Question

A Specimen made from a brittie material with a cross-section area of 0.004m was gradually loaded in tonsion until it yielded at a land of 280 KN and fractured Slightly after the yield point. If the specimon's material observed slashert clashe deformation until fracture, determine the material's toughness in terms of energy cidsor bed, in kj. Take E = 200 Gpa.​

Answer 1

the material's toughness, in terms of energy absorbed until fracture, is 98 kJ.

Step-by-step explanation:

Step 1: Calculate the strain at the yield point:

Given that the cross-sectional area of the specimen is 0.004 m², and it yielded at a load of 280 kN (280,000 N), we can calculate the stress at the yield point:

Stress = Load / Cross-sectional area

Stress = 280,000 N / 0.004 m²

Stress = 70,000,000 N/m² (or Pa)

Using Hooke's Law, we know that stress is equal to the modulus of elasticity (E) multiplied by the strain (ε) at the yield point:

Stress = E * Strain

70,000,000 N/m² = 200,000,000,000 N/m² * Strain (converting E from GPa to N/m²)

Strain = 70,000,000 / 200,000,000,000

Strain = 0.00035

Step 2: Calculate the strain energy up to the yield point:

Strain energy (U) = (1/2) * Stress * Strain * Volume

We can assume that the volume remains constant during the test, so:

Strain energy (U) = (1/2) * Stress * Strain * Cross-sectional area * Length

For simplicity, let's assume the length of the specimen is 1 meter (this assumption won't affect the toughness calculation since it is based on energy per unit volume). Substituting the values:

Strain energy (U) = (1/2) * 70,000,000 N/m² * 0.00035 * 0.004 m² * 1 m

Strain energy (U) = 98,000 J (Joules)

Step 3: Convert the strain energy to kilojoules (kJ):

Toughness = Strain energy / 1000

Toughness = 98,000 J / 1000

Toughness = 98 kJ