Chapter 5 Stress Strain Relation

Chapter 5 Stress Strain Relation 5.1 General Stress Strain sstem Parallelepiped, cube F 5.1.1 Surface Stress Surface stresses: normal stress shear stress, lim A 0 ( A ) F A lim F lim A 0 A A 0 F A lim A 0 ( A ) F A lim F lim A 0 A A 0 F A lim A 0 ( A ) F A lim F lim A 0 A A 0 F A 5-1

where F, F, F component of force vector F F acting in the direction of the -ais A area of the - face of the element A A subscripts : subscript indicates the direction of stress : 1st - direction of the normal to the face on which acts 2nd - direction in which acts general stress sstem: stress tensor ~ 9 scalar components [Re] Tensor ~ an ordered arra of entities which is invariant under coordinate transformation; includes scalars & vectors ~ 3 n 0th order 1 component, scalar (mass, length, pressure) 5-2

1st order 3 components, vector (velocit, force, acceleration) 2nd order 9 components (stress, rate of strain, turbulent diffusion) At three other surfaces, ' + ' + ' + ' + ' + ' + (5.1) Shear stress is smmetric. Shear stress pairs with subscripts differing in order are equal. [Proof] In static equilibrium, sum of all moments and sum of all forces equal ero for the element. 5-3

First, appl Newton's 2nd law F du m dt Then, consider torque (angular momentum), T d d d dω dt dt dt dt 2 T ( rmu) ( r mω) ( Iω) I where I moment of inertia 2 rm r radius of gration dω angular acceleration dt Thus, 2 dω T mr (A) dt Now, take a moment about a centroid ais in the -direction LHS T 2 2 2 ( ) ( ) ( ) dω RHS ρdvolr ρr dt 2 2 dω dt 2 ( ) 2 ρr After canceling terms, this gives 2ρr 2 dω dt dω dt 5-4

r lim 2 0,, 0 0 [Homework Assignment-Special work] Due: 1 week from toda 1. Make our own Stress Cube using paper bo. 5-5

5.1.2 Strain components Strain normal strain: shear strain: ε linear deformation γ angular deformation i) Displacement (translation): ξηζ,, (,, ) '( + ξ, + η, + ζ ) O O ii) Deformation: due to sstem of eternal forces OABC O' A' B' C ' (1) Deformation 1) Normal strain, ε ε change in length original length 5-6

OC ' ' OC ξ + + ξ + ( + ξ ) O'C' OC ξ ε lim lim 0 OC 0 ε η + + η + ( η ) O'A' OA + η lim lim 0 OA 0 ζ ε ~ ε is positive when element elongates under deformation 2) Shear strain, γ ~ change in angle between two originall perpendicular elements For plane C D O D ( ) ( ) γ lim θ + θ lim tanθ + tanθ c A c A, 0, 0 lim, 0 η ξ + η ξ ξ η + + + ξ < A E O E 5-7

γ η + ξ γ γ ζ η + ξ ζ + (5.4) [Re] displacement vs. deformation Motion translation rotation Deformation linear deformation angular deformation (2) displacement vector δ δ ξi + η j + ζk (3) Volume dilation e change of volume of deformed element original volume 5-8

ξ η ζ + d ( V ) + + e V ξ η ζ + + ε + ε + ε e ε + ε + ε (5.6) ξ η ζ e + + δ --- divergence (5.7) 5-9

5.2 Relations between Stress and Strain for Elastic Solids 5.2.1 Normal Stresses Hooke's law: stress is linear with strain E ε ε 1 E in which E Young's modulus of elasticit ε elongation in the dir. due to normal stress, dir. : ε E dir. : ε E Now, we have to consider other elongations because of lateral contraction of matter under tension. ε ' elongation in the dir. due to ε '' elongation in the dir. due to ' ε ε ε 5-10

Now, define ε ' nε n E ε '' nε n E (5.9) (5.10) where n Poisson's ratio [Re] Cork Thus, total strain ε is n 1 ε ε + ε' + ε'' E E E ( + ) n ( + ) ε 1 n( ) E + 1 ε n( + ) E (5.12) 5-11

5.2.2 Shear Stress ~ Hooke s law Gγ γ γ γ η ξ G + ζ η G + ξ ζ G + where G shear modulus of elasticit G E 2( 1+ n) (5.14) Volume dialation 1 e ε + ε + ε n E ( + ) 1 + n ( + ) E 1 + n E ( + ) 1 ( 1 2n)( + + ) E (5.15) arithmetic mean of 3 normal stresses 1 ( + + ) (5.16) 3 5-12

Combine Eqs. (5.12), (5.14) and (5.15) ne 2G ε + 1 2n (5.17) Therefore e 2G ε 3 e 2G ε 3 e 2G ε 3 (5.18) η ξ G + ζ η G + ξ ζ G + (5.19) [Proof] Derivation of Eqs. (5.17) & (5.18) 1 1 2 E (A) (5.15) e ( n)( + + ) 1 E (B) (5.12) ε n( + ) (5.14) G E 2(1 + n) E 2 G(1 + n) (C) 5-13

i) Combine (A) and (B) + ( ) ( ) ( ) n n 1 2 n e n + + + + ( 1+ 2n) ( 1 2n) E E 1 ε n( + ) E n 1+ n e + ε ( 1 2n) E E n ε + e 1+ n ( 1 2n) (D) Substitute (C) into (D) n 2G ε + e ( 1 2n) Eq. (5.17) ii) Subtract (5.16) from (5.17) n 1 2G ε + e ( 1 2n) 3 ( + + ) (E) Substitute (A) into (E); + + E e ( 1 2n) n 1 E RHS of ( E) 2G ε + e e ( 1 2n) 3 ( 1 2n) 1 n 2Gn 12G( ) + 1+ n 2 2 n Gε + e G ε 3 e ( 1 2n) 3 ( 1 2n) ( 1 2n) ( 1+ 2n) 1 ( 1 2 n) 1 2G ε 3 e 2G + ε e ( 1 2n) 3 5-14 Eq. (5.18)

5.3 Relations between Stress and Rate of Strain for Newtonian Fluids Eperimental evidence suggests that, in fluid, stress is linear with time rate of strain. stress t ( strain) Newtonian fluid (Newton's law of viscosit) [Cf] For solid, stress strain 5.3.1 Normal stress For solid, Eq. (5.18) can be used as Hookeian elastic solid: F e 2 ε 2 L 3 B analog, G Newtonian fluid: Ft e 2 t ε 2 L 3 (5.20) Ft Now set 2 L µ dnamic viscosit Time rate of strain Then, ε 2 e 2µ µ t 3 t (5.21) B the wa, 5-15

ε ξ, e δ ξ η ζ u, v, w ( ξηζ,, displacement) t t Therefore, ε ξ ξ u t t t e δ u v w q + + t t δ ξi+ η j+ ζk δ q ui + v j + wk t u v w q + + (5.22) (5.23) Eq. (5.21) becomes u 2 + 2µ µ 3 ( q) For compressible fluid, u 2 + 2µ µ 3 ( q) v 2 + 2µ µ 3 ( q) w 2 + 2µ µ 3 ( q) (5.24) 5-16

For incompressible fluid, de q 0 dt q 0 time rate of volume epansion0 Continuit Eq. Therefore, Eq. (5.24) becomes u + 2 µ v + 2 µ w + 2 µ 5.3.2. Shear stress B following the same analog µ G η ξ Ft η ξ + + L 2 t η t η ξ v u µ + µ + t t v t ξ u 5-17

v u µ + w v µ + u w µ + (5.25) [Appendi 1] µ v u + i), ' 5-18

ii), ' iii) composition Relation between thermodnamic pressure p and mean normal stress 1) Assume viscous effects are completel represented b the viscosit µ incompressible fluid 1 p ( + + ) (5.26) 3 ~ minus sign accounts for pressure (compression) for 2) For compressible fluid '( q) p + µ in which µ ' 2nd coefficient of viscosit associated solel with dilation bulk viscosit 5-19

Since, dilation effect is small for most cases ( q) µ ' 0 p For ero-dilation viscosit effects ( µ ' 0), (5.24) becomes u 2 p 2µ + µ 3 ( q) v 2 p 2µ + µ 3 ( q) w 2 p 2µ + µ 3 ( q) (5.29) Normal stress pressure Viscous effects Shear stresses in a real fluid µ v u + µ w v + u w µ + (5.30) For ero viscous effects ( µ 0) inviscid fluids in motion and for all fluids at rest p 0 5-20

[Appendi 2] Bulk viscosit and thermodnamic pressure Boundar-Laer Theor (Schlichting, 1979) pp. 61-63 '( q) p + µ If fluid is compressed, epanded or made to oscillate at a finite rate, work done in a thermodnamicall reversible process per unit volume is de W p q P dt ~ dissipation of energ where µ ' bulk viscosit of fluid that represents that propert which is responsible for energ dissipation in a fluid of uniform temperature during a change in volume at a finite rate second propert of a compressible, isotropic, Newtonian fluid [Cf] µ shear viscosit first propert µ ' 0, p µ ' 0, p Direct measurement of bulk viscosit is ver difficult. 5-21

[Appendi 3] Normal stress Normal stress pressure + deviation from it p + ' p + ' p + ' Thus, stress matri becomes p 0 0 ' 0 p 0 ' + 0 0 p ' Normal stresses are proportional to the volume change (compressibilit) and corresponding components of linear deformation, a, b, c. Thus, p+ λ ( a+ b+ c) + 2µ a p+ λ ( a+ b+ c) + 2µ b p+ λ ( a+ b+ c) + 2µ c where λ compressibilit coefficient 5-22

Homework Assignment # 3 Due: 1 week from toda 5-1. Verif Eq. (5-14) G E 2(1 + n). 5-3. Consider a fluid element under a general state of stress as illustrated in Fig. 5-1. Given that the element is in a gravit field, show that the equilibrium requirement between surface, bod and inertial forces leads to the equations g + + + ρ ρa + + + ρg ρa g + + + ρ ρa 5-23