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A tensile test is peformed on a brass specimen \(10 \mathrm{mm}\) in diameter using a gage length of \(50 \mathrm{mm}\) (see figure). When the tensile load \(P\) reaches a value of \(20 \mathrm{kN}\) the distance between the gage marks has increased by \(0.122 \mathrm{mm}\) (a) What is the modulus of elasticity \(E\) of the brass? (b) If the diameter decreases by \(0.00830 \mathrm{mm}\), what is Poisson's ratio?

Short Answer

Expert verified
(a) \(E = 104,379 \mathrm{MPa}\); (b) Poisson's ratio = 0.34.

Step by step solution

01

Understand the Problem

The problem involves finding two properties of a brass specimen under a tensile test. We first need to find the modulus of elasticity \(E\), and then Poisson's ratio. Modulus of elasticity relates stress with strain, while Poisson's ratio relates lateral and axial strains.
02

Calculate Stress

Stress \(\sigma\) is calculated by dividing the tensile load by the cross-sectional area of the specimen. The cross-sectional area \(A\) is given by \(A = \frac{\pi}{4}d^2\), where \(d = 10 \mathrm{mm}\). Thus:\[A = \frac{\pi}{4} \times (10)^2 = 78.54 \mathrm{mm}^2\]The stress \(\sigma\) is:\[\sigma = \frac{P}{A} = \frac{20000 \mathrm{N}}{78.54 \times 10^{-6} \mathrm{m}^2} = 254.65 \mathrm{MPa}\]
03

Calculate Strain

Strain \(\varepsilon\) is calculated as the change in gauge length divided by the original gauge length:\[\varepsilon = \frac{\Delta L}{L_0} = \frac{0.122 \mathrm{mm}}{50 \mathrm{mm}} = 0.00244\]
04

Calculate Modulus of Elasticity

Modulus of elasticity \(E\) is calculated using the formula \(E = \frac{\sigma}{\varepsilon}\):\[E = \frac{254.65 \mathrm{MPa}}{0.00244} = 104,379 \mathrm{MPa}\]
05

Calculate Lateral Strain

The lateral strain \(\varepsilon_l\) is calculated as the change in diameter divided by the original diameter:\[\varepsilon_l = \frac{\Delta d}{d_0} = \frac{0.0083 \mathrm{mm}}{10 \mathrm{mm}} = 0.00083\]
06

Calculate Poisson's Ratio

Poisson's ratio \(u\) is the negative ratio of lateral strain to axial strain:\[u = -\frac{\varepsilon_l}{\varepsilon} = -\frac{0.00083}{0.00244} = -0.34016\]The negative sign indicates the direction of deformation (contraction in diameter). Simplified, the Poisson's ratio is 0.34.

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Key Concepts

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

Tensile Test
A tensile test is a fundamental experiment that helps us understand how materials react when pulling forces are applied. This kind of test involves stretching a material, such as a brass specimen, until it deforms. We measure its ability to withstand these forces without breaking. The test gives us valuable information such as the material's strength and ductility.

The procedure involves clamping the specimen securely and applying a force along its length. In our case, we used a brass specimen 10 mm in diameter and with a gage length of 50 mm. We observed that the gage marks on the specimen elongated as the tensile load increased, providing us with crucial strain measurements.

Key aspects of tensile tests include:
  • Determining the stress-strain characteristics of a material.
  • Gaining insights into mechanical properties such as elasticity, yield strength, and tensile strength.
  • Characterizing materials for engineering applications, ensuring they meet required specifications.
Poisson's Ratio
Poisson's Ratio is an essential parameter in material science, which describes the ratio of transverse strain to axial strain in a stretched material. In simpler terms, when a material is compressed in one direction, it tends to expand in the perpendicular direction. Poisson's Ratio helps quantify this phenomenon.

In our given example, as the brass specimen was subjected to tensile loading, its length increased while the diameter slightly decreased. To find Poisson's Ratio, we observed the changes in the diameter and longitudinal dimensions. A decrease in diameter by 0.0083 mm for our specimen revealed how lateral strain occurred.

The calculation uses:
  • Lateral strain, which is the change in the diameter divided by the original diameter.
  • Axial strain, which denotes how much the length changed.
Poisson's Ratio (\( u \) ) is significant as it helps:
  • Understand material behavior under multi-axial loading conditions.
  • Predict how materials will deform, impacting the structural integrity of components.
Stress and Strain
Stress and strain are crucial concepts that describe how materials react under force. Stress refers to the internal forces per unit area within a material, and strain represents the deformation or displacement of the material. These concepts are deployed to assess how materials stretch, compress, or change shape.

In the context of the tensile test on our brass specimen, stress was calculated as the force applied divided by the cross-sectional area. With a tensile load of 20 kN and an area of 78.54 mm², we found a stress value of 254.65 MPa.

Strain, the measure of deformation, is the ratio of the change in length to the original length. In our example, the strain was derived from a 0.122 mm elongation over an original 50 mm length, resulting in a strain of 0.00244.

Understanding stress and strain is vital for:
  • Evaluating material performance in engineering applications.
  • Ensuring structures can withstand intended loads without failure.
  • Designing components that adhere to safety and durability standards.

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Most popular questions from this chapter

A steel pad supporting heavy machinery rests on four short, hollow, cast iron piers (see figure). The ultimate strength of the cast iron in compression is \(344.5 \mathrm{MPa}\). The outer diameter of the piers is \(d=114 \mathrm{mm}\) and the wall thickness is \(t=10 \mathrm{mm}\). Using a factor of safety of 4.0 with respect to the ultimate strength, determine the total load \(P\) that may be supported by the pad.

A bar of solid circular cross section is loaded in tension by forces \(P\) (see figure). The bar has length \(L=380 \mathrm{mm}\) and diameter \(d=6 \mathrm{mm} .\) The material is a magnesium alloy having modulus of elasticity \(E=42.7\) GPa. The allowable stress in tension is \(\sigma_{\text {allow }}=89.6 \mathrm{GPa}\), and the elongation of the bar must not exceed \(0.08 \mathrm{mm}\). What is the allowable value of the forces \(P ?\)

\(\text { A solid bar of circular cross section (diameter } d)\) has a hole of diameter \(d\) / 5 drilled laterally through the center of the bar (see figure). The allowable average tensile stress on the net cross section of the bar is \(\sigma_{\text {allow }}\) (a) Obtain a formula for the allowable load \(P_{\text {allow that }}\) that the bar can carry in tension. (b) Calculate the value of \(P_{\text {allow }}\) if the bar is made of brass with diameter \(d=45 \mathrm{mm}\) and \(\sigma_{\text {allow }}=83 \mathrm{MPa}\) (Hint: Use the formulas of Case \(15,\) Appendix D.)

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 torque \(T_{0}\) is transmitted between two flanged shafts by means of ten 20 -mm bolts (see figure and photo). The diameter of the bolt circle is \(d=250 \mathrm{mm}\) If the allowable shear stress in the bolts is \(85 \mathrm{MPa}\) what is the maximum permissible torque? (Disregard friction between the flanges)

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