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

wavelength variability is the right one

layers of the earth affect the waves and these effects are seen on instruiments such as seimometers