Answer:
0.00116
Explanation
The tensile modulus is the measure of the elastic deformation that a material has undergo.
While elastic strain is define as form of strain in which a body return to its original length after the removal of the force .
Strain s expressed as
[tex]e=\frac{F}{AE}[/tex]
Where e is the strain, A is the area, F is the applied force and E itensile modulus
From the data given,
Force, F=40,000
Area=πr^2
and the tensile modulus E=110GPa
If we insert values we arrive at
[tex]e=\frac{F}{(\pi d^{2}/2)E}\\e=\frac{40,000}{\pi (20*10^{-3}/2)^{2} *110*10^{9}} \\e=0.00116[/tex]
Hence the strain experienced by the material is 0.00116
(A) 0.00116
Tensile Modulus (γ) is the ratio of elastic stress (Eₓ) to the elastic strain (Eₙ). i.e
γ = Eₓ / Eₙ --------------------------(i)
Where;
Eₓ = Force(F) / Area(A)
Eₓ = F / A -----------------(ii)
From the question;
Force (F) applied = 40000N
The diameter (d) of the specimen = 20mm = 0.02m.
We can use that to calculate the cross-sectional area (A) of the brass as follows;
A = π x d² / 4 [Take π = 3.142]
=> A = 3.142 x (0.02)² / 4
=> A = 0.0003142 m²
Now substitute the values of F and A into equation (ii) to find the tensile/elastic stress as follows;
Eₓ = 40000 / 0.0003142
Eₓ = 127307447.49 N/m²
Going back to equation (i);
γ = Eₓ / Eₙ
Where;
γ = tensile modulus = 110Gpa = 110 x 10⁹Pa
Eₓ = tensile stress = 127307447.49 N/m²
Let's calculate the tensile strain (Eₙ) by substituting these values into equation (i) as follows;
γ = Eₓ / Eₙ
110 x 10⁹ = 127307447.49 / Eₙ
Eₙ = 127307447.49 / (110 x 10⁹)
Eₙ = 0.001157
Eₙ = 0.00116 [Tensile strain has no unit]
Therefore, the tensile strain experienced by the specimen is 0.00116
