1. 4. Properties of Materials Sediment (size) Physical States of Soil Concepts of Stress and Strain Normal and Shear Stress Additional Resistance Components

2. Sediment Sediments (rock fragments – debris – soil) at the core of geomorphological investigations Erosion  transport  deposition Landform creation

3. Sediment size Size of sediment (or range of sizes) control many phenomena such as –Rate of infiltration of water into soils –Volume of water contained in soils –Ease of erosion of earth surfaces –Stability of slopes –Etc…

4. Size measurements Size determined from laboratory analysis External diameter of particles being measured, directly of indirectly (e.g. sieve size analysis). Methods vary with range of particle sizes involved. Sediment size varies from microns to meters

5. Particle sizes Expressed either in mmor  (phi)  = - log 2 D (mm)  = - log 2 D (mm) if D = 8 mm,  = -3 (since 2 3 = 8) if D = 8 mm,  = -3 (since 2 3 = 8) D = 2 mm,  = -1 D = 2 mm,  = -1 D = 0.065 mm,  = 4.0 D = 0.065 mm,  = 4.0 since 2 -4 = 0.0625 since 2 -4 = 0.0625

6. Particle sizes - classification mm  (phi) Gravel> 2 mm 2 mm< -1.0 Sand0.065 – 2.0 mm-1.0 to 4.0 Silt0.0039-0.0654.0 to 8.0 Clay 8.0

9. Physical States of Soil (and dimensionless ratios) Physical States of Soil (and dimensionless ratios) Surficial materials composed of a combination of air, water, and solids (particles) Proportion of each phase varies in any sample but has significant impact on how and when movement will occur (e.g. surface erosion, mass movements…)

10. Physical States of soil…. V: volume W: weight ρ (rho): unit weight (mass density) V s : volume of solid V w : volume of water V v : volume of voids (air + water)

11. Porosity: n = V v /V Volume of voids/total volume Volume of voids/total volume Tends to increase as particle size decreases This controls in part infiltration of water, water movement in soils… Voids ratio: e = V v /V s volume of voids/volume of solids volume of voids/volume of solids

12. Moisture content: m = W w /W s Weight of water/weight of sediment Weight of water/weight of sediment Degree of saturation: S = V w /V v Volume of water/volume of voids Volume of water/volume of voids

13. Moisture content and stability Moisture content (e.g. presence or absence of saturation) affects the stability of a given volume of material Particularly important when dealing with fine material (e.g. clay particles smaller than 0.002 mm or two microns). Definition of “Atterberg Limits” to describe state of a soil

14. “Atterberg” Limits Series of tests that can be performed on the state of a soil “Limits” defined in relation to moisture content: –Liquid limit (L l ) –Plastic limit (P l ) –Shrinkage limit (S l ) (increasing moisture content from S l to L l )

15. Example of landslides where “liquid” limit has been reached (taking place in very fine clay deposit)…

16. Concepts of Stress and Strain Stress (σ or  ): Defined as the force (F) per unit area (A) acting on a plane σ = F/A Units for stress are Pascals (N/m 2 - Newtons per square meter) or dyn/cm 2. Units for stress are Pascals (N/m 2 - Newtons per square meter) or dyn/cm 2.

17. Stress and Strain Strain: Strain, represented by symbol ε (Greek letter epsilon) is defined as a ratio of lengths before and after deformation (measured by dividing the change in length - ∆l – by the original length – L): (measured by dividing the change in length - ∆l – by the original length – L): ε = ∆l / L So, forces exerted on an object can cause the object to deform.

18. The way in which a material exhibits strain when stressed is of importance in characterizing that material A good example of that concept is internal deformation and glacier flows (“plastic flow”…)

19. Normal and Shear Stresses Normal stress: Stress applied normal to the horizontal plane (σ) Shear stress: Force (per unit area) acting parallel to a given surface, symbolized by , is a shear stress and is also expressed in N/m 2. σ : normal stress σ : normal stress  : shear stress  : shear stress ε : strain ε : strain

21. Shear stress and normal stress for single particles  = ρ s g D sin θ σ = ρ s g D cos θ where ρ s is density of sediment (2650 kg m -3, g is acceleration due to gravity (9.81 m s - 2), D is particle size (m) and  is slope angle sin θcos θ θ = 1 deg0.01740.999 100.17360.984 100.17360.984 200.3420.939 200.3420.939 300.50.866 300.50.866 Therefore, as slope angle increases, driving force (  ) increases while σ decreases.

22. Definitions of normal and shear stresses when a volume of soil is concerned:

23. Shear strength Normal stress is only once component of resistance to movement Total resistance force is called shear strength (S t ) S t = f (c, σ, , S) Where c: cohesion of material σ: normal stress σ: normal stress  : friction angle  : friction angle S: degree of saturation S: degree of saturation

24. The ratio of shear stress (driving force) to shear strength (resisting force) will determine if a slope is stable or if mass movement will occur. When shear stress exceeds shear strength: unstable conditions and mass movements. When shear stress exceeds shear strength: unstable conditions and mass movements.

25. Cohesion - c Cohesion (if any) increases the strength. This can be measured and has the same units as stress. Coarse-grained material has no cohesion (sand, gravel…). Cohesive strength within fine-grained sediment (e.g. clay-rich soils posses a significant degree of cohesion) Coarse-grained material has no cohesion (sand, gravel…). Cohesive strength within fine-grained sediment (e.g. clay-rich soils posses a significant degree of cohesion)

26. Friction angle (  ): affects the stability of particles; a larger angle increases the stability (or resisting force). Often assumed to be around 32 degrees for sand and gravel particles

27. Moisture content and normal stress: the total normal stress includes two elements: –Effective normal stress (σ‘) –Pore pressure (  )

28. Effective normal stress (σ‘) Voids filled with air and/or water; pore pressure exists in these spaces. These spaces tend to support part of the normal stress. Normal stress modified to take into account the degree of saturation, i.e.: σ‘ = σ –  σ‘ = σ –  σ‘: effective normal stress σ: applied normal stress  : pore water pressure

29. Effective normal stress Takes the effects of pore water pressure into account when strength of soil is calculated Pressure exerted by water in soil pore spaces acts upon grains, tending to force them apart when saturated (positive pore water pressure) OR pull them together when the pores spaces have only a small amount of water in them (negative pore water pressure) pull them together when the pores spaces have only a small amount of water in them (negative pore water pressure)

30. Effective normal stress (cont.) Note that if  is negative - capillary attraction, then σ‘ > σ and vice-versa if the soil is fully saturated Shear strength (S t ) therefore defined as: S t = c + σ‘ tan  S t = c + σ‘ tan 

31. Shear strength…. Below the water table  is positive; σ‘ = σ –  and σ‘ < σ Above water table  is negative; σ‘ > σ and shear strength increases In dry soil, σ‘ = σ