Hysteretic model

Hysteretic models are mathematical models capable of simulating the complex nonlinear behavior characterizing mechanical systems and materials used in different fields of engineering, such as aerospace, civil, and mechanical engineering. Some examples of mechanical systems and materials having hysteretic behavior are:
* materials, such as steel, reinforced concrete, wood;
* structural elements, such as steel, reinforced concrete, or wood joints;
* devices, such as seismic isolators and dampers.

Hysteretic models may have a generalized displacement $u$ as input variable and a generalized force $f$ as output variable, or vice versa. In particular, in rate-independent hysteretic models, the output variable does not depend on the rate of variation of the input one.

Rate-independent hysteretic models can be classified into four different categories depending on the type of equation that needs to be solved to compute the output variable:
* Algebraic models;
* Transcendental models;
* Differential models;
* Integral models.

Algebraic models

In algebraic models, the output variable is computed by solving algebraic equations.

Bilinear model

Model formulation

In the bilinear model formulated by Vaiana et al. (2018), the generalized force at time ''t'', representing the output variable, is evaluated as a function of the generalized displacement as follows:
f_t=k_a\left(u_t-u_j\right)+k_b u_j+s_t f_0 \quad \text \quad u_t s_t

$f_t=k_b u_t+s_t f_0 \quad \text \quad u_t s_t>u_j s_t,$
where $k_a, k_b,$ and $f_0$ are the three model parameters to be calibrated from experimental or numerical tests, whereas $s_t$ is the sign of the generalized velocity at time $t$, that is, $s_t = \operatorname(\dot_t)$. Furthermore, $u_0$is an internal model parameter evaluated as:
$u_0=\frac,$
whereas $u_j$ is the internal variable:
$u_j=\frac$.

Hysteresis loop shapes

Figure 1.1 shows two different hysteresis loop shapes obtained by applying a sinusoidal generalized displacement having unit amplitude and frequency and simulated by adopting the Bilinear Model (BM) parameters listed in Table 1.1.

Matlab code

%

=
% June 2020
% Bilinear Model Algorithm
% Nicolò Vaiana, Research Fellow in Structural Mechanics and Dynamics, PhD
% Department of Structures for Engineering and Architecture
% University of Naples Federico II
% via Claudio, 21 - 80124, Napoli
%

=

clc; clear all; close all;

%% APPLIED DISPLACEMENT TIME HISTORY

dt = 0.001; %time step
t = 0:dt:1.50; %time interval
a0 = 1; %applied displacement amplitude
fr = 1; %applied displacement frequency
u = a0*sin((2*pi*fr)*t(1:length(t))); %applied displacement vector
v = 2*pi*fr*a0*cos((2*pi*fr)*t(1:length(t))); %applied velocity vector
n = length(u); %applied displacement vector length

%% 1. INITIAL SETTINGS
%1.1 Set the three model parameters
ka = 10.0; %model parameter
kb = 1.0; %model parameter
f0 = 0.5; %model parameter
%1.2 Compute the internal model parameters
u0 = f0/(ka-kb); %internal model parameter
%1.3 Initialize the generalized force vector
f = zeros(1,n);

%% 2. CALCULATIONS AT EACH TIME STEP

for i = 2:n
%2.1 Update the history variable
uj = (ka*u(i-1)+sign(v(i))*f0-f(i-1))/(ka-kb);
%2.2 Evaluate the generalized force at time t
if (sign(v(i))*uj)-2*u0 < sign(v(i))*u(i) && sign(v(i))*u(i) < sign(v(i))*uj
f(i) = ka*(u(i)-uj)+kb*uj+sign(v(i))*f0;
else
f(i) = kb*u(i)+sign(v(i))*f0;
end
end

%% PLOT
figure
plot(u,f,'k','linewidth',4)
set(gca,'FontSize',28)
set(gca,'FontName','Times New Roman')
xlabel('generalized displacement'), ylabel('generalized force')
grid
zoom off

Asymmetric bilinear model

Model formulation

In the asymmetric bilinear model formulated by Vaiana et al. (2020), the generalized force at time ''t'', representing the output variable, is evaluated as a function of the generalized displacement as follows:
f_t=k_a\left(u_t-u_j\right)+k_b u_j+s_t f_0 \quad \text \quad u_t s_t

$f_t=k_b u_t+s_t f_0 \quad \text \quad u_t s_t>u_j s_t,$
where $k_a=k_a^+(k_a^-), k_b=k_b^+(k_b^-),$ and $f_0 = f_0^+ (f_0^-)$ are the three model parameters of the generic loading (unloading) case to be calibrated from experimental or numerical tests, whereas $s_t$ is the sign of the generalized velocity at time $t$, that is, $s_t = \operatorname(\dot_t)$. Furthermore, $u_0$ is an internal model parameter evaluated as:
$u_0=\frac \quad\Biggl(\Biggr) \quad\text\quad s_t>0 \quad (s_t<0),$
whereas $u_j$ is the internal variable:
$u_j=\frac \quad\Biggl(\Biggr) \quad\text\quad s_t>0 \quad (s_t<0).$

Matlab code

%

=
% February 2021
% Asymmetric Bilinear Model Algorithm
% Nicolò Vaiana, Research Fellow in Structural Mechanics and Dynamics, PhD
% Department of Structures for Engineering and Architecture
% University of Naples Federico II
% via Claudio, 21 - 80124, Napoli
%

=

clc; clear all; close all;

%% APPLIED DISPLACEMENT TIME HISTORY

dt = 0.001; %time step
t = 0:dt:1.50; %time interval
a0 = 1; %applied displacement amplitude
fr = 1; %applied displacement frequency
u = a0*sin((2*pi*fr)*t(1:length(t))); %applied displacement vector
v = 2*pi*fr*a0*cos((2*pi*fr)*t(1:length(t))); %applied velocity vector
n = length(u); %applied displacement vector length

%% 1. INITIAL SETTINGS
%1.1 Set the six model parameters
kap = 5.0; %model parameter
kbp = 0.5; %model parameter
f0p = 0.75; %model parameter
kam = 15.0; %model parameter
kbm = 0.1; %model parameter
f0m = 0.25; %model parameter
%1.2 Initialize the generalized force vector
f = zeros(1,n);

%% 2. CALCULATIONS AT EACH TIME STEP

for i = 2:n
%2.1 Update the model parameters, the history variable, and the internal model parameter
if v(i)>0
ka = kap; kb = kbp; f0 = f0p;
else
ka = kam; kb = kbm; f0 = f0m;
end
uj = (ka*u(i-1)+sign(v(i))*f0-f(i-1))/(ka-kb);
if v(i)>0
u0 = ((kbp-kbm)*uj+f0p+f0m)/(2*(kap-kbm));
else
u0 = ((kbp-kbm)*uj+f0p+f0m)/(2*(kam-kbp));
end
%2.2 Evaluate the generalized force at time t
if (sign(v(i))*uj)-2*u0 < sign(v(i))*u(i) && sign(v(i))*u(i) < sign(v(i))*uj
f(i) = ka*(u(i)-uj)+kb*uj+sign(v(i))*f0;
else
f(i) = kb*u(i)+sign(v(i))*f0;
end
end

%% PLOT
figure
plot(u,f,'k','linewidth',4)
set(gca,'FontSize',28)
set(gca,'FontName','Times New Roman')
xlabel('generalized displacement'), ylabel('generalized force')
grid
zoom off

Animation

The following gif shows the nonlinear response of a single-degree-of-freedom (SDOF) mechanical system, with unit mass and asymmetric rate-independent hysteretic behavior, subjected to an external random force. To simulate its response, the following ABM parameters have been used: $k_a^+=50 \,\, (k_a^-=50), k_b^+=10 \,\, (k_b^-=1), f_0^+=1 \,\, (f_0^-=2)$.

Algebraic model by Vaiana et al. (2019)

Model formulation

In the algebraic model developed by Vaiana et al. (2019), the generalized force at time $t$, representing the output variable, is evaluated as a function of the generalized displacement as follows:
f_t=\beta_1 u_t^3 + \beta_2 u_t^5+k_b u_t+\left(k_a-k_b\right)\left frac-\frac\rights_t f_0 \quad \text \quad u_t s_t

$f_t=\beta_1 u_t^3+\beta_2 u_t^5 + k_b u_t+s_t f_0 \quad \text \quad u_t s_t>u_j s_t,$
where $k_a, k_b, \alpha$, $\beta_1$ and $\beta_2$ are the five model parameters to be calibrated from experimental or numerical tests, whereas $s_t$ is the sign of the generalized velocity at time $t$, that is, $s_t = \operatorname(\dot_t)$. Furthermore, $u_0$and $f_0$ are two internal model parameters evaluated as:
u_0=\frac\left left(\frac\right)^-1\right

f_0=\frac\left frac\right
whereas $u_j$ is the internal variable:
u_j=u_+s_t(1+2u_0)-s_t\left\lbrace \frac \left _-\beta_1 u_^3 - \beta_2 u_^5 - k_b u_-s_t f_0+(k_a-k_b) \frac\rightright\rbrace^.

Hysteresis loop shapes

Figure 1.2 shows four different hysteresis loop shapes obtained by applying a sinusoidal generalized displacement having unit amplitude and frequency and simulated by adopting the Algebraic Model (AM) parameters listed in Table 1.2.

Matlab code

%

=
% September 2019
% Algebraic Model Algorithm
% Nicolò Vaiana, Post-Doctoral Researcher, PhD
% Department of Structures for Engineering and Architecture
% University of Naples Federico II
% via Claudio, 21 - 80125, Napoli
%

=

clc; clear all; close all;

%% APPLIED DISPLACEMENT TIME HISTORY

dt = 0.001; %time step
t = 0:dt:1.50; %time interval
a0 = 1; %applied displacement amplitude
fr = 1; %applied displacement frequency
u = a0*sin((2*pi*fr)*t(1:length(t))); %applied displacement vector
v = 2*pi*fr*a0*cos((2*pi*fr)*t(1:length(t))); %applied velocity vector
n = length(u); %applied displacement vector length

%% 1. INITIAL SETTINGS
%1.1 Set the five model parameters
ka = 10.0; %model parameter
kb = 1.0; %model parameter
alfa = 10.0; %model parameter
beta1 = 0.0; %model parameter
beta2 = 0.0; %model parameter
%1.2 Compute the internal model parameters
u0 = (1/2)*((((ka-kb)/10^-20)^(1/alfa))-1); %internal model parameter
f0 = ((ka-kb)/2)*((((1+2*u0)^(1-alfa))-1)/(1-alfa)); %internal model parameter
%1.3 Initialize the generalized force vector
f = zeros(1, n);

%% 2. CALCULATIONS AT EACH TIME STEP

for i = 2:n
%2.1 Update the history variable
uj = u(i-1)+sign(v(i))*(1+2*u0)-sign(v(i))*((((sign(v(i))*(1-alfa))/(ka-kb))*(f(i-1)-beta1*u(i-1)^3-beta2*u(i-1)^5-kb*u(i-1)-sign(v(i))*f0+(ka-kb)*(((1+2*u0)^(1-alfa))/(sign(v(i))*(1-alfa)))))^(1/(1-alfa)));
%2.2 Evaluate the generalized force at time t
if (sign(v(i))*uj)-2*u0 < sign(v(i))*u(i) , , sign(v(i))*u(i) < sign(v(i))*uj
f(i) = beta1*u(i)^3+beta2*u(i)^5+kb*u(i)+(ka-kb)*((((1+2*u0+sign(v(i))*(u(i)-uj))^(1-alfa))/(sign(v(i))*(1-alfa)))-(((1+2*u0)^(1-alfa))/(sign(v(i))*(1-alfa))))+sign(v(i))*f0;
else
f(i) = beta1*u(i)^3+beta2*u(i)^5+kb*u(i)+sign(v(i))*f0;
end
end

%% PLOT
figure
plot(u, f, 'k', 'linewidth', 4)
set(gca, 'FontSize', 28)
set(gca, 'FontName', 'Times New Roman')
xlabel('generalized displacement'), ylabel('generalized force')
grid
zoom off

Transcendental models

In transcendental models, the output variable is computed by solving transcendental equations, namely equations involving trigonometric, inverse trigonometric, exponential, logarithmic, and/or hyperbolic functions.

Exponential models

Exponential model by Vaiana et al. (2018)

=Model formulation

=
In the exponential model developed by Vaiana et al. (2018), the generalized force at time $t$, representing the output variable, is evaluated as a function of the generalized displacement as follows:
f_t=-2\beta u_t+e^-e^ +k_b u_t-s_t\frac\left ^-e^\rightf_0 s_t \quad \text \quad u_t s_t < u_j s_t,

$f_t=-2\beta u_t+e^-e^+k_b u_t+f_0 s_t \quad \text \quad u_t s_t>u_j s_t,$
where $k_a, k_b, \alpha$ and $\beta$ are the four model parameters to be calibrated from experimental or numerical tests, whereas $s_t$ is the sign of the generalized velocity at time $t$, that is, $s_t = \operatorname(\dot_t)$. Furthermore, $u_0$and $f_0$ are two internal model parameters evaluated as:
$u_0=-\frac \ln\left(\frac\right),$

$f_0=\frac\left(1-e^\right),$
whereas $u_j$ is the internal variable:
u_j=u_+2u_0 s_t + \frac \ln\left frac \left(-2\beta u_+e^-e^+k_b u_+\frac s_t e^ + f_0 s_t-f_\right)\right

=Hysteresis loop shapes

=
Figure 2.1 shows four different hysteresis loop shapes obtained by applying a sinusoidal generalized displacement having unit amplitude and frequency and simulated by adopting the Exponential Model (EM) parameters listed in Table 2.1.

=Matlab code

=

%

=
% September 2019
% Exponential Model Algorithm
% Nicolò Vaiana, Post-Doctoral Researcher, PhD
% Department of Structures for Engineering and Architecture
% University of Naples Federico II
% via Claudio, 21 - 80125, Napoli
%

=

clc; clear all; close all;

%% APPLIED DISPLACEMENT TIME HISTORY

dt = 0.001; %time step
t = 0:dt:1.50; %time interval
a0 = 1; %applied displacement amplitude
fr = 1; %applied displacement frequency
u = a0*sin((2*pi*fr)*t(1:length(t))); %applied displacement vector
v = 2*pi*fr*a0*cos((2*pi*fr)*t(1:length(t))); %applied velocity vector
n = length(u); %applied displacement vector length

%% 1. INITIAL SETTINGS
%1.1 Set the four model parameters
ka = 5.0; %model parameter
kb = 0.5; %model parameter
alfa = 5.0; %model parameter
beta = 1.0; %model parameter
%1.2 Compute the internal model parameters
u0 = -(1/(2*alfa))*log(10^-20/(ka-kb)); %internal model parameter
f0 = ((ka-kb)/(2*alfa))*(1-exp(-2*alfa*u0)); %internal model parameter
%1.3 Initialize the generalized force vector
f = zeros(1, n);

%% 2. CALCULATIONS AT EACH TIME STEP

for i = 2:n
%2.1 Update the history variable
uj = u(i-1)+2*u0*sign(v(i))+sign(v(i))*(1/alfa)*log(sign(v(i))*(alfa/(ka-kb))*(-2*beta*u(i-1)+exp(beta*u(i-1))-exp(-beta*u(i-1))+kb*u(i-1)+sign(v(i))*((ka-kb)/alfa)*exp(-2*alfa*u0)+sign(v(i))*f0-f(i-1)));
%2.2 Evaluate the generalized force at time t
if (sign(v(i))*uj)-2*u0 < sign(v(i))*u(i) , , sign(v(i))*u(i) < sign(v(i))*uj
f(i) = -2*beta*u(i)+exp(beta*u(i))-exp(-beta*u(i))+kb*u(i)-sign(v(i))*((ka-kb)/alfa)*(exp(-alfa*(sign(v(i))*(u(i)-uj)+2*u0))-exp(-2*alfa*u0))+sign(v(i))*f0;
else
f(i) = -2*beta*u(i)+exp(beta*u(i))-exp(-beta*u(i))+kb*u(i)+sign(v(i))*f0;
end
end

%% PLOT
figure
plot(u ,f, 'k', 'linewidth', 4)
set(gca, 'FontSize', 28)
set(gca, 'FontName', 'Times New Roman')
xlabel('generalized displacement'), ylabel('generalized force')
grid
zoom off

References

{{reflist

Dynamical systems
Hysteresis
Materials science
Nonlinear systems