# Comparing structural bending strength

1. Nov 4, 2013

### nivek0078

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.

2. Nov 4, 2013

### nivek0078

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??

3. Nov 4, 2013

### nivek0078

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.

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?

4. Nov 5, 2013

### haruspex

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.