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oceanwalk
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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
[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
[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
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