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

Answer:

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 $d_{o}$ = 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