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How to solve a system of 8 equations having integrals!

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Abubakar Rashid
Abubakar Rashid 2022 年 11 月 29 日
コメント済み: Abubakar Rashid 2022 年 11 月 30 日
d=1.18; E=27000000; L=121.53;
I=(pi*d^4)/64;
x1=16.99; x2=34.16; x3=51.33; x4=68.49; x5=85.66; x6=102.83; x7=115.61; x8=120.12;
F1=49; F2=49; F3=49; F4=49; F5=49; F6=49; F7=49; F8=11;
K1=100; K2=100; K3=100; K4=100; K5=100; K6=100; K7=100; K8=100;
These are my moment equation, that are further used in the deflection (d1, d2, ...., d8) calculations.
M1 = @(x)(F1*x-K1*d1*x);
M2 = @(x)(F1*x+F2*(x-x1)-K1*d1*x-K2*d2*(x-x1));
M3 = @(x)(F1*x+F2*(x-x1)+F3*(x-x2)-K1*d1*x-K2*d2*(x-x1)-K3*d3*(x-x2));
M4 = @(x)(F1*x+F2*(x-x1)+F3*(x-x2)+F4*(x-x3)-K1*d1*x-K2*d2*(x-x1)-K3*d3*(x-x2)-K4*d4*(x-x3));
M5 = @(x)(F1*x+F2*(x-x1)+F3*(x-x2)+F4*(x-x3)+F5*(x-x4)-K1*d1*x-K2*d2*(x-x1)-K3*d3*(x-x2)-K4*d4*(x-x3)-K5*d5*(x-x4));
M6 = @(x)(F1*x+F2*(x-x1)+F3*(x-x2)+F4*(x-x3)+F5*(x-x4)+F6*(x-x5)-K1*d1*x-K2*d2*(x-x1)-K3*d3*(x-x2)-K4*d4*(x-x3)-K5*d5*(x-x4)-K6*d6*(x-x5));
M7 = @(x)(F1*x+F2*(x-x1)+F3*(x-x2)+F4*(x-x3)+F5*(x-x4)+F6*(x-x5)+F7*(x-x6)-K1*d1*x-K2*d2*(x-x1)-K3*d3*(x-x2)-K4*d4*(x-x3)-K5*d5*(x-x4)-K6*d6*(x-x5)-K7*d7*(x-x6));
M8 = @(x)(F1*x+F2*(x-x1)+F3*(x-x2)+F4*(x-x3)+F5*(x-x4)+F6*(x-x5)+F7*(x-x6)+F8*(x-x7)-K1*d1*x-K2*d2*(x-x1)-K3*d3*(x-x2)-K4*d4*(x-x3)-K5*d5*(x-x4)-K6*d6*(x-x5)-K7*d7*(x-x6)-K8*d8*(x-x7));
Below are my eight equations with defined boundaries for integrals. And it contains eight variables i.e. d1, d2, ...., d8 as well. I am fine with both numerical or analytical solution.
eq1 = d1 ==(1/(E*I))*(integral(@(x)(M1(x).*x),0,x1)+integral(@(x)(M2(x).*x),x1,x2)+integral(@(x)(M3(x).*x),x2,x3)+integral(@(x)(M4(x).*x),x3,x4)+integral(@(x)(M5(x).*x),x4,x5)+integral(@(x)(M6(x).*x),x5,x6)+integral(@(x)(M7(x).*x),x6,x7)+integral(@(x)(M8(x).*x),x7,x8));
eq2 = d2 ==(1/(E*I))*(integral(@(x)(M1(x).*0),0,x1)+integral(@(x)(M2(x).*(x-x1)),x1,x2)+integral(@(x)(M3(x).*(x-x1)),x2,x3)+integral(@(x)(M4(x).*(x-x1)),x3,x4)+integral(@(x)(M5(x).*(x-x1)),x4,x5)+integral(@(x)(M6(x).*(x-x1)),x5,x6)+integral(@(x)(M7(x).*(x-x1)),x6,x7)+integral(@(x)(M8(x).*(x-x1)),x7,x8));
eq3 = d3 ==(1/(E*I))*(integral(@(x)(M1(x).*0),0,x1)+integral(@(x)(M2(x).*0),x1,x2)+integral(@(x)(M3(x).*(x-x2)),x2,x3)+integral(@(x)(M4(x).*(x-x2)),x3,x4)+integral(@(x)(M5(x).*(x-x2)),x4,x5)+integral(@(x)(M6(x).*(x-x2)),x5,x6)+integral(@(x)(M7(x).*(x-x2)),x6,x7)+integral(@(x)(M8(x).*(x-x2)),x7,x8));
eq4 = d4 ==(1/(E*I))*(integral(@(x)(M1(x).*0),0,x1)+integral(@(x)(M2(x).*0),x1,x2)+integral(@(x)(M3(x).*0),x2,x3)+integral(@(x)(M4(x).*(x-x3)),x3,x4)+integral(@(x)(M5(x).*(x-x3)),x4,x5)+integral(@(x)(M6(x).*(x-x3)),x5,x6)+integral(@(x)(M7(x).*(x-x3)),x6,x7)+integral(@(x)(M8(x).*(x-x2)),x7,x8));
eq5 = d5 ==(1/(E*I))*(integral(@(x)(M1(x).*0),0,x1)+integral(@(x)(M2(x).*0),x1,x2)+integral(@(x)(M3(x).*0),x2,x3)+integral(@(x)(M4(x).*0),x3,x4)+integral(@(x)(M5(x).*(x-x4)),x4,x5)+integral(@(x)(M6(x).*(x-x4)),x5,x6)+integral(@(x)(M7(x).*(x-x4)),x6,x7)+integral(@(x)(M8(x).*(x-x4)),x7,x8));
eq6 = d6 ==(1/(E*I))*(integral(@(x)(M1(x).*0),0,x1)+integral(@(x)(M2(x).*0),x1,x2)+integral(@(x)(M3(x).*0),x2,x3)+integral(@(x)(M4(x).*0),x3,x4)+integral(@(x)(M5(x).*0),x4,x5)+integral(@(x)(M6(x).*(x-x5)),x5,x6)+integral(@(x)(M7(x).*(x-x5)),x6,x7)+integral(@(x)(M8(x).*(x-x5)),x7,x8));
eq7 = d7 ==(1/(E*I))*(integral(@(x)(M1(x).*0),0,x1)+integral(@(x)(M2(x).*0),x1,x2)+integral(@(x)(M3(x).*0),x2,x3)+integral(@(x)(M4(x).*0),x3,x4)+integral(@(x)(M5(x).*0),x4,x5)+integral(@(x)(M6(x).*0),x5,x6)+integral(@(x)(M7(x).*(x-x6)),x6,x7)+integral(@(x)(M8(x).*(x-x6)),x7,x8));
eq8 = d8 ==(1/(E*I))*(integral(@(x)(M1(x).*0),0,x1)+integral(@(x)(M2(x).*0),x1,x2)+integral(@(x)(M3(x).*0),x2,x3)+integral(@(x)(M4(x).*0),x3,x4)+integral(@(x)(M5(x).*0),x4,x5)+integral(@(x)(M6(x).*0),x5,x6)+integral(@(x)(M7(x).*0),x6,x7)+integral(@(x)(M8(x).*(x-x7)),x7,x8));

採用された回答

Torsten
Torsten 2022 年 11 月 29 日
d0 = ones(1,8);
d = fsolve(@fun,d0)
Equation solved. fsolve completed because the vector of function values is near zero as measured by the value of the function tolerance, and the problem appears regular as measured by the gradient.
d = 1×8
0.5346 0.5133 0.4823 0.4784 0.3431 0.2181 0.0778 0.0067
norm(fun(d))
ans = 1.4844e-15
function res = fun(D)
d=1.18;
E=27000000;
L=121.53;
I=(pi*d^4)/64;
x1=16.99; x2=34.16; x3=51.33; x4=68.49; x5=85.66; x6=102.83; x7=115.61; x8=120.12;
F1=49; F2=49; F3=49; F4=49; F5=49; F6=49; F7=49; F8=11;
K1=100; K2=100; K3=100; K4=100; K5=100; K6=100; K7=100; K8=100;
d1 = D(1);
d2 = D(2);
d3 = D(3);
d4 = D(4);
d5 = D(5);
d6 = D(6);
d7 = D(7);
d8 = D(8);
M1 = @(x)(F1*x-K1*d1*x);
M2 = @(x)(F1*x+F2*(x-x1)-K1*d1*x-K2*d2*(x-x1));
M3 = @(x)(F1*x+F2*(x-x1)+F3*(x-x2)-K1*d1*x-K2*d2*(x-x1)-K3*d3*(x-x2));
M4 = @(x)(F1*x+F2*(x-x1)+F3*(x-x2)+F4*(x-x3)-K1*d1*x-K2*d2*(x-x1)-K3*d3*(x-x2)-K4*d4*(x-x3));
M5 = @(x)(F1*x+F2*(x-x1)+F3*(x-x2)+F4*(x-x3)+F5*(x-x4)-K1*d1*x-K2*d2*(x-x1)-K3*d3*(x-x2)-K4*d4*(x-x3)-K5*d5*(x-x4));
M6 = @(x)(F1*x+F2*(x-x1)+F3*(x-x2)+F4*(x-x3)+F5*(x-x4)+F6*(x-x5)-K1*d1*x-K2*d2*(x-x1)-K3*d3*(x-x2)-K4*d4*(x-x3)-K5*d5*(x-x4)-K6*d6*(x-x5));
M7 = @(x)(F1*x+F2*(x-x1)+F3*(x-x2)+F4*(x-x3)+F5*(x-x4)+F6*(x-x5)+F7*(x-x6)-K1*d1*x-K2*d2*(x-x1)-K3*d3*(x-x2)-K4*d4*(x-x3)-K5*d5*(x-x4)-K6*d6*(x-x5)-K7*d7*(x-x6));
M8 = @(x)(F1*x+F2*(x-x1)+F3*(x-x2)+F4*(x-x3)+F5*(x-x4)+F6*(x-x5)+F7*(x-x6)+F8*(x-x7)-K1*d1*x-K2*d2*(x-x1)-K3*d3*(x-x2)-K4*d4*(x-x3)-K5*d5*(x-x4)-K6*d6*(x-x5)-K7*d7*(x-x6)-K8*d8*(x-x7));
res(1) = d1 -((1/(E*I))*(integral(@(x)(M1(x).*x),0,x1)+integral(@(x)(M2(x).*x),x1,x2)+integral(@(x)(M3(x).*x),x2,x3)+integral(@(x)(M4(x).*x),x3,x4)+integral(@(x)(M5(x).*x),x4,x5)+integral(@(x)(M6(x).*x),x5,x6)+integral(@(x)(M7(x).*x),x6,x7)+integral(@(x)(M8(x).*x),x7,x8)));
res(2) = d2 -((1/(E*I))*(integral(@(x)(M1(x).*0),0,x1)+integral(@(x)(M2(x).*(x-x1)),x1,x2)+integral(@(x)(M3(x).*(x-x1)),x2,x3)+integral(@(x)(M4(x).*(x-x1)),x3,x4)+integral(@(x)(M5(x).*(x-x1)),x4,x5)+integral(@(x)(M6(x).*(x-x1)),x5,x6)+integral(@(x)(M7(x).*(x-x1)),x6,x7)+integral(@(x)(M8(x).*(x-x1)),x7,x8)));
res(3) = d3 -((1/(E*I))*(integral(@(x)(M1(x).*0),0,x1)+integral(@(x)(M2(x).*0),x1,x2)+integral(@(x)(M3(x).*(x-x2)),x2,x3)+integral(@(x)(M4(x).*(x-x2)),x3,x4)+integral(@(x)(M5(x).*(x-x2)),x4,x5)+integral(@(x)(M6(x).*(x-x2)),x5,x6)+integral(@(x)(M7(x).*(x-x2)),x6,x7)+integral(@(x)(M8(x).*(x-x2)),x7,x8)));
res(4) = d4 -((1/(E*I))*(integral(@(x)(M1(x).*0),0,x1)+integral(@(x)(M2(x).*0),x1,x2)+integral(@(x)(M3(x).*0),x2,x3)+integral(@(x)(M4(x).*(x-x3)),x3,x4)+integral(@(x)(M5(x).*(x-x3)),x4,x5)+integral(@(x)(M6(x).*(x-x3)),x5,x6)+integral(@(x)(M7(x).*(x-x3)),x6,x7)+integral(@(x)(M8(x).*(x-x2)),x7,x8)));
res(5) = d5-((1/(E*I))*(integral(@(x)(M1(x).*0),0,x1)+integral(@(x)(M2(x).*0),x1,x2)+integral(@(x)(M3(x).*0),x2,x3)+integral(@(x)(M4(x).*0),x3,x4)+integral(@(x)(M5(x).*(x-x4)),x4,x5)+integral(@(x)(M6(x).*(x-x4)),x5,x6)+integral(@(x)(M7(x).*(x-x4)),x6,x7)+integral(@(x)(M8(x).*(x-x4)),x7,x8)));
res(6) = d6 -((1/(E*I))*(integral(@(x)(M1(x).*0),0,x1)+integral(@(x)(M2(x).*0),x1,x2)+integral(@(x)(M3(x).*0),x2,x3)+integral(@(x)(M4(x).*0),x3,x4)+integral(@(x)(M5(x).*0),x4,x5)+integral(@(x)(M6(x).*(x-x5)),x5,x6)+integral(@(x)(M7(x).*(x-x5)),x6,x7)+integral(@(x)(M8(x).*(x-x5)),x7,x8)));
res(7) = d7 -((1/(E*I))*(integral(@(x)(M1(x).*0),0,x1)+integral(@(x)(M2(x).*0),x1,x2)+integral(@(x)(M3(x).*0),x2,x3)+integral(@(x)(M4(x).*0),x3,x4)+integral(@(x)(M5(x).*0),x4,x5)+integral(@(x)(M6(x).*0),x5,x6)+integral(@(x)(M7(x).*(x-x6)),x6,x7)+integral(@(x)(M8(x).*(x-x6)),x7,x8)));
res(8) = d8 -((1/(E*I))*(integral(@(x)(M1(x).*0),0,x1)+integral(@(x)(M2(x).*0),x1,x2)+integral(@(x)(M3(x).*0),x2,x3)+integral(@(x)(M4(x).*0),x3,x4)+integral(@(x)(M5(x).*0),x4,x5)+integral(@(x)(M6(x).*0),x5,x6)+integral(@(x)(M7(x).*0),x6,x7)+integral(@(x)(M8(x).*(x-x7)),x7,x8)));
end
  3 件のコメント
Torsten
Torsten 2022 年 11 月 29 日
編集済み: Torsten 2022 年 11 月 29 日
The code solves your equations eq1,...,eq8 for d1,...,d8 using Newton's method (thus a numerical solution).
I think equations 1-8 are just linear equations in the unknowns d1,...,d8. Thus in principle you solve a linear system of equations of the form A*d = b.
But getting the entries of A and b from your equations would be quite tedious - so I solved it as if it were a nonlinear system.
A symbolic approach and EquationsToMatrix could be of help to recover A and b.
Abubakar Rashid
Abubakar Rashid 2022 年 11 月 30 日
Thank you Torsten, it helps a lot!

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その他の回答 (1 件)

Walter Roberson
Walter Roberson 2022 年 11 月 29 日
Do not use == with integral() . integral() is for numeric integration, and if you use numeric integration returning a numeric value then the right side of the == would be numeric, and the == would be for asking whether d1 is exactly the same value, bit-for-bit identical (except for different nans being considered the same, and negative 0 being treated the same as regular 0)
If you want to process this system numerically using fsolve() then you should change the NAME == EXPRESSION into NAME - (EXPRESSION) and then you would have the function return [eq1, eq2, eq3, eq4, eq5, eq6, eq7, eq8]
You could alternately switch to using the Symbolic Toolbox and use int() or vpaintegral() instead of integral, and try to solve() or vpasolve() the vector of equations (which can be written with == in that case.)
  4 件のコメント
Walter Roberson
Walter Roberson 2022 年 11 月 29 日
Torsten's response is good, but for efficiency I would change, for example,
res(2) = d2 -((1/(E*I))*(integral(@(x)(M1(x).*0),0,x1)+integral(@(x)(M2(x).*(x-x1)),x1,x2)+integral(@(x)(M3(x).*(x-x1)),x2,x3)+integral(@(x)(M4(x).*(x-x1)),x3,x4)+integral(@(x)(M5(x).*(x-x1)),x4,x5)+integral(@(x)(M6(x).*(x-x1)),x5,x6)+integral(@(x)(M7(x).*(x-x1)),x6,x7)+integral(@(x)(M8(x).*(x-x1)),x7,x8)));
to
res(2) = d2 -((1/(E*I))*(integral(@(x)(M2(x).*(x-x1)),x1,x2)+integral(@(x)(M3(x).*(x-x1)),x2,x3)+integral(@(x)(M4(x).*(x-x1)),x3,x4)+integral(@(x)(M5(x).*(x-x1)),x4,x5)+integral(@(x)(M6(x).*(x-x1)),x5,x6)+integral(@(x)(M7(x).*(x-x1)),x6,x7)+integral(@(x)(M8(x).*(x-x1)),x7,x8)));
which got rid of the integral(@(x)(M1(x).*0),0,x1) since we know the result is going to be 0 for the term.
Abubakar Rashid
Abubakar Rashid 2022 年 11 月 30 日
Thank you Walter for the quick feedback.

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