Download presentation

Presentation is loading. Please wait.

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 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

Similar presentations

Presentation is loading. Please wait....

OK

MECHANICAL PROPERTIES OF MATERIALS

MECHANICAL PROPERTIES OF MATERIALS

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on eye of tiger Ppt on bresenham's line drawing algorithm Ppt on water conservation download Ppt on printers and plotters Ppt on web browser security Ppt on id ego superego quiz Ppt on history of facebook Ppt on time division switching pdf Ppt on micro operations of an airport Convert pdf ppt to ppt online ticket