What is the deformation of a shaft under axial loading?

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SUMMARY

The deformation of a shaft under axial loading can be calculated using the formula deformation = PL/AE, where A is the cross-sectional area given by A = π/4(d²). In the discussion, the internal forces were determined as 6 kips for segment AB, resulting in a deformation of 0.0024 inches. For segment BC, the internal force was calculated as 2 kips, leading to a corrected deformation of 0.0026 inches, not 0.026 inches as initially stated. Segment CD experienced an internal load of 1 kip, resulting in a deformation of 0.042 inches.

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Homework Statement


http://prntscr.com/7rw2hj

Homework Equations


deformation = PL/AE
A= pi/4 (d^2)

The Attempt at a Solution


The displacement of the shaft is equal to the summation of the displacement of each individual piece. I started on end A, and got the internal force to be 6kips
then
deformation at AB ~ 6x50/(9pi/4 x 18x10^3)
= .0024 in

for the middle shaft i got the internal force to be 2 Kips ( this could be where i am wrong, I am not completely sure I am doing these FBD's right.

deformation at BC ~ 2x75/(pi x 18x10^3) ( the 4's cancel out)
= .026

internal load at CD is 1Kip

deformation at CD ~ 60/(pi/4 x 18x10^3)
= .042

for the last two calculations, i believe they are off by a power of 10...and i can't figure out why.
I have attached my FBD's also, thanks in advance
 

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You just did the math wrong; 2*75/(pi*18000)=.0026 not .026
 
thats ridiculous.
Did i have the fBD's correct?
 

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