1. VISCOSITY

2. Plastic deformation occurs by dislocation motions in crystallic structures.
For noncrystallic structures plastic deformations are due to viscous flow.
The characteristic property for viscous flow, viscosity, is a measure of a noncrystalline material’s resistance to deformation. L
Plate A F V
When a tangential force (F) acts on the plate, the plate moves with respect to the bottom.

3. The velocity of the liquid particles in each layer is a function of the distance L. Thus the rate at which the particles change their position is the measure of the rate of flow. dγ dL dV =
velocity gradient dt
rate of flow A dL dV
Newton expresses: F = η
Since τ =
F / A
η : coefficient of viscosity dt dγ
τ = η
τ = η dL dV
& 1 2

4. Unit of viscosity is Pa.s (Pascal-seconds) (N.s/m2)
The liquids that follow equations (1) & (2) are termed as Newtonian Liquids.
Newtonian liquids η

5. Viscosity also varies with temperature.
A: Constant
E: Energy of activation
R: Gas constant
T: Absolute temperature η 1
= A . e-E/RT
When solid particles are introduced into Newtonian liquids, viscosity increases.
η0: Coefficient of viscosity of the parent liquid.
Ø: Volume concentration of suspended particles
η = η0 (1+2.5 Ø)
η = η0 (1+2.5 Ø Ø2)

6. NON-NEWTONIAN MATERIALS
In certain materials, τ-dγ/dt does not obey the linearity described by Newton, i.e. Viscosity may vary with the rate of shear strain.
Dilatant: η increases with increasing dγ/dt or τ (clay)
Newtonian: (all liquids)
Pseudoplastic: η decreases with dγ/dt or τ (most plastics)
Newtonian η

7. The relationship between dγ/dt & τ can be described by the following general equation.η 1
= τn . dt dγ
If n=1 → Newtonian
n > 1 → Pseudoplastic
n < 1 → Dilatant

8. Fresh cement pastes & mortars, have highly concentrated solid particles in the liquid medium. Such a behaviour is described by Bingham’s equation. dt dγ
τ = τy + η dt dγ τ τy
(Upto τy there is no flow)

9. VISCOELASTICITY & RHEOLOGICAL CONCEPTS
Viscoelastic behaviour, as the name implies, is a combination of elasticity & viscosity.
Such a behaviour can be described by Rheological Models consisting of springs (for elasticity) & dashpots (for viscosity).

10. (Load) (Elastic) ε = σ/E (Viscous) dε/dt = σ/γ (Viscoelastic) t1 Time
Strain
Load t0
(Load)
(Elastic)
ε = σ/E
(Viscous)
dε/dt = σ/γ
(Viscoelastic)

11. Models to explain the viscoelastic behavior:
Maxwell Model
Kelvin Model
4-Element Model (Burger’s Model)

12. 1. Maxwell Model: A spring & a dashpot connected in series.σ
k = E
β = 1/η
A spring & a dashpot connected in series.
The stress on each element is the same:
σspring = σdashpot
However, the deformations are not the same:
εspring ≠ εdashpot

13. When the force (stress) is applied the spring responds immediately and shows a deformation εspring = σ/E
At the same time the dashpot piston starts to move at a rate βσ = σ/η
and the displacement of the piston at time t, is given by:
Therefore the total displacement becomes:

14. ε P t
εdashpot→ viscous permanent def.
εspring
εdashpot

15. Relaxation: An important mode of behavior for viscoelastic materials can be observed when a material is suddenly stroked to ε0 & this strain is kept constant.
Instead of this loading pattern strain is kept constant. ε0 ε t σ0 σ

16. Solving this differential equation Where : σ0 = Eε0
&
If ε is constant →
Solving this differential equation
Where :
σ0 = Eε0

17. ε
Relaxation time is a parameter of a viscoelastic material. ε0 t
If the body is left under a constant strain, the stress gradually disappears (relaxes). This pehenemenon can be observed in glasses & some ceramics. σ σ0
t=0
0.37σ0
trel t

18. 2. Kelvin Model:
Consists of a spring & a dashpot connected in parallel.
In this case the deformations are equal but the stresses are different.
εspring = εdashpot
σspring ≠ σdashpot
σ = σspring + σdashpot E
1/η σ

19. σ σ0 t ε t
(Delayed elasticity)

20. At each strain increment the spring will extend by σ/E so that a part of the load is taken over and the force on the piston decreases. Thus a final displacement is reached asymtotically and when the load is removed, there will be an asymtotic recovery until σ=0.
When a viscoelastic Kelvin Body is subjected to a constant stress, σ0, the response could be obtained by solving the differential equation.

21. is reached at time t=∞ retardation timeσ t
is reached at time t=∞
retardation time
When stressed the elastic deformation in the spring is retarded by the viscous deformation of the dashpot. ε
tret ε0
Retarded elastic strain (delayed elastic strain)
0.63ε0

22. 3. Burger’s Model:
The actual viscoelastic behavior of materials is very complex. The simplest models, Maxwell & Kelvin Models, explain the basic characteristics of viscoelastic behavior.
The Maxwell Model, for example, has a viscous character and explains the relaxation behavior of viscoelastic materials
The Kelvin Model on the other hand has a solid character and explains the retarded elasticity behavior.

23. However, none of the mentioned models completely explain the real behavior of viscoelastic materials.
There are other models with different E and η constants but they are rather complex.
One such model is given by BURGER, which consists of a Maxwell Model and Kelvin Model connected in series.

24. σ E1 E2 η1 η2 σ σ0 t
εvis
εvis+εret ε1 t ε

25. Kelvin (retarded elasticity)
Spring (elastic)
Dashpot (viscous)
Kelvin (retarded elasticity)
Most engineering materials show certain deviations from the behavior described by the 4-Element Model. Therefore the deformation equation is usually approximated as:
Instantaneous elastic
Retarded elastic
Viscous

26. Where “k, β & γ” are material constants & “α, n” are constants accounting for nonlinearity.

27. Example 1: For a certain oil, the experimentally determined shear stress, rate of flow data provided the following plot. Determine the viscosity of the oil.
dγ/dt (1/sec)
0.9
0.6
0.3 30 20 10
τ (Pa)

28. Example 2: When a concrete specimen of 75 cm in length is subjected to a 150 kgf/cm2 of constant compressive stress, the following data were obtained.
t (month) ε 1
0.0006
0.0007
Assume
where B is constant.
What will be the total deformation under 150 kgf/cm2 after 6 months?

29. Example 3: A glass rod of 2. 5 cm in diameter & 2
Example 3: A glass rod of 2.5 cm in diameter & 2.5 m in length, is subjected to a tansile load of °C.
Calculate the deformation of the rod after 100 hrs.
Determine trel (relaxation time)
What is the time during which the stress in the material would decay to 5% of its initial volume?
η=2x °C & E=1.55x105 kgf/cm2
Assume that the behavior of glass at this temperature can be approximated by a Maxwell Model.

30. For Normal Stresses & Strains the viscous behavior is described by dε/dt=σ/λ where λ is called “the Coefficient of Viscous Traction” & equals to “3η”.
η=2x1012 poise (1 poise = 1 dyne.sec/cm2) & (1 kgf = 106 dyne)
After 100 hrs the total strain is 3.4x10-7x100x60x60 =
Δ = x250 = 30.6 cm