Populating a vector between upper and lower bound?

Question

raymond bryant 2021-11-27

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1597004-populating-a-vector-between-upper-and-lower-bound

评论： raymond bryant 2021-11-27

I'm trying to simulate a cyclic loading in Matlab by varying strain between a positive upper bound and negative lower bound.

The code below stops at the upper bound, but ignores the lower bound. Ideally, I would be able to set however many loops I want and get values that incrementally alternate between e_max and -emax.

For the code below, the vector values should have gone up to 4 and down to -4 several times. Instead they continue down to -16 after hitting the upper bound.

e = [0 1]; %Initial strain values
e_max = 4; %Max strain values
e_t = 1; %Strain rate
delta_t = 1; %Time interval
for i = 2:1:24
    if e(i) < e_max &&  e(i) > e(i-1); %Increases strain as long as strain is below upper bound 
            e(end+1) = e(i) + e_t*delta_t;
    else e(i) > -e_max && e(i-1) > e(i); %Decreases strain as long as strain is below upper bound 
            e(end+1) = e(i) - e_t*delta_t;
    end
end
>> e
e =
     0     1     2     3     4     3     2     1     0    -1    -2    -3    -4    -5    
     -6    -7    -8    -9   -10   -11   -12   -13   -14   -15   -16

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Answer 1

Chunru 2021-11-27

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链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1597004-populating-a-vector-between-upper-and-lower-bound#answer_841384

e0 = 0;         % Initial strain values
e_max = 4;      % Max strain values
inc = 1;        % +ve increase, -ve decrease
e_t = 1;        % Strain rate
delta_t = 1;    % Time interval
n = 24;  
e = zeros(n, 1);
e(1) = e0;
for i = 2:1:n
    e(i) = e(i-1) + inc * e_t * delta_t;
    
    if e(i) >= e_max 
        e(i) = e_max;
        inc = -1;
    elseif e(i)<=-e_max
        e(i) = -e_max;
        inc = 1;
    end
end
e'
ans = 1×24
     0     1     2     3     4     3     2     1     0    -1    -2    -3    -4    -3    -2    -1     0     1     2     3     4     3     2     1

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raymond bryant 2021-11-27

e0 = 0;         % Initial strain values
e_max = 0.01;      % Max strain values
inc = 0.001;        % +ve increase, -ve decrease       
delta_t = 1;    % Time interval
n = 50;  
e = zeros(n, 1);
e(1) = e0;
for i = 2:1:n
    e(i) = e(i-1) + inc * delta_t;
    
    if e(i) >= e_max 
        e(i) = e_max;
        inc = -0.001;
    elseif e(i)<=-e_max
        e(i) = -e_max;
        inc = 0.001;
    end
end