Comparing structural bending strength

AI Thread Summary
To determine the outer diameter of a 2mm thick hollow rod that matches the structural bending strength of a solid rod with a 7mm diameter, the calculations involve using the moment of area (I) and stress equations. The initial calculations for the solid rod yield an area of 38.48 mm² and a moment of inertia of 117.85 mm⁴. The proposed outer diameter for the hollow rod is initially calculated as 4.03mm, but later corrections suggest it should be 7.0115mm⁴ when using the appropriate equations for hollow sections. Additionally, the weight per unit length comparison requires integrating the weight density into the calculations for both rod types. The discussion emphasizes the importance of correctly applying the moment of area formulas for both solid and hollow rods to ensure accurate results.
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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