8. Penulangan Tiang Pancang PERENCANAAN DAN PEMBAHASAN
85 t
ijin
= 0,65 √f’c = 0,65√300 = 11,258
t t
ijin
► OK tebal pile cap cukup
4. 4. 8. 4. Penulangan Tiang Pancang
Kondisi I Dua Tumpuan q = 0,25 x π x 0,35
2
x 2400 = 231 kgm L = 7 m
4a
2
+ 4aL - L
2
= 0 4a
2
+ 4a7 - 7
2
= 0 4a
2
+ 28a – 49 = 0 a
2
+ 7a - 12,25 = 0 a = 1,449 m
M
1
= M
2
= ½ . q . a
2
= ½ . 231 . 1,449
2
= 242,504 kgm Dmaks = ½ . q . L = ½ . 231 . 7 = 808,5 kg
Universitas Sumatera Utara
86 Kondisi II Satu Tumpuan
2a
2
+ 4aL - L
2
= 0 2a
2
+ 4a7 - 7
2
= 0 2a
2
+ 28a – 49 = 0 a
2
+ 14a – 24,5 = 0 a = 1,573 m
M
1
= M
2
= ½ . q . a
2
= ½ . 231 . 1,573
2
= 285,785 kgm
D
1
=
�.�
2
−2��� 2�−�
D
1
=
231.7
2
−2.1,573.231.7 27−1,573
D
1
= 574,159 kg Dari kedua kondisi, dipilih : M = 242,504 kgm; D = 808,5 kg
Dimensi tiang = Ø 30 cm D = 30 cm
Universitas Sumatera Utara
87 Berat jenis beton = 2400 kgm
3
F’c = 30 MPa Fy = 400 MPa
H = 300 mm P = 40 mm
Øtulangan = 20 mm Øsengkang = 8 mm
d = h – p – Øsengkang – Øtulangan = 300 – 40 – 8 – 10 = 242 mm d’ = p – Øsengkang – Øtulangan = 40 – 8 – 10 = 22 mm
Tulangan Memanjang Tiang Pancang
M = 242,504 kgm = 2,425 kNm ��
�. �
2
= 2,425
0,3. 0,242
2
= 138,026
�� �.�
2
= �. �. ��1 − 0,588. �.
�� �
′
�
1,38026 = �. 0,8.400 �1 − 0,588. �.
400 30
�
1,38026 = 320 � − 25088�
2
� = 0,00431 Pemeriksaan syarat rasio penulangan ρmin ρ ρmax
Universitas Sumatera Utara
88 ρmin = 1,4fy = 1,4400 = 0,0035
ρmin =
�.450 600+��
�
0,85.�′� 400
=
0,85.450 600+400
�
0,85.30 400
= 0,446 Maka dipakai ρ = 0,00431
As = ρ.b.d.
= 0,00431 . 300 . 242
= 312,906 mm
2
Digunakan tulangan 2D22 As = 760 mm
2
Cek Terhadap Tekuk
Dianggap kedua ujung sendi, diperoleh harga k = 1
r = 0,3 . h = 0,3 . 300 = 90 mm
K=
�.�� �
=
1.7000 90
= 77,78
K 20, maka kelangsingan diperhitngkan
Ig = 164 . π . D
4
= 164 . π . 300
4
= 397767857 mm
4
Ec = 4700. √f’c = 4700. √30 = 23500 MPa
EI =
��.��.0,4 1+�.�
=
23500.397767857.0,4 1+0,85.0,242
= 3,101 � 10
12
��
Universitas Sumatera Utara
89 Pu = 52,344 ton = 523,44 kN
Pcr =
�
2
. ��
�.��
2
=
�
2
. 3,101.10
12
1.7000
2
= 625107,872 �
�� = ��
�1 − �� � ����
= 1
�1 − 523,44
0,65.625107,872� = 1,0013
Mn = Cs . Mu = 1,0013 . 0,242504 = 2428192,55 Nmm
ea = MnPu = 2428192,55523440 = 4,639 mm
e = ea + h2 –d’ = 4,639 + 3002 – 22 = 132,639 mm
cb =
600.� ��+600
=
600.242 400+600
= 145,2 ��
a =
�� 0,85∗�
′
�∗�
=
523440 0,85.30.300
= 68,424 ��
ab = 0,85 . cb = 0,85 . 145,2 = 123,42
Jika a ab, dipakai rumus
�� = ��
′
= ��� − � + ��
2. �
�
. �
��� − �
′
= 523440132,639
− 242 + 132,639
2. 0,85.30.300 300242
− 22
= −867,263 ��
2
Digunakan As min 1 Ag = 0 ,01.14.π.300
2
= 707,143 mm
Universitas Sumatera Utara
90 Digunakan tulangan 4 D 22 As
terpasang
= 1521 mm
2
Penulangan Geser Tiang Pancang
Vu = 2760,7 kg = 27607 N
Vn = VuØ = 276070,6 = 46011,667 N
Vc = 16 . √f’c . b . d = 16 . √30 . 300 . 242 = 66274,429 N
Periksa v
u
fv
c
Vu = Vub.d = 27607300.242 = 0,380 MPa
Vc = Vcb.d = 16 . √f’c = 16 . √30 = 0,913
fVc = 0,6 . 0,913 = 0,548
v
u
fv
c
Þ dipakai tulangan praktis
Digunakan tulangan sengkang ø8 – 200.
4. 4. 8. 5. Penulangan Pile Cap