6-58 DESIGN OF LAMINATES 



The neutral axis of the transformed area can now be found as follows: 



* = y~n^ (6.39) 



where x = distance to neutral axis, in. 



A'j_ = Equivalent area of i-th lamina, in2 

 x^ = Distance to center of i-th lamina, in. 



Taking moments about the bottom face of the woven roving lamina: 



_ ( 0.0173 x 0.216 + O.Q285 x 0.178 + 3.11+8 x 1 x 0.07U ) 

 (0.0173 + 0.0285 + 0.11+8) 



0.0198 



x = = 0.102 in. from bottom face of woven roving lamina 



0.1938 



Obtaining the neutral axis by using equation 6. 24: 



= L E i A ?- x 3. (6.24) 



lEiAi 



(0 .016 x1. 96xl0 6 xO .216+0 .060x0 . 3 6x10 6 x0 .178+0 .Ili8xl.8lxl0 6 x3 .071+ 1 .. n nno ,„ 



X = ; ; ; 7 7 7 - U.UJ^ in. 



(0 .016x1. 96x106+0 .060x0 .86xl0 6 +0 .11+8x1. 81x10 6 J 



The results are the same. The transformed section method, however, gives the 

 designer a clearer indication of how the components of the laminate behave. 



The equivalent moment of inertia and section modulus of the composite section can now 

 be obtained. 



Moment of inertia: 



I 



b^ta.3 



tf- + A 'i x r 



(6.40) 



where b 1 = equivalent width for woven roving lamina 



I- = 1 - 08 g X °- 0163 + 0.0173X0.111,2 ♦ °- h75 g - 0603 + 0.0285x0.0762 + ^ + 0.11,8x0.0282 



= (3700xlO -9 + 2252xlO" 9 ) + (8552xlO -9 + l61+6xl0 -9 ) + (2701xlO -9 + 1160X10 -9 ) 

 = 781+9 x 10~ 9 = 0.00078 in^ 



