How to denote σ and this symbol ε

when I give σ as sigma = g(x) its showing error but when i give tau = f(x) its is showing write can anyone help me to figure it out please.

[Ameer Hamza]

Can you show your code?

[Avinash T]

[Avinash T]

This is my code I dont know how to type this in MATLAB

[Walter Roberson]

I do not recognize the bar to the left in this context, and I do not know what the (1) represents in this context. If it indicates first derivative then I would have expected sigma to have a single x instead of xx. The double xx suggests partial derivative with respect to x and then x but then I would expect '' or (2)

[Avinash T]

Then 1 represents sample 1. Can you help me how to denote those symbols

[Paul Hoffrichter]

>> when I give σ as sigma = g(x) its showing error but when i give tau = f(x) its is showing write

Can you show the code for g(x) and f(x) that you describe. Include the error and the result for tau.

[Avinash T]

I too face the same problem.

so what I tried is use live editor in MATLAB.

I dont whether it is right but I did like that and I got correct so please cross check it because I am begineer to MATLAB

syms l tau

x = l

sym x

f(x) = tau

Mr.Paul can you say the paper tittle and authour of the paper your reffering.

[Walter Roberson]

 sym x

You need

syms x

[Avinash T]

in your previous comment you asked for bar at the left it is cursor bro and I am sorry I didnt notice it and 1 represents the sample 1 and for sample 2 we represent sigma xx2

Walter bro can you help me how to write those symbol in this equation and in previous commnetts equations as notation in MATLAB

τ = f (x) and σ = g(x).

Shear-free condition at the right and left adherend edges ( x = 0 and L)

Answer 1

[Walter Roberson]

syms b h_1 h_2 I_1 p_1 S_1 sigma_xx t_0 y_1 
syms f(xi) g(zeta) M_1(x)
LHS = S_1 / (b*h_1) - M_1(x)*y_1 / I_1

LHS = 

RHS_1 = p_1

RHS_1 = 

RHS_2 = - 1/h_1 * int(f(xi), xi, 0, x)

RHS_2 = 

RHS_31 = M_1(x)

RHS_31 = 

RHS_32 = t_0*h_2*x

RHS_32 = 

RHS_33 = -int(int(g(zeta),zeta,0,xi),xi,0,x)

RHS_33 = 

RHS_34 = -h_1/2 * int(f(xi), xi, 0, x);
RHS_3 = 12*y_1/h_1^3 * (RHS_31 + RHS_32 + RHS_33 + RHS_34)

RHS_3 = 

RHS = RHS_1 + RHS_2 + RHS_3

RHS = 

equation = sigma_xx == (LHS == RHS)

equation = 

However, I was not able to duplicate the formula you posted. The formula you posted has a } between the double integral and the following single integral, but there is no { anywhere in the formula, so I have no idea what the } means in that context.

[Avinash T]

Thank alot for clarrifying my doubt bro

[Walter Roberson]

In cases like this, it just isn't worth packing long expressions into one line just to make shorter code. The way I broke up the code here made it easy to see the individual parts and how they went together, and made it easy to debug the code.

In practice you would put more semi-colons on so you did not display the individual parts such as RHS_33

[Avinash T]

Sure bro I will note down.

[Walter Roberson]

Also, I am not proud of

equation = sigma_xx == (LHS == RHS)

MATLAB is not designed to have three parts all == each other, but when I used

sigma_xx = LHS == RHS

then I could not get it to display sigma_xx in nice form .

[Avinash T]

ok bro I will try coding by splitting it