Presented By : Ketulkumar Amin Enroll No:130460119043

Elastic Constant

ELASTIC CONSTANTS IN ISOTROPIC MATERIALS 1.Elasticity Modulus (E) 2. Poisson’s Ratio (n) 3. Shear Modulus (G) 4. Bulk Modulus (K)

1. Modulus of Elasticity, E (Young’s Modulus) simple tension test 1. Modulus of Elasticity, E (Young’s Modulus) s Linear- elastic E e s = E e Units: E: [GPa]

Slope of stress strain plot (which is proportional to the elastic modulus) depends on bond strength of metal E= Adapted from Fig. 6.7, Callister 7e.

2. Poisson's ratio, n F eT eL n eT n = - eL eT : Transverse Strain simple tension test 2. Poisson's ratio, n eT eL n “n” is the ratio of transverse contraction strain to longitudinal extension strain in the direction of stretching force. Either transverse strain or longitudional strain is negative,  ν is positive eT n = - eL eT : Transverse Strain eL : Longitudional Strain Units: n: dimensionless

Virtually all common materials undergo a transverse contraction when stretched in one direction and a transverse expansion when compressed. In an isotropic material the allowable (theoretical) range of Poisson's ratio is from -1.0 to +0.5, based on the theory of elasticity. metals: n ~ 0.33 ceramics: n ~ 0.25 polymers: n ~ 0.40

3. Shear Modulus, G M t G t = G g g M Units: G: [GPa] simple torsion test t = G g M Units: G: [GPa]

4. Bulk Modulus, K P Initial Volume = V0 Volume Change = DV P P savg = σavg is the average of three stresses applied along three principal directions. D V V o K Units: K: [GPa]

Elastic Constants Stresses Strains s = E e t = G g savg = K D V o Normal Shear Volumetric Stresses Strains

Example: Uniaxial Loading of a Prismatic Specimen P=1000 kgf 9.9 cm Before 10 cm Determine E and n 10.4 cm 9.9 cm After

1000 P=1000kgf → σ= = 10kgf/cm2 10*10 E= σ ε = 10 0.04 = 250 kgf/cm2 Δd/2=0.05cm Δl/2=0.2cm E= σ ε = 10 0.04 = 250 kgf/cm2 10cm εlong= Δl l0 = =0.04 0.4 10 εlat= Δd d0 = = -0.01 -0.1 10 10cm ν = - -0.01 0.04 = 0.25

For an isotropic material the stress-strain relations are as follows:

RELATION B/W K & E Consider a cube with a unit volume σ σ causes an elongation in the direction CD and contraction in the directions AB & BC. 1 σ D C B A The new dimensions of the cube is : CD direction is 1+ε BC direction is 1-νε AB direction is 1-νε

ε is small, ε2 & ε3 are smaller and can be neglected. V0 = 1 Final volume Vf of the cube is now: (1+ε) (1-νε) (1-νε) = (1+ε) (1-2νε+μ2ε2) = 1 - 2νε + μ2ε2 + ε-2νε2 + μ2ε3 = 1 + ε - 2νε - 2νε2 + μ2ε2 + μ2ε3 ε is small, ε2 & ε3 are smaller and can be neglected. Vf = 1+ ε - 2νε → ΔV = Vf - V0 = ε (1-2ν) If equal tensile stresses are applied to each of the other two pairs of faces of the cube than the total change in volume will be : ΔV = 3ε (1-2ν)

Ξ + SΔV = 3ε (1-2ν) = ε (1-2ν) + ε (1-2ν) + ε (1-2ν) savg (σ+σ+σ)/3 σ E 3 (1-2ν) K = = = = DV/V0 3ε (1-2ν) 3ε (1-2ν) K = E 3 (1-2ν)

K = E 3 (1-2ν) The relation between K and E is : Moreover the relation between G and E is : G = E 2 (1+ν) The relation between G, E and K is : E 1 = + 9K 3G Therefore, out of the four elastic constants only two of them are independent.

For very soft materials such as pastes, gels, putties, K is very large Note that as K → ∞ → ν → 0.5 & E ≈ 3G If K is very large → ΔV/V0 ≈ 0 *No volume change For materials like metals, fibers & certain plastics K must be considered.

Modulus of Elasticity : High in covalent compounds such as diamond Lower in metallic and ionic crystals Lowest in molecular amorphous solids such as plastics and rubber.

Elastic Constants of Some Materials E(psi)x106 (GPa) G(psi)x106 (GPa) ν (-) Cast Iron 16 110 7.4 50 0.17 Steel 30 205 11.8 80 0.26 Aluminum 10 70 3.6 25 0.33 Concrete 1.5-5.5 10-40 0.62-2.30 4-15 0.2 Wood Long 1.81 12 Tang 0.10 0.7 0.11 0.7 0.03 0.2 ?

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