/*! 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 45 A circular specimen of \(\mathrm... [FREE SOLUTION] | 91Ó°ÊÓ

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A circular specimen of \(\mathrm{MgO}\) is loaded using a three-point bending mode. Compute the minimum possible radius of the specimen without fracture, given that the applied load is \(5560 \mathrm{~N}\left(1250 \mathrm{lb}_{\mathrm{f}}\right)\), the flexural strength is \(105 \mathrm{MPa}(15,000 \mathrm{psi})\), and the separation between load points is \(45 \mathrm{~mm}\) (1.75 in.).

Short Answer

Expert verified
Answer: The minimum possible radius of the MgO specimen under the given conditions is approximately 3.32 mm.

Step by step solution

01

Write down the given information

We are given the applied load \(F = 5560 \mathrm{~N}\), the flexural strength \(\sigma = 105 \mathrm{MPa}\), and the separation between load points \(L = 45 \mathrm{~mm}\). We will also convert the flexural strength to \(\mathrm{N/mm}^2\), so \(\sigma = 10^{3} \times 105 \mathrm{N/mm}^2\).
02

Write down the equation for bending stress in a three-point bending test

To find the minimum possible radius \(R\), we need to find the stress in the specimen. The equation for the bending stress in a three-point bending test is: \(\sigma = \frac{3FL}{2\pi R^3}\)
03

Substitute the given information into the equation

Substitute the given values for applied load, flexural strength, and separation between load points into the equation: \(10^{3} \times 105 = \frac{3 \times 5560 \times 45}{2\pi R^3}\)
04

Solve for the radius R

Now we need to solve the equation for the radius \(R\). First, simplify the equation: \(R^3 = \frac{3 \times 5560 \times 45}{2\pi \times 10^3 \times 105}\) Next, evaluate the right side of the equation: \(R^3 \approx 36.53\) Lastly, find the cube root of both sides: \(R \approx 3.32\) The minimum possible radius of the MgO specimen without fracture under the given conditions is approximately \(3.32 \mathrm{~mm}\).

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Key Concepts

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

MgO
Magnesium Oxide, often abbreviated as MgO, is a versatile material used in a variety of industries. It is known for its excellent thermal conductivity and electrical resistive properties. MgO is used in refractory materials due to its high melting point. It can withstand extreme temperatures, making it ideal for industrial applications.
In materials science, MgO is investigated for its structural properties, especially its mechanical strength. This strength makes it useful in applications requiring high durability like ceramics and insulators.
When subjected to mechanical loading, like in three-point bending tests, it is crucial to understand how MgO will behave under stress, ensuring that it meets the required performance standards without fracturing.
Flexural Strength
Flexural strength, also known as bending strength, is a measure of a material's ability to resist deformation under load. It indicates how much stress a material can handle before it cracks or breaks when bent.
  • The flexural strength of MgO is given as 105 MPa, which is quite significant, showing its ability to withstand strong forces without fracturing.
  • This property is essential for materials that encounter bending in their applications, such as in structural beams or slabs.
  • Understanding flexural strength helps engineers design components that are safe and effective under mechanical stress.
Measuring flexural strength involves applying a force to a material until it bends or breaks, helping to determine its point of failure.
Three-Point Bending
The three-point bending test is a method used to determine the flexural properties of a material. This method involves placing a specimen on two supports and applying a load at the midpoint between them.
This setup creates a bending moment that subjects the material to tensile stress on one side and compressive stress on the opposite side. The goal is to observe how the material deforms under these conditions.
  • Three-point bending is highly effective in evaluating materials like ceramics and metals, including MgO.
  • It allows the calculation of the flexural strength, helping to predict the performance of the material in real-world applications.
In the exercise, a circular specimen of MgO undergoes a three-point bending test to determine the minimum radius it must have to avoid fracture, given certain constraints.
Bending Stress
Bending stress is a critical factor in materials science and engineering. It refers to the internal stress induced in a material when subjected to a bending load. This stress can cause deformation and, potentially, structural failure.
Bending stress is calculated to ensure that materials can handle the loads they will encounter during use. The formula used in three-point bending is:\[\sigma = \frac{3FL}{2\pi R^3}\]
  • Where \(\sigma\) is the bending stress, \(F\) is the applied force, \(L\) is the span length, and \(R\) is the radius of the specimen.
  • This equation helps in determining the minimum dimensions that a sample should have.
By substituting the known values into this equation, we can calculate the bending stress and ensure it remains below the flexural strength limit, preventing material failure.

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Most popular questions from this chapter

(a) A three-point transverse bending test is onducted on a cylindrical specimen of aluminum oxide having a reported flexural strength of 300 MPa \((43,500 \mathrm{psi}) .\) If the specimen radius is \(5.0\) \(\mathrm{nm}(0.20 \mathrm{in} .)\) and the support point separation listance is \(15.0 \mathrm{~mm}\) (0.61 in.), would you expect he specimen to fracture when a load of \(7500 \mathrm{~N}\) \(1690 \mathrm{lb}_{\mathrm{f}}\) ) is applied? Justify your answer. b) Would you be \(100 \%\) certain of the answer in art (a)? Whv or whv not?

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