Question

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Problem 2: The state of stress at a point is given by 80 20 40 [a] 20 60 10|MPa 40 10 20 (a) Determine the strains using Young's modulus of 100 GPa and Poisson's ratio of 0.25 (b) Compute the strain energy density using these stresses and strains (c) Calculate the principal stresses and principal stress directions (can use the MATLAB command) (d) Write the strains calculated in (a) above in the form of strain matrixel and calculate the principal strains and principal strain directions using MATLAB. (e) Are the principal strain directions and principal stress directions in (c) the same? (f) Show that the principal stresses and principal strains satisfy the constitutive relations (g) Calculate the strain energy density using the principal stresses and strains, and show the strain energy density is the same as in (b)

Answer 1

a) Strain

$\sigma =\begin{bmatrix} 80 & 20 &40\\ 20&60 &10 \\ 40 &10 &20 \end{bmatrix}$

But standard representation of stress matrix is :

Ơi Try Ty:

Also,

Strain in x direction : $\varepsilon _{x} = \frac{\sigma _{x}}{E} - \mu\frac{\sigma _{y}}{E}-\mu\frac{\sigma _{z}}{E}$

80 103 0.25.60 2520 = 60 * 10-5

Similarly , $\varepsilon _{y} = \frac{\sigma _{y}}{E} - \mu\frac{\sigma _{x}}{E}-\mu\frac{\sigma _{z}}{E}$

$\therefore \varepsilon _{y} =\frac{60}{10^{5}} - 0.25\frac{80}{10^{5}}-0.25\frac{20}{10^{5}} = 35 * 10^{-5}$

Similarly, $\varepsilon _{z} = \frac{\sigma _{z}}{E} - \mu\frac{\sigma _{x}}{E}-\mu\frac{\sigma _{y}}{E}$

$\therefore \varepsilon _{z} =\frac{20}{10^{5}} - 0.25\frac{80}{10^{5}}-0.25\frac{60}{10^{5}} = -15 * 10^{-5}$

b) Strain Energy Density

$u_{x}= \frac{1}{2}\sigma_{x} *\varepsilon_{x}$

$u_{x}= \frac{1}{2}(80 *10^{6})*(60*10^{-5})= 24000 J/m^{3} = 24 KJ/m^{3}$

Similarly , $u_{y}= \frac{1}{2}\sigma_{y} *\varepsilon_{y}$

$u_{y}= \frac{1}{2}(60*10^{6})*(35*10^{-5})= 10500 J/m^{3} = 10.5 KJ/m^{3}$

Similarly, $u_{z}= \frac{1}{2}\sigma_{z} *\varepsilon_{z}$

$u_{z}= \frac{1}{2}(20*10^{6})*(-15*10^{-5})= -1500 J/m^{3} = -1.5 KJ/m^{3}$

C) Principal Stress

RESULTS Parameter Value Unit Principal stress-1 (01) Principal stress-2 (o) Principal stress-3 () Max shear stress-1 (Tmax Max shear stress-2 (τmax2) Max shear stress -3 (Tmax3) Von Mises stress (o) 110 50 25 MPa 30 95.4

d)

Shear Strain in x direction ; $\phi _{x}= \frac{\tau_{xy} }{G}$

where , $\tau_{xy} = Shear Stress\ on \ x\ plane\ in\ y \ direction$

But $G = \frac{E}{2(1+\mu )} = \frac{100}{2(1+.25 )} = 40 MPa$

$\therefore \phi _{xz}= 40/40 = 1 = \phi _{zx}$

Similarly , $\therefore \phi _{yz}= 10/40 = 0.4 = \phi _{zy}$

$\therefore \phi _{xy}= 20/40 = 0.5 =\phi _{yx}$

Now, Standard strain matrix representation

$\sigma =\begin{bmatrix} \varepsilon _{_{x}} & \frac{\phi_{xy}}{2} &\frac{\phi_{xz}}{2}\\ \frac{\phi_{yx}}{2}&\varepsilon _{_{y}} &\frac{\phi_{yz}}{2} \\ \frac{\phi_{zx}}{2} &\frac{\phi_{zy}}{2} &\varepsilon _{_{z}} \end{bmatrix}$

$\sigma =\begin{bmatrix} \60 * 10^{-5} & 0.25 &0.5\\ 0.25&35* 10^{-5}&0.2 \\ 0.5 &0.2 &-15* 10^{-5} \end{bmatrix}$