Soal :

Diketahui jari-jari suato terowongan lubang bukaan 17 meter, dimana batuan tersebut

adalah Breksi. Hasil pengujian dengan tegangan insitu σo adalah 40 kg/cm2. data-data

yang diperoleh :

No σc (mpa)1 502 13,83 18,94 30,95 14,16 13,67 26,48 37,69 41,1

Sedangkan dari uji geser θ adalah 68,7o dengan koreksi 2,73 mpa. Hitunglah tebal

daerah plastis ?

Jawab :

No σc (σc – σe)2

1 50 511,8002 13,8 184,3353 18,9 71,8594 30,9 12,4125 14,1 176,2796 13,6 189,8067 26,4 0,9558 37,6 10,5099 41,1 13,723Σ 246,4 1251,955

Σ σc = 246,4

Σe =

R1 = R (

λ =

=

Standar Deviasi

SD =

=

= 12,509

σc = σe - k . SD

dimana k : 0 sampai dengan 3 maka :

σc* (0) = 27,377 – (0 x 12,509) = 27,377σc* (0,1) = 27,377 – (0,1 x 12,509) = 26,1261σc* (0,2) = 27,377 – (0,2 x 12,509) = 24,8752σc* (0,3) = 27,377 – (0,3 x 12,509) = 23,6243σc* (0,4) = 27,377 – (0,4 x 12,509) = 22,3743σc* (0,5) = 27,377 – (0,4 x 12,509) = 21,1225σc* (0,6) = 27,377 – (0,6 x 12,509) = 19,8716σc* (0,7) = 27,377 – (0,7 x 12,509) = 18,6207σc* (0,8) = 27,377 – (0,8 x 12,509) = 17,3698σc* (0,9) = 27,377 – (0,9 x 12,509) = 16,1189σc* (1) = 27,377 – (1 x 12,509) = 14,8680σc* (1,1) = 27,377 – (1,1 x 12,509) = 13,6171σc* (1,2) = 27,377 – (1,2 x 12,509) = 12,3662σc* (1,3) = 27,377 – (1,3 x 12,509) = 11,1153σc* (1,4) = 27,377 – (1,4 x 12,509) = 9,8644σc* (1,5) = 27,377 – (1,5 x 12,509) = 8,6135σc* (1,6) = 27,377 – (1,6 x 12,509) = 7,3626σc* (1,7) = 27,377 – (1,7 x 12,509) = 6,1117σc* (1,8) = 27,377 – (1,8 x 12,509) = 4,8608σc* (1,9) = 27,377 – (1,9 x 12,509) = 3,6099σc* (2) = 27,377 – (2 x 12,509) = 0,359σc* (2,1) = 27,377 – (2,1 x 12,509) = 1,1081σc* (2,2) = 27,377 – (2,2 x 12,509) = -0,1428σc* (2,3) = 27,377 – (2,3 x 12,509) = -1,3937σc* (2,4) = 27,377 – (2,4 x 12,509) = -2,6446σc* (2,5) = 27,377 – (2,5 x 12,509) = -3,8955σc* (2,6) = 27,377 – (2,6 x 12,509) = -5,1464σc* (2,7) = 27,377 – (2,7 x 12,509) = -6,3937

σc* (2,8) = 27,377 – (2,8 x 12,509) = -7,6482σc* (2,9) = 27,377 – (2,9 x 12,509) = -8,8991σc* (3) = 27,377 – (3 x 12,509) = -10,1500

λ =

=

= tan2 (79,35)= 28,2788

R1 = R (

R1 (0) = 9 (

= 1,0429R1 (0,1) = 9 (

= 1,4334R1 (0,2) = 9 (

= 1,0449R1 (0,3) = 9 (

= 1,0465R1 (0,4) = 9 (

= 1,0482R1 (0,5) = 9 (

= 1,0500

R1 (0,6) = 9 (

= 1,0520R1 (0,7) = 9 (

= 1,0542R1 (0,8) = 9 (

= 1,0565R1 (0,9) = 9 (

= 1,0590R1 (1) = 9 (

= 1,0617R1 (1,1) = 9 (

= 1,0648R1 (1,2) = 9 (

= 1,0681R1 (1,3) = 9 (

= 1,0719R1 (1,4) = 9 (

= 1,0719

R1 (1,5) = 9 (

= 1,0811R1 (1,6) = 9 (

= 1,0867R1 (1,7) = 9 (

= 1,0939R1 (1,8) = 9 (

= 1,1028R1 (1,9) = 9 (

= 1,1101R1 (2) = 9 (

= 1,1265R1 (2,1) = 9 (

= 1,1566R1 (2,2) = 9 (

= -1,2430R1 (2,3) = 9 (

= -1,1463

R1 (2,4) = 9(

= -1,1202R1 (2,5) = 9(

= -1,1045R1 (2,6) = 9 (

= -1,0932R1 (2,7) = 9(

= -1,0844R1 (2,8) = 9(

= -1,0770R1 (2,9) = 9 (

= -1,0735

R1 (3) = 9 (

= -1,0678

Kesimpulan :

Untuk nilai k dari 0 sampai dengan 3

R1 < R , adalah daerah elastis

Nilai elastis maksimum adalah 1,1566