Largest mass for Ultimate Tensile Strength

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Homework Help Overview

The problem involves determining the maximum mass that a steel wire can support when a solid steel ball is hung from it. The context is within the subject area of material strength, specifically focusing on ultimate tensile strength and stress-strain relationships.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between ultimate strength, stress, and strain, with some attempting to rearrange equations to find the necessary variables. There is a suggestion to visualize the problem using a free body diagram. Questions arise about the definitions of ultimate strength and whether additional information is needed.

Discussion Status

The discussion has revealed some misunderstandings regarding the definitions of ultimate strength and stress. Participants are exploring different interpretations of the problem and clarifying concepts, with some expressing breakthroughs in understanding. No explicit consensus has been reached, but productive dialogue is ongoing.

Contextual Notes

Participants are working under the constraint of not providing full solutions and are encouraged to suggest approaches instead. There is uncertainty regarding the definitions of terms related to stress and strain, which is being actively discussed.

AJKing
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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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DEs not required. Do you know what Y represents?
 
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.
 
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]
 
Last edited:
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 :).
 

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