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A vibration isolation support consists of a rod \(A\) of radius \(R_{1}\) and a tube \(B\) of inner radius \(R_{2}\) bonded to an 80 -mm-long hollow rubber cylinder with a modulus of rigidity \(G=10.93\) MPa. Determine the required value of the ratio \(R_{2} / R_{1}\) if a 10 -kN force \(\mathbf{P}\) is to cause a \(2-\mathrm{mm}\) deflection of rod \(A\)

Short Answer

Expert verified
We are given: - The modulus of rigidity, \(G = 10.93 \text{ MPa} = 10.93 \times 10^6 \text{ Pa}\).- The length of the rubber cylinder, \(L = 80 \text{ mm} = 0.08 \text{ m}\).- The total force, \(P = 10 \text{ kN} = 10,000 \text{ N}\).- The desired deflection, \(\Delta = 2 \text{ mm} = 0.002 \text{ m}\).

Step by step solution

01

Identify Given Values

We are given: - The modulus of rigidity, \(G = 10.93 \text{ MPa} = 10.93 \times 10^6 \text{ Pa}\).- The length of the rubber cylinder, \(L = 80 \text{ mm} = 0.08 \text{ m}\).- The total force, \(P = 10 \text{ kN} = 10,000 \text{ N}\).- The desired deflection, \(\Delta = 2 \text{ mm} = 0.002 \text{ m}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Modulus of Rigidity
The modulus of rigidity is a key property of materials, especially when dealing with mechanical deformations. It is also known as the shear modulus, symbolized as \( G \), and measures how a material deforms under shear stress.
Specifically, it is defined as the ratio of shear stress to the shear strain: \[ G = \frac{\tau}{\gamma} \] Where \( \tau \) is the shear stress, and \( \gamma \) is the shear strain.
The modulus of rigidity provides insight into the stiffness of a material: a higher modulus indicates a stiffer material. In the context of vibration isolation, like in this problem, the modulus of rigidity helps us understand how much the material (rubber in this case) can be deformed before returning to its original shape.
  • Rubber, typically, has a low modulus, which means it's quite flexible and can absorb more vibration, making it ideal for vibration isolation.
  • It's important to choose a material with the right balance of stiffness and flexibility to ensure effective vibration isolation.
This is crucial when determining how much a part will deflect when a force is applied.
Deflection Calculation
In mechanics, deflection refers to the way a material bends or changes shape when a force is applied. Here, we calculate deflection to understand how the rubber cylinder behaves under a load of 10 kN.
To calculate deflection, you need to consider both material properties and geometry. For cylindrical objects like rods and tubes, these calculations often involve the modulus of rigidity and specific dimensions like radius and length.
The deflection \( \Delta \) can be determined by manipulating the formula for shear deformation: \[ \Delta = \frac{PL}{A_{shear} G} \] Where:
  • \( P \) is the applied force,
  • \( L \) is the length of the material,
  • \( A_{shear} \) is the area subjected to shear force,
  • and \( G \) is the modulus of rigidity.
The area \( A_{shear} \) depends on the geometry of the piece; for a hollow cylinder, it involves the inner and outer radii. In our problem, getting the right ratio between \( R_2 \) and \( R_1 \) is critical to achieving the desired deflection, especially given the material's inherent properties.
Mechanics of Materials
Mechanics of materials, also known as strength of materials, is the study of how different materials deform and bear loads. This domain of engineering is essential to designing and analyzing parts like the vibration isolation system in our exercise.
Here, we delve into how external forces, like the 10 kN load on the rod, affect the deformation and stress distribution in the vibration isolation setup.
  • This involves understanding stress-strain relationships, key for analyzing potential failure points and ensuring the design remains within safe limits.
  • Consideration of material properties like modulus of rigidity, elasticity, and yield strength is crucial.
Understanding these principles helps in predicting deflections accurately and choosing the right materials to prevent excessive deformation.
This ensures functionality and safety, making mechanics of materials a fundamental topic for students and practitioners in engineering contexts.

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Most popular questions from this chapter

The steel rails of a railroad track \(\left(E_{3}=200 \mathrm{GPa}, \alpha_{5}=11.7 \times 10^{-6} /^{\circ} \mathrm{C}\right)\) were laid at a temperature of \(6^{\circ} \mathrm{C}\). Determine the normal stress in the rails when the temperature reaches \(48^{\circ} \mathrm{C}\), assuming that the rails \((a)\) are welded to form a continuous track, \((b)\) are \(10 \mathrm{m}\) long with \(3-\) mm gaps between them.

An 18 -m-long steel wire of 5 -mm diameter is to be used in the manufacture of a prestressed concrete beam. It is observed that the wire stretches \(45 \mathrm{mm}\) when a tensile force \(\mathbf{P}\) is applied. Knowing that \(E=200 \mathrm{GPa}\), determine ( \(a\) ) the magnitude of the force \(\mathbf{P}\), (b) the corresponding normal stress in the wire.

The cylindrical rod \(A B\) has a length \(L=5 \mathrm{ft}\) and a 0.75 -in. diameter; it is made of a mild steel that is assumed to be elastoplastic with \(E=29 \times 10^{6} \mathrm{psi}\) and \(\sigma_{Y}=36 \mathrm{ksi} .\) A force \(\mathrm{P}\) is applied to the bar and then removed to give it a permanent set \(\delta_{P}\). Determine the maximum value of the force \(\mathbf{P}\) and the maximum amount \(\delta_{m}\) by which the bar should be stretched if the desired value of \(\delta_{P}\) is \((a) 0.1\) in. \((b) 0.2\) in.

A square yellow-brass bar must not stretch more than \(2.5 \mathrm{mm}\) when it is subjected to a tensile load. Knowing that \(E=105\) GPa and that the allowable tensile strength is \(180 \mathrm{MPa}\) determine \((a)\) the maximum allowable length of the bar, \((b)\) the required dimensions of the cross section if the tensile load is \(40 \mathrm{kN}\)

In many situations physical constraints prevent strain from occurring in a given direction. For example, \(\epsilon_{z}=0\) in the case shown, where longitudinal movement of the long prism is prevented at every point. Plane sections perpendicular to the longitudinal axis remain plane and the same distance apart. Show that for this situation, which is known as plane strain, we can express \(\sigma_{z}, \epsilon_{x},\) and \(\epsilon_{y}\) as follows: \\[ \begin{array}{l} \sigma_{z}=\nu\left(\sigma_{x}+\sigma_{y}\right) \\ \epsilon_{x}=\frac{1}{E}\left[\left(1-\nu^{2}\right) \sigma_{x}-\nu(1+\nu) \sigma_{y}\right] \\ \epsilon_{y}=\frac{1}{E}\left[\left(1-\nu^{2}\right) \sigma_{y}-\nu(1+\nu) \sigma_{x}\right] \end{array} \\]

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