 # Differential Equations Question

## Transcribed Text

Derive a frequency equation for a uniform beam of length L subjected to the specified boundary condition: fixed free. 7.3 Free Vibration of Uniform Beam Consider beam of length L bending as shown in Figure 7.3. The beam is assumed uniform so that density P cross- sectional area A and moment of inertia I are all constants. The vertical deflection of the beam is denoted by u(x,t) Furthermore, E is Young's modulus of elasticity, and the product El is the flexural rigidity. u Density p Flexural rigidity El Cross-sectional ar A u(x,t) x L Figure 7.3 Auniform beam in bending Application of Newton's second law, as well as the moment equation, leads to the equation of motion of the beam (7.14) Boundary Conditions There are three types of boundary conditions to consider for uniform beam. o Fixed (clamped) end u = 0 (deflection), u, =0 (slope of deflection) Free end =0 (bending moment), Elupor =0 (shear force) o Simply supported (hinged, pinned) end u=0 (deflection), Elu, =0 (bending moment) Note that Eq. (7 14) contains fourth-order spatial partial derivative, which would require four boundary conditions to determine the four unknown constants These are provided by the conditions listed above. As an example, for fixed- free (fixed at the left end, free at the right end) uniform beam of length L, the four conditions are (0,1)=0, 41(0,1)=0 Fixed at the left end Free at the right end Free Vibration of inned-Pinned Uniform Beam Consider the uniform beam shown in Figure 7.4. The equation of motion, Eq. (7.14), is conveniently written as El un =0 a = (7.15) Beam qquation PA Boundary conditions u(x,0) = f(x) u,(x,0)= g(x) Initial displacement Initial velocity Note that the boundary conditions on bending moment have been simplified since El const As it was the case with the elastic string, because geometry of the problem is simple, and the boundary conditions are homogeneous, the model can be solved by separation of variables The method allows u(x,1) to be expressed as u(x,t) X(x)T(t), the product of spatial function, and temporal function. EI,A x=0 x=L Figure 7.4 A pinned pinned uniform beam Substitution into Eq. (7.15), letting "dots" and "primes" denote derivatives with respect to and x. respectively, yields Divide by XT = x B const Two differential equations emerge from this last equation: (7.16) Using the same logic as before, imposing T(t) 0 to avoida trivial solution, the boundary conditions are interpreted as X(0)=0=X(L), X"(0)=0=X"(L) (7.17 The solutions of the ODEs in Eq. (7 16) depend on the nature of B and as such, all three possible cases must be tested Proceeding as in the case of the vibrating string, can easily be shown that B>0 is the only case that results in non trivial solution. Since B is positive, can be written as B =22 where w#0. With this, the spatial ODE becomes x10-yix-0.1-(a) with general solution , X(x) =c cosyx +c2sinyx +c3 cosh yx FCA sinhyx Applying the boundary conditions at x=0 to this solution yields X(0)=G+G= 15 9999 (x"0)=-1219-9)=0 19-9=0 Update the spatial solution as X(x) Applying the boundary conditions at x=L, we find x(L)=c2sinyL+c4sinhyL=0 c,sinyL+c_sinhyL=0 X"(L)=r²(-c2sinyL+c4sinhyL)=0 Adding these last two equations, we have sinhyLw0 2c, sinh yL= => c4=0 With this, the spatial solution is updated as X(x)=c2sinyx Since 9=03=C4 =0, unen C2 +0, otherwise the solution becomes trivial. Therefore, by either of the two equations in Eq. (7.18), sinyL=0 which is known as the characteristic equation, or the frequency equation, and whose solutions are known as the characteristic values. Because y depends on index n, more accurate notation for the spatial function is nax n=1,2,3, These are the characteristic functions or natural modes of vibration. Recall y (a) where the correct notation for y is 7n. As a result, wn n=1,2,3, 3.48 which are the beam natural frequencies. We next consider the temporal ODE in Eq. (7.16), T+WOT-0. With the correct notation in place, the solution to this ODE is Therefore, u(x,t) n-] n=] Letting An = 22,nc5,n and Bn =c2ndone the above simplifies to . (7.19) N-1 The initial conditions are used next to find the coefficients An and Bn, Applying u(x,0)= f(x) to Eq. (7.19), we nzx n=l This is the Fourier sine series of f(x), Section 6.1, hence nax, Finally, application of u,(x,0) g(x) leads to nox "=] L This the Fourier sine series of g(x) hence Solving for Bn,

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