3.Normal load:It is denoted as M): N=m9(e2- 2 at the foot of the blade (x=//10): N(t/10)#12,000daN 4.Longitudinal modulus of elasticity: Using the relation of Section 3.3.1: Ee ErVr+EmVm with (Section 1.6):E,=86,000 MPa;E=4,000 MPa. E=53,200 MPa 5.Section of the spar of the blade made of glass/epoxy: The longitudinal rupture tensile stress of the unidirectional is Oe nupture 1700 MPa With a factor of safety of 6,the admissible stress at a section S(x)becomes 。==1700=283MPa S(x) 6 then: S(ax)=Na) S(x)= 20(-x m at the foot of the blade: S(/10)=4.24cm 2003 by CRC Press LLC3. Normal load: It is denoted as N(x): at the foot of the blade (x = l/10): 4. Longitudinal modulus of elasticity: Using the relation of Section 3.3.1: with (Section 1.6): Ef = 86,000 MPa; Em = 4,000 MPa. 5. Section of the spar of the blade made of glass/epoxy: The longitudinal rupture tensile stress of the unidirectional is With a factor of safety of 6, the admissible stress at a section S(x) becomes then: at the foot of the blade: N x( ) Fc q # dFc x
Ú d cos x
Ú mw2 x dx x
Ú = = N x( ) mw2 2 -----------
2 x2 = ( ) – N( )
/10 # 12,000 daN E
= EfVf + EmVm E
= 53,200 MPa s
rupture # 1700 MPa s N x( ) S x( ) ----------- 1700 6 === ----------- 283 MPa S x( ) N x( ) s = ----------- S x( ) mw2 2s -----------
2 x2 = ( ) – S( )
/10 4.24 cm2 = TX846_Frame_C18a Page 351 Monday, November 18, 2002 12:40 PM © 2003 by CRC Press LLC