In this code how can I put multiple values of "betasqr" for example 0.2, 0.4, 0,6 and subsequently multiple values of "sbar" for example 1/0.02, 1/0.04, 1/0.06 (contd in des)

% (contd.) to obtan 9 values of Wbar, Fbar, fbar and Ybar respectively corresponding to "betasqr" = 0.2 ,"sbar" = 1/0.02 ; "betasqr" = 0.2 ,"sbar" = 1/0.04 and similarly for other combinations of "betasqr" and "sbar".

betasqr = 0.2;

gamma = 0.3;

psi = 0.001;

mu = 1.0;

sbar = 1/0.02;

a = 2;

% Steps for finding dimensionless load carrying capacity (Wbar) of the bearing

hbar = @(x) a - a.*x + x;

A = @(x) 4 .* (hbar(x) .^ 3) + sbar .* (hbar(x) .^ 4) - sbar*betasqr*gamma .* (hbar(x) .^ 2) .* sqrt(x .* (1-x)) ;

B = @(x) (1 + sbar .* hbar(x)) .* (1 - betasqr * sqrt(x .* (1-x)));

G = @(x) 12 * psi + A(x) ./ B(x);

Z = @(x) 12 .* hbar(x) + 6*sbar .* (hbar(x) .^ 2) - 2*sbar*betasqr*gamma .* sqrt(x .* (1-x)) ;

E = @(x) Z(x) ./ (1 + sbar .* hbar(x));

C = @(x) E(x) ./ G(x);

IntEbyG = integral(C,0,1);

D = @(x) 1 ./ G(x);

IntGinv = integral(D,0,1);

Q = IntEbyG/IntGinv;

Wdash = @(x) x.*(E(x)-Q) ./ G(x);

IntWdash = integral(Wdash,0,1);

Wbar = (mu/12) - IntWdash

% Steps for finding dimensionless frictional force (Fbar) of the slider

I = @(x) hbar(x) .* (2 + sbar .* hbar(x)) .* (E(x) - Q );

J = @(x) 2 .* G(x) .* (1 + sbar .* hbar(x)) .* (1 - betasqr * sqrt(x .* (1-x)));

Fdash = @(x) (sbar ./ (1 + sbar .* hbar(x))) + I(x) ./ J(x);

Fbar = integral(Fdash,0,1)

% Steps for finding coefficient of friction (fbar)

fbar = Fbar/Wbar % Coefficient of friction

% Steps for finding dimensionless form of x coordinate of center of pressure(Ybar)

L = @(x) (x.^2).*(E(x)-Q) ./ G(x);

IntL = integral(L,0,1);

Ybar = ((mu/24)-(0.5*IntL)) / Wbar

Answer 1

[Dyuman Joshi]

Define dbar as follows, and run a for loop 9 times

dbar = 0.2*(1:9)

dbar = 1×9

0.2000 0.4000 0.6000 0.8000 1.0000 1.2000 1.4000 1.6000 1.8000

sbar = 0.1./dbar

sbar = 1×9

0.5000 0.2500 0.1667 0.1250 0.1000 0.0833 0.0714 0.0625 0.0556