Value of first derivative of Shape function at -1,0 and +1 using for loop?? Error using sym/diff (line 69) The second argument must be a variable or a nonnegative integer specifying the number of differentiations. Error in practise3 (line 29) N1=dif
3 ビュー (過去 30 日間)
古いコメントを表示
%%Shape function for axial effect
N1epsn=0.5*epsn^2-0.5*epsn; (ERROR HERE)
N2epsn=-1*epsn^2+1;
N3epsn=0.5*epsn^2+0.5*epsn;
for epsn=-1:1:1
%%First Derivatives of Shape Function
N1=diff(N1epsn,epsn)
N2=diff(N2epsn,epsn)
N3=diff(N3epsn,epsn)
end
5 件のコメント
Walter Roberson
2018 年 10 月 9 日
Which variable does the -1 designate, or how many derivatives should be taken?
Notice that you are inside a for loop that is changing the value of the second parameter to diff and that you start at -1 which is confusing us.
採用された回答
KSSV
2018 年 10 月 9 日
%%Shape function for axial effect
syms epsn
N1epsn=0.5*epsn^2-0.5*epsn;
N2epsn=-1*epsn^2+1;
N3epsn=0.5*epsn^2+0.5*epsn;
for epsn=-1:1:1
%%First Derivatives of Shape Function
N1=subs(diff(N1epsn),epsn) ;
N2=subs(diff(N2epsn),epsn) ;
N3=subs(diff(N3epsn),epsn) ;
end
You should use double to convert sym to double.
5 件のコメント
その他の回答 (0 件)
参考
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!