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Presentation on theme: "VISCOSITY."— Presentation transcript:


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. dL dV = velocity gradient dt rate of flow A dL dV Newton expresses: F = η Since τ = F / A η : coefficient of viscosity dt 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)

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 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 τ = τy + η dt τ τy (Upto τy there is no flow)

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 time
σ 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

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