6-52 



DESIGN OF LAMINATES 



In obtaining the deflection of a laterally loaded member, the effect of shear may be 

 appreciable depending upon the material used and the depth to span ratio. When considering 

 the shear effect in fiberglass reinforced laminates, the shear modulus, G, is important. 

 The shear modulus for a composite laminate will vary with the different types of reinforce- 

 ments used and an assumed value based on any one of the reinforcements may not be correct. 

 Therefore, when the accuracy of the deflection is important, tests on the particular com- 

 posite laminate in question should be made. 



For a uniformly loaded simply supported beam the total deflection for flexure and shear 

 deformation is (14): 



d-4-^* 



381i EI 



pLf 



Sag 



(6. 28) 



where d = deflection, in. 



p = load, lbs. per in. 



L = span, in. 



E = flexural modulus of elasticity, psi 



I = moment of inertia, in. 



A = shear area, in. 2 



G = shear modulus, psi 



The above equation assumes uniform shear distribution across the section and can be 

 considered approximately correct for sections with thin deep webs where all the shear is 

 taken by the web, similar to an I-beam. 



For rectangular sections where the shear is taken by the entire section and the shear 

 distribution across the section is non-uniform with a maximum at the neutral axis, equa- 

 tion 6.28 becomes: 



d =4r^ 



im ei 



pi/ 



■Bag 



(6. 28a) 



DESIGN EXAMPLE 6-18. BENDING OF A MAT OR ISOTROPIC LAMINATE 



The fixed ended beam indicated in Fig. 6-34 is made of mat reinforcement and is 12 in. 

 long and 2 in. wide. What should the thickness of the laminate be if it is to support a uni- 

 form load of 10 lbs. per linear in. and have a factor of safety of 4 on the ultimate strength? 

 Also compute the beam deflection. 



