Elastic Limit and Constants MCQ Quiz in मल्याळम - Objective Question with Answer for Elastic Limit and Constants - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Explanation:

For static equilibrium τ_xy = τ_yx, τ_xz = τ_zx, τ_zy = τ_yz resulting in six independent scalar quantities. These six scalars can be arranged in a 3 x 3 matrix, giving us a stress tensor.

\(\sigma = {\sigma _y} = \left[ {\begin{array}{*{20}{c}} {{\sigma _x}}&{{\tau _{yx}}}&{{\tau _{zx}}}\\ {{\tau _{xy}}}&{{\sigma _y}}&{{\tau _{zy}}}\\ {{\tau _{xz}}}&{{\tau _{yz}}}&{{\sigma _z}} \end{array}} \right]\;\)

So total of 9 components are required in stress tensor for defining that stress. i.e. σ_x, σ_y, σ_z, τ_xy, τ_yx, τ_xz,τ_zx, τ_zy, τ_yz.

But number of independent components will be 6 only i.e. σ_x, σ_y, σ_z, τ_xy (= τ_yx ), τ_xz (= τ_zx), τ_zy (= τ_yz).

Mistake PointsPlease read the question properly, when it will be asked about independent stress it will be 6, but when it will ask about the number of stress components then it will be 9.

Concept:

The relation between E, G, μ, and k will be,

E = 2G (1 + μ),

E = 3K (1 – 2μ)

Where E = Young’s modulus, G = Modulus of rigidity, μ = Poisson ratio, K = Bulk modulus of elasticity.

Calculation:

Given:

E = 2G

→ 2G = 2G (1 + μ)

→ μ = 0

Now

E = 3K (1 – 2μ)

E = 3K (1 – 2 × 0)

→ E = 3K

K = E/3

Concepts:

• Isotropic Material: Isotropic materials have identical material properties in all directions at every given point. When a specific load is applied at any point of an isotropic materials, it will exhibit the same strength, stress, strain, young’s modulus and hardness in the x, y or z-axis direction. Examples: metals, glasses.
• Almost every metal exhibit anisotropic behaviour at micro level because all single crystal systems are anisotropic with respect to mechanical properties. Although there are a few exceptional cases where metals exhibit isotropic behaviour at micro level. But due to random localized distribution of anisotropic behaviour at micro level, they nullify each other and the net effect at macro level is isotropic behaviour.

∴ Statement II is incorrect.

• The number of independent elastic constants for isotropic materials is 2. It means to fully define the elastic behaviour of an isotropic material only 2 elastic constants are good enough. Because among modulus of elasticity (E), modulus of rigidity (G), bulk modulus (K) and Poisson’s ratio (μ), if any two constants are known for any linear elastic and isotropic material than rest two can be derived.

∴ Statement I is correct.

Important Points:

Number of independent elastic constant for isotropic material is 2.

Number of independent elastic constant for orthotropic material is 9.

Number of independent elastic constant for anisotropic material is 21.

Explanation:

Poisson's ratio is the ratio of transverse strain to longitudinal strain

\(\rm{\nu = - \frac{{{\epsilon_{lateral}}}}{{{\epsilon_{longitudinal}}}}}\)

Most materials have Poisson's ratio values ranging between 0.0 and 0.5 (mostly 0.33).

A perfectly incompressible material has a value of 0.5.

Stainless steel: 0.30–0.31,

Steel: 0.27–0.30,

Cast iron:0.21–0.26,

Cork: 0.0,

Rubber: 0.5.

∴ Poisson’s ratio of rubber lies between 0.45 to 0.50.

Additional Information

For a rigid body, the value of Poisson’s ratio is zero.

A zero Poisson’s ratio means that there is no transverse deformation resulting from an axial strain.

• Most materials have Poisson's ratio values ranging between 0.0 and 0.5.
• A perfectly incompressible material deformed elastically at small strains would have a Poisson's ratio of exactly 0.5.
• Most steels and rigid polymers when used within their design limits (before yield) exhibit values of about 0.3, increasing to 0.5 for post-yield deformation which occurs largely at constant volume.
• Rubber has a Poisson ratio of nearly 0.5.
• Cork's Poisson ratio is close to 0, showing very little lateral expansion when compressed.

Explanation: 

The relation between Young’s modulus (E), rigidity modulus (G), and poison’s ratio (μ) is given by

E = 2G (1 + μ)

\(\frac{G}{E} = \frac{1}{2(1\ +\ μ )}\)

We know that The value of Poisson's ratio lies in the range of 0 to 0.5.

At μ = 0,

\(\frac{G}{E} = \frac{1}{2}\)

At μ = 0.5,

\(\frac{G}{E} = \frac{1}{2(1\ +\ 0.5 )}\)

\(\frac{G}{E} = \frac{1}{3}\)

Hence, the G/E value will be less than 1/2 and more than 1/3. 

Explanation:

Isotropic Material: 

• Isotropic materials have identical material properties in all directions at every given point.
• When a specific load is applied at any point of isotropic materials, it will exhibit the same strength, stress, strain, young’s modulus, and hardness in the x, y, or z-axis direction.
• For an isotropic, homogeneous, and elastic material obeying Hook's law, the number of independent elastic constants is 2.

Examples: metals, glasses.

Anisotropic material

• A material is said to be anisotropic if it has different properties in each direction. They show non- homogeneous behavior.
• There are 21 independent elastic constants required to define the stress-strain relationship for an anisotropic material.

Orthotropic material

• A material is said to be orthotropic if it has three different properties in three mutually perpendicular directions.
• There are 9 independent elastic constants required to define the stress-strain relationship for orthotropic material.

Concept:

E = 3K (1 - 2ν)

where, E = Modulus of elasticity, K = Bulk modulus, v = poisson's ratio.

Calculation:

Given:

E = E GPa, v = 0.5, and P =  p MPa.

Placing these values in the above equation

E = 3K (1 - 2ν)

K = ∞

The Bulk Modulus of Elasticity is given by:

\(K = \frac{P}{{\frac{{ - {\rm{Δ }}V}}{V}}}\)

we have Pressure and Volume value given as finite so to make K = ∞, Change in volume, ΔV = 0.

ΔV = 0 signifies the block is perfectly incompressible.

Important Points

Other relationships between various elastic constants are:

E = 2G (1 + ν)

\({{E}} = \frac{{9{{KG}}}}{{3{{K}}\; + \,{{G}}\;}}\)

K = Bulk modulus, G = Shear modulus, or Modulus of rigidity.

Explanation:

Isotropic Material: 

• Isotropic materials have identical material properties in all directions at every given point.
• When a specific load is applied at any point of isotropic materials, it will exhibit the same strength, stress, strain, young’s modulus, and hardness in the x, y, or z-axis direction.

Examples: metals, glasses.

• Almost every metal exhibits isotropic behaviour at the micro-level because all single-crystal systems are isotropic with respect to mechanical properties.
• There are a few exceptional cases where metals exhibit anisotropic behaviour at the micro-level.
• Random localized distribution of anisotropic behaviour at the micro-level, nullify each other and the net effect at the macro level is isotropic behaviour.

For an isotropic, homogeneous, and elastic material obeying Hook's law, the number of independent elastic constants is 2.

It means to fully define the elastic behaviour of isotropic material only 2 elastic constants are good enough.

E = 2G (1 + μ) = 3K(1 - 2μ)

As seen in the above equations there are a total of 4 elastic constants.

If we know two of them, then we can find other constants also.

∴ There are 2 independent elastic constants required to define the stress-strain relationship for isotropic material. 

Important PointsIn the homogenous isotropic, elastic material, there are only two independent elastic constants. These constants may be any two of the E, G, ν, and K.

Additional Information

Orthotropic material 

• A material is said to be orthotropic if it has three different properties in three mutually perpendicular directions.
• There are 9 independent elastic constants required to define the stress-strain relationship for orthotropic material.

Anisotropic material

• A material is said to be anisotropic if it has different properties in each direction.
• There are 21 independent elastic constants required to define the stress-strain relationship for an anisotropic material.

Concept:

by Hooke's law,

\(E=\frac{\sigma }{\epsilon }\)

Where,

E = Young’s modulus

σ = uniaxial stress

ϵ = strain.

Explanation:

The young’s modulus of high tensile steel and mild steel is the same i.e 200 GPa

∴ \(\frac{{{E}_{high~tensile~steel}}}{{{E}_{mild~steel}}}=\frac{200}{200}=1\)

Important points:

Ecast iron = ½  of Esteel

Ealuminium = ⅓ of Esteel

Explanation:

Composite Beam is the one in which the beam is made up of two or more material and rigidly connected together in such a way that they behave as one piece.

In the composite beam, there is a common neutral axis through the centroid of the equivalent homogeneous section.