Soil and Rock

Soil and Rock

اسلاید 1: Soil and RockSoil and rock are the principle components of many construction projects. Knowledge of their properties, characteristics, and behavior is important to those associated with the design or construction of projects.

اسلاید 2: Soil and Rock

اسلاید 3: Soil and Rock

اسلاید 4: Steel and concrete are construction materials that are basically homogeneous in composition. As such, their behavior can be predicted. Soil and rock are just the opposite. By nature they are heterogeneous. In their natural state, they are rarely uniform. Soil and Rock

اسلاید 5: Soil and rock are heterogeneous. They are rarely uniform and work processes are developed by comparison to a similar type material with which previous experience has been gained. To accomplish this, soil and rock types must be classified. Soil and Rock

اسلاید 6: GRADATIONSoil gradation is the distribution, in percent (%) by weight, of individual particle sizes.

اسلاید 7: SOIL TYPESORGANIC SOILSWill usually have to remove before building.

اسلاید 8: SOIL TYPESBulky shaped soil grainsNON-COHESIVE

اسلاید 9: SOIL TYPESSmall grained  #200 Mesh sieve Platy shaped soil grainsCOHESIVE

اسلاید 10: SOIL LIMITSAtterburg LimitsLL - Liquid limitPL - Plastic limitPI - Plasticity Index

اسلاید 11: SOIL LIMITSStages of ConsistencyMoisture content decreasing

اسلاید 12: SOIL LIMITSLL - Liquid limitis the water content of a soil when it passes from the plastic to liquid state.

اسلاید 13: SOIL LIMITSLL - Liquid limitNon-cohesive or sandy soils have low LLs -- less than 20.Clay soils have LLs ranging from 20 to 100.

اسلاید 14: SOIL LIMITSPL - Liquid limitis the lowest water content at which a soil remains plastic.1/8 inch diameter thread

اسلاید 15: SOIL LIMITSPI - Plastic IndexPI = LL - PLThe higher the PI the more clay that is present in the soil.

اسلاید 16: Volumetric Measure Bank cubic yards (bcy) Loose cubic yards (lcy)Compacted cubic yards (ccy)bcylcyccy

اسلاید 17: COMPACTIONEach soil has its particular optimum moisture content (OMC) at which a corresponding maximum density can be obtained for a given amount of compactive input energy.

اسلاید 18: COMPACTION

اسلاید 19: PROCTOR TESTStandard Proctor orAASHTO T-99Soil sample 1/30 cubic foot3 layersCOMPACTION

اسلاید 20: COMPACTIONPROCTOR TESTModified Proctor orAASHTO T-180Soil sample 1/30 cubic foot5 layers

اسلاید 21: COMPACTION SPECIFICATIONSTypically specifications give an acceptable range of water content, OMC ± 2% for example.

اسلاید 22: COMPACTION SPECIFICATIONSThe specification also sets a minimum density, 95% of max. dry density for a specific test126.4

اسلاید 23: Must work in the box.

اسلاید 24: Soil Weight-Volume RelationshipsEqu. 4.1Equ. 4.3Equ. 4.2

اسلاید 25: Water Content =

اسلاید 26: Water Content = = 0.18 or 18%

اسلاید 27: Soil Weight-Volume RelationshipsDry weight is related to unit weight by water content,and when you move rock and dirt the only thing that stays constant is the weight of the solid particles.

اسلاید 28: Soil Weight-Volume Relationshipsis the weight of the solid particles.When you move rock and dirt the only thing that stays constant

اسلاید 29: EXERCISE unit weight () of 94.3 pcf water content () of 8%.The excavated material has a

اسلاید 30: EXERCISEdry unit weight (d) of 114 pcfwater content () of 12%.The embankment will be compacted to

اسلاید 31: EXERCISEThe net section of the embankment is 113,000 cy.How many cubic yards of excavation will be required to construct theembankment?

اسلاید 32: As material is moved from the excavation to the compacted fill the only constant is the weight of the solid particles (d).EXERCISE

اسلاید 33: EXERCISE Step 1Weight of the solid particles which make up the embankment (fill). a dry unit weight (d) of 114 pcf 113,000 cy embankmentConversion factor cy to ft3

اسلاید 34: EXERCISE Step 2Use relationship d -  to calculate the dry unit weight of the excavated material. a unit weight () of 94.3 pcf a water content () of 8% d = 87.31 pcf

اسلاید 35: EXERCISE Step 3Calculate the weight of the solid particles which make up the excavation (cut).

اسلاید 36: EXERCISE Step 4The weights must be equal therefore:=Conversion factors cancel out.

اسلاید 37: EXERCISE Step 4The weights must be equal therefore: x =x = 147,535 cy excavated material required

اسلاید 38: EXERCISECheck the water requirements.Will a water truck be needed on the job or will it be necessary to dry the material?

اسلاید 39: Water ?Water content () is? x d = weight of water/cf

اسلاید 40: Step 1 Water from CutVol. Cutdconversion factor()

اسلاید 41: Step 1 Water from Cut = 27,825,120 lb waterdelivered with the borrow material

اسلاید 42: Step 2 Water needed at the FillVol. Embdconversion factor()

اسلاید 43: Step 2 Water needed at the Fill = 41,737,680 lb waterneeded at the fill

اسلاید 44: Step 3 Water DeficiencyNeeded at the fill41,737,680 lb Delivered w/ cut27,825,120 lbWater deficiency13,912,560 lb

اسلاید 45: PE 2 Step 4 Convert to GallonsWater deficiency 13,912,560 lbWater weights 8.33 lb/galNeed to add 1,670,175 gallons

اسلاید 46: Step 5 Gallons per cyWater deficiency 1,670,175 galVolume of cut147,535 cy Need 11.3 gal/cy

اسلاید 47: Adding WaterUsing sprinklers to add moisture to a foundation fill.

اسلاید 48: Reducing MoistureDisking a heavy clay fill to reduce moisture.