Elastic Behaviour of Unidirectional Lamina PDF

Elastic Behaviour of Unidirectional Lamina Chapter 2 : ”Linear Elastic Stress-Strain Characteristic of Fiber-reinforced Materials”. 1 Outline Stress Strain Linear Elastic Stress-Strain relationship 3-D Orthotropic (princ...

Elastic Behaviour of Unidirectional Lamina Chapter 2 : ”Linear Elastic Stress-Strain Characteristic of Fiber-reinforced Materials”. 1 Outline Stress Strain Linear Elastic Stress-Strain relationship 3-D Orthotropic (principal directions or on-axis) 3-D Transversely Isotropic 3-D Isotropic Stress-Strain Relationship Including the Effect of Free Thermal Strains and the Effect of Free Moisture Strains 2 Stress Is a measure of internal forces within a body. F Together with strain are the key variables for determination of stiffness and strength. A There is no direct measurement of stress. To get stress: – Applied force using stress analysis – Measure displacement and using stress analysis F – Measure strain using stress-strain relationship Force F Stress = = Stress nothing to do with material Area A properties 3 Stress Two types of stress – Normal Stress (σ) σx=Normal stress acting on the element face with its outward normal in the x-direction Sign convention Tension – Tension : Positive + + σ – Compression : Negative - – Shear Stress (τ) Compression τxy =Shear stress acting on the x - σ face in the y-direction y Sign convention – Positive when acting on a τxy (+) positive face and directed toward a positive axis Or acting x on a negative face and directed toward a negative axis Sign of shear stress = (Sign of face)*(Sign of force) 4 Strain Relative displacement of different Before loading points on a material. Together with stress are the key L variables for determination of stiffness and strength. Can be measured using strain After loading gages. L + ∆L Strain nothing to do with material properties ∆L Strain Strain = Gage L 5 Strain (cont.) Two types of strain: Normal Strain and Shear Strain Normal Strain (ε) – Associated with changes in the length of an infinitesimal element. – Rectangular element after deformation remains rectangular although its length and width may change. – No distortion produced by normal strain. – ∆u =relative infinitesimal displacement along x-axis – ∆v =relative infinitesimal displacement along y-axis ∆u ∂u y ε x = lim ∆x→0 = ∆x ∂x ∆v ∆v ∂v ε y = lim ∆y →0 = ∆y ∆y ∂y x ∆x ∆u 6 Strain (cont.) Shear Strain (γ) – Associated with changes in the angle of an infinitesimal element – Rectangular element will be distorted into a parallelogram. – Equivalent with stretching one diagonal and compressing the other. – Distortion is measured by the change of angle. – Engineering Shear Strain can be calculated as: ∂v ∂u γ xy = a + b where a ≅ tan a ≅ and b ≅ tan b ≅ ∂x ∂y ∂v ∂u γ xy = + ∂x ∂y γ xy = Engineering Shear Strain. Positive shear strain negative shear strain 7 Normal and Shear Strains 8 Stress-Strain Relation Force Balance Geometry And And Equilibrium Relative displacement Stress Strain Material Properties 9 Linear-Elastic Elasticity: – Full reversibility. We can load,unload, and reload a material without incurring any permanent strain or hystersis – Material response to loading is instantaneous. No time lag and no time or rate dependency. Orthotropic Material/Specially Orthotropic Material: – Three perpendicular planes of material symmetry: 9 independent constants Transversely Isotropic Material: – One plane of material isotropy in an orthotropic body: 5 independent constants Isotropic Material: – All planes in an orthotropic body are identical: 2 independent constants 10 Property plus direction E1, E2, E3 E1 ,E2=E3 E1=E2=E3=E Orthotropic Transversely Isotropic Isotropic 11 Prepreg- Building Block 12 Analysis of Composite Materials Micromechanics focuses on the stresses, strains, damage, etc., happening at the level of individual fibres and the fibre/matrix interface Macromechanics focuses on the analysis at the level of individual lamina Structural mechanics involves analysis at the structural scale 13 Hook’s law Orthotropic: Three perpendicular planes of material symmetry Principal Material Coordinate System or On-axis 1-2-3 1-axis aligned with the fiber direction 2 and 3-axis are perpendicular to the fiber direction (matrix or transvers direction) 3 (matrix direction) 2 (matrix direction) 1 (fiber direction) 14 15  Hook’s law- Orthotropic Materials (On-axis) ε1   S11 S12 S13 0 0 0  σ 1  ε    σ   2   S 21 S 22 S 23 0 0 0  2  ε 3   S31 S32 S33 0 0 0  σ 3   =   γ 23   0 0 0 S 44 0 0  τ 23  γ 13   0 0 0 0 S55 0  τ 13       γ 12   0 0 0 0 0 S 66  τ 12  S=Compliance Matrix 1 − ν 21 − ν 31 S11 = S12 = S13 = E1 E2 E3 − ν 12 1 − ν 32 S21 = S22 = S23 = Nine independent Constants: E1 E2 E3 − ν 13 − ν 23 1 E1 , E2 , E3,ν12 ,ν13 ,ν23,G12 ,G13 ,G23 S31 = S32 = S33 = E1 E2 E3 1 1 1 S44 = S55 = S66 = G23 G13 G12 16 Hook’s law- Orthotropic Materials Compliance matrix is symmetric. Therefore: Sij = S ji Since Ei is different from Ej , then ν ij ≠ ν ji For example: ν 12 ν 21 S12 = S21 ⇒ = and E1 ≠ E2 ⇒ ν 12 ≠ ν 21 E1 E2 ν 12 ν 21 ν 13 ν 31 ν 23 ν 32 = = = E1 E2 E1 E3 E2 E3 17 Stiffness Matrix If we want to calculate the stresses from given strains, we can obtain: {ε } = [S ]{σ } [S ]−1{ε } = [S ]−1 [S ]{σ } [S ]−1{ε } = [I ]{σ } [S ] {ε } = {σ } −1 Inverse of Compliance matrix is called Stiffness Matrix (Modulus Matrix) and denoted by C. [C ] = [S ]−1 {σ } = [C ]{ε } 18 Stiffness Matrix σ 1  C11 C12 C13 0 0 0   ε1  σ    2  C21 C22 C23 0 0 0   ε 2  σ 3  C31 C32 C33 0 0 0   ε 3   =   τ 23   0 0 0 C44 0 0  γ 23  τ 13   0 0 0 0 C55 0  γ 13       τ 12   0 0 0 0 0 C66  γ 12  C=Stiffness Matrix S 22 S 33 − S 23 S 23 S13 S 23 − S12 S 33 C11 = C12 = S S S 33 S11 − S13 S13 S S − S13 S 22 C 22 = C13 = 12 23 S S S S −S S S S − S 23 S11 C33 = 11 22 12 12 C23 = 12 13 S S 1 1 1 C 44 = C 55 = C 66 = S 44 S 55 S 66 S = S11 S 22 S 33 − S11 S 23 S 23 − S 22 S13 S13 − S 33 S12 S12 + 2 S12 S 23 S13 19 Hook’s law- Transversely Isotropic Transversely isotropic materials lies between isotropic and orthotropc materials. Material properties in the 2-direction is identical to the material properties in the 3-direction (The material is said to be isotropic in the 2-3 plane). Therefore: E2 = E3 ν 12 = ν 13 G12 = G13 And more importantly 3 (matrix direction) E2 2 (matrix direction) G23 = 2(1 + ν 23 ) 1 (fiber direction) The unidirectional composite laminates are transversely isotropic. 20 Hook’s law- Transversely Isotropic (On-axis) ε1   S11 S12 S13 0 0 0  σ 1  ε    σ  Five independent  2   S 21 S 22 S 23 0 0 0  2  ε 3   S31 S32 S33 0 0 0  σ 3  Constants:  =   E1 , E2 ,ν12 ,ν23 γ 23   0 0 0 S 44 0 0  τ 23  γ 13   0 0 0 0 S55 0  τ 13       G12 γ 12   0 0 0 0 0 S 66  τ 12  and 3 (matrix direction) 1 − ν 12 S11 = S12 = S13 = E1 E1 2 (matrix direction) 1 − ν 23 S22 = S33 = S23 = E2 E2 1 (fiber direction) 1 2(1 + ν 23 ) 1 S44 = = S55 = S66 = G23 E2 G12 21 Hook’s law- Isotropic Materials ε1   S11 S12 S13 0 0 0  σ 1  ε    σ   2   S 21 S 22 S 23 0 0 0  2  Two independent ε 3   S31 S32 S33 0 0 0  σ 3   =   Constants: E and ν γ 23   0 0 0 S 44 0 0  τ 23  γ 13   0 0 0 0 S55 0  τ 13       γ 12   0 0 0 0 0 S 66  τ 12  E1 = E 2 = E3 = E ν 12 = ν 13 = ν 23 = ν E G12 = G13 = G23 = G = 2(1 + ν ) 1 S11 = S 22 = S33 = E ν S12 = S 21 = S13 = S31 = S 23 = S32 = − E 1 2(1 + ν ) S 44 = S55 = S 66 = = G E 22 Hook’s law- Isotropic Materials σ 1  C11 C12 C13 0 0 0   ε1  σ    ε   2  C21 C22 C23 0 0 0  2  σ 3  C31 C32  = C33 0 0 0   ε 3    {σ } = [C ]{ε } τ 23   0 0 0 C44 0 0  γ 23  τ 13   0 0 0 0 C55 0  γ 13       τ 12   0 0 0 0 0 C66  γ 12  For isotropic material: C11 = C 22 = C 33 = (1 − ν )E (1 + ν )(1 − 2ν ) νE C12 = C 21 = C13 = C 31 = C 23 = C 32 = (1 + ν )(1 − 2ν ) E C 44 = C 55 = C 66 = G = 2(1 + ν ) 23 Typical Engineering Material Properties Graphite-Polymer Glass-Polymer Aluminum Composite Composite E1 155 GPa 50 GPa 72.4 GPa E2 12.1 GPa 15.2 GPa 72.4 GPa E3 12.1 GPa 15.2 GPa 72.4 GPa ν23 0.458 0.428 0.3 ν13 0.248 0.254 0.3 ν12 0.248 0.254 0.3 G23 3.2 GPa 3.28 GPa 27.8 GPa G13 4.4 GPa 4.7 GPa 27.8 GPa G12 4.4 GPa 4.7 GPa 27.8 GPa α1 -0.018X10-6 /°C 6.34X10-6 /°C 22.5X10-6 /°C α2 24.3X10-6 /°C 23.3X10-6 /°C 22.5X10-6 /°C α3 24.3X10-6 /°C 23.3X10-6 /°C 22.5X10-6 /°C β1 146X10-6 /%M 434X10-6 /%M 0 β2 4770X10-6 /%M 6320X10-6 /%M 0 β3 4770X10-6 /%M 6320X10-6 /%M 0 24  Example: [C] and [S] matrices for graphite-polymers Composite  6.45 − 1.6 − 1.6 0 0 0   − 1.6 82.6 − 37.9 0 0 0     − 1.6 − 37.9 82.6 0 0 0  [S ] =   (TPa )−1 TPa = terraPasca ls = 1012 Pa  0 0 0 312 0 0  S11 = 6.45 (TPa ) = 6.45 × 10−12 (Pa ) −1 −1  0 0 0 0 227 0     0 0 0 0 0 227   158 5.64 5.64 0 0 0 5.64 15.51 7.21 0 0 0   5.64 7.21 15.51 0 0 0 GPa = gigaPascals = 109 Pa [C ] =   GPa  0 0 0 3. 2 0 0  C11 = 158 (GPa ) = 158 × 109 (Pa )  0 0 0 0 4.4 0     0 0 0 0 0 4.4  25 Interpretation of Stress-Strain Relations Simple algebraic relations between 12 quantities σ 1 , σ 2 , σ 3 , τ 23 , τ 13 , τ 12 ε 1 , ε 2 , ε 3 , γ 23 , γ 13 , γ 12 Given any six quantities, the other six can be determined. The 12 stresses and strains should be considered in pairs (σ 1 − ε 1 ) (τ 23 − γ 23 ) (σ 2 − ε 2 ) (τ 13 − γ 13 ) (σ 3 − ε 3 ) (τ 12 − γ 12 ) Physically one can only describe either a stress or a strain from each pair, but not both. 26 Interpretation of Stress-Strain Relations There is one exception: Doing experiment to determine the material properties. For example, to determine E1 from test, both σ1 and ε1 have to be known. (σ 1 − ε 1 ) ⇒ E1 27 Example A 50-mm cube of graphite-reinforced materials subjected to 125 KN compressive force to the fiber direction (direction 2). Calculate the change of 50-mm dimensions for three cases: A) The cube is free to expand or contract B) The cube is constrained against expansion in the 3 direction. C) The cube is constrained against expansion in the 1 direction. 28 Example-Part A A 50-mm cube of graphite-reinforced materials subjected to 125 KN compressive force to the fiber direction (direction 2). Calculate the change of 50-mm dimensions for three cases: A)The cube is free to expand or contract Assume the compressive force is uniformly distributed then P − 125000 N σ2 = = = −50 MPa ∆ 1 ∆ 3 (0.05mm) (0.05mm) Cube is free from stress on the other faces. Thus σ 2 = −50 MPa σ 1 = σ 3 = τ 12 = τ 13 = τ 23 = 0 Since all stress components are defined, nothing can be said regarding strain components. They have to be calculated (one from each pair can be defined) 29 Example-Part A. ε1   S11 S12 S13 0 0 0  σ 1  ε    2   S 21 S 22 S 23 0 0 0  σ 2  ε 3   S31 S32 S33 0 0 0  σ 3   =   γ 23   0 0 0 S 44 0 0  τ 23  γ 13   0 0 0 0 S55 0  τ 13       γ 12   0 0 0 0 0 S 66  τ 12  ε 1   S11 S12 S13 0 0 0  0  ε    σ   2   S 21 S 22 S 23 0 0 0  2  ε 3   S31 S32 S33 0 0 0   0   =   γ 23   0 0 0 S 44 0 0  0  γ 13   0 0 0 0 S55 0  0       γ 12   0 0 0 0 0 S 66   0  ε 1 = S12σ 2 ε 2 = S 22σ 2 ε 3 = S 23σ 2 γ 12 = γ 23 = γ 13 = 0 30 Example- Part A By definition µε = 10 −6 m / m = 10 −6 mm / mm  ε 1   6.45 − 1.6 − 1.6 0 0 0   0   ε  − 1.6 82.6 − 37.9 0 0 0  − 50    2     ε 3  − 1.6 − 37.9 82.6 0 0 0   0  −1     =  × 10 −12 (Pa )   × 10 (Pa ) 6 γ 23   0 0 0 312 0 0   0  γ 13   0 0 0 0 227 0   0        γ 12   0 0 0 0 0 227   0  ε 1 = S12σ 2 = (−1.6 × 10 −12 )(−50 × 10 6 ) = 80 × 10 −6 m / m = 80 × 10 −6 mm / mm = 80 µ mm / mm = 80 µε ε 2 = S 22σ 2 = (82.6 × 10 −12 )(−50 × 10 6 ) = −4130 × 10 −6 m / m = −4130 µmm / mm = −4130 µε ε 3 = S 23σ 2 = (−37.9 × 10 −12 )(−50 × 10 6 ) = 1893 × 10 −6 m / m = 1893 µmm / mm = 1893 µε 31  Example- Part A Displacement δ ∆1 ε1 = δ∆1 = ∆1ε 1 = ∆1 S12σ 2 = (0.05)(−1.6 × 10 −12 )(−50 × 10 6 ) = 4 × 10 −6 m = 0.004 mm ∆1 δ ∆2 ε2 = δ∆ 2 = ∆ 2 ε 2 = ∆ 2 S 22σ 2 = (0.05)(82.6 × 10 −12 )(−50 × 10 6 ) = −207 × 10 −6 m = −0.207 mm ∆2 δ ∆3 ε3 = δ∆ 3 = ∆ 3ε 3 = ∆ 3 S 23σ 2 = (0.05)(−37.9 × 10 −12 )(−50 × 10 6 ) = 94.6 × 10 −6 m = 0.0946 mm ∆3 32 Example-Part B B)The cube is constrained against expansion in the 3 direction. It means the displacement in the 3 direction is zero δ∆ 3 = 0 ⇒ ε 3 = 0 Therefore the normal stress in the 3 direction is unknown.(one from each pair can be defined). The presence of rollers at the cube boundary means there is no friction which means no shear force at the boundary. Therefore, our six component that are known σ 2 = −50 MPa σ 1 = ε 3 = τ 12 = τ 13 = τ 23 = 0 Then: ε 1   S11 S12 S13 0 0 0  0  ε    2   S 21 S 22 S 23 0 0 0  σ 2  ε 1 = S12σ 2 + S13σ 3 0   S 31 0  σ 3   = S 32 S 33 0 0   ε 2 = S 22σ 2 + S 23σ 3 γ 23   0 0 0 S 44 0 0  0  γ 13   0 0 0 0 S 55 0  0  0 = S 23σ 2 + S 33σ 3      γ 12   0 0 0 0 0 S 66   0  33  Example-Part B ε 1 = S12σ 2 + S13σ 3 From third equation: σ3 = − S 23 σ2 S 33 ε 2 = S 22σ 2 + S 23σ 3 Solving the first and  S13 S 23   ε 1 = S12σ 2 + S13σ 3 =  S12 −  σ 2 0 = S 23σ 2 + S 33σ 3 second equations  S 33   S S  ε 2 = S 22σ 2 + S 23σ 3 =  S 22 − 23 23  σ 2  S 33   S13 S 23  δ∆ 1 = ε 1 ∆ 1 = ∆ 1  S12 −  σ 2 = (0.05)(−2.33 × 10 −12 )(−50 × 10 6 ) = 5.83 × 10 −6 = 0.00583mm  S 33   S S  δ∆ 2 = ε 2 ∆ 2 = ∆ 2  S 22 − 23 23  σ 2 = (0.05)(65.3 × 10 −12 )(−50 × 10 6 ) = −163.3 × 10 −6 m = −0.1633mm  S 33  δ∆ 3 = 0 S 23 σ3 = − σ 2 = −22.9 MPa S 33 34  Example-Part C C)The cube is constrained against expansion in the 1 direction. It means the displacement in the 1 direction is zero δ∆1 = 0 ⇒ ε 1 = 0 Therefore the normal stress in the 1 direction is unknown.(one from each pair can be defined). Therefore, σ 2 = −50 MPa ε 1 = σ 3 = τ 12 = τ 13 = τ 23 = 0 Then: 0   S11 S12 S13 0 0 0  σ 1  ε   0  σ 2  0 = S11σ 1 + S12σ 2  2   S 21 S 22 S 23 0 0   ε 3   S 31 S 32 S 33 0 0  0  0  ε 2 = S12σ 1 + S 22σ 2  =   γ 23   0 0 0 S 44 0 0  0  ε 3 = S13σ 1 + S 23σ 2 γ 13   0 0 0 0 S 55 0  0       γ 12   0 0 0 0 0 S 66   0  35  Example-Part C : 0 = S11σ 1 + S12σ 2 S12 From first equation: σ1 = − σ 2 ε 2 = S12σ 1 + S 22σ 2 S11  S S  ε 2 = S12σ 1 + S 22σ 2 =  S 22 − 12 12  σ 2 ε 3 = S13σ 1 + S 23σ 2 Solving the second  S11  and third equations  S S  ε 3 = S13σ 1 + S 23σ 2 =  S 23 − 12 13  σ 2  S11  and S12 δ∆1 = 0 and σ 1 = − σ 2 = −12.4 MPa S11  S S  δ∆ 2 = ε 2 ∆ 2 = ∆ 2  S 22 − 12 12  σ 2 = (0.05)(82.2 ×10 −12 )(−50 ×106 ) = −206 ×10 −6 m = −0.206mm  S11   S12 S13  δ∆ 3 = ε 3∆ 3 = ∆ 3  S 23 −  σ 2 = (0.05)(−38.2 ×10 −12 )(−50 ×106 ) = 95.6 ×10 −6 = 0.0956mm  S11  36 Thermal Strains Deformation due to heating or cooling independent of any applied load. Unlike isotropic materials, thermal expansions are different at different directions. The strains are strictly dilatational, not distortional (Dilatational identified by change in volume, whereas distortional identified with a change of shape. Free thermal strains can be written as (no shear strain): ε1T (T , Tref ) = α1∆T ε 2T (T , Tref ) = α 2 ∆T ε 3T (T , Tref ) = α 3∆T ∆T = T − Tref T is the temperature at the state of interest, Tref is the reference temperature which is often taken to be the stress-free processing conditions (curing temperature) α1, α2 , and α3 are the coefficients of thermal expansion CTE at different directions 37 Thermal Strains-Example A 50-mm cube of graphite-reinforced materials is heated 50C. Calculate the change of 50-mm dimensions 38 Thermal Strains-Example A 50-mm cube of graphite-reinforced materials is heated 50C. Calculate the change of 50-mm dimensions δ∆ 1 = ∆ 1α 1 ∆T = (50)(−0.18 × 10 −6 )(50) = −0.000045mm δ∆ 2 = ∆ 2α 2 ∆T = (50)(24.3 × 10 −6 )(50) = 0.0608mm δ∆ 3 = ∆ 3α 3 ∆T = (50)(24.3 × 10 −6 )(50) = 0.0608mm The dimensional change in the 1 direction is very small and compressive due to the negative value of thermal expansion coefficient because of graphite fibers. ∆T = T − Tref 39 Total Strains Total strains can Mechanical strains, the key be calculated as: strains in the stress-strain relations, are given as:  ε 1  ε 1mech + α1∆T  ε 1mech   ε 1 − α1∆T   ε   mech   mech    ε + α ∆T ε  2   2 ε − α ∆T   2  2 2  2  ε 3  ε 3mech + α 3 ∆T  ε mech  ε 3 − α 3 ∆T   =   3mech  =   γ  23   γ mech 23  γ 23   γ 23  γ 13   γ 13mech  γ 13mech   γ 13       mech    γ 12   γ 12 mech  γ 12   γ 12   40 Hook’s Law- Mechanical and Thermal Effects  ε 1 − α1∆T   S11 S12 S13 0 0 0  σ 1  ε − α ∆T    2 2   S 21 S 22 S 23 0 0 0  σ 2  ε 3 − α 3 ∆T   S31 S32 S33 0 0 0  σ 3   =    γ 23  0 0 0 S 44 0 0  τ 23   γ 13  0 0 0 0 S55 0  τ 13        γ 12   0 0 0 0 0 S 66  τ 12  σ 1  C11 C12 C13 0 0 0   ε 1 − α1∆T  σ    2  C21 C22 C23 0 0 0  ε 2 − α 2 ∆T  σ 3  C31 C32 C33 0 0 0  ε 3 − α 3 ∆T   =   τ 23   0 0 0 C44 0 0   γ 23  τ 13   0 0 0 0 C55 0   γ 13       τ 12   0 0 0 0 0 C66   γ 12   41 Thermal Strains-Example A 50-mm cube of graphite-reinforced materials is heated ΔT. There is no constraints. Calculate the mechanical stresses. T = Tref ∆T = T − Tref 42 Thermal Strains-Total Strain-Example A 50-mm cube of graphite-reinforced materials is heated ΔT. There is no constraints and no mechanical stress. Then total strain will be ε 1 = α 1 ∆T ε 2 = α 2 ∆T ε 3 = α 3 ∆T γ 12 = γ 23 = γ 13 = 0 The mechanical strain can be calculated as ε 1mech   ε 1 − α 1 ∆T  α 1 ∆T − α 1 ∆T  0  mech        ε 2  ε 2 − α 2 ∆T  α 2 ∆T − α 2 ∆T  0 ε mech  ε 3 − α 3 ∆T  α 3 ∆T − α 3 ∆T  0  3mech  =  = =   γ 23   γ 23   0  0 γ 13mech   γ 13   0  0  mech        γ 12   γ 12   0  0 Therefore there will be no mechanical stresses. 43 Thermal Strains-Example A 50-mm cube of graphite-reinforced materials is heated ΔT=50C. There is constraints all around. Calculate the mechanical stresses. T = Tref ∆T = T − Tref 44 Thermal Strains-Example A 50-mm cube of graphite-reinforced materials is heated ΔT=50C. There is constraints all around. Then total strain will be ε1 = 0 ε 2 = 0 ε 3 = 0 γ 12 = γ 23 = γ 13 = 0 The mechanical strain can be calculated as ε 1mech   ε 1 − α 1 ∆T  − α 1 ∆T  0.90   mech    − α ∆T  − 1215  ε 2   ε 2 − α 2 ∆T   2    ε mech  ε 3 − α 3 ∆T  − α 3 ∆T  − 1215  3mech  =  = = µε γ  23   γ 23   0   0  γ 13mech   γ 13   0  0   mech        γ 12   γ 12   0  0  Therefore stresses will be σ 1  C11 C12 C13 0 0 0   − α 1 ∆T  − 13.55 σ    2  C 21 C 22 C 23 0 0 0  − α 2 ∆T  − 27.6    σ 3  C 31 C 32 C 33 0 0 0  − α 3 ∆T  − 27.6   =  = MPa τ 23   0 0 0 C 44 0 0   γ 23   0  τ 13   0 0 0 0 C 55 0   γ 13   0         τ 12   0 0 0 0 0 C 66   γ 12   0  45 Moisture Strains Deformation due to absorption of moisture (expansion) Free moisture strains can be written as (no shear strains): ε1M (M , M ref ) = β1∆M ε 2M (M , M ref ) = β 2 ∆M ε 3M (M , M ref ) = β 3∆M where ∆M = M − M ref M is the moisture percentage at the state of interest, Mref is the reference moisture concentration which is often taken to be zero. β1, β2, and β3 are the coefficients of moisture expansion at different directions (Table 2.1 page 64). 46 Moisture Strains-Example A 50-mm cube of graphite-reinforced materials has absorbed 0.5% moisture. Calculate the change of 50-mm dimensions. M = M ref ∆M = M − M ref 47 Moisture Strains-Example A 50-mm cube of graphite-reinforced materials has absorbed 0.5% moisture. Calculate the change of 50-mm dimensions δ∆1 = ∆1 β1 ∆M = (50)(146 × 10 −6 )(0.5) = 0.00365mm δ∆ 2 = ∆ 2 β 2 ∆M = (50)(4770 × 10 −6 )(0.5) = 0.1193mm δ∆ 3 = ∆ 3 β 3 ∆M = (50)(4770 × 10 −6 )(0.5) = 0.1193mm The dimensional changes in the 2 and 3 directions are significant and re-straining them leads to large moisture-induced stresses. 48  Total Strains Total strains can Mechanical strains, the key be calculated as: strains in the stress-strain relations, are given as:  ε1   ε1mech + α1∆T + β1∆M  ε 1mech   ε1 − α1∆T − β1∆M   ε   mech   mech    ε + α ∆ T + β ∆M ε  2   2 ε − α ∆ T − β ∆M  2  2 2 2  2 2   ε 3  ε 3mech + α 3∆T + β 3∆M  ε mech  ε 3 − α 3∆T − β 3∆M   =   mech  =  3  γ  23   γ mech 23  γ 23   γ 23  γ 13   γ 13mech  γ 13mech   γ 13       mech    γ  12    γ 12 mech  γ 12   γ 12  49 Hook’s Law- Mechanical, Thermal, and Moisture Effects  ε1 − α1∆T − β1∆M   S11 S12 S13 0 0 0  σ 1  ε − α ∆T − β ∆M   S S22 S23 0 0 0  σ   2 2 2   21  2  ε 3 − α 3∆T − β 3∆M   S31 S32 S33 0 0 0  σ 3   =    γ 23  0 0 0 S44 0 0  τ 23   γ 13  0 0 0 0 S55 0  τ 13        γ 12  0 0 0 0 0 S66  τ 12  σ 1  C11 C12 C13 0 0 0   ε1 − α1∆T − β1∆M  σ  C 0  ε 2 − α 2 ∆T − β 2 ∆M   2   21 C22 C23 0 0   σ 3  C31 C32 C33 0 0 0  ε 3 − α 3∆T − β 3∆M   =   τ 23   0 0 0 C44 0 0  γ 23  τ 13   0 0 0 0 C55 0  γ 13       τ 12   0 0 0 0 0 C66   γ 12  50 Moisture Strains-Example A 50-mm cube of graphite-reinforced materials has absorbed 0.5% moisture but has experienced no temperature change.There is constraints all around. Calculate the mechanical stresses. M = M ref ∆M = M − M ref 51 Moisture Strains-Example A 50-mm cube of graphite-reinforced materials has absorbed 0.5% moisture but has experienced no temperature change.. There is constraints all around. Then total strain will be ε1 = 0 ε 2 = 0 ε 3 = 0 γ 12 = γ 23 = γ 13 = 0 The mechanical strain can be calculated as ε 1mech   ε 1 − β1 ∆M  − β1 ∆M  − 73   mech        ε 2  ε 2 − β 2 ∆M  − β 2 ∆M  − 2380 ε mech  ε 3 − β 3 ∆M  − β 3 ∆M  − 2380  3mech  =  = = µε γ  23   γ 23   0   0  γ 13mech   γ 13   0  0   mech        γ 12   γ 12   0  0  Therefore stresses will be σ 1  C11 C12 C13 0 0 0   − β1 ∆M  − 38.4 σ    − β ∆M  − 54.6  2  C 21 C 22 C 23 0 0 0  2    σ 3  C 31 C 32 C 33 0 0 0  − β 3 ∆M  − 54.6  =  = MPa τ 23   0 0 0 C 44 0 0  γ 23   0  τ 13   0 0 0 0 C 55 0   γ 13   0         τ 12   0 0 0 0 0 C 66   γ 12   0  52