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Linear Time History Analysis

Linear time history analysis is used to calculate the transient linear structure responses to dynamic loads or base acceleration using modal superposition.

The dynamic equation of motion of structure subjected to dynamic loads is

Mu¨+Cu˙+Ku=pf(t)\mathbf{M}\ddot{\mathbf{u}} + \mathbf{C}\dot{\mathbf{u}} + \mathbf{Ku} = \mathbf{p}f(t)

If the excitation is base acceleration

p=Mv\mathbf{p} = \mathbf{Mv}

where v\mathbf{v} is an influence vector that represents the displacement of the masses resulting from static application of a unit base displacement defined by the base excitation direction and the force due to the base acceleration is

f(t)=u¨g(t)f(t) = {\ddot{u}}_{g}(t)

To use the results (mode shapes) from modal dynamic analysis, the nodal displacements, velocities, and accelerations can be expressed in modal coordinates as

u=Φqu˙=Φq˙u¨=Φq¨\begin{aligned} \mathbf{u} = \boldsymbol{\Phi q} \\ \dot{\mathbf{u}} = \boldsymbol{\Phi}\dot{\mathbf{q}} \\ \ddot{\mathbf{u}} = \boldsymbol{\Phi}\ddot{\mathbf{q}}\end{aligned}

Then setting

m^i=φiTMφik^i=φiTKφic^i=φiTCφip^i=φiTp\begin{aligned}{\widehat{m}}_{i} &= {\boldsymbol{\varphi}_{i}}^{T}\mathbf{M}\boldsymbol{\varphi}_{i} \\ {\widehat{k}}_{i} &= {\boldsymbol{\varphi}_{i}}^{T}\mathbf{K}\boldsymbol{\varphi}_{i} \\ {\widehat{c}}_{i} &= {\boldsymbol{\varphi}_{i}}^{T}\mathbf{C}\boldsymbol{\varphi}_{i} \\ {\widehat{p}}_{i} &= {\boldsymbol{\varphi}_{i}}^{T}\mathbf{p}\end{aligned}

gives

m^q¨+c^q˙+k^q=p^if(t)\widehat{m}\ddot{\mathbf{q}} + \widehat{c}\dot{\mathbf{q}} + \widehat{k}\mathbf{q} = {\widehat{p}}_{i}f(t)

This gives a single degree of freedom problem that can be solved using any of the direct numerical analysis methods such as Newmark or central differences (Newmark is used in GSA). There are m such equations that are corresponding to each of the modes from the modal dynamic analysis. Superimposing the responses from each of the one degree of freedom problem the total responses of the structure can be calculated from

u=∑i=1mφiqiu˙=∑i=1mφiq˙iu¨=∑i=1mφiq¨i\mathbf{u} = \sum_{i = 1}^{m}\boldsymbol{\varphi}_{i}\mathbf{q}_{i}\qquad\dot{\mathbf{u}} = \sum_{i = 1}^{m}\boldsymbol{\varphi}_{i}{\dot{\mathbf{q}}}_{i}\qquad\ddot{\mathbf{u}} = \sum_{i = 1}^{m}\boldsymbol{\varphi}_{i}{\ddot{\mathbf{q}}}_{i}