The Behavior of Structures Composed of Composite Materials Part 7 pot

170 BEAM CLAMPED AT x = 0, FREE AT x = L BEAM SIMPLY SUPPORTED AT EACH END These equations are quite general for a beam of uniform flexural stiffness subjected to any concentrated load P acting at For instance, Equation (4.60) can be used for a clamped-free beam with a load P acting at the tip of x = L by letting Also because the equations evolve from linear theory, superposition can be used if there are two concentrated loads at two different locations, but care must be used to accurately depict regions to the left and right of each load to insure correct solutions. After the solutions are found, Equations (4.14) and (4.17) are used to obtain bending moments and shear resultants, and Equation (4.26) is used to determine stresses everywhere. 4.4 Solutions by Green’s Functions 171 From Section 4.3, consider a composite beam subjected to a unit concentrated load, i.e., P = 1. By way of example, consider the beam simply supported at each end given by Equations (4.62) and (4.63). In this case, Note that is the deflection to the left of the load, to the right of the load. Green’s function and can be defined as the deflection at x due to a unit load at It can be reasoned that any distributed load q(x ) is in fact an infinity of concentrated loads, which can be summed to obtain the solution of the response to a distributed load. Because of the infinity of concentrated loads, the infinite summation can be replaced by an integration such that the following is correct. For a beam simply supported at each end Equations (4.66) and (4.67) would be inserted for and and other analogous expressions from Section 4.3 would be used for beams with other boundary conditions. As an example, solving Equations (4.68), (4.66) and (4.67) with uniform load over the entire length, results in which is the solution obtained through solving the governing differential equation (4.20) and satisfying the appropriate boundary conditions, with the explicit solution shown in 172 Equation (4.49). Likewise, using the Green’s function approach discussed above for a uniform load results in Equation (4.29) for the clamped-clamped beam, and Equation (4.37) for the cantilevered beam. Thus, with the use of Green’s functions one uses an integral equation where the Green’s functions satisfy the boundary conditions, rather than solving the governing differential equation and matching boundary conditions. This alternative approach 1. 2. is general for any governing differential equation such as a beam, plate or shell. can save great computational difficulty for complicated problems. As an example of the latter, consider the following problem (Figure 4.7): To solve the problem above, developing a solution requires solving the governing differential equations and matching boundary conditions involves dividing the beam into six sections, wherein twenty-four boundary value constants must be solved for, and particular solutions obtained for each of the distributed loads. As a result, w ( x ) is determined everywhere after all constants are solved for simultaneously. Using the Green’s function approach, suppose that one hypothesizes that the maximum deflection and bending moment occurs in the region One obtains and for the clamped-simply supported beam from Equations (4.58) and (4.59). Then for is given by w ( x ) 173 To check that the maxima occur in this region, one could easily investigate adjoining regions, i.e., and Having obtained Equation (4.78), maximum deflections and stresses are determined straightforwardly as shown before. 4. 5 Composite Beams of Continuously Varying Cross-Section There are many applications in which beams can be better designed and utilized through having a variable cross-section rather than a constant cross-section discussed in Sections 4.1 through 4.4. In that case, the and are functions of x, the length dimension, i.e., and In that case, it is straightforward to derive the governing equation of beam bending, analogous to the derivation leading to Equation (4.20). The result is where here I ( x ) is a continuously varying function of x, and the mid-surface of the beam is z = 0. Depending on what that function of x is, the solution could be rather complicated and/or tedious. For these types of problems, Galerkin’s method is well suited. No attempt is made here to provide a detailed comprehensive introduction of Galerkin’s method, which is well treated in numerous other texts. However, consider an ordinary differential equation (although the method is equally useful for partial differential equations as well as nonlinear equations). where L is any differential operator, and q(x ) is a forcing function. Boundary conditions must be homogeneous; if not, a transformation of variables must be made first to attain homogeneous boundary conditions. In Galerkin’s method one assumes a complete set of coordinate functions n = 1, 2, 3, , N, which satisfies the prescribed homogeneous boundary conditions. Although these functions need not be an orthogonal set, if they are, the procedure is much simpler. Therefore, assume 174 where the are constants. By using only the first N terms, an error can be defined from Equations (4.73) and (4.72), as If is sufficiently small, then of Equation (4.73) is considered to be a satisfactory approximation to w ( x ) . So can be viewed as an error function, and the task is to select proper values of to minimize In the Galerkin method, this is done by the following orthogonality condition: This is equivalent to minimizing the mean square error and insures that will converge to w ( x ) in the mean. From above, this is equivalent to stating: Therefore, Equation (4.76) is a set of N algebraic equations through which one can determine the to minimize This is a powerful technique. As an example, consider a tapered beam subjected to a lateral distributed load q ( x ) , as developed by R.L. Daugherty when he taught a course using the first edition of this text. Assume the variable stiffness can be written as: Equation (4.71) can then be written as where primes denote differentiation with respect to x. Hence, 175 If the beam is s i mply supported, i.e. , at x=0, L, coordinate functions can be chosen as follows: These are a complete set of functions, which satisfy the boundary conditions. Thus, and from Equation (4.74) From Equations (4.75) and (4.76), Galerkin’s procedure requires To continue the example, let f ( x ) = 1 + ( x/ L ) , then Equation (4.83) becomes: 176 So Equation (4.82) is, for For n = m Thus, Equation (4.82) becomes for n = m If N is taken as 3 for example, the final set of three non-homogeneous algebraic equations are written as follows to obtain and The first, second and third equations are for m = 1, 2, 3, respectively. For 177 To complete the problem for any distributed load, q(x ) , is straightforward. To know what value of N to take can only be determined by solving the problem with one higher integer and determining that the previous approximation is sufficient. One thing to remember is that the result of this method is the determination of an approximate deflection w(x). To determine stresses requires solving for since the bending stresses are proportional to Taking derivatives of an approximative function causes an increase in the error through differentiation. Hence, for stress critical structural members, N must be determined by suitably approximate the maximum stress. This may require a higher value of N than that required to achieve a certain accuracy in maximum deflection. 4.6 Rods Consider the rod shown in Figure 4.1 and 4.8 subjected to a uniform tensile axial load in the x -direction. One wishes to determine the in-plane mid-plane displacement and the axial stresses in each lamina The assumptions are: a . Individual plies (laminae) are perfectly bonded (no slip). 178 b. c. d. e. f. g. h. The thickness dimensions of the rod are small compared to the length of the rod. Displacements and strains are small compared to all beam/rod dimensions. A balanced, mid-plane symmetric laminate is used, i.e., The loading is static. For simplicity of example hygrothermal effects are ignored. The loading is in the x-z plane. Since b << L, strains in the y -direction are ignored. The governing equation for axial equilibrium is easily shown to be as follows, where here an axial component has been added to (4.5) where from Figure 4.8, p ( x ) is the axial distributed force per unit area, which can be a function of x. The mid-plane strain-axial displacement equation is again seen to be The constitutive equation can be given by the following for each ply, where is the ply stiffness in the x -direction of the kth lamina, and is equivalent to if Poisson’s ratio effects are ignored. It has been shown in many texts that can be written as follows: where all quantities have been defined previously. *Note that as alternative to using Equations (4.87) and (4.88) to obtain the stress in the k th lamina, one could use Equation (4.19) where of course because with a beam The integrated axial stress resultant is seen to be, as in (4.3) Utilizing the above equations the governing differential equation is found to be 179 If the rod is held fixed at x = 0 in Figure 4.8, then the boundary conditions are: The solution is found straightforwardly to be and the stress in the kth lamina is 4.7 Vibration of Composite Beams In Sections 4.1 through 4.5 the problems studied have concentrated on (1) finding the maximum deflection in composite beams to insure that they are not too large for a deflection-limited or stiffness-critical structure, and (2) determining the maximum stresses in the beam structure for those structures, which are strength critical. However, there are two other ways in which a structure can become damaged or useless; one is through a dynamic response to time-dependent loads, and the other is through the occurrence of an elastic instability (buckling). In the former, dynamic loading on a structure can vary from a recurring cyclic loading of the same repeated magnitude, such as a structure supporting an unbalanced motor that is turning at a specified number of revolutions per minute (for example), to the other extreme of a short time, intense, nonrecurring load, termed shock or impact loading, such as a bird striking an aircraft component during flight. A continuous infinity of dynamic loads exists between these extremes of harmonic oscillation and impact. [...]... are written on the dynamic response of composite structures to time-dependent loads, but that is beyond the scope of this text There are a number of texts dealing with dynamic response of isotropic structures However, one common thread to all dynamic responses are the natural frequencies of vibration and their associated mode shapes Mathematically, any continuous structure has an infinity of natural frequencies... and the rotatory inertia respectively defined as: where is the mass density of each ply, e.g lb These equations can be simplified to the following coupled equations: Making use of the constitutive equations of (2.66), and the strain displacement relations, the following dynamic equations can be obtained in terms of the displacements: 186 If the loads are static, these equations become Note, in the. .. above the subscript T denotes thermal terms, and the subscript m denotes moisture effects represents the axial moment distribution Solving the first equation above for the function of which is then substituted into the second equation, the result is: One sees immediately that the reduced bending stiffness is It is seen that if the beam is mid-plane symmetric, the last term on the right hand side of Equation... Hence, the natural frequencies and the buckling load are the eigenvalues; while the vibrational mode and the buckling mode are the eigenfunctions In analyzing any structure, one should therefore determine four things: the maximum deflection, the maximum stresses, the natural frequencies (if there is any dynamic loading to the structure, or nearby to the structure), and the buckling loads (if there are... nonlinear theory will not be included herein because it is covered in so many other texts [7, 8], except to say that for a beam the strain-displacement relation is modified to include a non-linear term as seen below 198 The result of including the terms to predict the advent or inception of buckling for the beam is the following: where there is a coupling between the in-plane load P and the lateral... beam theory if one uses the relationship one can conclude that then from above where is a negative number associated with the bottom of lamina 1, i.e., the lower face of the beam Now for the specific case of (a constant) 189 and the stresses are Now as an example problem consider the mid-plane asymmetric beam subjected only to a static hygrothermal loading, and no mechanical loads From above, the governing... every other structure are identical So as long as one is using linear theory, the two effects obviously are superposed as seen above But the important thing to remember is that if one has the thermal effect solution to any solid mechanics (linear, problem, then through superposition one has the solution of the hygrothermal problem for the same structure and boundary conditions Finally, consider the following... coefficients of thermal expansion are given by The thermal loadings are as follows Finally, the deformations for this problem are 193 4.10 Advanced Beam Theory Including Transverse Shear Deformation Effects The effects of transverse shear deformation are discussed now so that a comparison can be made with the analyses of Section 4.8, which did not include these effects The development of this section... the in-plane action is entirely independent of the lateral action Yet it is well known and often observed that in-plane loads (through buckling) do cause lateral deflections, which are usually disastrous The answer to the paradox is that up to this point in this chapter only linear elasticity theory has been used, and the physical event of buckling is a nonlinear theory problem The development of the. .. equation (4. 97) is only applicable over a portion of this range, i.e., it applies to those mode shapes in which the classical beam equations apply For an isotropic single-layer beam, the equation breaks down when the half wavelength becomes close to the height, h, of the beam, because then transverse shear deformation effects become important, and the classical theory of Equations (4.94) and (4. 97) yields . beam, the equation breaks down when the half wavelength becomes close to the height, h, of the beam, because then transverse shear deformation effects become important, and the classical theory of. terms of the displacements: 186 If the loads are static, these equations become Note, in the above the subscript T denotes thermal terms, and the subscript m denotes moisture effects. represents the. conclude that then from above where is a negative number associated with the bottom of lamina 1, i.e., the lower face of the beam. Now for the specific case of (a constant). 189 and the stresses