The change in diameter of a large steel bolt is carefully measured as the nut is tightened. knowing that e 5 29 3 10 6 psi and n 5 0.30, determine the internal force in the bolt if the diameter is observed to decrease by 0.5 3 10 23 in. beer. mechanics of materials (mechanical engineering) (page 110). mcgraw-hill education. kindle edition.

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.