Mechanical engineeringLesson 4 of 5 in the course Strength of materials

Hooke's Law: Springs, Elasticity, and Material Response

5 flashcards · 5 quiz questions · 5 match pairs · 5 fill-in sentences · English

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Summary

Hooke's law describes how an elastic object responds to a small force. It is a foundational model for understanding springs and many engineering materials.

For a spring displaced a distance xx from its relaxed position, the magnitude of the force needed to hold it there is

Here, FsF_s is the applied force, xx is the extension or compression, and kk is the spring constant, or stiffness. A larger kk means more force is required for the same displacement. Rearranging gives x=Fs/kx=F_s/k.

All flashcards

What does the slope of an applied-force-versus-displacement graph represent for a linear spring?
The spring constant kk, which measures stiffness.
How is elastic potential energy related to the displacement of an ideal linear spring?
It is Uel=rac12kx2U_{el}=rac{1}{2}kx^2.
What does a stiffness matrix represent in linear elasticity?
A linear mapping from a displacement vector to a force vector, including directional coupling.
What distinguishes elastic recovery from linear proportionality?
A material can recover its shape while its stress–strain response is not linear.
How does the axial stiffness of a uniform elastic rod depend on its dimensions?
It is k=AE/Lk=AE/L, where AA is cross-sectional area, EE is Young's modulus, and LL is length.

Quiz questions

  1. For a torsional spring with angular displacement θ\theta, which equation gives its restoring torque?Answer: τ=−kθ\tau=-k\theta
  2. Which pair gives the two independent elastic moduli used to describe an isotropic linear-elastic material?Answer: Bulk modulus KK and shear modulus GG
  3. How many independent elastic coefficients describe a general anisotropic material in three-dimensional linear elasticity?Answer: 21
  4. A uniform rod has length LL, area AA, and Young's modulus EE. What is its equivalent axial spring stiffness?Answer: k=AE/Lk=AE/L
  5. For an ideal mass–spring oscillator with mass mm and stiffness kk, which expression gives its frequency?Answer: f=12πkmf=\frac{1}{2\pi}\sqrt{\frac{k}{m}}

Based on the Wikipedia article Hooke's law, licensed CC BY-SA 4.0. Adapted into a summary, flashcards and quizzes.

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