/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 82 A vibration isolation unit consi... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A vibration isolation unit consists of two blocks of hard rubber with a modulus of rigidity \(G=19\) MPa bonded to a plate \(A B\) and to rigid supports as shown. Denoting by \(P\) the magnitude of the force applied to the plate and by \(\delta\) the corresponding deflection, determine the effective spring constant, \(k=P / \delta,\) of the system.

Short Answer

Expert verified
The effective spring constant is \(k = \frac{GA}{t}\).

Step by step solution

01

Understand the given variables

The exercise involves a vibration isolation unit with two blocks of hard rubber. The modulus of rigidity is given as \(G = 19\) MPa. The force applied to the plate is \(P\), and the deflection caused by this force is \(\delta\). Our goal is to find the effective spring constant \(k\) of this system using the formula \(k = \frac{P}{\delta}\).
02

Relate shear modulus to force and deflection

The modulus of rigidity \(G\) relates the shear stress to the shear strain. In this context, shear stress is \(\tau = \frac{P}{A}\), where \(A\) is the area, and shear strain is \(\gamma = \frac{\delta}{t}\), where \(t\) is the thickness of the rubber. Thus, \(G = \frac{\tau}{\gamma} = \frac{P}{A} \div \frac{\delta}{t} = \frac{Pt}{A\delta}\). Rearrange for \(P\): \(P = \frac{GA\delta}{t}\).
03

Substituting into the spring constant formula

Replace \(P\) in the spring constant formula \(k = \frac{P}{\delta}\) with the expression from Step 2: \(k = \frac{GA\delta}{t \cdot \delta} = \frac{GA}{t}\). Here, \(A\) is the area of one plate and \(t\) is the thickness of one rubber block.
04

Calculate the effective spring constant

Using the expression for \(k\), we find \(k = \frac{GA}{t}\). Given \(G = 19\) MPa, \(A\) and \(t\) depend on the specific dimensions of the rubber blocks, which should be provided as part of the diagram or additional information in the problem.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Modulus of Rigidity
The modulus of rigidity, also known as the shear modulus, is a fundamental property of materials that helps describe their ability to resist deformation under shear stress. It is a measure of the material's stiffness when subjected to shear forces. In mathematical terms, the modulus of rigidity is defined as the ratio of shear stress to shear strain: \[ G = \frac{\tau}{\gamma} \]- \(G\) represents the modulus of rigidity.- \(\tau\) is the shear stress.- \(\gamma\) is the shear strain.In this exercise, we were given a modulus of rigidity of 19 MPa for hard rubber. This number tells us how stiff the rubber is when shear forces are applied, providing insight into how effectively the material can isolate vibrations. Remember, the higher the modulus of rigidity, the stiffer the material, meaning more force is needed for the same amount of deformation.
This concept is crucial in the field of mechanical engineering, particularly in designing systems that can manage or mitigate the transmission of vibrations.
Shear Stress
Shear stress occurs when a force is applied parallel or tangentially to the surface of a material, causing it to deform. It's basically the force distributed over an area, leading to a sliding failure along the surface of the material. The formula for shear stress is:\[ \tau = \frac{P}{A} \]- \(\tau\) stands for shear stress.- \(P\) is the force applied.- \(A\) is the area over which the force is distributed.In the context of vibration isolation units, calculating shear stress is important as it determines the amount of force that the material can endure before it begins to deform. For example, in this exercise, the force \(P\) applied to the plate causes shear stress in the rubber blocks, resulting in deformation.
Understanding shear stress helps in the design and selection of materials that can effectively absorb and dampen vibrations, protecting structures or devices from potential damage.
Spring Constant
The spring constant, represented by \(k\), is a measure of a spring's stiffness. This concept extends to vibration isolation systems, where materials or combinations behave like springs. The spring constant is defined using Hooke's Law for simple springs: \[ k = \frac{P}{\delta} \]- \(k\) is the spring constant.- \(P\) represents the applied force.- \(\delta\) is the resulting deflection.In the problem, the effective spring constant of the vibration isolation unit was calculated as \(k = \frac{GA}{t}\). The spring constant indicates how much force is required to deflect the system by a certain amount.
A higher spring constant means the system is stiffer, requiring more force to achieve the same level of deflection. Understanding how to calculate the spring constant in these systems is vital for designing mechanisms that need controlled motion or specific vibration reduction.
Shear Strain
Shear strain is the deformation measure resulting from shear stress, representing the angular distortion of the material. It is expressed as the ratio of the change in shape to the original shape under an applied force, without changing its volume. Mathematically, shear strain is defined as:\[ \gamma = \frac{\delta}{t} \]- \(\gamma\) denotes shear strain.- \(\delta\) is the lateral or tangential deflection.- \(t\) is the original thickness of the material.In this context, shear strain informs us about how much a material will deform when subjected to a force, which is critical when working with vibration isolators. The ability of a material to undergo deformation without failure determines its effectiveness in absorbing and dissipating energy.
A thorough understanding of shear strain helps in selecting materials that accommodate the required deformation levels in various applications, ensuring the longevity and performance of the vibrational isolation system.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Two steel bars \(\left(E_{s}=200 \mathrm{GPa} \text { and } \alpha_{s}=11.7 \times 10^{-6} /^{\circ} \mathrm{C}\right)\) are used to reinforce a brass bar \(\left(E_{b}=105 \mathrm{GPa}, \alpha_{b}=20.9 \times 10^{-6} /^{\circ} \mathrm{C}\right)\) that is subjected to a load \(P=25\) kN. When the steel bars were fabricated, the distance between the centers of the holes that were to fit on the pins was made \(0.5 \mathrm{mm}\) smaller than the \(2 \mathrm{m}\) needed. The steel bars were then placed in an oven to increase their length so that they would just fit on the pins. Following fabrication, the temperature in the steel bars dropped back to room temperature. Determine \((a)\) the increase in temperature that was required to fit the steel bars on the pins, ( \(b\) ) the stress in the brass bar after the load is applied to it.

A standard tension test is used to determine the properties of an experimental plastic. The test specimen is a \(\frac{2}{8}\) -in.- diameter rod and it is subjected to an 800 -lb tensile force. Knowing that an elongation of 0.45 in. and a decrease in diameter of 0.025 in. are observed in a 5 -in. gage length, determine the modulus of elasticity, the modulus of rigidity, and Poisson's ratio for the material.

Denoting by \(\epsilon\) the "engineering strain" in a tensile specimen, show that the true strain is \(\epsilon_{t}=\ln (1+\epsilon)\)

Two tempered-steel bars, each \(\frac{3}{16}\) in. thick, are bonded to a \(\frac{1}{2}\) -in. mild-steel bar. This composite bar is subjected as shown to a centric axial load of magnitude \(P\). Both steels are elastoplastic with \(E=29 \times 10^{6} \mathrm{psi}\) and with yield strengths equal to \(100 \mathrm{ksi}\) and \(50 \mathrm{ksi},\) respectively, for the tempered and mild steel. The load \(P\) is gradually increased from zero until the deformation of the bar reaches a maximum value \(\delta_{m}=0.04\) in. and then decreased back to zero. Determine \((a)\) the maximum value of \(P\), (b) the maximum stress in the tempered-steel bars, \((c)\) the permanent set after the load is removed.

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.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.