Plot summation series | For Loop | Creep Strain

Question

mpz 2022-4-10

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1692875-plot-summation-series-for-loop-creep-strain

评论： VBBV 2022-4-10

Hi,

Need some assistance in plotting the below equation. I don't know why it is not coming out right. Below is my code and the plot below shows the expected results. Appreciate any feedback.

See picture at the bottom for equation to plot (strain vs time)

t=0:10000 (in seconds)

sigma0=100 MPa

D0= 360e-6 MPA^-1

D1=11.05e-6 MPa^-1; tr1=10 s;

D2=12.27e-6 MPa^-1; tr2=100 s;

D3=17.35e-6 MPa^-1; tr3=1000 s;

D4=21.63e-6 MPa^-1; tr4=10000 s;

D5=13.13e-6 MPa^-1; tr5=100000 s;

D6=41.78e-6 MPa^-1; tr6=1000000 s;

r in the summation series goes from r=1 to r=6 representing the D1,D2,D3,D4,D5,D6

clear all;clc
%% Linear Creep Example 
s=100; % constant tensile stress, (MPa)
t=linspace(0,10000,11); % duration of applied stress on specimen (seconds)
D=[11.05e-6 12.27e-6 17.35e-6 21.63e-6 13.13e-6 41.78e-6]; % Prony series coefficients, (MPA^-1)
ta=[10 100 1000 10000 100000 1000000]; % seconds
% D1=11.05e-6;ta1=10;
% D2=12.27e-6;ta2=100;
% D3=17.35e-6;ta3=1000;
% D4=21.63e-6;ta4=10000;
% D5=13.13e-6;ta5=100000;
% D6=41.78e-6;ta6=1000000;
% strn=s*(D(1)*(1-exp(-t(1)/ta(1))) + D(2)*(1-exp(-t(1)/ta(2))) + D(3)*(1-exp(-t(1)/ta(3)))...
%     + D(4)*(1-exp(-t(1)/ta(4))) + D(5)*(1-exp(-t(1)/ta(5))) + D(6)*(1-exp(-t(1)/ta(6))));
strn=zeros(1,11);
for i=1:11
    for n=1:6
     strn(i)=s*(D(n)*(1-exp(-t(i)/ta(n))));
    end
end
plot(t,strn)

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Answer 1

VBBV 2022-4-10

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1692875-plot-summation-series-for-loop-creep-strain#answer_938975

编辑：VBBV 2022-4-10

clear all;clc

%% Linear Creep Example

s=100; % constant tensile stress, (MPa)

t=linspace(0,10000,11); % duration of applied stress on specimen (seconds)

D=[11.05e-6 12.27e-6 17.35e-6 21.63e-6 13.13e-6 41.78e-6]; % Prony series coefficients, (MPA^-1)

ta=[10 100 1000 10000 100000 1000000]; % seconds

% D1=11.05e-6;ta1=10;

% D2=12.27e-6;ta2=100;

% D3=17.35e-6;ta3=1000;

% D4=21.63e-6;ta4=10000;

% D5=13.13e-6;ta5=100000;

% D6=41.78e-6;ta6=1000000;

% strn=s*(D(1)*(1-exp(-t(1)/ta(1))) + D(2)*(1-exp(-t(1)/ta(2))) + D(3)*(1-exp(-t(1)/ta(3)))...

% + D(4)*(1-exp(-t(1)/ta(4))) + D(5)*(1-exp(-t(1)/ta(5))) + D(6)*(1-exp(-t(1)/ta(6))));

strn=zeros(1,11);

for i=1:11

for n=1:6

strn(n,i)=(D(n)*(1-exp(-t(i)/ta(n)))); % multiply the s outside the summation

end

Strn(i) = s*sum(strn(:,i)); % multiply with s here

end

plot(t,Strn)

Multiply with s outside of the summation.

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mpz 2022-4-10

@VBBV @Akira Agata you both got it right. @VBBV you got it right first.

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Answer 2

Akira Agata 2022-4-10

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1692875-plot-summation-series-for-loop-creep-strain#answer_938980

How about the following?

s = 100; % constant tensile stress, (MPa)

t = linspace(0, 10000)'; % duration of applied stress on specimen (seconds)

D = [11.05e-6 12.27e-6 17.35e-6 21.63e-6 13.13e-6 41.78e-6]; % Prony series coefficients, (MPA^-1)

ta = [10 100 1000 10000 100000 1000000]; % seconds

dim = 2;

strn = s*sum(D.*(1-exp(-1*t./ta)), dim);

figure

plot(t, strn)

xlabel("Time [sec]")

ylabel("Strain \epsilon(t)")

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mpz 2022-4-10

s = 100; % constant tensile stress, (MPa)
t = linspace(0, 10000)'; % duration of applied stress on specimen (seconds)
D = [11.05e-6 12.27e-6 17.35e-6 21.63e-6 13.13e-6 41.78e-6]; % Prony series coefficients, (MPA^-1)
ta = [10 100 1000 10000 100000 1000000]; % seconds
dim = 2;
strn = s*Do+s*sum(D.*(1-exp(-1*t./ta)), dim);
figure
plot(t, strn)
xlabel("Time [sec]")
ylabel("Strain \epsilon(t)")

@Akira Agata and @VBBV I believe this should solve my problem

VBBV 2022-4-10

Yes,by adding an offset Do, your code seems to produce what you have shown in the figure.

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