Perhitungan Struktur Rumah Tinggal

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PERHITUNGAN STRUKTUR

RUMAH TINGGAL

a

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Rencana Desain Pembangunan RUMAH TINGGAL

Luas Tanah : 297 m2

Luas Total Bangunan : 280 m2

Luas Lantai 1 : 116 m2

Building Coverage Ratio (BCR) : 40 %

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RENCANA PERHITUNGAN STRUKTUR DAN KONSTRUKSI 1. Preliminary Design Elemen Struktur

Pendimensian Pelat

Asumsi :

Fc’ : 30 Mpa Fy’ : 240 Mpa

a. Menentukan Panjang Bersih

Ly = 5000 Lx = 3000

β = Ly = 5000 = 1.6667 Pelat dua arah

Lx 3000

b. Menentukan tebal Pelat

Berdasarkan RSNI 2002 h = Lny x 0.8 + Fy / 1500 36 + 9β h = 5000 x 0.8 + 240 / 1500 36 + (9 * 1.6667) h = 94.117 mm 0.8 + Fy / 1500 _{36 + 9β}

Tebal pelat yang diambil = h = 100 mm = 10 cm

Pendimensian Balok

Balok l = 5000 mm

Berdasarkan RSNI 2002 (untuk balok 2 tumpuan)

h = 1/16 l

h = 1/16 * 5000 mm = 312.5 mm

5000

30

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Diambil h = 350 mm (asumsi)

B = 0.5 x h = 175 mm Digunakan Balok Ukuran 175 / 350 mm

Pendimensian Kolom

Kolom A ( dari lantai 3 ke atap ) Beban mati atap :

Berat sendiri pelat (0.1 x 2400) = 240 kg / m2

Utilitas = 25 kg / m2 Berat finishing (0.02 x 2400) = 48 kg / m2 a Dead Load (DL) = 313 kg / m2 q terfaktor = 1.2 DL = 1.2 x 313 = 375.6 kg / m2 Beban kolom = 3 x 5 x 375.6 = 5634 kg = 56340 N Σ DI = 56340 N A > _{40 % x 0.85 x ƒc’}Σ DI A > _{40 % x 0.85 x 30’}56340 A > 5523.529 mm2 Asumsi _{b = h A = h}2 h = √5523.529 = 74.32 mm b = h > 74.32 mm

Diambil asumsi ukuran kolom 150 x 150 mm

Kolom B ( dari lantai 2 ke lantai 3 ) Beban kolom :

Berat sendiri pelat ( 0.1 x 2400 x 5 x 3 ) = 3600 kg / m2 Berat sendiri balok 175 / 350 ( 0.175 x 0.35 x 2400 x 5 ) = 735 kg / m2 Berat sendiri balok 175 / 350 ( 0.175 x 0.35 x 2400 x 3 ) = 441 kg / m2

Berat finishing (0.02 x 2400) = 48 kg / m2

Berat kolom atas ( 0.15 x 0. 15 x 2400 x 3 ) = 162 kg / m2

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DL = 5028 kg / m2 Σ DI = 5028 kg / m2 = 50280 N A > _{40 % x 0.85 x ƒc’}Σ DI A > _{40 % x 0.85 x 30’}50280 A > 4929.41 mm2 Asumsi _{b = h A = h}2 h = √4929.41 = 70.20 mm b = h > 70.20 mm

Kolom C ( dari lantai 1 ke lantai 2 ) Beban kolom :

Berat sendiri pelat ( 0.1 x 2400 x 5 x 3 ) = 3600 kg / m2 Berat sendiri balok 175 / 350 ( 0.175 x 0.35 x 2400 x 5 ) = 735 kg / m2

Berat sendiri balok 175 / 350 ( 0.175 x 0.35 x 2400 x 3 ) = 441 kg / m2

Berat finishing (0.02 x 2400) = 48 kg / m2

Berat kolom atas lantai 3 ( 0.15 x 0. 15 x 2400 x 3 ) = 162 kg / m2 Berat kolom atas lantai 2 ( 0.15 x 0. 15 x 2400 x 4 ) = 216 kg / m2

Berat spesi (2 x 21) = 42 kg / m2 a DL = 5244 kg / m2 Σ DI = 5244 kg / m2 = 52440 N A > _{40 % x 0.85 x ƒc’}Σ DI A > _{40 % x 0.85 x 30’}52440 A > 5141.176 mm2

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Asumsi _{b = h A = h}2

h = √5141.176 = 71.70 mm b = h > 71.70 mm

2. Pembebanan Pelat Lantai

1. Beban Mati / Dead Load (DL)

Berat sendiri pelat t = 100 mm (0.1 x 2400) = 240 kg / m2

Berat finishing (0.1 x 24) = 2.4 kg / m2

DL = 242.4 kg / m2

2. Beban hidup / Living Load (LL)

Untuk bangunan rumah tinggal LL = 150 kg / m2

3. Perhitungan Momen Rencana Terfaktor

Berdasarkan RSNI 2002

qu = 1.2 DL + 1.6 LL

= 1.2 (242.4) + 1.6 (150)

= 290.88 + 240 = 530.88 kg / m2

4. Perhitungan Momen Lentur Pelat

Diambil panel terbesar

Lx = 3000 = 3 m ; Ly = 5000 = 5 m

β = Ly = 5000 = 1.6667 Pelat dua arah

Lx 3000 Dimana nilai x : Mlx = _{- Mtx 35} Mly = _{- Mty 35} Mlx = - Mtx = 0.001 * qu * Lx2 * x = 0.001 * 530.88 * (3)2 * 35

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= 167.2272 Kg.m Mlx = - Mtx = 0.001 * qu * Lx2 * x = 0.001 * 530.88 * (3)2 * 35 = 167.2272 Kg.m 5. Penulangan Pelat Arah x : Fy = 240 Mpa Tulangan lapis (jd) I = d1 = 100 - 20 = 80 mm As = Mn = 1672272 = 120.968 mm2 Φ x ƒy x jd 0.8 x 240 x 0.9 x 80 Pakai Φ 8 : As = ¼ π D2 = ¼ x 3.14 x 0.82 = 0.5 cm2 S = _1.209680.5 X 100 = 4.333 cm Ambil S = 40 cm Digunakan Penulangan Φ 8 – 400 mm

Cek terhadap ρmin pelat !

ρmin = 0.002 Amin = ρmin x b x h = 0.002 x 100 x 9 = 1.8 cm2 = 180 mm2

S = 0.5 X 100 = 27.7777 cm 1.8 Digunakan Penulangan Φ 8 – 250 mm Arah y : fy = 240 Mpa Tulangan lapis (jd) II = d2 = 80 - 8 = 72 mm As = Mn = 1672272 = 134.41 mm2 Φ x ƒy x jd 0.8 x 240 x 0.9 x 72 Pakai Φ 8 : As = ¼ π D2 = ¼ x 3.14 x 0.82 = 0.5 cm2

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S = _1.34410.5 X 100 = 37.199 cm Ambil S = 35 cm Digunakan Penulangan Φ 8 – 350 mm PENULANGAN PORTAL Penulangan Balok Diketahui : Dimensi Balok = 175 x 350 mm Dimensi Kolom = 150 x 150 mm Selimut Beton = 1 / 10 h = 1 / 10 (350) = 35 mm d = h – sel. beton = 350 – 35 = 315 mm

Penulangan Tarik Balok Terlentur

(dari : Struktur Beton Bertulang Istimawan)

Tumpuan Mu = 814.4 kg-m = 8144000 Nmm As = Mu = 8144000 = 149.67 mm2 Φ x γ x d x ƒy 0.8 x 0.9 x 315 x 240 Asumsi tulangan 2 D 12 2 x (1/4) x π x (122) = 226.1946 mm2 Kontrol : Ratio penulangan : ρ = As = 226.1946 = 4.10 x 10-3 b x d 175 x 315 0.75 ρ = 0.75 x 0.85 x ƒc’ x β1 600 ƒy 600 + ƒy 0.75 ρ = 0.75 x 0.85 x 30 x 0.85 600 240 600 + 240 0.75 ρ = 0.04821 ρmin = 1.4 = 1.4 = 0.00583 ƒy 240

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Syarat ρ < 0.75 x ρb

0.00410 < 0.04821 OK Kedalaman balok tegangan beton tekan :

a = As x ƒy = 226.1946 x 240 = 12.1650 mm

(0.85 x ƒc’ x b) (0.85 x 30 x 175)

Panjang lengan momen kopel dalam :

Z = D - 1 ₂ a = 315 - 1 ₂ 12.1650 = 308.9175 mm

Momen tahanan (momen dalam) ideal :

Mn = As x fy x z = 226.1946 x 240 x 308.9175 = 16770112 Nmm Mr = Φ Mn = 0.8 x 16770112 = 13416089 Nmm

Mu < Mr = 8144000 Nmm < 13416089 Nmm OK Pada tumpuan digunakan tulangan 2 D 12

Lapangan Mu = 1407.3 Kg-m = 14073000 Nmm As = Mu = 14073000 = 258.542 mm2 Φ x γ x d x ƒy 0.8 x 0.9 x 315 x 240 Asumsi tulangan 3 D 12 3 x (1/4) x π x (122) = 339.12 mm2 Kontrol : Ratio penulangan : ρ = As = 339.12 = 6.15 x 10-3 b x d 175 x 315 0.75 ρ = 0.75 x 0.85 x ƒc’ x β1 600 ƒy 600 + ƒy 0.75 ρ = 0.75 x 0.85 x 30 x 0.85 ₂₄₀ _{600 + 240}600 0.75 ρ = 0.04821 ρmin = 1.4 _ƒy = ₂₄₀1.4 = 0.00583 Syarat ρmin < ρ < 0.75 x ρb 0.00583 < 0.00615 < 0.04821 OK

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Kedalaman balok tegangan beton tekan :

a = As x ƒy = 339.12 x 240 = 18.2383 mm

(0.85 x ƒc’ x b) (0.85 x 30 x 175)

Panjang lengan momen kopel dalam :

Z = D - 1 a = 315 - 1 18.2383 = 305.8808 mm

2 2

Momen tahanan (momen dalam) ideal :

Mn = As x fy x z = 339.12 x 240 x 305.8808 = 24895271 Nmm Mr = Φ Mn = 0.8 x 16770112 = 19916216 Nmm

Mu < Mr = 14073000 Nmm < 19916216 Nmm _OK Pada tumpuan digunakan tulangan 3 D 12

Penulangan Geser Balok Terlentur (dari : Struktur Beton Bertulang Istimawan)

Tumpuan

Vu = 2891.4 kg = 28914 N

Vc = 1/6 √ƒc’ * bw * d = 1/6 √30 * 175 * 315 = 50318.1 N Pemeriksaan perlu tidaknya sengkang :

½ Φ * Vc = ½ * 0.6 * 50318.1 = 15095.43 N Vu = 28914 N > ½ Φ Vc = 15095.43 N Tulangan sengkang diperlukan Perencanaan Sengkang : 7560000

Vs = Vu - Vc = 28914 - 50318.1 = -2128.1

Φ 0.6

Asumsi dengan tulangan Φ 8 Av = 2 As = 100 mm2

s = Av . ƒy . d = 100 * 240 * 315 = 3552.464 mm

bw -2128.1

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b 175 Pada tumpuan digunakan tulangan Φ 8 – 400

Kontrol : Vs = Av . ƒy . d = 100 * 240 * 315 = 30240 N s 250 Φ (Vc + Vs) = 0.6 (50318.1 + 30240) = 48334.86 N Vu = 28914 N < Φ (Vc + Vs) = 48334.86 N OK Lapangan L = 5 m = 5000 mm Vu = 722.85 kg _{= 7228.5 N (jarak ¼ x 5000 = 1250 mm)} Vc = 1/6 √ƒc’ * bw * d = 1/6 √30 * 175 * 315 = 50318.1 N Pemeriksaan perlu tidaknya sengkang :

½ Φ * Vc = ½ * 0.6 * 50318.1 = 15095.43 N Vu = 722.85 N < ½ Φ Vc = 15095.43 N Tulangan sengkang tidak diperlukan Penulangan Torsi Balok Terlentur

(dari : Struktur Beton Bertulang Istimawan) 5.477

Tumpuan

Tu = 0.28 kg-m = 2800 N Σ x2y = 1752 * 500 = 15312500 m m3 Pemeriksaan perlu tidaknya sengkang :

Φ (1/4 √ƒc’) Σ x2y = 0.6 (1/4 √30) 15312500 = 12579984 Nmm Tu = 2800 Nmm < Φ (1/4 √ƒc’) Σ x2y = 12579984 Nmm

Tulangan sengkang tidak diperlukan Cat : Tidak diperlukan tulangan Torsi

PENULANGAN KOLOM

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Fc’ = 30 Mpa B = 150 mm H = 150 mm Fy = 240 Mpa Pu = 3412.19 kg = 341219 N Mu = 814.4 kg-m = 8144000 Nmm Pn = Pu = 341219 = 524952.3 N Φ 0.65 Mn = Mu = 8144000 = 12529230 N Φ 0.65 Pn x 0.7 = 524952.3 x 0.7 = 0.8367 ƒc . b . h 0.65 30 * 150 * 150 0.65 Mn x 0.7 = 12529230 x 0.7 = 0.1332 ƒc . b . h2 0.65 30 * 150 * 1502 0.65 G = h – (2.40 + 2.10 + 13) _h = 150 – (2.40 + 2.10 + 13) ₁₅₀ = 0.2466 m = ƒy = 240 = 16 0.85 * ƒc’ 0.85 * 30 ρtperlu = 0.1307 ₁₆ = 0.00816 Ast = ρtperlu . b . h = 0.00816 . 150 . 150 = 183.6 mm2 Dipakai tulangan 4 D 10 = 314 m2

Penulangan Geser Kolom (tulangan Tumpuan, tulangan Lapangan)

Gaya Geser

Data ETABS :

Gaya geser lapangan = 40.24 kg Gaya geser tumpuan atas = 40.24 kg Gaya geser tumpuan bawah = 40.24 kg

bw . d Ag 6 =

(

1 + 0.3 * 341219

)

(

√30

)

150 . 130 (150)2 6 = 98767.5 N Φ Vc = 0.75 * 98767.5 = 74075.625 N Φ Vc = 74075.625 > Vu kritis = 402.4 N OK Tulangan geser minimum

Selimut beton = 1 / 10 h = 20 mm d = 150 – 20 = 130 mm

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Pakai sengkang Φ 10 = Av = 2 Φ 10 = 157 mm2

s = Av . ƒy . d _ΦVc = 157 * 240 * 130 _74075.625 = 66.127 mm

s = 66.127 mm > ½ 130 = 65 mm Maka dipakai tulangan Φ 10 - 65

150 . 130 = 98767.5 N Φ Vc = 0.75 * 98767.5 = 74075.625 N Φ Vc = 74075.625 > Vu kritis = 402.4 N OK Tulangan geser minimum

Selimut beton = 1 / 10 h = 20 mm d = 150 – 20 = 130 mm Pakai sengkang Φ 10 = Av = 2 Φ 10 = 157 mm2 s = Av . ƒy . d = 157 * 240 * 130 = 66.127 mm ΦVc 74075.625 s = 66.127 mm > ½ 130 = 65 mm Maka dipakai tulangan Φ 10 - 65

PENULANGAN PONDASI Penulangan Pondasi

Diketahui :

Kolom 150 / 150 mm

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P = 218 N H = 330 N M = 99 Nmm

Tebal pelat pondasi 250 mm, cover 75 mm Tebal Tanah diatas pondasi = 1500 mm fc’ = 30 Mpa fy = 240 Mpa σ t = 0.1 Mpa γ tanah = 18 KN / m3 = 1800 N / m3 γ beton = 24 KN / m3 = 2400 N / m3 1. Pembebanan Pondasi : Berat tanah = 1.5 x 1800 = 2700 N Berat pondasi = 0.25 x 2400 = 600 N Berat kolom = 52440 N P total = 55740 N (R) A perlu = P tot = 55740 = 557400 mm2 σ t 0.1

diambil Pondasi B x L = 1200 x 1200 mm2 , A pakai = 1440000 mm2 σ r = _{1200 x 1200}55740 = 0.0387 Mpa < σ r = 0.1 Mpa OK 2. Syarat Geser e = M + H . h = 99 + 330 * 1750 = 10.362 mm R 55740 e = 10.362 mm < L / 6 = 200 mm, maka : q max = R + 6 . M B . L B . L2 q max = 55740 + 6 . 99 = 0.039 N / mm2 12002 12003 q min = R - 6 . M B . L B . L2

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q max = 55740 ₁₂₀₀2 - ₁₂₀₀6 . 99 3 = 0.0387 N / mm2

Geser satu arah d = 150 – 75 = 75 mm s1 = s2 = ½ (1200 – 150) – 75 = 450 mm Vc = 1/6 √ƒc’ B . d = 1/6 √30 . 1200 . 75 = 0.912 . 1200 . 75 = 82080 N ΦVc = 0.75 Vc = 61560 N Va = q max . B.s = 0.039 . 1200 . 450 = 21060 N Va = 21060 N < ΦVc = 61560 N OK

Geser dua arah

Bo = (C1 + d + C2 + d) = (150 + 75 + 150 + 75) = 450 mm

Bc = C_C1 = 150 = 1

2 150

α s = 40 (untuk kolom interior)

Vu = q max (B . L – (C1 + d)(C2 + d)) = 0.039 (12002 – 2252) = 54185.625 N Vc = 1

(

1 + 2

)

√ƒc’ . bo . d = 92340 N 6 Bc Vc = 1 √ƒc’ . bo . d = 61560 N 3 Maka Vc = 92340 N Φ Vc = 0.75 Vc = 69255 N Φ Vc = 69255 N > Vu = 54185.625 N

OK

3. Tulangan angker / stek As min = 0.005 . A kolom

= 0.005 . ( 150 x 150) = 112.5 mm2

As stek = 112.5 mm2 db stek perlu = √112.5 / 4 * ¼ π = 5.985

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4. Transfer gaya pada dasar kolom Kuat tumpu dasar kolom

P = 218 N

Φ Pnb 2 = Φ (0.85 . ƒcp . Ap) = 17850000 N P _{= 218 N < Φ Pnb 2 = 17850000 N OK} Kuat tumpu permukaan pondasi

A2 = 12002 = 1440000 mm2 A1 = 1502 = 22500 mm2 √ A2 / A1 = 1440000 / 22500 = 64 > 2 P _{= 218 N < Φ Pnb 2 = 17850000 N OK} 5. Penulangan pondasi Mu = q x (s + s) + b x 2 x s 2 3 Mu = 0.039 x (450 + 450) + 1200 x 2 x 450 = 360017.55 N 2 3 As = _{Φ . γ d . ƒy}Mu = 26.667 mm2 As min = 0.0018 * 250 * 1200 = 540 mm2 A perlu = 450 mm2_{6 D 10 = 471.238} PERENCANAAN TANGGA

Tangga direncanakan menggunakan konstruksi kayu (pengembangan selanjutnya akan dilakukan dengan memperhatikan kondisi finansial yang tersedia).

+ 6.20 Lantai 3

+ 3.20 Lantai 2

+ 0.00 Lantai 1 + 4.70

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Syarat tangga :

- Sudut Kemiringan demi kenyamanan adalah 25° hingga 35° - Tinggi tanjakan (Optrade) : 16 – 20 cm

- Lebar tanjakan (Antrade) : 25 – 30 cm Rencana pendimensian tangga :

Asumsi Lebar tanjakan yang diambil adalah : 25 cm

Jumlah tanjakan (n) = L / 25 = 200 / 25 = 8 buah tanjakan

Dengan begitu total jumlah tanjakan hingga lantai berikutnya adalah 18 buah (seperti pada gambar rencana tangga).

Tinggi tanjakan Tangga lantai 1 ke lantai 2 (o) = 320 / 18 = 17.77 cm Tinggi tanjakan Tangga lantai 2 ke lantai 3 (o) = 300 / 18

= 16.66 cm Jumlah Injakan = 18 buah

Syarat :

Lebar langkah = 20 + A < 45 – 65 cm = 20 + 25 < 45 – 65 cm

= 45 _{< 45 – 65 cm OK}

Pembebanan Pada Pelat Tangga 1. Beban Mati (DL)

Berat sendiri pelat (0.15 x 2400 x sec 37) = 450.769 kg/ m2

1.00 2.00 Lantai atas Lantai bawah 1 2 3 4 5 6 7 8 17 16 15 14 13 12 11 10 9 18

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Berat sendiri anak tangga = (0.15 x 0.166 x 2400) = 59.76 kg/ m2 DL = 510.529 kg/ m2

2. Beban Super Dead (SDL)

B. S Ubin (t = 0.5 cm) 0.005 x 1800 = 12 kg/ m2 B. S Spesi (t = 3 cm) 0.03 x 2100 = 63 kg/ m2 B. S Sandaran = 20 kg/ m2 Pasir (t = 3 cm) 0.03 x 1800 = 54 kg/ m2 SDL = 149 kg/ m2 3. Beban Hidup (LL) LL = 150 kg/ m2 qu = 1.2 DL + 1.2 SDL + 1.6 LL = 1.2 (510.529) + 1.2 (149) + 1.6 (150) = 1031.434 kg/ m2 Pembebanan Bordes 1. Beban Mati (DL)

Berat sendiri pelat (0.15 x 2400) = 360 kg/ m2 DL = 360 kg/ m2 2. Beban Super Dead (SDL)

B. S Ubin (t = 0.5 cm) 0.005 x 1800 = 12 kg/ m2 B. S Spesi (t = 3 cm) 0.03 x 2100 = 63 kg/ m2 B. S Sandaran = 20 kg/ m2 Pasir (t = 3 cm) 0.03 x 1800 = 54 kg/ m2 SDL = 149 kg/ m2 4. Beban Hidup (LL) LL = 150 kg/ m2 qu = 1.2 DL + 1.2 SDL + 1.6 LL = 1.2 (360) + 1.2 (149) + 1.6 (150) = 850.8 kg/ m2

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Dari hasil ETAB didapat data sebagai berikut : Pada pelat tangga

Momen M11 = 257.717 kg . m Momen M22 = 627.443 kg . m Pada pelat bordes

Momen M11 = 522.904 kg . m Momen M22 = 764.004 kg . m Penulangan Tangga : Penulangan Arah X : Mu = 2577170 Nmm d = dx – 20 = 150 – 20 = 130 Nmm A perlu = Mu = 2577170 = 114.775 mm2 ΣΦ. Jd . ƒy 0.8 (0.9 * 130) 240 As min = 0.0018 . 1200 . 130 = 280.8 mm2 As perlu = 114.775 mm2 < As min = 280.8 mm2 Diameter tulangan Φ 10 = 79 mm2 280.8 / 79 = 3.55 4 buah Dengan jarak : 1200 – (20 + 8 + (3.10) + 8 + 20) = 278.5 300 mm 4 Dipakai tulangan (4 Φ 10 – 300) As = 236 mm2 Penulangan Arah Y : Mu = 6274430 Nmm d = dx – 20 = 150 – 20 = 130 Nmm A perlu = Mu = 6274430 = 279.310 mm2 ΣΦ. Jd . ƒy 0.8 (0.9 * 130) 240 As min = 0.0018 . 1800 . 130 = 421.2 mm2

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As perlu = 279.310 mm2 < As min = 421.2 mm2 Diameter tulangan Φ 10 = 79 mm2 421.2 / 79 = 5.33 6 buah Dengan jarak : 1800 – (20 + 8 + (6.10) + 8 + 20) = 280.6607 250 mm 6 Dipakai tulangan (4 Φ 10 – 250) As = 471 mm2 Penulangan Bordes Penulangan Arah X : Mu = 5229040 Nmm d = dx – 20 = 150 – 20 = 130 Nmm A perlu = _{ΣΦ. Jd . ƒy}Mu = _{0.8 (0.9 * 130) 240}5229040 = 232.774 mm2 As min = 0.0018 . 1800 . 130 = 421.2 mm2 As perlu = 232.774 mm2 < As min = 421.2 mm2 Diameter tulangan Φ 10 = 79 mm2 421.2 / 79 = 5.33 6 buah Dengan jarak : 1800 – (20 + 8 + (6.10) + 8 + 20) = 280.6607 250 mm 6 Dipakai tulangan (4 Φ 10 – 250) As = 471 mm2 Penulangan Arah Y : 22464 Mu = 760040 Nmm d = dx – 20 = 150 – 20 = 130 Nmm A perlu = Mu = 760040 = 340.101 mm2 ΣΦ. Jd . ƒy 0.8 (0.9 * 130) 240 As min = 0.0018 . 1200 . 130 = 280.8 mm2

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As perlu = 340.101 mm2 < As min = 280.8 mm2 Diameter tulangan Φ 10 = 79 mm2 280.8 / 79 = 3.55 4 buah Dengan jarak : 1200 – (20 + 8 + (4.10) + 8 + 20) = 276 250 mm 4 Dipakai tulangan (4 Φ 10 – 250) As = 471 mm2