Dynamic Analysis Method

Dynamic Analysis Method

In this section, the dynamic problem analysis method with a direct time integration method applied is described. As presented below, with this development code, it is possible to perform time history response analysis by an implicit or explicit method.

Formulation of the implicit method

Focusing on dynamic problems, the direct time integration method was applied to solve the following equation motion indicated:

$$\begin{equation} M( t + \Delta t ) \ddot{U} (t + \Delta t) + C( t + \Delta t ) \dot{U}(t + \Delta t) + Q( t + \Delta t ) = F( t + \Delta t ) \label{eq:2.5.1} \end{equation}$$

where $M$ and the mass matrix, $C$ is the damping matrix, and $Q$ is the internal stress vector, and $F$ is the external force vector. This software does not consider the changes in mass; thus, the mass matrix is non-linear and constant regardless of deformation.

The displacement within time increment $\Delta t$, and the change in speed and acceleration are approximated with the Newmark-$\beta$ method, as expressed in Eq.$\eqref{eq:2.5.2}$ and Eq.$\eqref{eq:2.5.3}$:

$$\begin{equation} \dot{U}(t + \Delta t) = \frac{\gamma}{\beta \Delta t} \Delta U( t + \Delta t ) - \frac{\gamma - \beta}{\beta} \dot{U}( t ) - \Delta t \frac{\gamma - 2\beta}{2\beta} \ddot{U}(t) \label{eq:2.5.2} \end{equation}$$

$$\begin{equation} \ddot{U}(t + \Delta t) = \frac{1}{\beta \Delta t^2}\Delta U(t + \Delta t) - \frac{1}{\beta \Delta t} \dot{U}(t) - \frac{1 - 2\beta}{2\beta} \ddot {U}(t) \label{eq:2.5.3} \end{equation}$$

where $\gamma$ and $\beta$ are the parameters of the Newmark-$\beta$ method.

if $\gamma$ and $\beta$ have the following values, it coincides with the linear acceleration method or the trapezoidal rule.

As it is already known, when $\gamma$ and $\beta$ are substituted into the following values, it will match the linear acceleration method, or trapezoid rule.

• $\gamma=\displaystyle \frac{1}{2}$, $\beta=\displaystyle \frac{1}{6}$ (Linear acceleration method)
• $\gamma=\displaystyle \frac{1}{2}$, $\displaystyle \beta=\frac{1}{4}$ (Trapezoid rule)

If Eq.$\eqref{eq:2.5.2}$ and Eq.$\eqref{eq:2.5.3}$ are substituted into Eq.$\eqref{eq:2.5.1}$, the following equation is obtaind:

$$\begin{align} \nonumber \left( \frac{1}{\beta \Delta t^2} \mathbf{M} + \frac{\gamma}{\beta \Delta t} C + K \right) \Delta U ( t + \Delta t ) &= F ( t + \Delta t ) - Q ( t + \Delta t ) \\ \nonumber &+ \frac{1}{\beta \Delta t} M \dot{U} ( t ) + \frac{1 - 2\beta}{2\beta} M \ddot{U} ( t ) \\ &+ \frac{\gamma - \beta}{\beta} C \dot{U} (t) + \Delta t \frac{\gamma - 2\beta}{2 \beta} C \ddot{U}(t) \label{eq:2.5.4} \end{align}$$

$K_L$ is linear stiffness matrix for linear problem; thus, $Q ( t + \Delta t ) = K_L U (t + \Delta t)$. If this equation is substituted into the equation above, the following equation is obtained:

$$\begin{equation} \left\lbrace M \left( - \frac{1}{(\Delta t)^2 \beta} U(t) - \frac{1}{(\Delta t)\beta} \dot{U}(t) - \frac{1-2\beta}{2\beta} \ddot{U}(t) \right) \\\ \qquad\qquad\qquad\qquad + C \left( - \frac{\gamma}{(\Delta t)\beta} U(t) + \left(1-\frac{\gamma}{\beta}\right) \dot{U}(t) + \Delta t \frac{2\beta - \gamma}{2\beta} \ddot{U}(t) \right) \right \rbrace \\\ + \left \lbrace \frac{1}{(\Delta t)^2\beta} M + \frac{\gamma}{(\Delta t)\beta} C + K_L \right \rbrace U(t+\Delta t) = F(t+\Delta t) \label{eq:2.5.5} \end{equation}$$

In the portion, where the acceleration is specified as a geometric boundary condition, the displacement of the following equation is obtained from Eq.$\eqref{eq:2.5.2}$.

$$\begin{equation} u_{is} (t+\Delta{t}) = u_{is} (t) + \Delta t\, \dot{u}_{is}(t) + (\Delta t)^2 \left(\frac{1}{2} -\beta \right) {\ddot{u}}_{is}(t) + (\Delta t)^2 \beta \ddot{u}_{is}(t + \Delta t) \label{eq:2.5.6} \end{equation}$$

Similarly, if the speed is specified, the displacement of the following equation is obtained from Eq.$\eqref{eq:2.5.6}$:

$$\begin{equation} u_{is}(t+\Delta t) = u_{is}(t) + \Delta t \frac{\gamma-\beta}{\gamma}\dot{u}_{is} + (\Delta t)^2 \frac{\gamma-2\beta}{2\gamma} \ddot{u}_{is} + \Delta t \frac{\beta}{\gamma} \dot{u}_{is}(t+\Delta t) \label{eq:2.5.7} \end{equation}$$

Where

• $u_{is}(t+\Delta t)$ is the nodal displacement at $t + \Delta t$,
• $\dot{u}_{is}(t+\Delta t)$ is the nodal speed at time $t + \Delta t$
• $\ddot{u}_{is}(t+\Delta t)$ is the nodal acceleration at time $t + \Delta t$
• $i$ is the nodal degree-of-freedom (DOF)
• $s$ is the Node number.

Furthermore, the mass and damping terms were handled as follows:

(1) Handling of mass term

The mass matrices are handled as concentrated mass matrices.

(2) Handling of damping terms

The damping terms are handled as Rayleigh damping expressed in Eq.$\eqref{eq:2.5.8}$.

$$\begin{equation} C = R_m M + R_k K_L \label{eq:2.5.8} \end{equation}$$

where $R_m$ and $R_k$ are parameters of Rayleigh damping.

Formulation of Explicit Method

The explicit method is based on the motion equation at time $t$ expressed in the following equation:

$$\begin{equation} M \ddot{U}(t) + C (t) \dot{U}(t) + Q(t) = F(t) \label{eq:2.5.9} \end{equation}$$

where the displacement at time $t + \Delta t$ and that at time $t - \Delta t$ are expressed by the Taylor expansion at time $t$. If it is expanded up to the secondary term with $\Delta t$, it becomes

$$\begin{equation} U(t+\Delta{t}) = U(t)+\dot{U}(t)(\Delta{t}) +\frac{1}{2!}\ddot{U}(t)(\Delta{t})^2 \label{eq:2.5.10} \end{equation}$$

$$\begin{equation} U(t-\Delta{t})=U(t)-\dot{U}(t)(\Delta{t}) +\frac{1}{2!}\ddot{U}(t)(\Delta{t})^2 \label{eq:2.5.11} \end{equation}$$

From the difference and sum of Eq.$\eqref{eq:2.5.3}$ and Eq.$\eqref{eq:2.5.4}$, the following equation is obtained:

$$\begin{equation} \dot{U}(t)=\frac{1}{2\Delta{t}} (U(t+\Delta{t})-U(t-\Delta{t})) \label{eq:2.5.12} \end{equation}$$

$$\begin{equation} \ddot{U}= \frac{1}{(2\Delta{t})^2} (U(t+\Delta{t})-2U(t)+U(t-\Delta{t})) \label{eq:2.5.13} \end{equation}$$

If Eq.$\eqref{eq:2.5.12}$ and Eq.$\eqref{eq:2.5.13}$ are substituted into Eq.$\eqref{eq:2.5.9}$, the following equation is obtained:

$$\begin{align} \left( \frac{1}{\Delta t^{2}} M + \frac{1}{2\Delta t} C \right) U ( t + \Delta t ) &= F(t) - Q(t) \nonumber \\\ & - \frac{1}{\Delta t^{2}} M [2 U(t) - U( t - \Delta t)] - \frac{1}{2\Delta t} C U(t - \Delta t) \label{eq:2.5.14} \end{align}$$

For linear problems, specifically, $Q(t)=K_L U(t)$, the above equation becomes

$$\begin{align} \left( \frac{1}{\Delta t^{2}} M + \frac{1}{2\Delta t} C \right) U( t + \Delta t ) & = F(t) - K_L U(t) \nonumber \\\ & - \frac{1}{\Delta t^{2}} M[2U(t) - U(t - \Delta t)] - \frac{1}{2\Delta t} C U(t - \Delta t) \label{eq:2.5.15} \end{align}$$

In this case, if mass matrix $M$ is set as a concentrated mass matrix, and the damping matrix as the proportional damping matrix $C=R_m M$, Eq.$\eqref{eq:2.5.15}$ eliminates the requirement of solving operations for simultaneous equations.

Therefore, from Eq.$\eqref{eq:2.5.15}$, $U(t+\Delta t)$ can be determined by the following equation:

$$\begin{equation} U(t + \Delta t) = \frac{1}{\left(\cfrac{1}{\Delta t^2}M + \cfrac{1}{2\Delta t}C\right)} \\ \left \{ F(t) - Q(t) - \frac{1}{\Delta t^2} M [2U(t) - U(t-\Delta t)] - \frac{1}{2\Delta t} C(t-\Delta t) U \right \} \label{eq:2.5.17} \end{equation}$$