MECHANICAL PROPERTIES OF SOLIDS-QnA

Q1: What is elasticity?

Elasticity is the property of a material by which it regains its original shape and size after the removal of the deforming force, provided the deformation is within the elastic limit.

Q2: Define stress.

Stress is the internal restoring force developed per unit area inside a material when an external force is applied.

Q3: What is strain?

Strain is the ratio of change in dimension to the original dimension of a body due to applied stress.

Q4: State Hooke’s law.

Hooke’s law states that stress is directly proportional to strain within the elastic limit of a material.

Q5: What is Young’s modulus?

Young’s modulus is the ratio of longitudinal stress to longitudinal strain in a material.

Q6: Write SI unit of stress.

The SI unit of stress is pascal (Pa).

Q7: Is strain dimensional or dimensionless?

Strain is dimensionless as it is a ratio of two similar quantities.

Q8: Define elastic limit.

Elastic limit is the maximum stress a material can withstand without undergoing permanent deformation.

Q9: What is bulk modulus?

Bulk modulus is the ratio of volume stress to volume strain in a material.

Q10: Define Poisson’s ratio.

Poisson’s ratio is the ratio of lateral strain to longitudinal strain.

Q11: What type of stress is produced in stretching a wire?

Tensile stress is produced when a wire is stretched.

Q12: Name the stress involved in twisting a rod.

Shear stress is involved in twisting a rod.

Q13: What is elastic fatigue?

Elastic fatigue is the reduction in elastic strength due to repeated loading and unloading of a material.

Q14: What does the slope of stress–strain graph represent?

The slope of the stress–strain graph in elastic region represents Young’s modulus.

Q15: Which material is more elastic: rubber or steel?

Steel is more elastic than rubber because it has a higher Young’s modulus.

Q1: Why are railway tracks provided with gaps?

Railway tracks are provided with gaps to allow thermal expansion during high temperatures and prevent buckling of rails.

Q2: State two factors affecting elongation of a wire.

Elongation depends on the applied force and the original length of the wire. It also depends inversely on cross-sectional area and Young’s modulus.

Q3: Why is steel used in construction of bridges?

Steel has high strength and elasticity, allowing it to bear heavy loads without excessive deformation, making it suitable for bridges.

Q4: What is the significance of elastic limit in design?

Structures are designed to operate below elastic limit to avoid permanent deformation and ensure safety.

Q5: Define shear modulus and write its unit.

Shear modulus is the ratio of shear stress to shear strain. Its SI unit is pascal.

Q6: Why is bulk modulus of liquids large?

Liquids are nearly incompressible, so a large pressure produces very small volume change, resulting in high bulk modulus.

Q7: State two applications of elasticity.

Elasticity is used in shock absorbers and suspension bridges to absorb energy and maintain structural stability.

Q8: Why does a stretched wire store energy?

Energy is stored because work is done against internal restoring forces during stretching.

Q9: What is the relation between stress and strain in elastic region?

Stress is directly proportional to strain in the elastic region, obeying Hooke’s law.

Q10: Why should machines work within elastic limit?

Working within elastic limit ensures the material regains original shape and prevents permanent damage.

Q1: Explain stress–strain curve for a ductile material.

The stress–strain curve initially shows a linear elastic region followed by yield point, plastic deformation region, ultimate stress, and finally fracture, indicating ductile behavior.

Q2: Derive expression for elongation of a wire under load.

Elongation is derived using Young’s modulus by relating stress and strain, showing elongation proportional to force and length, and inversely proportional to area and Young’s modulus.

Q3: Explain elastic potential energy in a stretched wire.

Elastic potential energy is the work done in stretching a wire within elastic limit and is stored as recoverable energy proportional to square of extension.

Q4: Discuss Poisson’s ratio and its significance.

Poisson’s ratio explains lateral contraction during stretching and helps understand material deformation behavior under stress.

Q5: What is elastic fatigue? Explain with example.

Elastic fatigue occurs due to repeated cyclic loading, such as in bridge cables, leading to gradual weakening of material.

Q1: Explain elastic behaviour of materials with stress–strain curve.

The elastic behaviour of materials describes the ability of a material to regain its original shape and size after the removal of an applied deforming force, provided the deformation remains within certain limits. This behaviour is clearly explained with the help of a stress–strain curve, which shows the relationship between the applied stress and the resulting strain.

When a gradually increasing force is applied to a material, stress increases and produces a corresponding strain. In the initial portion of the stress–strain curve, stress is directly proportional to strain. This straight-line region represents the elastic region and obeys Hooke’s law. In this region, the material behaves like an ideal elastic body. The slope of this linear portion gives Young’s modulus, which is a measure of the stiffness of the material. A larger slope indicates a stiffer material that resists deformation more strongly.

The point up to which the material shows this linear behaviour is known as the elastic limit. If the applied stress is removed before crossing this limit, the material completely returns to its original dimensions, showing perfect elastic behaviour. Beyond the elastic limit, the stress–strain curve becomes non-linear and the material enters the plastic region.

In the plastic region, even a small increase in stress produces a large strain, and the material no longer returns to its original shape when the load is removed. This results in permanent deformation. As stress increases further, the material reaches the ultimate stress, which is the maximum stress it can withstand. Beyond this point, the material begins to fracture and finally breaks at the breaking point.

Thus, the stress–strain curve effectively explains elastic behaviour and helps in determining safe working limits of materials used in engineering and construction.

Q2: Describe Young’s modulus and its importance in engineering.

Young’s modulus is a fundamental elastic constant that measures the stiffness of a solid material. It is defined as the ratio of longitudinal stress to the corresponding longitudinal strain produced within the elastic limit of the material. In simple terms, it tells us how much a material resists deformation when it is stretched or compressed along its length. Mathematically, Young’s modulus is expressed as

\[ \begin{aligned} Y &= \frac{\text{Longitudinal stress}}{\text{Longitudinal strain}} \\ &= \frac{F/A}{\Delta L/L} \end{aligned} \]

Here, \(F\) is the applied force, \(A\) is the cross-sectional area of the material, \(L\) is the original length, and \(\Delta L\) is the change in length. A high value of Young’s modulus indicates that the material is very stiff and undergoes very little deformation for a given stress, whereas a low value implies that the material is easily deformable.

The importance of Young’s modulus in engineering lies in its direct role in material selection and structural design. Engineers use this quantity to predict how materials will behave under load and to ensure that structures do not deform excessively during use. For example, materials with high Young’s modulus, such as steel, are preferred in bridges, buildings, and machine components where rigidity and dimensional stability are crucial. On the other hand, materials with lower Young’s modulus are chosen where flexibility is required, such as in springs or shock-absorbing components.

Young’s modulus also helps engineers calculate elongation, stress distribution, and safety margins in structures. By designing components to operate well within the elastic limit, engineers ensure that structures return to their original shape after loading, thereby enhancing safety, durability, and reliability in practical applications.

Q3: Explain applications of elasticity in daily life.

Elasticity plays a vital role in daily life by allowing materials to deform under applied forces and return to their original shape when the forces are removed. This property ensures safety, comfort, and efficiency in many objects and structures we use every day. Elastic behaviour is especially important in situations where materials are repeatedly subjected to forces and must respond without permanent damage.

One of the most common applications of elasticity is in buildings and bridges. Materials such as steel and concrete are chosen because they can withstand large stresses while undergoing only small elastic deformations. This elastic response allows structures to bear loads, vibrations, and even mild earthquakes without collapsing. Engineers design structures so that stresses remain within the elastic limit, ensuring that they return to their original shape after the load is removed.

Elasticity is also crucial in vehicles. Springs and shock absorbers in cars, buses, and trains work on the principle of elasticity. When a vehicle passes over an uneven road, these components compress and expand elastically, absorbing shocks and providing a smooth and comfortable ride. Without elastic suspension systems, vehicles would experience severe jolts, leading to discomfort and mechanical damage.

In daily tools and devices, elasticity is widely used. Rubber bands, springs in watches, weighing machines, and mattresses rely on elastic behaviour for proper functioning. Elasticity is equally important in sports equipment such as balls, rackets, and trampolines, where energy is stored during deformation and released to produce motion. Thus, elasticity is an essential physical property that quietly supports convenience, safety, and performance in everyday life.

Q4: Discuss bulk modulus and its applications.

Bulk modulus is an important elastic constant that describes how a material responds to uniform pressure applied from all directions. It measures the resistance offered by a substance to changes in its volume when subjected to compressive stress. In other words, bulk modulus tells us how incompressible a material is. Materials with a high bulk modulus undergo very small volume changes even under large pressures, while those with a low bulk modulus are easily compressible.

\[ \begin{aligned} K &= \frac{\text{Volume stress}}{\text{Volume strain}} \\ &= -\frac{\Delta P}{\Delta V / V} \end{aligned} \]

In this expression, \( \Delta P \) represents the change in pressure applied to the material, \( V \) is the original volume, and \( \Delta V \) is the resulting change in volume. The negative sign indicates that an increase in pressure causes a decrease in volume. Bulk modulus has the same SI unit as pressure, which is the pascal. Solids and liquids generally have large bulk moduli, while gases have comparatively smaller values because they are highly compressible.

The applications of bulk modulus are significant in both science and engineering. In hydraulic systems such as hydraulic presses and brakes, liquids with a high bulk modulus are used so that pressure applied at one point is transmitted efficiently throughout the fluid. This ensures smooth and effective operation of these devices. Bulk modulus is also important in understanding the behavior of materials under high pressure, such as in deep-sea exploration and geophysical studies where rocks experience enormous compressive forces.

In construction and mechanical design, knowledge of bulk modulus helps engineers select suitable materials that can withstand compressive loads without significant volume change. Thus, bulk modulus plays a crucial role in ensuring stability, safety, and reliability in practical applications involving pressure and compression.

Q5: Explain importance of elastic constants in material selection.

Elastic constants play a crucial role in material selection because they quantitatively describe how a material responds to different types of deformation under applied forces. These constants, namely Young’s modulus, bulk modulus, and shear modulus, help engineers and scientists predict whether a material will stretch, compress, or change shape safely when used in real-life applications. Choosing materials without understanding their elastic constants can lead to excessive deformation, structural failure, or reduced service life.

Young’s modulus is particularly important when rigidity and dimensional stability are required. It helps determine how much a material will elongate or compress under a given load. Materials with high Young’s modulus, such as steel, are preferred in buildings, bridges, and machine parts where minimal deformation is essential. This relationship can be expressed as

\[ \begin{aligned} Y &= \frac{\text{Longitudinal stress}}{\text{Longitudinal strain}} \end{aligned} \]

Bulk modulus becomes significant in applications involving uniform pressure, such as hydraulic systems, underwater structures, and high-pressure vessels. A material with a high bulk modulus undergoes very little change in volume under pressure, ensuring efficient transmission of force and structural stability. Similarly, shear modulus is vital when materials are subjected to twisting or shear forces, as in shafts, bolts, and mechanical joints. It helps engineers ensure that components do not undergo excessive angular deformation during operation.

The importance of elastic constants also lies in ensuring safety and durability. Engineers design structures so that stresses remain well within elastic limits, allowing materials to return to their original shape after unloading. By comparing elastic constants, materials can be selected to balance strength, flexibility, and resistance to deformation. Thus, elastic constants serve as reliable guides for choosing suitable materials that meet functional requirements while maintaining safety and efficiency.