Question

A load of 120 KN is applied on a bar of 200mm diameter .The bar which is 400mm long is elongated by 0.7. Determine the modulus of elasticity of bar material if poisson's ratio is 0.3​

Answer 1

Let the number = X

• Applied on 200 =x

• Determine = 400 mm

• 0.3 and 0.4 point 0.10.5 x coconut Karenge X2 barabar Mein square to barabar mein
• 600 answers OK

Answer 2

Concept

The elasticity of a material is determined by its modulus of elasticity, commonly referred to as elastic modulus or simply modulus. A material's resistance to elastic, or non-permanent, deformation is measured by its elastic modulus.

Given

A load of $120 KN$ is applied on a bar of $200mm$ diameter .The bar which is $400mm$ long is elongated by $0.7$.

To Find

We have to find the modulus of elasticity of bar material.

Solution

According to the problem,

Poisson's ratio = dtrans/daxial.

$0.3=ddiameter/0.7$

⇒Δdiameter=$0.7*0.3=0.21$

Modulus of elasticity= stress/strain

$=\frac{120000}{\pi*(0.1)^{2} } \frac{0.4}{0.0007} =\frac{36000000000*7}{22*7}$$=1636363636.63$

Hence, Modulus of elasticity is $1636363636.63$

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