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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:
Hysteretic models may have a generalized displacement as input variable and a generalized force 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.[2][3]
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:
In algebraic models, the output variable is computed by solving algebraic equations.
In the bilinear model formulated by Vaiana et al. (2018),[4] the generalized force at time t, representing the output variable, is evaluated as a function of the generalized displacement as follows:
where and are the three model parameters to be calibrated from experimental or numerical tests, whereas is the sign of the generalized velocity at time , that is, . Furthermore, is an internal model parameter evaluated as:
whereas is the internal variable:
.
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. File:Bilinear Model.tif
| (a) | 10.0 | 1.0 | 0.5 |
| (b) | 10.0 | -1.0 | 0.5 |
% =========================================================================================
% 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
In the asymmetric bilinear model formulated by Vaiana et al. (2020),[3] the generalized force at time t, representing the output variable, is evaluated as a function of the generalized displacement as follows:
where and are the three model parameters of the generic loading (unloading) case to be calibrated from experimental or numerical tests, whereas is the sign of the generalized velocity at time , that is, . Furthermore, is an internal model parameter evaluated as:
whereas is the internal variable:
% =========================================================================================
% 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
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: .

In the algebraic model developed by Vaiana et al. (2019),[5] the generalized force at time , representing the output variable, is evaluated as a function of the generalized displacement as follows:
where , and are the five model parameters to be calibrated from experimental or numerical tests, whereas is the sign of the generalized velocity at time , that is, . Furthermore, and are two internal model parameters evaluated as:
whereas is the internal variable:
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.File:Figure NV 3.tif
| (a) | 10.0 | 1.0 | 10.0 | 0.0 | 0.0 |
| (b) | 10.0 | 1.0 | 10.0 | 0.2 | 0.2 |
| (c) | 10.0 | 1.0 | 10.0 | −0.2 | −0.2 |
| (d) | 10.0 | 1.0 | 10.0 | −1.2 | 1.2 |
% =========================================================================================
% 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
In transcendental models, the output variable is computed by solving transcendental equations, namely equations involving trigonometric, inverse trigonometric, exponential, logarithmic, and/or hyperbolic functions.
In the exponential model developed by Vaiana et al. (2018),[4] the generalized force at time , representing the output variable, is evaluated as a function of the generalized displacement as follows:
where and are the four model parameters to be calibrated from experimental or numerical tests, whereas is the sign of the generalized velocity at time , that is, . Furthermore, and are two internal model parameters evaluated as:
whereas is the internal variable:
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.File:Figure NV 2.tif
| (a) | 5.0 | 0.5 | 5.0 | 0.0 |
| (b) | 5.0 | −0.5 | 5.0 | 0.0 |
| (c) | 5.0 | 0.5 | 5.0 | 1.0 |
| (d) | 5.0 | 0.5 | 5.0 | −1.0 |
% =========================================================================================
% 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