By Christian Decolon (Auth.)
This ebook offers the foundation for calculations of composite constructions, utilizing continuum mechanics to facilitate the therapy of extra complex theories. A composite constitution combines conventional fabrics (such as concrete) with new fabrics (such as excessive functionality fibres) to discover and advance new structures.
The writer bargains with person layers in laminate composites, discussing the elemental legislation that govern combos.
· steered for either scholar use
· a scientific, compact presentation in one volume
· Covers the governing equations of composite beams, plates and buildings
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Extra info for Analysis of Composite Structures
When E f >> E m , the expression for E 2 becomes: Em E2 = ~ . l-V/ From the calculations above it may be noted that the two Young's moduli E I and E 2 are given by expressions analogous to those encountered when the equivalent stiffnesses of two springs in parallel and in series are calculated. 6. 3. ~n. ~n = 0"6 , Gm where G f and G m represent the shear moduli of the fibres and the matrix. The displacement in the Xl direction of the upper plane of the composite with respect to the lower plane is" u=uf +um .
E4 = 0, E2 = $260"6 , E5 = 0, E3 = $360. 6, E6 = $660. 6. The layer undergoes unit extensions in the x I , x 2 and x a directions, and an angular distortion in the two directions xt and x2. For ct = 0 or a = g-, 2 the elastic compliances S i 6 , $26 and $36 are zero. The linear expansions are therefore zero. The corresponding strains are represented by the following figures. 7. 8. 8. Reduced stiffness matrix for the orthotropic layer In this paragraph, we will describe the Q and Q of an orthotropic layer with through thickness normal stress tr 3 zero.
1 l O'l rt O'l rc Orl rt O'l rc 1 E l l -" - - ~ " O'l rt O'l rc The constants F 2 , F22, F 3 and F33 are obtained from tensile and compression tests in the X 2 and X3 directions. - In the case of a shear failure in the plane (X 2, X 3), there are two equations: F4~4r + F44~2r --F4G4r + F44~:r = l, = l, with ~4r and - ~ 4 r being the shear stress. From this: Failure criteria 61 F,=O, F~ ~,1 In an analogous manner we obtain the values of F 5, F55 , F 6 and F66. - The coupling term Fl2 can be obtained from a biaxial test but this is very difficult to perform and in the absence of experimental results the following expression for Fl2 can be used: l /712= 24~lrt~lrc~2rt~:rc The expressions for F23 and /731 are analogous to that for F12.