## Compute modulus of toughness resilience and modulus Assignment Help

Assignment Help: >> Toughness - Compute modulus of toughness resilience and modulus

A steel specimen of 10 millimeter diameter and 50 millimeter gauge length was tested in tension and given observations were recorded.

Load at upper yield point = 20600N; Load at lower yield point = 19650 N; Maximum load = 35550 N

Gauge length after fracture = 62.43 N

Compute modulus of toughness's resilience and modulus. Also compute % elongation.  E = 210 × 103 N/mm2.

Solution

Area of cross-section of specimen,

A0 = (π /4)d2

A0 = (π/4) (10)2 = 78.57 mm2

∴    Yield strength,

σY  = 19650/ 78.57= 250 N/mm2

Ultimate tensile strength σu = 35550/78.57 = 452.5 N/mm2

Strain at fracture or % elongation = εf = (62.43 - 50)/ 50 = 0.25 or 25%

∴    Modulus of resilience = σ2Y/2E

= ((250)2)/ 2 × 210 × 103 N-mm/mm3

= 148.8 × 10-3  N-mm/mm3

Now compare along with illustration .1 to notice that for higher yield strength modulus of resilience is larger.

Modulus of toughness = ((σu   + σY )/2)   εf

= ((452.5 + 250)/2) × 0.25 N-mm/mm3

= 87.8 N-mm/mm3

% elongation = 25% 