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    What Is Hooke's Law for Stress and Strain

    By Ramraj Saini
    25 Nov'22  5 min read
    What Is Hooke's Law for Stress and Strain
    Synopsis

    Hooke’s Law is an important concept of the NCERT class 11th Physics. Hooke’s law explains the behaviour of the elastic material with experiments giving mathematical equations and graphical representation. Questions in Boards and JEE-Mains are frequent from Hooke’s Law. Read on to know more about Hooke’s Law and its working principle.

    What Is Hooke's Law for Stress and Strain
    Synopsis

    Hooke’s Law is an important concept of the NCERT class 11th Physics. Hooke’s law explains the behaviour of the elastic material with experiments giving mathematical equations and graphical representation. Questions in Boards and JEE-Mains are frequent from Hooke’s Law. Read on to know more about Hooke’s Law and its working principle.

    Stress and Strain relationship shows a similar pattern when we study the behaviour of springs and their elasticity. Hooke’s Law gives the mathematical relationship between the stress and strain in a spring system. According to Hooke's Law, for small deformations, the stress and strain are almost proportional to each other, showing a linear relationship

    In simpler terms, Hooke’s Law states that the strain caused in the material is directly proportional to the stress applied, and the stress applied must be in the elastic limit of the material.

    On applying the stress, atoms and molecules in the elastic material get deformed and when the stress is removed, atoms and molecules get back to their initial position.

    Hooke’s law is represented by:

    F = - kx

    Where, F represents the force exerted, x represents the extension in length and k is the proportionality constant which is also known as the spring constant having unit N/m.

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    Hooke’s Law Experiment

    Hooke’s experiment using the spring system before giving the stress and strain relationship. The below figure shows the experiment apparatus where stress is applied on the spring in increasing order.

    Hookes-Law-Experiment

    In the above figure, When no load is applied to the spring, it is in a stable condition. When a load of 1 N is applied, it gets elongated to a length of x and when the load of 2N is applied, it gets elongated to the 2x length. Spring constants are different for different springs which depends on the material of the spring.

    In the above figure, 3 instances are shown which are :

    Stable condition of the spring in which the spring is elongated to a length of x under a load of 1 N, and the spring elongated to a length of 2x under a load of 2 N. Spring constant of the spring can be calculated if these values are substituted in the Hooke’s law equation.

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    Young’s Modulus

    Modulus of elasticity is represented by Young’s Modulus. It represents the resilience of a material when mechanical force is applied. Young's Modulus is denoted by Y and is measured for the mechanical properties of rods, wires etc. The relationship between Stress and Strain is represented by Young's Modulus and Mathematically it is represented by:

    Young’s Modulus= Stress / Strain

    This ratio tells us how easily a material can be stretched and deformed. Here stress is the amount of force applied on the material per unit area and strain is the extension in length.

    The important property of the Young’s Modulus is that it applies to linear objects in the elastic range only. Formula for Young’s Modulus can be used to solve many numerical problems.

    Stress-Strain Curve for Low Carbon Steel

    One of the important curves in the NCERT syllabus of Hooke’s Law is the stress-strain curve for low carbon steel. As from the figure below showing the stress and strain relationship, it can be seen that the material exhibits the elastic behaviour till the yield strength point, after which the material loses elasticity and the relationship between stress and strain is not linear. After elasticity, the material exhibits plasticity.

    Stress-Strain-Curve-for-Low-Carbon-Steel_rl14jGT

    When the stress and strain are proportional, from the origin to the limit nearing yield strength, the straight line relationship implies that the material follows Hooke’s law and the material is in the elastic range. Once the stress is extended beyond the elastic limit the material loses its elasticity and exhibits plasticity. The area under the curve which is formed by the stress-strain relationship from the origin to the proportional limit belongs to the elastic range. The area under the curve from a proportional limit to the fracture point shown on the graph belongs to the plastic range.

    The ultimate strength of the material is assessed by the ordinary value of the stress-strain curve calculated from origin to rupture point.

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    Applications of Hooke’s Law

    Hooke’s law provides a mathematical relationship between stress and strain which is used in the mechanical industry and various devices. Below are some of the important applications:

    1. Hooke’s Law is used as a fundamental principle behind the manometer which is used to measure the pressure difference, it is used in the spring scale and the balance wheel of a clock.

    2. Hooke's law has applications in seismology and molecular mechanics.

    Important Questions

    Following are some of the important questions and their answers in brief related to Hooke’s Law that are included in the Board exams.

    Do all materials follow Hooke’s Law ?

    Hooke’s Law is applied to the objects which are elastic in nature. Beyond the elastic limit, Hooke’s law doesn't apply to the objects.

    What is the nature of graphs of Hooke’s Law?

    Hooke’s Law as shown in the above graph shows a linear relationship. According to Hooke’s law, the restoring force is proportional to the displacement of the object.

    Beyond which limit, Hooke’s Law fails?

    Hooke’s law is applied to perfectly elastic materials and does not apply beyond the elastic limit of materials. Hence beyond the elastic limit, Hooke's Law fails.

    What is the use of Hooke’s Law?

    Hooke’s Law is used in multiple industries as it helps us understand the stretchability of an object when it is stretched or compressed.

    List out the various types of strain

    Following are the types of strains:

    1. Longitudinal Strain

    2. Volumetric Strain

    3. Shear Strain

    JEE Mains Previous Year Questions

    Q-1 (JEE Main 2022)

    The elastic behaviour of material for linear stress and linear strain, is shown in the figure. The energy density for a linear strain of 5 \times 10^{-4}\: \: is________\mathrm{kJ} / \mathrm{m}^{3}. Assume that material is elastic upto the linear strain of 5 \times 10^{-4}.

    Q-1%20(1)

    Answer = 25

    Solution:

    \mathrm{Energy\: density= \frac{1}{2}\times stress\times strain }

    \mathrm{= \frac{1}{2}\left ( strain \right )^{2}\gamma }

    \mathrm{\because slope\: of\: graph= \frac{1}{\gamma }= \frac{10^{-10}}{20} }

    \mathrm{\therefore \; Y= 20\times 10^{+10} }

    \mathrm{Energy\: density= \frac{1}{2}\times \left ( 5\times 10^{-4} \right )^{2}\times \left ( 20\times 10^{10} \right )}

    \mathrm{= 250\times 10^{2}}

    \mathrm{Energy\, density= 25\, \frac{kJ}{m^{3}}}

    Concepts Used:

    • Concept to calculate slope of a straight line graph

    • stress - strain relation of hooke’s law

    • Concept of relation between energy density, stress and strain

    Q-2 (JEE Main 2022)

    A metal wire of length 0.5 \mathrm{~m} and cross-sectional area 10^{-4} \mathrm{~m}^{2} has breaking stress 5 \times 10^{8} \mathrm{Nm}^{-2}. A block of is attached at one end of the string and is rotating in a horizontal circle. The maximum linear velocity of block will be__________ \mathrm{ms}^{-1}

    Answer = 50

    Solution:

    \mathrm{\ell=0.5m}

    \mathrm{A=10^{-4}m^{2}}

    \mathrm{Stress=5\times 10^{8}}

    Q-2%20(1)

    \mathrm{T =\frac{m v^2}{\ell} }

    \mathrm{\text { Stress } =\frac{I}{A}=\frac{m v^2}{\ell A} }

    \mathrm{5 \times 10^8 =\frac{10 v^2}{(0.5) \times 10^{-4}} }

    \mathrm{2.5 \times 10^3 =V^2 }

    \mathrm{V^2 =2500 }

    \mathrm{V =50 \mathrm{~m} / \mathrm{s} }

    The maximum linear velocity will be \mathrm{50 \mathrm{~m} / \mathrm{s} }

    Concept used

    • Centrifugal force

    • Concept to calculate stress

    Q-3 (JEE Main 2021)

    Wires W_{1}$ and $W_{2} are made of same material having the breaking stress of 1.25 \times 10^{9} \mathrm{~N} / \mathrm{m}^{2} . \mathrm{W}_{1}$ and $\mathrm{W}_{2} have cross-sectional area of 8 \times 10^{-7} \mathrm{~m}^{2}$ and $4 \times 10^{-7} \mathrm{~m}^{2}, respectively. Masses of 20 \mathrm{~kg}$ and $10 \mathrm{~kg} hang from them as shown in the figure. The maximum mass that can be placed in the pan without breaking the wires is _______kg. (Use \: \: g=10 \mathrm{~m} / \mathrm{s}^{2} )

    q-3%20(1)

    Answer = 40 Kg

    Solution:

    ans-3

    T_{1}= T_{2}+200\\ ......(1)

    T_{2}= mg+100\\ .......(2)

    A_{1}= 8\times10^{-7}m^{2}\\

    A_{2}= 4\times 10^{-7}m^{2}\\

    Breaking\: Stress = \frac{F}{A}\\

    For wire 1

    Breaking\: Stress = \frac{T_{1}}{A_{1}}= \frac{T_{2}+200}{A_{1}}= \frac{mg+300}{A_{1}}\\

    1.25\times10^{9}=\frac{10m+300}{8\times 10^{-7}}\\

    10\times10^{2} = 10m+300\\

    700= 10m\\

    m=70kg\\

    Maximum mass in pan wire 1 can beat is 70kg

    For wire 2

    Breaking \: Stress= \frac{T_{2}}{A_{2}}= \frac{mg+100}{4\times 10^{-7}}\\

    1.25\times 10^{9}\times 4\times 10^{-7}= 10m+100\\

    500-100= 10m\\

    m= 40kg

    Maximum mass in pan wire 2 can bear is 40kg

    Concepts Used:

    • Concept to find out tension

    • Concept of stress and breaking stress

    We have discussed in brief about the spring force formula and the Hooke’s law of the spring along with the graphical representation of the spring force equation. Hooke’s law gives a better understanding of the concept of spring and its elastic behaviour and this concept is helpful in understanding the working principle of the mechanical devices.

    • Chemistry
    • Applied and interdisciplinary physics
    • Materials
    • Physics
    • Materials science
    • Chemical product engineering
    • Building engineering
    • Physical quantities
    • Classical mechanics
    • Mechanics

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