3.2 Hooke s law anisotropic elasticity Robert Hooke ( ) Most general relationship

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1 3.2 Hooke s law anisotropic elasticity Robert Hooke ( ) Most general relationship σ = C ε + C ε + C ε + C γ + C γ + C γ yy 13 zz 14 xy 15 xz 16 yz σ = C ε + C ε + C ε + C γ + C γ + C γ yy yy 23 zz 24 xy 25 xz 26 yz σ = C ε + C ε + C ε + C γ + C γ + C γ zz yy 33 zz 34 xy 35 xz 36 yz σ = C ε + C ε + C ε + C γ + C γ + C γ xy yy 43 zz 44 xy 45 xz 46 yz σ = C ε + C ε + C xz yy 53 zz 54 xy 55 xz 56 yz yz yy 63 zz 64 xy 65 xz 66 yz { } [ ]{} ε + C γ + C γ + C γ σ = C ε + C ε + C ε + C γ + C γ + C γ or σ = C ε or σ = C ε ij ijkl kl What is the physical meaning of C 16?

2 Symmetry and coordinate transformation Combining with energy relations U0 = C11 C12 yy C66 yz 2 ε + ε ε + + γ U 0 = σ = C ε + C ε + C ε + C γ + C γ + C γ ε U γ 0 yz yy 13 zz 14 xy 15 xz 16 yz = σ = C ε + C ε + C ε + C γ + C γ + C γ yz yy 63 zz 64 xy 65 xz 66 yz U U ε γ γ ε = C61 = C16 yz yz What is needed to make C a fourth order tensor with transformation? C = C n n n n ijkl i' j ' k ' l ' i' i j ' j k ' k l ' l Principal coordinate system for C? Why symmetric?

3 3.3 Isotropic Hooke s law In isotropic material may as well use coordinate system of principal strains U0 = C11ε 1 + C22ε2 + C33ε3 + 2C12εε C13εε C23εε Down to six constants, then isotropy means C11 = C22 = C33 = C1 C12 = C13 = C23 = C2 May re-write as Lamé coefficients Gabriel Lamé ( ) 1 U0 = λ ε1+ ε2 + ε 3 + µ ε1 + ε2 + ε3 2 λ = C µ = G = C C ( ) ( ) ( ) 2 1 2

4 Transformation to general coordinates In terms of strain tensor invariants 1 U = λ + G I 2GI 2 I = ε + ε + ε I = ε ε + ε ε + ε ε ε ε ε yy zz 2 yy yy zz zz xy xz yz U0 = λ ( ε + εyy + εzz) + G( ε + εyy + ε yy + 2εxy + 2εxz + 2ε yz) 2 Differentiating for stresses σ = λi + 2Gε σ = λi + 2Gε σ = λi + 2Gε 1 yy 1 yy zz 1 zz Alternately σ = (1 ν) ε + ν ε + ε (1 + ν)(1 2 ν) σ = 2Gε σ = 2Gε σ = 2Gε ( ) yy zz xy xy xz xz yz yz How do we see from these equations that the principal stress coordinate system is also the principal strain system?

5 We like the inverted form Thomas Young ( ), Simeon Denis Poisson ( ) Similarly for other components Coefficient relationships 1 σ xy ε = ( σ νσ yy νσ zz ) ε xy = 2G G(3λ + 2 G) λ = ν = λ+ G 2 + ( λ G) ν 3νK λ = = G = K = (1 + ν )(1 2 ν) 1+ ν 2(1 + ν) 3(1 2 ν) What happens when Poisson s ratio is above 0.5 or below -1?

6 xample A bar with cross-sectional area of one square inch is enclosed in a rigid enclosure except for one side. It s Young modulus is one million psi. Calculate the force needed to apply a strain of 0.1% for Poisson s ratios of 0.3, 0.4, 0.49, (1 ν ) ε σ = (1 ν) ε + ν ( εyy + εzz ) = σ = (1 + ν )(1 2 ν) (1 + ν)(1 2 ν) 1000(1 ν ) = (1 + ν)(1 2 ν) ν Force, lb ,100

7 Finally stress-strain equations Three dimensional Plane stress σ = ν ε + ν ε + ε σ ( )( ) ( 1 ) ( ) 1+ ν 1 2ν yy zz xy = 2Gε xy σ = ε νε σ νε ε 1 ν + = 1 ν + ( ) ( ) 2 yy yy 2 yy Why did we opt for Young and Poisson rather than Lamé?

8 Problem 3.5 For an isotropic elastic medium subjected to hydrostatic state of stress σ = σ yy = σzz = p and no shear stresses. Show that for this state p=-ke, where e= I and is 1 = ε + εyy + ε K = /3(1 [ 2) ν ] zz the bulk modulus We have Adding the three equations We also had σ = p = λe+ 2Gε σ = p= λe+ 2Gε yy yy σ = p= λe+ 2Gε zz ( λ+ G) zz p = λ + λ 1 2ν ν = G = λ 2 3ν 2G 1+ ν λ+ = λ = K 3 3ν 2G 3 e

9 3.4 quations of thermoelaticity for isotropic materials Strain due to thermal expansion ε ' = ε' = ε' = α T ε ' = ε ' = ε ' = 0 yy zz xy xz yz Mechanical strain ε " = ε α T ε" = ε α T ε" = ε α T yy yy zz zz ε" = ε ε" = ε ε" = ε xy xy xz xz yz yz Stresses are due only to mechanical part of strain, so we can have stresses without displacements.

10 Reading assignment Sections 3,4-5: Question: What are the minimal conditions on the elastic tensor C for the principal directions to be the same for stresses and strains? Source: Page11.htm