SUPPLEMENTAL PROBLEM (Please detach and include with your homework submission. S

Question

SUPPLEMENTAL PROBLEM (Please detach and include with your homework submission. Show all work) Fill obtained from a local borrow area (G. 2.70) was compacted in accordance with the following end-product specifications: (1 ) RC 95 % (standard Proctor) (2) wept-2% w wage + 1 % A standard Proctor compaction test was conducted on the soil, yielding the following results: 110 106 102 98 94 90 12 14 16 18 20 222426 Moisture Content, w (%) (a) Fill in the table below and plot the Zero Air Voids (ZAV Curve on the graph above (2 points). 24 26 20 (b) Calculate the degree of saturation of the compacted soil at w- Wa (2 points). (c) Draw the acceptable zone of compaction on the graph (2 points)

Explanation / Answer

Solution:

here each smallest square in the graph is = 0.5 units of moisture content along X-axis

and = 1 unit of dry density along Y-axis.

so for w=20 , Yd = 103 pcf

w= 22, Yd= 98.5 pcf

w=24, Yd= 95 pcf

w=26, Yd= not plotted on graph, can be found by extrapolation

(y-y1)/ (y2-y1) = (x-x1)/(x2-x1)

(26-24)/(22-24) = (x- 95)/ (98.5 -95)

x= 91.5 = Yd

we know, G*w = e*S and for zero air voids,the soil is completely saturated or S=1

also Yd = G*Yw / 1+e , e= void ratio and Yw = unit weight of water =62.4 pcf

thus for each corresponding water content we get a value of e., and again values of Yd

G*w= e

G=2.7 ,thus the values are,

w=0.24, e= 0.648 , Yd= 102

w=0.22, e= 0.594 , Yd= 106

w= 0.18, e= 0.486 , Yd= 113.4

w=0.16, e= 0.432 , Yd= 118

If the w and Yd values are plotted on he graph it would yield a straight line which is the zero air voids line.

Ans b- From the graph it is evident that wopt = 18.5 which is the peak of the curve

from the values above , we need to find value of e at w=18.5

so, 18.5 -18 / 22 -18 = e - 0.486 / 0.594 - 0.486 (3-pt straight line formula)

e=0.4995= 0.5

G*w =e*S

2.7*0.185 = 0.5*S

S= 1

Ans d-

relative compaction= compaction in field(Yd) / compaction in lab (Yd)

Y (field) = weight / volume = 2.38 / (34.58/ 123)

=118.93 pcf

thus, Yd= Y / 1+w = 101.3

thus RC= (96 / 101.3) = 94.76 % = 95% , thus conforming with the standard proctor test

based on sand cone test,

Yd= G*Yw / 1+e

101.3 = 2.7 * 62.4 / 1+e

e= 0.663

G*w =e*S

2.7 * 0.174 = 0.663 * S

S= 70.84%

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