Tugas 3 Pondasi Dangkal Persegi Panjang Cpt.docx

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TUGAS 3 PERENCANAAN PONDASI DANGKAL PERSEGI PANJANG BEBAN SIMETRIS DUA ARAHDENGAN DATA TANAH CONE PENETRATION TEST (CPT)

1. Diketahui: Data CPT sample 2 (s1) Masjid SMA Sholahudin Malang. 𝛾𝑏 = 1,40 π‘‘π‘œπ‘›/π‘š3 𝛾𝑀 = 1,00 π‘‘π‘œπ‘›/π‘š3 π›Ύπ‘ π‘Žπ‘‘ πΏπ‘’π‘šπ‘π‘’π‘›π‘” = 1,80 π‘‘π‘œπ‘›/π‘š3 π›Ύπ‘π‘’π‘‘π‘œπ‘› π‘π‘’π‘Ÿπ‘‘π‘’π‘™π‘Žπ‘›π‘” = 2,40 π‘‘π‘œπ‘›/π‘š3 π›Ύπ‘π‘’π‘‘π‘œπ‘› π‘‘π‘Žπ‘˜ π‘π‘’π‘Ÿπ‘‘π‘’π‘™π‘Žπ‘›π‘” = 1,80 π‘‘π‘œπ‘›/π‘š3 𝐷𝑓 = 1,20 π‘š 𝐷𝑀 = 1,60 π‘š 𝐻 = 2,80 π‘š π‘žπ‘ = 20 π‘˜π‘”/π‘π‘š2 π‘ž01 = 2600 π‘˜π‘”/π‘š3 π‘ž02 = 400 π‘˜π‘”/π‘π‘š3 𝑒π‘₯ = 0,40 π‘š 𝑒𝑦 = 0,50 π‘š 𝑁 = 20 π‘‘π‘œπ‘› 𝑀𝐿 = 5,00 π‘‘π‘œπ‘›. π‘š 𝑀𝐡 = 3,00 π‘‘π‘œπ‘›. π‘š 𝐴 = 10 π‘π‘š2 𝑇 = 0,50 π‘š

2. Ditanya: Rencanakan pondasi dangkal persegi panjang eksentris 2 arah.

3. Penyelesaian: 3.1.Dicoba:

𝐿 = 5,50 π‘š 𝐡 = 3.60 π‘š 𝐿𝐴 = 1,00 π‘š 𝐿𝐡 = 0,70 π‘š 𝑑1 = 0,60 π‘š 𝑑2 = 0,30 π‘š 𝐿𝐾 = 0,10 π‘š

3.2.Gambar Data Tanah CPT

Gambar1. Data CPT Tanah dan Rencana Pondasi Skala Panjang 1m = 1cm

Gambar2. Tampak Atas Sakala Panjang 1m = 1cm

Karena M.A.T merpakan kasus c dimana M.A.T dibawah DP maka berat volume yang dipakai adalah Ξ³rt. 𝑧

𝛾 β€² = π›Ύπ‘ π‘Žπ‘‘ πΏπ‘’π‘šπ‘π‘’π‘›π‘” βˆ’ 𝛾𝑀

π›Ύπ‘Ÿπ‘‘ = 𝛾 β€² + (𝐡) . (𝛾𝑏 βˆ’ 𝛾 β€² )

𝛾′ = 1,80 βˆ’ 0,80

π›Ύπ‘Ÿπ‘‘ = 0,80 + (3.6) . (1,40 βˆ’ 0,80)

𝛾 β€² = 0,80 π‘‘π‘œπ‘›/π‘š3

π›Ύπ‘Ÿπ‘‘ = 0,867 π‘‘π‘œπ‘›/π‘š3

0,4

a. Menghitung qn (Tekanan Pondasi Netto) 1. Berat pondasi

= 21,103 ton

2. Berat tanah urug

= 11,256 ton

3. Beban q01 = penutup lantai granit (PMI1970)

= 1,986 ton

4. Beban q02 = beban berguna masjid (PMI1970) = 7,920 ton 5. Beban akibat konstruksi (N)

= 20 ton

6. Beban lantai kerja

= 3,564 ton Ξ£V

Uraian Perhitungan Ξ£V 1. Menghitung Berat Pondasi

=65,829 ton

+

i. Volume Plat Pondasi Rata

Gambar3. Pondasi Plat Rata Skala Panjang 1m=1cm

𝑉𝑖 = 𝐡. 𝐿. 𝑑2 𝑉𝑖 = 3,60π‘₯5,50π‘₯0,30 = 5,94 π‘š3 π΅π‘’π‘Ÿπ‘Žπ‘‘ 𝑖 = 𝑉𝑖π‘₯π›Ύπ‘π‘’π‘‘π‘œπ‘› π‘π‘’π‘Ÿπ‘‘π‘’π‘™π‘Žπ‘›π‘” π΅π‘’π‘Ÿπ‘Žπ‘‘ 𝑖 = 5,940π‘₯2,0 = 14,256 π‘‘π‘œπ‘›

ii. Volume Limas Terpancung

Gambar 4. Pondasi Limas Terpancung Skala Panjang 1m =1cm

1

𝑉𝑖𝑖 = 3 (𝑑1 βˆ’ 𝑑2)((𝐿𝐴. 𝐿𝐡) + (𝐿. 𝐡) + √(𝐿𝐴. 𝐿𝐡) + (𝐿. 𝐡)) 1

𝑉𝑖𝑖 = 3 (0,6 βˆ’ 0,3)((1π‘₯0,7) + (5,5π‘₯3,6) + √(1π‘₯0,7) + (5,5π‘₯3,6)) 𝑉𝑖𝑖 = 2,503 π‘š3 π΅π‘’π‘Ÿπ‘Žπ‘‘ 𝑖𝑖 = 𝑉𝑖𝑖. π›Ύπ‘π‘’π‘‘π‘œπ‘› π‘π‘’π‘Ÿπ‘‘π‘’π‘™π‘Žπ‘›π‘” π΅π‘’π‘Ÿπ‘Žπ‘‘ 𝑖𝑖 = 2,503π‘₯2,4 = 6,007 π‘‘π‘œπ‘›

iii. Volume Kolom Pendek

Gambar 5. Kolom Pendek Skala Panjang 1m = 2 cm

T kolom Pendek

= 𝐷𝑓 βˆ’ 𝑑1 βˆ’ 𝐿𝐾 = 1,20 βˆ’ 0,60 βˆ’ 0,30 = 0,50 π‘š

𝑉𝑖𝑖𝑖 = 𝐿𝐴. 𝐿𝐡. 𝑇 𝑉𝑖𝑖𝑖 = 1,00π‘₯0,70π‘₯0,50 𝑉𝑖𝑖𝑖 = 0,350 π‘š3 π΅π‘’π‘Ÿπ‘Žπ‘‘ 𝑖𝑖𝑖 = 𝑉𝑖𝑖𝑖. π›Ύπ‘π‘’π‘‘π‘œπ‘› π‘π‘’π‘Ÿπ‘‘π‘’π‘™π‘Žπ‘›π‘” π΅π‘’π‘Ÿπ‘Žπ‘‘ 𝑖𝑖𝑖 = 0,350π‘₯2,4 = 0,84 π‘‘π‘œπ‘› Jadi, berat pondasi = π΅π‘’π‘Ÿπ‘Žπ‘‘ 𝑖 + π΅π‘’π‘Ÿπ‘Žπ‘‘ 𝑖𝑖 + π΅π‘’π‘Ÿπ‘Žπ‘‘ 𝑖𝑖𝑖 = 14,256 + 6,007 + 0,840 = 21,103 π‘‘π‘œπ‘›

2. Menghitung Tanah Urug

Gambar 6. Tanah Urug Skala Panjang 1m = 1cm

𝑉 π‘’π‘Ÿπ‘’π‘”π‘Žπ‘› = (𝐡. 𝐿. 𝐷𝑓) βˆ’ (𝑉 𝑖 + 𝑉𝑖𝑖 + 𝑉𝑖𝑖𝑖 + (𝐡. 𝐿. 𝐿𝐾)) 𝑉 π‘’π‘Ÿπ‘’π‘”π‘Žπ‘› = (3,6π‘₯5,5π‘₯1,2) βˆ’ (5,94 + 2,503 + 0,35 + (3,6.5,5.0,1)) 𝑉 π‘’π‘Ÿπ‘’π‘”π‘Žπ‘› = 12,987 π‘š3 π΅π‘’π‘Ÿπ‘Žπ‘‘ = π‘‰π‘’π‘Ÿπ‘’π‘”π‘Žπ‘› . π›Ύπ‘Ÿπ‘‘ π΅π‘’π‘Ÿπ‘Žπ‘‘ = 12,987 . 0,867 π΅π‘’π‘Ÿπ‘Žπ‘‘ = 11,256 π‘‘π‘œπ‘›

3. Menhitung q01= Beban Penutup Lantai Granit 2600 π‘˜π‘”/π‘š3 = 2,6 π‘˜π‘”/π‘š3 Tebal lantai = 4 π‘π‘š = 0,04 π‘š π‘ž01 = 0,04π‘₯2,6 = 0,104 π‘˜π‘”/π‘š2 Akibat π‘ž01 = π‘ž01 ((𝐡. 𝐿) βˆ’ (𝐿𝐴. 𝐿𝐡)) π‘ž01 = 0,104((3.6π‘₯5,5) βˆ’ (1π‘₯0,7)) = 1,986 π‘‘π‘œπ‘›

4. Menghitung q02= Beban Berguna Masjid (PMI1970) 400 kg/m2 = 0,4 ton/m2 Akibat π‘ž02 = π‘ž02 (𝐡. 𝐿) π‘ž02 = 0,4 (3,6π‘₯5,5) = 7,920 π‘‘π‘œπ‘› 5. Beban Akibat Konstruksi (N) = 20 ton

6. Menghitung Beban Lantai Kerja = 𝐡. 𝐿. π‘‘π‘’π‘π‘Žπ‘™ 𝐿𝐾. π›Ύπ‘π‘’π‘‘π‘œπ‘› π‘‘π‘Žπ‘˜ π‘π‘’π‘Ÿπ‘‘π‘’π‘™π‘Žπ‘›π‘”

Beban lantai kerja

= 3,6π‘₯5,5π‘₯0,1π‘₯1,8 = 3,564 π‘‘π‘œπ‘› b. Menghitung qmax dan qmin Untuk

menghitung

qmax

dan

qmin

diperlukan

nilai

Re

ex/B=0.4/3,6=0,1 sehingga Re=0,6 (Grafik !. Pembebanan Eksentris pada Pondasi Memanjang) 𝐡 β€² = 𝐡 βˆ’ (2. 𝑒π‘₯)

𝐿′ = 𝐿 βˆ’ (2. 𝑒𝑦)

𝐡 β€² = 3,60 βˆ’ (2π‘₯0,4)

𝐿′ = 5,5 βˆ’ (2π‘₯0,5)

𝐡 β€² = 2,80 π‘š

𝐿′ = 4,50 π‘š

1) Menghitung qmax dan qmin Arah L 𝛴𝑉

𝑀𝐿

π‘žπ‘šπ‘Žπ‘₯ = 𝐡′ .𝐿′ + 𝐡′ .𝐿′2 65,829

5

𝛴𝑉

𝑀𝐿

π‘žπ‘šπ‘–π‘› = 𝐡′ .𝐿′ βˆ’ 𝐡′ .𝐿′2 65,829

5

π‘žπ‘šπ‘Žπ‘₯ = 2,8π‘₯4,5 + 2.8.4,52

π‘žπ‘šπ‘–π‘› = 2,8π‘₯4,5 βˆ’ 2.8.4,52

π‘žπ‘šπ‘Žπ‘₯ = 5,754 π‘‘π‘œπ‘›/π‘š2

π‘žπ‘šπ‘–π‘› = 4,695 π‘‘π‘œπ‘›/π‘š2

Gambar 7.Diagram qmax dan qmin Arah L Skala Panjng 1m = 1cm Skala Tegagan 15 ton/m2 = 1cm

2) Menghitung qmax dan qmin Arah B π‘žπ‘šπ‘Žπ‘₯ = π‘žπ‘šπ‘Žπ‘₯ =

𝛴𝑉.𝑅𝑒 𝐡′ .𝐿′

𝑀𝐿

+ 𝐿′ .𝐡′2

65,829π‘₯0,6 2,8π‘₯4,5

π‘žπ‘šπ‘–π‘› = 3

+ 4,5.2,82

π‘žπ‘šπ‘Žπ‘₯ = 3,645 π‘‘π‘œπ‘›/π‘š

2

π‘žπ‘šπ‘–π‘› =

𝛴𝑉.𝑅𝑒 𝐡′ .𝐿′

𝑀𝐿

βˆ’ 𝐿′ .𝐡′2

65,829π‘₯0,6 2,8π‘₯4,5

3

βˆ’ 4,5.2,82

π‘žπ‘šπ‘–π‘› = 2,624 π‘‘π‘œπ‘›/π‘š2

Gambar 8. Diagram qmax dan qmin Arah B Skala Panjang 1cm = 1m Skala Tegangan 15ton/m2 = 1cm

c. Menghitung qs A = 10 cm2 Sf = 3,00 karena bangunan monumental (factor aman yang disarankan , Reese & o’neil 1989) π‘žπ‘

20

π‘žπ‘  = 𝐴.𝑆𝐹 = 10π‘₯3 = 0,667

π‘˜π‘”β„ π‘‘π‘œπ‘›β„ π‘π‘š2 = 6,67 π‘š2

Gambar 9. Diagram qs Skala Panjang 1m = 1cm Skala Tegangan 15ton/m2 = 1cm

Selisih antara qs dan qmax adalah: π‘π‘’π‘Ÿπ‘ π‘’π‘›π‘‘π‘Žπ‘ π‘’ = π‘π‘’π‘Ÿπ‘ π‘’π‘›π‘‘π‘Žπ‘ π‘’ =

(π‘žπ‘ βˆ’π‘žπ‘šπ‘Žπ‘₯) π‘žπ‘šπ‘Žπ‘₯

100%

(6,67βˆ’5,754) 5,754

π‘π‘’π‘Ÿπ‘ π‘’π‘›π‘‘π‘Žπ‘ π‘’ = 15,870%

100% < 20% (Aman)

1.1.Perhitungan Penurunan a. Menghitung Tegangan Efektif Tanah (po’)

Gambar 10. Diagram Penyebaran Tegangan Tanah Skala Panjang 1m = 1cm

1) π‘ƒπ‘œβ€² 1 = (π‘žπ‘šπ‘Žπ‘₯. 𝑧1) + (𝛾𝑏. 𝑧) + ((π›Ύπ‘ π‘Žπ‘‘ π‘™π‘’π‘šπ‘π‘’π‘›π‘” βˆ’ 𝛾𝑀). 𝑧1) π‘ƒπ‘œβ€² 1 = (5,754.1,4) + (1,4.0,4) + ((1,8 βˆ’ 1). 1,4) π‘ƒπ‘œβ€² 1 = 9,415 π‘‘π‘œπ‘›β„π‘š2 2) π‘ƒπ‘œβ€² 2 = π‘ƒπ‘œβ€² 1 + ((π›Ύπ‘ π‘Žπ‘‘ π‘™π‘’π‘šπ‘π‘’π‘›π‘” βˆ’ 𝛾𝑀). (𝑧2 βˆ’ 𝑧1)) π‘ƒπ‘œβ€² 2 = 9,415 + ((1,8 βˆ’ 1). (3,4 βˆ’ 1,4)) π‘ƒπ‘œβ€² 2 = 11,015 π‘‘π‘œπ‘›β„π‘š2 b. Menghitung Tegangan Tanah (πˆπ’—) 𝛴𝑣

𝜎1 = (𝐿+𝑍1)(𝐡+𝑍1) 65,829

𝜎1 = (5,5+1,4)(3,6+1,4) 𝜎1 = 1,9081 π‘‘π‘œπ‘›β„π‘š2 𝛴𝑣

𝜎2 = (𝐿+𝑍2)(𝐡+𝑍2)

65,829

𝜎2 = (5,5+3,4)(3,6+3,4) 𝜎2 = 1,0566 π‘‘π‘œπ‘›β„π‘š2 π‘žπ‘1 = π‘žπ‘1 =

π‘…π‘Žπ‘‘π‘Žβˆ’π‘Ÿπ‘Žπ‘‘π‘Ž π‘˜π‘’π‘‘π‘Žπ‘™π‘Žπ‘šπ‘Žπ‘› 𝐷𝑓 𝑠.𝑑 βˆ†β„Ž1 4 30+25+25+30 4

π‘žπ‘1 = 27,308

π‘žπ‘2 = π‘žπ‘2 =

π‘˜π‘”β„ π‘‘π‘œπ‘›β„ π‘π‘š2 = 273,08 π‘š2

π‘…π‘Žπ‘‘π‘Žβˆ’π‘Ÿπ‘Žπ‘‘π‘Ž π‘˜π‘’π‘‘π‘Žπ‘™π‘Žπ‘šπ‘Žπ‘› βˆ†β„Ž2 𝑠𝑑 𝑇𝐾 13 30+30+30+(7π‘₯20)+25+110+220 13

π‘žπ‘2 = 76,5

π‘˜π‘”β„ π‘‘π‘œπ‘›β„ π‘π‘š2 = 765 π‘š2

300

𝑐1 = 2.π‘ƒπ‘œβ€²1 π‘₯π‘žπ‘1 𝑐1 =

300 2π‘₯9,415

π‘₯273,08

𝑐1 = 4350,65 300

𝑐2 = 2.π‘ƒπ‘œβ€²2 π‘₯π‘žπ‘2 300

𝑐1 = 2π‘₯11,015 π‘₯765 𝑐1 = 10417

c. Menghitung Besar Penurunan (𝝆) 𝜌1 =

βˆ†β„Ž1 𝑐1

𝑙𝑛.

1,4

π‘ƒπ‘œ β€² 1+πœŽπ‘£1 π‘ƒπ‘œβ€²1

𝜌1 = 4350,65 𝑙𝑛.

9,415+1,908 1,908

𝜌1 = 0,000059 π‘š = 0,0059 π‘π‘š

𝜌2 =

βˆ†β„Ž2 𝑐2

𝑙𝑛.

3,4

π‘ƒπ‘œ β€² 2+πœŽπ‘£2 π‘ƒπ‘œβ€²2

𝜌1 = 10417,58 𝑙𝑛.

11,015+1,0566 11,015

𝜌1 = 0,000012 π‘š = 0,0012 π‘π‘š

Jadi, pondasi akan mengalami penurunan sedalam penurunan total. Penurunan total = 𝜌1 + 𝜌2 = 0,0059 + 0,0012 = 0,0072 π‘π‘š

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