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Question 16.9: Design diagram of uniform cantilevered beam with intermediat......

Design diagram of uniform cantilevered beam with intermediate support is shown in Fig. 16.17. Parameters of beam are the total length l, mass per unit length m, flexural stiffness EI=const. Derive the frequency equation (Babakov 1965).

16.17
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The origin of the coordinates is placed on the extreme left support, the x axis is directed to the right. Initial parameters and conditions for intermediate support at x=l / 2  and for free end are shown in Fig. 16.17. According to (16.28), the total expression for vertical displacements is

\begin{aligned}& Y(x)=Y_0 S(k x)+\theta_0 \frac{1}{k} T(k x)+\frac{M_0}{k^2 E I} U(k x)+\frac{Q_0}{k^3 E I} V(k x) \\& \theta(x)=Y_0 k V(k x)+\theta_0 S(k x)+\frac{M_0}{k E I} T(k x)+\frac{Q_0}{k^2 E I} U(k x) \\& M(x)=E I Y_0 k^2 U(k x)+E I \theta_0 k V(k x)+M_0 S(k x)+\frac{Q_0}{k} T(k x) \\& Q(x)=E I Y_0 k^3 T(k x)+E I \theta_0 k^2 U(k x)+M_0 k V(k x)+Q_0 S(k x)\end{aligned}          ( 16.28)

Y(x)=\frac{M_0}{k^2 E I} U(k x)+\frac{Q_0}{k^3 E I} V(k x)+\frac{R}{k^3 E I} V\left[k\left(x-\frac{l}{2}\right)\right]

The structure of the terms which take into account the reaction R and shear force Q_0  are same, i.e., they contain the same coefficient \frac{1}{k^3 E I} , and the same function V; the difference is that the function V for the reaction R has a shifted argument x-l / 2 . It means that the last term should be take into account only for x>l / 2 .

At the intermediate support we have

Y\left(\frac{l}{2}\right)=\frac{M_0}{k^2 E I} U\left(\frac{k l}{2}\right)+\frac{Q_0}{k^3 E I} V\left(\frac{k l}{2}\right)=0             (a)

The total expression for bending moments and shear are

\begin{aligned}& M(x)=M_0 S(k x)+\frac{Q_0}{k} T(k x)+\frac{R}{k} T\left[k\left(x-\frac{l}{2}\right)\right] ; \\& Q(x)=M_0 k V(k x)+Q_0 S(k x)+R S\left[k\left(x-\frac{l}{2}\right)\right]\end{aligned}

so at the free end we have

\begin{aligned}& M(l)=M_0 S(k l)+\frac{Q_0}{k} T(k l)+\frac{R}{k} T\left(k \frac{l}{2}\right)=0 \\& Q(l)=M_0 k V(k l)+Q_0 S(k l)+R S\left(\frac{k l}{2}\right)=0\end{aligned}            (b)

The unknowns of equations (a) and (b) are M_0, Q_0 , and R. The frequency equation we get from a condition of the nontrivial solution of homogeneous algebraic equations (a) and (b) is

D=\left|\begin{array}{ccc}\frac{1}{k^2 E I} U\left(\frac{\lambda}{2}\right) & \frac{1}{k^3 E I} V\left(\frac{\lambda}{2}\right) & 0 \\S(\lambda) & \frac{1}{k} T(\lambda) & \frac{1}{k} T\left(\frac{\lambda}{2}\right) \\k V(\lambda) & S(\lambda) & S\left(\frac{\lambda}{2}\right)\end{array}\right|=0, \quad \lambda=k l

The frequency equation in the expanded form is

U_{\lambda / 2} T_\lambda S_{\lambda / 2}+V_{\lambda / 2} T_{\lambda / 2} V_\lambda-S_\lambda T_{\lambda / 2} U_{\lambda / 2}-S_{\lambda / 2} S_\lambda V_{\lambda / 2}=0

The frequencies equation in the elementary functions becomes

\sin \lambda \cosh \lambda-\sinh \lambda \cos \lambda=\sinh \lambda-\sin \lambda

To determine the interval in which the roots of the equation will be located, the following theorem can be used (Babakov 1965, pages 134,279).

If a linear constraint is imposed on a system with n-degrees of freedom, then the frequencies of the modified system with n-1  degrees of freedom are located between the frequencies of an initial system.

This theorem may be extended to structures with distributed parameters. For given Example 16.9, the clamped-free beam of length l should be treated as initial structure. For this beam the three eigenvalues are k l=1.875,4.694,7.855, \ldots  (Table A.24). Therefore, the first and second eigenvalues \lambda_1  and \lambda_2  of cantilevered beams with additional constraint (modified system) satisfied to conditions

1.875 \leq \lambda_1 \leq 4.694 \leq \lambda_2 \leq 7.855

This wide range should be narrowed down. The definition of the boundaries of the refined range is based on a change in the sign of the determinant D. The refined interval for the first eigenvalue is 3.1<\lambda_1<3.2 . The first frequency vibration is \omega_1=k_1^2 \sqrt{\frac{E I}{m}}=\frac{\lambda_1^2}{l^2} \sqrt{\frac{E I}{m}} . Corresponding first mode of vibration is shown in Fig. 16.17.

Initial parameters method may be successfully applied to beam with elastic support, beam with distributed mass together with lumped mass, etc.

Table A.24 Frequency equation, eigenvalue and nodal points for one-span uniform beams,   k^4=\frac{m \omega^2}{E I}
# Type of beam Frequency equation n Eigenvalue   \lambda_n Nodal points \xi=x / l  of mode shape X
1 Pinned-pinned \sin k_n l=0 1 3.14159265   \begin{aligned} & 0 ; 1.0 \\ & 0 ; 0.5 ; 1.0 \\ & 0 ; 0.333 ; 0.667 ; 1.0\end{aligned}
2 6.28318531
3 9.42477796
2 Clamped-clamped \cos k_n l \cosh k_n l=1 1 4.73004074   \begin{aligned} & 0 ; \quad 1.0 \\ & 0 ; 0.5 ; 1.0 \\ & 0 ; 0.359 ; 0.641 ; 1.0 \\ & \end{aligned}
2 7.85320462
3 10.9956079
3 Pinned-clamped \tan k_n l-\tanh k_n l=0 1 3.92660231   \begin{array}{llll} 0 ; & 1.0 & \\ 0 ; & 0.440 ; & 1.0 & \\ 0 ; & 0.308 ; & 0.616 ; & 1.0 \end{array}
2 7.06858275
3 10.21017612
4 Clamped-free \cos k_n l \cosh k_n l=-1 1 1.87510407   \begin{aligned} & 0 \\ & 0 ; \quad 0.774 \\ & 0 ; \quad 0.5001 ; \quad 0.868 \\ & \end{aligned}
2 4.69409113
3 7.85475744
5 Free-free \cos k_n l \cosh k_n l=1 1,2 0   \begin{aligned} & \text { Rigid-body modes } \\ & 0.224 ; \quad 0.776 \\ & 0.132 ; \quad 0.500 ; \quad 0.868 \end{aligned}
3 4.73004074
4 7.85320462
6 Pinned-free \tan k_n l-\tanh k_n l=0 1 0 Rigid-body mode    \begin{array}{lll} 0 ; & 0.736 & \\ 0 ; & 0.446 ; & 0.853 \end{array}
2 3.92660231
3 7.06858275

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