Solving an equation and using the answer for other equations

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Finlay Brierton 2020-1-7

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评论： Cameron B 2020-1-7

Hi,

I am trying to write some code for my thesis and am struggling to convert an equation I derived into matlab. The code relates to steel member selection for varyingnd portal fram dimensions. Y - R is the target bending capacity of the beam. I have attached my written work and attempted coding, thanks.

w = 8.37;
L =  30;
h = 6.5;
theta = 12;
N = L/2 + h/tand(theta);
syms m
eqn = m^2/2*w + m*(5*N/3 - L/3) + w*L^2/8 == 0
M = solve(eqn,m)
Y = -(w/2)*(-m/w)^2 + w*L^2/8;
R = -(m^2/w) - m*N;
Mr = Y - R

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Answer 1

Cameron B 2020-1-7

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w = 8.37;
L =  30;
h = 6.5;
theta = 12;
N = L/2 + h/tand(theta);
syms m
eqn = m^2/(2*w) + m*(5*N/3 - L/3) + w*L^2/8;
M = solve(eqn,m);
m1 = double(M);
Y = -(w/2)*(-m1/w).^2 + w*L^2/8;
R = -(m1.^2/w) - m1.*N;
Mr = Y - R

This gives both roots of m and gives Mr = 294.8667 and 22178.87.

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Finlay Brierton 2020-1-7

Also I am not looking to solve the bending moment capacity for a portal frame of variables: w, L, h, theta.

Once I require Mr, will insert the values of Mr from the steel eurocodes and the program will suggest suitable steel section based on the calculated value.

thanks

Cameron B 2020-1-7

format longG %formats your answer how you want it

If you want the smallest room of m, then there are two ways of thinking about it. The first is picking the smallest of the two numbers which would be:

m1 = min(double(M));

However, the above line is biased towards negative values. If you wanted the smallest absolute value of the two, you would do this:

m1 = min(abs(double(M)));

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