Bab IV Perencanaan Tangga

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BAB IV

PERENCANAAN TANGGA

1. Data Bahan Tangga

a. Ruang Yang Dipakai

 Panjang : 400 cm

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 Tinggi antar lantai : 400 cm  Tinggi bordes : 200 cm b. Perhitungan Ukuran Tangga

 Perencanaan tinggi optride (O) : 18 cm

 Jumlah Optride : 200/18 = 11 buah  Lebar Antride (A) : 28 cm

Syarat A + 2 x O = 60 s/d 65 28 + 2 x 18 = 64 …… OK c. Perhitungan Tangga dan Bordes

 Jumlah Antride : 11 – 1 = 10 buah  Lebar Bordes : 100 cm  Panjang Tangga : 300 cm

 Sudut kemiringan tangga ( α ) = tan-1 ( 200/290 ) = 27°  Tebal bordes = 14 cm  0,15 m  Tebal tangga = 15 cm  0,15 m

 Tebal rata-rata = (i / 2) . sin  = (0,28/2) . sin 27 = 0,070 m

 Tebal total = 0,15+ 0,070 = 0,220 m

2. Analisa Pembebanan Pelat Tangga a. Beban Mati

 berat sendiri pelat tangga : 0,22 x 2400 = 528 kg/m2

 berat spesi : 2 x 21 = 42 kg/m2  berat tegel : 2 x 24 = 48 kg/m2  berat handrill : = 10 kg/m2 + WD = 628 kg/m2 = 6,28 kN/m² b. Beban Hidup (WL) = 250 kg/m2

c. Beban Berfaktor (Wu)

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B C

A 3. Analisa Statika

Untuk mempermudah perhitungan momen dan gaya – gaya dalam, tumpuan pada tangga dianggap jepit dan tumpuan pada balok bordes dianggap sendi dengan menganggap tangga dan bordes sebagai elemen.

- Stiffness Factor (SF) KBC = 4 EI L = 4 EI 1,00

=

4,00 EI KBA = 4 EI L = 4 EI 3,00 = 1,33 EI - Distributie Factor (DF) μ BC = 4 5,33= 0,75 μ CB : μ CB = 0,75 : 0,25 ( OK = 1 ) μ BA = 1,33 5,33= 0,25 Perhitungan Momen q1 = 1153,6 kg/m2 M BC = -M CB = 1/12 x q x L2 = 1/12 x 1153,6 x 1,002 = 96,133 Kgm q2= q cos.27 = 1153,6 cos.27 = 1294,715 Kg/m 2 M BA = M AB = 1/12 x ql x L2 = 1/12 x 1294,715 x 3,002 = 971 Kgm 1,00 _3,00

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METODE CROSS JOINT C B A ANGGOTA CB BC BA AB FD -0.75 -0.25 MF 96.133 -96.133 971 -971 MD 0 -656.15025 -218.71675 0 MI -328.075125 0 0 -109.358375 MD 0 0 0 0 JUMLAH -231.94 -752.28 752.28 -1080.358

Hasil Perhitungan Cross Tangga

BIDANG MOMEN

752.28 231.94 1080.358 231.94 752.28 q1 = 1153,6 kg/m2 q2 = 1294,715 KGM 752.28 1080.358

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150 cm 4. Perhitungan pelat tangga

Direncanakan :

- Mutu baja ( fy ) = 400 mpa - Mutu beton ( fc’) = 35 mpa - Bj beton = 2400 kg/m2 - Dimensi pelat Ly = 355 cm Lx = 150 cm 𝜷 =𝑳𝒚 𝑳𝒙= 355 150= 2,36 Beban Berfaktor (qu) qu = 1,2. WD + 1,6. WL = 1,2 x 628 + 1,6 x 250 = 1153,6 kg/m2

 Perhitungan momen

(PBI`71 bab 13 tabel 13.3.1) Mulx = 0,001 × qu × Lx2 × x1 = 0,001 × 1153,6 × 1,52 × 97 = 252 kgm ≈ 2,52 ×106 Nmm Muly = 0,001 × qu × Lx2 × x2 = 0,001 × 1153,6 × 1,52 × 36 = 94 kgm ≈ 9,4 ×105 Nmm Muty = - 0,001 × qu × Lx2 × x3 = - 0,001 × 1153,6 × 1,52 ×122 = - 317 kgm ≈ - 3,17 × 106 Nmm Mlx = 252 kgm Mly = 94 kgm Mty = 317 kgm

 Perhitungan tulangan pelat tangga

Direncanakan - Tulangan Utama = 10 mm - Decking (d’) = 20 mm - Tebal Pelat (h) = 150 mm 150 cm 355 cm m 355 cm m

Tangga

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- b = 1000

- fy = 400 mpa

- fc’ = 35 mpa

f’c = 35 MPa > 30 MPa, maka β = 0,88

Dx = h – d - ½ Ø = 150 – 20 - ½.10 = 125 mm Dy = h – d – Ø - ½ Ø = 150 – 20 – 10 - ½.10 = 115 mm 0035 , 0 400 4 , 1 4 , 1 min    fy  (SNI 03-2847-2002 pasal 12/12.5) 𝝆𝒃 = [𝟎, 𝟖𝟓 𝒙 𝒇 ′_{𝒄 𝒙 𝜷} 𝒇𝒚 ] 𝒙 [ 𝟔𝟎𝟎 𝟔𝟎𝟎 𝒙 𝒇𝒚] 039 , 0 400 600 600 400 88 , 0 35 85 , 0 _           x x x b  ρmax = 0,75 x ρb = 0,75 x 0,039 = 0,029

 Penulangan lapangan arah X

Mu = 252 kgm ≈ 2,52 ×106 _Nmm

𝑴𝒏 =

𝑴𝒖 ∅ Mn =  Mulx = 8 , 0 10 52 , 2  6 = 3,15 × 106 Nmm 𝑹𝒏 = 𝑴𝒏 𝒃 . 𝒅𝒙𝟐 2 dx b Mn RnLx   = ₂ 6 125 1000 10 15 , 3   = 0,20

𝒎 =

𝒇𝒚 𝟎,𝟖𝟓 .𝒇𝒄′

=

400 0,85 .35

=

13,45 𝝆𝟎 = 𝟏 𝒎 𝒙 [𝟏 − √𝟏 − 𝟐.𝒎 𝒙𝑹𝒏 𝒇𝒚 ] h d’ dy _dx

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      _ _    400 20 , 0 45 , 13 2 1 1 45 , 13 1 0 x lx  = 0,0005 < max

< min, mka dipakai min = 0,0035 As = min . b . dx

As = 0,0035 . 1000 . 125 = 437,5 mm2 Ø 10 – 175 ( 449 mm2 )

 Penulangan lapangan arah Y Mu = 94 kgm ≈ 9,4 ×105_Nmm Mn =  Muly = 8 , 0 10 9,4 5 = 11,75×105 𝑹𝒏 = 𝑴𝒏 𝒃 . 𝒅𝒚𝟐 2 dy b Mn RnLy   = ₂ 5 115 1000 10 75 , 11   = 0,0888

𝒎 =

𝒇𝒚 𝟎,𝟖𝟓 .𝒇𝒄′

=

400 0,85 .35

=

13,45 𝝆𝟎 = 𝟏 𝒎 𝒙 [𝟏 − √𝟏 − 𝟐.𝒎 𝒙𝑹𝒏 𝒇𝒚 ]       _ _    400 888 , 0 45 , 13 2 1 1 45 , 13 1 0 x x ly  = 0,00225 < max

< min, mka dipakai min = 0,0035 As = min . b . dy

As = 0,0035 . 1000 . 115 = 402,5 mm2 Ø 10 – 175 ( 449 mm2 )

 Penulangan Tumpuan Arah Y Mu = 317 kgm ≈ 3,17 × 106 _Nmm Mn =  Muty = 8 , 0 10 17 , 3  6 = 3,9625×106

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𝑹𝒏 = 𝑴𝒏 𝒃 . 𝒅𝒚𝟐 2 dy b Mn Rntx   = ₂ 6 115 1000 10 9625 , 3   = 0,3

𝑚 =

𝑓𝑦 0,85 .𝑓𝑐′

=

400 0,85 .35

=

13,45 𝝆𝟎= 𝟏 𝒎 𝒙 [𝟏 − √𝟏 − 𝟐.𝒎 𝒙𝑹𝒏 𝒇𝒚 ]       _ _    400 3 , 0 45 , 13 2 1 1 45 , 13 1 0 x ty  = 0,00075 < max

>min, mka dipakai  = 0,0035 As =  . b . dy

As = 0,0035 . 1000 . 115 = 402,5 mm2 _{Ø 10 – 175 ( 449 mm}2₎

Dari perhitungan tulangan di atas di peroleh Tulangan lapangan arah X = Ø 10 – 175 ( 449 mm2 ) Tulangan lapangan arah Y = Ø 10 – 175 ( 449 mm2 ) Tulangan tumpuan arah Y = Ø 10 – 175 ( 449 mm2 )

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5. Perhitungan Plat Bodres



Perhitungan Balok Bordes

h = 30 cm b = 15 cm d’ = 4 cm f’c = 35 MPa fy = 400 MPa Ø tulangan = 12 mm Sengkang = 8 mm

d = h – d’ – Ø sengkang – ½.Ø tul . utama = 300 – 40 – 8 – ½ . 12 = 245 mm

Pembebanan pada balok bordes ( tabel 2 . 1 PPIUG 1989 ) a. Akibat beban mati (qD)

 Berat sendiri = 0,15 x 0,30 x 2400 = 108 kg/m  Berat dinding = 0,15 x 3 x 1700 = 765 kg/m  Berat plat bordes = 0,15 x 0,3 x 2400 = 360 kg/m +

qD = 1233 kg/m b. Beban hidup = 3 x 250 = 1050 kg/m

c. Beban Berfaktor (Wu) Wu = 1,2 . WD + 1,6 . WL = 1,2 . 1233 + 1,6 . 1050 = 3159,6 kg/m Analisa statika M lapangan = 1/12 . Wu . l2 = 1/12 ×3159,6 × 32 = 2369,7 kgm M tumpuan = -1/24 . Wu. L2 = -1/24 × 3159,6 × 32

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= -1184,85 kgm



Perhitungan Tulangan Lentur Balok Bordes

 Tulangan Lapangan

Mu = 2369,7 kgm = 23,697 × 106 _N/mm

f’c = 35 MPa > 30 MPa, maka β = 0,88               fy x fy x xfc b 600 600 ' 85 , 0   039 , 0 400 600 600 400 88 , 0 35 85 , 0 _           x x x b  0035 , 0 400 4 , 1 4 , 1    fy MIN  MAX = 0,75 x ρb = 0,75 x 0,039 = 0,0292 45 , 13 35 85 , 0 400 ' 85 , 0    x fc fy m Mn=  Mu = 8 , 0 10 × 23,697 6 = 29,621×106 2 bxd Mn Rn = ₂ 6 245 150 10 29,621×  = 3,289          fy mxRn x m 2 1 1 1



0,0087 400 289 , 3 45 , 13 2 1 1 45 , 13 1 _          x x x < max

> min, mka dipakai min = 0,0087 As = min . b. d As = 0,0087 x 150 x 245 = 319,725 mm2 Di pakai tulangan Ø 12 mm = ¼ . π x 122 = 113,04 mm2 Jumlah tulangan = 319,725 113,04 = 2,828 ≈ 3 buah As yang timbul = 3. ¼ .π. d2 = 339,12 mm2_{> As ... Aman !}

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Dipakai tulangan 3 Ø 12 mm

 Tulangan Tumpuan

Mu = 1184,85 Kg m = 11,85 × 106_N/mm

f’c = 35 MPa > 30 MPa, maka β = 0,88

              fy x fy x xfc b 600 600 ' 85 , 0   039 , 0 400 600 600 400 88 , 0 35 85 , 0 _           x x x b  0035 , 0 400 4 , 1 4 , 1 _ _  fy MIN  MAX = 0,75 x ρb = 0,75 x 0,039 = 0,0292 45 , 13 35 85 , 0 400 ' 85 , 0    x fc fy m Mn=  Mu = 8 , 0 10 11,85x 6 = 14,8125 x 106 Nmm 2 bxd Mn Rn = ₂ 6 245 150 10 14,8125 x x = 1,645          fy mxRn x m 2 1 1 1



0,00423 400 645 , 1 45 , 13 2 1 1 45 , 13 1 _          x x x < max

 > min, mka dipakai min = 0,00423 As = min . b. dx As = 0,00423 x 150 x 245 = 155,45 mm Di pakai tulangan Ø 12 mm = ¼ . π x 122 = 113,04 mm2 Jumlah tulangan = 155,45 113,04 = 1,37 ≈ 2 buah As yang timbul = 2. ¼ .π. d2 = 226,08 mm2 > As ... Aman !

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Dipakai tulangan 2 Ø 12 mm



Perhitungan tulangan geser balok bordes Wu = 3159,6 kg/m = 31,596 kN/m Vu dukungan = ½ . Wu . L Vu dukungan = ½ ×31,596 ×3 = 47,394 kN = 47394 N Maka Vu / = 47,394 / 0,8 = 59,242 kN = 59242 N d = h – deking - Av – ½ As = 300 – 40 – 8 – ½.12 = 245 mm Vc = 1/ 6 . b.d. √f'c . = 1/6 . 150 . 245. √35 . = 36.235,98 N Ø Vc = 0,75 . Vc = 0,75 x 36.235,98 = 27177 N

Vu >  Vc , maka dibutuhkan tulangan geser Vs = Vu -  Vc = 59242 – 27177 = 32065 N Vs = 32065/0,6 = 53441.67 N Vs = 𝐴𝑣.𝑓𝑦.𝑑 𝑠 ≤ ( 2 3) √𝑓𝑐′. 𝑏𝑤. 𝑑 (2 3) √35 × 150 × 245 = 144944 𝑁

Vs < Vs max Ukuran penampang memenuhi

Perencanaan tulangan geser

Digunakan tulangan geser 2 kaki diameter 8 mm Av = 2 x ¼  d2

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Vs = 𝑨𝒗.𝒇𝒚.𝒅 𝒔

s =

𝑨𝒗.𝒇𝒚.𝒅 𝑽𝒔 s = 𝟏𝟎𝟎,𝟒𝟖×𝟒𝟎𝟎.×𝟐𝟒𝟓 𝟓𝟑𝟒𝟒𝟏.𝟔𝟕 s = 184,257 mm S max = ½ d = 122,5 mm

Digunakan sengkang diameter 8 mm jarak 120 mm Kontrol kecukupan tulangan geser :

Vu = 47394 N Vs = 𝐴𝑣.𝑓𝑦.𝑑 𝑠 = 100,48 ×400×245₁₂₀ = 82058,67 N Vn = Vc + Vs = 36.235,98 + 82058,67 = 118294,65 N  Vn = 0,6 x 118294,65 = 70976,78 N Vu = 47394 N  Vn ≥ Vu 70976,78 N > 47394 N Okey

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Tulangan Lapangan Tulangan Tumpuan

6. Analisa Pembebanan Pelat Bordes

a. Beban Mati

 berat sendiri pelat tangga : 0,15 x 2400 = 528 kg/m2  berat spesi : 2 x 21 = 42 kg/m2  berat tegel : 2 x 24 = 48 kg/m2  berat handrill : = 10 kg/m2 + WD = 628 kg/m2 = 6,28 kN/m² b. Beban Hidup (WL) = 250 kg/m2

c. Beban Berfaktor (Wu)

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100 cm m

100 cm m 7. Perhitungan kebutuhan tulangan pelat bordes

Direncanakan :

- Mutu baja ( fy ) = 400 mpa - Mutu beton ( fc’) = 35 mpa - Bj beton = 2400 kg/m2 - Dimensi pelat Ly = 300 cm Lx = 100 cm 𝜷 =𝑳𝒚 𝑳_𝒙 = 300 100= 3

 Perhitungan momen

(PBI`71 bab 13 tabel 13.3.1) Mulx = 0,001 x qu x Lx2 x x1 = 0,001 x 1153,6 x 12 x 63 = 72,6768 kgm ≈ 726768 Nmm Muly = 0,001 x qu x Lx2 x x2 = 0,001 x 1153,6 x 12 x 13 = 15 kgm ≈ 15000 Nmm Muty = - 0,001 x qu x Lx2 x x3 = - 0,001 x 1153,6 x 12 x125 = -144,2 kgm ≈ - 1442000 Nmm Mlx = 726768 Nmm Mly = 15000 Nmm Mty = 1442000 Nmm

 Perhitungan tulangan pelat lantai

Direncanakan - Tulangan Utama = 10 mm - Decking (d’) = 20 mm - Tebal Pelat (h) = 150 mm - b = 1000 - fy = 400 mpa - fc’ = 35 mpa

f’c = 35 MPa > 30 MPa, maka β = 0,88

h d’ dy _dx 300 cm 300 cm

Tangga

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Dx = h – d - ½ Ø = 150 – 20 - ½.10 = 125 mm Dy = h – d – Ø - ½ Ø 150 – 20 – 10 - ½.10 = 115 mm 0035 , 0 400 4 , 1 4 , 1 min    fy  (SNI 03-2847-2002 pasal 12/12.5) 039 , 0 400 600 600 400 88 , 0 35 85 , 0 _           x x x b  MAX = 0,75 x ρb = 0,75 x 0,039 = 0,0292 45 , 13 35 85 , 0 400 ' 85 , 0    x fc fy m

 Penulangan lapangan arah X

Mu = 72,6768 kgm ≈ 726768 _Nmm

𝑴𝒏 =

𝑴𝒖 ∅ Mn =  Mulx = 8 , 0 726768 = 908460 Nmm 𝑹𝒏 = 𝑴𝒏 𝒃 . 𝒅𝒙𝟐 2 dx b Mn RnLx   = ₂ 125 1000 908460 x = 0,058

𝒎 =

𝒇𝒚 𝟎,𝟖𝟓 .𝒇𝒄′

=

400 0,85 .35

=

13,45 𝝆𝟎 = 𝟏 𝒎 𝒙 [𝟏 − √𝟏 − 𝟐.𝒎 𝒙𝑹𝒏 𝒇𝒚 ]       _ _    400 058 , 0 45 , 13 2 1 1 45 , 13 1 0 x lx  = 0,00146 < max

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< min, mka dipakai min = 0,0035 As = min . b . dx

As = 0,0035 . 1000 . 125 = 437,5 mm2 _{Ø 10 – 175 ( 449 mm}2₎

 Penulangan lapangan arah Y Mu = 15 kgm ≈ 15000 Nmm Mn =  Muly = 8 , 0 15000 = 18750 Nmm 𝑹𝒏 = 𝑴𝒏 𝒃 . 𝒅𝒚𝟐 2 dy b Mn RnLy   = ₂ 115 1000 18750 x = 0,0014

𝒎 =

𝒇𝒚 𝟎,𝟖𝟓 .𝒇𝒄′

=

400 0,85 .35

= 13,45

𝝆𝟎 ₌𝟏 𝒎 𝒙 [𝟏 − √𝟏 − 𝟐.𝒎 𝒙𝑹𝒏 𝒇𝒚 ]       _ _    400 0014 , 0 45 , 13 2 1 1 45 , 13 1 0 x ly  = 0,000035 < max

< min, mka dipakai min = 0,0035 As = min . b . dy

As = 0,0035 . 1000 . 115 = 402.5 mm2 Ø 10 – 175 ( 449 mm2 )

 Penulangan Tumpuan Arah Y Mu = 144,2 kgm ≈ 1442000 Nmm Mn =  Muty = 8 , 0 1442000 = 1802500 Nmm 𝑹𝒏 = 𝑴𝒏 𝒃 . 𝒅𝒚𝟐

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2 dy b Mn Rntx   = ₂ 115 1000 1802500 x = 0,136

𝑚 =

𝑓𝑦 0,85 .𝑓𝑐′

=

400 0,85 .35

=

13,45 𝝆𝟎= 𝟏 𝒎 𝒙 [𝟏 − √𝟏 − 𝟐.𝒎 𝒙𝑹𝒏 𝒇𝒚 ]            400 136 , 0 45 , 13 2 1 1 45 , 13 1 0 x ty  = 0,00034 < max

>min, mka dipakai min = 0,0035 As =  . b . dy

As = 0,0035 . 1000 . 115 = 402.5 mm2 Ø 10 – 175 ( 449 mm2 )

Dari perhitungan tulangan di atas di peroleh Tulangan lapangan arah X = Ø 10 – 175 ( 449 mm2 ) Tulangan lapangan arah Y = Ø 10 – 175 ( 449 mm2 ) Tulangan tumpuan arah Y = Ø 10 – 175 ( 449 mm2 )

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Detail Penulangan Plat Tangga

300 100

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