STRUKTUR BETON
Farhan Fadhilah
2204101010139
01
Diketahui Fy = 400 Mpa f’c = 30 Mpa b = 400 mm d = 600 mm h = 700 mm Ditanya:
Cb …?
N…?
As…?
Nt…?
β₁ sesuai standar terbaru SNI 2847-2019 β₁ = 0,85 – 0,05
= 0,85 – 0,05 0,846
• Cb = d = . 600 = 360 mm
• � = β1 . Cb = 0,846 (360) = 304,56 mm
• NDC = 0,846 . f’c . � . b
= 0,846 (30) (304,56) (400) = 3.106.512 N 310,6512 Ton
• As.fy = NDC As =
= 7.766,28 mm2 Butuh 8D36 - Tabel A - 4
• Mn = NDC (d - )
= 3.121.200 (600 - ) = 1.395.176.400 Nmm Kapasitas = 139,5 tm
• b =
=
= 0,323
Kondisi Seimbang
Cek
• b =
= 0,0323
• NT = As . Fy
= 7.766,28 (400) = 3.106.512 N = 310,6512 t
Kondisi Under
= b = 0,0323
As. fy = =
=
= =
= =
Kondisi Over
• As = 91690,9 Butuh 9D36
NDC = NT
0,85 . f’c . . b = As . fs � 0,85 . f’c . . b = As . Es . εs � 0,85 . f’c ( . c) b = As . Es . �₁ . 0,003
0,85 (30) (400) (0,846) c = 9.160,9 (200.000) () 0,003 8.692,2 c2 = -5.496.540 c + 3.267.924.000
8.692,2 c2 + 5.496.540 c – 3.267.924.000 = 0
• X12 = C12= C12=
C1 = 376,939 C2= -1013,908
• � = �₁ . c
= 0,846 (376,939) = 318,890 mm
• εs = ( 0,003 = () 0,003 = 0,00177
• fs = Es . εs
= 200.000 (0,00177) = 355,061
• Nd = 0,85 f’c . . b �
= 0,85 (30) (319,988) (400) = 3.263.883,447 N
• Nt = As . fs
= 9160,9 (355,061) = 3.252.682,739 N
• Mn = Ndc
= 3.252.682,739 ( 600 - ) = 1.432.985.644 Nm Kapasitas 143,298 tm
Seimbang Under Over
Cb (mm) 360 235,915 376,939
As (mm2) 7.766,28 5.089,4 9160,9
� (mm) 304,56 199,584 318,890
Mn ( tm ) 139,08 101,83 143,298
02
Buktikan bahwa balok pada gambar telah cukup memenuhi persyaratan SK SNI T-15-1991-03.
Beban mati merata = 12 kN/m (diluar berat sendiri), beban hidup merata = 12kN/m, beban hidup terpusat= 54 kN (ditengah bentang). Mutu bahan ; f’c = 30 MPa, fy = 400 Mpa. Pembuktian dilakukan dengan cara yang ditimbulkan oleh beban rencana (beban terfaktor) Mu. Jika Mr Mu maka balok akan memenuhi persyaratan.
• b = = = 0,0325
• As = 0,5 x Pb x b x d = (0,5)(0,0325)(300)(650) = 3169,968 mm2
• =
=
= 0,0162
Dari tabel A6 didapat 0,75 Pb = 0,0244 , karena 0,162 < 0,0244
Dapat dipastikan tulangan baja sudah luluh.
min = = 0,0035 0,0162
• � = =
= 165,75 mm
• z = (d - ) = (650 - ) = 569,125
Berdasarkan pada tulangan baja
• Mn = As fy z
= (3169,968)(400)( 569,125 x 10 -6 ) = 721.643.385,9 x 10 -6
= 721,643 KNm
Menghitung Mu
o Berat sendiri balok = (0,7)(0,3)(23) = 4,83 o Beban mati = 12 KN/m
o Total beban mati merata = 16,83 KN/m
o Beban mati merata terfaktor = (1,2)(16,83) 20,196 KN/m o Beban hidup merata terfaktor = (1,6)(12) 19,2 KN/m o Beban hidup terpusat terfaktor = (1,6)(54) 86,4 KN/m
• Wu = 20,196 + 19,2 39,396 KN/m
• Pu = 86,4 KN/m
• Mu = Wu. + Pu . �
= (39,396) (8) + (86,4) (8) = 315,168 + 172,8
= 487,968 kNm
Mr = 577,314 > Mu = 487,968 ( Memenuhi Syarat )
03
Penyelesaian:
As = 4824 mm2
= = = 0,0322
(0,75 b = 0,0241 maks )
NT = NDC
0,85 f’c . � . b = As . fs 0,85 f’c . � . b = As . Es . εs 0,85 f’c b (β₁.c) = As . Es .( ) 0,003
0,85 (20) (300) (0,85) c = 4824.200.000( ) 0,003 4.335 c = 2.894.400 ( )
4.335 c2 + 2.894.400 c – 1.477.200.000 = 0 c12 = - 2.894.400
=
C1 = 333,460 C2 = - 1001,141 Hitung Mr dari balok dengan d= 500 mm, b=300 mm, As= 6D32 dan mutu bahan : kuat beton 20 Mpa, tulangan baja fy= 300 MPa
a = β₁.c = 0,85 (333,46)
= 283,441
εs = ( ) 0,003 = ( ) 0,003 = 0,001498
fs = Es . εs
= 200.000 (0,001498) = 299,6578
z = d - = 500 -
= 358,2794 mm
Mn = As . fs . z . 10-6
= 4824 (299,6578) (358,2794) = 517,910510 kNm
Mr = Mn
= (0,8) 517,91051 = 414,3284 kNm
NT = As . fs
= 4824 (299,6578) = 1.445.549,23
NDC = 0,85 . f’c . �. b
= 0,85 (20) (283,441) (300) = 1.445.549,1
04
Rencanakan suatu balok persegi beton bertulang untuk menahan momen beban kerja mati 100 kNm (termasuk berat sendiri ) ditambah beban momen guna hidup 200 kNm. Pertimbangkan arsitektur butuh lebar balok 320 mm,dan tinggi di cari .gunakan beton bahan f`c = 20 Mpa dan baja fy = 400 MpaDiketahui:
Beban kerja mati = 100 kNm Beban kerja hidup = 200 kNm b = 320 mm
f’c = 20 Mpa = 400 Mpa
Total momen rencana :
= 1,2 () + 1,6 ( )
=1,2 (100) + 1,6(200) = 440 kNm
Untuk = 20 Mpa dan = 400 Mpa, diperoleh nilai dari tabel : = 0,02168
= 0,01355 ( penampang terkendali tarik) = 4,09962 Mpa
= = = 107.327.021
Untuk b = 320 mm ,maka d = 579,1346 mm Sehingga
= b d
= 0,01355 x 320 x 579,1346 = 2511,1276
Gunakan 4D29, = 2642,1 (ditabel A-4 & A-38)
• � = = 194,272058 mm
• c = = = 228, 555 mm
• = = 0,3946 > 0,375
(penampang pada daerah transisi) = 0,65 + 0,2 [ - = 0,65 + 0,25 [ -
= 0,867 = )
= 0,867(2642,1)(400)(579,1346-194,272/2) = 441.645.812,2 Nmm
= 441,64 kNm >
441,64 > 440...(ok)
Kapasitas > beban
h perlu = 579,1346 + (29) + 10 + 40 =643,63 mm gunakan h = 650 mm
tinggi balok efektif actual
d = 650 -40 -10 - (24) = 585,5 > 579,1346 nilai akhir = = 0,0141
rasio = = 1,83… ( rasio baik )
05
b = 300 mm As’ = 3D16 603,2 mm2
d = 500 mm As = 4D28 2463 mm2
d’ = 50 mm ES = 200.000 MPa
fy = 300 MPa 1. fc’ = 20 MPa
2. fc’ = 35 MPa Untuk f’c = 20 MPa
Asumsikan bahwa tegangan baja tarik dan tekan sudah luluh
NT = As . fy
= 2463 (300)
= 738.900
NDS = A’s . fy
= 603,2 (300)
= 180.960
• =
=
= 109,4
• C =
=
= 128,70
• = 0,003 = 0,008654
= 0,003
= 0,001834
Mn = 0,85 f’c ab (d-d’)
= 0,85 (20) (109,4) (300) + 603,2 (300) (500- 50)
= 329.882.684
Untuk f’c = 35 MPa
� =
=
= 62,5142 mm
c =
=
= 77,178 mm
’ = 0,003
= 0,001056 0,001056 = 0,0015
• NDC + NDS = NT
0,85 f’c.a.b+As’ ( ( 0,003 . 200.000 – 0,85 f’c ) = As. Fy 0,85 (35) (β1.c) (300) + 603,2 ( ( 600 – 0,85 (35) ) = 2463 (300) 0,85 (35) (0,81) (300) c + 603,2 ( 600 ( - 0,85 (35) ) = 2463 (300)
7229,25 C2 + (-394925) c – 18096000 = 0 C2 -54,6288 - 2503,164 = 0 Menggunakan rumus abc
• X12 = C12 = C12 = C12 =
C1 = 84,31651 C2 = -29,68771
= β1 . c
= 0,81 (84,31651) = 68,29637
s = 0,003 ( β1 . ) = 0,01536 Fs = Es . = 3072,022 = 0,003 () = 0,001221 Fs’ = .
= 244,1978
NDC = 0,85 . f’c . a . b
= 0,85 (35) (68,29637) (300) = 609545,1
NDS = As’ (fs’ – 0,85 f’c)
= 603,2 (244,1978 – 0,85 (35)) = 129354,9
NDC + NDS = NT
609545,1 + 129354,9 = As . Fy 738900 = 738900 (ok)
