Question

Chapter 06, Practice Problem 6.35 A cylindrical metal specimen having an original diameter of 11.72 mm and gauge length of 51.3 mm is pulled in tension until fracture occurs. The diameter at the point of fracture is 6.58 mm, and the fractured gauge length is 72.7 mm. Calculate the ductility in terms of (a) percent reduction in area (percent RA), and (b) percent elongation (percent EL).

Answer 1

Initial area of the specimen

`{\left( {{\rm{Area}}} \right)_{{\rm{initial x-section}}}} = \pi {\left( {\frac{{{d_0}}}{2}} \right)^2}`

Where,
`{d_i}` is the initial diameter of the metal specimen.

Substituting
`11.72{\rm{ mm}}` for
`{d_i}`

`\begin{array}{c}\\{\left( {{\rm{Area}}} \right)_{{\rm{initial x-section}}}} = \pi {\left( {\frac{{11.72{\rm{ mm}}}}{2}} \right)^2}\\\\ = 107.881{\rm{ m}}{{\rm{m}}^2}\\\end{array}`

Final minimum area of the specimen

`{\left( {{\rm{Area}}} \right)_{{\rm{final minimum}}}} = \pi {\left( {\frac{{{d_f}}}{2}} \right)^2}(Area)`

Where
`{d_f}` is final minimum diameter.

Substituting
`6.58{\rm{ mm}}` for
`{d_f}`

`\begin{array}{c}\\{\left( {{\rm{Area}}} \right)_{{\rm{final minimum}}}} = \pi {\left( {\frac{{6.58{\rm{ mm}}}}{2}} \right)^2}\\\\ = 34.00491{\rm{ m}}{{\rm{m}}^2}\\\end{array}`

Percentage reduction in the area.

`\% {\rm{RA}} = \frac{{{{\left( {{\rm{Area}}} \right)}_{{\rm{initial x-section}}}} - {{\left( {{\rm{Area}}} \right)}_{{\rm{final minimum}}}}}}{{{{\left( {{\rm{Area}}} \right)}_{{\rm{initial x-section}}}}}} * 100%`

Substituting
`107.881{\rm{ m}}{{\rm{m}}^2} for {\left( {{\rm{Area}}} \right)_{{\rm{initial x-section}}}}` and
`34.00491{\rm{ m}}{{\rm{m}}^2}` for
`{\left( {{\rm{Area}}} \right)_{{\rm{final minimum}}}}`

`\begin{array}{c}\\\% {\rm{RA}} = \frac{{107.881{\rm{ m}}{{\rm{m}}^2} – 34.00491{\rm{ m}}{{\rm{m}}^2}}}{{107.881{\rm{ m}}{{\rm{m}}^2}}} * 100\\\\ = 0.684792* 100\\\\ = 68.47925\% \\\end{array}`

(b)

The expression for the percent increase in the gauge length of the specimen.

`\% {\rm{EL}} = \frac{{{l_f} - {l_i}}}{{{l_i}}} * 100%`

Substituting
`72.7{\rm{ mm}}` for
`{l_f}` and
`51.3{\rm{ mm}}`for
`{l_i}`

`\begin{array}{c}\\\% {\rm{EL}} = \frac{{72.7{\rm{ mm}} - 51.3{\rm{ mm}}}}{{51.3{\rm{ mm}}}} * 100\\\\ = 0.417154* 100\\\\ = 41.7154\% \\\end{array}`