Applications to Nonlinear Solid Mechanics

Abstract

Solid mechanics is essential and important to a variety of structures, which are widely used in almost all engineering areas, e.g., bridge structures, architecture structures, ship structures, mechanical structures, aerospace structures, rail structures, etc.

Appendices

Appendix 9.1 Wavelet Numerical Integration Method with Modified Scaling Function

Consider the integral

$$g(x) = \int\limits_{0}^{x} {f(y)dy} ,\quad x \in [0,1].$$

(9.121)

When the approximation scheme, Eq. (5.6), is substituted into the preceding formula, one gets

$$g(x) \approx \tilde{g}(x) = \sum\limits_{k = 0}^{{2^{j} }} {f_{n,k} \tilde{\theta }_{n,k} (x)} {,}\quad x \in [0, \, 1],$$

(9.122)

in which

$$\tilde{\theta }_{n,k} (x) = 2^{ - j} \left\{ {\begin{array}{*{20}l} {\sum\limits_{{l = 1 - \tilde{N} + M_{1} }}^{ - 1} {u_{0,k} (\frac{l}{{2^{n} }})\theta |_{{M_{1} - l}}^{{2^{n} x - l + M_{1} }} + \theta |_{{M_{1} - k}}^{{2^{n} x - k + M_{1} }} } } \hfill & {k \in [0, \, r - 1]} \hfill \\ {\theta |_{{M_{1} - k}}^{{2^{n} x - k + M_{1} }} } \hfill & {k \in [r, \, 2^{n} - r]} \hfill \\ {\sum\limits_{{l = 2^{n} + 1}}^{{2^{n} - 1 + M_{1} }} {u_{{1,2^{n} - k}} (\frac{l}{{2^{n} }})\theta |_{{M_{1} - l}}^{{2^{n} x - l + M_{1} }} } + \theta |_{{M_{1} - k}}^{{2^{n} x - k + M_{1} }} } \hfill & {k \in [2^{n} - r + 1, \, 2^{n} ]} \hfill \\ \end{array} } \right.,$$

(9.123)

where $\theta |_{y}^{x} = \theta (x) - \theta (y)$, and the integral of the scaling function is

$$\theta (x) = \int\limits_{0}^{x} {\varphi (y)dy} .$$

(9.124)

In accordance with the normalization condition, the compact support of the scale function is $\left[ {0,\tilde{N}} \right]$. From Sect. 2.5.2, we obtain that the integral in Eq. (9.124) of the scale function has properties

$$\theta \left( x \right) = \left\{ {\begin{array}{*{20}l} {0,} \hfill & {x \le 0} \hfill \\ {1,} \hfill & {x \ge \tilde{N}} \hfill \\ \end{array} } \right..$$

(9.125)

Appendix 9.2 An Analytical Solution of Elastic Line Equation of Flexible Rod

For such case, when $\alpha = 0$ in Eq. (9.45), the governing equation for the bending problems of the cantilevered rod or beam can be written as

$$n\frac{{d^{2} \theta }}{{ds^{2} }} = - \lambda \cos \theta (\frac{d\theta }{{ds}})^{1 - n} ,$$

(9.126)

where the curvature is κ = dθ/ds and θ is the bending angle. Then, the governing equation can then be rewritten into

$$n\kappa^{n} d\kappa = - \lambda \cos \theta d\theta .$$

(9.127)

By integrating the preceding equation once and considering the boundary condition at the free end, $\kappa (1) = 0$, we get the following integration:

$$\frac{n}{n + 1}\kappa^{n + 1} = \lambda \sin \theta_{L} - \lambda \sin \theta ,$$

(9.128)

where θ_L is the bending angle at the free end.

Substitution of the expression of the curvature into Eq. (9.128) leads to

$$(\sin \theta_{L} - \sin \theta )^{ - 1/(n + 1)} d\theta = \left( {\frac{n + 1}{n}\lambda } \right)^{ - 1/(n + 1)} ds.$$

(9.129)

In accordance with the non-extensibility assumption in Euler–Bernoulli beam theory, integration of Eq. (9.129) from the fixed end to the free end yields

$$\lambda = \frac{n}{n + 1}\left[ {\int\limits_{0}^{{\theta_{L} }} {g(\theta )d\theta } } \right]^{n + 1} ,$$

(9.130)

where $g(\theta ) = (\sin \theta_{L} - \sin \theta )^{{ - \frac{1}{n + 1}}}$. This equation provides an analytical relation between the bending angle, θ_L, and the normalized load, λ.

Further, the curvature, κ, can then be expressed as a function of the bending angle, θ_L, as follows:

$$\kappa = \frac{{\int_{0}^{{\theta_{L} }} {g(\theta )d\theta } }}{Lg(\theta )}.$$

(9.131)

Due to the geometric relationship

$$dx = \cos \theta ds = \cos \theta \frac{d\theta }{\kappa },$$

(9.132)

substituting the curvature of Eq. (9.131) into Eq. (9.132) and integrating the induced equation from the fixed end, we obtain the following formula:

$$x = \frac{{\int_{0}^{\theta } {\cos \theta g(\theta )d\theta } }}{{\int_{0}^{{\theta_{L} }} {g(\theta )d\theta } }}.$$

(9.133)

Further, the axial displacement of the free end is formulated by

$$\delta_{h} = 1 - \frac{{\int_{0}^{{\theta_{L} }} {\cos \theta g(\theta )d\theta } }}{{\int_{0}^{{\theta_{L} }} {g(\theta )d\theta } }}.$$

(9.134)

With the similar steps, the deflection of every point of the rod can be obtained as

$$y = \frac{{\int_{0}^{\theta } {\sin \theta g(\theta )d\theta } }}{{\int_{0}^{{\theta_{L} }} {g(\theta )d\theta } }}.$$

(9.135)

Then, the deflection of the free end is denoted by $\delta_{v} = y(\theta_{L} )$.

Appendix 9.3 Approximate Theoretical Analyses on the Nonlinear Free Vibration of Beams

When the beam is undergoing large amplitude vibration, the following nonlinear strain–displacement relation is employed [8]:

$$\varepsilon_{x} = u_{,x} + 0.5w_{,x}^{2} - zw_{,xx} ,$$

(9.136)

where u, w, and z are the axial displacement, the transverse displacement or deflection, and the distance from the neutral plane along the thickness direction of the beam, respectively.

Based on Eq. (9.136), the strain energy and kinetic energy of the elastic beam are written by the form [8]

$$U = \frac{1}{2}\int_{0}^{l} {[EA(u_{,x}^{2} + u_{,x} w_{,x}^{2} + \frac{1}{4}w_{,x}^{4} ) + EIw_{,xx}^{2} ]dx} ,$$

(9.137)

$$T = \int_{0}^{l} {mw_{,t}^{2} dx} /2,$$

(9.138)

where $E,A,I$, and $m$ are the elastic modulus, area of cross section, inertia moment of cross section, and mass per unit length of the beam, respectively. Here, the axial motion is disregarded in calculating kinetic energy.

Assume

$$u = u_{0} (t)\sin (2\pi x/l),$$

(9.139)

$$w = w_{0} (t)\sin (\pi x/l),$$

(9.140)

which satisfy the kinematic boundary conditions of a simply supported beam with immovable ends. Then, substituting them into Eqs. (9.137) and (9.138), the Lagrange dynamic equation of the free vibration of beams can be written as [8]

$$m\ddot{w}_{0} + \alpha w_{0} + \beta w_{0}^{3} = 0,$$

(9.141)

where parameters $\alpha = \pi^{4} EI/l^{4}$ and $\beta = \pi^{4} EA/4l^{4}$.

The energy conservation equation [8] can be further obtained by multiplying both sides of Eq. (9.141) by $\dot{w}_{0}$ and integrating with respect to time such that

$$m\dot{w}_{0}^{2} + \alpha w_{0}^{2} + \beta w_{0}^{4} /2 = T + U = {\text{constant}}{.}$$

(9.142)

Assume that the nonlinear free vibration of the elastic beam satisfies the single harmonic response relationship $w_{0} = W_{0} \cos (\omega t)$, where $W_{0}$ and $\omega$ are dealt with by the amplitude and characteristic frequency of the nonlinear free vibration of a beam, respectively. Further, the dynamic equation (9.141) and the energy Eq. (9.127), respectively, yield [8].

$$\omega /\omega_{0} = \sqrt {1 + (W_{0} /\xi )^{2} /4} ,$$

(9.143)

$$\omega /\omega_{0} = \sqrt {1 + (W_{0} /\xi )^{2} /8} ,$$

(9.144)

where $\omega_{0} = \sqrt {\alpha /m}$ is the natural frequency of the linear free vibration of the same beam, and parameter $\xi = \sqrt {I/A}$ is the radius of gyration of the beam’s cross section.

It is obvious that these two relations of Eqs. (9.143) and 9.144) are conflicted, which means that on the basis of assumption of the harmonic oscillations, such induced results cannot satisfy the equation of motion and the energy balance equation simultaneously. Singh et al. [26] pointed out that Eqs. (9.143) and (9.144) are, respectively, the upper and lower bounds on the actual solution, as displayed in Fig. 9.35.

In addition, the governing equation of motion can also be solved by using the perturbation method without the assumption of harmonic response, in which the following relation is obtained [8]:

$$\omega /\omega_{0} = \sqrt {1 + \frac{3}{16}(W_{0} /\xi )^{2} } .$$

(9.145)

As pointed out previously, such perturbation solution is suitable only to the small initial amplitude.