Solid Mechanics - Force in a bolted assembly

[oceanwalk]

A bolted assembly consists of a steel bolt A, a brass tube B and a nut C. The nut is turned so that it just secures the tube, and then is tightened one extra turn. Determine the resulting force Fs (in kN) in the steel bolt. The initial length of the tube is L = 158 mm, and the screw pitch is 1mm (the nut moves 1mm along the bolt for each turn). For the steel bolt, Es = 200 GPa, As = 28 mm2. For the brass tube, Eb = 100GPa, Ab = 142 mm2.
https://www.vista.ubc.ca/webct/RelativeResourceManager/Template/RspQ-ps-221-3-v1/ps-221-3-q19-1.jpg

Homework Equations

[F][/b]+[F][/s] = F (not sure about this because it's a bolt and not something placed between the nut like the tube)
[δ][/s]= [δ][/b]=δ = 1mm
[σ][/s] = [F][/s]/ [A][/s] = [E][/s]δ/L

The Attempt at a Solution

[σ][/s] = [E][/s]δ/L = 200*(1/158) = 1.2658GPa
[F][/s] = [σ][/s] * [A][/s] = (1.2658*10^9) * (28*10^-6) = 35kN

 

[nvn]

oceanwalk: What do mean by all the brackets and such? To correct typos, you can hit the Edit button (for 24 hours).

[F][/b] + [F][/s] = F

Incorrect. Try again.

[delta][/s] = [delta][/b] = delta = 1 mm

Incorrect. Try again.