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3 years ago

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A cylindrical specimen of brass that has a diameter of 20 mm, a tensile modulus of 110 GPa, and a Poisson’s ratio of 0.35 is pul

led in tension with force of 40,000 N. If the deformation is totally elastic, what is the strain experienced by the specimen?
(A) 0.00116
(B) 0.00029
(C) 0.00463
(D) 0.01350

2 answers:

kupik [55]3 years ago

8 0

<h2>Answer:</h2>

(A) 0.00116

<h2>Explanation:</h2>

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

stealth61 [152]3 years ago

7 0

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

$e=\frac{F}{AE}$

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

$e=\frac{F}{(\pi d^{2}/2)E}\\e=\frac{40,000}{\pi (20*10^{-3}/2)^{2} *110*10^{9}} \\e=0.00116$

Hence the strain experienced by the material is 0.00116

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