F 2 v 2 1 Icos22

This relationship is plotted as GU/GLTin Fig. 2.34. Unlike isotropic materials, stress T12 causes a strain e1 in the 1 -direction e, = - m,t!2/EL (2-99) and a strain e2 in the 2-direction e2 = -m2T,2/EL (2-100) in which (Fig. 2.34)

m2 = sin 2a j vLT+ | - 1 -sin2 a + 2 vLT+ | - ^ j | (2-101)

These values might correspond, for example, to a square-weave or symmetrical satin-weave fabric-reinforced construction.

As an example of the application of the foregoing equations, the tensile stress <7\ acting on the small plate at the top of Fig. 2.34 is 10,000 psi, the shear stress t12 is 4000 psi, and the angle a is 30°. Then from Fig. 2.34,

E}/El = 0.367 or f, = 0.367 x 5,000,000 = 1,830,000 psi Gn/GLT = 0.81 or 612 = 0.81 x 550,000 = 445,000 psi v12 = -0.0286 m, = 4.66 m2=4.98

Then, strains caused by Oi are

Total strains, therefore, are

712= -2.04 x 10"3 e, = 1.72 x 10"3 e2 = -3.82 x 10"3

Problems involving Fig. 2.35 can be solved in an analogous manner.

It must be kept in mind that Eqs. 2-92, 2-94, 2-96, 2-98, and 2-101 are valid and useful if the fibers and the resin behave together in accordance with the assumptions upon which their derivation is based. If only the values of ET, GLT, and vLT are available, the intermediate values of G12, Vi2, and the values of m, and m2 can be estimated by means of these equations.

Composite Plates

Fibrous reinforced plates in practice are often made up of several layers, and the individual layers may be of different construction, such as mat, fabric, or roving. Furthermore, the various layers may be oriented at different angles with respect to each other in order to provide the best combination to resist some particular loading condition. Outside loads or stresses applied to a composite plate of this type result in internal stresses which are different in the individual layers. External direct e, = 10,000/1,830,000 = 5.45 x 10~3 e2 = - (-0.0286) 5.45 x 10"3 = 0.16 x 10"3 7l2 = -4.66x 10,000/5,000,000 = -9.32 x 10"3

and strains caused by t12 are

= -4.66x4,000/5,000,000 = -3.73 x 10"3 e2 = -4.98x4,000/5,000,000 = -3.98 x 10"3

Composite panel with layers a and b of different orthotropic materials oriented at arbitrary angles a and P with respect to applied stresses o,, o2 and y12

M> = ¿ti + *bz t = ta + ti stresses may result not only in internal direct stresses but in internal shear stresses, and external shear stresses may result in internal direct stresses as well as internal shear stresses.

Fig. 2.36 depicts a small composite plate made up of materials a and b having principal longitudinal and transverse directions Ln and Ta, and L[, and T¡„ respectively. Several layers of each are present but their total thicknesses are tn and t!n respectively, and the overall thickness is t. Outside stresses CTj, <72, ar>d Ti2 are applied in the 1 and 2 directions, as shown. The 1-direction makes an angle a with Ln3 and a reverse angle [5 with Lb. The angle a is considered to be positive and the angle [5 negative.

The internal stresses (Tlrt, <J2a, t12«, and Glb¡ (T2fo b >n the individual layers can be found by observing that the sums of the internal stresses in the 1 and 2 directions must equal the external stresses in these directions, and that the strains must be the same in all layers. These relationships may be written in the following forms:

Strains and stresses are induced in each layer. Because the layers are firmly bonded together the strains are the same in the a and b layers, and are equal to the strains in the whole plate:





fi a




~ m2b

<*2b ELb

Solution of the foregoing Eqs. 2-109 to 2-117 leads to the following simultaneous equations:

/»„(T10 + Ano2a + AnT12o - -L [ 21 - V216 p - rn,b J1 ) (2-118)

^21^10 + A22a2a + A12t12o = ( - v12b ^ + p- - m2b ^ I (2-119)

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