# A cylindrical bar of metal having a diameter of 15.7 mm and a length of 178 mm is deformed...

## Question:

A cylindrical bar of metal having a diameter of 15.7 mm and a length of 178 mm is deformed elastically in tension with a force of 49100 N. Given that the elastic modulus and Poisson's ratio of the metal are 67.1 GPa and 0.34 respectively, Determine the following:

(a) The amount by which this specimen will elongate (in mm) in the direction of the applied stress. The entry field with incorrect answer now contains modified data.

(b) The change in diameter of the specimen (in mm). Indicate an increase in diameter with a positive number and a decrease with a negative number.

## Poisson's Ratio:

Poisson's Ratio is defined as the ratio of lateral strain to linear strain. The diameter of the cylindrical bar will reduce only when an axial tensile force is applied at the cross-sectional area of the bar. Strain in lateral direction is defined as the ratio of change in length of the specimen to the original length. Strain in longitudinal direction is defined as the ratio of change in diameter of the specimen to the original diameter. The force which produces an longitudinal strain can be calculated by using the formula given below.

{eq}\varepsilon_l = \dfrac{F}{AE} {/eq}

Where,

• {eq}\varepsilon_l {/eq} is the longitudinal strain.
• F is the applied tensile force.
• E is the modulus of elasticity.
• A is the area of cross-section.

We're given following information in the problem:

• The diameter of the cylindrical bar, {eq}d = 15.7\ \rm{mm} {/eq}
• The length of the cylindrical bar, {eq}L = 178\ \rm{mm} {/eq}
• The tensile force applied is, {eq}F = 49100\ \rm{N} {/eq}
• The Young's Modulus is, {eq}E = 67.1\ \rm{GPa} = 67100\ \rm{MPa} {/eq}
• The Poisson's ratio is, {eq}\mu = 0.34 {/eq}

Cross-sectional area of the cylindrical bar is, {eq}A_c = \dfrac{\pi}{4}(15.7\ \rm{mm})^2 = 193.59\ \rm{mm^2} {/eq}

The tensile stress acting on the cylindrical bar is, {eq}\sigma = \dfrac{49100\ \rm{N}}{193.59\ \rm{mm^2}} = 253.63\ \rm{MPa} {/eq}

(a) The amount by which this specimen will elongate in the direction of the applied stress is:

Linear strain:

{eq}\epsilon_l = \dfrac{\sigma}{E}\\ \dfrac{\delta L}{178\ \rm{mm}} = \dfrac{253.63\ \rm{MPa}}{67100\ \rm{MPa}} \delta L = 0.673\ \rm{mm} {/eq}

(b) The change in diameter of the specimen is:

{eq}\mu = - \dfrac{\left(\dfrac{\Delta D}{D}\right)}{\left(\dfrac{\Delta L}{L}\right)}\\ 0.34 = - \dfrac{\left(\dfrac{\Delta D}{15.7\ \rm{mm}}\right)}{\left(\dfrac{0.673\ \rm{mm}}{178\ \rm{mm}}\right)}\\ \Delta D = - 0.02 \rm{mm} {/eq}

Negative sign shows that the diameter of the bar decreases when the tensile load is applied at the cross-section. 