Question

A cylindrical specimen of steel has an original diameter of 12.8 mm. It is tested in tension its engineering fracture strength is found to be 460 MPa. If the cross-sectional diameter at fracture is 10.7 mm, determine (max. pts. 8): a. The ductility in terms of percent reduction in area b. The true stress at fracture

Answer 1

a) The ductility = -30.12%

the negative sign means reduction

Therefore, there is 30.12% reduction

b) the true stress at fracture is 658.26 Mpa

Step-by-step explanation:

Given that;

Original diameter
`d_(o)` = 12.8 mm

Final diameter
`d_(f)` = 10.7

Engineering stress
`\alpha _(E)` = 460 Mpa

a) determine The ductility in terms of percent reduction in area;

Ai = π/4(
`d_(o)` )² ; Ag = π/4(
`d_(f)` )²

% = π/4 [ ( (
`d_(f)` )² - (
`d_(o)` )²) / ( π/4 (
`d_(o)` )²) ]

= ( (
`d_(f)` )² - (
`d_(o)` )²) / (
`d_(o)` )² × 100

we substitute

= [( (10.7)² - (12.8)²) / (12.8)² ] × 100

= [(114.49 - 163.84) / 163.84 ] × 100

= - 0.3012 × 100

= -30.12%

the negative sign means reduction

Therefore, there is 30.12% reduction

b) The true stress at fracture;

True stress
`\alpha _(T)` =
`\alpha _(E)` ( 1 +
`E_(E)` )

`E_(E)` is engineering strain

`E_(E)` = dL / Lo

= (do² - df²) / df² = (12.8² - 10.7²) / 10.7² = (163.84 - 114.49) / 114.49

= 49.35 / 114.49

`E_(E)` = 0.431

so we substitute the value of
`E_(E)` into our initial equation;

True stress
`\alpha _(T)` = 460 ( 1 + 0.431)

True stress
`\alpha _(T)` = 460 (1.431)

True stress
`\alpha _(T)` = 658.26 Mpa

Therefore, the true stress at fracture is 658.26 Mpa