1. Soil and Rock
Soil 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.

2. Soil and Rock

3. Soil and Rock

4. Soil and Rock
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.

5. Soil and Rock
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.

6. GRADATION
Soil gradation is the distribution, in percent (%) by weight, of individual particle sizes.

7. SOIL TYPES
ORGANIC SOILS
Will usually have to remove before building.

8. SOIL TYPES
NON-COHESIVE
Bulky shaped soil grains

9. SOIL TYPES COHESIVE Platy shaped soil grains Small grained
 #200 Mesh sieve

10. SOIL LIMITS
Atterburg Limits
LL - Liquid limit
PL - Plastic limit
PI - Plasticity Index

11. SOIL LIMITS
Stages of Consistency
Moisture content decreasing

12. SOIL LIMITS LL - Liquid limit
is the water content of a soil when it passes from the plastic to liquid state.

13. SOIL LIMITS LL - Liquid limit
Non-cohesive or sandy soils have low LLs -- less than 20.
Clay soils have LLs ranging from 20 to 100.

14. SOIL LIMITS PL - Liquid limit
is the lowest water content at which a soil remains plastic.
1/8 inch diameter thread

15. SOIL LIMITS PI - Plastic Index PI = LL - PL
The higher the PI the more clay that is present in the soil.

16. Volumetric Measure Bank cubic yards (bcy) Loose cubic yards (lcy)
Compacted cubic yards (ccy)
lcy
ccy
bcy

17. COMPACTION
Each 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.

18. COMPACTION

19. COMPACTION PROCTOR TEST Standard Proctor or AASHTO T-99
Soil sample 1/30 cubic foot
3 layers

20. COMPACTION PROCTOR TEST Modified Proctor or AASHTO T-180
Soil sample 1/30 cubic foot
5 layers

21. COMPACTION SPECIFICATIONS
Typically specifications give an acceptable range of water content, OMC ± 2% for example.

22. COMPACTION SPECIFICATIONS
The specification also sets a minimum density, 95% of max. dry density for a specific test
126.4

23. Must work in the box.

24. Soil Weight-Volume Relationships
Equ. 4.1
Equ. 4.2
Equ. 4.3

25. Water Content
 =

26. Water Content
 = = 0.18 or 18%

27. Soil Weight-Volume Relationships
Dry 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.

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

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

30. dry unit weight (d) of 114 pcf water content () of 12%.
EXERCISE
The embankment will be compacted to
dry unit weight (d) of 114 pcf
water content () of 12%.

31. How many cubic yards of excavation will be required to construct the
EXERCISE
The net section of the embankment is 113,000 cy.
How many cubic yards of excavation will be required to construct the
embankment?

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

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

34. EXERCISE Step 2
Use 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 =  pcf

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

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

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

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

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

40. Step 1 Water from Cut
Vol. Cut d
conversion factor
()

41. Step 1 Water from Cut = 27,825,120 lb water
delivered with the borrow material

42. Step 2 Water needed at the Fill
Vol. Emb d
conversion factor
()

43. Step 2 Water needed at the Fill
= 41,737,680 lb water
needed at the fill

44. Step 3 Water Deficiency Needed at the fill 41,737,680 lb
Delivered w/ cut 27,825,120 lb
Water deficiency 13,912,560 lb

45. PE 2 Step 4 Convert to Gallons
Water deficiency 13,912,560 lb
Water weights lb/gal
Need to add 1,670,175 gallons

46. 11.3 gal/cy Step 5 Gallons per cy Need Water deficiency 1,670,175 gal
Volume of cut 147,535 cy
Need
11.3 gal/cy

47. Adding Water
Using sprinklers to add moisture to a foundation fill.

48. Reducing Moisture
Disking a heavy clay fill to reduce moisture.