Comparing structural bending strength

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Homework Help Overview

The discussion revolves around determining the outer diameter of a hollow rod that would provide the same structural bending strength as a solid rod, while also comparing the weight per unit length of both rods. The materials involved are stainless steel, with specific properties provided, including Young's modulus and weight density.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to clarify the meaning of the 2mm thickness and its relevance to the hollow rod's dimensions. They also question how weight density factors into the calculations. Subsequent participants provide calculations for the solid rod's area and moment of inertia, while also exploring the necessary equations for the hollow rod. There are inquiries about the correctness of the approach and the values derived for the outer diameter and moment of inertia.

Discussion Status

Participants are actively sharing their calculations and seeking validation for their approaches. Some have provided partial solutions and are discussing the implications of their findings, while others are questioning the methods and values used, particularly in relation to the weight per unit length comparison.

Contextual Notes

There are indications of confusion regarding the definitions and implications of certain parameters, such as the thickness of the hollow rod and the application of weight density in the context of the problem. Additionally, there is a correction noted regarding the moment of inertia values used in the calculations.

nivek0078
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Hello,

Not sure how to go about solving this problem:

What outer diameter must a 2mm thick hollow rod have in order to have the same structural bending strength as a solid rod with diameter of 7mm? Also compare the weight per unit length of the two rods. Assume that material is stainless steel (E=180MPa) having weight density of 7.78 x 10^4N/m^3.

Does the 2mm thick refer to the length of the hollow rod? Where does weight density play into the equation?

Given equations:
stress = Mc/I

Bending rigidity -> ymax= cFl^3/EI

Thank you in advance for any information you can provide.
 
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Ok so this is what I have so far:

for the soild bar
find area
A=1/4(pie)(dia)^2 -> =38.48mm^2

find moment
I=Pie/64(D^4) -> = 117.85mm^4

Find stress
stress=My/I where M=bending moment, y=neutral axis, I=moment
-> =M(3.5)/117.85 -> stress=.0297(M)
for this equation I don't know how to find M

So to have the outer diameter of the hollow rod
find the area
A=pie((Do^2)-(Di^2)) where Do is outer dia and Di is inner dia

solution for outer diameter is Do=4.03mm

Is this the correct approach for that part of the problem??
 
Still having some issues with how to compare the weight per unit length of the two rods. I'm given E=108x10^3MPa, weight density= 7.78x10^4N/m^3 for both rods, calculated I=117.85 for solid bar. Using the equation Ymax= cFl^3/EI where c=constant, l=1.

So please advise on how to proceed here.

Also a fix to the above post.
using the I value from the solid and plugging that value into this equation for the hollow bar I=(pi/64)((Do^4)-(Di^4)) Do=7.0115mm^4 instead of the 4.03 value, is this correct?
 
You have that for a solid rod diameter D the 2nd moment of area is ##\frac{\pi D^4}{64}##. For a hollow rod, just treat it as the difference of two solid rods. I.e. subtract the 2d MoA of the missing part from what you would get if the rod were solid.
 

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