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VISCOSITY.  Plastic deformation occurs by dislocation motions in crystallic structures.  For noncrystallic structures plastic deformations are due to.

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Presentation on theme: "VISCOSITY.  Plastic deformation occurs by dislocation motions in crystallic structures.  For noncrystallic structures plastic deformations are due to."— Presentation transcript:

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. 1 Since τ =F / A velocity gradient = dL dV dt dγdγ rate of flow A dL dV  Newton expresses: F = η τ = η dL dV & dt dγdγ τ = η 2 η : coefficient of viscosity

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

5  Viscosity also varies with temperature. 1) η = η 0 (1+2.5 Ø) 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. 2) η = η 0 (1+2.5 Ø Ø 2 ) η 0 : Coefficient of viscosity of the parent liquid. Ø: Volume concentration of suspended particles

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. Newtonian η Dilatant: η increases with increasing dγ/dt or τ (clay) Newtonian: (all liquids) Pseudoplastic: η decreases with dγ/dt or τ (most plastics)

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

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

9  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). VISCOELASTICITY & RHEOLOGICAL CONCEPTS

10 (Load) t1t1 Time Strain Load Strain Time t0t0 t0t0 t0t0 t0t0 t1t1 t1t1 t1t1 (Elastic) ε = σ/E (Viscous) dε/dt = σ/γ (Viscoelastic)

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

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

13  When the force (stress) is applied the spring responds immediately and shows a deformation ε spring = 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 ε dashpot → viscous permanent def. ε spring ε dashpot

15  Relaxation: An important mode of behavior for viscoelastic materials can be observed when a material is suddenly stroked to ε 0 & this strain is kept constant. Instead of this loading pattern strain is kept constant. ε0ε0 ε t t σ0σ0 σ

16 & If ε is constant → Solving this differential equation Where : σ 0 = Eε 0

17 Relaxation time is a parameter of a viscoelastic material. t=0 t rel t 0.37σ 0 σ0σ0 σ t ε0ε0 ε  If the body is left under a constant strain, the stress gradually disappears (relaxes). This pehenemenon can be observed in glasses & some ceramics.

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.  ε spring = ε dashpot  σ spring ≠ σ dashpot  σ = σ spring + σ dashpot E 1/ η σ σ

19 (Delayed elasticity) t t ε σ0σ0 σ

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, σ 0, the response could be obtained by solving the differential equation.

21 σ t ε t ret ε0ε0 Retarded elastic strain (delayed elastic strain) 0.63ε 0 is reached at time t=∞ retardation time  When stressed the elastic deformation in the spring is retarded by the viscous deformation of the dashpot.

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 σ E1E1 E2E2 η1η1 η2η2 σ σ0σ0 t ε vis ε vis +ε ret ε1ε1 ε1ε1 t ε

25 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) τ (Pa)

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

29 Example 3: A glass rod of 2.5 cm in diameter & 2.5 m in length, is subjected to a tansile load of °C. a) Calculate the deformation of the rod after 100 hrs. b) Determine trel (relaxation time) c) What is the time during which the stress in the material would decay to 5% of its initial volume? η=2x °C & E=1.55x10 5 kgf/cm 2 Assume that the behavior of glass at this temperature can be approximated by a Maxwell Model.

30 a) 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=2x10 12 poise (1 poise = 1 dyne.sec/cm 2 ) & (1 kgf = 10 6 dyne) After 100 hrs the total strain is 3.4x10 -7 x100x60x60 = Δ = x250 = 30.6 cm


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