Elastic Limit and Constants MCQ Quiz in मल्याळम - Objective Question with Answer for Elastic Limit and Constants - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക
Last updated on Mar 10, 2025
Latest Elastic Limit and Constants MCQ Objective Questions
Top Elastic Limit and Constants MCQ Objective Questions
Elastic Limit and Constants Question 1:
If for a given material E = 2G, then the bulk modulus K will be
Answer (Detailed Solution Below)
Elastic Limit and Constants Question 1 Detailed Solution
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
Elastic Limit and Constants Question 2:
If the stress acting on a point is in the three dimensions, then what is the number of components in a stress tensor required for defining that stress?
Answer (Detailed Solution Below)
Elastic Limit and Constants Question 2 Detailed Solution
Explanation:
For static equilibrium τxy = τyx, τxz = τzx, τzy = τyz 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. σx, σy, σz, τxy, τyx, τxz,τzx, τzy, τyz.
But number of independent components will be 6 only i.e. σx, σy, σz, τxy (= τyx ), τxz (= τzx), τzy (= τyz).
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.
Elastic Limit and Constants Question 3:
Statement (I): There are two independent elastic constants for an isotropic material.
Statement (II): All metals at micro-level are isotropic
Answer (Detailed Solution Below)
Elastic Limit and Constants Question 3 Detailed Solution
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.
Elastic Limit and Constants Question 4:
_______ .has the highest value of Poisson's ratio
Answer (Detailed Solution Below)
Elastic Limit and Constants Question 4 Detailed Solution
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.
Elastic Limit and Constants Question 5:
The ratio of modulus of rigidity and modulus of elasticity (G/E) for any elastic isotropic material is:
Answer (Detailed Solution Below)
Elastic Limit and Constants Question 5 Detailed Solution
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.
Elastic Limit and Constants Question 6:
How many Independent elastic constants of a non-homogeneous, non-isotropic material will be?
Answer (Detailed Solution Below)
Elastic Limit and Constants Question 6 Detailed Solution
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.
Elastic Limit and Constants Question 7:
A block of volume V mm3 is subjected to hydrostatic pressure p MPa. Modulus of elasticity is E GPa and Poisson’s ratio v = 0.5. Which statement is true about the block?
Answer (Detailed Solution Below)
Elastic Limit and Constants Question 7 Detailed Solution
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.
Elastic Limit and Constants Question 8:
The independent elastic constants for a homogeneous and isotropic material are
Answer (Detailed Solution Below)
Elastic Limit and Constants Question 8 Detailed Solution
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.
Elastic Limit and Constants Question 9:
The ratio of Young’s modulus of high tensile steel to that of mild steel is about:
Answer (Detailed Solution Below)
Elastic Limit and Constants Question 9 Detailed Solution
Concept:
by Hooke's law,
\(E=\frac{\sigma }{\epsilon }\)
Where,
E = Young’s modulus
σ = uniaxial stress
ϵ = strain.
Sr no. |
Materials |
Young's modulus(E) in GPa |
1. |
Soft Rubber |
10 |
2. |
Diamond |
1200 |
3. |
Steel |
200 |
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
Elastic Limit and Constants Question 10:
Beams composed of more than one material, rigidly connected together so as to behave as one piece, are known as
Answer (Detailed Solution Below)
Elastic Limit and Constants Question 10 Detailed Solution
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.
While the compound beam is a beam of different element assembled together to form a single unit.