T k

A,S,

-A3E,

A*Et

y Neutral

Ei = 5 0 X 10" psi Et = 3.0 X 10" p«i £. = 1.0 X 10" psi E. = 5.0 X 10" psi B. = 3.0 X 10* psi m'

Neutral

«■ -- 40,000 psi = 25,000 psi a i = 5,000 psi tr. = 40,000 psi = 25,000 psi

of the span would be WL/4. Setting this equal to 3,500 in.-lb gives the load W as 1400 lb. Shear V is W/2 or 700 lb. Using this value, the shear stress intensity at various horizontal planes in the beam may be computed by means of Eq. 2-128.

Plane Layers Eft; y'

Q"

Plane Layers Eft; y'

Q"

b-b

1

0.2 x 5 x 106

0.285"

0.285 x 106

1150 psi

rl

0.2 x 5 x 106

0.285

c-c i

0.326 X 106

1315 psi

[2

+ 0.1 x3x 106

0.135

[4

0.2 x 5 x 106

0.215

d-d

0.324 x 106

1310 psi

l5

+ 0.1 x 3 x 106

0.365

These would be the critical planes because they represent planes between layers of different materials, and consequently the resin alone would largely carry the stress. The shear stress at the neutral axis would be slightly higher and might or might not represent the critical plane, depending upon the structure of the material in layer 3.

Structural Sandwiches

In usual construction practice, a structural sandwich is a special case of a laminate in which two thin facings of relatively stiff, hard, dense, strong material are bonded to a thick core of relatively lightweight material considerably less dense, stiff, and strong than the facings.

With this geometry and relationship of mechanical properties, the facings are subjected to almost all of the stresses in transverse bending or in axial loading, and the geometry of the arrangement provides high stiffness combined with lightness because the stiff facings are at maximum distance from the neutral axis, similar to the flanges of an I-beam. The continuous core takes the place of the web of an I-beam or box beam, it absorbs most of the shear, and it also stabilizes the thin facings against buckling or wrinkling under compressive stresses. The bond between core and facings must resist shear and any transverse tensile stresses set up as the facings tend to wrinkle or pull away from the core.

Stiffness

For an isotropic material with modulus of elasticity E, the bending stiffness factor EI of a rectangular beam b wide and h deep is

In a rectangular structural sandwich of the same dimensions as above whose facings and core have moduli of elasticity Ey and Ec, respectively, and a core thickness C, the bending stiffness factor EI is

This equation is exact if the facings are of equal thickness, and approximate if they are not, but the approximation is close if facings are thin relative to the core.

If, as is usually the case, Ec is much smaller than Ej-, the last term in the equation can be ignored.

For unsymmetrical sandwiches with different materials or different thicknesses in the facings, or both, the more general equation for £EI given in the previous section may be used.

In many isotropic materials the shear modulus G is high compared to the elastic modulus E, and shear distortion of a transversely loaded beam is so small that it can be neglected in calculating deflection. In a structural sandwich the core shear modulus Gc is usually so much smaller than Ej- of the facings that shear distortion of the core may be large and therefore contribute significandy to the deflection of a transversely loaded beam. The total deflection of a beam is therefore composed of two factors: the deflection caused by bending moment alone, and the deflection caused by shear, that is

w here

8 = total deflection 8m = moment deflection 8g = shear deflection

Under transverse loading, bending moment deflection is proportional to the load and the cube of the span and inversely proportional to the stiffness factor EI. Shear deflection is proportional to the load and span and inversely proportional to a shear stiffness factor N whose value for symmetrical sandwiches is

where

Gc = core shear modulus The total deflection may therefore be written g = KJNL? + KsjA/L (2_m

The values of K,„ and Ks depend on the type of load. Values for several typical loading conditions are given as shown in Table 2.4.

Stresses in Sandwich Beams

The familiar equation for stresses in an isotropic beam subjected to bending

must be modified for sandwiches to the form

where y = distance from neutral axis to fiber at y

Ey = elastic modulus of fiber at y El = stiffness factor

For a symmetrical sandwich the stress in the outermost facing fiber is found by setting

Y = h/2 Ey= Ef and the stress in the outermost core fiber by setting y = c/2 Ey= Ec

The mean stress in the facings of a symmetrical sandwich can be found from a =

bt(h+c) where t = facing thickness

Similarly the general equation for the shear stresses in a laminate (see preceding section)

can be used for any sandwich. For the symmetrical sandwich the value of T can be closely approximated by

Axially-Loaded Sandwich

Edge-loaded sandwiches such as columns and walls are subject to failure by overstressing the facings or core, or by buckling of the member as a whole. Direct stresses in facings and core can be calculated by assuming that their strains are equal, so that

where

Of oc

A,r total load facing stress core stress cross-sectional area of facings cross-sectional area of core

Usually the elastic modulus Ec of the core is so small that the core carries little of the total load, and die equation can be simplified by ignoring the last term, so that for a sandwich b wide with facings t thick,

The column buckling load of a sandwich L long simply supported at the ends is given by

This variation of the Euler equation takes into account the low shear stiffness of the core.

For wall panels held in line along their vertical edges an approximate buckling formula is

provided the length L of the panel is at least as great as the width b and provided the second term in the bracket of the denominator is not greater than unity.

Filament-Wound Shells, Internal Hydrostatic Pressure

Basic Equations

Cylindrically symmetric shells are considered which are of the form r=r (z) in a system of cylindrical polar coordinates (r, 6, z). Inextensible fibers are wound on and bonded to this shell in such a way that at any point on it equal numbers of fibers are inclined at angles a and n-a to the line of latitude (z=constant) passing through that point. At a point (r, 0, z) of the shell there are n1} nu ... np fibers, per unit length measured perpendicular to the length of the fibers, with positive inclinations CLj, a2, ... o.p; and an equal number of fibers with inclinations K-a.\, tc-ol ... n-a.p, to the line of latitude passing through (r, 6, z). The number n\, n2 ... np, and angles oci, a2 ... o.p are independent of 6 and, since r is a function of z, they may be regarded as functions of z only.

The shell is subjected to internal hydrostatic pressure P and all resulting forces are carried by the fibers, which are considered to constitute an undeformable membrane. T\ and T2 are the normal components of stress in the latitudinal and longitudinal directions at a point (r, 8, z). Because of cylindrical symmetry, 7"! and T2 are independent of 6; and because there are equal numbers of fibers inclined at a, and n - a,- (i = 1,2, ... p), the shearing components of stress are zero.

Let Ti, ■■■ Tp be the tensions at (r, 9, z) in the fibers inclined at a1? (x2 ... ap. Then because of symmetry, X\ (i=l, 2, ...p) must also be the tensions in the fibers inclined at n-O^. The number of fibers with inclination a; per unit length measured along a line of longitude (9 = constant) is n, cos a,. Resolving the tensions in the fibers in the latitudinal direction the latitudinal tension Ti is obtained

cos2a¡

Similarly, the number of fibers inclined at a, per unit length measured along a line of latitude (z = const) in w, sin a„ and resolving the tensions in fibers parallel to the longitudinal direction, the longitudinal tension T2 is obtained

sin2o/

If ki and k2 are the curvatures of the membrane in the latitudinal and longitudinal directions

dz d and

where T is the distance from the equator, measured along a line of longitude.

The equations of equilibrium for the membrane are k:T2 = :hP equilibrium in direction normal to membrane /c^, + kj2 = P equilibrium along a line of longitude (2-146)

Solving for Ti and T2 and combining

n,cosz

0 0

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