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q=
L1
L2
D
Tebal (L1)
= 2.2
γ
= 1537 kg/m3
ø
=
Tebal (L2)
= 2.5
γsat
= 1953 kg/m3
ø
=
c
= 0
ø
=
γ
= 1987 kg/m3
1400
kg/m2
m 29
o
m 30
o
kg/m2 31
o
¨ Keterangan : 1. Lapisan 1
: γ1 ø1
2. Lapisan 2
: γSat 2 ø2 γ2'
= 1537 kg/m3 =
29
= 1953 kg/m3 =
30 o = γSat 2 - γw = 1953 -
3. Lapisan 3
: γSat 3 ø2 γ3'
o
1000 =
953
kg/m3
987
kg/m3
= 1987 kg/m3 =
31 o = γSat 2 - γw = 1987 -
1000 =
¨ Menentukan koefisien tekanan tanah aktif dan pasif 1. Lapisan 1 (L1) Ka1 = Tan2 (45 - Ф/2)
Kp1
= Tan2 (45 + Ф/2)
= Tan2 ( 45 = 0.35
-
29 /2 )
= Tan2 ( 45 = 2.88
2. Lapisan 2 (L2) Ka2 = Tan2 (45 - Ф/2) = Tan2 ( 45 = 0.33
-
Kp2 30 /2 )
= Tan2 ( 45 = 0.32
-
Kp3 31 /2 )
29 /2 )
= Tan2 (45 + Ф/2) = Tan2 ( 45 = 3.00
3. Lapisan 3 (D) Ka3 = Tan2 (45 - Ф/2)
+
+
30 /2 )
= Tan2 (45 + Ф/2) = Tan2 ( 45 = 3.12
+
31 /2 )
¨ Menghitung tekanan tanah untuk setiap lapisan 1. Untuk kedalaman Z = 0.00 m σv0 = q =
1400
σh0 = P1 = =
kg/m2 q . Ka1
485.76
2. Untuk kedalaman Z
kg/m2 =
L1
=
2.20
m
σv1 = q + γ1 . L1 = = σh2 =
1400 + 1537.00 x 4781.40 kg/m2 P2 = σv1 . Ka2 = =
4781.40 1593.80
3. Untuk kedalaman Z σv2 = = =
2.20
=
x 0.33 2 kg/m L2
=
2.50
1400 + ( 1537.00 x 7163.90 kg/m2
2.20
m
q + γ1 . L1 + γ2' . L2
σ'h 3A = P3A= σv2 . Ka2 = 7163.90 = 2387.97 σh 3B = P3B = σv2 . Ka3
x 0.33 2 kg/m
7163.90 2293.16
x 0.32 2 kg/m
= =
¨ Menghitung kedalaman L3 P3B L3 =
)+(
953.00 x
2.50
)
L3
=
γ3' (Kp3 - Ka3) 2293.16 3.12 -
=
987.00 x ( = 0.83 m
0.32
)
¨ Menghitung luasan diagram tekanan tanah diatas dredge line A 1 = P 1 . L1 = 485.76 x 2.20 1 2
A2 =
1/2 . (P2 - P1 ). L1
3
A3 =
P 2 . L2
4
A4 =
1/2 .( P3A -P2 ). L2
5
A5 =
1/2 . P3B . L3
•
P
= = =
= 1068.6800 kg/m = 1/2 x 1108.04 x = 1218.84 kg/m = 1593.80 x 2.50 = 3984.50 kg/m = 1/2 x 794.17 x = 992.71 kg/m = 1/2 x 2293.16 x = 950.06 kg/m
2.20
2.5 0.83
A1 + A 2 + A 3 + A 4 + A 5 1068.68 8214.79
+ 1218.84 kg/m
+
3984.50
+
992.71
+
950.06
¨ Menghitung letak P dari L3 (z) / titik berat dari tekanan ACDE • ΣMc = P . Ž ΣMc ž = P (A • ΣMc = 1)(1/2. L1 + L2 + L3) + (A2)(1/3. L1+ L2 + L3) + (A3)(1/2. L2 + L3) + (A4)(1/3.L2+L3) + A5 . (2/3 .L3) = 1068.680 ( 1/2 2.20 + 2.50 + 0.83 ) + 1218.84 ( 1/3 2.20 + 2.50 + 3984.50 ( 1/2. 2.50 + 0.83 ) + 992.71 ( 1/3 2.50 + 0.83 + 950.06 (2/3. 0.83 ) = 19318.73 kg/m2 ΣMc 19318.73 ž = = = 2.35 m • P 8214.79 ¨ Menghitung P5 • P5 = [(q + γ1.L1 + γ2'.L2)Kp3]+ (γ3'.L3(Kp3 - Ka3)) + 953.00 x 2.50 )) x 3.12 ] =[( 1400 + 1537.00 x 2.20 0.83 ( 3.12 0.32 ) + 987.00 x 2293.16 = 22380.28 + = 24673.43 kg/m2
•
A1'
=
P5 γ3'(Kp-Ka)
=
24673.43 987.00 x ( 2.80
)
•
A2'
=
8.P γ3'(Kp-Ka)
=
65718.32 987.00 x ( 2.80
)
=
8.92
=
23.75
•
•
A3'
A4'
6.P(2.z.γ3'(Kp-Ka)+P5) γ3'2(Kp-Ka)2 49288.7 ((2 x 2.35 x 987.00 x ( 2.80 = 987.00 2 x ( 2.80 )2 1,857,693,919.54 = 7,658,973.72 = 242.55 =
P(6.z.P5 + 4.P) γ3'2(Kp-Ka)2 8214.79 (6 x 2.35 x 24673.43 + 4 x = 987.00 2 x ( 2.80 )2 3,129,887,134.32 = 7,658,973.72 = 408.66
) + 24673.43
=
8214.79 )
¨ Menghitung kedalaman L4 L44 + A1'.L43 - A2'.L42 - A3'.L4 - A4' = 0 L44 +
8.92
L43
-
23.75
L42 -
242.55
L4 -
408.66
=
0
Dengan cara coba-coba, maka L4 : Dianggap L4 (m) 5.64000 5.56350 5.56327
Hasil Persamaan 79.326 0.235 0.00307
¨ Menghitung P4 P4 = P5 + γ3' L4 (Kp-Ka)3 = 24673.43 + 987 = 40069.70 kg/m2 ¨ Menghitung P3 P3 = γ3' L4 (Kp-Ka) = 987 x 5.56 ( 2 = 15396.27 kg/m
x
5.56
2.80
Maka L4
(
2.80
)
¨ Menghitung L5 (P3 x L4) - 2.P L5 = P3 + P4 5.56 ) - 2 x ( 15396.27 x = 15396.27 + 40069.70 69224.01 = 55465.96 = 1.25 m
8214.79
)
=
5.56
m
¨ Kedalaman D aktual dan teoritis Dteorotis = L3 + L4 = =
0.83 6.39
+ m
5.56
Kedalaman dinaikkan sebesar 30% (1/3) dari D sebenarnya. Drencana
= 1.3Dteorotis = 1.3 x = 8.31
Drencana
= =
### 6.39 m
( 0.30 x 6.39 8.31 m
)
+
6.39
¨ Menghitung nilai z' (menentukan dimensi sheetpile) 2xP 0.5 Ö z' = γ3'(Kp-Ka)3 = =
16429.58 2767.49 2.44 m
0.5
¨ Menghitung momen maksimum Mmax
= P(z+z') - ((0.5.γ3'.z'2.(Kp-Ka)).(z'/3) = 8214.79 ( 2.35 + 2.44 ) - ( 0.5 x x 2.8039 x 0.8122 ) = 39334.26 6671.84 32662.42 = Kg.m
Digunakan tekanan ijin sebesar w = PZ-27
M = σijin
32662.42 17200000
162.3 x 10-5 m3
= =
172
987
MN/m2 =
0.0018990
x
2.44
2 17200000 kg/m
= 189.90 x 10-5 m3
skala = 5000 p1 = p2 =
0.097 0.319
p3a =
0.478
p3b = p5 =
0.459 4.935
p4 = p3 =
8.014 3.079
Mmax = 6.5324831 p = 1.6429581
950.06
.L2+L3) + A5 . (2/3 .L3) + 0.83 ) 0.83 )
L1 g'1 C f
= = = =
7.2 1554 0 28
m kg/m^3 kg/m^2 °
D g2 C f
= = = =
? 1965 5832 0
m kg/m^3 kg/m^2 °
l1= 1m l2= 6.2 m
g'2
=
965
a).Menghitung Koefisien tekanan Tanah - Lapis 1 Ka1 = Tan^2 = Tan^2 = 0.36 kg/m^3
- Lapis 2 Ka2 = = =
Tan^2 Tan^2 1.000
b).Menghitung tekanan Tanah - Lapis 1 s1 = ( g'1 = ( 1554 = 4039.531 - Lapis 2 s2 = = =
( g'1 ( 1554 11188.800
c).Menghitung L2 L2 = s2 g'2 ( Kp2 = 11188.800 965 ( = #DIV/0! m d).Hitung resultan gaya P di atas titik balik Lapis
Bentuk
1
2
∑
P=
jarak garis kerja P ke E z = ∑ M
= =
P #DIV/0! #DIV/0! #DIV/0! m
e).Menghitung L3 L3^3 + 1.5
L3
=
L3^2
L3^3
+
1.5
L3^3
+
#DIV/0!
2.2
L3^2
m
f).Menghitung s 7 s7 = g'2 ( Kp2 = 965 ( = 0.000 g).Menghitung kedalaman D Dteori = L2 + = ### + = ### m Dactual
= = =
1.3 1.4 ### ### m
h).Menghitung gaya anker F = P - 1/2 = #DIV/0! = #DIV/0!
i).Menghitung z' z' =Kedalaman dari permukaan tanah 1/2 Ka1 g'1 z'^2 0.5 0.361 1554 280.523 z'^2 z'
=
6.47
m
#DIV/0!
j).Menghitung momen maksimum. Mmax = - 1/6 = 0.167 = #DIV/0! k).Modulus penampang yang diperlukan pakai type ASTM-328 sall = 172 mN/m^2 S
= = =
Mmax sall #DIV/0! 17200000 #DIV/0!
Perencanaan angkur a g'1 Ka1 f Kp1 H/u
= = = = = >
1 1554 0.361 28 2.770 1,5-2
m kg/m^3
H = a + H/u = a/h + 1.5 = 1 h h = B = H = a + = 1 + = 1.5
0.5 h 0.5
° diambil
1.5
0.5
+
®
1 m 0.5 h 0.5 m
-Menghitung besar sudut f1 dan f2 - f1 = 45 + f / 2 = 45 + 28 / 2 = 59 ° - f2 = 45 f / 2 = 45 - 28 / 2 = 31 ° -Menghitung pajang a' Tan f1 = L1 + a'
L2
Tan 59
=
7.2
a'
=
#DIV/0! 1.66
a'
=
#DIV/0! m
-Menghitung pajang b' Tan f2 = H b'
+ ### a'
1 h
=
1.5
-
0.5
®
h
=
1 1
=
a'
Tan 31
=
b'
=
1.5 0.601
b'
=
2.5
b'
= =
+
1.5 b'
m
### + 2.5 ### m
Jadi jarak plat angkur dari turap minimal
= ### m
METODE OVESEN DAN STROMANN(1972) -Menghitung beban ultimate(Pult) P'ult = 1 g'1 H^2 2 w
= = =
Kp1 Sin d'
H 1.5
- Ka1 Cos f
)
g'1
t 1 2331
1554 kg/m
+ 0.5 g'1 0.5 g'1
=
w
=
2331
=
( Kp1 Cos d'
1.503
+ 0.5 0.5
H^2 H^2 1554 1554
Ka1 Sin f
2.25 2.25
0.361
Sin 28
P'ult
=
= =
1 2
=
=
( Kp1 Cos d'
H^2
0.5 1554 2.25 11621.549 kg
- jika pasir lepas, - jika pasir Padat,
P'us =
g'1
Cov = Cov =
14 19
(
Cov + Cov +
1 H h
(
14 14
+ +
11246.66
)
1 1.5 1
(
6.3
- Ka1 Cos f
-
0.361 Cos 28
P'ult
)
11621.549
kg
-Menghitung beban yang di izinkan(Pall) FS yang digunakan untuk metode ovesen dan stromann(1972) = Pall = P'us = 11246.6608 = 5623.330 kg/m FS 2 -Menghitung jarak antar angkur S' = Pall F = 5623.330 #DIV/0! = #DIV/0! m S' H
+
B h
=
### 1.5
+
1 1
)
= #DIV/0!
2
)
Be
= Pult = = =
P'us Be 11246.66 0 0 kg
Koefisien tekanan Tanah ( (
45 45
-
( (
f 28
/ /
2 2
) )
) )
Kp1 = = =
Tan^2 Tan^2 2.77
( (
45 45
+ +
( (
f 28
/ /
2 2
) )
) )
( (
45 45
-
( (
f 0
/ /
2 2
) )
) )
Kp2 = = =
Tan^2 Tan^2 1.000
( (
45 45
+ +
( (
f 0
/ /
2 2
) )
) )
tekanan Tanah L1 ) Ka1 1554 7.2 ) 4039.531 kg/m^2
0.361
L1 ) Ka2 1554 7.2 ) 11188.800 kg/m^2
1.000
s2 - Ka2 ) 11188.800 1.000 -
1.000
)
an gaya P di atas titik balik dan letak garis kerjanya. Bentuk
P=
Gaya (s1*L1)/2=
14542.313
(s2*L2)/2=
#DIV/0!
#DIV/0!
Lengan
Momen
(L1/3)+L2= #DIV/0!
#DIV/0!
(L2)*2/3=
#DIV/0!
M=
#DIV/0!
#DIV/0!
L3^2
( l2
+
L2
)
-
3
L3^2
(
+
### )
-
3
=
0
6.2
L3^2
-
#DIV/0!
P
[
(
#DIV/0!
L1 g'2
[ ( 965 (
#DIV/0!
- Ka2 ) 1.000 kg/m^2
L3 1.000
)
2.2
kedalaman D L3 2.2
1.4 (
Dteori )
g'2 ( Kp2 - 0.5 965 kg
Ka2 ) L3^2 ( 1.000 -
aman dari permukaan tanah F = 0 z'^2 #DIV/0! #DIV/0!
#DIV/0! = 0
=
0
1.000
+ L2 ) ( ( Kp2 - Ka2 )
)
4.8
z
+
l1
)
]
=
0
7.2 + ### ) 1.000 - 1.000
( )
#DIV/0! +
1
]
momen maksimum. Ka1 g'1 z'^3 + F ( z' 0.167 0.361 1554 270.470 + #DIV/0! kg.m
-
l1 ) #DIV/0!
(
6.47
ampang yang diperlukan mN/m^2
=
#DIV/0! 17200000 #DIV/0! m^3/m
172000 kN/m^2 =
=
#DIV/0! x 10^5
17200000
m^3/m
kg/m^2
-
1
)
1
m
METODE GHALY(1997)
METODE NEELY,STUART,DAN GRAHAM(1
-Menghitung beban ultimate(Pult) A = B x h = 1 x 1 = 1 m^2 28 f = ° H = 1.5 m F = #DIV/0! kg
-Menghitung beban ultimate(Pult) H = 1.5 = h 1 B h
5.4 Tan f
x
(
H^2 A
)
^0,28
=
5.4 0.532
x
(
2.25 1
)
^0,28
=
29707.956
kg
Pult =
-Menghitung beban yang di izinkan(Pall) FS yg digunakan untuk metode ghaly(1997) Pall = Pult = 29707.9562 FS 3 -Menghitung jarak antar angkur S' = Pall F = 9902.652 #DIV/0! = #DIV/0! m
=
x
x
g'1 A
H
1554
1
= 3 9902.652 kg/m
1.5
=
1 1
=
Pult = = =
Myg g'1 13 1554 30303
-Menghitung beban yang di izinkan(Pall FS yg digunakan untuk metode neely,stua Pall = Pult FS -Menghitung jarak antar angkur S' = Pall F = 20202 #DIV/0! = #DIV/0!
=
0
,STUART,DAN GRAHAM(1973)
eban ultimate(Pult) 1.5
1
h^2 B Fs 1 1 1.5 1554 kg
eban yang di izinkan(Pall) n untuk metode neely,stuart and graham(1973) = 30303 = 20202 kg/m 1.5
rak antar angkur
20202 #DIV/0! m
=
1.5
L1
L2
D
Tebal (L1)
= 1.8
m
γ
= 1543 kg/m3
ø
=
Tebal (L2)
= 5.5
γ
= 1543 kg/m3
ø
=
c
=
ø γsat
=
30
o
m 30
o
4795 kg/m2 0
o
= 1979 kg/m3
¨ Keterangan : 1. Lapisan 1
: γ1 ø1
2. Lapisan 2
: γSat 2 ø2 γ2'
= 1543 kg/m3 =
30
= 1543 kg/m3 =
30 o = γSat 2 - γw = 1543 -
3. Lapisan 3
: γSat 3 ø2 γ3'
o
1000 =
543
kg/m3
979
kg/m3
= 1979 kg/m3 =
o 0 = γSat 2 - γw
= 1979 -
1000 =
¨ Menentukan koefisien tekanan tanah aktif dan pasif 1. Lapisan 1 (L1) Ka1 = Tan2 (45 - Ф/2) = Tan2 ( 45 = 0.33 2. Lapisan 2 (L2)
-
Kp1 30 /2 )
= Tan2 (45 + Ф/2) = Tan2 ( 45 = 3.00
+
30 /2 )
Ka2 = Tan2 (45 - Ф/2) = Tan2 ( 45 = 0.33
Kp2
-
30 /2 )
= Tan2 ( 45 = 3.00
3. Lapisan 3 (D) Ka3 = Tan2 (45 - Ф/2) = Tan2 ( 45 = 1.00
Kp3
-
0
/2 )
=
30 /2 )
+
0
/2 )
m
γ1 . L1
= 1543.00 x = 2777.40
1.80 kg/m2
σh1 = P1 = σv1 . Ka1 = 2777.40 = 925.80 2. Untuk kedalaman Z σv2 = = =
+
= Tan2 (45 + Ф/2) = Tan2 ( 45 = 1.00
¨ Menghitung tekanan tanah untuk setiap lapisan 1. Untuk kedalaman Z = L1 = 1.80 σv1
= Tan2 (45 + Ф/2)
=
x 0.33 2 kg/m L2
=
0.9258
5.50
0.4629
m
γ1 . L1 + γsat2 . L2 ( 1543.00 x 1.80 11263.90 kg/m2
)
+
( 1543.00 x
σ'h 2A = P2A= σv2 . Ka2 = 11263.90 = 3754.63 σh 2B = P2B = σv2 . Ka3
x 0.33 2 kg/m
11263.90 11263.90
x 1.00 2 kg/m
= =
5.50
3.7546
¨ Menghitung luasan diagram tekanan tanah diatas dredge line A1 = 1/2 . P1 . L1 = 1/2 x 925.80 2 3
A2 =
P 1 . L2
4
A3 =
1/2 .( P2a -P1 ). L2
•
P
= A1 + A 2 + A 3 = 833.22 = 13704.41
)
1.8773
x
= 833.22 kg/m = 925.80 x 5.50 = 5091.90 kg/m = 1/2 x 2828.83 x = 7779.29 kg/m
+ 5091.90 kg/m
¨ Menghitung momen statis terhadap E • ΣMc = P . Ž
+
1.80
5.5
7779.29 13.704412 6.8522058
ΣMc P • ΣMc = (A1)(1/3. L1 + L2) + (A2)(1/2. L2) + (A3)(1/3. L2 ) = 833.220 ( 1/3 1.80 + 5.50 ) + 5091.90 = 33347.40 kg/m2 ž
•
ž
=
=
ΣMc = P
33347.40 13704.41
=
2.43
¨ Menghitung P6 • P6 = 4c - (γ.L1 + γ2.L2) = 4 ( 4795 )-( ( 1543.00 x = 7916.10 kg/m2
( 1/2 5.50 )+
7779.29
( 1/3
m
1.80
) + ( 1543.00 x 5.50 )) 7.9161
¨ Kedalaman D aktual dan teoritis Dteorotis = P6D2 + 2 P6 D (L1+L2-l1) - 2 P (L1+L2-l1-z)
Daktual
=
7916.10
D2 +
2
7916.10
= =
7916.10 0.996
D + 101326.08 D m (dari kalkulator) 2
= 1.3Dteorotis = 1.3 x = 1.294
D( 1.80 + 5.50 -
0.9 ) - 2 13704.41
108721.67
0.996 m
¨ Menghitung Nilai F F = P - P6 D = 13704.41 7916.10 kg/m = 5823.49
0.996 5.8235
¨ Menghitung momen maksimum turap ½ P1 L1 - F + P1 (z-L1) + ½ Ka1 γ1 (z-L1)2 = 0 misal z - L1 = x ½ P1 L1 - F + P1 (x) + ½ Ka1 γ1 (x) 2 = 0 ½ x 925.80 x
1.8
-
5823.49
+
257.17 X2 + 925.80 X -4990.27 = X = 2.959 (dari kalkulator)
925.80 X + ½ x 0.33 x 1543 X2 = 0 0
Mmax = - ½ P1 L1 (x + L1/3) + F (x + lz) - P1 x/2 - ½ Ka1 γ1 x 2 (x/3) = - ½ x 925.80 x 1.8 x ( = -2965.13 + 22470.783 = 15915.9978 kg/m
2.959 + 0.6 )+ 5823.492 x ( 2.959 + 1369.56 2220.10 7.9579988869
¨ PERENCANAAN TULANGAN DAN PERENCANAAN DIMENSI Direncanakan : ϕ Tulangan Pokok, d = 16 mm Selimut Beton, sb = 40 mm
0.9
b h F'c Fy Momen Ultimate
= = = = = =
800 mm 250 mm 37 Mpa 370 Mpa 15915.9978 kg.m 159159978 N.mm
1. Menghitung tinggi efektif (d) • d = h sb = =
250 202
-
- 1/2 x
ϕp
- /2 x
16
1
40
2. Menghitung Koefisien Tahanan Penampang Mu • Rn = ϕ x b x d2 159159978 = 0.8 x 800 x 40804 = 6.0947 •
m
Fy 0.85 x F'c 370 = 0.85 x 37 = 11.7647 =
3. Rasio Penulangan •
ρ min
= = =
• ρ perlu = = =
1.4 Fy 1.4 370 0.0038 1 x m 1 x 11.765 0.0185
1
1
1 -
0.85 x F'c ρ • balance = Fy 0.85 x 37 = 370 = 0.0447
x
2 . m . Rn fy
1 -
β
x 0.85
• ρ max = 0.75 x ρ balance = 0.75 x 0.0447 = 0.0335
x x
2x
11.765 x 370
600 600 + Fy 600 600 + 370
6.095
Syarat :
ρ min < ρ perlu < ρ max 0.0038 < 0.0185 < 0.0335 Sehingga digunakan rasio (ρ perlu = 0.0185 ) 4. Luas Tulangan • As = ρ x b = 0.0185 x 800 = 2986.5750 mm2
x x
d 202
• Ast = ¼ x π x ϕ2 = ¼ x π x 16 = 201.0619 mm2 5. Jumlah Tulangan As • n = Ast 2986.5750 = 201.0619 = 14.854 buah ≈
15 buah
6. Jarak Antar Tulangan b • s = 15 1 800 = 14 = 57.14 mm ≈
57 mm
1. menghitung stability number Sn = C 1.25 (ᵧ.L1 + ᵧ. L2) = 0.5321 2. Menghitung nilai ᾳ ᾳ = L1 + L2 L1 + L2 + Dact = 0.8494 3. Menghitung flexibility numberv ( P ) modulus elastisitas E = 25742.96 MN/m2 = 25.74296 Kpa momen inersia I = 0 1 1 I = bh3 = 800 150 3 = = 12 12 225000000.00 maka P 10.9 x
10-7 (L1 + L2 + Dact)4 EI = 0.0016187894 log p = -2.79080965
0.0002
4. Menentukan Md dari hasil plot grafik ( dlm buku braja M das bab 9 hal 484 ) , maka Md= 1 Mmaks Md = 1 ### Md= #REF! ¨ Perencanaan angkur
Dalam merencanakan letak dan posisi antar angkur digunakan metode tegangan angkur, direncanakan jarak ½ L1 dari permukaan tanah. Dik : γ1 = 1543 kg/m3 ø =
30
0
Direncanakan pelat angkur bujur sangkar (B=h) teng 1962 menghasilkan metode perhitungan tahanan ultim dari tanah-tanah berbutir yang dekat dari permukaan tanah, yaitu: H/h > 1.2 - 2.0, digunakan H/h = H = H = H = H/h = 1.5 = h = H = H = =
1/2 1/2 0.9 0.9 0.9 0.9
L1 + 1/2 h 1.8 + 0.5 h + 0.5 h /h + 0.5 /h + 0.5 m
1/2 L1 + 1/2 h 1/2 1.8 + 0.5 1.35 m
1.5
h =
0.9
Be = 0.19 (H + h) + B = 0.19 ( 1.35 + 0.9 ) + 0.9 = 1.33 m Pult = P x Be = 13704.41 x = 18192.61 kg Pall = = = S' = =
Pult Fs 18192.61 2 9096.30 Kg Pall F 9096.30
1.33
0.9
b=h=
0.9
5823.49 = 1.56 m ≈
2
m
jadi , jarak spasi angkur adalah =
2
m
Menghitung momen ultimate (Pult) menggunakan metode Ovsen dan Stromen Diketahui : H = 1.35 m γ = 1543 kg/m2 ᶲ b=h ᵧ beton t Ka Kp
= = = = = =
Penyelesain : W = H.t.ᵧ beton = 1.35 x 0.15 = 486.000 kg/m
30 0.9 2400 0.15 0.333 3.000
x
o m kg/m2 m
2400
Kp sin ᵹ' = W + 1/2 ˠH^2 . Ka sin ᶲ /1/2 ˠH^2 = 486.000 + 1406.059 0.167 = 0.512 Berdasarkan grafik 9.39 (b) (Braja M. Das hal 849 ) , sehingga Pult = 1/2 ˠH^2 . (Kp cos ᵹ' + Ka cos ᶲ ) = 1406.06 x 3.689 = 5186.5 kg/m
3.4 Kp cos ᵹ'
Diasumsikan tanah pasir dgn ᶲ = 30 adalah dense sand (Cov = Pu's = Cov + 1 P' ult Cov + H/h = 20 5186.494 21.500 = 4824.65 kg/m
19
¨ Perencanaan penempatan angkur 1. Menghitung besar sudut Ф1 = 45 + Ф /2 = 45 + 30 /2 =
60 0
Ф2 = =
45 45 -
=
30 0
Ф /2 30 /2
2. Menghitung a' L1 + L2 + D Tan Ф1 =
Tan Ф1 = Tan 60
a'
8.59 a' a' = 4.962 m =
3. Menghitung a' H b' 1.35 Tan 30 = b' b' = 2.338 m Tan Ф2 =
a' + b' = =
4.96 7.30
+ m
2.34
( 1/3
5.50
13704.41 (
)
1.80 + 5.50 -
lz =
0.9 -
2.43 )
0.9
0.9 ) - 925.80 x
1.479
- ½ x
0.33 x
1543 x
2.959 2 x
0.986
0.0002
ngkur, direncanakan jarak
erhitungan tahanan ultimate
1.45
P=
L1
L2
L3
31 ton
P=
jenis tanah tebal lapisan, l1
= =
5
berat isi basah, γ Berat isi jenuh, γsat
=
16.5
kg/m3
=
-
kg/m3
Kohesi, c
=
-
kg/m2
Sudut gesek, ø jenis tanah tebal lapisan, l2
= = =
32 7
m
berat isi basah, γ Berat isi jenuh, γsat
=
-
kg/m3
=
16.7
kg/m3
Kohesi, c
=
12
kg/m2
Sudut gesek, ø jenis tanah tebal lapisan, l1
= = =
22
o
13
m
berat isi basah, γ
=
-
kg/m3
Berat isi jenuh, γsat
=
19.2
kg/m3
Kohesi, c
=
39
kg/m2
Sudut gesek, ø
m
o
o = eo = 0.7 Cc = 0.4 Lapisan Batuan
1. Perhitungan pondasi tiang tunggal Direncanakan menggunakan tiang HP 250 x 85 dik : P σbaja d1 d2 w W
= = = = = =
31 ton
=
310
kn
62000 kN/m2 254 mm 260 mm 14.4 mm 125 kg/m
¨ Luas penampang yang diperlukan P 310 A = = = σbaja 62000 ¨ Luas profil baja
0.005
m2
140 ton
Ap1 = (2 . d2 . w) + ((d1- 2.w) . w) = ( 2 x 260 x 14.4 ) + = =
10730.88 mm 0.0107
(( 254 -
m2
¨ Luas disekitar profil baja Ap2 = (d2 - w) x (d1 - 2w) = ( 260 - 14.4 ) x ( 254 = =
2 14.4 )
55309.12 mm2 0.0553
m2
¨ Luas Penampang Ap = Ap1 + Ap2 = 0.0107 + =
2 14.4 )
2
0.0660
m
0.0553 2
¨ Keliling p = 2 (d1 + d2) = 2 x ( 254 + 260 ) = 1028 mm = 1.028 m 2. Menghitung tahanan ujung (Qp) a) Metode mayerhof -lapisan 3 Lempung Qp = Nc* . Cu . Ap = 9 . Cu . Ap = 9 x 39 x 0.0660 = 23.18 Kg b) Metode Vesic -lapisan 3 Lempung Qp = Ap . qp
= Ap . Cu. Nc*
4 π (In . Irr + 1) + 1 3 2 4 = ( ln 100 + 1 )+ 3 = 10.04
Nc * =
Qp = Ap . Cu. Nc* = 0.0660 x 39 = 25.87 Kg Tabel kesimpulan nilai Qp No Lapisan Mayerhof
Vesic
x
π 2
+ 1
10.04
Rata - rata
14.4 )
3
3
23.18
25.87
3. Menghitung tekanan friksi (Qs) -Lapisan Pasir Qs = P . L1 . Fav L' = 20 D = 20 x 0.254 = 5.08 k =
1.65
(penampang H, hal. 570)
σv' = (γ1 . L1) = ( 16.5 x =
82.5
5
)
kg/m2
δ' = 0.8 ø = 0.8 x = 25.6
32
Fav = k . σv' tan δ = 1.65 x 82.5 = 65.2202 kg/m
x 0.479
2
Qs = P . L1 . Fav = 1.028 x 5 = 335.23 kg
x
65.2202
-lapisan Lempung Qs = P . L2 . Fav L' = 20 D = 20 x 0.254 = 5.08 k =
1.65
(penampang H, hal. 570)
σv' = (γ2' . L2) = ( -983 x =
-6883.1 kg/m
δ' = 0.8 ø = 0.8 x = 17.6
22
Fav = k . σv' tan δ
7 2
)
24.52
=
1.65 x
-6883.1 x 0.317
= -3602.69 kg/m2 Qs = P . L2 . Fav = 1.028 x 7 = -25924.95 kg
x -3602.69
P=
L1
L2
L3
8
ton
P=
jenis tanah tebal lapisan, l1
= =
6.5
berat isi basah, γ Berat isi jenuh, γsat
=
1750
kg/m3
=
-
kg/m3
Kohesi, c
=
120
kg/m2
Sudut gesek, ø jenis tanah tebal lapisan, l2
= = =
26
o
5.4
m
berat isi basah, γ Berat isi jenuh, γsat
=
-
kg/m3
=
1960
kg/m3
Kohesi, c
=
1800
kg/m2
Sudut gesek, ø jenis tanah tebal lapisan, l1
= = =
10
o
16
m
berat isi basah, γ
=
-
kg/m3
Berat isi jenuh, γsat
=
1970
kg/m3
Kohesi, c
=
1850
kg/m2
Sudut gesek, ø
=
9
m
o
Lapisan Batuan jumlah tiang kelompok =
2
x
2
1. Perhitungan pondasi tiang tunggal Direncanakan menggunakan tiang HP 200 x 53 dik : P σbaja d1 d2 w W
= = = = = =
8
ton
=
80
kn
62000 kN/m2 204 mm 207 mm 11.3 mm 125 kg/m
¨ Luas penampang yang diperlukan P 80 A = = = σbaja 62000 ¨ Luas profil baja
0.001
m2
90 ton
Ap1 = (2 . d2 . w) + ((d1- 2.w) . w) = ( 2 x 207 x 11.3 ) + =
6728.02
mm
=
0.0067
m2
(( 204 -
¨ Luas disekitar profil baja Ap2 = (d2 - w) x (d1 - 2w) = ( 207 - 11.3 ) x ( 204 =
2 11.3 )
35499.98 mm2
=
0.0355
m2
¨ Luas Penampang Ap = Ap1 + Ap2 = 0.0067 + = ¨ Q ijin Qijin = = =
2 11.3 )
2
0.0422
m
0.0355 2
Ap x σbaja 0.0422 x 62000 2618.1360 kN
Qijin 2618.14
> kN >
P 80
kN
Oke
¨ Keliling p = 2 (d1 + d2) = 2 x ( 204 + 207 ) = 822 mm = 0.822 m 2. Menghitung tahanan ujung (Qp) a) Metode mayerhof -lapisan 3 Lempung Qp = Nc* . Cu . Ap = 9 . Cu . Ap = 9 x 1850 x 0.0422 = 703.096 Kg b) Metode Vesic -lapisan 3 Lempung Qp = Ap . qp
= Ap . Cu. Nc*
4 π (In . Irr + 1) + 1 3 2 4 = ( ln 100 + 1 )+ 3
Nc * =
π 2
+ 1
11.3 )
=
10.04
Qp = Ap . Cu. Nc* = 0.0422 x 1850 = 784.68 Kg Tabel kesimpulan nilai Qp No Lapisan Mayerhof 3 3 703.10
x
Vesic 784.68
3. Menghitung tekanan friksi (Qs) -Lapisan Pasir Qs = P . L' . Fav L' = 15 D = 15 x 0.204 = 3.06 k =
1.65
(penampang H, hal. 570)
σv' = γ . L' = 1750 x 3.06 =
5355
kg/m2
δ'1 = 0.8 ø1 = 0.8 x = 20.8
26
δ'2 = 0.8 ø2 = 0.8 x = 8
10
Z = 0 sd L' Fav1 = k . (σv'/2) tan δ1 5355.00 = 1.65 x 2
x 0.38
= 1678.19 kg/m2 Qs1 = P . L' . Fav1 = 0.822 x 3.06 x = 4221.19 kg Z = L' sd L1 Fav2 = k . σv' tan δ1 = 1.65 x 5355 = 3356.39 kg/m
2
Qs2 = P . (L1 - L') . Fav2
x
1678.19
0.38
10.04
Rata - rata 743.89
= =
0.822 x ( 6.5 9490.79 kg
Z = L1 sd L2 Fav3 = k . σv' tan δ2 = 1.65 x 5355 = 1241.78 kg/m
3.06 )x
x 0.141
2
Qs3 = P . L2 . Fav3 = 0.822 x 5.4 x = 5512.03 kg
1241.78
Maka : Qs = Qs1 + Qs2 + Qs3 = 4221.19 + 9490.79 = 19224.01 kg -lapisan Lempung σv1 = (γ1 . L1) = ( 1750 x =
3356.39
+
5512.03
6.5 )
11375 kg/m2
σv2 = (γ1 . L1) + (γ2' . L2) = ( 1750 x 6.5 ) + ( 960 x =
16559 kg/m2
σv3 = (γ1 . L1) + (γ2' . L2) + (γ3' . L3) = ( 1750 x 6.5 ) + ( 960 x =
L1
5.4 )
21409 kg/m
5.4 ) + ( 970 x
A1 A2
L2
A3 A4
L3
σo =
A5
A2
+
5
)
2
A3
+
A4
+
A5
A2 = = =
σv1 x L2 11375 x 5.4 61425 kg/m
A3 = = =
1/2 x (σv2 - σv1) x 1/2 x 5184 x 5.4 13996.8 kg/m
A4 = = =
σv2 x L3 16559 x 5 82795 kg/m
A5 = = =
1/2 x (σv3 - σv2) x 1/2 x 4850 x 5 12125 kg/m
σo = 61425
= =
L2 13996.8 5.4
+
+ + +
L3 82795 5
+
12125
16379.02 kg/m2
Metode λ Cu = =
(
Cu2
x
(
1800
x
L2 ) + ( Cu3 L2 + L3 5.4 ) + ( 1850 5.4 + 5
x
L3 )
x
5
)
= 1824.04 kg/m2 Untuk nilai λ dari tabel 11.9 L = 17 m λ = 0.173 Fav = =
λ . (σv' + 2.cu) 0.173 x ( 16379.02 +
2 1824.04 )
= 3464.69 kg/m2 Qs = P . L . Fav = 0.822 x 16.9 x = 48130.75 kg Metode α Cu 1800 = Pa 10000
=
3464.69
Cu Pa
0.180
=
1850 10000
0.183 Maka α1 = Maka α2 =
0.872 0.786
`
Fav = α1 x Cu + α2 x Cu = 0.872 x 1800 + = 3023.70 kg/m
0.786 x
2
Qs = P . L . Fav = 0.822 x 16.9 x = 42004.64 kg
3023.70
Qs rata -rata Lempung Qs α + Qs λ σo = 2 42004.64 + 48130.75 = 2 = 45067.69 kg
1850
=
0.185
Qs Total Qs = = =
Qspasir + QsLempung 19224.01 + 45067.69 64291.70 kg
4. Menghitung Dukung Ijin Tiang Qu = Qp + Qs = 743.89 + 64291.70 = 65035.59 kg Qall
= = =
Qu Fs 65035.59 2 32517.80 Kg
Qytd = P + (Ap . L . γbaja) = 8000 + ( 0.0422 x 16.9 x = 13602.1776 Kg Syarat =
Qall 32517.80
)
> Qytd Kg > 13602.1776 Kg
5. Menghitung Penurunan Tiang Tunggal Perhitungan Empiris Dik : D = 204 mm = Qytd = 13602.1776 Kg L = 16.9 m = =
Aman
20.4
cm
1690
cm
422.28
cm2
Ap
=
0.0422
m
Ep
=
200000
Mpa = 2000000.00 kg/cm2
2
= S
7850
= = =
D 100 20.4 100 0.23 S
0.23
20000000000 kg/m2
Qytd . L Ap . Ep 13602.1776 x 1690 + 422.28 x 2000000.00 cm +
< cm
Qytd 112408.71
Kg
7850
Aman
)
7. Menghitung Penurunan Tiang Kelompok Dik : s = 0.23 cm (penurunan tiang tunggal) Bg = 81.6 cm D = 20.4 cm Sg
=
s
=
0.23
= Sg 0.46
0.46 < cm
