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A steel rod has a radius of 10 mm and a length of 1.0 m. A force stretches it along its length and produce a strain of 0.16%. Young’s modulus of the steel is 2.0 x 10¹¹ Nm⁻². What is the magnitude of the force stretching the rod?
The expression of stress, strain, and Young's modulus is as follows: Stress = Y × Strain where, - Y = 2.0 × 10¹¹ N/m² - Strain = 0.16% = 0.0016 Stress can also be defined as follows: Stress = Force / Area Cross-sectional area of the rod A = πr² = π × (10 × 10⁻³)² = π × 10⁻⁴ m² Put it into the equatiRead more
The expression of stress, strain, and Young’s modulus is as follows:
Stress = Y × Strain
where,
– Y = 2.0 × 10¹¹ N/m²
– Strain = 0.16% = 0.0016
Stress can also be defined as follows:
Stress = Force / Area
Cross-sectional area of the rod
A = πr² = π × (10 × 10⁻³)² = π × 10⁻⁴ m²
Put it into the equation for stress:
Y × Strain = Force / Area
Force =
Force = Y × Strain × A
Put the values:
Force = (2.0 × 10¹¹) × (0.0016) × (π × 10⁻⁴)
Force = 100.5 × 10³ N ≈ 100 kN
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A cube is subjected to a uniform volume compression. If the side of the cube decreases by 2% the bulk strain is
The bulk strain can be described as the fractional alteration in the volume of a body. In this case of uniform compression for the cube, this is computed with: Bulk Strain = 3 × Linear Strain Since the length of the cube has been compressed by 2%, then this gives a value of: Linear Strain = ΔL / L =Read more
The bulk strain can be described as the fractional alteration in the volume of a body. In this case of uniform compression for the cube, this is computed with:
Bulk Strain = 3 × Linear Strain
Since the length of the cube has been compressed by 2%, then this gives a value of:
Linear Strain = ΔL / L = 0.02
This therefore, implies that,
Bulk Strain = 3 × 0.02 = 0.06
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Minimum and maximum values of possion’s ratio for a metal lies between
Most metals have a Poisson's ratio in the range of 0 to 0.5. A value of 0 indicates no lateral contraction when stretched, and 0.5 represents a material that maintains constant volume under deformation. This is theoretically true for perfectly incompressible materials.
Most metals have a Poisson’s ratio in the range of 0 to 0.5. A value of 0 indicates no lateral contraction when stretched, and 0.5 represents a material that maintains constant volume under deformation. This is theoretically true for perfectly incompressible materials.
See lessThe work done per unit volume in deforming a body is given by
Work done per unit volume in deforming a body is given by the formula: Work done per unit volume = 1/2 × Stress × Strain This is derived from the area under stress-strain curve for elastic deformation, which forms a triangle. Click here for more: https://www.tiwariacademy.com/ncert-solutions/class-1Read more
Work done per unit volume in deforming a body is given by the formula:
Work done per unit volume = 1/2 × Stress × Strain
This is derived from the area under stress-strain curve for elastic deformation, which forms a triangle.
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If in a wire of Young’s modulus Y, longitudinal strain X is produced, then the value of potential energy stored in its unit volume will be
Formula: The potential energy per unit volume of a taut wire may be given as follows: Energy per unit volume = ½ × Stress × Strain For a wire: Tension = Stress Stress = Y × Strain Substituting Stress to above equation: Potential Energy per unit volume = 1/2 x (Yx X) × X= 0.5Yx² Click for more: hRead more
Formula:
The potential energy per unit volume of a taut wire may be given as follows:
Energy per unit volume = ½ × Stress × Strain
For a wire:
Tension = Stress
Stress = Y × Strain
Substituting Stress to above equation:
Potential Energy per unit volume = 1/2 x (Yx X) × X= 0.5Yx²
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