4

rorb V. Clb t2b J

Solution of these equations and reference to Eqs. 2-109 to 2-111 show that

°io = 0it> = cr, = 19,200 psi a2a = a2b = = 9.600 psi

This proves what might have been expected intuitively; because of symmetry with respect to the 1-2 directions chosen; the internal direct stresses Oi„ o^ (T2(„ and a2h are equal to the imposed stresses Oj and a2, and there is no internal shear stress.

The same result is found for Case 2. In this balanced fabric m\ = m2 = 0 at 45°, there is no shear distortion caused by direct stress, and shear therefore is zero. In Case 3:

The values of m\y and m2b are negative because the 30° angles of orientation of the longitudinal direction Lb of layers b is measured in the negative direction whereas it is positive for the a-layers.

Equations 22 become

Aua]a + Ana2o + o = -L - v21b -P- + 0| <arb V Mb ^26 /

0 + 0 + /l33r12fl = f - m16 ^ - m2b +0] 'o'b V. Hb tLb )

The first two of these equations are exactly like the first two equations for Cases 1 and 2 and show that the internal direct stresses are equal to the imposed, that is

0) a = b = 0) = 19,200 psi 02a = 02b = 02 = 9,600 psi

The third equation, however, is not equal to zero, and its solution, together with equation 19c, shows that r^a = 6750 psi r12b = 6750 psi

Appreciable shear stresses are set up within the body of the cylinder wall when layers are oriented as in Case 3, even though no shear forces are applied to the cylinder itself. The shear stresses in layers b are oriented in the direcdon opposite to the shear stresses in layers a.

The difference in the shear stresses between the two layers must be taken up by shear in the adhesive bond between them, that is, in the layer of resin that holds the fiber-reinforced layers together. The difference is

This shear stress in the resin bonding the layers together is therefore seen to be high.

In Cases 1 and 2 the orientadon of the fibers with respect to the 1-2 directions chosen resulted in zero shear stresses associated with those directions, whereas in Case 3 the shear stresses were not zero. In all three cases, symmetry of the fiber orientations with respect to the stress directions resulted in internal stresses equal to the external stresses. These are special cases. In the more general case the internal direct stresses in the individual layers are not necessarily equal to the external direct stresses, nor are they the same in the various layers. Furthermore, even symmetrical Case 3 leads to internal shear stresses when external shear stresses are absent. In the more general case it is still more true that internal shear stresses may be appreciable, or they may be absent, depending upon the magnitude of the external stresses and the orientation of the 1-2 directions with respect to the external stresses.

A more general case in shown in Fig. 2.37b in which the same cylinder is chosen as in Fig. 2.37a except that torsional effect equal to a twisting couple of 25,000 in.-lb has been added. The construction of the wall has also been changed. Layers a of unbalanced material having the properties of Fig. 2.34 are a total of 0.13 in. thick, and are oriented at 15° to the circumferential direction as shown. Layers b, of balanced material having the properties of Fig. 2.35, are a total of 0.07 in. thick and are oriented at 45° as shown. Referring to Fig. 2.35, the properties are found to be o-l aye rs

6-layers to a

Gi2a

0.703 x 5x 106

Vl26 G 12b

Solving for the various constants and substituting in Fqs. 2-118 to 2-120

11.241 2ct1o - 5.6641 o2a - 4.046U12o = +190,180 -5.6641 CJio + 23.2030cJ2o - 4.5156T12o = -21,150 -4.0461 <7lb - 4.5156cj2o+ 26.4668T12o = +16,190

The solution of the forgoing simultaneous equations leads to the following results for stresses in the »-layers:

When these results are employed with Eqs. 2-109 to 2-111 it is found that stresses in the ¿-layers are:

<7lb = 15,700 psi d2b = 17,800 psi r12b = -6,150 psi

Bending of Beams and Plates

Plates and beams of fibrous glass reinforced plastics may be homogeneous and isotropic or composite and nonisotropic depending upon their structure. Mat-reinforced plates may be considered to be essentially isotropic and the usual engineering formulas may be applied. Composite structures require suitably modified formulas but otherwise the procedures for computing bending stresses, stiffness, and bending shear stresses are essentially the same as for isotropic materials. The differences and similarities may be brought out by considering two beams of identical overall dimensions, one isotropic and the other composite. Two such cross sections are shown in Fig. 2.38. For each cross section it is necessary to know the stiffness factor EI to compute deflection, the section modulus to compute bending stresses, and the statical moments of portions of the cross section to compute shear stresses. For isotropic materials (a) the neutral axis of a rectangular cross section is at middepth, and the familiar formulas are bcP

y 6M

Bendinq stress = a = M4 = -r-o for outermost fiber I uu

Shear stres = -rr = 7r, for maximum shear at the neutral axis o/ 2 ba

For composite materials the neutral axis is not necessarily at middepth of a rectangular section, and it must first be found.

in which Eb A„ x, are the modulus of elasticity; cross-sectional area (bdi); and distance from some reference line, such as the bottom of the cross section, to the center of gravity of any particular layer.

in which E, and I, are, for any particular layer, the modulus of elasticity and the moment of inertia about the neutral axis.

Bending stress a = MEyy/Ej

in which y is the distance from the neutral axis to any point, and Ey is the modulus of elasticity of the layer at that point. The maximum bending stress does not necessarily occur at the outermost (top or bottom) fiber, as it does in isotropic materials.

in which V is the total shear on the cross section, t is the shear stress intensity along some horizontal plane, and Q' is the weighted statical moment, E^Ay about the beam's neutral axis, of the pordon of the cross section between the horizontal plane in quesdon and the outer edge (top or bottom) of the cross section.

An example of the forgoing is illustrated in Fig. 2.38c in which a composite beam is made up of five layers having three different moduli of elasticity, and three different strengths, as shown.

The neutral axis, found by applying Eq. 2-125 is 0.415 in. from the bottom of the cross section. Distance from the neutral axis to the centers of the individual layers are computed, and the stiffness factor EI calculated by means of Eq. 2-126. This is found to be

Bending stresses are next computed for the top and bottom edges of the cross section and for the outer edge of each layer, that is, the edge of each layer farther from the neutral axis. From these, the bending moment the cross section is capable of carrying can be computed. This may be done for example by applying a bending moment M of one in.-lb and computing the unit bending stresses. These unit bending stresses divided into the strengths of the individual layers give a series of calculated resisting moments, the smallest of which is the maximum bending moment the beam is capable of carrying without exceeding the strength of any portion of the cross section.

For a unit bending moment M= 1 in.-lb, <5y = from Eq. 2-127 Plane y Ey cyin.lb a = M

Gy/in.lb

0-0

0.385 in.

5 x 106

11.1 psi

40,000/11.1

= 3,600 in.

-lb

b-b

0.185 in.

3 x 106

3.19 psi

25,000/3.19

= 7,800 in.

-lb

c-c

0.085 in.

1 x 10G

0.49 psi

5,000/0.49

= 10,200 in.

-lb

d-d

0.115 in.

1 x 106

0.66 psi

5,000/0.66

= 7,600 in.

-lb

e-e

0.315 in.

5 x 106

9.07 psi

40,000/9.07

= 4,400 in.

-lb

f-f

0.415 in.

3 x 106

7.16 psi

25,000/7.16

= 3,500 in.

-lb

If, for example, the beam were a simple beam carrying a load W on a 10-in. span, as shown in Fig. 2.38, the bending moment at the center

Fig Lire 2.38 Cross section of: (a) isotropic beam (b) composite beam made in layers of different materials, and (c) composite beam having properties (where all E are x 106 psi) E, = 5, E2 = 3, E3 = 1, E4 = 5, and E5 = 3; also (where o are 103 psi) o, = 40, o2 25, o3 = 5, o4 = 40, and o5 = 25

' <f

A.S

Neutral axis

0 0

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