A 2 m length of an aluminum pipe of 240 mm outer diameter and 10 mm wall thickness is used as a short column to carry a 640? centric axial load. Knowing that? = 73? P? and? = 0.33, determine (a) the change in length of the pipe, (b) the change in its outer diameter, (c) the change in its wall thickness.

A) Change in length of pipe = - 2.4265 x 10^ (-3) m

B) Change in pipe outer diameter = 9.6096 x 10^ (-2) mm

C) change in thickness = 4.004 x 10^ (-3) mm

Explanation:

Load = 640KN or 640,000N

E = 73 GPa or 73 x 10^ (9) Pa

and v = 0.33

A) Inner Diameter of rod (di) = do - 2t

= 0.24m - (2x0.01) = 0.24 - 0.02 = 0.22m

So, area of cross section = (π/4) [ (do) ^2 - (di) ^2] = (π/4) [ (0.24) ^2 - (0.22) ^2] = 7.226 x 10^ (-3) m^2

Now change in pipe length;

ΔL = - PL/EA

= - [640,000 x 2] / [ (73 x 10^ (9)) x (7.226 x 10^ (-3) ] =

- [1280000]/[527.498 x 10^ (6) ] =

-2.4265 x 10^ (-3) m

B) let's first calculate the strain in longitudinal direction;

ε = (δL) / L = [-2.4265 x 10^ (-3) ]/2

= - 1.2133 x 10^ (-3)

For change in diameter;

Δdo = do (εLAT)

εLAT = - v (ε)

So v = 0.33 from the question, thus

εLAT = - (0.33 x - 1.2133 x 10^ (-3)) = 4.004 x 10^ (-4)

Thus; Δdo = 240 x 4.004 x 10^ (-4) =

9.6096 x 10^ (-2) mm

C) for change in thickness;

Δt = t (εLAT) = 10 x 4.004 x 10^ (-4) = 4.004 x 10^ (-3) mm