Radius of Specimen from Flexural Strength

In summary: So in summary, the minimum possible radius of the specimen without fracture is 10.24 mm, given that the applied load is 4000 N, the flexural strength is 65 MPa, and the separation between load points is 35mm.
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Homework Statement



A circular specimen of glass is loaded using the three-point bending method. Compute the minimum possible radius of the specimen without fracture, given that the applied load is 4000 N the flexural strength is 65 MPa, and the separation between load points is 35mm.

Homework Equations



Flexural Strength = (FL)/(pi*R3) where F is the load at fracture, L is the distance between support points (in mm), and R is the radius of the specimen.


The Attempt at a Solution




Flexural Strength = (FL)/(pi*R3)

so

R = cube root of ((4000N)(0.035m)/(pi*6.5*10^7))
R = 8.818 mm

This is not one of the answers given. What am I doing wrong? The only thing I can think of is that the load given is NOT the fracture load, however I don't know how/from where to get that load.
 
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  • #2


your first step should be to double check your equations and make sure they are correct. In this case, your equation for flexural strength is incorrect. The correct equation is:

Flexural Strength = (3FL)/(2pi*R3)

You also need to convert the load from N to MPa, as the flexural strength is given in MPa. So the correct equation becomes:

Flexural Strength = (3FL)/(2pi*R3)

F = 4000N = 4N

Flexural Strength = (3*4N*0.035m)/(2pi*R3)

= 0.0857 MPa

Solving for R:

R = cube root of ((3*4000N*0.035m)/(2pi*0.0857MPa))

R = 10.24 mm

This is the minimum possible radius of the specimen without fracture.
 

FAQ: Radius of Specimen from Flexural Strength

1. What is the radius of a specimen in regards to flexural strength?

The radius of a specimen refers to the distance from the center to the outer edge of a cylindrical or circular sample that is being tested for flexural strength.

2. How does the radius of a specimen affect its flexural strength?

The radius of a specimen can significantly impact its flexural strength. Generally, a larger radius will result in a higher flexural strength, as the larger surface area can distribute the load more evenly and reduce the stress on the material.

3. Is there an ideal radius for testing flexural strength?

There is no one ideal radius for testing flexural strength, as it can vary depending on the material being tested and the purpose of the test. However, it is important to ensure that the radius is large enough to accurately represent the behavior of the material in real-world applications.

4. How is the radius of a specimen measured?

The radius of a specimen can be measured using a variety of tools, such as a caliper or micrometer. It is important to measure the radius at multiple points around the specimen to ensure accuracy.

5. Can the radius of a specimen affect the validity of flexural strength results?

Yes, the radius of a specimen can greatly impact the validity of flexural strength results. If the radius is too small, it can lead to localized stress concentrations and inaccurate results. It is important to carefully select the appropriate radius for testing to ensure reliable and valid results.

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