6-32 DESIGN OF LAMINATES 



1 _ n.o62$ + 0.8705 + Q.QQU 5 

 F 2 !j5b0 17900 IhOOO 



from which ? 2 = 13170 psi = F^at 15° 

 (b) The value of E can be found by referring to Fig. 6-9. From the graph for fl 



the ratio at 15° is equal to 0.8U5 

 Then E l5 



E L 



= 0.8U5 or Ex^o = E L x 0.8U5 = 1.U0 x 10 6 x 0.8U5 



F -15° = 1.183 x 10 6 p 



si 



(e) With the required F and E values known at 15 degrees the maximum stress that each 



lamina of the composite can withstand is now found by means of equations 6 . 14a and 6 .14b: 



(6 .14a) 

 (6. 14b) 



or 



If the cloth laminate ultimate tensile stress of 13, 170 psi is used as the controlling 

 stress, the mat laminate would attain a value of 9017 psi. This value exceeds the allowable 

 ultimate and therefore failure would occur. 



The cloth laminate attains a safe stress of 9639 psi since the mat laminate tensile 

 ultimate controls. Consequently these values are the values to be used, namely: 



f c l = 9639 psi 

 f m = 6600 psi 



The total load on the composite laminate is then found by means of equation 6.15; 



P = A m x f m + A cl x f cl 



P = 9639 x 0.125 + 6600 x 0.375 



= 1205 + 825 = 2030 lbs. per in. of width 



