contoh-soal-dan-jawaban-teknik-sipil.doc

contoh-soal-dan-jawaban-teknik-sipil.doc - CONTOH SOAL DAN...

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CONTOH SOAL DAN JAWABAN TEKNIK SIPIL Sumber : Buku paket Mekanika Tanah 1(Hary C) Contoh Soal 3.1 : Pada kondisi di lapangan, tanah mempunyai volume = 10 cm 3 dan berat basah tanah = 18 gr. Berat tanah kering oven = 16 gr, jika berat jenis tanah (Gs) = 2,71. Hitung kadar air, berat volume basah, berat volume kering, angka pori, porositas dan derajat kejenuhannya. (dianggap berat volume air = 1 gr/cm 3 ). Penyelesaian : a. kadar air ( w ) = % 5 , 12 % 100 16 16 18 x W W W W W s s s w b. berat volume basah ( ) = 3 / 8 , 1 10 18 cm gr V W c. berat volume kering ( d ) = 3 / 6 , 1 10 16 cm gr V W s d. angka pori ( e ) = s v V V 3 90 , 5 1 71 , 2 16 . cm Gs W V w s s 3 10 , 4 90 , 5 10 cm V V V s v 69 , 0 90 , 5 10 , 4 s v V V e e. porositas ( n ) = 41 , 0 69 , 0 1 69 , 0 1 e e f. derajat kejenuhan ( S r ) = v w V V 3 2 1 16 18 cm W V w w w % 49 % 100 10 , 4 2 r S Contoh Soal 3.2 : Suatu tanah mempunyai nilai e = 0,75, w = 22 % dan Gs = 2,66. Hitung porositas, berat volume basah, berat volume kering dan derajat kejenuhan. Gunakan sistem BS (satuan Inggris). Penyelesaian : a. porositas ( n ) = 43 , 0 75 , 0 1 75 , 0 1 e e
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b. berat volume basah ( ) = 3 / 7 , 115 75 , 0 1 4 , 62 . 66 , 2 . 22 , 0 1 1 . . 1 ft lb e Gs w w c. berat volume kering ( d ) = 3 / 9 , 94 75 , 0 1 4 , 62 * 66 , 2 1 . ft lb e Gs w d. derajat kejenuhan ( S r ) = % 78 100 75 , 0 66 , 2 * 22 , 0 . e Gs w Contoh Soal 3.3 : Data dari pengujian di laboratorium pada benda uji jenuh menghasilkan angka pori e = 0,45 dan berat jenis Gs = 2,65. Untuk keadaan ini, tentukan berat volume basah ( b ) dan kadar airnya ( w ). Penyelesaian : Benda uji dalam kondisi jenuh. Jadi, seluruh ruang pori terisi dengan air. 45 , 0 s v V V e Tapi V v dan V s belum diketahui, pada Gambar C 1.3 , dengan menganggap V s = 1, maka untuk kondisi jenuh : V v = V w = e.V s = e % 17 65 , 2 45 , 0 s w W W w V = V s + e.V s = 1 + (0,45 x 1) = 1,45 m 3 jadi, tanah ini mempunyai W s = V s .Gs. w = 1 x 2,65 x 1 = 2,65 ton berat volume basah ( b ) = W w = V w . w = 0,45 x 1 = 0,45 ton 2,14 t/m 3 dan kadar air ( w ) W = W s + W w = 2,65 + 0,45 = 3,1 ton = 17 %. air butira n V v = e.V s V s = 1 W s W w Gambar C 1.3
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3 / 14 , 2 45 , 1 1 , 3 m t V W b Contoh Soal 3.4 : Dilakukan uji batas susut pada suatu tanah dimana mineral lempung yang paling dominan dikandungnya adalah Illite. Hasil pengujian yang didapat adalah : m 1 = 44,6 gr Vi = 16,2 cm 3 m 2 = 32,8 gr V f = 10,8 cm 3 Hitunglah batas susut tanah tersebut.
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