Structural analysis of nonlocal nanobeam via FEM using equivalent nonlocal differential model

Abstract

In the present work, we developed a new FEM framework to simulate the mechanical responses of the Euler–Bernoulli beam with a two-phase local/nonlocal mixed model. The shape function is a fifth-order polynomial and constitutive boundary conditions (CBCs) are treated as external loads. The main advantages of the present model are the efficient of convergence, simplicity of expressions, and the flexibility on handling various boundary conditions as well as the external loads. Several numerical tests, including static bending, free vibration, and elastic bulking, are carried out to validate the FEM framework. The results showed the complete agreement with the exact solution from Laplace transformation and indicated a simple and reliable scheme to deal with complicated nanosystems.

Appendices

A List of the terms in the weak form of the present model

Some terms in the the weak form of the present model from the theorem of divergence (integration by parts for one-dimension case) are listed here:

$$\begin{aligned}&\begin{aligned} \int _0^L w^{(6)}\delta v \mathrm {d}x&= \int _0^L \{(w^{(5)}\delta v)^\prime -w^{(5)}\delta v^\prime \} \mathrm {d}x \\&=-\int _0^L w^{(3)}\delta v^{(3)}\mathrm {d}x+w^{(3)}\delta v^{\prime \prime }\vert _0^L-w^{(4)}\delta v^\prime \vert _0^L+w^{(5)}\delta v\vert _0^L \end{aligned} \end{aligned}$$

(41)

$$\begin{aligned}&\begin{aligned} \int _0^L w^{(4)}\delta v\mathrm {d}x&=\int _0^L\{(w^{(3)}\delta v)^\prime -w^{(3)}\delta v^\prime \}\mathrm {d}x \\&=\int _0^Lw^{\prime \prime }\delta v^{\prime \prime }\mathrm {d}x\\&\quad -w^{\prime \prime }\delta v^\prime \vert _0^L + w^{(3)}\delta v\vert _0^L \end{aligned} \end{aligned}$$

(42)

$$\begin{aligned}&\begin{aligned} \int _0^L w^{\prime \prime }\delta v\mathrm {d}x&=\int _0^L \{(w^\prime \delta v)-w^\prime \delta v^\prime \}\mathrm {d}x \\&=\int _0^L w \delta v^{\prime ^\prime } \mathrm {d}x - w \delta v^\prime \vert _0^L + w^\prime \delta v \vert _0^L \end{aligned} \end{aligned}$$

(43)

$$\begin{aligned}&\begin{aligned} \int _0^L q^{\prime \prime } \delta v \mathrm {d}x =&\int _0^L\{ \left( q^\prime \delta v\right) ^\prime -q^\prime \delta v^\prime \}\mathrm {d}x \\&= \int _0^L q \delta v^{\prime \prime }\mathrm {d}x-q\delta v^\prime \vert _0^L + q^\prime \delta v\vert _0^L \end{aligned} \end{aligned}$$

(44)

$$\begin{aligned}&\begin{aligned} \int _0^L q^\prime \delta v^\prime \mathrm {d}x&= \int _0^L \{(q \delta ^\prime )^\prime -q\delta v^{\prime \prime }\}\mathrm {d}x \\&= -\int _0^L q\delta v^{\prime \prime } \mathrm {d}V+q\delta v^\prime \vert _0^L . \end{aligned} \end{aligned}$$

(45)

B Shape and geometric matrices of the present beam element

The explicit expression of the shape function and the ith geometric matrix of the fifth-order polynomial are listed (over the internal $[0,l_\mathrm {e}]$):

$$\begin{aligned} w = {\mathbf {N}} \cdot {\mathbf {w}} = \left[ N_1\ N_2\ N_3\ N_4\ N_5\ N_6\right] \begin{bmatrix} w_1 \\ w^\prime _1 \\ w^{\prime \prime }_1 \\ w_2 \\ w^\prime _2 \\ w^{\prime \prime }_2 \end{bmatrix} \end{aligned}$$

(46)

$$\begin{aligned} N_1= & {} \frac{\left( l_\mathrm {e}-x\right) ^3 \left( l_\mathrm {e}^2+3l_\mathrm {e} x+6 x^2\right) }{l_\mathrm {e}^5} \end{aligned}$$

(47a)

$$\begin{aligned} N_2= & {} \frac{x \left( l_\mathrm {e}-x\right) ^3 \left( l_\mathrm {e}+3 x\right) }{l_\mathrm {e}^4} \end{aligned}$$

(47b)

$$\begin{aligned} N_3= & {} \frac{x^2 \left( l_\mathrm {e}-x\right) ^3}{2l_\mathrm {e}^3} \end{aligned}$$

(47c)

$$\begin{aligned} N_4= & {} \frac{x^3 \left( 10l_\mathrm {e}^2-15l_\mathrm {e}x+6 x^2\right) }{l_\mathrm {e}^5} \end{aligned}$$

(47d)

$$\begin{aligned} N_5= & {} -\frac{x^3 \left( 4 l_\mathrm {e}-3 x\right) \left( l_\mathrm {e}-x\right) }{l_\mathrm {e}^4} \end{aligned}$$

(47e)

$$\begin{aligned} N_6= & {} \frac{x^3 \left( l_\mathrm {e}-x\right) ^2}{2 l_\mathrm {e}^3} \end{aligned}$$

(47f)

$$\begin{aligned} w^{\prime }=\; & {} {\mathbf {B}}^1 \cdot {\mathbf {w}} = [B^{(1)}_1\ B^{(1)}_2\ B^{(1)}_3\ B^{(1)}_4\ B^{(1)}_5\ B^{(1)}_6]\begin{bmatrix} w_1 \\ w^\prime _1 \\ w^{\prime \prime }_1 \\ w_2 \\ w^\prime _2 \\ w^{\prime \prime }_2. \end{bmatrix} \end{aligned}$$

(48)

$$\begin{aligned} B^{(1)}_1&= -\frac{30 x^2 \left( l_\mathrm {e}-x\right) ^2}{l_\mathrm {e}^5} \end{aligned}$$

(49a)

$$\begin{aligned} B^{(1)}_2&= \frac{\left( l_\mathrm {e}-3 x\right) \left( l_\mathrm {e}-x\right) ^2 \left( l_\mathrm {e}+5 x\right) }{l_\mathrm {e}^4} \end{aligned}$$

(49b)

$$\begin{aligned} B^{(1)}_3&= \frac{x \left( 2 l_\mathrm {e}-5 x\right) \left( l_\mathrm {e}-x\right) ^2}{2 l_\mathrm {e}^3} \end{aligned}$$

(49c)

$$\begin{aligned} B^{(1)}_4&= \frac{30 x^2 \left( l_\mathrm {e}-x\right) ^2}{l_\mathrm {e}^5} \end{aligned}$$

(49d)

$$\begin{aligned} B^{(1)}_5&= -\frac{x^2 \left( 6 l_\mathrm {e}-5 x\right) \left( 2 l_\mathrm {e}-3 x\right) }{l_\mathrm {e}^4} \end{aligned}$$

(49e)

$$\begin{aligned} B^{(1)}_6&= \frac{x^2 \left( 3 l_\mathrm {e}-5 x\right) \left( l_\mathrm {e}-x\right) }{2 l_\mathrm {e}^3}. \end{aligned}$$

(49f)

$$\begin{aligned} w^{\prime \prime } = {\mathbf {B}}^2 \cdot {\mathbf {w}} = [B^{(2)}_1\ B^{(2)}_2\ B^{(2)}_3\ B^{(2)}_4\ B^{(2)}_5\ B^{(2)}_6]\begin{bmatrix} w_1 \\ w^\prime _1 \\ w^{\prime \prime }_1 \\ w_2 \\ w^\prime _2 \\ w^{\prime \prime }_2. \end{bmatrix} \end{aligned}$$

(50)

$$\begin{aligned} B^{(2)}_1&= -\frac{60 x \left( l_\mathrm {e}-2 x\right) \left( l_\mathrm {e}-x\right) }{l_\mathrm {e}^5} \end{aligned}$$

(51a)

$$\begin{aligned} B^{(2)}_2&= -\frac{12 x \left( 3 l_\mathrm {e}-5 x\right) \left( l_\mathrm {e}-x\right) }{l_\mathrm {e}^4} \end{aligned}$$

(51b)

$$\begin{aligned} B^{(2)}_3&= \frac{\left( l_\mathrm {e}-x\right) \left( l_\mathrm {e}^2-8l_\mathrm {e} x+10 x^2\right) }{l_\mathrm {e}^3} \end{aligned}$$

(51c)

$$\begin{aligned} B^{(2)}_4&= \frac{60 x \left( l_\mathrm {e}-2 x\right) \left( l_\mathrm {e}-x\right) }{l_\mathrm {e}^5} \end{aligned}$$

(51d)

$$\begin{aligned} B^{(2)}_5&= -\frac{12 x \left( 2 l_\mathrm {e}-5 x\right) \left( l_\mathrm {e}-x\right) }{l_\mathrm {e}^4} \end{aligned}$$

(51e)

$$\begin{aligned} B^{(2)}_6&= \frac{x \left( 3 l_\mathrm {e}^2-12 l_\mathrm {e} x+10x^2\right) }{l_\mathrm {e}^3}. \end{aligned}$$

(51f)

$$\begin{aligned} w^{\prime \prime \prime } = {\mathbf {B}}^3 \cdot {\mathbf {w}} = [B^{(3)}_1\ B^{(3)}_2\ B^{(3)}_3\ B^{(3)}_4\ B^{(3)}_5\ B^{(3)}_6]\begin{bmatrix} w_1 \\ w^\prime _1 \\ w^{\prime \prime }_1 \\ w_2 \\ w^\prime _2 \\ w^{\prime \prime }_2 \end{bmatrix} \end{aligned}$$

(52)

$$\begin{aligned} B^{(3)}_1&= -\frac{60 \left( l_\mathrm {e}^2-6l_\mathrm {e} x+6 x^2\right) }{l_\mathrm {e}^5} \end{aligned}$$

(53a)

$$\begin{aligned} B^{(3)}_2&= -\frac{12 \left( 3 l_\mathrm {e}^2-16 l_\mathrm {e} x+15x^2\right) }{l_\mathrm {e}^4} \end{aligned}$$

(53b)

$$\begin{aligned} B^{(3)}_3&= \frac{-9 l_\mathrm {e}^2+36 l_\mathrm {e} x-30 x^2}{l_\mathrm {e}^3} \end{aligned}$$

(53c)

$$\begin{aligned} B^{(3)}_4&= \frac{60 \left( l_\mathrm {e}^2-6 l_\mathrm {e} x+6 x^2\right) }{l_\mathrm {e}^5} \end{aligned}$$

(53d)

$$\begin{aligned} B^{(3)}_5&= -\frac{12 \left( 2 l_\mathrm {e}^2-14 l_\mathrm {e} x+15 x^2\right) }{l_\mathrm {e}^4} \end{aligned}$$

(53e)

$$\begin{aligned} B^{(3)}_6&= \frac{3 \left( l_\mathrm {e}^2-8 l_\mathrm {e} x+10 x^2\right) }{l_\mathrm {e}^3}. \end{aligned}$$

(53f)

C Table of coefficients of CBCs at both sides of the beam

$$\begin{aligned} c^\mathrm {L}_1&= -\frac{60 EI \kappa \left( 12 \kappa ^2 (\xi -1)+l_\mathrm {e}^2 \xi +6 \kappa l_\mathrm {e} (\xi -1)\right) }{l_\mathrm {e}^5} \end{aligned}$$

(54a)

$$\begin{aligned} c^\mathrm {L}_2&= -\frac{12 EI \kappa \left( 30 \kappa ^2 (\xi -1)+3 l_\mathrm {e}^2 \xi +16 \kappa l_\mathrm {e} (\xi -1)\right) }{l_\mathrm {e}^4} \end{aligned}$$

(54b)

$$\begin{aligned} c^\mathrm {L}_3&= -\frac{EI \left( 60 \kappa ^3 (\xi -1)+9 \kappa l_\mathrm {e}^2 \xi +l_\mathrm {e}^3 \xi +36 \kappa ^2 l_\mathrm {e} (\xi -1)\right) }{l_\mathrm {e}^3} \end{aligned}$$

(54c)

$$\begin{aligned} c^\mathrm {L}_4&= -c^\mathrm {L}_1 = \frac{60 EI \kappa \left( 12 \kappa ^2 (\xi -1)+l_\mathrm {e}^2 \xi +6 \kappa l_\mathrm {e} (\xi -1)\right) }{l_\mathrm {e}^5} \end{aligned}$$

(54d)

$$\begin{aligned} c^\mathrm {L}_5&= -\frac{24 EI \kappa \left( 15 \kappa ^2 (\xi -1)+l_\mathrm {e}^2 \xi +7 \kappa l_\mathrm {e} (\xi -1)\right) }{l_\mathrm {e}^4} \end{aligned}$$

(54e)

$$\begin{aligned} c^\mathrm {L}_6&= \frac{3 EI \kappa \left( 20 \kappa ^2 (\xi -1)+l_\mathrm {e}^2 \xi +8 \kappa l_\mathrm {e} (\xi -1)\right) }{l_\mathrm {e}^3} \end{aligned}$$

(54f)

$$\begin{aligned} c^\mathrm {L}_0&= \kappa ^2 \left( -q + q^\prime \kappa \right) . \end{aligned}$$

(54g)

$$\begin{aligned} c^\mathrm {R}_1&= -\frac{60 EI\kappa \left( 12 \kappa ^2 (\xi -1)+l_\mathrm {e}^2 \xi +6 \kappa l_\mathrm {e} (\xi -1)\right) }{l_\mathrm {e}^5} \end{aligned}$$

(55a)

$$\begin{aligned} c^\mathrm {R}_2&= -\frac{24 EI\kappa \left( 15 \kappa ^2 (\xi -1)+l_\mathrm {e}^2 \xi +7 \kappa l_\mathrm {e} (\xi -1)\right) }{l_\mathrm {e}^4} \end{aligned}$$

(55b)

$$\begin{aligned} c^\mathrm {R}_3&= -\frac{3 EI\kappa \left( 20 \kappa ^2 (\xi -1)+l_\mathrm {e}^2 \xi +8 \kappa l_\mathrm {e} (\xi -1)\right) }{l_\mathrm {e}^3} \end{aligned}$$

(55c)

$$\begin{aligned} c^\mathrm {R}_4&= \frac{60 EI\kappa \left( 12 \kappa ^2 (\xi -1)+l_\mathrm {e}^2 \xi +6 \kappa l_\mathrm {e} (\xi -1)\right) }{l_\mathrm {e}^5} \end{aligned}$$

(55d)

$$\begin{aligned} c^\mathrm {R}_5&= -\frac{12 EI\kappa \left( 30 \kappa ^2 (\xi -1)+3 l_\mathrm {e}^2 \xi +16 \kappa l_\mathrm {e} (\xi -1)\right) }{l_\mathrm {e}^4} \end{aligned}$$

(55e)

$$\begin{aligned} c^\mathrm {R}_6&= \frac{EI\left( 60 \kappa ^3 (\xi -1)+9 \kappa l_\mathrm {e}^2 \xi +l_\mathrm {e}^3 \xi +36 \kappa ^2 l_\mathrm {e} (\xi -1)\right) }{l_\mathrm {e}^3} \end{aligned}$$

(55f)

$$\begin{aligned} c^\mathrm {R}_0&= \kappa ^2 \left( q + q^\prime \kappa \right) . \end{aligned}$$

(55g)

D Table of locations and weights with Gaussian quadrature

Some low-order quadrature rules are tabulated below (over interval $[-1,1]$) (Table 2).

Table 2 Weights and points of Gaussian quadrature

where 7-points quadrature is only necessary for calculating $\int _V{\mathbf {N}}^\mathrm {T}{\mathbf {N}}\mathrm {d}V$

E Lagrange multiplier method for linear constrain in FEM

The solution of the differential equation with constraints is equivalent to finding the stationary points of the stationary points of :

$$\begin{aligned} \begin{bmatrix} {\mathbf {K}}^\mathrm {int}-{\mathbf {K}}^\mathrm {ext} &{} {\mathbf {G}}^\mathrm {T} \\ {\mathbf {G}} &{} {\mathbf {0}} \end{bmatrix} \begin{bmatrix} {\mathbf {u}} \\ \varvec{\lambda } \end{bmatrix} = \begin{bmatrix} {\mathbf {f}}^\mathrm {ext}-{\mathbf {f}}^\mathrm {int} \\ {\mathbf {r}}_0. \end{bmatrix} \end{aligned}$$

(56)

The value of $\lambda _I$ is the react-force or moment at boundaries.

F Derivation of the exact solution to free vibration analysis of the present nonlocal Euler–Bernoulli beam with Laplace transform

Ignoring the axial force P and performing Laplace transformation on Eq. (5) , one gets

$$\begin{aligned} {\mathcal {L}}\left( M^{\prime \prime }\right) =-\omega ^2{\mathcal {L}}(W) \end{aligned}$$

(57)

Applying the Laplace transformation to both sides of Eq. (7) and taking into account the convolution theorem, one can derive

$$\begin{aligned} -\frac{{\mathcal {L}}(M)}{EI} = \frac{(1-\xi )\kappa ^2s^2-1}{\kappa ^2s^2-1}{\mathcal {L}} \left( W^{\prime \prime }\right) +\frac{\xi C_0}{2(\kappa s -1)}, \end{aligned}$$

(58)

where

$$\begin{aligned} C_0 = \int _VW^{\prime \prime }\mathrm {e}^{-\frac{\eta }{\kappa }}\mathrm {d}\eta . \end{aligned}$$

(59)

According to the derivative property on Laplace transformation, one can obtain

$$\begin{aligned} {\mathcal {L}}\left( M^{\prime \prime }\right)&= s^2{\mathcal {L}}(M)-sC_1-C_2 \end{aligned}$$

(60a)

$$\begin{aligned} {\mathcal {L}}\left( W^{\prime \prime }\right)&= s^2{\mathcal {L}}(W)-sC_3-C_4 \end{aligned}$$

(60b)

in which

$$\begin{aligned} {\left\{ \begin{array}{ll} C_1=M(0) \\ C_2=M^\prime (0) \\ C_3=W(0) \\ C_4=W^\prime (0) \\ . \end{array}\right. } \end{aligned}$$

(61)

Combining Eqs. (57), (58), and (60), and taking into account

$$\begin{aligned} s^6-\frac{1}{\kappa ^2(1-\xi )}s^4- \frac{m_0\omega ^2}{EI(1-\xi )}s^2+ \frac{m_0\omega ^2}{EI\kappa ^2(1-\xi )}= \left( s^2-a^2\right) \left( s^2-b^2\right) \left( s^2-c^2\right) , \end{aligned}$$

(62)

where

$$\begin{aligned} a&= \sqrt{ \root 3 \of {\frac{\sqrt{12p^3+81q^2}-9q}{18}}-\frac{a_0}{3}- \frac{2}{\root 3 \of {12\sqrt{12p^3+81q^2}-108q}}p } \end{aligned}$$

(63a)

$$\begin{aligned} b&= \sqrt{ \frac{1+\mathrm {j}\sqrt{3}}{\root 3 \of {12\sqrt{12p^3+81q^2}}-108q}p- \frac{1-\mathrm {j}}{2} \root 3 \of {\frac{\sqrt{12p^3+81q^2}-9q}{18}} -\frac{a_0}{3} } \end{aligned}$$

(63b)

$$\begin{aligned} b&= \sqrt{ \frac{1-\mathrm {j}\sqrt{3}}{\root 3 \of {12\sqrt{12p^3+81q^2}}-108q}p- \frac{1+\mathrm {j}}{2} \root 3 \of {\frac{\sqrt{12p^3+81q^2}-9q}{18}} -\frac{a_0}{3}. } \end{aligned}$$

(63c)

The parameters $a_0$, p and q are listed as follows:

$$\begin{aligned} a_0&= -\frac{1}{\kappa ^2(1-\xi )} \end{aligned}$$

(64a)

$$\begin{aligned} p&= - \frac{m_0\omega ^2}{EI(1-\xi )}-\frac{a_0^2}{3} \end{aligned}$$

(64b)

$$\begin{aligned} q&= \frac{2a_0^3}{27}+\frac{a_0^2m_0\omega ^2}{3EI(1-\xi )}+\frac{m_0\omega }{EI\kappa ^2(1-\xi )}. \end{aligned}$$

(64c)

One obtains Eq. (62)

$$\begin{aligned}&\begin{aligned} {\mathcal {L}}(M)=&\frac{2 \kappa ^2(1-\xi )\left[ C_{1} {~s}^{5}+C_{2} {~s}^{4}\right] -\left( 2 C_{1}+\kappa \omega ^{2}\left( 2 C_{3} m_{0} \kappa (1-\xi )+C_{0} m_{2} \xi \right) \right) {s}^{3}}{2 \kappa ^{2}(1-\xi )\left( {s}^2-a^2\right) \left( {s}^{2}-b^{2}\right) \left( {s}^{2}-c^{2}\right) }\\&+\frac{\left( 2 C_{2}+\omega ^2\left( 2 C_{4} m_{0} \kappa ^{2}(1-\xi )+C_{0} m_{2} \xi \right) \right) {s}^{2}+m_{0}\left( 2 C_{3}+C_{0} \kappa \xi \right) s+m_{0} \omega ^{2}\left( 2 C_{4}+C_{0} \xi \right) }{2 \kappa ^{2}(1-\xi )\left( {s}^{2}-a^{2}\right) \left( {s}^{2}-b^{2}\right) \left( {s}^{2}-c^{2}\right) } \end{aligned} \end{aligned}$$

(65a)

$$\begin{aligned}&\begin{aligned} {\mathcal {L}}(W)=&\frac{2 E I \kappa ^{2}(1-\xi )\left[ C_{3} {~s}^{5}+C_{4} {~s}^{4}\right] -\left( E I_{2}\left( 2 C_{3}+C_{0} \kappa \xi \right) +2 \kappa ^{2}\left( C_{1}-C_{3} m_{2} \omega ^{2}\right) \right) {s}^{3}}{2 E I \kappa ^{2}(1-\xi )\left( {s}^{2}-a^{2}\right) \left( {s}^{2}-b^{2}\right) \left( {s}^{2}-c^{2}\right) }\\&+\frac{\left( E I_{2}\left( 2 C_{4}+C_{0} \xi \right) +2 \kappa ^{2}\left( C_{2}-C_{4} m_{2} \omega ^{2}\right) \right) {s}^{2}+2\left( C_{1}-C_{3} m_{2} \omega ^{2}\right) s+2\left( C_{2}-C_{4} m_{0} \omega ^{2}\right) }{2 E I_{2} \kappa ^{2}(1-\xi )\left( {s}^{2}-a^{2}\right) \left( {s}^{2}-b^{2}\right) \left( {s}^{2}-c^{2}\right) }. \end{aligned} \end{aligned}$$

(65b)

For the nonlocal beam with clamped–clamped boundary conditions, we can obtain from Eq. (62) and Eq. (65)

$$\begin{aligned}&\begin{aligned} M =&C_1 L_5(x)+C_2 L_4(x) - \frac{2C_1}{2\kappa ^2(1-\xi )}L_3(x)-\frac{C_2}{\kappa ^2(1-\xi )}L_2(x)\\&+ \frac{m_0\omega ^2C_0\xi }{2\kappa (1-\xi )}L_1(x)+\frac{m_0\omega ^2C_0\xi }{2\kappa ^2(1-\xi )}L_0(x) \end{aligned} \end{aligned}$$

(66a)

$$\begin{aligned} W&= \frac{C_2L_0(x)+C_1L_1(x)}{EI\kappa ^2(1-\xi )} - \frac{EIC_0\xi +2\kappa ^2 C_2}{2EI\kappa ^2(1-\xi )}L_2(x) - \frac{EI C_0 \xi + 2\kappa C_1}{2EI\kappa ^2 (1-\xi )}L_3(x), \end{aligned}$$

(66b)

where $L_i(x),\ i=0,1,\ldots ,5$ are defined as

$$\begin{aligned} L_0(x)&= \frac{bc(b^2-c^2)\sinh (ax)+ac(c^2-a^2)\sinh (bx)+ab(a^2-b^2)\sinh (cx)}{abc(a^2-b^2)(b^2-c^2)(c^2-a^2)} \end{aligned}$$

(67a)

$$\begin{aligned} L_1(x)&= \frac{(b^2-c^2)\cosh (ax)+(c^2-a^2)\cosh (bx)+(a^2-b^2)\cosh (cx)}{(a^2-b^2)(b^2-c^2)(c^2-a^2)} \end{aligned}$$

(67b)

$$\begin{aligned} L_2(x)&= \frac{a^2(b^2-c^2)\sinh (ax)+b^2(c^2-a^2)\sinh (bx)+c^2(a^2-b^2)\sinh (cx)}{(a^2-b^2)(b^2-c^2)(c^2-a^2)} \end{aligned}$$

(67c)

$$\begin{aligned} L_3(x)&= \frac{a^2(b^2-c^2)\cosh (ax)+b^2(c^2-a^2)\cosh (bx)+c^2(a^2-b^2)\cosh (cx)}{(a^2-b^2)(b^2-c^2)(c^2-a^2)} \end{aligned}$$

(67d)

$$\begin{aligned} L_4(x)&= \frac{a^3(b^2-c^2)\sinh (ax)+b^3(c^2-a^2)\sinh (bx)+c^3(a^2-b^2)\sinh (cx)}{(a^2-b^2)(b^2-c^2)(c^2-a^2)} \end{aligned}$$

(67e)

$$\begin{aligned} L_5(x)&= \frac{a^4(b^2-c^2)\cosh (ax)+b^4(c^2-a^2)\cosh (bx)+c^4(a^2-b^2)\cosh (cx)}{(a^2-b^2)(b^2-c^2)(c^2-a^2)}. \end{aligned}$$

(67f)

Combining the Eqs. (59) and (66) as well as clamped boundary at the both sides of the beam, we can obtain

$$\begin{aligned}&A_{10}^\mathrm {CC}C_0+A_{11}^\mathrm {CC}C_1+A_{12}^\mathrm {CC}C_3=0 \end{aligned}$$

(68a)

$$\begin{aligned}&A_{20}^\mathrm {CC}C_0+A_{21}^\mathrm {CC}C_1+A_{22}^\mathrm {CC}C_3=0 \end{aligned}$$

(68b)

$$\begin{aligned}&A_{30}^\mathrm {CC}C_0+A_{31}^\mathrm {CC}C_1+A_{32}^\mathrm {CC}C_3=0, \end{aligned}$$

(68c)

where $A_{ij}^\mathrm {CC}$ are listed as follows:

$$\begin{aligned}&\begin{aligned} A_{10}^\mathrm {CC}=&\frac{\xi (a^3(c^2\kappa ^2-1)(b^2\kappa ^2-1)(b^2-c^2)(2a\kappa \cosh (aL)+(1+a^2\kappa ^2)\sinh (aL)}{2\kappa (a^2\kappa ^2-1)(b^2\kappa ^2-1)(c^2\kappa ^2-1)} \\&+\frac{b^3(c^2-a^2)(c^2\kappa ^2-1)(a^2\kappa ^2-1)(2b\kappa \cosh (bL)+b^2\kappa ^2+1)\sinh (bL)}{2\kappa (a^2\kappa ^2-1)(b^2\kappa ^2-1)(c^2\kappa ^2-1)} \\&+\frac{(a^2-b^2)c^3(a^2\kappa ^2-1)(b^2\kappa ^2-1)(2c\kappa \cosh (cL)+(1+c^2\kappa ^2)\sinh (cL))}{2\kappa (a^2\kappa ^2-1)(b^2\kappa ^2-1)(c^2\kappa ^2-1)} \\&+\frac{2(a^2-b^2)(a^2-c^2)(b^2-c^2)\mathrm {e}^{\frac{L}{\kappa }}\kappa (\kappa ^2(c^2+b^2(1-c^2\kappa ^2)+a^2(b^2\kappa ^2-1)(c^2\kappa ^2-1))(1-\xi )-1)}{2\kappa (a^2\kappa ^2-1)(b^2\kappa ^2-1)(c^2\kappa ^2-1)} \end{aligned} \end{aligned}$$

(69a)

$$\begin{aligned}&\begin{aligned} A_{11}^\mathrm {CC} =&\frac{(a^2(b^2-c^2)(\cosh (aL)+a\kappa \sinh (aL))+b^2(c^2-a^2)(\cosh (bL)+b\kappa \sinh (bL)))}{EI\kappa } \\&+ \frac{c^2(a^2-b^2)(\cosh (cL)+c\kappa \sinh (cL))}{EI\kappa } \end{aligned} \end{aligned}$$

(69b)

$$\begin{aligned}&\begin{aligned} A_{12}^\mathrm {CC} =&\frac{(a(b^2-c^2)(a\kappa \cosh (aL)+\sinh (aL))+b(c^2-a^2)(b\kappa \cosh (bL)+\sinh (bL)))}{EI\kappa } \\&+\frac{c(a^2-b^2)(c\kappa \cosh (cL)+\sinh (cL))}{EI\kappa } \end{aligned} \end{aligned}$$

(69c)

$$\begin{aligned}&\begin{aligned} A_{20}^\mathrm {CC} =\;&\xi a^2(b^2-c^2)(\cosh (aL)+a\kappa \sinh (aL))/2 + \xi b^2(c^2-a^2)(\cosh (bL)+b\kappa \sinh (bL))/2+ \\&+\xi c^2(a^2-b^2)(\cosh (cL)+c\kappa \sinh (cL))/2 \end{aligned} \end{aligned}$$

(69d)

$$\begin{aligned}&\begin{aligned} A_{21}^\mathrm {CC} =\;&a(b^2-c^2)(a^2\kappa ^2-1)\sinh (aL)/EI + b(c^2-a^2)(b^2\kappa ^2-1)\sinh (bL)/EI \\&+ c(a^2-b^2)(c^2\kappa ^2-1)\sinh (cL)/EI \end{aligned} \end{aligned}$$

(69e)

$$\begin{aligned}&\begin{aligned} A_{22}^\mathrm {CC} =\;&(b^2-c^2)(a^2\kappa ^2-1)\cosh (aL)/EI + (c^2-a^2)(b^2\kappa ^2-1)\cosh (bL)/EI \\&+ (a^2-b^2)(c^2\kappa ^2-1)\cosh (cL)/EI \end{aligned} \end{aligned}$$

(69f)

$$\begin{aligned}&\begin{aligned} A_{30}^\mathrm {CC} =\;&\xi a(b^2-c^2) (\sinh (aL)+a\kappa \cosh (aL))/2 +\xi b(c^2-a^2) (\sinh (bL)+b\kappa \cosh (bL))/2 \\&+ \xi c(a^2-b^2) (\sinh (cL)+c\kappa \cosh (cL))/2 \end{aligned} \end{aligned}$$

(69g)

$$\begin{aligned}&\begin{aligned} A_{31}^\mathrm {CC} =\;&\xi (b^2-c^2)(a^2\kappa ^2-1)\cosh (aL)/EI + \xi (c^2-a^2)(b^2\kappa ^2-1)\cosh (bL)/EI \\&+ \xi (a^2-b^2)(c^2\kappa ^2-1)\cosh (cL)/EI \end{aligned} \end{aligned}$$

(69h)

$$\begin{aligned}&\begin{aligned} A_{32}^\mathrm {CC} =\;&bc(b^2-c^2)(a^2\kappa ^2-1)\sinh (aL)/(abEI) + ac(c^2-a^2)(b^2\kappa ^2-1)\sinh (bL)/(abEI) \\&+ ab(a^2-b^2)(c^2\kappa ^2-1)\sinh (cL)/(abEI). \end{aligned} \end{aligned}$$

(69i)

None-zero solution of homogeneous linear Eq. (68) requires the coefficient matrix of $C_0$, $C_1$, and $C_2$ singular, e.g.,

$$\begin{aligned} \begin{vmatrix} A_{10}^\mathrm {CC}&A_{11}^\mathrm {CC}&A_{12}^\mathrm {CC} \\ A_{20}^\mathrm {CC}&A_{21}^\mathrm {CC}&A_{22}^\mathrm {CC} \\ A_{30}^\mathrm {CC}&A_{31}^\mathrm {CC}&A_{32}^\mathrm {CC} \end{vmatrix}=0 \end{aligned}$$

(70)

The eigenvalues $\omega$ can be thus obtained numerically.