Stress and Strain Questions

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Very Short Answers Questions [1 Mark Questions]

Ques. Elastic limit is the point

1. Up to which stress is proportional to strain
2. Up to which if the load is removed, original volume and shapes are regained
3. At which elongation takes place without application of additional load
4. None of the above

Ans. The correct answer is b. Up to which if the load is removed, original volume and shapes are regained

Explanation: The elastic limit is the point at which any material acts like it is elastic i.e. it regains its original shapes and volume when the load is removed.

Ques. The dimensional formula of stress is

1. ML^-1T^-2
2. ML^-1T
3. MLT^-2
4. M^-1LT^-2

Ans. The correct answer is a. ML^-1T^-2

Explanation: Stress is given by

Stress = Force/Area

The dimensional formula of force is [MLT^-2]

The dimension formula of the area is [L²]

Therefore the dimensional formula of stress is, [stress] = [MLT^-2]/[L²]

⇒ [Stress] = [ML^-1T^-2]

Ques. Which of the following have well-defined geometrical external shapes?

1. Gases
2. Liquids
3. Amorphous solids
4. Crystalline solids

Ans. The correct answer is d. Crystalline solids

Explanation: Because the atoms and molecules are arranged in a definite geometrical repeating manner throughout the body of the crystal, crystalline solids have well-defined geometrical exterior shapes.

Ques. The lateral strain is

1. The ratio of the length of the body to the tensile force applied on it
2. The ratio of deformation in the area to the original area
3. The ratio of axial deformation to the original length
4. The strain at right angles to the direction of applied load

Ans. The correct answer is d. The strain at right angles to the direction of applied load

Explanation: The lateral strain is the strain at right angles to the applied load direction. The longitudinal strain is followed by the lateral strain.

Ques. What is the stress-strain curve?

1. It is the difference between stress and strain
2. It is the relationship between stress and strain
3. It is the percentage of stress and stain
4. None of the above

Ans. The correct answer is a. It is the relationship between stress and strain

Explanation: The stress-strain curve shows the relationship between stress and strain on a graph. It indicates the change in stress as a function of strain.

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Short Answers Questions [2 Marks Questions]

Ques. Define Stress.

Ans. When a deforming force is applied to a body, then the restoring forces are developed inside the body. The restoring force developed per unit area in a body is called Stress.

Stress = Restoring force/Area

At equilibrium, restoring force is equal to the external deforming force, therefore

Stress = Deforming force/Area

Ques. Define Strain.

Ans. The strain is the relative change in the dimensions of a body resulting from external forces. It is given by the ratio of change in configuration to the original configuration.

Strain = (Change in configuration)/(Original configuration)

It is a dimensionless quantity and has no unit.

Ques. What is Hooke’s Law?

Ans. According to Hooke’s Law, stress is directly proportional to strain i.e.

Stress/Strain = Constant

This constant is known as the elastic modulus or modulus of elasticity.

Hooke’s law i.e. proportionality of stress and strain is obeyed only up to a limited range of force for a given material.

Ques. What is elasticity?

Ans. The ability of a material to resist distorting influences and return to its normal shape and size when that influence or force is removed is referred to as elasticity.

When sufficient loads are applied to solid objects, they will deform. If the material is elastic, the object will return to its original shape and size after removal.

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Long Answers Questions [3 Marks Questions]

Ques. What are the different types of Stress?

Ans. Stress is mainly of three types

• Longitudinal or Normal Stress: When two equal and opposite forces act perpendicular to the surface of an object, then the length of the object may increase or decrease depending on the type of applied force. In this case, the object is under Normal stress. When the force applied increases the length of the body then the stress is known as tensile stress, and when the force applied decreases the length of the body then the stress is known as compressive stress.
• Shear or Tangential Stress: When two equal and opposite forces act tangent to the surface of opposite ends of the object, then one face of the object is displaced with respect to the other face. In this case, the object is under shearing or tangential stress. It is defined as the force acting tangent (parallel) to the surface, divided by the area of the surface in which it acts.
• Volumetric or Bulk Stress: When an object is immersed in a fluid, the fluid exerts uniform pressure everywhere on the surface of the object, resulting decrease in the volume of the object. The stress is now a uniform pressure on all sides of the object. Hence the stress is known as Bulk stress. It is defined as the force acting perpendicular to the surface of the object per unit area of the object.

Ques. What is the difference between stress and strain?

Ans. The following are the differences between stress and strain

Ques. A spherical ball contracts in volume by 0.05%, when subjected to a normal uniform pressure. Calculate the volume strain produced in the spherical ball.

Ans. Let the initial volume of the spherical ball be V₀

When normal uniform pressure is applied, the spherical ball contracts in volume by 0.05% of its original volume i.e.

Change in volume (ΔV) = 0.05% of V₀

⇒ ΔV = (0.05/100)V₀

Volume strain is given by

Volume strain = (Change in volume)/(Original volume) = ΔV/V₀

⇒ Volume strain = [(0.05/100)V₀]/V₀ = 5 x 10^-4

Hence the volume strain produced in the spherical ball is 5 x 10^-4

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Very Long Answers Questions [5 Marks Questions]

Ques. A steel rod 2.0 m long has a cross-sectional area of 0.30 cm². The rod is now hung by one end from a support structure, and a 550 kg milling machine is hung from the rod’s lower end. Determine the stress, strain, and the elongation of the rod.

Ans. Given

• Original length of the rod, L₀ = 2 m
• Area of the cross-section of the rod, A = 0.30 cm² = 0.30 x 10^-4 m²

Since a load of mass 550 kg is attached to the lower end of the rod, therefore force acting on the rod is

F = 550 x 9.8 = 5390 N

The stress produced on the steel rod is given by

Stress = Force(F)/Area(A)

On substituting the values, we get

Stress = 5390/(0.30 x 10^-4)

⇒ Stress = 1.8 x 10⁸ Pa

Also stress on the rod is given by

Stress = Y x strain

Where Y is Young’s modulus for steel rod i.e. Y = 2 x 10¹¹ Pa

⇒ Strain = Stress/Y

⇒ Strain = (1.8 x 10⁸)/(2 x 10¹¹)

⇒ Strain = 9 x 10^-4

Also, strain is the ratio of the change in length (elongation ΔL) of the rod to the original length (L0) i.e.

Strain = ΔL/L₀

Therefore, elongation of the rod is

ΔL = Strain x L₀

⇒ ΔL = 9 x 10^-4 x 2 = 18 x 10^-4 m

⇒ ΔL = 1.8 mm

Ques. Draw the stress-strain curve. What are the various regions on a stress-strain curve?

Ans. The Stress-strain curve is given below

The Stress-strain curve

The various regions on a stress-strain curve are

• Proportional limit: It is the part of the stress-strain curve when Hooke's law is followed. The stress-strain ratio provides us with a proportionality constant known as Young's modulus in this limit. The proportional limit of the graph is represented by point OA.
• Elastic limit: It is the point on the graph when the material returns to its original configuration once the load acting on it has been removed entirely. Beyond this limit, the material does not return to its initial position and begins to exhibit plastic deformation.
• Yield point: The yield point is the point at which a material begins to deform plastically. Permanent plastic deformation occurs after the yield point is crossed. There are two yield points: Upper and lower yield points.
• Ultimate stress point: It is a point that represents the maximum amount of stress that a material can withstand before breaking. 
• Fracture or breaking point: The breaking down of the material occurs at this point on the stress-strain curve.

Ques. A 5 m long aluminium wire (Y = 7 x 10¹⁰ N/m²) of diameter 3 mm supports a 4 kg mass. In order to have the same elongation in copper wire (Y = 12 x 10¹⁰ N/m²) of the same length under the same weight, what will be the diameter of the copper wire?

Ans. The ratio of the tensile stress to that of the tensile strain gives Young’s modulus of a material i.e.

Y = Stress/Strain

But Stress = Force(F)/Area(A) = F/πr²

Where r is the radius of the cross-section of the material.

Also, Strain = Elongation(ΔL)/Original length(L₀)

⇒ Y = (F/πr²)/(ΔL/L₀)

⇒ Elongation, ΔL = (FL₀)/(πr²Y)

But, radius of the cross-section, r = d/2

Where d is the diameter of the cross-section.

⇒ Elongation, ΔL = (4FL₀)/(πd²Y)

Since elongation, force, and original length of both the wire is same, then the above equation becomes

\(\frac{1}{d_{aluminium}^2Y_{aluminium}}=\frac{1}{d_{copper}^2Y_{copper}}\)

⇒ d_copper = \((\frac{d_{aluminium}^2Y_{aluminium}}{Y_{copper}})^{1/2}\)

Given

• Diameter of aluminum wire, d_aluminum = 3 mm = 3 x 10^-3 m
• Young’s modulus of aluminum, Y_aluminum = 7 x 10¹⁰ N/m²
• Young’s modulus of copper, Y_copper = 12 x 10¹⁰ N/m²

On substituting the values, the required diameter of the copper wire can be calculated as

⇒ d_copper = \((\frac{(3 \times 10^{-3})^2 \times (7 \times 10^{10})}{(12 \times 10^{10})})^{1/2}\)

⇒ d_copper = 2.29 x 10^-3 m = 2.29 mm

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