Poisson’s Ratio & Relation between the Elastic Modulii14 min read

POISSON’S RATIO

Let us assume If a body is subjected to a load ( Tensile), Due to this load the dimension of this body changes. its length increased while the area of the body decreased. The Ratio of this change in the length to the original length is known as Linear or Primary Strain and the Ratio of the corresponding cross-section width and original width is called secondary or transverse strain. The linear strain will be tensile and transverse strain will be a compressive strain.

“The ratio of Lateral or transverse strain to the linear strain known as Poisson’s Ratio.”

i.e. Poisson’s ratio,

 \boxed{ \mu = \frac{\mbox{Lateral Or Transverse Strain }}{\mbox{Linear Or Primary Strain }} = \frac{1}{m}}

where m is a constant and its value varies between 3 to 4 for different materials.

Relation Between Young’s Modulus “E” & Modulus of Rigidity “C”

ABCD is a solid cube subjected to a shearing force F. Let τ be the shear stress produced in the faces BC and AD due to this shear force. The complementary shear stress consequently produced in the faces BA and CD be τ too. By the shearing force, the cube deformed WXYZ TO WX’Y’Z as W moved to W’ and Y moved to Y’.relation EC

 \mbox{Shear strain} = \phi = \frac{YY'}{YZ}

Also

 \mbox{ Shear Strain } = \frac{\tau}{C}

 \therefore \frac{YY'}{YZ} = \frac{\tau}{C}      ……(1)

On the diagonal WY’, draw a perpendicular YN from Y.

Now diagonal strain

 \Rightarrow \frac{NY'}{WN} = \frac{NY'}{WY}     ….(2)

 NY' = YY' \cos 45^{\circ} = \frac{YY'}{\sqrt{2}}

“∠WY’Z is assume to be equal to ∠WYZ because YY’ is very small”

and               WY = YZ \times \sqrt{2}

Putting the value of WY in equation (2), we get

 \mbox{Diagonal Strain} = \frac{YY'}{\sqrt{2}YZ \times \sqrt{2}} = \frac{YY'}{2 YZ}

But               \frac{YY'}{YZ} = \frac{\tau}{C}

 \therefore \mbox{ Diagonal strain} = \frac{\tau}{2C} = \frac{ \sigma_n}{2C}    ….(3)

where  \sigma_n is the normal stress due to shear stress τ. The net strain in the direction of diagonal WY

 = \frac{\sigma_n}{E} + \frac{\sigma_n}{mE}

” The diagonal WY and XZ have normal tensile and compressive stress  \sigma_n , respectively”

 = \frac{\sigma_n}{E} \left [1 + \frac{\sigma_n}{mE} \right ]       …….(4)

Comparing equation (3) and (4), we get

⇒        \frac{\sigma_n}{2C} = \frac{\sigma_n}{E}\left [ 1 + \frac{1}{m} \right ]

 \boxed{E = 2C \left [ 1 + \frac{1}{m} \right ]}

Relation Between Young’s Modulus “E” & Bulk Modulus of Elasticity “K”

if a solid cube is subjected to normal comressive stress  \sigma_n on all 6 faces. The direct compressive strain in each axis be  \frac{\sigma_n}{E} and lateral tensile strain in other axis be  \frac{\sigma_n}{mE} .Relation EK

∴ Net compressive stress in each axis

⇒           = \frac{\sigma_n}{E} - \frac{\sigma_n}{mE} - \frac{\sigma_n}{mE}

⇒           = \frac{\sigma_n}{E} \left[1- \frac{2}{m} \right]

Volumetric strain ( e_v ) in each axis will be,

⇒        e_v = 3 \times \mbox{linear strain} = 3 \times \frac{\sigma_n}{E} \left[ 1 - \frac{2}{m} \right]

But              e_v = \frac{\sigma_n}{K}

⇒  ∴             \frac{\sigma_n}{K} = \frac{3 \sigma_n}{E} \left[ 1 - \frac{2}{m} \right]

OR

⇒            \boxed{ E = 3K \left[ 1 - \frac{2}{m} \right] }

Relation Between Modulus of Rigidity “C” & Bulk Modulus of Elasticity “K”

We have now two equations now,

 E = 2C \left[ 1 + \frac{1}{m} \right]    …(1)

and

 E = 3K \left[ 1- \frac{2}{m}\right]     …(2)

from equation (1), we have

 m = \frac{2C}{E-2C}

Putting the value of m into equation (2), we have

 E = 3K \left[ 1- \frac{2}{2C/ \left( E- 2C\right)}\right]

⇒      E = 3K \left[ 1 - \frac{E-2C}{C}\right]

⇒      \frac{E}{3K} = \frac{C - E + 2C}{C} = \frac{-E +3C}{C}

⇒      \frac{E}{3K} = 3 - \frac{E}{C}

⇒      \frac{E}{3K} + \frac{E}{C} = 3

⇒      \frac{EC + 3KE}{3KC} =3

⇒      EC + 3 KE = 9 KC

⇒     E \left( C + 3K\right) = 9C

⇒     \boxed{E = \frac{9KC}{3K +C}}

 

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