Two tubes (one solid, one hollow) concentrically arranged as shown attached to the ground with a known mass on top. Both have equal areas of cross section.
Which tube sees more load? More deflection? Calculate stresses and strains in both. What if the mass on top was rigid? The mass is not attached, just placed.
Key things to recognize:
d = FL/AE (where d is displacement, F is the force imparted by the block, A is the cross-sectional area of the axially loaded body, E is the elastic modulus of the axially loaded body.)
Since d_al = d_st,
(F_al * L) / (A * E_al) = (F_st * L) / (A * E_st)
A and L can be canceled to get,
(F_al) / (E_al) = (F_st) / (E_st)
We recognize that E_al < E_st, thus F_st > F_al.
stress_st = F_st/A
stress_al = F_al/A
strain = (d_0 - d)/d_0
Summary:
The steel tube sees more load.
Deflection is the same in both tubes.
(not sure what is meant by what if the mass on top was rigid, I assumed it was rigid)
Hello,
I think the question is just focused on two steel tubes. One hollow and one solid, hence I feel the difference will be related to the moment of inertia. What do you think? @Mark_Washington
My 2 cents:
Case 1: When mass is not rigid (assume extremely heavy jelly), the aluminum would deflect more since young’s modulus is lower and deflection = PL/AE, everything else being the same . Aluminum would deflect 3 times as much as steel would. Here, the force on the aluminum would be much higher than steel and also the stress. Perhaps, as the aluminum compresses, all the weight of the jelly “m” would be on the aluminum and none on the steel?
Case 2: When the mass is rigid, displacement of both would be the same. The displacement would be limited by how much the stiffer material (steel) would deflect. Not sure what the distribution of the weight would be between the aluminum and steel, but maybe assume half and half?
@Mark_Washington @Chirag_Shah @ronron
Any critiques?
