Largest mass for Ultimate Tensile Strength

In summary, the conversation discusses a problem involving a solid steel ball hanging from a steel wire and asks for the maximum radius and mass of the ball that the wire can bear. The equation Y = σ/ε is mentioned, with the suggestion to arrange it by knowns and unknowns. The idea of using differential equations is brought up, but it is determined that they are not necessary. The conversation then shifts to discussing the concept of ultimate strength, which is the stress associated with failure. The problem is solved correctly using this understanding.
  • #1
AJKing
104
2
Vibrations and Waves, A.P. French, questions 3-9
Please /do not/ provide full solutions. Just suggestions.

Homework Statement



A solid steel ball is hung at the bottom of a steel wire of length 2m and radius 1mm, the ultimate strength of steel is 1.1E9 N/m2. What are the radius and mass of the biggest ball the wire can bear?

Homework Equations



[itex]Y = \frac{\sigma}{\epsilon}[/itex] (line)

[itex]U = \frac{\sigma_u}{\epsilon_u}[/itex] (point below line)

[itex]m = \frac{4}{3}\pi*r^3 \rho[/itex]

[itex]\rho \approx 7850 Kg/m^3[/itex] (Internet)

The Attempt at a Solution



For the sake of sanity, I arrange eq.2 by knowns and unknowns.

[itex]\frac{U*A}{g*l_0} = \frac{m}{\Delta l} [/itex]

I must find Δ l

I consider two points on the Y line:

[itex]Y = \frac{\sigma_u}{\epsilon_1}[/itex]

[itex]Y = \frac{\sigma_2}{\epsilon_u}[/itex]

both can be written as

[itex]Y = C\frac{\sigma_u}{\epsilon_u} = C U[/itex]

[itex]C = Y/U[/itex]

Then I stop and think about how this will lead me in circles.

------

I have a suspicion that the solution to my problem requires differential equations. Anyone second that notion?
 
Last edited:
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  • #2
DEs not required. Do you know what Y represents?
 
  • #3
You know the ultimate stress the wire is capable of sustaining. You also know the size of the wire. Can you use this information to work back and figure out the amount of mass required to produce this stress? Draw a free body diagram if it helps you to visualize the situation.
 
  • #4
SteamKing said:
You know the ultimate stress the wire is capable of sustaining. You also know the size of the wire. Can you use this information to work back and figure out the amount of mass required to produce this stress? Draw a free body diagram if it helps you to visualize the situation.

Y is available in the text.

[strike]The ultimate stress σu. Do I really know this without knowing the ultimate strain εu?[/strike]

Have I been wrong this entire time? Is ultimate strength just the stress associated with failure?
Suggesting the following is incorrect:

[itex]U = \frac{\sigma_u}{\epsilon_u}[/itex]
 
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  • #5
Holy hell.

Yes. I was incorrect about ultimate strength!
Break through in understanding here folks.

Ultimate strength is the stress of failure.

Question solved correctly.

Our discussion in the other forum makes more sense now SteamKing - cheers :).
 

Related to Largest mass for Ultimate Tensile Strength

1. What is the definition of Ultimate Tensile Strength?

Ultimate Tensile Strength (UTS) is the maximum amount of stress that a material can withstand before breaking or fracturing.

2. How is the largest mass for Ultimate Tensile Strength determined?

The largest mass for Ultimate Tensile Strength is determined through testing and analysis of different materials under various tensile loads until the breaking point is reached. The mass of the material that can withstand the highest tensile load before breaking is considered to have the highest UTS.

3. What factors affect the largest mass for Ultimate Tensile Strength?

The largest mass for Ultimate Tensile Strength can be affected by several factors, including the type and composition of the material, its density, and any external forces or environmental conditions that it may be exposed to.

4. Why is the largest mass for Ultimate Tensile Strength important?

The largest mass for Ultimate Tensile Strength is important because it indicates the strength and durability of a material. Materials with high UTS are more likely to withstand stress and tension, making them suitable for use in structural and load-bearing applications.

5. How is the largest mass for Ultimate Tensile Strength used in engineering and manufacturing?

The largest mass for Ultimate Tensile Strength is a crucial factor in material selection for engineering and manufacturing purposes. It helps engineers and manufacturers determine which materials are suitable for specific applications and can withstand the required amount of stress and tension.

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