{"id":2271,"date":"2023-05-03T04:28:25","date_gmt":"2023-05-03T04:28:25","guid":{"rendered":"https:\/\/vartmaaninstitutesirsa.com\/?page_id=2271"},"modified":"2024-03-09T06:52:51","modified_gmt":"2024-03-09T06:52:51","slug":"chapter-9-mechanical-properties-of-solids","status":"publish","type":"page","link":"https:\/\/vartmaaninstitutesirsa.com\/index.php\/chapter-9-mechanical-properties-of-solids\/","title":{"rendered":"Chapter 9 Mechanical Properties of Solids"},"content":{"rendered":"

Chapter 9 Mechanical Properties of Solids<\/h1>\n
\n

Chapter\u20139: Mechanical Properties of Solids:<\/span><\/strong> Elasticity,<\/span><\/strong> Stress strain<\/span><\/strong> relationship,<\/span><\/strong> Hooke’s<\/span><\/strong> law,<\/span><\/strong> \u00a0<\/span><\/strong>Young\u2019s<\/span><\/strong> modulus,<\/span><\/strong> bulk<\/span><\/strong> modulus,<\/span><\/strong> shear<\/span><\/strong> modulus<\/span><\/strong> \u00a0<\/span><\/strong>of<\/span><\/strong> rigidity<\/span><\/strong> (qualitative<\/span><\/strong>\u00a0\u00a0\u00a0\u00a0<\/span><\/strong>idea<\/span><\/strong>\u00a0\u00a0\u00a0\u00a0<\/span><\/strong>only),<\/span><\/strong> Poisson\u2019s ratio; elastic energy.<\/span><\/strong><\/p>\n

1 Solid:<\/span><\/strong><\/p>\n

The material having definite shape and definite volume is called solid or rigid body.\u00a0<\/span><\/p>\n

Some terms related to solids:<\/span><\/strong><\/p>\n

Deforming Force:-<\/span><\/strong>\u00a0<\/span><\/em><\/strong><\/p>\n

An external force which changes the form or configuration (shape, size or volume) of a rigid body called deforming force.<\/span><\/p>\n

Restoring Force:-<\/span><\/strong>\u00a0<\/span><\/em><\/strong><\/p>\n

An opposing force which comes in the body when a deforming force is applied on the body is called restoring force.\u00a0<\/span><\/p>\n

Solid basically may be divided into two parts:<\/span><\/p>\n

Elastic Solids:-<\/span><\/strong><\/p>\n

If a solid regains its original position after removal of deforming force then the solid is called elastic solid and the property by virtue of which it regains its original position is called elasticity<\/span>.<\/span><\/em><\/p>\n

Plastic Solids:-<\/span><\/strong><\/p>\n

If a solid does not regains its original position after removal of deforming force then the solid is called plastic solid and the property by virtue of which it does not regain its original position called plasticity.<\/span><\/p>\n

Perfectly Elastic Body:-<\/span><\/strong><\/p>\n

A body is said to be perfectly elastic if it completely regains its original position after removal of deforming force, then the body is said to be perfectly elastic body. In reality no any material regains its original position so we can say that nobody is perfectly elastic. A\u00a0<\/span>quartz crystal<\/span><\/strong>\u00a0nearly approach to perfectly elastic body.<\/span><\/p>\n

Perfectly Plastic Body:-<\/span><\/strong><\/p>\n

A body is said to be perfectly plastic if it does not regains its original position, even small after removal of deforming force. Since every material partially regains its original position so we, can say that no anybody is perfectly plastic body.\u00a0<\/span>Paraffin wax, wet clay<\/span><\/strong>\u00a0are nearly perfectly bodies.<\/span><\/p>\n

2 Elastic Behavior of Solids in terms of Inter Molecular Forces:<\/span><\/strong><\/p>\n

\"MechanicalIn solids, atoms and molecules are joined like by springs due to inter molecular force of attraction between atoms or molecules. When we apply external force, then the atoms and molecules of body get stretched and a restoring force is develops in the molecules which tends to brings the molecules to its original position. Due to this phenomenon solids shows elasticity.\u00a0<\/span><\/p>\n

3 Stress:-<\/span><\/strong><\/p>\n

We know that, when a deforming force is applied on the bodies then restoring force are developed in the body; the restoring force developed per unit area in the body is called stress.<\/span><\/p>\n

i.e.<\/span>\u00a0\u00a0\u00a0\u00a0<\/span>\"\"<\/p>\n

Unit:-\u00a0<\/span>\"\"\u00a0and in\u00a0<\/span>\"\"\u00a0\u00a0<\/span>\u27f6<\/span>\"\"\u00a0its unit are same as pressure.<\/span><\/p>\n

Dimensional formula<\/span>\"\",\u00a0<\/span><\/p>\n

Stress is a scalar quantity.<\/span><\/p>\n

4 Types of Stress:-<\/span><\/strong><\/p>\n

A Normal Stress or Longitudinal Stress:-<\/span><\/strong><\/p>\n

It is defined as the restoring force developed perpendicular to the surface per unit area.<\/span><\/p>\n

\"Tensile\u00a0Normal stress is of two types:\u00a0<\/span><\/p>\n

(i)\u00a0<\/span><\/em>Tensile Stress<\/span>\u00a0<\/span><\/em>(ii)\u00a0<\/span><\/em>Compressive Stress<\/span>.<\/span><\/em><\/p>\n

I.<\/span><\/strong>\u00a0\u00a0<\/span><\/strong>Tensile Stress:<\/span><\/strong>\u00a0<\/span><\/em><\/strong><\/p>\n

When deforming force increase the length of body, then the restoring force developed per unit area perpendicular to the cross-section of the body is called tensile stress.<\/span>\u00a0\u00a0\u00a0<\/span><\/em>\"\"<\/p>\n

II.<\/span><\/strong>\u00a0\u00a0\u00a0<\/span><\/strong>Compressive Stress:-<\/span><\/strong>\u00a0<\/span><\/em><\/strong><\/p>\n

\"BulkWhen deforming force decrease the length of the body, then the restoring force developed per unit area perpendicular to cross-section of the body is called compressive stress.<\/span><\/em><\/p>\n

B Tangential Stress:\u00a0<\/span><\/strong><\/p>\n

If restoring force acts tangentially on the body, then the stress produced in the body is called tangential stress.<\/span><\/p>\n

\"\"\u00a0<\/span><\/p>\n

C. Bulk Stress or Volume Stress or Hydraulic Stress<\/span><\/strong>:<\/span><\/em><\/strong><\/p>\n

When an object is immersed in a fluid or liquid then force acting\u00a0<\/span>\u22a5<\/span>\u00a0to the surface of the body is called bulk stress or volume stress.<\/span><\/p>\n

\"\"<\/p>\n

\uf0d8<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>Stress changes the configuration of the body.<\/span><\/p>\n

Question 1: A lead of 4.0 kg is suspended from a ceiling through a steel wire of radius 2.0 mm. Find the tensile stress developed in the wire when equilibrium is achieved. Take\u00a0<\/span><\/strong>\"\"<\/p>\n

Sol:<\/span>\u00a0\u00a0\u00a0<\/span>Stress is restoring force per unit area of cross-section. <\/span><\/p>\n

Tension in the wire is<\/span>\u00a0\u00a0\u00a0<\/span>\"\"<\/p>\n

The area of cross section is\u00a0<\/span><\/p>\n

\u00a0<\/span>\"\"<\/p>\n

\"\"<\/p>\n

Thus, the tensile stress developed<\/span><\/p>\n

\u00a0<\/span>\"\"<\/p>\n

\"\"<\/p>\n

Question2:\u00a0<\/span><\/em><\/strong>A rectangular bar having a cross-sectional area of 70m<\/span><\/strong>2<\/span><\/strong>\u00a0has a tensile force\u00a0<\/span><\/strong><\/p>\n

Of 14 kN applied to It. Determine the stress in the bar.<\/span><\/strong><\/p>\n

Solution:<\/span><\/em><\/strong>\u00a0\u00a0\u00a0\u00a0<\/span><\/em><\/strong>Cross-sectional area\u00a0<\/span>A<\/span><\/em>\u00a0= 70 m<\/span>2<\/sup><\/span>\u00a0<\/span>\u00a0\u00a0\u00a0<\/span><\/p>\n

Tensile force\u00a0<\/span>F<\/span><\/em>\u00a0= 14 kN = 14\u00a0<\/span>\uf0b4<\/span>\u00a010<\/span>3<\/sup><\/span>\u00a0N<\/span><\/p>\n

Stress in the bar =\u00a0<\/span>\"\"=\u00a0<\/span>200 N<\/span><\/strong><\/p>\n

5 Strain:<\/span><\/strong><\/p>\n

It is defined as the ratio of change in the configuration to the original configuration.<\/span><\/em><\/p>\n

i.e.\u00a0<\/span>\"\"<\/p>\n

*<\/span>\u00a0\u00a0<\/span>it has no unit and Dimensional formula.<\/span><\/p>\n

Strain are of three types:-<\/span><\/strong><\/p>\n

I. Longitudinal Strain:-<\/span><\/strong>\u00a0<\/span><\/em><\/strong><\/p>\n

It is defined as the ratio of change in length to the original length of the body. It produces due to normal stress.<\/span><\/p>\n

i.e.<\/span>\u00a0\u00a0<\/span>\"\"<\/p>\n

Or<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>\"\"\u00a0<\/span><\/p>\n

II.<\/span><\/strong>\u00a0\u00a0\u00a0<\/span><\/strong>Volumetric Strain:-<\/span><\/strong>\u00a0<\/span><\/em><\/strong><\/p>\n

It is defined as the ratio of change in the volume to the original volume when a deforming force is applied on the body. This strain produces in the body when body is under Bulk Stress.<\/span><\/p>\n

Volumetric<\/span>\"\"\u00a0<\/span>\"\"\u00a0<\/span><\/p>\n

\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>Here\u00a0<\/span>\"\"sign indicate that volume decrease when deforming force is applied on the body.<\/span><\/p>\n

III. Shear -Strain:-<\/span><\/strong>\u00a0<\/span><\/em><\/strong><\/p>\n

The tangential force applied per unit area of the body is equal to shear stress. When shear stress act on the body, then shape of the body changes . So shear strain may be defined as the ratio of relative displacement between two parallel plans to the distance between parallel plans.<\/span><\/p>\n

\"\"<\/p>\n

Question3:\u00a0<\/span><\/em><\/strong>A wire of length 2.5 m has a percentage strain of 0.12% under a tensile\u00a0<\/span><\/strong><\/p>\n

Force . Determine the extension in the wire.<\/span><\/strong><\/p>\n

Solution:<\/span><\/em><\/strong>\u00a0<\/span><\/em>\u00a0<\/span>Original length of wire\u00a0<\/span>L<\/span><\/em>\u00a0= 2.5 m<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

\u00a0<\/span>Strain<\/span>\u00a0<\/span>=0. 12 %\u00a0<\/span>\u00a0<\/span><\/p>\n

Strain =\u00a0<\/span>\"\"=\u00a0<\/span>\"\"<\/p>\n

\uf044<\/span>L<\/span><\/em>=<\/span>\u00a0\u00a0<\/span>\"\"\u00a0= 3\u00a0<\/span>\uf0b4<\/span>\u00a010<\/span>\uf02d<\/span>3<\/span>\u00a0m<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

Extension =<\/span>3 mm<\/span><\/strong>.<\/span><\/p>\n

Question 4:<\/strong>\u00a0\u00a0\u00a0\u00a0<\/strong>A\u00a0<\/strong>\"\"\u00a0long copper wire of cross sectional area\u00a0<\/strong>\"\"is stretched by a force of\u00a0<\/strong>\"\"\u00a0<\/strong>\"\".\u00a0<\/strong><\/p>\n

Stress will be-\u00a0<\/strong><\/p>\n

\u00a0<\/strong>\"\"\u00a0\u00a0<\/strong><\/p>\n

(B)\u00a0<\/strong>\"\"\u00a0<\/strong><\/p>\n

(C)\u00a0<\/strong>\"\"<\/p>\n

(D) None<\/strong><\/p>\n

Sol:\u00a0 (C) Stress is restoring force per unit area of cross-section.<\/p>\n

Stress = \"\"<\/p>\n

\"\"<\/p>\n

Question 5:<\/strong>\u00a0\u00a0\u00a0<\/strong>A copper rod 2m long is stretched by 1mm. Strain will be<\/strong>\u00a0\u00a0<\/strong><\/p>\n

\u00a0<\/strong>(A)\u00a0<\/strong>\"\", volumetric<\/strong>\u00a0\u00a0\u00a0<\/strong><\/p>\n

(B)\u00a0<\/strong>\"\", volumetric<\/strong>\u00a0\u00a0\u00a0<\/strong><\/p>\n

(C)\u00a0<\/strong>\"\", longitudinal<\/strong>\u00a0\u00a0\u00a0<\/strong><\/p>\n

(D)\u00a0<\/strong>\"\", volumetric<\/strong><\/p>\n

Sol:\u00a0 (C) Strain \"\"<\/p>\n

\"\"\u00a0Longitudinal<\/p>\n

Question 6:<\/strong>\u00a0\u00a0<\/strong>Find the stress on a bone (1 cm in radius and 50 cm long) that supports a mass of 100kg. Find the strain on the bone if it is compressed 0.15 mm by this weight. Find the proportionality constant C for this bone.\u00a0<\/strong><\/p>\n

\u00adSol: Stress is restoring force per unit area of cross-section.<\/p>\n

Strain is equal to change in length per unit length.<\/p>\n

\"\"<\/p>\n

So Stress = \"\"<\/p>\n

\"\"<\/p>\n

\u00a0 Strain = \"\"<\/p>\n

\u00a0Since \"\"<\/p>\n

\"\"<\/p>\n

6.<\/span><\/strong>\u00a0\u00a0<\/span><\/strong>Elastic Limit<\/span><\/strong>:<\/span><\/em><\/strong><\/p>\n

The maximum value of stress with in which the body regains its original shape and size, after that value body gets permanently deformed is called elastic limit.<\/span><\/em><\/p>\n

Hooks Law:<\/span><\/strong><\/p>\n

\"HooksAccording to hooks law, the extension produced in a wire is directly proportional to applied load, which is modified by\u00a0<\/span>Thomas Young<\/span><\/strong>, so modified Hooks law states that: within elastic limit, the stress is directly proportional to strain.\u00a0<\/span><\/p>\n

i.e.<\/span>\u00a0<\/span>\"\"<\/p>\n

Or<\/span>\u00a0\u00a0\u00a0<\/span>\"\"<\/p>\n

Or<\/span>\u00a0<\/span>\"\"<\/p>\n

Here the ratio of stress to the strain is called coefficient of elasticity and its value depends upon the nature of the material and the manner in which it is determined.\u00a0<\/span><\/p>\n

7 Variation of strain with stress or stress-strain relationship:\u00a0<\/span><\/strong><\/p>\n

\"stress-strainWhen a wire is stretched, then it is found that for a small value of load (stress), there is a (strain) extension in the wire. When load is removed then wire comes in its original position up to\u00a0<\/span>\"\", this\u00a0<\/span>\"\"\u00a0is called\u00a0<\/span>proportionality limit.<\/span><\/em><\/strong><\/p>\n

After this value up to\u00a0<\/span>\"\"\u00a0wire regain its original position, the value\u00a0<\/span>\"\"\u00a0is called\u00a0<\/span>elastic limit<\/span><\/em><\/strong>.\u00a0<\/span><\/em>Thus elastic limit is the maximum stress up to which body shows elastic behavior and after that body have some permanent strain in the wire. The point\u00a0<\/span>\"\"\u00a0is called\u00a0<\/span>elastic limit<\/span><\/em><\/strong>\u00a0or\u00a0<\/span>yield point<\/span><\/em><\/strong>.<\/span>\u00a0\u00a0<\/span>The wire shows plastic behavior between<\/span>\"\"\u00a0to\u00a0<\/span>\"\"\u00a0at\u00a0<\/span>\"\"\u00a0(maximum plastic behavior) the wire breaks, this stress required to break the wire is called\u00a0<\/span>breaking stress.<\/span><\/em><\/strong><\/p>\n

\"\"8<\/span><\/strong>\u00a0\u00a0<\/span><\/strong>Modulus of Elasticity:<\/span><\/strong><\/p>\n

It is defined as the ratio of stress to the strain within the elastic limit.\u00a0<\/span><\/p>\n

\u00a0<\/span>i.e.<\/span>\"\"<\/p>\n

Unit:-\u00a0<\/span>\"\"\u00a0and dimensional formula\u00a0<\/span>\"\"<\/p>\n

There are three types of modulus of elasticity<\/span><\/p>\n

1. Young Modulus (<\/span><\/strong>\"\"):-\u00a0<\/span><\/strong><\/p>\n

It is defined as the ratio of tensile stress to the longitudinal strain<\/span><\/p>\n

\"\"<\/p>\n

Suppose\u00a0<\/span>\"\"\u00a0amount of force is applied on a wire of length\u00a0<\/span>\"\"\u00a0and area\u00a0<\/span>\"\"\u00a0then young modulus of the wire is<\/span>\u00a0\u00a0<\/span><\/p>\n

\u00a0<\/span>\"\"\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

Or\u00a0<\/span>\u00a0<\/span>\"\"<\/p>\n

\uf0d8<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>Larger the value of young modulus, larger the force required to change the length of the wire.<\/span><\/p>\n

\uf0d8<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>Young modulus of an object depends upon the nature of the materials.<\/span><\/p>\n

\uf0d8<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>Young modulus does not depend upon the dimensions of the object.<\/span><\/p>\n

\uf0d8<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>Young modulus of a material decrease with increase in temperature.<\/span><\/p>\n

\uf0d8<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>Young modulus of a perfectly rigid body is infinite.<\/span><\/p>\n

Question 7:<\/span><\/strong>\u00a0\u00a0<\/span><\/strong>Two wires of equal cross section but one made of steel and the other of copper are joined end to end. When the combination is kept under tension, the elongations in the two wires are found to be equal. Find the ratio of the lengths of the two wires. <\/span><\/strong><\/p>\n

Young\u2019s modulus of steel<\/span><\/strong>\"\"\u00a0<\/span><\/strong>\"\"\u00a0<\/span><\/strong><\/p>\n

Sol: The wires joined together have same stress and same elongation. Ratio of stress and young\u2019s modulus is strain. As young\u2019s modulus for steel and copper is different, strains of the wires will be different.<\/span><\/p>\n

Let the original lengths of the steel wire and the copper wire be\u00a0<\/span>\"\"\u00a0and\u00a0<\/span>\"\"\u00a0respectively and the elongation in each wire be<\/span>\u00a0\u00a0<\/span>.<\/span>\u00a0\u00a0<\/span>Than<\/span>\u00a0\u00a0<\/span>\"\"\u00a0<\/span><\/p>\n

And<\/span>\u00a0\u00a0<\/span>\"\"\u00a0<\/span><\/p>\n

Dividing (ii) by (i<\/span>\"\"\u00a0<\/span><\/p>\n

Question 8: A solid cylindrical steel column is 4.0 m long and 9.0 cm in diameter. What will be decrease in length when carrying a load of 80000 kg?\u00a0<\/span><\/strong>\"\"<\/p>\n

Sol: The stress will be equal to load per unit cross section. Strain is the ratio of stress and young\u2019s modulus.<\/span><\/p>\n

Let us first calculate the cross-sectional area of column <\/span><\/p>\n

=\u00a0<\/span>\"\"<\/p>\n

\u00a0<\/span>Then, from\u00a0<\/span>\"\"\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

we have\u00a0<\/span>\"\"\u00a0<\/span><\/p>\n

Illustration 9: A load of 4.0 kg is suspended from a ceiling through a steel wire of length 20 m and radius 2.0 mm. It is found that the length of the wire increases by 0.031 mm as equilibrium is achieved. Find Young\u2019s modulus of steel. Take\u00a0<\/span><\/strong>\"\"<\/p>\n

Sol: The longitudinal stress<\/span>\"\"<\/p>\n

\u00a0<\/span>The longitudinal strain =\u00a0<\/span>\"\"\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

Thus\u00a0<\/span>\"\"\u00a0<\/span>\"\"<\/p>\n

2. Bulk Modulus (<\/span><\/strong>\"\"):-\u00a0<\/span><\/strong><\/p>\n

It is defined as the ratio of normal stress to the volumetric strain\u00a0<\/span><\/em><\/p>\n

i.e.<\/span>\u00a0\u00a0\u00a0<\/span>\u00a0<\/span>\"\"<\/p>\n

\"BulkSuppose a body of volume\u00a0<\/span>\"\"\u00a0and area<\/span>\"\", again suppose that\u00a0<\/span>\"\"\u00a0is the amount of force applied on the body per unit area due to which volume changes\u00a0<\/span>\"\"\u00a0to\u00a0<\/span>\"\".<\/span><\/p>\n

\u00a0<\/span>So bulk modulus of the body<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

\"\"\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

Or<\/span> \u00a0\u00a0<\/span>\"\"<\/p>\n

Here\u00a0<\/span>\"\"\u00a0sign shows that volume decreases when we apply stress.<\/span><\/p>\n

\uf0d8<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>Compressibility:-<\/span><\/strong><\/p>\n

The reciprocal of Bulk modulus of a material is called compressibility.<\/span><\/p>\n

\u00a0<\/span>i.e.<\/span>\"\"<\/p>\n

\uf0d8<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>Bulk modulus for a perfectly rigid body and ideal liquid is \u221e.<\/span><\/em><\/p>\n

\uf0d8<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>Bulk modulus of solids is larger than liquid and gases so solids are less compressible as compared to solids.<\/span><\/p>\n

Question 10: Find the decrease in the volume of a sample of water from the following data. Initial volume\u00a0<\/span><\/strong>\"\",initial pressure=<\/span><\/strong>\"\",final pressure\u00a0<\/span><\/strong>\"\", compressibility of water=\u00a0<\/span><\/strong>\"\"<\/p>\n

Sol: Using the formula for bulk modulus deduce the value for decrease in volume.<\/span><\/p>\n

The change in pressure\u00a0<\/span><\/p>\n

=\u00a0<\/span>\"\"<\/p>\n

Compressibility =<\/span>\"\"<\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

Thus the decrease in volume is 0.45 cm<\/span>3<\/sup><\/span><\/p>\n

    \n
  1. Shear Modulus (<\/strong>\"\"):-<\/strong><\/li>\n<\/ol>\n

    It is defined as the ratio of tangential stress to the shear strain.<\/span><\/p>\n

    \"ShearSuppose a cube of sides<\/span>\"\", whose lower face is fixed and a tangential force\u00a0<\/span>\"\"\u00a0is applied on the upper face whose area is\u00a0<\/span>\"\".<\/span><\/p>\n

    So<\/span>\u00a0\u00a0\u00a0\u00a0<\/span>tangential stress\u00a0<\/span>\"\"<\/p>\n

    Now due to this stress, the shear strain produced in the cube is shear strain<\/span><\/p>\n

    \u00a0<\/span>\"\"<\/p>\n

    So<\/span>\u00a0\u00a0\u00a0<\/span>shear modulus\u00a0<\/span>\"\"\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

    Or<\/span>\u00a0\u00a0<\/span>\"\"\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

    Or<\/span>\u00a0\u00a0\u00a0\u00a0<\/span>\"\"<\/p>\n

    \uf0d8<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>Shear modulus for ideal liquid is zero.<\/span><\/p>\n

    \uf0d8<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>A solid have all types of modulus but liquid and gases have only bulk modulus.<\/span><\/p>\n

    Question 11: A box shaped piece of gelatin dessert has a top area of\u00a0<\/span><\/strong>\"\"and a height of 3cm. When a shearing force of 0.50 N is applied to the upper surface, the upper surface displaces 4 mm relative to the bottom surface. What are the shearing stress, the shearing strain and the shear modulus for the gelatin?<\/span><\/strong>\u00a0\u00a0<\/span><\/strong><\/p>\n

    Sol:\u00a0<\/span>\"\"<\/p>\n

    \"\"<\/p>\n

    \u00a0\u00a0<\/span>\"\"<\/p>\n

    \"\"<\/p>\n

    \"\"\u00a0<\/span><\/p>\n

    \"Poisson\u2019s9\u00a0<\/span><\/em><\/strong>Poisson\u2019s Ratio:<\/span><\/strong><\/p>\n

    When we stretch a wire, then the length of the wire increases and diameter of the wire decreases.<\/span>\u00a0\u00a0<\/span>As shown in figure. Here the ratio of increases in length to the original length is called\u00a0<\/span>longitudinal strain<\/span><\/em>\u00a0and the ratio of decreases in radius to the original radius is called\u00a0<\/span>lateral strain<\/span><\/em>. <\/span>Hence the ratio of lateral strain to the longitudinal strain of the wire is called poisons ratio (<\/span><\/em>\"\").<\/span><\/em>i.e.<\/span><\/p>\n

    \"\"<\/p>\n

    \"\"\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

    O r<\/span><\/em>\u00a0 \u00a0 \u00a0 \u00a0 <\/span><\/em>\"\"<\/p>\n

    Here\u00a0<\/span>\"\"\u00a0sign shows that diameter of the bar decreases when it is stretched.<\/span><\/p>\n

    Poisson\u2019s ratio is a unit less and dimensionless quantity.<\/span><\/p>\n

    \uf0d8<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>In practical, the value of poisons ratio lies between<\/span><\/em>\u00a0\u00a0\u00a0<\/span><\/em>0<\/span><\/em>\u00a0\u00a0\u00a0<\/span><\/em>to<\/span><\/em>\u00a0\u00a0<\/span><\/em>0.5.<\/span><\/em><\/p>\n

    10<\/span><\/em><\/strong>\u00a0\u00a0\u00a0<\/span><\/em><\/strong>Some Other Elastic Effects<\/span><\/strong>:<\/span><\/em><\/strong><\/p>\n

      \n
    • Elastic After Effect:<\/span><\/strong>–<\/span><\/em><\/strong><\/li>\n<\/ul>\n

      The delay in regaining the original state of a body on removal of deforming force is called elastic after effect.<\/span><\/p>\n

        \n
      • Elastic Fatigue:<\/span><\/strong>\u00a0<\/span><\/em><\/strong><\/li>\n<\/ul>\n

        The loss of strength of the material due to repeated strains on the material is called elastic fatigue.<\/span>\u00a0<\/span><\/em><\/strong>e.g.: A wire breaks when we bend it again and again due to elastic fatigue.<\/span><\/p>\n

          \n
        • Elastic Hysteresis<\/span><\/strong>:<\/span><\/em><\/strong>\u00a0<\/span><\/li>\n<\/ul>\n

          Hysteresis means delay; hence elastic hysteresis means that stress-strain curve is not retraced on reversing the strain is known as elastic hysteresis.<\/span><\/p>\n

          11.\u00a0<\/span><\/em><\/strong>Types of Materials:<\/span><\/strong><\/p>\n

          a. Ductile Materials:-<\/span><\/strong>\u00a0<\/span><\/em><\/strong>The materials which have large plastic ranges of extension are called Ductile Materials. e.g.- Gold, Silver, Copper have large plastic range before breaking point so can be easily extended.<\/span><\/p>\n

          b. Brittle Materials<\/span><\/strong>:-<\/span><\/em><\/strong>The materials which have very small plastic ranges are called brittle materials. The breaking point of these materials is near to plastic range. Such materials break as soon as the stress is increased beyond elastic.<\/span><\/p>\n

          c. Elastomers:<\/span><\/strong>\u00a0\u00a0<\/span><\/em><\/strong>The materials which shows a very large strain when a small stress is applied are called elastomers. These materials after removal of stress regain their original size. e.g.- rubber and does not obey hooks law.<\/span><\/p>\n

          \u00a0<\/span><\/strong><\/p>\n

          12 Elastic Potential Energy of a Stretched Wire:<\/span><\/strong><\/p>\n

          Suppose a force\u00a0<\/span>\"\"\u00a0is applied on a wire of length<\/span>\"\", increases its length form\u00a0<\/span>\"\"\u00a0to\u00a0<\/span>\"\".<\/span><\/p>\n

          Due to this force average internal force induced in the wire is<\/span><\/p>\n

          \u00a0<\/span>\"\"<\/p>\n

          \"ElasticNow the work done on the wire is\u00a0<\/span>\"\"<\/p>\n

          \"\"<\/p>\n

          Or<\/span>\u00a0<\/span>\"\"<\/p>\n

          This work is stored in the wire in form of elastic potential energy.<\/span>\u00a0<\/span><\/strong><\/p>\n

          So<\/span>\u00a0<\/span>\"\"<\/p>\n

          Or<\/span>\u00a0<\/span>\"\"<\/p>\n

          Or<\/span>\u00a0<\/span>\"\"<\/p>\n

          Now elastic potential energy per unit volume of the wire is\u00a0<\/span><\/p>\n

          \"\"<\/p>\n

          Also\u00a0<\/span>\u00a0<\/span>\"\"<\/p>\n

          \u27f9<\/span>\u00a0<\/span>\"\"<\/p>\n

          Question 12:A steel wire of length 2.0 m is stretched through 2.0 mm. The cross-sectional area of the wire is<\/span><\/strong>\u00a0\u00a0\u00a0<\/span><\/strong>\"\".Calculate the elastic potential energy stored in the wire in the stretched condition. Young modulus of steel<\/span><\/strong>\"\".<\/span><\/strong>\u00a0\u00a0<\/span><\/strong><\/p>\n

          Sol: We know the formula to find the elastic potential energy stored per unit volume of the wire. Calculate the volume of the wire and find the energy stored in the entire wire.<\/span><\/p>\n

          The strain in the wire<\/span><\/p>\n

          \u00a0<\/span>\"\"<\/p>\n

          The stress in the wire<\/span><\/p>\n

          \u00a0<\/span>=\u00a0<\/span>\"\"<\/p>\n

          \"\"<\/p>\n

          The volume of the wire<\/span><\/p>\n

          \u00a0<\/span>=\u00a0<\/span>\"\"<\/p>\n

          \"\"<\/p>\n

          The elastic potential energy stored\u00a0<\/span><\/p>\n

          \"\"<\/p>\n

          \"\"\u00a0<\/span><\/p>\n

          13 Application of Elasticity:<\/span><\/strong><\/p>\n

            \n
          1. The metallic part of machinery is never subjected to a stress beyond elastic limit because a stress beyond elastic limit will permanently deform that metallic part.<\/li>\n
          2. Bridges are declare unsafe after long use:<\/span><\/strong>\u00a0<\/span><\/li>\n<\/ol>\n

            Due to continuous motion of traffic over the bridge, the bridge loses its elasticity due to which they<\/span>\u00a0\u00a0\u00a0<\/span>produce large strain even for small stress and may breaks, hence bridge becomes unsafe.<\/span><\/p>\n

              \n
            1. \u00a0<\/span>The radius of steel rope may be easily calculated by elasticity use to lift a weight safely.\u00a0<\/span>e.g.- to lift a weight of \"\" the radius required rope can be calculated as. Elastic limit of steel is\u00a0<\/span>\"\"\u00a0<\/span><\/li>\n<\/ol>\n

              \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>Let\u00a0<\/span>\"\"\u00a0be the radius of the required rope then<\/span><\/p>\n

              \u00a0 \u00a0 \u00a0 \u00a0<\/span>\"\"<\/p>\n

              \u27f9<\/span>\u00a0<\/span>\"\"<\/p>\n

              Or<\/span>\u00a0<\/span>\"\"<\/p>\n

              Thus the radius of steel rope used crane to lift maximum load of\u00a0<\/span>\"\"\u00a0should be about 3cm. In practice the radius of wire is used larger for safety.<\/span><\/p>\n

                \n
              1. To Estimate the maximum height of a mountain:<\/li>\n<\/ol>\n

                The elastic behavior of earth helps us to calculate the maximum height of a mountain.<\/span><\/p>\n

                \u00a0\u00a0<\/span>The pressure at the base of the mountain<\/span><\/p>\n

                \u00a0<\/span>\"\"\u00a0<\/span><\/p>\n

                and the elastic limit of typical rock is\u00a0<\/span><\/p>\n

                \"\"<\/p>\n

                The stress must be less than the elastic limit so<\/span>\u00a0<\/span><\/p>\n

                \u00a0<\/span>\"\"<\/p>\n

                \u00a0 <\/span>\"\"<\/p>\n

                or<\/span>\u00a0<\/span>\u00a0\u00a0<\/span>\"\"<\/p>\n

                It should be noted that the height of Mount Everest is nearly 9Km.<\/span><\/em><\/p>\n

                  \n
                • The girder are generally formed of I shaped because the stress is mostly concentrated on top and bottom flanges and practically there is no stress in the central part of the girder.<\/span><\/li>\n
                • A hollow cylinder is stronger than a solid one of same material due to elasticity.<\/span><\/li>\n<\/ul>\n

                  14<\/span><\/strong>\u00a0\u00a0<\/span><\/strong>Factor Affecting Elasticity:<\/span><\/strong><\/p>\n

                  The elasticity of a material is affected by the following factors.<\/span><\/p>\n

                    \n
                  • Effect of Hammering and Rolling:<\/span><\/strong>\u00a0<\/span><\/li>\n<\/ul>\n

                    Hammering and Rolling breaks the crystal grains into small part due to which elasticity increases.<\/span><\/p>\n

                      \n
                    • Effects of Annealing<\/span><\/strong>:\u00a0<\/span><\/em><\/li>\n<\/ul>\n

                      Annealing means heating and cooling a materials gradually which forms the large crystal grains. Hence elasticity of the materials decreases.<\/span><\/p>\n

                        \n
                      • Effect of Temperature<\/span><\/strong>:<\/span><\/em><\/li>\n<\/ul>\n

                        \u00a0<\/span><\/em>The elasticity of material decreases with increases in temperature and increases with decreases in temperature.<\/span><\/p>\n

                          \n
                        • Effect of Impurity in the Materials<\/span><\/strong>:\u00a0<\/span><\/li>\n<\/ul>\n

                          The elasticity of a material can be increased or decreased by adding impurities in materials depending on the nature of the material.<\/span><\/p>\n

                          To free download pdf of class 11 physics notes of Chapter 10 Mechanical Properties of Fluids click on the link given below<\/h3>\n

                          Chapter 10 Mechanical Properties of Fluids<\/h2>\n

                          \u00a0<\/span><\/p>\n

                          \n

                          1<\/strong><\/p>\n

                          \u00a0<\/em><\/strong><\/p>\n<\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

                          Chapter 9 Mechanical Properties of Solids Chapter\u20139: Mechanical Properties of Solids: Elasticity, Stress strain relationship, Hooke’s law, \u00a0Young\u2019s modulus, bulk modulus, shear modulus \u00a0of rigidity (qualitative\u00a0\u00a0\u00a0\u00a0idea\u00a0\u00a0\u00a0\u00a0only), Poisson\u2019s ratio; elastic energy. 1 Solid: The material having definite shape and definite volume is called solid or rigid body.\u00a0 Some terms related to solids: Deforming Force:-\u00a0 An external […]<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"_uag_custom_page_level_css":"","site-sidebar-layout":"default","site-content-layout":"default","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"default","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"set","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"var(--ast-global-color-4)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"class_list":["post-2271","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_hostinger_reach_plugin_has_subscription_block":false,"_hostinger_reach_plugin_is_elementor":false,"uagb_featured_image_src":{"full":false,"thumbnail":false,"medium":false,"medium_large":false,"large":false,"1536x1536":false,"2048x2048":false},"uagb_author_info":{"display_name":"sandeep.soni484@gmail.com","author_link":"https:\/\/vartmaaninstitutesirsa.com\/index.php\/author\/sandeep-soni484gmail-com\/"},"uagb_comment_info":0,"uagb_excerpt":"Chapter 9 Mechanical Properties of Solids Chapter\u20139: Mechanical Properties of Solids: Elasticity, Stress strain relationship, Hooke’s law, \u00a0Young\u2019s modulus, bulk modulus, shear modulus \u00a0of rigidity (qualitative\u00a0\u00a0\u00a0\u00a0idea\u00a0\u00a0\u00a0\u00a0only), Poisson\u2019s ratio; elastic energy. 1 Solid: The material having definite shape and definite volume is called solid or rigid body.\u00a0 Some terms related to solids: Deforming Force:-\u00a0 An external…","_links":{"self":[{"href":"https:\/\/vartmaaninstitutesirsa.com\/index.php\/wp-json\/wp\/v2\/pages\/2271","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/vartmaaninstitutesirsa.com\/index.php\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/vartmaaninstitutesirsa.com\/index.php\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/vartmaaninstitutesirsa.com\/index.php\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/vartmaaninstitutesirsa.com\/index.php\/wp-json\/wp\/v2\/comments?post=2271"}],"version-history":[{"count":5,"href":"https:\/\/vartmaaninstitutesirsa.com\/index.php\/wp-json\/wp\/v2\/pages\/2271\/revisions"}],"predecessor-version":[{"id":4682,"href":"https:\/\/vartmaaninstitutesirsa.com\/index.php\/wp-json\/wp\/v2\/pages\/2271\/revisions\/4682"}],"wp:attachment":[{"href":"https:\/\/vartmaaninstitutesirsa.com\/index.php\/wp-json\/wp\/v2\/media?parent=2271"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}