Question

A piece of copper having a rectangular cross-section of \[{\mathbf{15}}.{\mathbf{2mm}}{\text{ }}{\mathbf{x}}{\text{ }}{\mathbf{19}}.{\mathbf{1mm}}\] is pulled in tension with\[{\mathbf{44}},{\text{ }}{\mathbf{500}}{\text{ }}{\mathbf{N}}{\text{ }}{\mathbf{force}}\], producing only elastic deformation. Calculate the resulting strain? (Modulus of elasticity of copper,).

Answer 1

Hint: When an elastic material is deformed due to an external force, it experiences internal resistance to the deformation and restores it to its original state if the external force is no longer applied.

Complete step by step solution:
There are various types of elastic modulus:

Young’s modulus of elasticity: Young’s modulus of elasticity is a mechanical property that measures the stiffness of a solid material. It defines the relationship between stress (force per unit area) and strain (proportional deformation) in a material.

Shear modulus of elasticity: The shear modulus of elasticity is concerned with the deformation of a solid when it experiences a force parallel to one of its surfaces while its opposite face experiences an opposing force (such as friction).

Bulk modulus of elasticity: The bulk modulus of elasticity of a substance is a measure of how resistant to compression that substance is. It is defined as the ratio of the infinitesimal pressure increase to the resulting relative decrease of the volume.

Given:
Length of the copper \[ = {\text{ }}{\mathbf{19}}.{\mathbf{1}}{\text{ }}{\mathbf{mm}}\]\[ = {\text{ }}{\mathbf{1}}.{\mathbf{9}}{\text{ }}{\mathbf{x}}{\text{ }}{\mathbf{1}}{{\mathbf{0}}^{ - 3}}{\text{ }}{\mathbf{m}}\]
Breadth of the copper \[ = {\text{ }}{\mathbf{1}}5.2{\text{ }}{\mathbf{mm}}\] \[ = {\text{ }}{\mathbf{1}}5.2{\text{ }}{\mathbf{x}}{\text{ }}{\mathbf{1}}{{\mathbf{0}}^{ - 3}}{\text{ }}{\mathbf{m}}\]
Area of copper piece
$Area = Length\,\times\,\,breadth$
$ = 19.1\times\,{10^{ - 3}}$$\times\,\,15.2\times\,{10^{ - 3}}$
$ = 2.9\times\,{10^{ - 4\,}}\,{m^2}$
Tension on copper, \[{\mathbf{T}}{\text{ }} = {\text{ }}{\mathbf{44500}}{\text{ }}{\mathbf{N}}\]

Modulus of elasticity of copper,
As we know that:
\[{\mathbf{Modulus}}{\text{ }}{\mathbf{of}}{\text{ }}{\mathbf{elasticity}}{\text{ }} = {\text{ }}{\mathbf{stress}}{\text{ }}/{\text{ }}{\mathbf{strain}}\;\]
$\eta = \dfrac{{\dfrac{f}{a}}}{{strain}}$
$strain = \dfrac{f}{{a\eta }}$
$ = \dfrac{{44500}}{{2.9\times\,{{10}^{ - 4}}}}\times\,42\times\,{10^9}$
$ = 3.65\times{10^{ - 5}}$

Therefore, the correct answer is $3.65 \times 10^{-5}$.

Note: Stress: It is the ratio of the internal force (F) produced when the substance is deformed, to the area (A) over which this force acts and the unit of stress is N/m2.
Type of stress:
1. Normal stress
2. Tangential stress

Strain: It is the ratio of the change in size or shape to the original size or shape. It has no dimensions; it is just a number.

Types of strain:
1. Longitudinal strain
2. Volumetric strain