DESIGN OF LAMINATES 6-29 



In Design Example 6-10 the composite laminate is made of 3 separate materials, mat, 

 cloth and woven roving. These three materials each have their own value for the modulus 

 of elasticity, E. The value of E for the composite laminate is not equal to any of the indi- 

 vidual lamina values but is a function of these values and the individual lamina areas. 

 equation 6. 16 expresses this relationship. 



i = n 



E x A= y E ± A ± (6. 16) 



i = 1 



where E = composite modulus of elasticity, psi 



A = composite area, sq. in. 



E^ = modulus of elasticity of i-th lamina 



Aj_ = area of i-th laiaina 



I ■ 



means summation of all laminae 



DESIGN EXAMPLE 6-11. MODULUS OF ELASTICITY OF COMPOSITE LAMINATE 



For the laminate in Design Example 6-10 find the modulus of elasticity, E of the 

 composite laminate. 



From Design Example 6-10 the following values are: 



^¥i 



EA 



i = 1 

 = 0.90 x 10 6 x 0.25 + 1.U0 x 10 6 x 0.0625 + 1.3U x 10 6 x 0.125 



= 0.225 x 10 6 + 0.0875 x 106 + 0.1675 x 10 6 



EA = 0.U8 x 106 lbs. 



Ar°a of composite = (0.25 + 0.0625 + 0.125) x 1 = C.U375 sq.in. = 7/l6 sq.in. 



Therefore 



E -y -Ml = Q- 1 * 9 * 1Q6 = 1.10 x 106 psi (6. 16a) 



*- A O.U375 



