∅

) . 𝑏𝑤. 𝑑 = (

6

− 𝑉𝑐 =

𝐴𝑣.𝑓𝑦.𝑑 𝑉𝑠

=

√25 ) . 200.342 6

64,041

− 57 = 28,388 𝐾𝑁 = 28388 𝑁

0,75

1 4

2 𝜋102 .240.342 28388 𝑑

S= 454,174 mm >

4

= 57000 𝑁 = 57 𝐾𝑁

= 454,174

342

=

2

= 171

Dipakai Jarak 171 mm Jadi, Dipakai Sengkang∅10 − 171 mm

c. Penulangan Geser Balok Lantai 3 Skema Gaya Geser rencana :

Vujung = Vub-(

𝑉𝑢𝑏−𝑉𝑢𝑏𝑡𝑒𝑟𝑝𝑎𝑘𝑎𝑖 𝑑

𝑥 𝑙𝑛)

65,854 −64,740

Vujung = 65,854 -(

0,342

𝑥4,25)

Vujung = 52,010 KN Penulangan Geser Dalam Daerah Sendi Plastis Vs =

𝑉𝑢𝑏𝑡𝑒𝑟𝑝𝑎𝑘𝑎𝑖 Ø

64,740

=

0,75

= 86,320 𝐾𝑁 = 86320 𝑁

Dipakai Sengkang ∅ 10 𝑚𝑚 S=

𝐴𝑣.𝑓𝑦.𝑑 𝑉𝑠

=

1 4

2 𝜋102 .240.342 86320

S= 149,364mm >

𝑑 4

=

= 149,364

342 4

= 85,5

Dipakai Jarak 85,5 mm Jadi, Dipakai Sengkang∅10 − 85,5 mm

Penulangan Geser Diluar Daerah Sendi Plastis 𝑙𝑛−2ℎ

Vub’ = Vujung + Vub’ = 52,010 +

𝑙𝑛

. (𝑉𝑢𝑏 𝑚𝑎𝑘𝑠𝑖𝑚𝑢𝑚 − 𝑉𝑢𝑗𝑢𝑛𝑔)

4,250−2.0,4 4,250

. (65,854 − 52,010 )

Vub’ = 63,248 KN √𝑓𝑐′

Vc = ( Vs = S=

𝑉𝑢𝑏′ ∅

) . 𝑏𝑤. 𝑑 = (

6

− 𝑉𝑐 =

𝐴𝑣.𝑓𝑦.𝑑 𝑉𝑠

=

√25 ) . 200.342 6

63,248

− 57 = 27,330 𝐾𝑁 = 27330 𝑁

0,75

1 4

2 𝜋102 .240.342 27330

S= 451,756mm >

𝑑

=

4

= 57000 𝑁 = 57 𝐾𝑁

= 451,756

342 2

= 171

Dipakai Jarak 171 mm Jadi, Dipakai Sengkang∅10 − 171 mm

d. Penulangan Geser Balok Lantai 4 Skema Gaya Geser rencana :

Vujung = Vub-(

𝑉𝑢𝑏−𝑉𝑢𝑏𝑡𝑒𝑟𝑝𝑎𝑘𝑎𝑖 𝑑 73,857−72,621

Vujung = 73,857 -(

0,342

𝑥 𝑙𝑛) 𝑥4,25)

Vujung = 58,497 KN Penulangan Geser Dalam Daerah Sendi Plastis Vs =

𝑉𝑢𝑏𝑡𝑒𝑟𝑝𝑎𝑘𝑎𝑖 Ø

=

72,621 0,75

= 96,828𝐾𝑁 = 96828 𝑁

Dipakai Sengkang ∅ 10 𝑚𝑚

S=

𝐴𝑣.𝑓𝑦.𝑑

=

𝑉𝑠

1 4

2 𝜋102 .240.342 96828 𝑑

S= 133,154 mm >

4

= 133,154

342

=

4

= 85,5

Dipakai Jarak 85,5 mm Jadi, Dipakai Sengkang∅10 − 85,5 mm

Penulangan Geser Diluar Daerah Sendi Plastis 𝑙𝑛−2ℎ

Vub’ = Vujung + Vub’ = 58,497 +

𝑙𝑛

. (𝑉𝑢𝑏 𝑚𝑎𝑘𝑠𝑖𝑚𝑢𝑚 − 𝑉𝑢𝑗𝑢𝑛𝑔)

4,250−2.0,4 4,250

. (73,857 − 58,497 )

Vub’ = 70,965 KN √𝑓𝑐′

Vc = ( Vs = S=

) . 𝑏𝑤. 𝑑 = (

6

𝑉𝑢𝑏′ ∅

− 𝑉𝑐 =

𝐴𝑣.𝑓𝑦.𝑑 𝑉𝑠

=

√25 ) . 200.342 6

70,965

− 57 = 37,620 𝐾𝑁 = 37620 𝑁

0,75

1 4

2 𝜋102 .240.342 37620

S= 342,719 mm >

𝑑 4

=

= 57000 𝑁 = 57 𝐾𝑁

= 342,719

342 2

= 171

Dipakai Jarak 171 mm Jadi, Dipakai Sengkang∅10 − 171 mm Penulangan lapangan melintang B-B 1). Momen lapanan Rencana Nilai nilai dari momen lapangan rencana dapat dilihat dari table 2). Penulangan lapangan Data-data perhitungan : L

= 4250 mm

Ln

= 4250 mm

Lebar efektif untuk lantai 1,2,3 dan 4 1

1

Beff ≤ 4L= 4 4250 = 1062,5 mm Beff ≤ 16 hf + bw = 16. 120 + 200 = 2120 mm Beff ≤ bw + ln = 200 + 4250 = 4450 mm Diambil Beff = 1062,5

a. Perhitungan Balok Lantai 1 Mu = 25,620 KNm Asumsi letak garis netral di dalam sayap a ≤ hf dan c ≤ hf Batasan tulangan 𝜌 min = 𝜌 min =

√𝑓 ′ 𝑐 4.𝑓𝑦

√25

= 4.240 = 0,0052

1,4 1,4 = = 0,0058 𝑓𝑦 240

As min = ρmin . bw . d = 0,0052 . 200 . 342 = 355,68 mm2 𝐴𝑚𝑎𝑘𝑠 = 0,75 . ( 𝐴𝑚𝑎𝑘𝑠 = 0,75 . (

0,85. 𝛽. 𝑓𝑐 ′ 600 . ) 𝑏𝑤. 𝑑 𝑓𝑦 600 + 𝑓𝑦

0,85.0,85.25 600 . ) 200.342 240 600 + 240

𝐴𝑚𝑎𝑘𝑠 = 2757,756 𝑚𝑚2 𝑀𝑢 𝑎 = 0,85. 𝑓𝑐 ′ . 𝑎. 𝑏 (𝑑 − ) ∅ 2 25,620. 106 𝑎 = 0,85.25. 𝑎. 1062,5 (342 − ) 0,8 2 𝑎 32,02.106 = 22567,5 𝑎 (342 − ) 2 𝑎2 − 684 𝑎 + 2837,709 = 0

𝑅𝑢𝑚𝑢𝑠 𝑎𝑏𝑐 =

𝑎12 =

−𝑏 ± √𝑏 2 − 4𝑎𝑐 2𝑎

684 ± √(−684)2 − 4. (1). (2837,709) 2. (1)

𝑎1 = 679,825𝑚𝑚 𝑎2 = 4,174 𝑚𝑚 Ambil a yang terkecil a = 4,174 mm 𝑐=

𝑎 4,174 𝑚𝑚 = = 4,910 < ℎ𝑓 = 120 𝑚𝑚 (𝑎𝑠𝑢𝑚𝑠𝑖 𝑏𝑒𝑛𝑎𝑟) 𝛽1 0.85

Cc = Ts

𝐴𝑠 =

0,85. 𝑓𝑐 ′ . 𝑐. 𝑏𝑒𝑓𝑓 0,85.25.4,910.1062,5 = = 461,910 𝑚𝑚2 𝑓𝑦 240

Dipakai 2 D 16 (As = 402,2 mm2)

b. Perhitungan Balok Lantai 2 Mu = 23,868 KNm Asumsi letak garis netral di dalam sayap a ≤ hf dan c ≤ hf Batasan tulangan 𝜌 min = 𝜌 min =

√𝑓 ′ 𝑐 4.𝑓𝑦

√25

= 4.240 = 0,0052

1,4 1,4 = = 0,0058 𝑓𝑦 240

As min = ρmin . bw . d = 0,0052 . 200 . 342 = 355,68 mm2 𝐴𝑚𝑎𝑘𝑠 = 0,75 . ( 𝐴𝑚𝑎𝑘𝑠 = 0,75 . (

0,85. 𝛽. 𝑓𝑐 ′ 600 . ) 𝑏𝑤. 𝑑 𝑓𝑦 600 + 𝑓𝑦

0,85.0,85.25 600 . ) 200.342 240 600 + 240

𝐴𝑚𝑎𝑘𝑠 = 2757,756 𝑚𝑚2 𝑀𝑢 𝑎 = 0,85. 𝑓𝑐 ′ . 𝑎. 𝑏 (𝑑 − ) ∅ 2 23,868. 106 𝑎 = 0,85.25. 𝑎. 1062,5 (342 − ) 0,8 2 𝑎 29,83.106 = 22567,5 𝑎 (342 − ) 2 𝑎2 − 684 𝑎 + 2643,624 = 0 −𝑏 ± √𝑏 2 − 4𝑎𝑐 𝑅𝑢𝑚𝑢𝑠 𝑎𝑏𝑐 = 2𝑎 𝑎12 =

684 ± √(−684)2 − 4. (1). (2643,624) 2. (1)

𝑎1 = 680,112 𝑚𝑚 𝑎2 = 3,887 𝑚𝑚 Ambil a yang terkecil a = 3,887 mm

𝑐=

𝑎 3,887 𝑚𝑚 = = 4,572 < ℎ𝑓 = 120 𝑚𝑚 (𝑎𝑠𝑢𝑚𝑠𝑖 𝑏𝑒𝑛𝑎𝑟) 𝛽1 0.85

Cc = Ts 𝐴𝑠 =

0,85. 𝑓𝑐 ′ . 𝑐. 𝑏𝑒𝑓𝑓 0,85.25.4,572.1062,5 = = 430,113 𝑚𝑚2 𝑓𝑦 240

Dipakai 2 D 16 (As = 402,2 mm2)

c. Perhitungan Balokk Lantai 3 Mu = 23,172 KNm Asumsi letak garis netral di dalam sayap a ≤ hf dan c ≤ hf Batasan tulangan 𝜌 min = 𝜌 min =

√𝑓 ′ 𝑐 4.𝑓𝑦

√25

= 4.240 = 0,0052

1,4 1,4 = = 0,0058 𝑓𝑦 240

As min = ρmin . bw . d = 0,0052 . 200 . 342 = 355,68 mm2 0,85. 𝛽. 𝑓𝑐 ′ 600 𝐴𝑚𝑎𝑘𝑠 = 0,75 . ( . ) 𝑏𝑤. 𝑑 𝑓𝑦 600 + 𝑓𝑦 𝐴𝑚𝑎𝑘𝑠 = 0,75 . (

0,85.0,85.25 600 . ) 200.342 240 600 + 240

𝐴𝑚𝑎𝑘𝑠 = 2757,756 𝑚𝑚2 𝑀𝑢 𝑎 = 0,85. 𝑓𝑐 ′ . 𝑎. 𝑏 (𝑑 − ) ∅ 2 23,172. 106 𝑎 = 0,85.25. 𝑎. 1062,5 (342 − ) 0,8 2 𝑎 28,96.106 = 22567,5 𝑎 (342 − ) 2 𝑎2 − 684 𝑎 + 2566,522 = 0

𝑅𝑢𝑚𝑢𝑠 𝑎𝑏𝑐 =

𝑎12

−𝑏 ± √𝑏 2 − 4𝑎𝑐 2𝑎

684 ± √(−684)2 − 4. (1). (2566,522) = 2. (1)

𝑎1 = 680,226 𝑚𝑚

𝑎2 = 3,773 𝑚𝑚 Ambil a yang terkecil a = 3,773 mm 𝑐=

𝑎 3,773 𝑚𝑚 = = 4,438 < ℎ𝑓 = 120 𝑚𝑚 (𝑎𝑠𝑢𝑚𝑠𝑖 𝑏𝑒𝑛𝑎𝑟) 𝛽1 0.85

Cc = Ts 0,85. 𝑓𝑐 ′ . 𝑐. 𝑏𝑒𝑓𝑓 0,85.25.4,438.1062,5 𝐴𝑠 = = = 417,507 𝑚𝑚2 𝑓𝑦 240 Dipakai 2 D 16 (As = 402,2 mm2)

d. Perhitungan Balokk Lantai 4 Mu = 29,468 KNm Asumsi letak garis netral di dalam sayap a ≤ hf dan c ≤ hf Batasan tulangan 𝜌 min = 𝜌 min =

√𝑓 ′ 𝑐 4.𝑓𝑦

√25

= 4.240 = 0,0052

1,4 1,4 = = 0,0058 𝑓𝑦 240

As min = ρmin . bw . d = 0,0052 . 200 . 342 = 355,68 mm2 0,85. 𝛽. 𝑓𝑐 ′ 600 𝐴𝑚𝑎𝑘𝑠 = 0,75 . ( . ) 𝑏𝑤. 𝑑 𝑓𝑦 600 + 𝑓𝑦 𝐴𝑚𝑎𝑘𝑠 = 0,75 . (

0,85.0,85.25 600 . ) 200.342 240 600 + 240

𝐴𝑚𝑎𝑘𝑠 = 2757,756 𝑚𝑚2 𝑀𝑢 𝑎 = 0,85. 𝑓𝑐 ′ . 𝑎. 𝑏 (𝑑 − ) ∅ 2 29,468 . 106 𝑎 = 0,85.25. 𝑎. 1062,5 (342 − ) 0,8 2 𝑎 36,83.106 = 22567,5 𝑎 (342 − ) 2 𝑎2 − 684 𝑎 + 3263,985 = 0

𝑅𝑢𝑚𝑢𝑠 𝑎𝑏𝑐 =

−𝑏 ± √𝑏 2 − 4𝑎𝑐 2𝑎

𝑎12 =

684 ± √(−684)2 − 4. (1). (3263,985) 2. (1)

𝑎1 = 679,194 𝑚𝑚 𝑎2 = 4,805 𝑚𝑚 Ambil a yang terkecil a = 4,805 mm 𝑐=

𝑎 4,805 𝑚𝑚 = = 5,653 < ℎ𝑓 = 120 𝑚𝑚 (𝑎𝑠𝑢𝑚𝑠𝑖 𝑏𝑒𝑛𝑎𝑟) 𝛽1 0.85

Cc = Ts 0,85. 𝑓𝑐 ′ . 𝑐. 𝑏𝑒𝑓𝑓 0,85.25.5,653.1062,5 𝐴𝑠 = = = 531,808 𝑚𝑚2 𝑓𝑦 240 Dipakai 3 D 16 (As = 603,2 mm2)