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Chapter 1: Problem 8

An elastomeric bearing pad consisting of two steel plates bonded to a chloroprene elastomer (an artificial rubber) is subjected to a shear force \(V\) during a static loading test (see figure). The pad has dimensions \(a=125 \mathrm{mm}\) and \(b=240 \mathrm{mm},\) and the elastomer has thickness \(t=50 \mathrm{mm} .\) When the force \(V\) equals \(12 \mathrm{kN}\), the top plate is found to have displaced laterally \(8,0 \mathrm{mm}\) with respect to the bottom plate. What is the shear modulus of elasticity \(G\) of the chloroprene?

Short Answer

Expert verified
The shear modulus of elasticity \( G \) is 2500 kPa.

Step by step solution

01

Understanding the Problem

We are given the dimensions of an elastomeric bearing pad, a force causing a lateral displacement, and need to find the shear modulus of elasticity of the chloroprene elastomer. The relevant formula for calculating shear modulus is derived from the relationship between shear stress, shear strain, and shear modulus.
02

Formula for Shear Modulus

The shear modulus of elasticity, denoted as \( G \), is defined by the relation:\[ G = \frac{ \text{Shear Stress} }{ \text{Shear Strain} } \]Where shear stress \( \tau \) is force per area \( \tau = \frac{V}{A} \), with \( A = a \times b \), and shear strain \( \gamma \) is lateral displacement over thickness \( \gamma = \frac{\Delta x}{t} \).
03

Calculate Shear Stress

First, we calculate the area \( A \) of the pad:\[ A = a \times b = 125 \text{ mm} \times 240 \text{ mm} = 30000 \text{ mm}^2 \] Convert to square meters:\[ A = 30000 \times 10^{-6} \text{ m}^2 = 0.03 \text{ m}^2 \]Now, calculate the shear stress \( \tau \):\[ \tau = \frac{12 \text{ kN}}{0.03 \text{ m}^2} = \frac{12000 \text{ N}}{0.03 \text{ m}^2} = 400000 \text{ Pa} \]
04

Calculate Shear Strain

Shear strain \( \gamma \) is the lateral displacement divided by thickness:\[ \gamma = \frac{8 \text{ mm}}{50 \text{ mm}} = \frac{8}{50} = 0.16 \]
05

Compute Shear Modulus of Elasticity

Finally, compute the shear modulus \( G \) using the shear stress and shear strain:\[ G = \frac{\tau}{\gamma} = \frac{400000 \text{ Pa}}{0.16} = 2500000 \text{ Pa} \]Convert \( G \) from Pascals to kilopascals for a more conventional unit:\[ G = 2500 \text{ kPa} \]
06

Conclusion

The shear modulus of elasticity \( G \) of the chloroprene elastomer is calculated to be 2500 kPa.

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Key Concepts

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

Elastomeric Bearing Pad
An elastomeric bearing pad plays a crucial role in structural systems, helping to absorb loads and accommodate movements. It consists of layers of elastomer (a flexible material like rubber) often bonded to steel plates.
Elastomeric bearing pads are used extensively in bridges and buildings for the redistribution of loads, allowing for controlled movement and rotations.
Their ability to mitigate stress and dampen vibrations makes them vital in engineering, enhancing stability while extending the structure's lifespan.
Chloroprene Elastomer
Chloroprene elastomer is a type of synthetic rubber, chemically known as polychloroprene. It's highly valued for its excellent mechanical properties and resilience, making it ideal for elastomeric bearing pads.
  • Durable and flexible, it absorbs energy effectively, enduring repetitive loads without losing performance.
  • Resistant to weathering, oils, and many chemicals, chloroprene elastomer performs well in harsh environmental conditions.
Its use in structures ensures reliability and adaptability, crucial for ensuring steady support and motion control.
Shear Stress
Shear stress is a measure of the force per unit area within materials, like elastomeric pads, due to applied loads. When a force slides one part of a material parallel to an adjacent part, shear stress occurs. It's crucial for understanding how materials deform and behave under load.
  • Shear stress (\( \tau \)) can be calculated using the formula \( \tau = \frac{F}{A} \), where \( F \) is the force applied, and \( A \) is the area.
  • In the exercise, a force of 12 kN acts on a pad area of 0.03 m虏, resulting in a shear stress of 400,000 Pa.
Understanding shear stress helps engineers design materials that can withstand forces without failure.
Shear Strain
Shear strain is the deformation measure when materials undergo stress, defined by the change in shape without a change in volume. It explains how much a material is distorted due to applied shear forces.
  • Shear strain (\( \gamma \)) can be expressed as \( \gamma = \frac{\Delta x}{t} \), where \( \Delta x \) is the displacement, and \( t \) is the material's original thickness.
  • In the described exercise, an 8 mm lateral shift over a 50 mm thick pad concluded in a shear strain of 0.16.
Comprehending shear strain is vital for ensuring structural integrity when designing load-bearing components.

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Most popular questions from this chapter

A pressurized circular cylinder has a sealed cover plate fastened with steel bolts (see figure). The pressure \(p\) of the gas in the cylinder is \(1900 \mathrm{kPa}\), the inside diameter \(D\) of the cylinder is \(250 \mathrm{mm}\), and the diameter \(d_{B}\) of the bolts is \(12 \mathrm{mm}\). If the allowable tensile stress in the bolts is \(70 \mathrm{MPa}\) find the number \(n\) of bolts needed to fasten the cover.

A suspender on a suspension bridge consists of a cable that passes over the main cable (see figure) and supports the bridge deck, which is far below. The suspender is held in position by a metal tie that is prevented from sliding downward by clamps around the suspender cable Let \(P\) represent the load in each part of the suspender cable, and let \(\theta\) represent the angle of the suspender cable just above the tie. Finally, let \(\sigma_{\text {allow }}\) represent the allowable tensile stress in the metal tie. (a) Obtain a formula for the minimum required crosssectional area of the tie. (b) Calculate the minimum area if \(P=130 \mathrm{kN}\) $$ \theta=75^{\circ}, \text { and } \sigma_{\text {allow }}=80 \mathrm{MPa} $$

A force \(P\) of \(70 \mathrm{N}\) is applied by a rider to the front hand brake of a bicycle ( \(P\) is the resultant of an evenly distributed pressure). As the hand brake pivots at \(A,\) a tension \(T\) develops in the 460 -mm long brake cable \(\left(A_{e}=1.075 \mathrm{mm}^{2}\right)\) which elongates by \(\delta=0.214 \mathrm{mm} .\) Find normal stress \(\sigma\) and strain \(\varepsilon\) in the brake cable.

A high-strength steel bar used in a large crane has diameter \(d=50 \mathrm{mm}\) (see figure). The steel has modulus of elasticity \(E=200 \quad\) GPa and Poisson's ratio \(v=0.3 .\) Because of clearance requirements, the diameter of the bar is limited to \(50.025 \mathrm{mm}\) when it is compressed by axial forces. What is the largest compressive load \(P_{\max }\) that is permitted?

A circular aluminum tube with a length of \(L=420 \mathrm{mm}\) is loaded in compression by forces \(P\) (see figure). The hollow segment of length \(L / 3\) had outside and inside diameters of \(60 \mathrm{mm}\) and \(35 \mathrm{mm}\), respectively. The solid segment of length \(2 L / 3\) has a diameter of \(60 \mathrm{mm} .\) A strain gage is placed on the outside of the hollow segment of the bar to measure normal strains in the longitudinal direction. (a) If the measured strain in the hollow segment is \(\varepsilon_{h}=470 \times 10^{-6},\) what is the strain \(\varepsilon_{s}\) in the solid part? (Hint: The strain in the solid segment is equal to that in the hollow segment multiplied by the ratio of the area of the hollow to that of the solid segment.) (b) What is the overall shortening \(\delta\) of the bar? (c) If the compressive stress in the bar cannot exceed \(48 \mathrm{MPa},\) what is the maximum permissible value of load \(P ?\)
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