Finding unknown in MATLAB

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Berke ERKAN
Berke ERKAN 2021-3-6
评论： Walter Roberson 2021-3-7
I would like to find the unknown value d in this code. Unfortunately, something wrong between line 19 and line 24 because the answer printed as ans = 0. I don't know why?
d answer should be approximately 4.33832
format short
%from table A-2
lbf = 4.45; %Pound-force to Newton 1 pound-force = 4.45 Newton
inch = 25.4; %inch to mm, 1 inch = 25.4 millimeter
%-----------------------------------
%Assumptions
Zie = 0.15; %Minimized space usage
F.S = 1.2; %Factor of Safety
%--------------------------------------------------------------------------
D = 1.25*inch, fprintf('mm') %Mean Diameter of the Coil
Req_valve_lift = 0.9*inch, fprintf('mm')
Fpreload = 50*lbf, fprintf ('N')
Fmax = 150*lbf, fprintf ('N')
%--------------------------------------------------------------------------
%Table 10-4
syms d
Ssy = (1974)*(d^0.108);
Kb = (4*((1.25*inch)/d)+2)/(4*((1.25*inch)/d)-3);
(Ssy*0.56)/F.S == (Kb*((8*Fmax*D)/(pi*d^3)));
sol=solve((Ssy*0.56)/F.S == (Kb*((8*Fmax*D)/(pi*d^3))),d);

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Walter Roberson
Walter Roberson 2021-3-6
编辑：Walter Roberson 2021-3-6
Ran in:
Q = @(v) sym(v);
format short
%from table A-2
lbf = Q(4.45); %Pound-force to Newton 1 pound-force = 4.45 Newton
inch = Q(25.4); %inch to mm, 1 inch = 25.4 millimeter
%-----------------------------------
%Assumptions
Zie = Q(0.15); %Minimized space usage
F.S = Q(1.2); %Factor of Safety
%--------------------------------------------------------------------------
D = Q(1.25)*inch, fprintf('mm') %Mean Diameter of the Coil
D = 
mm
Req_valve_lift = Q(0.9)*inch, fprintf('mm')
Req_valve_lift = 
mm
Fpreload = Q(50)*lbf, fprintf ('N')
Fpreload = 
N
Fmax = Q(150)*lbf, fprintf ('N')
Fmax = 
N
%--------------------------------------------------------------------------
%Table 10-4
syms d
Ssy = Q(1974)*(d^Q(0.108));
Kb = (Q(4)*((Q(1.25)*inch)/d)+Q(2))/(Q(4)*((Q(1.25)*inch)/d)-Q(3));
eqn = (Ssy*Q(0.56))/F.S == (Kb*((Q(8)*Fmax*D)/(Q(pi)*d^Q(3))));
eqn
eqn = 
sol = solve(eqn,d, 'returncondition', true);
condch = children(sol.conditions);
P = sym2poly(lhs(condch{1}));
z1 = roots(P);
mask = isAlways(subs(condch{2} & condch{3},sol.parameters,z1));
useful_z1 = z1(mask);
d_solutions = subs(sol.d, sol.parameters, useful_z1)
d_solutions = 
vpa(d_solutions)
ans = 
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Berke ERKAN
Berke ERKAN 2021-3-7
I am using R2020a.
Firstly, my main goal was to code an emulator calculator that can be calculate spring diameter which is "d" for this question. I edited the question just after the submitted here. I added the expected "d" value which is 4.33832.
Walter Roberson
Walter Roberson 2021-3-7
The change for R2020a turned out to be small. The results are the same as above, though.
You can test for yourself that d around 4.33832 is not a solution by substituting it into eqn
Q = @(v) sym(v);
format short
%from table A-2
lbf = Q(4.45); %Pound-force to Newton 1 pound-force = 4.45 Newton
inch = Q(25.4); %inch to mm, 1 inch = 25.4 millimeter
%-----------------------------------
%Assumptions
Zie = Q(0.15); %Minimized space usage
F.S = Q(1.2); %Factor of Safety
%--------------------------------------------------------------------------
D = Q(1.25)*inch, fprintf('mm') %Mean Diameter of the Coil
Req_valve_lift = Q(0.9)*inch, fprintf('mm')
Fpreload = Q(50)*lbf, fprintf ('N')
Fmax = Q(150)*lbf, fprintf ('N')
%--------------------------------------------------------------------------
%Table 10-4
syms d
Ssy = Q(1974)*(d^Q(0.108));
Kb = (Q(4)*((Q(1.25)*inch)/d)+Q(2))/(Q(4)*((Q(1.25)*inch)/d)-Q(3));
eqn = (Ssy*Q(0.56))/F.S == (Kb*((Q(8)*Fmax*D)/(Q(pi)*d^Q(3))));
eqn
sol = solve(eqn,d, 'returncondition', true);
condch = children(sol.conditions);
P = sym2poly(lhs(condch(1)));
z1 = roots(P);
mask = isAlways(subs(condch(2) & condch(3),sol.parameters,z1));
useful_z1 = z1(mask);
d_solutions = subs(sol.d, sol.parameters, useful_z1)
vpa(d_solutions)

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