It sounds to me like 2mm thick refers to the distance from the outer diameter to the inner diameter. The weight density looks like it comes into comparing the weight per unit length of the two rods.
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 lenght 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 lenght 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.
Thank you SlipEternal that makes more sense now. So far I've found the area of the solid rod to be A=1/4(pie)(dia)^2 = 38.48.
How do I find the stress or force using this equation F= stress(Area). No force was given in the problem?
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?
Area is calculated by radius, not diameter. I am not sure if that will change any of your results. It sounds like you want . In other words, you want . Solving for D, you get . I am not sure where you are getting your values.
Thank you for responding again. This is a site i used to get the equation for my numbers. How do you calculate the maximum bending stress of a hollow tube? - Yahoo! Answers
where there equations are for the solid I=pi(D^4)/64 and for the hollow I= pi((Do^4)-(Di^4))/64. Your approach seems more true is your D value the outer diameter or radius?
The outer diameter is mm the inner diameter is mm. To calculate weight per unit length, calculate the cross-sectional area. For the solid tube, it is . Multiply that by the density: .
For the hollow tube, it would be .
Note: For each, I divide by 4 since .
Note also, I did not use E at all, so it is possible I did this wrong. I just calculated the diameter needed for the I values to be the same. It is possible that does not imply the same bending strength. The formulas from that forum were for bending stress, not bending strength. Since you don't know the length, maybe that doesn't tell you the bending strength. I don't know enough about it.
Thanks for your help again! You really make the problem easy to understand. I was having a really hard time just finding the right information to use. I'm pretty sure you are doing it correctly. The weight per unit length is a force value. so you consider equal I values.
Where the deflection equation is
Ymax = cFl^3/EI
where c = constant, F= weight per unit length, l = length 1
So doing the math you find that the deflection of the hollow rod is more that the solid which makes sense.