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

Stress-Strain curve for some materials:

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

Concept:

Hooke’s law: As per Hooke’s law, stress is proportional to strain.

i.e. σ ∝ ξ or, σ = ξ E

Where, E = Modulus of Elasticity

For Hooke’s law to be valid:

(a) The material should be homogenous.

(b) The material should be isotropic.

(c) The material should behave in a linearly elastic manner.

Thus for a plane bar with Area ‘A’, Length ‘L’ and Modulus of Elasticity ‘E’,

\(\sigma = \frac{P}{A}\;and\;\xi = \frac{\delta }{\ell }\)

As per Hooke’s law,

∵ σ = ξ E

\(\frac{P}{A} = \frac{\delta }{\ell } \times E \Rightarrow \delta = \frac{{P\ell }}{{AE}}\)

Calculation:

Take moment about A,

T₁a + T₂b = Pl.            (1)

\({\left( {\delta l} \right)_1} = \frac{{{T_1}l}}{{AE}}\) and \({\left( {\delta l} \right)_2} = \frac{{{T_2}l}}{{AE}}\)

By similar triangles,

\(\frac{{{{(\delta l)}_1}}}{a} = \frac{{{{(\delta l)}_2}}}{b}\)

\(\frac{{{T_1}l}}{{AEa}} = \frac{{{T_2}l}}{{AEb}}\)                  (2)

Solving (1) and (2), 

T₁ = \(\frac{{Pla}}{{{a^2} + {b^2}}}\)

T₂ = \(\frac{{Plb}}{{{a^2} + {b^2}}}\)

Concept:

Hooke’s law: As per Hooke’s law, stress is proportional to strain.

i.e. σ ∝ ξ or, σ = ξ E

Where, E = Modulus of Elasticity

For Hooke’s law to be valid:

(a) The material should be homogenous.

(b) The material should be isotropic.

(c) The material should behave in a linearly elastic manner.

Thus for a plane bar with Area ‘A’, Length ‘L’ and Modulus of Elasticity ‘E’,

\(\sigma = \frac{P}{A}\;and\;\xi = \frac{\delta }{\ell }\)

As per Hooke’s law,

∵ σ = ξ E

\(\frac{P}{A} = \frac{\delta }{\ell } \times E \Rightarrow \delta = \frac{{P\ell }}{{AE}}\)

Calculation:

Point B will move downwards by the amount of elongation of CD and BC.

As per Hooke’s law,

\(\delta = \frac{{PL}}{{AE}}\)

FBD of the bar

From the FBD drawn above there is compressive force in member CD and tensile in BC. Hence there will be compression of CD and expansion of BC.

\({\delta _{BC}} = \frac{{PL}}{{AE}} = \frac{{P\left( {L/2} \right)}}{{\frac{\pi }{4}{d^2} \times E}} = \frac{{2\;PL}}{{\pi {d^2}E}}\)

\({\delta _{CD}} = \frac{{ - PL}}{{{A_{CD}}E}} = \frac{{ - PL}}{{\frac{\pi }{4}{{\left( {2d} \right)}^2}E}} = \frac{{ - PL}}{{\pi {d^2}E}}\)     ...(Negative sign indicates compression)

\(\therefore {\delta _B} = {\delta _{BC}} + {\delta _{CD}} = \frac{{2\;PL}}{{\pi {d^2}E}} + \left( {\frac{{ - PL}}{{\pi {d^2}E}}} \right) = \frac{{PL}}{{\pi {d^2}E}}\)

Concept:

The material will expand or contract depending on the material's thermal expansion coefficient.

As long as the material is free to move, the material can expand or contract freely without generating stresses.

Once this material is fixed between two rigid supports, thermal stresses can be created.

The total elongation (δ ) due to thermal stress is calculated as δ  = LαΔT 

As per Hooke’s law,

∵ σ = ξ E

\(\frac{P}{A} = \frac{\delta }{\ell } \times E \Rightarrow \delta = \frac{{P\ell }}{{AE}}\)

Calculation:

The ends A and C are fixed and hence there will be no net elongation.

The increase in length due to temperature rise will be compressed by the reaction developed at A and C. Let the reaction be R_A and R_C respectively.

Drawing the FBD:

∴ For equilibrium, R_A = R_C

Elongation produced due to temperature rise ,  \({δ _1} = {\ell _{AB}}α {\rm{\Delta }}T + {\ell _{BC}}α {\rm{\Delta }}T\)

= 500 × 12 × 10^-6 × 50 + 500 × 12 × 10^-6 × 50

∴ δ₁ = 0.6 mm

Compression due to reaction developed

\({δ _2} = \frac{{{R_A}{L_{AB}}}}{{{A_1}E}} + \frac{{{R_A}{L_{BC}}}}{{A_2\;E}}\; = \frac{{{R_A} \times 500}}{{100 \times 2 \times {{10}^5}}} + \frac{{{R_A} \times 500}}{{400 \times 2 \times {{10}^5}}}\)  = (3.125 × 10^-5)R_A

Since the supports are fixed and non-yielding, the magnitude of δ₁ and δ₂ should be the same.

⇒ 0.6 = (3.125 × 10^-5) R_A

⇒ R_A = 19.2 kN = 19200 N

∴ Maximum stress will occur in the bar with lesser area

Concept:

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

Shear strain \(\left( \gamma \right)=\frac{τ }{G}\)

Given:

E = 2 × 10⁵ MPa, μ = 0.3, τ  = 20 MPa

\(\therefore G=\frac{2\times {{10}^{5}}}{2\left( 1+0.3 \right)}=0.769\times {{10}^{5}}\)

F_th = F_bulcking

\(\alpha {\rm{\Delta }}TEA = \frac{{{\pi ^2}EI}}{{L_e^2}}\)

\(\alpha {\rm{\Delta }}TE.\frac{\pi }{4}{d^2} = \frac{{{\pi ^2}E.\pi {d^4}}}{{64\;L_e^2}}\)

\({\rm{\Delta }}T = \left( {\frac{{{\pi ^2}}}{{16E\alpha L_e^2}}} \right){d^2}\)

Concept:

If ϵ_x, ϵ_y, ϵ_z are the longitudinal stress along x-axis, y-axis and z-axis respectively.

\({e_x} = \frac{{{\sigma _x}}}{E} - \mu \times \frac{{{\sigma _y}}}{E} - \mu \times \frac{{{\sigma _z}}}{E}\)

\({e_y} = \frac{{{\sigma _y}}}{E} - \mu \times \frac{{{\sigma _x}}}{E} - \mu \times \frac{{{\sigma _z}}}{E}\)

\({e_x} = \frac{{{\sigma _z}}}{E} - \mu \times \frac{{{\sigma _x}}}{E} - \mu \times \frac{{{\sigma _y}}}{E}\)

Calculation:

\({{\rm{\sigma }}_{\rm{y}}} = \frac{{{{\rm{P}}_{\rm{x}}}}}{{{{\rm{A}}_{\rm{x}}}}} = \frac{{10 \times {{10}^3}}}{{50 \times 10}} = 20~{\rm{N}}/{\rm{m}}{{\rm{m}}^2}\)

\(\therefore {\epsilon_{\rm{y}}} = \frac{{20}}{{2 \times {{10}^5}}} - 0.35 \times \frac{{66.67}}{{2 \times {{10}^5}}} - 0.35 \times \frac{{20}}{{2 \times {{10}^5}}} = - 5.16725 \times {10^{ - 5}}\)