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Soal . Diketahui struktur Dinding Penahan Tanah seperti tergambar. Periksalah Stabilitas Struktur tersebut terhadap Guling dan Keruntuhan Geser pada dasar pondasi.



0.45 m



q = 50 kN/m2



γ = 16 kN/m3 Ø = 30o C = 0 kN/m2



5.5 m



γ c = 23,50 kN/m3



1.25 m



0.6 m 1m



0.6m



1.54 m



γ = 19 kN/m3 Ø = 360 C = 40 kN/m2



Jawaban : A. Tekanan Tanah Lateral Rankine I. Tekanan Tanah Aktif a) Akibat Beban Luar Merata, q = 50 kN/m2 0≤ Z ≤ 5.5 m ( γ = 16 kN/m3, Ø = 30o, C = 0 ) Koefisien Tek. Tanah Aktif Rankine ( Ka ) Ka = tan2 (45o - Ø/2) = tan2 (45o - 30o/2) = 0.333 Z= 0 → σv = q = 50 kN/m2 ; σa = ka. σv = 0.333 *50 = 16.65 kN/m2



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Z = 5.5 m → σv = q = 50 kN/m2 ; σa = ka. σv = 0.333 *50 = 16.65 kN/m2 b) Akibat Berat Tanah 0≤ Z ≤ 5.5 m ( γ = 16 kN/m3, Ø = 30o, C = 0 ) Z= 0 → σv = γ .z = 16 * 0 = 0 kN/m2 ; σa = ka. σv = 0.333 *0 = 0 kN/m2 Z = 5.5 m → σv = γ .z = 16 * 5.5 = 8.8 kN/m2 ; σa = ka. σv = 0.33*8.8 = 2.93 kN/m2 II. Tekanan Tanah Pasif 0≤ Z ≤ 1.25 m ( γ = 19 kN/m3, Ø = 36o, C = 40 kn/m2 ) Koefisien Tekanan Tanah Pasif Rankine (kp) Kp = tan2 (45o + Ø/2) = tan2 (45o + 36o/2) = 3.85 Z= 0 → σv = γ .z = 19 * 0 = 0 kN/m2 ; σp = kp. σv + 2 C2√kp= 0 + 2(40) √3.85 = 0 + 156.97 = 156.97 kN/m2 Z = 1.25 m → σv = γ .z = 19 * 1.25 = 23.75 kN/m2 ; σp = kp. σv + 2 C2√kp σp = 3.85 * 23.75 + 2(40) √3.85 = 91.44 + 156.97 = 248.41 kN/m2 Gambar Diagram Tekanan Lateral 0.45 A 2



1 tanah



x1 x2 x3 Pa1 3 w2



w1



Pa Pa2



W3 Pp1 Pp Pp2



4



2.75



beton



Za 5.5/3



Zp C



x4 1m



W4 0.6 m



B 1.54 m



B. Gaya Lateral Rankine 1. Gaya Aktif ( Pa ) Pa = Pa1 + Pa2 = 16.65 * 5.5 + 0.5 * 2.93 * 5.5 = 91.575+8.057 = 99.63 kN/m’ Posisi Gaya Aktif ( Za ) Za = (Pa1 * 5.5/2 + Pa2 * 5.5/3)/Pa = (251.83 + 14.77)/99.63 = 2.68 m 2. Gaya Pasif ( Pp )



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Pp = Pp1 + Pp2 = (156.97 * 1.25 + 0.5 * 91.44 * 1.25) = 196.21 + 57.15 = 253.36 kN/m’ Posisi Gaya Pasif ( Zp ) Zp = (Pp1* 1.25/2 + Pp2 * 1.25/3)/Pp = (122.63 + 23.8)/253.36 = 0.58 m 



Periksa Struktur Terhadap Guling



FK = ΣMR/ ΣMo ≥ 1.5 ΣMo ( Jumlah Momen-momen Penyebab Guling ) ΣMo = Pa * Za = 99.63 * 2.68 = 267.01 kn - m ΣMR ( Jumlah Momen-momen Penahan Guling ) Tabel Momen Penahan Guling sbb : No. Ele. 1 2 3 4



Luas Penamp Elemen (m2) A1=1.54*4.9 = 7.55 A2=0.45*4.9=2.21 A3=0.5*0.15*4.9=0.37



A4=0.6*3.14=1.88



Berat Elemen ( kN/m’) W1=7.55*16= 120.74 W2=2.21*23.50= 51.82 W3=0.37*23.50= 8.64 W4=1.88*23.50= 44.27 ΣV =225.47



Jarak ke ttk C(m) X1=2.37 X2=1.38 X3=1.1 X4=1.57



Momen thd. Ttk C (kN-m) M1=120.74*2.37= 286.15 M2=51.82*1.38 = 71.25 M3=8.64*1.1= 9.50 M4=44.27*1.57= 69.50 ΣMR = 436.40



FK = (ΣMR + Pp * Zp )/ ΣMo = (436.4 + 253.36*0.58)/267.01 = 2.2 > 1.5 ( aman) 



Periksa Struktur Terhadap Geser



FK = ΣFR / ΣFd ≥ 1.5 ΣFd (Jumlah Gaya-gaya Penyebab Geser) ΣFd = Pa = 99.63 kN/m’ ΣFR ( Jumlah Gaya-gaya Penahan Geser) ΣFR = B*C2 + ΣV tan Ø2 + Pp = 3.14*40 + 225.47*tan 36º + 253.36 = 543.14 kN FK = ΣFR / ΣFd = 543.14/99.63 = 5.45 ≥ 1.5 ( Aman )



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