A prismatic bar made from an unknown but homogeneous elastic material is tested in tension at constant strain rate.  The gauge length and diameter of the bar are 100 mm and 10 mm respectively.  The force-extension graph resulting from the test is linear until the bar fractures suddenly at a load of 2.23 kN, and the extension at this point is 2.233 mm.  Just before fracture a reduction of 0.9% in the bar’s diameter was measured.  Determine:

(i)   the fracture stress;

(ii)  he fracture strain;

(iii) Young’s modulus;

(iv) Poisson’s ratio;

What type of response is exhibited by the material?

A prismatic bar made from an unknown but homogeneous elastic material is tested in tension at constant strain rate.  The gauge length and diameter of the bar are 100 mm and 10 mm respectively.  The force-extension graph resulting from the test is linear until the bar fractures suddenly at a load of 2.23 kN, and the extension at this point is 2.233 mm.  Just before fracture a reduction of 0.9% in the bar’s diameter was measured.  Determine:

(i)   the fracture stress;

Sol) Stress = Force / Area

Force = 2.23kN = 2230 N, Radius R = 5mm = 5 x 10^-3 m

Area = π R² =  3.14 x (5 x 10^-3)² = 7.85 x 10^-5 m²

Fracture Stress = 2230 / 7.85 x 10^-5 = 2.84 x 10⁷ N/m²

(ii)  the fracture strain;

Sol) Longitudinal strain = change in length / Original length = dL/L

Initially diameter d = 10mm or R = 5mm, length of bar L = 100mm

Thus volume of bar = π R²L

Now reduction in bar’s diameter before fracture limit = 0.9%

Thus diameter of bar at fracture limit = d’ = 10 x 99.1/100 = 9.91mm

Or R’ = 9.91/2 = 4.955mm

Length length of the bar at fracture limit = L’

Now it is such that the volume is constant thus

π R²L = π R’²L’

or R²L = R’²L’

or 5² x 100 = 4.955² x L’

or L’ = 101.824 mm

Thus dL = change in length = dL =  L’ – L = 101.824 – 100 = 1.824mm

Thus frature strain = dL /L = 1.824/100 = 1.824 x 10^-2

(iii) Young’s modulus;

Youngs modulus = Y = Stress/Strain

From part (i) Fracture Stress = 2.84 x 10⁷ N/m²

Now as calculated in part (ii), strain = 1.824/100 = 1.824 x 10^-2

Thus, Y = Stress/Strain = 2.84 x 10⁷ / 1.824 x 10^-2

Or Y =1.557 x 10⁹ N/m²

(iv)  Poisson’s ratio;

It is another kind of modulus of elasticity associated with the longitudinal stress and strain. When a rod or a wire is subjected to a tensile stress, its length increases in the direction of the tensile force. At the same time the length perpendicular to the tensile force decreases.

Now, for a cylindrical rod, the length increases and the diameter decreases when the rod is stretched

The fractional change in the transverse length (diameter) is proportional to the fractional change in the longitudinal length (length of the cylinder). The constant of proportionality is called Poisson’s ratio. Thus poisson’s ratio is

σ = - (Δd/d)/(ΔL/L)

The negative sign ensures that the σ is positive.

Now change in diameter Δd = d’ – d =  9.91 – 10 = -0.09mm

[ Note: value of d’ = 9.91mm is calculated as above in part (i)]

Also change in length of the bar = ΔL = L’ – L = 101.824 – 100 = 1.824mm

Thus Δd/d = -0.09 / 10 = -0.009

And ΔL/L = 1.824 / 100 = 1.824 x 10^-2

Hence σ =- (Δd/d)/(ΔL/L) =  - (-0.009) /(1.824 x 10^-2) =  0.493

What type of response is exhibited by the material?

Ans) Material is Brittle in nature.

Metal exhibits elastic behaviour or response until the fracture point that corresponds to force of 2.23 kN, Its proportional limit, elastic limit and the fracture limit is the same (or so closely lying that it can assumed to be same).

Thus the material breaks soon after the elastic limit is crosses, it is BRITTLE