Jawaban Tugas03 - BETON I

Soal 1: Tentukan nilai momen nominal (Mn) balok T seperti pada gambar! Digunakan beton mutu fc’ =20 MPa dan mutu baja fy = 240 MPa dengan tulangan (As) adalah 12φ25 = 5.890 mm2.

bw = 400 mm. hf = 100 mm. be = 1.200 mm. d = 500 mm. fc’ = 20 MPa = 20 N/mm2. fy = 240 MPa = 240 N/mm2. Penyelesaian: Anggap persyaratan lebar efektif dipenuhi, periksa posisi sumbu: C = 0,85 . fc’ . be . a C = 0,85 . (20 N/mm2) . (1.200 mm) . a. C = (20.400 N/m) . a. Keseimbangan gaya horizontal: C = T = As . fy C = (5.890 mm2) . (240 N/mm2). C = 1.413.600 N. Maka: (20.400 N/mm) . a = 1.413.600 N → a = (1.413.600 N) / (20.400 N/mm). Maka, (20.400 N/mm) . a = 1.413.600 N → a = 69,294 mm.

a = 69,294 mm < hf = 100 mm → Analisis Balok Persegi. Mn = As . fy . (d – a/2) Mn = (5.890 mm2) . (240 N/mm2) . (500 mm – (69,294 mm /2)). Mn = 657.823.000,8 Nm.

Soal 2: Suatu balok T, lebar flens 800 mm, tebal 160 mm dan lebar balok 400 mm dicor monolit dengan badan balok. Balok tersebut memikul momen akibat beban mati sebesar 60 tm dan beban hidup sebesar 80 tm. Jarak garis sumbu tulangan dari bidang atas adalah 950 mm. Mutu beton fc’= 25 MPa dan baja fy = 400 MPa. Hitunglah luas tulangan balok T tsb.! Penyelesaian: bw = 400 mm. be = 800 mm. hf = 160 mm. MDL = 60 tm. MLL = 80 tm. d = 950 mm. fc’ = 25 MPa = 25 N/mm2 → β1 = 0,85. fy = 400 MPa = 400 N/mm2. Menghitung Momen berfaktor (Mu): Mu = 1,2.MDL + 1,6.MLL Mu = 1,2.(60 tm) + 1,6.(80 tm). Mu = 200 tm. Menghitung Momen Nominal (Mn): φ = 0,80 → Lentur. Mn = Mu / φ Mn = 200 tm / 0,80. Mn = 250 tm.

Menghitung nilai Mf dan bandingkan dengan nilai Mn: Asf = {0,85 . fc’ . hf . (be - bw)} / fy Asf = {0,85 . (25 N/mm2) . (160 mm) . (800 mm - 400 mm)} / (400 N/mm2). Asf = (1.360.000 N) / (400 N/mm2). Asf = 3.400 mm2. Mf = Asf . fy . (d – hf/2) Mf = (3.400 mm2) . (400 N/mm2) . (950 mm – 160 mm/2). Mf = 1.183.200.000 Nmm. Mf = 118,320 tm < Mn = 250 tm → Analisis Balok T. Menentukan Momen pada Web: Mw = Mn - Mf Mw = (250 tm) – (118,320 tm) Mw = 131,680 tm. Menentukan nilai k yang diperlukan:

k = 1− 1−

2 . Mw 0,85 . fc' . bw . d 2

k = 1− 1−

2 . (131,680 tm) . (10 7 ) = 0,1896. 0,85 . (25 N/mm 2 ).(400 mm) . (950 mm) 2

Menentukan nilai Asw yang diperlukan: Asw = Mw / {fy . ( d - k/2)} Asw = {(131,680 tm) . (107)} / {(400 N/mm2) . ( 950 mm - 0,1896/2)} Asw = 3.465,609 mm2. Menentukan nilai Asf yang diperlukan: Asf = 3.400 mm2. Luas Total Tulangan Tarik: As = Asf + Asw As = (3.400 mm2) + (3.465,609 mm2) = 6.865,609 mm2.

Pilih tulangan dengan syarat: Ast ≥ As Diambil 9 buah tulangan ulir dengan diameter 32 mm. Ast = 9 . (1/4 . π .(32 mm)2). Ast = 7.238,2295 mm2 > As = 6.865,609 mm2 → OKE! Periksa Pembatasan Luas Tulangan Maksimum:  0,85 . fc' . 1   600  !b =  ⋅  fy    600 + fy   0,85 . (25 N/mm 2 ) . (0,85)    600  ⋅   = 0,0271 !b =  2 2  (400 N/mm )    600 + (400 N/mm ) 

ρf = Asf / (bw . d) ρf = (3.400 mm2) / (400 mm × 950 mm). ρf = 0,00895. ρb = (bw / be) . ( !b +ρf) ρb = (400 mm / 800 mm) . (0,0271 + 0,00895) ρb = 0,0180 Asb = !b . bw . d Asb = 0,0271 . (400 mm) . (950 mm). Asb = 10.298 mm2. Asmaks = Asmaks (web) + Asmaks (flens) Asmaks = (0,75 . Asb) + (0,75 . Asf) Asmaks = (0,75 . 10.298 mm2) + (0,75 . 3.400 mm2) Asmaks = 0,75 . (10.298 mm2 + 3.400 mm2) Asmaks = 10.273,5 mm2.

Periksa Pembatasan Luas Tulangan Minimum:

ρmin = 1,4 / fy ρmin = 1,4 / (400 N/mm2) ρmin = 0,0035. Asmin = ρmin . be . d Asmin = 0,0035 . (800 mm) . (950 mm). Asmin = 2.660 mm2. (Asmin = 2.660 mm2) < (Ast = 7.238,2295 mm2) < (Asmaks = 10.273,5 mm2) → OKE! Syarat daktilitas sudah terpenuhi. Sketsa Penulangan Balok T:

Soal 3: Sebuah balok pada struktur portal dengan panjang bentang bersih 5 m mempunyai hasil analisa gaya lintang seperti pada gambar di bawah ini, rencanakan dan gambarlah tulangan geser balok tersebut! Data: a. fyh = 350 MPa, fc’ = 28 MPa., bw = 30 cm, d = 45 cm, Nu = 2,50 ton (tekan). b. fyh = 400 MPa, fc’ = 35 MPa., bw = 35 cm, d = 50 cm, Nu = 2,00 ton (tarik)

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Penyelesaian: a. fyh = 350 MPa, fc’ = 28 MPa., bw = 30 cm, d = 45 cm, Nu = 2,50 ton (tekan). Tahap 1: Gaya geser pada muka tumpuan: Ru = +15,50 ton = +155,0 kN. Gaya geser pada jarak d dari muka tumpuan: Vu =

2,50 m − 0,45 m x (+15,50 ton) = +12,71 ton. 2,50 m

Vu = +127,1 kN. Tahap 2: Kekuatan geser yang diberikan beton: fc’ = 28 Mpa = 28 N/mm2. Nu = +2,50 ton = +25,0 kN = +25.000 N (tekan).

bw = 30 cm = 300 mm Æ b = bw = 30 cm = 300 mm d = 45 cm = 450 mm. Asumsi: 1 lapis. d = 0,9 . h h = d / 0,9 h = (450 mm) / 0,9 h = 500 mm. Ag = b . h Ag = luas penampang kotor Ag = 300 mm × 500 mm Ag = 150.000 mm2.

Nu   1  Vc =  fc'  ⋅ bw ⋅ d ⋅ 1 +  6   14 ⋅ Ag    (+25.000 N) 1   Vc =  (28 N/mm2 )  ⋅ (300 mm) ⋅ (450 mm) ⋅ 1 + 6   14 ⋅ (300 mm × 500 mm)  Vc = 120.476,1758 N = 120,5 kN.

φ.Vc = 0,75 x 120,5 = 90,375 kN. ½ . φ.Vc = 45,1875 kN.

Karena Vu > φ.Vc maka diperlukan tulangan sengkang.

Tahap 3: Penulangan geser pada daerah yang perlu tulangan geser: Kekuatan geser yang diberikan tulangan baja: fyh = 350 Mpa = 350 N/mm2. fc’ = 28 Mpa = 28 N/mm2. bw = 30 cm = 300 mm. d = 45 cm = 450 mm. Vu = +127,1 kN.

φ.Vc = 90,375 kN. 39s = Vu − 39c = (+127,1 kN) − (90,375 kN) = 36,725 kN. 1  1  3⋅ fc' ⋅ bw ⋅ d  = 0,75 ⋅  (28 N/mm 2 ) ⋅ (300 mm) ⋅ (450 mm) . 3  3  1  3⋅ fc' ⋅ bw ⋅ d  = 178.588,2135 N = 178,60 kN. 3  φ.Vs < φ.(1/3.√fc’.bw.d) Dicoba φ = 8 mm ( Luas satu kaki = 50,286 mm2 ). Jadi Av = 2 x 50,286 = 100,572 mm2. Jarak spasi tulangan sengkang (α α = 90o): Untuk sengkang vertikal: S=

S=

Av ⋅ f yh ⋅ d 39s (100,572 mm 2 ) ⋅ (350 N/mm 2 ) ⋅ (450 mm) (36,725 × 103 N)

S = 431,316 mm

Dipakai S = 200 mm. Jadi dipakai tulangan geser φ8-200. Penulangan geser pada daerah yang cukup tulangan geser minimum: fyh = 350 Mpa = 350 N/mm2. fc’ = 28 Mpa = 28 N/mm2. bw = 30 cm = 300 mm. d = 45 cm = 450 mm. Syarat tulangan geser minimum jika: 1 ⋅ 3 ⋅ Vc < Vu ≤ 3 ⋅ Vc 2 1  Karena 39s < 3 ⋅  3 fc' ⋅ bw ⋅ d  maka: Smax =

d 450 = = 225 mm 2 2

Dipakai S = 200 mm. Av =

bw ⋅ S (300 mm) ⋅ (200 mm) = = 57,143 mm 2 2 3 ⋅ fy 3 ⋅ (350 N/mm )

Luas 1 kaki tulangan sengkang = (57,143 mm2)/2 = 28,5715 mm2. Dipakai φs = 8 mm (As = 50,286 mm2).

Batasan daerah tulangan sengkang: Ru = +155,0 kN. Ln = 5 m = 5.000 mm.

φ.Vc = 90,375 kN. ½ . φ.Vc = 45,1875 kN. x1 =

x2 =

Ru − 39c (155 kN) − (90,375 kN) ⋅ (0,5 ⋅ Ln) = ⋅ (2.500 mm) = 1.042,34 mm. Ru 155 kN Ru −

1 ⋅ 39c (155 kN) − (45,1875 kN) 2 ⋅ (0,5 ⋅ Ln) = ⋅ (2.500 mm) = 1.771,17 mm. Ru (155 kN)

Daerah perlu tulangan sengkang: x1 = 1.042,34 mm Daerah tulangan sengkang minimum: x2 – x1 = (1.771,17 mm) – (1.042,34 mm) = 728,83 mm. Daerah tidak perlu tulangan sengkang: (0,5.Ln) – x2 = (2.500 mm) – (1.771,17 mm) = 728,83 mm.

b. fyh = 400 MPa, fc’ = 35 MPa., bw = 35 cm, d = 50 cm, Nu = 2,00 ton (tarik) Tahap 1: Gaya geser pada muka tumpuan: Ru = +15,50 ton = +155,0 kN. Gaya geser pada jarak d dari muka tumpuan: Vu =

2,50 m − 0,50 m x (+15,50 ton) = +12,4 ton. 2,50 m

Vu = +124 kN. Tahap 2: Kekuatan geser yang diberikan beton: fc’ = 35 Mpa = 35 N/mm2. Nu = -2,00 ton = -20,0 kN = -20.000 N (tarik).

bw = 35 cm = 350 mm Æ b = bw = 35 cm = 350 mm d = 50 cm = 500 mm. Asumsi: 1 lapis.

d = 0,9 . h h = d / 0,9 h = (500 mm) / 0,9 h = 555,556 mm. Ag = b . h Ag = luas penampang kotor. Ag = 350 mm × 555,556 mm Ag = 194.444,444 mm2.

 0,3 ⋅ Nu  1  Vc =  fc'  ⋅ bw ⋅ d ⋅ 1 +  Ag  6     0,3 ⋅ (−20.000 N) 1   Vc =  (35 N/mm2 )  ⋅ (350 mm) ⋅ (500 mm) ⋅ 1 + 6   (350 mm × 555,556 mm)  Vc = 167.227,8552 N = 167,228 kN.

φ.Vc = 0,75 x (167,228 kN) = 125,421 kN. ½ . φ.Vc = 62,7105 kN. Karena ½ . φ.Vc < Vu < φ.Vc maka cukup tulangan geser minimum.

Tahap 3: Kekuatan geser yang diberikan tulangan baja: fyh = 400 Mpa = 400 N/mm2. fc’ = 35 Mpa = 35 N/mm2. bw = 35 cm = 350 mm. d = 50 cm = 500 mm. Vu = +124 kN.

φ.Vc = 125,421 kN. 39s = Vu − 39c = (+124 kN) − (125,421 kN) = -1,421 kN. 1  1  3 ⋅  fc' ⋅ bw ⋅ d  = 0,75 ⋅  (35 N/mm2 ) ⋅ (350 mm) ⋅ (500 mm) . 3  3  1  3 ⋅  fc' ⋅ bw ⋅ d  = 258.828,4905 N = 258,829 kN. 3  φ.Vs < φ.(1/3.√fc’.bw.d)

Penulangan geser pada daerah yang cukup tulangan geser minimum: fyh = 400 Mpa = 400 N/mm2. fc’ = 35 Mpa = 35 N/mm2. bw = 35 cm = 350 mm. d = 50 cm = 500 mm. Syarat tulangan geser minimum jika: 1 ⋅ 3 ⋅ Vc < Vu ≤ 3 ⋅ Vc 2 1  Karena 39s < 3 ⋅  3 fc' ⋅ bw ⋅ d  maka: Smax =

d 500 = = 250 mm 2 2

Dipakai S = 200 mm. Av =

bw ⋅ S (350 mm) ⋅ (200 mm) = = 58,3333 mm 2 2 3 ⋅ fy 3 ⋅ (400 N/mm )

Luas 1 kaki tulangan sengkang = (58,3333 mm2)/2 = 29,1667 mm2. Dipakai φs = 8 mm (As = 50,286 mm2).

Batasan daerah tulangan sengkang: Ru = +155,0 kN. Ln = 5 m = 5.000 mm.

φ.Vc = 125,421 kN. ½ . φ.Vc = 62,7105 kN. x1 =

x2 =

Ru − 39c (155 kN) − (125,421 kN) ⋅ (0,5 ⋅ Ln) = ⋅ (2.500 mm) = 477,081 mm. Ru 155 kN Ru −

1 ⋅ 39c (155 kN) − (62,7105 kN) 2 ⋅ (0,5 ⋅ Ln) = ⋅ (2.500 mm) = 1.488,540 mm. Ru (155 kN)

Karena

1 ⋅ 3 ⋅ Vc < Vu ≤ 3 ⋅ Vc , maka cukup tulangan geser minimum. 2

Daerah tulangan sengkang minimum: x = x2 = 1.488,540 mm. Daerah tidak perlu tulangan sengkang: (0,5.Ln) – x2 = (2.500 mm) – (1.488,540 mm) = 1.011,46 mm.