Transcript Document
Hygrothermal behavior of composite laminates • Composite materials can absorb moisture and expand in the same way that they expand due to temperature. • We can use the same kind of analysis to deal with stresses that arise from both. • In this course we will focus on thermal behavior. • The effect of temperatures on composite laminates is more pronounced than on metals because of large disparity in coefficient of thermal expansion. Effect of mismatch • Graphite/epoxy typical thermal expansion coefficients are 𝛼1 = 0.02 × 10−6 ∘𝐶 , 𝛼2 = 22.5 × 10−6 ∘𝐶. • A 200𝑜 𝐶 drop in temperature will produce a strain of -0.0045 in the transverse direction, and almost nothing in the fiber direction. • In a 0/90 laminate the fibers in one ply will not allow the matrix in the other ply to shrink by more than a fraction of the -0.0045 strain. • This will eat up a substantial part of the load carrying capacity. • For cryogenic conditions (e.g. liquid hydrogen tanks) the temperature drop is more than doubled, and the laminate can fail just due to thermal loads. • What can we do? Thermal deformations • If ply can expand or contract due to hygrothermal effects, only strains, but no stresses. • Stress free expansion strain in a unidirectional layer (Fig. 3.2) Strains and stresses • “Free” strains 1F 1 1 F T 2 2 2 C F 0 0 12 • Hooke’s law when total strains are different from free strains. 1 Q11 Q12 2 Q12 Q22 0 0 12 0 1t 1F 0 2t 2F Q66 12t Interaction between plies • 0/90/45 laminate • Constraining effect of adjacent laminates (0/90)s laminate • Stress-free expansion of layers • Constrained deformation Residual strains • Free strains in x-y system (Checks?) xF m 21F n 2 2F m 21 n 2 2 F 2 F 2 2 F 2 n m n m y 1 2 1 2 T F 2mn( F F ) 2mn( ) 1 2 1 2 xy m cos , n sin • In vector form 𝜺𝐹𝑇 = 𝜶𝛥𝑇 • What is the coefficient of thermal expansion in the 45-deg direction? • The actual strains in the ply due to thermal loads are denoted as 𝜺𝑁 for “non-mechanical” • The “residual strains” or the mechanical strains of non-mechanical origin are 𝜺𝑀 = 𝜺𝑁 − 𝜺𝐹 Residual stresses • Residual strains in k-th layer 𝜺𝑅 = 𝜺0𝑁 + 𝑧 𝑘 𝑁 𝜿 − 𝐹 𝜺𝑘 • Residual stresses in k-th layer xR Q11 Q12 R y Q12 Q22 R Q Q 26 xy 16 xN xF Q16 x0 N Q26 y0 N z( k ) yN yF N F Q66 xy0 N xy xy • What terms will be the same in two layers, under what conditions? Thermal loads • If a laminate is subjected to temperature change without any loads then the stress and moment resultants should be zero. N xR h /2 xR R R N y y dz 0 N R h /2 R xy xy M xR h /2 xR R R M y y zdz 0 M R h /2 R xy xy • Substituting from previous slide for in-plane equation for symmetric laminate 0N F N x h /2 x Nx 0N F N A y Q y dz N y 0 N h /2 F N N xy xy xy • Loads needed to produce same strains without temperatures Thermal loads for a single isotropic layer • For a single isotropic layer 𝜀1𝐹 = 𝜀2𝐹 = 𝛼𝛥𝑇 E E • The matrix A 0 1 1 2 E A 1 2 0 • Thermal loads 2 E 1 2 0 0h G T 1 ETh N N A T 1 1 0 0 • Justify letter by letter! With bending • Thermal moment resultants F x h /2 F N M Q y zdz h /2 F xy • Calculating strains and curvatures under thermal loads 𝐀𝜺0𝑁 + 𝐁𝜿𝑁 = 𝐍𝑁 𝐁𝜺0𝑁 + 𝐃𝜿𝑁 = 𝐌𝑁 With Tsai-Pagano Material constants • Thermal loads N xT K K 2 cos 2 h /2 1 T N 0.5 K K cos 2 y 1 Tdz 2 h /2 N T K sin 2 xy 3 • Thermal material constants K1 U1 U 4 1 2 U 2 1 2 K 2 U 2 1 2 U1 2U 3 U 4 1 2 K 3 U 2 1 2 2 U 3 U 5 1 2 • Any checks that we can apply?