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

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

Explanation:

Given that;

Original diameter = 12.8 mm

Final diameter = 10.7

Engineering stress   = 460 Mpa

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

Ai = π/4( )²  ; Ag = π/4( )²

% = π/4 [ ( ( )² - ( )²) / ( π/4  ( )²) ]

= ( ( )² - ( )²) / ( )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   = ( 1 +   )

 is engineering strain

 = dL / Lo

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

= 49.35 / 114.49  

= 0.431

so we substitute the value of  into our initial equation;

True stress   = 460 ( 1 +  0.431)

True stress   = 460 (1.431)

True stress   = 658.26 Mpa

Therefore, the true stress at fracture is 658.26 Mpa

b. check your rearview mirror

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