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انثى عدد المساهمات : 1
البلد : oman
تاريخ التسجيل : 15/05/2015

مُساهمةموضوع: ممكن مساعدة في كتابة discussion??   الجمعة مايو 15, 2015 8:49 am

سلام عليكم

ه>ا التقرير تبع تجربة Atterberg Limit Test

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INTRODUCTION:
In the early 1990s, a Swedish scientist named Atterberg developed a method to describe the consistency of fine-grained soils with varying moisture contents. Atterberg limits are defined as the water corresponding to different behavior conditions of fine-grained soil (silts and clays). The four states of consistency in Atterberg limits are liquid, plastic, semisolid and solid. The dividing line between liquid and plastic states is the liquid limit; the dividing line between plastic and semisolid states is the shrinkage limit. If a soil in the liquid state is gradually dried out, it wills past through the liquid limit, plastic state, plastic limit, semisolid state and shrinkage limit and reach the solid stage. The liquid, plastic and shrinkage limits are therefore quantified in terms of the water content at which a soil changes from the liquid to the plastic state. The difference between the liquid limit and plastic limit is the plasticity index. Because the liquid limit and plastic limit are the two most commonly used Atterberg limits, the following discussion is limited to the test procedures and calculation for these two laboratory tests.

The liquid limit is that moisture content at which a soil changes from the liquid state to the plastic state. It along with the plastic limit provides a means of soil classification as well as being useful in determining other soil properties.


OBJECTIVE:
To determine the liquid limit of a soil.
To determine the liquid limit of a soil.

THEORY:
m = Ww/Ws=(Wwet – Wdry)/(Wdry – Wcantainer)
m = Ww/Ws x 100






EQUIPMENTS:
Casagrande apparatus
Glass plate
Balance
Spatula
Containers
Soil
Oven












PRODEURE:
Liquid Limit
We obtain a soil sample and place in the brass cup of the liquid limit device, cut a standard groove from the back of the cup to the front, and count the number of drops of the device that are required to close the groove. If the number of drops is less than 20 or more than 30, the water content of the soil sample is adjusted and the procedure repeated until the number of drops is between 20 and 30 drops, at which time the water content of the soil is determined. The liquid limit is computed using an equation involving the number of drops required to close the groove and the associated water content of the soil. Two such determinations are done, and an average value is taken as the liquid limit.

We obtain a soil sample and place in the cup of the liquid limit device, cut a standard groove from the back of the cup to the front by using grooving tool, and count the number of drops of the device that are required to determine the correct rate to rotate the crank so that the cup drops approximately two times per second. If the number of drops is less than 20 or more than 30, the water content of the soil sample is adjusted and the procedure repeated until the number of drops is between 20 and 30 drops, at which time the water content of the soil is determined. The liquid limit is computed using an equation involving the number of drops required to close the groove and the associated water content of the soil. Two such determinations are done, and an average value is taken as the liquid limit.



Plastic Limit
Roll another portion of the soil sample between the fingers and a roughed glass rolling surface until a 1/8-inch diameter thread is obtained. Break the resulting specimen into smaller pieces, compress and repeat the rolling process to a diameter of 1/8-inch. This procedure is repeated until the mass crumbles and can no longer be made into threads. The plastic limit is now assumed to be reached and the water content.







OBSERVATION AND CALCULATION:

Plastic limit 1 2
Container No. 5 7
Mass of dry soil + container (m2)            g 50.051 45.756
Mass of dry soil + container (m3)            g 48.734 64.486
Mass of container (m3)                            g 41.69 57.64
Mass of moisture                                     g 1.317 1.27
Mass of dry soil                                       g 7.044 6.846
Moisture content   % 18.7 18.6
Average 18.65

Liquid limit 3 4 5
Container No. 10 9 8
Number of bumbs 43 28 17
Mass of wet soil + container (m2)           g 96.172 72.985 79.746
Mass of dry soil + container (m3)           g 93.810 71.513 77.418
Mass of container                                    g 82.32 64.717 66.704
Mass of dry moisture                              g 2.362 1.472 2.328
Mass of dry soil                                       g 11.49 9.796 10.714
Moisture content % 20.56 21.7 21.7






SAMPLE CALCULATION:
Plastic limit:
Mwet of soil    = ( Mw + container ) – ( Mcontainer )
                  = 50.051 – 41.69 = 8.361 g
Mwet of soil  = ( Mw + container) – ( Mcontainer )
               = 65.756 - 57.64 = 8.11 g
Mdry    = ( Mw + container ) – ( Mcontainer )
         = 48.374 – 41.69 = 7.044 g
Dry mass of soil = ( dry of soil + container) – (mass of container)
                           = 64.486 – 57.64 = 6.846 g
Mass of moisture  = ( Mwet of soil  ) – ( MDry )
                             = 8.361 – 7.044 = 1.317 g
Mass of moisture   = ( Mwet of soil ) – ( MDry )
                              = 8.116 – 6.846 = 1.27 g
Moisture content %  = ( Mwet – Mdry ) / MDry x100
                                 = 1.317/7.044x100
                                 = 18.7 %
Moisture content %  = ( Mwet – Mdry ) / Mass Dry x 100
                                 = 1.27 / 6.846 x 100
                                 = 18.6 %





Liquid limit:
Wet mass of soil = (mass wet of soil + container) – (Mcontainer)
                           = 96.172 – 82.32 = 13.852 g
Wet mass of soil = ( mass wet of soil + container) – (Mcontainer)
                           = 72.985 – 64.717 = 8.268 g
Wet mass of soil = (mass wet of soil + container) – (mass of container)
                           = 79.746 – 66.704 = 13.042 g
Dry mass of soil (10) = (dry mass of soil + container) – (mass of container)
                                 = 93.810 – 82.32 = 11.49 g
Dry mass of soil (9) = (dry mass of soil + container) – (mass of container)
                                = 71.513 – 64.717 = 6.796 g
Dry mass of soil (Cool = (dry mass of soil + container) – (mass of container)
                                = 13.042 – 10.714 = 2.328 g
Mass of moisture (10) = (Mwet of soil – Mdry of soil)
                                    = 13.852 – 11.49 = 2.362 g
Mass of moisture (9) = (Mwet of soil – Mdry of soil)
                                  = 8.268 – 6.796 = 1.472 g
Mass of moisture (Cool = (Mwet of soil – Mdry of soil)
                                  = 13.042 – 10.714 = 2.328 g
Moisture content % = (mass wet – Mdry) / dry mass x 100
                                = 2.362/11.49 x 100
                                = 20.6 %
Moisture content % = (mass wet – Mdry) / dry mass x 100
                                = 1.472 / 6.796 x 100
                                = 21.7 %
Moisture content % = (mass wet – Mdry) / dry mass x 100
                               = 2.328 / 10.714 x 100
                               = 21.7 %

DISCUSSION:












CONCLUSION:
After we have done the experiment (Liquid Limit and Plastic Limit Tests), we found that the plastic index of the soil is equal to 27 and the liquid limit of the soil is equal to 50.3 (obtained from fall-cone method). From the chart we found that the type of soil was falling under Inorganic silts of compressibility and organic silts.
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