Strength of Materials Test 1

Concept:

1) In Isotropic materials properties in different directions at a point do not vary (e.g. metals & glasses).

For Isotropic materials number of Independent elastic constant in the elastic constant matrix is 2. In such materials E_x, E_y and E_z are all same. Similarly G and μ along all directions are also same.

Hence the independent elastic constants are either E and G or E and μ or G and μ  and the third always depends on the other two defined by the relation :

\(G = \frac{E}{{2\left( {1 + \mu } \right)}}\)

Elastic constant Matrix for isotropic materials:

\(\left[ {\begin{array}{*{20}{c}}{{\varepsilon _x}}\\{{\varepsilon _y}}\\{{\varepsilon _z}}\\{{\gamma _{xy}}}\\{{\gamma _{yz}}}\\{{\gamma _{zx}}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{\frac{1}{E}}&{ - \frac{\mu }{E}}&{ - \frac{\mu }{E}}&0&0&0\\{ - \frac{\mu }{E}}&{\frac{1}{E}}&{ - \frac{\mu }{E}}&0&0&0\\{ - \frac{\mu }{E}}&{ - \frac{\mu }{E}}&{\frac{1}{E}}&0&0&0\\0&0&0&{\frac{1}{G}}&0&0\\0&0&0&0&{\frac{1}{G}}&0\\0&0&0&0&0&{\frac{1}{G}}\end{array}} \right]\;\left[ {\begin{array}{*{20}{c}}{{\sigma _x}}\\{{\sigma _y}}\\{{\sigma _z}}\\{{\tau _{xy}}}\\{{\tau _{yz}}}\\{{\tau _{zx}}}\end{array}} \right]\)

Glass is an example of isotropic material. Hence the number of independent elastic constants is 2.

2) In Orthotropic materials properties in different directions are different and the normal strain does not depend on the shear strain. e.g. Wood

For Orthotropic materials number of Independent elastic constant in the elastic constant matrix is 9. These are \({E_x},\;{E_y},\;{E_z},\;{\mu _x},\;{\mu _y},\;{\mu _z},\;{G_{xy}},\;{G_{xz}},\;{G_{yz}}\).

3) In Anisotropic materials properties in different directions are different and the normal strain depends on the shear strain. The number of independent elastic constants for Anisotropic material is 21. The zero terms in the above matrix won’t be zero.

Important Point:

Concept:

Mathematical definition of strain:

Let u, v, and w are the displacements in x, y, and z-direction respectively.

Then Normal strain is given as:

\({{\epsilon }_{x}}=\frac{\delta u}{\delta x};~{{\epsilon }_{y}}=\frac{\delta v}{\delta y};~{{\epsilon }_{z}}=\frac{\delta w}{\delta z}\)

The Shear strain is given as:

i. \({{\gamma }_{xy}}=\frac{\delta v}{\delta x}+\frac{\delta u}{\delta y}\)

ii. \({{\gamma }_{yz}}=\frac{\delta w}{\delta y}+\frac{\delta v}{\delta z}\)

iii. \({{\gamma }_{zx}}=\frac{\delta u}{\delta z}+\frac{\delta w}{\delta x}\)

Given:

u = 5xy, v = 2xy²

\({{\gamma }_{xy}}=\frac{\partial v}{\partial x}+\frac{\partial u}{\partial y}=\frac{\partial \left( 2x{{y}^{2}} \right)}{\partial x}+\frac{\partial \left( 5xy \right)}{\partial y}\)

γ_xy = 2y² + 5x

Concept:

True Strain: It is given as

\({\varepsilon _t} =\ln\frac{\ell }{{{\ell _0}}}\)

Where, ℓ = final length, ℓ₀ = initial length

Calculation:

Let the initial length be ‘ℓ’ and final length be ℓ’.

Since volume remains constant for the specimen.

∴ Initial volume = Final volume

\(\Rightarrow \frac{\pi }{4}{\left( {2d} \right)^2} \times \ell = \frac{\pi }{4} \times {\left( {1.5\;d} \right)^2} \times \ell {\rm{'}}\)

\(\Rightarrow \frac{{\ell '}}{\ell } = \frac{4}{{2.25}}\)

From the definition of true strain