Normal Strain

 

 

*Example:  A brass, circular bar with a diameter, d, of 10 mm is subjected to an axial load,

P, of 25 kN.  The gage length, L, between points A and B on the bar (shown below) is 50 mm. 

As a result of the axial load the bar extends 0.152 mm.  Poisson’s ratio, ν, for brass  is 0.34.

Assume response of the bar is in the elastic range.  Find the modulus of elasticity, E, and

the change in diameter of the bar, ∆d.

 

 

 

                      

 

 

 

Strategy:  Calculate the axial stress, σ, , and the axial strain,ε, then apply Hooke’s Law, 

σ  =  .  Next calculate the lateral strain using Poisson’s ratio.  Use it to calculate the

change in length of the diameter.

 

 

                                                 

Axial stress:       σ  =  P/A   =  25000 / [π(10)2/4]  =  318.3 MPa

 

Axial strain:    ε  =  ∆L / L  =  152/50  =  0.00304    (result for axial strain)

 

From Hooke’s Law:     E  =  σ/ε  =  318.3/0.00304  =  104,700 N/mm2    (result)

 

 

As the bar elongates the diameter decreases a small amount.  The strain related to the

decrease in diameter is   – ν ε  =  – (0.34)(0.00304)  =  – 0.001034

 

 

The reduced diameter, ∆d  =  – ν ε d  =    – 0.001034)(10)  =  – 0.01034 mm     (result)

 

 

 


 

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