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.