# A stainless steel rod with a circular cross section has a length of 100 mm and a diameter of 2.0...

## Question:

A stainless steel rod with a circular cross section has a length of 100 mm and a diameter of 2.0 mm. It is deformed elastically in compression by a force of 500 N applied parallel to its length. The material has a Young?s modulus of 200 GPa, and a shear modulus of 77 GPa. Calculate (i) the amount by which the rod will decrease in length, and (ii) the amount by which the rod will increase in diameter.

## Poisson's Ratio:

Poisson's Ratio is defined as the ratio of lateral strain to linear strain. Strain in the lateral direction is defined as the ratio of change in length of the specimen to the original length. Strain in the longitudinal direction is defined as the ratio of change in the diameter of the specimen to the original diameter. The relationship between the Poisson's ratio, Young's modulus of elasticity and the shear modulus is given by:-

{eq}E = 2G (1 + \mu) {/eq}

Where

• {eq}\mu {/eq} is the Poisson's ratio.

We're given the following information in the problem:

• The diameter of the stainless steel rod, {eq}d = 2\ \rm{mm} {/eq}
• The length of the stainless steel rod, {eq}L = 100\ \rm{mm} {/eq}
• The compressive force applied on the stainless steel rod is, {eq}F = 500\ \rm{N} {/eq}
• The Young's Modulus is, {eq}E = 200\ \rm{GPa} = 200000\ \rm{MPa} {/eq}
• The shear modulus is, {eq}G = 77\ \rm{GPa} = 77000\ \rm{MPa} {/eq}

Cross-sectional area of the cylindrical bar is, {eq}A_c = \dfrac{\pi}{4}(2\ \rm{mm})^2 = 3.14\ \rm{mm^2} {/eq}

Poisson's ratio:

{eq}E = 2G(1 + \mu)\\ 200000\ \rm{MPa} = 2*(77000\ \rm{MPa})(1 + \mu)\\ \mu = 0.3 {/eq}

(i) The amount by which the rod will decrease in length:

{eq}\delta L = -\dfrac{PL}{AE}\\ \delta L = - \dfrac{(500\ \rm{N})( 100\ \rm{mm})}{(3.14\ \rm{mm^2})(200000\ \rm{MPa})} \\ \delta L = - 0.0796\ \rm{mm} {/eq}

The negative change in length shows the decrement of the length.

(ii) The change in diameter of the rod is:

{eq}\mu = - \dfrac{\left(\dfrac{\Delta D}{D}\right)}{\left(\dfrac{\Delta L}{L}\right)}\\ 0.3 = - \dfrac{\left(\dfrac{\Delta D}{2\ \rm{mm}}\right)}{\left(\dfrac{-0.0796\ \rm{mm}}{100\ \rm{mm}}\right)}\\ \Delta D = 0.0005\rm{mm} {/eq}

The Positive sign shows that the diameter of the rod increases when the compressive load is applied at the cross-section.