6-54 DESIGN OF LAMINATES 



Ultimate flexural strength, F fe = 20500 psi Table 5-9 



Flexural modulus of elasticity, E = 0.86 x 10 6 psi Table 5-10 



Ultimate shear strength, perpendicular, F s = 9900 psi Table 5-lh 



By use of the standard beam formula, the required section modulus, Z is obtained: 



f = HZ = M (6.31) 



b I Z 



where f, = bending stress, psi 



y = distance from neutral axis to the outermost fiber, in. 



I = moment of inertia, in.k 



Z = section modulus, in.^ 



Rearranging the terms in the formula the required section modulus is: 



Z = J- (6.31a) 



Z = l ^° = 0.023h in. 3 

 20500 



The section modulus, Z, for the outermost fiber of the beam is obtained from: 



I _ bt 2 (6.32) 



y 



z = - " -5" 



where J, the moment of inertia is: 



t = bt3 (6.23) 



b = width, in. 



t = thickness, in. 



Therefore the required depth or thickness of the laminate is: 



/ 6 x 

 t = v — 



(6.32a) 



t . / 6 * °' 02 ^ - 0.265 in. 



