6-150 



Determine constants: 



DESIGN OF LAMINATES 



x = 1 - a TL a LT = 1 " 0-20 2 = 0.96 

 A = E T a + 2 \G TL = 1.96xl0 6 x0.20 + 2(u.96)x0.52xl0 6 



= 0.392 x 10 6 + 0.9981 x 10 6 = 1.39 x 10 6 



(6.45) 



0.7596 



(6.53a) 



E = 0.5 x 0.965 = 0.U8 (6.54a) 



Referring to Fig. 6-46, for the value of (3 a = 0.48 and fc = 0.76, the value of C a is 

 found to be 13.7. 



and k„ _ L a 



C. 



3^ 



J T 



1 



13.7 

 3x0.96 



-.1 



6 



1.70xl0 c 



1.96x10 



A 



h. 76 x 0.965 = h.59 (6.56a) 



and q cr = *&> £ - ^ xl. 9 6x10^x0. g 2 = ^ ^ 

 a 2 1*00 



(6. 55a) 



This value is less than the ultimate parallel shear stress; 



F su = 105C0 psi 



(Table 5-14) 



DESIGN EXAMPLE 6-23. CRITICAL SHEAR BUCKLING OF A MAT 



OR ISOTROPIC PANEL 



Compute the critical shear stress for a mat laminate with similar dimensions as those 

 given in Design Example 6-22. 



E T = E L = 0.86 x 10 6 

 G TL = O.liO x 10 6 



(Table 5-10) 

 (Table 5-14) 



J TL 



c^T = 0.37 (Average of Tables 5-8 and 5-13) 



X = 1 - 0.37 2 =0.863 



A = 0.86x106x0.37 + 2x0. 86x0. 1^0x10° = 1.006 x 10 6 

 „ _ 1.006xl0 6 



0.86xl0 6 



= 1.17 



(6.45) 

 (6.53a) 



