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// SciLAB plotear un circulo

xc=3; yc=-3; r=6.5;n=50;k=0:n;fi=2*%pi*(k/n);x=xc+r*cos(fi);y=yc+r*sin(fi);plot(xc,yc,'or',x,y)

% igual en MATLAB, solo que no hac falta definir pi=%pi

xc=3; yc=-3; r=6.5;n=50;k=0:n;fi=2*%pi*(k/n);x=xc+r*cos(fi);y=yc+r*sin(fi);plot(xc,yc,'or',x,y)--------------------------------------------en SCILAB

function salida = oval(C)N=1000;c1=C(2);b=C(1);pi=%pi;B=sqrt(45^2 - b^2);m=-abs(b/B);ab=atan(m);xt=B-c1*cos(ab);yt=-c1*sin(ab);//

salida=[b,c1]//pausedisp([b,c1,ab,m,xt,yt]);xclickendfunction

bs=fsolve([0,61],oval)

-->bs=fsolve([0,61],oval)

column 1 to 5

0.0D+00 6.1D+01 0.0D+00 0.0D+00 - 1.6D+01

column 6

0.0D+00

-----------------------------------------------------------------------

en MATLAB, 1 definimos la funcion

function [salida] = oval(C)%UNTITLED Summary of this function goes here% Detailed explanation goes here

syms b c1N=1000;C=[b,c1] %c1=C(1,2);b=C(1,1);B=sqrt(45^2 - b^2);m=-abs(b/B);ab=atan(m);xt=B-c1*cos(ab);yt=-c1*sin(ab);salida=[b,c1]disp([b,c1,ab,m,xt,yt])

end

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bs=fsolve([0,61],oval)

no sale las parametriza............... y no m itera los valores iniciales

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Dedicated to my recently deceased father.

Literature:

'Theory of Elasticity 1934 Timoshenko-Goodier'

'Mechanics of materials 5th ed. Timoshenko-Gere'

'Steel Structures: Behavior and LRFD. Vinnakota'

'UNE-ENV 1993-1-1:1996: Eurocode 3'

It is a cantilever beam AB with distributed torque loads. The beam is attached at point A.

Given:

**L=5 % [mts]**

**t=0.03 % [mts] 03 cm**

**b=0.4 % [mts] 40 cm**

**hA=0.5 % [mts] 50 cm**

**hB=0.15 % [mts] 15 cm**

**G=70000000 % [kN/m^2]**

**t0=18 % [kN*m/m] t(x)=t0*(L-x)**

**SF=2**

**tau_faliure_admisible=275 % 550 * 2 [MPa]**

See figures for details (x-axis and geometry)

![enter image description here][1]![enter image description here][2]![enter image description here][3]

(1) Obtain the symbolic expresion in terms of (**L ,x, $h_A$, $h_B$, and b**) for Shear Stress $\tau_(max)$

HINT: Use the approximation given by Saint Venant in paragraph 98. Theory of elasticity for laminated cross-sections.

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$\mathcal{T} (x)= G*I_t(x)*\frac{\partial \varphi_x(x)}{\partial x}-E*I_w*\frac{\partial^3\varphi_x }{\partial x^3}$

Only uniform twist on this model will be used

$\mathcal{T} (x)= G*I_t(x)*\frac{\partial \varphi_x(x)}{\partial x}$

According with Venant, to make your life easier:

$\tau_(max) (x)=\frac{\mathcal{T} (x) * e_(max)}{I_t(x)}$

$I_t(x)=\frac{1}{3}*\sum b_i*e_1^3$

(2) Obtain the numerical value in MPa for Shrear Strain $\tau_(max)$

(3) CHALLENGE. Plot $\tau_(max)$ vs length with datas provided !

Remember: **SF=2 & tau_faliure_admisible=275**

(4) Well, solution it is not working..

so we will put a reinforcement plate...

![enter image description here][4]![enter image description here][5]![enter image description here][6]

Calculate length **L*** and thickness **t*** minimum of the reinforcement plateto solve the problem & represent the solution tau_max vs length.

.

.

.

.

**SOL**

(1)

$\tau _(max)=\frac{54*(L-x)}{t^2*(h_A+b-\frac{h_A-h_B}{L}*x)}$

(2)

333.33 MPa

(3)

![enter image description here][7]

(4)

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minimal solution

**L*** must be >1.30 mts

**t*** must be > 2.5 cm

![enter image description here][8]

[1]: https://edxuploads.s3.amazonaws.com/13714977634379855.jpg[2]: https://edxuploads.s3.amazonaws.com/13714979347837238.jpg[3]: https://edxuploads.s3.amazonaws.com/13714980014937602.jpg[4]: https://edxuploads.s3.amazonaws.com/13715003696944239.jpg[5]: https://edxuploads.s3.amazonaws.com/1371500513563529.jpg[6]: https://edxuploads.s3.amazonaws.com/13715005983997798.jpg[7]: https://edxuploads.s3.amazonaws.com/13714999148364526.jpg[8]: https://edxuploads.s3.amazonaws.com/13715013607826315.jpg