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14. Taty Yuniarti_UAS Bu Hetty
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UAS STABILITAS BANGUNAN AIR NAMA : TATY YUNIARTI NIM. 95014318 DOSEN IR. SRI HETTY SUSANTIN, M.ENG PRORAM STUDI MAGISTER PENGELOLAAN SUMBER DAYA AIR INSTITUT TEKNOLOGI BANDUNG 2015
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Stabilitas Bangunan Air
Transcript of 14. Taty Yuniarti_UAS Bu Hetty
UASSTABILITAS BANGUNAN AIR
NAMA : TATY YUNIARTINIM. 95014318
DOSENIR. SRI HETTY SUSANTIN, M.ENG
PRORAM STUDI MAGISTER PENGELOLAAN SUMBER DAYA AIRINSTITUT TEKNOLOGI BANDUNG
2015
Problem 1 - Lateral Pressure
Dik:
A
17.4
φ = 26
c = 14.36
zH = 6 m
ket: tensile crack terjadi pada kedalaman zc
BDit:a) Kedalaman zcb) Tekanan aktif rankine dan lokasinya sebelum tensile crack terjadic) Tekanan aktif rankine dan lokasinya sesudah tensile crack terjadi
Jawab:
>> Ka =
= 0.390
a) Kedalaman zc
= 2 x 14.36
17.4 x 0.62
= 2.64 m
b)sebelum tensile crack
>> Tekanan tanah lateral
= 0.00 - 17.946248
ϒ = kN/m3
o
kN/m3
tg2 (45+φ/2)
�_ − √ ℎ�=�_�.�_�=�.�.�_� 2� (�_�)
� = /(_� 2� )�√��
= -17.95
= 40.76 - 17.946248= 22.82
>> Diagram tekanan tanah lateral
-17.95
I ket: (+)
zc = 2.641
H-Zc = 3.36
II
>> GayaPa1 = 0.5 x -17.95 x 2.64
= -23.702253 kN/m
Pa2 = 0.5 x 3.36 x 22.82= 38.3173728 kN/m
>> Tekanan tanah aktif rankine (Pa) sebelum tensile crackPa = Pa1 + Pa2
= -23.702253 + 38.317373= 14.6151199 kN/m
>> Lokasi Pa sebelum tensile crack
= -23.7 x (( 2/3 x 2.64 ) +38.3 x ( 1/3 x 3.36 )
14.61512
= -121.34 + 42.9014.61512
= -5.3675404 m
kN/m2
kN/m2
(tanda negatif menunjukan bahwa lokasi berada di bawah titik B)
�=(∑▒ 〖����� �� ������〗 ) /�_�
�_ − √ ℎ�=�_�.�_�=�.�.�_� 2� (�_�)
c)sesudah tensile crack
>> Tekanan tanah lateral
= 22.82 - 17.946248
= 4.87
= 40.76 - 17.946248
= 22.82
>> Diagram tekanan tanah lateral
AI ket:
(+)zc = 2.641
c 4.87
H-Zc = 3.36 III
B>> Gaya
Pa1 = 0.5 x 17.95 x 3.36= 30.1364905 kN/m
Pa2 = 3.36 x 4.87= 16.3617646 kN/m
>> Tekanan tanah aktif rankine (Pa) setelah tensile crackPa = Pa1 + Pa2
= 30.1364905 + 16.361765= 46.4982551 kN/m
>> Lokasi Pa setelah tensile crack
= 30.1 x 1/3 x 3.36 +16.4 x 1/2 x 3.36
(tanda negatif menunjukan bahwa lokasi berada di bawah titik B)
kN/m2
kN/m2
�_ − √ )ℎ�=�_�.�_�=�.�−��.�_� 2� (�_�
�=(∑▒ 〖����� �� ������〗 ) /�_�
�_ − √ ℎ�=�_�.�_�=�.�.�_� 2� (�_�)
46.498255
= 33.74 + 27.4846.498255
= 1.31647542 m (tanda negatif menunjukan bahwa lokasi berada di atas titik B)
ket: tensile crack terjadi pada kedalaman zc
22.82
3.36 +
(tanda negatif menunjukan bahwa lokasi berada di
22.82
(tanda negatif menunjukan bahwa lokasi berada di
Problem 2 - Cantilever Retaining wall
Dik:x1 = 0.25 m α
A
16
φ1 = 32
c1 = 0
H = 6 m
B
18 D 1.5 m
φ2 = 15 x5 = 0.5 m
c2 = 45 Cx3 x2 = x4 =
1.2 m 0.5 m 3 m
Dit:SF guling dan SF geser dengan teori rankinecatatan: Drainase berfungsi maksimal dan tekanan tanah lateral pasif dipertimbangkan
Jawab:
active earth pressure coefficient (Ka)
28 30 32 34 36 38 400 0.361 0.333 0.307 0.283 0.26 0.283 0.2175 0.366 0.337 0.311 0.286 0.262 0.24 0.219
10 0.38 0.35 0.321 0.294 0.27 0.246 0.22515 0.409 0.373 0.341 0.311 0.283 0.258 0.23520 0.461 0.414 0.374 0.338 0.306 0.277 0.2525 0.573 0.494 0.434 0.385 0.343 0.307 0.275
>> Koefisieni = 1/3 φ1 cos i = 0.984807753
= 1/3 32 cos I ^2 = 0.9698463104= 10.67 ≈ 10 cos φ2 = 0.9659
cos φ2 ^2= 0.9330 Dari tabel didapat nila Ka, untuk
= 10φ1 = 32Ka1 = 0.321
= 0.984807753 1.17670.7929
Kp = 1.46156
>> Tekanan tanah lateral aktif
= 0.00 - 0.00= 0.00
= 30.82 - 0.00= 30.82
= 33.38 - 0.00= 33.38
ϒ1 = kN/m3
o
kN/m3
ϒ2 = kN/m3
o
kN/m3
α (deg) φ'(deg)
I =α
kN/m2
kN/m2
kN/m2
�_ 1 1.0 1− _1 √ 1 )ℎ�=�_�.�_� =� .�_� 2� (�_�
�_ 1 1 1− _1 √ 1 )ℎ�=�_�.�_� =� .�.�_� 2� (�_�
�_ 1 1. 5) 1− _1 √ 1 )ℎ�=�_�.�_� =� (�+� .�_� 2� (�_�
� cos _�=���� 〖� +√( 〖���〗 ^2 − � 〖���〗^2 2) � 〗 /(cos √〖�− ( 〗〖��� ^2 − � 〖�
〗�� ^2 ) � 〗 2)
>> Tekanan tanah lateral pasif
= 39.46 + 192.25= 231.72
>> Diagram tekanan tanah lateral
A
6
I Pa
30.82B
IIPp
III 1.5C 33.38
231.7
>> Resultante Gaya-Gaya
aktifPa1 = 0.5 x 30.82 x 6.00
= 92.448 kN/mPa2 = 30.82 x 1.50
= 46.224 kN/mPa3 = 0.50 x ( 33.38 - 30.82 ) x 1.5
= 1.926 kN/m
Pa = 92.448 + 46.22 + 1.9= 140.60 kN/m
pasif
Pp = 0.5 x 231.72 x 1.50= 173.787132 kN/m
Resultante= 0
Pp-Pa = 173.787132 - 140.60= 33.19 kN/m
>> Jarak titik tangkap
aktifterhadap titik c
= 92.448 x (( 1/3 x 6.00 ) + 1.5 ) +
46.22 x ( 1/2 x 1.50 ) +
1.93 x ( 1/3 x 1.50 ) +
140.60
= 323.568 + 34.67 + 0.96140.60
= 2.554795 m
kN/m2
∑H
�_ 2 2 2+ _2 √ 2 )ℎ�=�_�.�_� =� .�.�_� 2� (�_�
�_ =(∑▒(� 〖����� ��������〗 ) )/�_�
pasifterhadap titik c
= 173.7871 x ( 1/3 x 1.50 ) 173.787132
= 0.50 m
>> Gaya Vertikal α 10x1 = 0.25 m
W3 T
= 24 tan 10 = T
3
0.176326981 = T
3B1 T = 0.5289809
6B2 W4
W2x
1 W1D 1.5
B3 x5 = 0.5 m
x3 x2 = x4 =
1.2 m 0.5 m 3 mBerat DPT
B1 = alas x tinggi x
= 0.25 x 6 x 24= 36
B2 = 1/2 x alas x tinggi x
= 1/2 x 0.25 x 6 x 24= 18.00
B3 = alas x tinggi x
= 4.70 x 0.5 x 24= 56.4
Berat tanah
W1 = alas x tinggi x
= 1.2 x 1 x 18= 21.6
W2 = 1/2 x alas x tinggi x
= 1/2 x 0.04166667 x 1 x 18= 0.75
W3 = 1/2 x alas x tinggi x
= 1/2 x 3 x 0.5289809421 x 16= 25.39109
W4 = alas x tinggi x
= 3 x 6 x 16= 288
γbeton kN/m3
γbeton
kN/m2
γbeton
kN/m2
γbeton
kN/m2
γ2
kN/m2
γ2
kN/m2
γ1
kN/m2
γ2
kN/m2
�_ =(∑▒(� 〖����� ��������〗 ) )/�_�
>> Perhitungan Momen terhadap titik C
Gaya Vertikal Lengan Gaya Horisontal lengan Momen Tahan Momen GulingW1 21.6 1 Pa 140.60 2.55 21.60 359.199W2 0.75 1.166667 Pp 173.787132 0.50 0.88 86.893566146W3 25.3911 6.676327 169.52W4 288 3.5 1008.00B1 36 3.5 126.00B2 18.0 2.5 45.00B3 56.4 0.25 14.10
jml 446.141 33.19 3.0547945 1385.09 446.092566146
>> Stabilitas geser
δ'=k1.φ2 c'a =k2c'2 10= 0.1763269807
= 446.1411 x 0.17632698 + 45 x 4.7
33.189132= 8.742823 > 1.5 ( aman terhadap geser)
>> Stabilitas guling
SF guling = Momen tahanMomen guling
= 1385.09419446.092566
= 3.10494792 > 2 (ok)
2/3φ2 =〖��〗 =(∑▒_����� 〖� )(�_2+� .� 〗 )/(∑ )▒�
� 2/3 =�� _2�
Problem 3- Cantilever Sheet Pile
Dik Asand
16 L1= 1 m
φ1 = 30
c1 = 0B
sand
18
9.81 φ1' = 30 L2= 7 m
c1' = 0
C DL (Dredge Line)clay
φ2' = 0 D?
c2' = 40E
Dit:a) Hitung kedalaman penetrasi (D teori)?b) Jika Dact = 0.3 D teori, hitung panjang turap?c) Momem maksimum?
Jawaba) D teori
sand clay
>> Ka1 = Ka2 =
= 0.333 = 1.0
Kp1 = Kp2 =
= 3.000 = 1.0
>> Tekanan tanah lateral aktif
= 0.00 - 0= 0.00
= 5.333 - 0
= 5.333
Titik CDi kanan -> aktif
(lap.I) = 24.44 - 0= 24.44
(lap. 2) = 73.33 - 80= -6.67
Di kiri ->pasif
= 0.00 + 80= 80.00
Titik EDi kanan -> pasif
= ( 16.00 + 57.33 + 8.19 D) 1.0 + 80= 153.33 + 8.19 D
Di kiri -> aktif
= 8.19 D - 80= 8.19 D - 80
ϒ1 = kN/m3
o
kN/m3
ϒ1sat = kN/m3
ϒw = kN/m3 o
kN/m3
o
kN/m3
tg2 (45-φ1/2) tg2 (45-φ2/2)
tg2 (45+φ1/2) tg2 (45+φ2/2)
kN/m2
kN/m2
kN/m2
kN/m2
kN/m2
�_ 1 1.0 1− _1 √ 1 )ℎ�=�_�.�_� =� .�_� 2� (�_�
�_ 1 1 1 1− _1 √ 1 )ℎ�=�_�.�_� =� .� .�_� 2� (�_�
�_ 1 1= 1 1 1^′ 2 1− _1 ℎ�� =�_�.�_� (� .� +� .� )�_� 2�√ 1 )(�_�
�_ 2 2= 1 1 1^′ 2 2− _2 ℎ�� =�_�.�_� (� .� +� .� )�_� 2�√ 2 )(�_�
�_ 1 .ℎ 2+ _2 √ 2 )ℎ��=�_�.�_� =� .�_� 2� (�_�
�_ 1= 1 1 1^′ 2 γ^′ ) 2+ _2 √ 2 )ℎ��=�_�.�_� (� .� +� .� +� �_� 2� (�_�
�_ 1 ^′ 2− _2 √ 2 )ℎ��=�_�.�_� =� .�.�_� 2� (�_�
>> Digram tekanan tanah lateral
Asand
L1= 1 m
I
B 5.33 sand
L2= 7 mII Pa = 106.885
III86.67
F C 24.44 DL (Dredge Line)clay
Pc
D?G
DZ Pe
HE 233.33
>> Gaya- gaya lateral diatas Dredge LinePa1 = 0.5 x 5.33 x 1.00
= 2.67 kN/m
Pa2 = 5.33 x 7.00= 37.33 kN/m
Pa3 = 0.5 x 19.11 x 7.00= 66.885 kN/m
>> Tekanan tanah aktif (Pa) Pa = Pa1 + Pa2 + Pa3
= 2.666667 + 37.333333 + 66.885= 106.885 kN/m
>> Lokasi Pa
= 2.67 x ( 1/3 x 1.00 ) + 7 +37.33 x ( 1/2 x 7.00 ) 66.9 x ( 1/3 x 7.00 )
106.9
= 19.56 + 130.7 + 156.1106.885
= 2.865577 m
>> Gaya- gaya lateral dibawah Dredge Line
Superposisi di titik C == 80.0 - -6.67= 86.7 kN/m
Pc = 86.67 x D= 86.67 D
Superposisi di titik E=
= ( 153.3 + 8.19 D ) - ( 8.19 D - 80.0 )= 233.3 + 0= 233.3 kN/m
Pe = 0.5 x 320 x z= 160 z
�=(∑▒ 〖����� �� ������〗 ) /�_�
�_ ℎ��− _� ℎ� 2�
�_ ℎ��− _ℎ� ��
>> = 0Pa + Pe - Pc = 0
106.9 + 160 z - 86.67 D = 0z = 86.67 D - 106.9
160z = 0.541688 D - 0.668031
>> = 0
= 0
106.9 2.865577 + D + 160 z 1/3 z - 86.67 D 1/2D = 0
306.28722 + 106.9 D + 53.333333 - 43.335 = 0
306.28722 + 106.885 D + 53.333333 0.541688 D - 0.6680313 43.335 = 0
306.28722 + 106.885 D + 53.333333 0.293425 0.7237284 D+ 0.446266 - 43.3 = 0
306.28722 + 106.885 D + 15.649352 38.59885 D+ 23.80 - 43.335 = 0
-27.68565 68.28615 D + 330.09 = 0
dari rumus diatas di dapatkan nilai D teori sebesar = 4.8899 m
sehingga nilai z dapat dihitung:z = 0.541688 D - 0.6680313
= 2.648798 - 0.6680313= 1.980766 m
b) D aktual dan panjang turapD aktual = 1.3 x D teori
= 1.3 x 4.8899= 6.35687 m
Panjang turap yang dibutuhkan = L1 +L2 +D aktual= 1 + 7 + 6.35687= 14.35687 m
c) Momen Maksimum
>> overbudden pressure
= ( 16 x 1 )+ ( 8.19 x 7 )
= 73.33
>> Momen Maksimum
= 106.885 106.885 + 2.865577 - 1/2 106.88586.67 86.67
= 106.885 4.099 - 0.62
= 438.1022 - 65.907484= 372.1947 kN
∑H
∑ME
z2 D2
)2 - D2
D2- D2
D2- D2
D2 +
kN/m2
�� ) (1⁄3 ) (1⁄2 )(�+� +�� � −�� �
�=γ 1+γ^′ 21� �
� =_��� [(�� � /( _2 ) )−(1/2 _� 4� −� +� � � /( _2 ))]_� 4� −�
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