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,

σ_Y  = 19650/ 78.57= 250 N/mm²

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

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

      ∴    Modulus of resilience = σ²_Y/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 = ((σ_u   + σ_Y )/2)   ε_f    

= ((452.5 + 250)/2) × 0.25 N-mm/mm³

= 87.8 N-mm/mm³

% elongation = 25%