1. The problem statement, all variables and given/known data 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) 2. Relevant 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) 3. 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.