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Physics Part 2

Mechanical Properties of Solids

Mechanical Properties of Solids

Action PlanNotesQuestionsMock-Test
Study Action Plan (Sorted by Difficulty)
Yield Strength, Breaking Stress and Safety Factor
Accuracy: 20%
Total Attempted5
Total Correct1
Total Incorrect4
Difficulty Level:
High
Stress and Strain
Accuracy: 40%
Total Attempted15
Total Correct6
Total Incorrect9
Difficulty Level:
High
Elastic Potential Energy
Accuracy: 45%
Total Attempted11
Total Correct5
Total Incorrect6
Difficulty Level:
High
Shear Modulus and Shearing Deformation
Accuracy: 50%
Total Attempted8
Total Correct4
Total Incorrect4
Difficulty Level:
Medium
Stress-Strain Curve
Accuracy: 50%
Total Attempted10
Total Correct5
Total Incorrect5
Difficulty Level:
Medium
Elastic Moduli
Accuracy: 53%
Total Attempted15
Total Correct8
Total Incorrect7
Difficulty Level:
Medium
Bulk Modulus and Compressibility
Accuracy: 55%
Total Attempted11
Total Correct6
Total Incorrect5
Difficulty Level:
Medium
Young’s Modulus and Longitudinal Deformation
Accuracy: 69%
Total Attempted13
Total Correct9
Total Incorrect4
Difficulty Level:
Medium
Poisson's Ratio
Accuracy: 75%
Total Attempted4
Total Correct3
Total Incorrect1
Difficulty Level:
Low
Applications of Elastic Behavior
Accuracy: 82%
Total Attempted17
Total Correct14
Total Incorrect3
Difficulty Level:
Low
Hooke's Law
Accuracy: 92%
Total Attempted13
Total Correct12
Total Incorrect1
Difficulty Level:
Low

Learning Objectives

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Learning Objectives

  • Understand the definitions and differences between stress and strain, including tensile, compressive, shearing, and hydraulic stress.
  • Calculate stress and strain in various materials using Hooke's Law and identify the modulus of elasticity as the proportionality constant.
  • Differentiate between Young's modulus, shear modulus, and bulk modulus, and their relevance to the elastic behavior of materials under different types of stress.
  • Analyze stress-strain curves to identify regions of elastic and plastic deformation, yield points, and ultimate tensile strength.
  • Explain Poisson's ratio and its significance in describing the deformation characteristics of materials.
  • Determine the elastic potential energy stored in materials when work is done against inter-atomic forces during deformation.
  • Apply knowledge of elastic behavior to practical engineering problems, such as the design of beams, columns, and bridges.
  • Define bulk modulus and compressibility, and calculate the effects of hydraulic stress on volume strain.
  • Solve problems involving Young’s modulus and longitudinal deformation, including series-wire and load-extension scenarios.
  • Examine shearing stress, shearing strain, angular displacement, and shear modulus in practical applications.
  • Evaluate yield strength, breaking stress, and safety factors in the design of ropes and cables for safe load management.

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Revision Notes & Summary

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Chapter Notes

Stress and Strain

  • Stress is defined as the restoring force per unit area. It is given by the formula Stress=FA\text{Stress} = \frac{F}{A} where FF is the force applied and AA is the area.
  • Strain is the fractional change in dimension, defined as the ratio of change in dimension to the original dimension.
  • Types of stress:
    • Tensile Stress: Associated with stretching.
    • Compressive Stress: Associated with compression.
    • Shearing Stress: Caused by forces applied parallel to the surface.
    • Hydraulic Stress: Caused by forces applied perpendicular to the surface.

Hooke's Law

  • States that for small deformations, stress is directly proportional to strain: Stress∝Strain\text{Stress} \propto \text{Strain}
  • The proportionality constant is the modulus of elasticity.

Elastic Moduli

  • Young's Modulus (Y): Describes tensile or compressive elasticity. Y=Tensile StressLongitudinal StrainY = \frac{\text{Tensile Stress}}{\text{Longitudinal Strain}}
  • Shear Modulus (G): Describes shear elasticity. G=Shearing StressShearing StrainG = \frac{\text{Shearing Stress}}{\text{Shearing Strain}}
  • Bulk Modulus (B): Describes volumetric elasticity. B=−pΔV/VB = -\frac{p}{\Delta V/V}

Stress-Strain Curve

  • Illustrates the relationship between stress and strain.
  • Key points:
    • Proportional Limit: End of linear elasticity.
    • Yield Point: Stress at which material begins to deform plastically.
    • Ultimate Tensile Strength: Maximum stress material can withstand.
    • Fracture Point: Point of material failure.

Poisson's Ratio

  • Ratio of lateral strain to longitudinal strain in a stretched wire.
  • Poisson's Ratio (ν\nu) is a dimensionless quantity.

Elastic Potential Energy

  • Energy stored in a material when work is done against inter-atomic forces during deformation.
  • For a wire, U=12×Stress×Strain×VolumeU = \frac{1}{2} \times \text{Stress} \times \text{Strain} \times \text{Volume}

Applications of Elastic Behavior

  • Essential in engineering design for structures like beams, columns, and bridges.
  • Considerations include material strength, Young's modulus, and safety factors.

Bulk Modulus and Compressibility

  • Bulk Modulus (B): Ratio of hydraulic stress to volume strain.
  • Compressibility (k): Reciprocal of bulk modulus, indicating how compressible a material is.

Young’s Modulus and Longitudinal Deformation

  • For elongation/compression of wires: Y=FLAΔLY = \frac{FL}{A\Delta L}

Shear Modulus and Shearing Deformation

  • Describes the response of a material to shearing stress: G=FAθG = \frac{F}{A\theta}

Yield Strength, Breaking Stress and Safety Factor

  • Yield Strength: Stress at which material begins to deform plastically.
  • Ultimate Tensile Strength: Maximum stress material can withstand before breaking.
  • Safety Factor: Ratio of breaking stress to actual stress applied, ensuring safety in design.
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Practice Test – MCQs, True/False

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Multiple Choice Questions

A.

2×10−42 \times 10^{-4}

B.

3×10−43 \times 10^{-4}

C.

1×10−41 \times 10^{-4}

D.

4×10−44 \times 10^{-4}
Correct Answer: C

Solution:

The fractional change in volume ΔVV\frac{\Delta V}{V} is given by ΔVV=−pB\frac{\Delta V}{V} = \frac{-p}{B}. Substituting the given values: ΔVV=−3×107150×109=1×10−4\frac{\Delta V}{V} = \frac{-3 \times 10^7}{150 \times 10^9} = 1 \times 10^{-4}.

Chapter Concept:

Stress and Strain

A.

2.0%

B.

1.0%

C.

0.5%

D.

0.25%
Correct Answer: B

Solution:

Using the definition of Poisson's ratio, which is the negative ratio of lateral strain to longitudinal strain, the longitudinal strain can be calculated as 0.5% / 0.25 = 2.0%. Therefore, the length of the rod increases by 2.0%.

Chapter Concept:

Poisson's Ratio

A.

2.5×107 N/m22.5 \times 10^7 \text{ N/m}^2

B.

2.5×106 N/m22.5 \times 10^6 \text{ N/m}^2

C.

2.5×105 N/m22.5 \times 10^5 \text{ N/m}^2

D.

2.5×108 N/m22.5 \times 10^8 \text{ N/m}^2
Correct Answer: A

Solution:

Tensile stress is calculated as force divided by area: σ=FA=50002×10−4=2.5×107 N/m2\sigma = \frac{F}{A} = \frac{5000}{2 \times 10^{-4}} = 2.5 \times 10^7 \text{ N/m}^2.

Chapter Concept:

Hooke's Law

A.

6.25×10−56.25 \times 10^{-5}

B.

7.85×10−57.85 \times 10^{-5}

C.

5.00×10−55.00 \times 10^{-5}

D.

4.00×10−54.00 \times 10^{-5}
Correct Answer: A

Solution:

Shearing strain Θ\Theta is given by Θ=FA⋅G\Theta = \frac{F}{A \cdot G}. The cross-sectional area A=π⋅(0.1)2=0.0314A = \pi \cdot (0.1)^2 = 0.0314 m². Substituting the values: Θ=50000.0314⋅80×109=6.25×10−5\Theta = \frac{5000}{0.0314 \cdot 80 \times 10^9} = 6.25 \times 10^{-5}.

Chapter Concept:

Stress and Strain

A.

Steel

B.

Rubber

C.

Copper

D.

Aluminium
Correct Answer: B

Solution:

Rubber is an example of an elastomer, which does not obey Hooke's Law as it can be stretched to large strains without a proportional increase in stress.

Chapter Concept:

Hooke's Law

A.

Elastic deformation

B.

Plastic deformation

C.

Hydraulic deformation

D.

Shear deformation
Correct Answer: B

Solution:

Plastic deformation is characterized by a permanent change in shape, meaning the material does not return to its original dimensions after the load is removed.

Chapter Concept:

Stress-Strain Curve

A.

1.2

B.

1.5

C.

2.0

D.

1.0
Correct Answer: B

Solution:

The ratio of the Young's modulus of steel to that of copper is calculated using the formula YsYc=(Ls/As)(Lc/Ac)=(4.7/3.0×10−5)(3.5/4.0×10−5)=1.5\frac{Y_s}{Y_c} = \frac{(L_s/A_s)}{(L_c/A_c)} = \frac{(4.7/3.0 \times 10^{-5})}{(3.5/4.0 \times 10^{-5})} = 1.5.

Chapter Concept:

Young’s Modulus and Longitudinal Deformation

A.

FA\frac{F}{A}

B.

YΔLLY \frac{\Delta L}{L}

C.

FA=YΔLL\frac{F}{A} = Y \frac{\Delta L}{L}

D.

ΔLL=FYA\frac{\Delta L}{L} = \frac{F}{Y A}
Correct Answer: C

Solution:

The tensile stress applied is given by the relation FA=YΔLL\frac{F}{A} = Y \frac{\Delta L}{L}, where FF is the force applied, AA is the cross-sectional area, YY is Young's modulus, and ΔLL\frac{\Delta L}{L} is the strain.

Chapter Concept:

Elastic Moduli

A.

θ\theta remains the same

B.

θ\theta doubles

C.

θ\theta halves

D.

θ\theta quadruples
Correct Answer: D

Solution:

The equation G=FAθG = \frac{F}{A\theta} can be rearranged to θ=FAG\theta = \frac{F}{AG}. If FF is doubled and AA is halved, then θ\theta becomes 2FA2G=4FAG\frac{2F}{\frac{A}{2}G} = 4\frac{F}{AG}, hence θ\theta quadruples.

Chapter Concept:

Shear Modulus and Shearing Deformation

A.

400 MPa400 \text{ MPa}

B.

200 MPa200 \text{ MPa}

C.

100 MPa100 \text{ MPa}

D.

300 MPa300 \text{ MPa}
Correct Answer: A

Solution:

Maximum stress without permanent deformation is given by σ=Y⋅ϵ=200×109×0.002=400×106 N/m2=400 MPa\sigma = Y \cdot \epsilon = 200 \times 10^9 \times 0.002 = 400 \times 10^6 \text{ N/m}^2 = 400 \text{ MPa}.

Chapter Concept:

Hooke's Law

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True or False

Correct Answer: False

Solution:

Hooke's Law is valid for small deformations, where stress is directly proportional to strain. This is evident in the linear region of the stress-strain curve.

Chapter Concept :

Hooke's Law

Correct Answer: True

Solution:

The bulk modulus measures a material's response to uniform pressure, which results in a change in volume. This property is applicable to solids, liquids, and gases.

Chapter Concept :

Elastic Moduli

Correct Answer: True

Solution:

Poisson's ratio is indeed defined as the ratio of lateral strain (change in diameter) to longitudinal strain (change in length) in a stretched wire, as described in the excerpt.

Chapter Concept :

Poisson's Ratio

Correct Answer: True

Solution:

In Hooke's Law, the modulus of elasticity (also known as Young's modulus for tensile/compressive stress) is the proportionality constant between stress and strain.

Chapter Concept :

Hooke's Law

Correct Answer: True

Solution:

Hooke's Law is defined as the proportional relationship between stress and strain for small deformations, with the modulus of elasticity as the proportionality constant.

Chapter Concept :

Hooke's Law

Correct Answer: True

Solution:

The bulk modulus BB is defined as B=−p/(ΔV/V)B = -p/(\Delta V/V), where pp is the pressure and ΔV/V\Delta V/V is the volume strain. The negative sign indicates that an increase in pressure results in a decrease in volume, ensuring BB is positive.

Chapter Concept :

Bulk Modulus and Compressibility

Correct Answer: False

Solution:

Young's modulus is specifically applicable to solids because it describes the relationship between tensile stress and longitudinal strain, which are not applicable to gases.

Chapter Concept :

Stress and Strain

Correct Answer: False

Solution:

The stress-strain curve is linear only up to the proportional limit, not the yield point. Beyond the proportional limit, the material may still return to its original dimensions when the load is removed, but the relationship is no longer linear.

Chapter Concept :

Stress-Strain Curve

Correct Answer: True

Solution:

Young's modulus, denoted by YY, is indeed the ratio of tensile (or compressive) stress to the longitudinal strain, applicable to solids as they have definite shapes and sizes. It is a measure of the stiffness of a solid material.

Chapter Concept :

Young’s Modulus and Longitudinal Deformation

Correct Answer: False

Solution:

Young's modulus is relevant only for solids since it measures the ability of a solid material to withstand changes in length when under lengthwise tension or compression.

Chapter Concept :

Elastic Moduli

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