solution :
write the expression for the lateral strain in the bolt.
ɛy = δ y/d
Here, δy and d are the decrease in diameter and original diameter respectively.
Substitute -0.5 × 10-3 in for δy and 2.5 in for d.
ɛy =
= -2.0 x 10-4
Write the expression for the Poisson’s ratio.
V = - ɛy/ ɛx
Substitute -2 x 10-4 for ɛy and 0.3 for v..
0.3 = - (2.0 x 10-4)/0.3
ɛx = 2.0 x 10-4/0.3
= 6.67 x 10-4
Write the expression for strain in -direction.
ɛx = σx/ E - vσx/E
Substitute 0 for σy, 29 x 106 psi for and 6.67 x 10-4 for ɛx.
6.67 x 10-4 = σx/29 x 106 psi
σx = 6.67 x 10-4 (29 x 106)
= 19.34 x 103 psi
the expression for the cross-section area of the bolt.
A =
Here, d is the diameter of a steel bolt.
Substitute 2.5 in for d.
A =
= 4.906
Consider the axial stress.
σx =
Substitute 19.34 x 103 psi for σx and 4.906 in2 for A.
19.34 x 10 psi =
F = 19.34 x 103 (4.906)
= 95 x 103
The initial force in the bolt is 95 x 103 ib.
