In summary, a steel bolt with a nominal diameter of 20 mm and a pitch of 2.5 mm is used as a spacer in conjunction with an aluminum tube with dimensions of 40 mm OD by 22 mm ID. The distance between the two plates is 0.35 m, and the nut is pulled snug and then given a one-third additional turn. Neglecting the deformation of the plates, the resulting stress in the bolt is 241 MPa and -86.3 MPa in the tube. This is determined by calculating the elongation of the bolt and tube and using their respective moduli of elasticity, with the assumption that the bolt is a solid shank for its full length.
hatchelhoff: Close, but currently incorrect. Show the steps you used to obtain your post 38 formula, so I can find where your algebra mistake occurred.
hatchelhoff: Your first equation in post 40 is currently incorrect. Show the steps you used to obtain your first equation in post 40, so I can find where your algebra mistake occurred.
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#42
hatchelhoff
65
0
((LF2*a2*e2)/L2)+((Lf1*a1*e1)/L1) = a1e1 + a2e2 (From Post 34)
sub LF2 = Lf1-2*tf3
((Lf1-2*tf3)*a2*e2)/L2)+((Lf1*a1*e1)/L1) = a1e1 + a2e2
Let
X = Lf1
T = tf3
A = a1*e1
B = a2*e2
hatchelhoff: The second equation (line 3) in post 42 is correct. The last equation in post 42 is currently incorrect. Check your work to go from the second equation to the last equation, and try again.
hatchelhoff: Excellent work. That is correct. Use six decimal places for all numerical values. Continue.
#48
hatchelhoff
65
0
nvn thanks very much for your help over the past few weeks
I now have the correct answers
Lf1 = 349.587
Lf2 = 329.587
Stress1 = 240.85 MPa
Stress2 = -86.325 MPa.