Determine modulus of resilience:
A steel specimen of 10 mm diameter & 50 mm gauge length was tested in tension and following observations were recorded.
Load at upper yield point = 20600 N
Load at lower yield point = 19650 N
Maximum load = 35550 N
Gauge length after fracture = 62.43 N
Determine modulus of resilience and modulus of toughness. Also compute % elongation. E = 210 × 10^{3} N/mm^{2} .
Solution
Area of cross-section of specimen,
A_{0} = (π/4 ) d^{ 2}
A _{0} = (π /4 )(10) ^{2} = 78.57 mm2
∴ Yield strength,
σ_{ Y} = 19650 /78.57 = 250 N/mm^{2}
Ultimate tensile strength, σ _{u} = 35550 /78.57 = 452.5 N/mm^{2}
Strain at fracture or % elongation =ε _{f }= 62.43 - 50 = 0.25 or 25%
∴ Modulus of resilience =
= (250)^{2} /2 × 210 × 10^{3} N-mm/mm^{3}
= 148.8 × 10^{- 3} N-mm/mm^{3}
(i) (Compare with Example 1 to note that for higher yield strength modulus of resilience is larger).
Modulus of toughness = (s_{u} +s_{Y }/2) ε _{f}
= (452.5 + 250 /2 )× 0.25 N-mm/mm3
= 87.8 N-mm/mm^{3}
% elongation = 25%