maple2015

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These are questions asked by maple2015

 

Dear Friends
Is there a way to solve a complicated integration in less possible time?

Thanks

_________________________________________________________________________________
 

restart;
Digits := 100:
tm := time():
with(LinearAlgebra):

m := 6:
a := 0.1:
b := 10*a:
E := 1:
h := 1:
nu := 0.3:

w := (r-b)^2*(r-a)^2*add(add(W[n, i]*r^n*t^(i-n), n = 0 .. i), i = 0 .. m):
ur := -z*(diff(w, r)):
ut := -z*(diff(w, t))/r:
er := diff(ur, r)+(1/2)*(diff(w, r))^2:
et := ur/r+(diff(ut, t))/r+(diff(w, t))^2/(2*r^2):
grt := diff(ut, r)-ut/r+(diff(ur, t))/r+(diff(diff(w, t), r))/r:
u := -(1/2)*E*(2*er*et*nu+er^2+et^2)/(nu^2-1)+(1/2)*E*grt^2/(2*(1+nu)):

PI := int(int(int(u*r, z = -(1/2)*h .. (1/2)*h), t = 0 .. 2*Pi), r = a .. b)-0.5*P*(int(int(r*(diff(w, r))^2, r = a .. b), t = 0 .. 2*Pi)):

Time = time()-tm;

Hi

Please download and check the attached file.
It seems when you run the code more than one time, various results are obtained each time.

What is the reason? How it can be fixed?

Thanks


 

restart; Digits := 20; tm := time(); with(LinearAlgebra); m := 6; a := .1; b := 10*a; E := 1; h := 1; nu := .3; ur := -w*z+u0; u0 := 0; ut := add(add(T[n, i]*r^n*t^(i-n), n = 0 .. i), i = 0 .. m); w := (r-b)^2*(r-a)^2*add(add(W[n, i]*r^n*t^(i-n), n = 0 .. i), i = 0 .. m); er := diff(ur, r); et := ur/r+(diff(ut, t))/r; ert := 1/2*(diff(ut, r)-ut/r+(diff(ur, t))/r); u := -(1/2)*E*(2*er*et*nu+er^2+et^2)/(nu^2-1)+2*E*ert^2/(2+2*nu); N := sum(i+1, i = 0 .. m); PI := int(int(int(u*r, z = -(1/2)*h .. (1/2)*h), t = 0 .. 2*Pi), r = a .. b)-.5*P*(int(int(r*(diff(w, r))^2, r = a .. b), t = 0 .. 2*Pi)); s1 := seq(indets(add(add(T[n, i], n = 0 .. i), i = 0 .. m))[k] = c[k], k = 1 .. N); s2 := seq(indets(add(add(W[n, i], n = 0 .. i), i = 0 .. m))[k] = c[k+N], k = 1 .. N); PI := subs(s1, s2, PI); for k to 2*N do diff(PI, c[k]); if % = 0 then ex := `union`({}, {k}) else eq[k] := % end if end do; NE := seq(ex[j], j = 1 .. numelems(ex)); M := GenerateMatrix([`$`(eq[j], j = 1 .. 2*N)], [`$`(c[j], j = 1 .. 2*N)])[1]; M := DeleteColumn(DeleteRow(M, NE), NE); Determinant(M); 12*fsolve(%, P = 0 .. 1)*(-nu^2+1)*a^2/(E*h^3); Time = time()-tm

Time = 85.363

(1)

``


 

Download Stability.mw

# Bending Moment Envelope Curve for Isostatic Two-Span Beam

restart:

with(plots):

with(Optimization):
PD:= proc (L1, L2, L3, N)

local l, R, Y, M, M1, M2, V:

R[1]:= piecewise(x <= L1, (L1-x)/L1, 0):

l[1]:=L1:

l[2]:=L2:

l[3]:=L3:

solve([add(R[i], i = 1 .. 3) = 1, add(R[j]*add(l[i], i = 1 .. j), j = 2 .. 3) = x], [R[2], R[3]]):

R[2]:=rhs(%[1][1]):

R[3]:= rhs(`%%`[1][2]):

M1:=piecewise(y <= add(l[i], i = 1 .. 2), R[1]*y, `and`(add(l[i], i = 1 .. 2) < y, y <= add(l[i], i = 1 .. 3)), R[1]*y+R[2]*(y-add(l[i], i = 1 .. 2))):

M2:=piecewise(y <= x, 0, x-y):

for Y to N do eval(M1+M2, y = Y*add(l[j], j = 1 .. 3)/N):

M[Y]:=Maximize(abs(%),x=0..add(l[j],j=1..3))[1]

end do:

pointplot(`<,>`(seq(i*add(l[j], j = 1 .. 3)/N, i = 1 .. N)), `<,>`(seq(M[i], i = 1 .. N)), color = red, symbol = asterisk):

display(%, axis = [gridlines = [10, color = black]], size = [700, "golden"], axesfont = [Times, 16])

end proc:

Hi

I use some commands like the evalf[10]

But they don't work for the definite double integral in attached file.

It leads to "Error, (in type/algfun) too many levels of recursion"

Please help me to find the answer

Thank You Very Much

Double-Int.mw

 

Hi

I have two equations as follows:

The goal is finding the parameter 'phi'. This parameter is a positive real numeric constant.

I uploaded two files that they are included two methods to solve the problem.

1.mw

2.mw

Is method in the first file mathematically logical? If it is a correct method, why the command fsolve dosent work?

In file 2, we have 2 equations with further indeterminantes. The constant 'phi' must be minimum possible amount. How we can use the commands like the Minimize in Optimization? Please hint me.

Moreover, if there is a method to solve this problem please help me to know.

Thank you very much 

 

 

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