Tugas 1 434m5c

TASK 1 SI – 5123 PERILAKU TANAH 8.8 The pressure versus void ratio data determined from a consolidation test on an undisturbed clay specimen are as follows: Pressure (kPa) 20 40 80 160 320 640

Void Ratio 0.864 0.853 0.843 0.830 0.785 0.696

Pressure (kPa) 1280 320 80 20 0

Void Ratio 0.602 0.628 0.663 0.704 0.801

(a) Plot the pressure versus void ratio curve on both arithmetic and semilogarithmic graphs (b) Determine the equations for the virgin compression curve and for the rebound curve for unloading starting at 1280 kPa (c) What are the corresponding modified compression and recompression indices for this soil? (d) Estimate the stress to which this clay has been preconsolidated (After A Casagrande) Solution

1

1

0.8

0.8

Void Ratio

Void Ratio

(a) Arithmetic graph and Semilogarithmic graph

0.6 0.4 0.2

0.6 0.4 0.2

0

0 0

500 1000 Pressure (kPa)

1500

1

10

Arithmetic graph

100 1000 Pressure (kPa)

10000

Semilogarithmic graph

(b)

Compression Index (Cc)

Recompression Index (Cr)

0.8

Void Ratio

Void Ratio

1

y = -0.132ln(x) + 1.5473

0.6 0.4 0.2 0 1

10

100 1000 Pressure (kPa)

10000

0.72 0.7 0.68 0.66 0.64 0.62 0.6 0.58

y = -0.025ln(x) + 0.7741 1

10

100 1000 Pressure (kPa)

10000

-

Compression Index e1 = 0.785 σ1 = 320 kPa e2 = 0.602 σ2 = 1280 kPa Cc =

-

𝑒1 − 𝑒2 0.785 − 0.602 = = 0.304 1280 𝜎′ 𝑙𝑜𝑔 320 𝑙𝑜𝑔 2 𝜎′1

Recompression Index e1 = 0.602 σ1 = 1280 kPa e2 = 0.704 σ2 = 20 kPa

Cr =

𝑒1 − 𝑒2 0.602 − 0.704 = = 0.056 20 𝜎′2 𝑙𝑜𝑔 𝑙𝑜𝑔 1280 𝜎′1

(c) -

Modified Compression Index 𝐶𝑐𝜀 =

-

𝐶𝑐 1 + 𝑒𝑜

=

0.304 1 + 0.864

= 0.163

Modified Recompression Index 𝐶𝑟𝜀 =

𝐶𝑟 1 + 𝑒𝑜

=

0.056 1 + 0.864

= 0.03

Void Ratio

(d) Preconsolidation Pressure from the semilogarithmic plot 𝜎′𝑐 = 280 kPa

Pressure (kPa)

8.9 A building is to be constructed on a stratum of the clay 7 m thick for which consolidation data are given in Problem 8.8. The average existing effective overburden pressure on this clay stratum is 126 kPa. The average applied pressure on th clay after construction of the building is 285 kPa. (a) Estimate the decrease in the thickness of the clay stratum caused by full consolidation under the building load. (b) Estimate the decrease in thickhness due to the building load if the clay had never been preconsolidation under a load greater than the existing overburden (c) Show on the e versus log plot of Problem 8.8 the values of e used for making the estimates in parts (a) and (b). (After A, Casagrande) Solution (a) Diketahui σ’vo = 126 kPa σ’p = 280 kPa ∆σ = 285 kPa 𝜎′

280𝑘𝑃𝑎

𝑂𝐶𝑅 = 𝜎′ 𝑃 = 126 𝑘𝑃𝑎 = 2.2 > 1 Over Consolidation 𝑣𝑜

σ’vo + ∆σ > σ’p

OC – NC, maka: 𝑆𝑐 = 𝐶𝑟

7

𝑆𝑐 = 0.056

1 + 0.864 𝑆𝑐 = 0.0723 + 0.184 𝑆𝑐 = 0.257 𝑚

𝑙𝑜𝑔

280 126

𝐻𝑜 1 + 𝑒𝑜

+ 0.304

𝑙𝑜𝑔

𝜎′𝑃 𝜎′𝑉𝑂

7 1 + 0.864

+ 𝐶𝑐

𝑙𝑜𝑔

𝐻𝑜 1 + 𝑒𝑜

𝑙𝑜𝑔

𝜎′𝑣𝑜 + ∆𝜎 𝜎′𝑝

126 + 285 280

(b) Diketahui σ’vo = 126 kPa σ’p = 126 kPa (asumsi) ∆σ = 285 kPa 𝜎′

126𝑘𝑃𝑎

𝑂𝐶𝑅 = 𝜎′ 𝑃 = 126 𝑘𝑃𝑎 = 1 Normally Consolidation 𝑣𝑜

𝑆𝑐 = 𝐶𝑐

𝑆𝑐 = 0.304

7 1 + 0.864

𝑆𝑐 = 0.586 𝑚

𝑙𝑜𝑔

126 + 285 126

𝐻𝑜 1 + 𝑒𝑜

𝑙𝑜𝑔

𝜎′𝑣𝑜 + ∆𝜎 𝜎′𝑝

(c) Over Consolidation – Normally Consolidation

∆e2

Void Ratio

∆e1

Pressure (kPa)

-

Dari grafik didapat ∆e1 = 0.02 dan ∆e2 = 0.05 Dari perhitungan: ∆e 1+e0 ∆e 1+e0

= 0.010

∆e1 = (1+0.864) × 0.010 = 0.019

= 0.027

∆e2 = (1+0.864) × 0.027 = 0.050

Normally Consolidation - Dari perhitungan: ∆e 1+e0

= 0.0837

∆e = (1+0.0837) × 0.0837 = 0.091

8.12 The following consolidation test data were obtained from undisturbed San Francisco Bay mud. For this clay, LL = 85, PL = 38, s = 2.70 Mg/m3, and n = 105.7%. Initially, the specimen height was 2.54 cm and its volume was 75.14 cm3. Plot the data as percent consolidation versus log pressure. Evaluate the preconsolidation pressure and the modified virgin compression index. Stress (kPa) 0 5 10 20 40 80 160 320 640 160 40 5

Dial Reading (mm) 12.700 12.352 12.294 12.131 11.224 9.053 6.665 4.272 2.548 2.951 3.533 4.350

Void Ratio 2.765 2.712 2.703 2.679 2.541 2.211 1.849 1.486 1.224 1.285 1.374 1.499

Solution Stress (kPa)

Dial Reading (mm)

Void Ratio

Strain 1

Strain 2

%Strain

0

12.7

2.765

5 10 20 40 80 160 320

12.352 12.294 12.131 11.224 9.053 6.665 4.272

2.712 2.703 2.679 2.541 2.211 1.849 1.486

640 160

2.548 2.951

1.224 1.285

40

3.533

1.374

0 0.014 0.016 0.023 0.059 0.147 0.243 0.340 0.409 0.393 0.369

0 0.014 0.016 0.022 0.058 0.144 0.238 0.332 0.400 0.384 0.361

0.000 1.389 1.623 2.262 5.880 14.536 24.045 33.576 40.449 38.846 36.518

5

4.35

1.499

0.336

0.329

33.250

Ccε

𝜎′𝑐 -

Strain 1 𝜺=

∆𝒆 𝒆𝟎 − 𝒆𝒊 = 𝟏 + 𝒆𝟎 𝟏 + 𝒆𝟎

Example: ∆e e0 − ei 2.765 − 2.712 ε= = = = 0.014 1 + e0 1 + e0 1 + 2.765 -

Strain 2 𝜺=

𝑹𝟎 − 𝑹𝒊 𝟐𝟓. 𝟒 𝒎𝒎

Example : R0 − Ri 12.700 − 12.352 ε= = = 0.014 25.4 mm 25.4 mm -

% Strain %𝑺𝒕𝒓𝒂𝒊𝒏 = (

𝒔𝒕𝒓𝒂𝒊𝒏 𝟏 − 𝒔𝒕𝒓𝒂𝒊𝒏 𝟐 ) × 𝟏𝟎𝟎% 𝟐

Example : %Strain = ( -

strain 1 − strain 2 0.14 − 0.14 ) × 100% = ( ) × 100% = 1.39 2 2

Modified Compression Index 𝐶𝑐𝜀 =

𝜀2 − 𝜀1 0.4045 − 0.0588 = 0.287 𝜎 = 640 𝑙𝑜𝑔 𝜎′2 𝑙𝑜𝑔 ′1 40

-

Compression Index

-

𝐶𝑐 = (1 + 𝑒0 )𝐶𝑐𝜀 = (1 + 2.765) × 0.287 = 1.08 Preconsolidation Pressure from the semilogarithmic plot 𝜎′𝑐 ≈ 36 kPa

8.13 Plot the data of Problem 8.12, on a void ratio versus log pressure graph. Evaluate the preconsolidation pressure and the virgin compression index. Do these values agree with what you found in Problem 8.12? Comments?

Void Ratio

Solution

Pressure (kPa)

-

Compression Index e1 = 2.541 σ1 = 40 kPa e2 = 1.224 σ2 = 640 kPa Cc =

-

𝑒1 − 𝑒2 2.541 − 1.224 = = 1.09 640 𝜎′ 𝑙𝑜𝑔 𝑙𝑜𝑔 2 40 𝜎′1

Preconsolidation Pressure from the semilogarithmic plot 𝜎′𝑐 ≈ 38 kPa