Mechanics of Materials: Deformation of a Hollow Cylinder

[Taiki_Kazuma]

Homework Statement

Hollow Circular tube of Length (L) 600 mm is compressed by forces P (axially).
Outside diameter (d2) is 75 mm.
Inside diameter (d1) is 63 mm.
Modulus of Elasticity (E) is 73 GPa
Poisson's ratio (v) is 0.33.
axial strain (ε) is 781 x 10^-6

Find shortening of tube (δ). (This was calculated to be 0.469 mm)
Find % change in cross sectional area. (Answer is -0.081%)
Find volume change of the tube. (Answer is -207 mm³)

Homework Equations

δ = εL
δ = L' - L
A = ¹/₄ π d²
v = -^ε'/_ε (ε' is lateral strain)
σ = Eε = ^P/_A (σ is stress)

The Attempt at a Solution

Using equation 1, the tube shortens 0.469 mm.

I figured I should calculate the lateral strain (ε') first.
Using equation 4, ε' = 258 x 10^-6

Then, using equation 1 (and 2) laterally for d₂, I get d'₂ is 75.02 mm.

Similarly using equation 1 and 2 for d₁, I get d'₁ is 62.98 mm. (I assumed that the walls would expand which is why d₁ to d'₁ decreased).

Now, calculating the area difference I used the following equation 3:
A = ¹/₄ π (d₁²-d₁²) = 1300 mm²
A' = ¹/₄ π (d'₁²-d'₁²) = 1304.5 mm²

(A' - A) / A = 0.346% *Should be -0.081%*

I also tried to calculate Volume change, but received the incorrect answer. I believe my issue with Volume is the same reason for missing Area. Please let me know what I'm missing.

 

[billy_joule]

Homework Statement

Hollow Circular tube of Length (L) 600 mm is compressed by forces P (axially).
Outside diameter (d2) is 75 mm.
Inside diameter (d1) is 63 mm.
Modulus of Elasticity (E) is 73 GPa
Poisson's ratio (v) is 0.33.
axial strain (ε) is 781 x 10^-6

Find shortening of tube (δ). (This was calculated to be 0.469 mm)
Find % change in cross sectional area. (Answer is -0.081%)
Find volume change of the tube. (Answer is -207 mm³)

The second answer should ring alarm bells right off the bat. Things bulge during axial compression, not contract, so we expect an increase in CSA.
I think your working is correct.

 

[Chestermiller]

What was the load P?

Chet

 

[Chestermiller]

The two diameters should increase in the compressed state. They should each increase by 0.0258 % (as you showed). The cross sectional area should increase by twice this percent, or 0.0516%. So the length decreases by 0.0781%, and the area increases by 0.0516%. So the volume decreases by 0.0781-0.0516=0.0265%. This percent of the volume translates into volume decrease of 207 mm³.

With an increase of 0.0258% for each of the diameters, what is the new cross sectional area? With a decrease of 0.0718% in length, what is the new length? What is the new volume? What is the decrease in volume? How does this compare with the 207 mm³?

Chet