Question

2. A consolidation test was made on a sample of saturated clay in a circular ring with a height of 1.22 inches and an area of 4.91 square inches. Primary consolidation under the addition of each load was essentially complete at the end of 24 hours. The dial readings corresponding to the end of primary consolidation under each load increment are provided below. The initial void ratio was 1.70 Dial Reading (inch x 10 Pressure (tsf) 0.000 0.064 0.131 0.264 0.531 1.062 2.120 4.250 8.540 0 67 105 211 382 730 1320 1975 2640 (a) From the dial gauge readings, compute the void ratios after each increment of load. Plot the void ratio versus log(pressure) curve. Estimate the preconsolidation pressure of the clay (b) The dial readings observed during the consolidation of the clay specimen for the load increment from 0.531 tsf to 1.062 tsf are provided below. Compute the value of the coefficient of consolidation (c) using the log-time-fitting procedure. Dial Reading in x 104 382.0 423.0 434.0 442.6 452.8 465.0 484.0 511.0 Time (min) ng Time (min) 0.00 0.10 0.25 0.50 in x 1 15 30 60 135 240 1180 1600 2625 548.5 594.1 636.3 670.1 688.9 727.0 728.4 730.3 2.0 4.0 8.0 (c) If the clay sample was taken from a layer of clay 10 ft thick that was drained on the top and bottom, how many days would it take this layer to reach 80% consolidation on average if using the cv value estimated using the log-time-fitting procedure 2(b) HINTS You can find how to estimate the Cv using the log-time method in the lecture note p36. 2(C) · To estimate time of 80% degree of consolidation, you need to use time factor (TV) for U-80%. You can get this information in lecture note p42.

Answer 1

(a) For computing the void ratio for each increment load follow the following steps

> Find dial change delta H

> Find specimen height H = H₁ + Delta H

> Find Height of voids = H-H_s

> Calculate void ratio e = (H-H_s)/H_s

The height of solids Hs is given by,

Md 103 1. 169cm-0.4605in GApw 2.78 31.671

by completing the table

FINAL DIAL DIAL CHANGE SPECIMEN HEIGHT OF APPLIED PRESSURE tsf VOID RATIO READING VOIDS H-H3 0.75952 0.75282 0.74902 0.73842 0.72132 0.68652 0.62752 0.56202 0.49552 HEIGHT H inch x10- inch 1.22 1.2133 1.2095 1.1989 1.1818 1.147 1.088 1.0225 0.956 1.649 1.635 1.627 1.604 1.566 1.491 1.363 1.221 1.076 0.064 0.131 0.264 0.531 1.062 2.12 4.25 8.54 67 105 211 382 730 1320 1975 2640 0.0067 0.0038 0.0106 -0.0171 0.0348 0.059 0.0655 -0.0665

Sample calculation: delta H = 67-105 = -38 x 10^-4 in

The height of specimen, H = H₁ +/- delta H = 1.2133.0038 = 1.2095 in

Height of voids = H - Hs = 1.2095 - 0.46048 = 0.74902 in

void ratio, e = (H-Hs)/Hs = 0.74902/0.46048 = 1.6267

From this we can draw the graph e v/s lop P

For finding out the pre-consolidatiom pressure, you have to follow the following steps

> The initial portion of the curve is curved which represents recompression curve of a remolded specimen and the lower portion is straight which is the virgin curve representing Cc

> The point A is selected, the point of maximum curve and a horizontal line AB is drawn

> A tangent AC is drawn at A

> Angle ABC is bisected by AD

> The straight portion of the virgin curve is extended to meet at AD, this point represents pre-consolidation pressure in x-axis

e-log P 10 1 0.1 0.01 log p

Pre-consolidation Pressure = 1.2 tsf

(b) For finding out Cv, we need to plot square root time v/s dial gauge reading

9-2 1 1-9 5 2-7 0 3 8 2 3 4-4 8 7-2 3 6 1-2 1-1 , 1

From the graph for obtaining T₉₀ We have to follow the following steps

> The straight portion (line A) is produced back ro meet the ordinate ar Rc

> From Rc another line B is drawn such that it's abscissa at every point is 1.15 times that of line A

> The intersection of Line with consolidation curve gives squae root T₉₀

_{sqrt T 30 0 10 50 sqrt Tgo Rc 420 드500 600 800}

Sqrt (T₉₀) = 7.6 minutes

.'. T = 57.76 minutes

now, Cv is given by the equation,

Cu = (T,.)90012 G, T90

Tv - Time factor corresponding to 90% consolidation = 0.848

d - Depth of drainage path = 0.95/2 (double drainage)

0.848 * (0.4752) = 3.312 * 10-3in2/min 57.76 T90

(c)

G, = Tso

0.567512) T80 3.312 103in2 /min -

from this, we get T₈₀= 616304.34 minutes = 427.98 years

* The values taken the graph may vary according to the accuracy.