Aerospace Structural Composites I Homework 6. Anirban Chaudhuri

Transcription

1 Aerospace Structural Composites I Homework 6 Anirban Chaudhuri. The unidirectional Boron/Epoxy layer of Example 6.. (using E = 4GPa, E = 8.5GPa, G = 5.59GPa, =.3, t =.5mm) is to be loaded by combined axial and shear stresses with xy r. x Determine the strength of this layer based on the Tsai-Hill failure criterion as a function of the fiber orientation in the range o 9 o, and plot the strength for r =.5 and r =.. Typical strength properties of Boron/Epoxy layers are X t = 6MPa, X c = 5MPa, Y t = 6MPa, Y c = MPa, and S = 67MPa. Solution: The stresses in the principal direction can be found by using the equation (). m n mn x n m mn y mn mn m n xy () where, m=cos and n=sin. We have y = and xy = r x. So the principal stresses are given below. ( m rn ) x ( n rm ) x mn( r ) x () Tsai-Hill criterion is given in equation (3) [Eq. 6.. in book] (3) X Y X S t t t For the case of compressive loading the X t and Y t are replaced by X c and Y c respectively. The values found in equation () are substituted in equation (3) for both tensile and compressive loading case and the strengths are plotted for r =.5 and r =.. The strength ( x ) in each case is found by solving the equation (3) (after substituting the values) in MATLAB. The MATLAB code used for this purpose is given in Appendix I.

2 4 x 8 Strength under tensile loading for r =.5.5 x 9 Strength under compressive loading for r =.5 x 8 6 x (deg) (deg) (a) Tensile loading (b) Compressive loading Fig. : Strength for r =.5 under tensile and compressive loading x 8 Strength under tensile loading for r =..5 x 9 Strength under compressive loading for r = x 5 4 x (deg) (deg) (a) Tensile loading (b) Compressive loading Fig. : Strength for r =. under tensile and compressive loading. Derive the strength coefficients F, G, H, P, Q, R, L, M, and N for Hoffman s strength criterion y of Eq. (6..3). Plot the failure envelope and show the load limits under biaxial load. for Glass/Epoxy with the following properties. x

3 Elastic properties: E = 5.6e6psi, E =.e6psi, G =.6e6psi, =.6, t =.mm. Strength properties: Xt = 54e3psi, X c = 88.5e3psi, Y t = 4.5e3psi, Y c = 7.e3psi, and S =.4e3psi. Solution: The starting general equation for Hoffman criterion is given in equation (4). F( ) G( ) H( ) P Q R L M N (4) The coefficients are evaluated by matching criterion to results obtained under simple load conditions. The equations obtained for simple tensile and compressive tests are given in equation (5). ( F H) Xt PXt ( F H) Xc PXc ( F G) Yt QYt ( F G) Yc QYc ( H G) Yt RYt ( H G) Yc RYc (5) The failure stresses in the and 3 directions are both equal to Y t and Y c. The equations in equation (5) can be easily solved simultaneously to get the values of the coefficients as given below. F H, G X X YY X X P, Q R X X Y Y t c t c t c t c t c (6) Now considering simple shear tests it can be shown that LS3, or L S 3 MS, or M S NS, or N S 3 3 (7) Substituting the values of the coefficients obtained in equations (6) and (7) in equation (4) and using a plane stress criterion with 3 = 3 = 3 = we get, XX t c XX t c YY t c Xt Xc Yt Yc S (8)

4 Equation (8) is exactly the same as given in equation 6..3 in the book. To plot the failure envelope for Hoffman criteria is set to and then equation (8) is plotted using MATLAB (details are given in the MATLAB code provided in Appendix II). Now for a unidirectional ply aligned along the x-direction we can assume the biaxial stress criteria is equivalent to / =.. This is the equation of a straight line passing through the origin and it is also plotted on the failure envelope and the failure region is shown in figure 3. The safe region is shown as the shaded (gray) region in figure 3. Fig. 3: Hoffman criterion failure envelope and the failure region for the given biaxial loading APPENDIX I MATLAB code for problem % Anirban Chaudhuri % Aerospace Structural Composites I % HW 6 prob close all;clc;clear all;format short;format compact; % Initial Constants for the given materials E = 4e9; % Youngs Modulus for material E = 8.5e9; % Youngs Modulus for material G = 5.59e9; % Shear Modulus for the given materials

5 nu =.3; t =.5; % Poisson's Ratio for the given materials % individual ply thickness (m) (CAN BE ANY RANDOM NUMBER) theta = linspace(,9); th = theta.*pi./8; % Strength properties of Boron/Epoxy layers Xt = 6e6; Xc = 5e6; Yt = 6e6; Yc = e6; S = 67e6; r =.; % Distribution of strength of this layer based on Tsai-Hill criterion as a % function of the fiber orientation m = cos(th); n = sin(th); % For Tensile loading var_intert = (/Xt^)*(m.^+*m.*n.*r).^+(/Yt^)*(n.^-*m.*n.*r).^- (/Xt^)*(m.^+*m.*n.*r).*(n.^-*m.*n.*r)+(/S^)*(-m.*n+(m.^- n.^).*r).^; sigxt = sqrt(./var_intert); figure() plot(theta,sigxt,'linewidth',) xlabel('\theta (deg)') ylabel('\sigma_x ') title('strength under tensile loading for r =.') % For Compressive loading var_interc = (/Xc^)*(m.^+*m.*n.*r).^+(/Yc^)*(n.^-*m.*n.*r).^- (/Xc^)*(m.^+*m.*n.*r).*(n.^-*m.*n.*r)+(/S^)*(-m.*n+(m.^- n.^).*r).^; sigxc = sqrt(./var_interc); figure() plot(theta,sigxc,'linewidth',) xlabel('\theta (deg)') ylabel('\sigma_x ') title('strength under compressive loading for r =.') APPENDIX II MATLAB code for problem % Anirban Chaudhuri % Aerospace Structural Composites I % HW 6 prob close all;clc;clear all;format short;format compact; % Initial Constants for the given materials E = 5.6e6; % Youngs Modulus for material E =.e6; % Youngs Modulus for material G =.6e6; % Shear Modulus for the given materials

6 nu =.6; % Poisson's Ratio for the given materials t =.; % individual ply thickness (mm) theta = linspace(,9); th = theta.*pi./8; % Strength properties of Boron/Epoxy layers Xt = 54e3; Xc = 88.5e3; Yt = 4.5e3; Yc = 7.e3; S =.4e3; r =.; % The constants in the Hoffman criteria are evaluated below and the values are directly entered in the ezplot command C = /(Xt*Xc); C = /(Yt*Yc); C3 = (/Xt-/Xc); C4 = (/Yt-/Yc); C5 = /S^; figure() hold on; % Plotting the failure envelope ezplot('7.3373e-*x^ e-*x*y+.995e-8*y^-4.859e-6*x+.6374e-4*y- ',[-Xc-35e3,Xt+35e3,-Yc-5e3,Yt+5e3]) % Plotting the straight line sigma =.*sigma for the biaxial loading % This is true only for the case when the plies are aligned in the % x-direction ezplot('y-.*x',[-xc-35e3,xt+35e3,-yc,yt+5e3])