A unique and comprehensive approach to investigate the transverse free vibration of non-uniform and functionally graded Euler–Bernoulli beams

Abstract

This study investigates the transverse free vibration of general type of non-uniform beams with general non-classical boundary conditions at both ends based on the exploitation of Cauchy’s formula for iterated integrals. A review of related sources shows that despite of many studies and various methods used to investigate the transverse vibrations of non-uniform and inhomogeneous beams, no solutions applicable for free vibrations in such beams with general boundary conditions were proposed. In this article, Hamilton's principle is utilized to derive the partial differential equation governing the transverse vibration of the inhomogeneous non-uniform beam as well as the general corresponding boundary conditions. The effect of the boundary conditions, mass object, eccentricity and aspect ratios on the natural dimensionless transverse frequencies has been analyzed. The flowcharts of analyzing frequency parameters and mode shapes are also given in details to state the proposed method visually. The performance of the proposed method has been verified through numerical examples available in the published literature. All results exhibit that the proposed method is capable to analyze the free vibration of general type of non-uniform and axially FG Euler–Bernoulli beams with various general boundary conditions. Utilizing the proposed method and their related equations, it is possible to analyze the vibrations of different types of homogeneous uniform, homogeneous non-uniform, functionally graded uniform, and functionally graded non-uniform beams.

Appendices

Appendix A

Components of the $a$ matrix,$a_{ij}$ $\left( {i = 1 - 4, j = 1 - 4} \right)$ in Eq. (19)

$$\begin{gathered} a_{11} = 0,\;a_{12} = 1,\;a_{13} = \hat{G}_{1L} ,\;a_{14} = \gamma_{0} \hat{G}_{1L} + \hat{G}_{2L} , \hfill \\ a_{21} = 1,\;a_{22} = - \gamma_{0} ,\;a_{23} = \hat{G}_{3L} ,\;a_{24} = \gamma_{0} \hat{G}_{3L} + \hat{G}_{4L} \hfill \\ a_{31} = 6 + \hat{G}_{2R} + \left( {\gamma + 3} \right)\hat{G}_{1R} , \hfill \\ a_{32} = 3\left[ {2 + \hat{G}_{2R} + \hat{G}_{1R} \left( {\gamma + 2} \right)} \right], \hfill \\ a_{33} = 6\left[ {\hat{G}_{1R} \left( {\gamma + 1} \right) + \hat{G}_{2R} } \right],\;a_{34} = 6\left( {\gamma \hat{G}_{1R} + \hat{G}_{2R} } \right), \hfill \\ a_{41} = \left( {\gamma + 3} \right)\hat{G}_{3R} + \hat{G}_{4R} - 6\left( {\gamma - 1} \right), \hfill \\ a_{42} = 3\left[ {\left( {\gamma + 2} \right)\hat{G}_{3R} + \hat{G}_{4R} - 2\gamma } \right], \hfill \\ a_{43} = 6\left[ {\left( {\gamma + 1} \right)\hat{G}_{3R} + \hat{G}_{4R} } \right],\; a_{44} = 6\left( {\gamma \hat{G}_{3R} + \hat{G}_{4R} } \right) \hfill \\ \end{gathered}$$

(A.1)

The column vector $b,$ $b_{i} \left( {i = 1 - 4} \right)$ in Eq. (19)

$$\begin{gathered} b_{1} = 0, \hfill \\ b_{2} = 0, \hfill \\ b_{3} = \left( {\gamma \hat{G}_{1R} + \hat{G}_{2R} } \right)\left( {6B_{2} - 12B_{1} - \beta^{4} B_{3} } \right) - 3\left[ {\beta^{4} \left( {\hat{G}_{1R} B_{5} + 2B_{6} } \right) - 2\hat{G}_{1R} B_{4} } \right] \hfill \\ b_{4} = \left( {\gamma \hat{G}_{3R} + \hat{G}_{4R} } \right)\left( {6B_{2} - 12B_{1} - \beta^{4} B_{3} } \right) - 3\left[ {\beta^{4} \left( {\hat{G}_{3R} B_{5} - 2\gamma B_{6} + 2B_{7} } \right) - 2\hat{G}_{3R} B_{4} } \right] \hfill \\ \end{gathered}$$

(A.2)

Solution to the unknowns $C_{i} \left( {i = 1 - 4} \right)$ utilizing Cramer’s rule

$$C_{1} = \frac{{{\text{det}}}}{\zeta }\left[ {\begin{array}{*{20}c} {\begin{array}{*{20}c} 0 & {a_{12} } \\ 0 & {a_{22} } \\ \end{array} } & {\begin{array}{*{20}c} {a_{13} } & {a_{14} } \\ {a_{23} } & {a_{24} } \\ \end{array} } \\ {\begin{array}{*{20}c} {b_{3} } & {a_{32} } \\ {b_{4} } & {a_{42} } \\ \end{array} } & {\begin{array}{*{20}c} {a_{33} } & {a_{34} } \\ {a_{43} } & {a_{44} } \\ \end{array} } \\ \end{array} } \right] = \frac{{b_{3} A_{1} - b_{4} A_{2} }}{\zeta }$$

(A.3)

$$C_{2} = \frac{{{\text{det}}}}{\zeta }\left[ {\begin{array}{*{20}c} {\begin{array}{*{20}c} {a_{11} } & 0 \\ {a_{21} } & 0 \\ \end{array} } & {\begin{array}{*{20}c} {a_{13} } & {a_{14} } \\ {a_{23} } & {a_{24} } \\ \end{array} } \\ {\begin{array}{*{20}c} {a_{31} } & {b_{3} } \\ {a_{41} } & {b_{4} } \\ \end{array} } & {\begin{array}{*{20}c} {a_{33} } & {a_{34} } \\ {a_{43} } & {a_{44} } \\ \end{array} } \\ \end{array} } \right] = - \frac{{b_{3} A_{3} - b_{4} A_{4} }}{\zeta }$$

(A.4)

$$C_{3} = \frac{{{\text{det}}}}{\zeta }\left[ {\begin{array}{*{20}c} {\begin{array}{*{20}c} {a_{11} } & {a_{12} } \\ {a_{21} } & {a_{22} } \\ \end{array} } & {\begin{array}{*{20}c} 0 & {a_{14} } \\ 0 & {a_{24} } \\ \end{array} } \\ {\begin{array}{*{20}c} {a_{31} } & {a_{32} } \\ {a_{41} } & {a_{42} } \\ \end{array} } & {\begin{array}{*{20}c} {b_{3} } & {a_{34} } \\ {b_{4} } & {a_{44} } \\ \end{array} } \\ \end{array} } \right] = \frac{{b_{3} A_{5} - b_{4} A_{6} }}{\zeta }$$

(A.5)

$$C_{4} = \frac{{{\text{det}}}}{\zeta }\left[ {\begin{array}{*{20}c} {\begin{array}{*{20}c} {a_{11} } & {a_{12} } \\ {a_{21} } & {a_{22} } \\ \end{array} } & {\begin{array}{*{20}c} {a_{13} } & 0 \\ {a_{23} } & 0 \\ \end{array} } \\ {\begin{array}{*{20}c} {a_{31} } & {a_{32} } \\ {a_{41} } & {a_{42} } \\ \end{array} } & {\begin{array}{*{20}c} {a_{33} } & {b_{3} } \\ {a_{43} } & {b_{4} } \\ \end{array} } \\ \end{array} } \right] = - \frac{{b_{3} A_{7} - b_{4} A_{8} }}{\zeta }$$

(A.6)

where coefficients $A_{1}$ to $A_{8}$ and $\zeta$ are given by Eqs. (A.7) to (A.11), respectively.

$$A_{1} = {\text{det}}\left[ {\begin{array}{*{20}c} {a_{12} } & {a_{13} } & {a_{14} } \\ {a_{22} } & {a_{23} } & {a_{24} } \\ {a_{42} } & {a_{43} } & {a_{44} } \\ \end{array} } \right],\quad A_{2} = {\text{det}}\left[ {\begin{array}{*{20}c} {a_{12} } & {a_{13} } & {a_{14} } \\ {a_{22} } & {a_{23} } & {a_{24} } \\ {a_{32} } & {a_{33} } & {a_{34} } \\ \end{array} } \right]$$

(A.7)

$$A_{3} = {\text{det}}\left[ {\begin{array}{*{20}c} {a_{11} } & {a_{13} } & {a_{14} } \\ {a_{21} } & {a_{23} } & {a_{24} } \\ {a_{41} } & {a_{43} } & {a_{44} } \\ \end{array} } \right], A_{4} = {\text{det}}\left[ {\begin{array}{*{20}c} {a_{11} } & {a_{13} } & {a_{14} } \\ {a_{21} } & {a_{23} } & {a_{24} } \\ {a_{31} } & {a_{33} } & {a_{34} } \\ \end{array} } \right]$$

(A.8)

$$A_{5} = {\text{det}}\left[ {\begin{array}{*{20}c} {a_{11} } & {a_{12} } & {a_{14} } \\ {a_{21} } & {a_{22} } & {a_{24} } \\ {a_{41} } & {a_{42} } & {a_{44} } \\ \end{array} } \right], A_{6} = {\text{det}}\left[ {\begin{array}{*{20}c} {a_{11} } & {a_{12} } & {a_{14} } \\ {a_{21} } & {a_{22} } & {a_{24} } \\ {a_{31} } & {a_{32} } & {a_{34} } \\ \end{array} } \right]$$

(A.9)

$$A_{7} = {\text{det}}\left[ {\begin{array}{*{20}c} {a_{11} } & {a_{12} } & {a_{13} } \\ {a_{21} } & {a_{22} } & {a_{23} } \\ {a_{41} } & {a_{42} } & {a_{43} } \\ \end{array} } \right], A_{8} = {\text{det}}\left[ {\begin{array}{*{20}c} {a_{11} } & {a_{12} } & {a_{13} } \\ {a_{21} } & {a_{22} } & {a_{23} } \\ {a_{31} } & {a_{32} } & {a_{33} } \\ \end{array} } \right]$$

(A.10)

and

$$\zeta = {\text{det}}\left[ {\begin{array}{*{20}c} {\begin{array}{*{20}c} {a_{11} } & {a_{12} } \\ {a_{21} } & {a_{22} } \\ \end{array} } & {\begin{array}{*{20}c} {a_{13} } & {a_{14} } \\ {a_{23} } & {a_{24} } \\ \end{array} } \\ {\begin{array}{*{20}c} {a_{31} } & {a_{32} } \\ {a_{41} } & {a_{42} } \\ \end{array} } & {\begin{array}{*{20}c} {a_{33} } & {a_{34} } \\ {a_{43} } & {a_{44} } \\ \end{array} } \\ \end{array} } \right]$$

(A.11)

$$h_{1} \left( X \right) = \frac{2}{\zeta }\left\{ {\left[ {\left( {\gamma \hat{G}_{3R} + \hat{G}_{4R} } \right)A_{2} - \left( {\gamma \hat{G}_{1R} + \hat{G}_{2R} } \right)A_{1} } \right]X^{3} + 3\left[ {\left( {\gamma \hat{G}_{1R} + \hat{G}_{2R} } \right)A_{3} - \left( {\gamma \hat{G}_{3R} + \hat{G}_{4R} } \right)A_{4} } \right]X^{2} + 6\left[ {\left( {\gamma \hat{G}_{3R} + \hat{G}_{4R} } \right)A_{6} - \left( {\gamma \hat{G}_{1R} + \hat{G}_{2R} } \right)A_{5} } \right]X + 6\left[ {\left( {\gamma \hat{G}_{1R} + \hat{G}_{2R} } \right)A_{7} - \left( {\gamma \hat{G}_{3R} + \hat{G}_{4R} } \right)A_{8} } \right]} \right\}.$$

(A.12)

$$h_{2} \left( X \right) = \frac{1}{\zeta }\left[ {\left( {\hat{G}_{1R} A_{1} - \hat{G}_{3R} A_{2} } \right)X^{3} + 3\left( {\hat{G}_{3R} A_{4} - \hat{G}_{1R} A_{3} } \right)X^{2} + 6\left( {\hat{G}_{1R} A_{5} - \hat{G}_{3R} A_{6} } \right) X + 6\left( {\hat{G}_{3R} A_{8} - \hat{G}_{1R} A_{7} } \right)} \right].$$

(A.13)

$$h_{3} \left( X \right) = \frac{{\beta^{4} }}{\zeta }\left[ { - \left( {A_{1} + A_{2} \gamma } \right)X^{3} + 3\left( {A_{3} + A_{4} \gamma } \right)X^{2} - 6\left( {A_{5} + A_{6} \gamma } \right)X + 6\left( {A_{7} + \gamma A_{8} } \right)} \right].$$

(A.14)

$$h_{4} \left( X \right) = \frac{{\beta^{4} }}{\zeta }\left( {A_{2} X^{3} - 3A_{4} X^{2} + 6A_{6} X - 6A_{8} } \right)$$

(A.15)

Appendix B

See Table 23

Table 23 Simulation of classical boundary conditions C: Clamped, S: Simply, G: Guided and F: Free