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stira [4]
3 years ago
15

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
Engineering
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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pashok25 [27]

Answer:

Modulus of  elasticity is dependent on the elemental constitution of  1080  steel and various heat treatment don't affect this elemental composition  but only affects the mechanical properties(like strength , hardness, ductility, malleability) of the 1080 steel that are affected by plastic deformation of the 1080 steel

Explanation:

Generally the underlying physical reason why when we conduct various heat treatments on 1018 steel we expect the modulus of elasticity to stay about the same for every heat treatment is  

 

Modulus of  elasticity is dependent on the elemental constitution of  1080  steel and various heat treatment don't affect this elemental composition  but only affects the mechanical properties(like strength , hardness, ductility, malleability) of the 1080 steel that are affected by plastic deformation of the 1080 steel

6 0
3 years ago
(20pts) Air T[infinity] = 10 °C and u[infinity] = 100 m/s flows over a flat plate. Assume that the density of air is 1.0 kg/m3 a
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Answer:

attached below

Explanation:

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3 years ago
Air is compressed isothermally from 13 psia and 55°F to 80 psia in a reversible steady-flow device. Calculate the work required,
solong [7]

Answer:64.10 Btu/lbm

Explanation:

Work done in an isothermally compressed steady flow device is expressed as

Work done = P₁V₁ In { P₁/ P₂}

Work done=RT In { P₁/ P₂}

where P₁=13 psia

          P₂= 80 psia

Temperature =°F Temperature is convert to  °R

T(°R) = T(°F) + 459.67

T(°R) = 55°F+ 459.67

=514.67T(°R)

According to the properties of molar gas, gas constant and critical properties table, R  which s the gas constant of air is given as 0.06855 Btu/lbm

Work = RT In { P₁/ P₂}

0.06855 x 514.67 In { 13/ 80}

=0.06855 x 514.67 In {0.1625}

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8 0
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The water in a large lake is to be used to generate electricity by the installation of a hydraulic turbine-generator at a locati
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Answer:

a) 0.76

b) 0.80

c) 1964 kW

Explanation:

GIVEN DATA:

\dot m = 5000 kg/s

Assume Mechanical energy at exist is negligible

A) Take lake bottom as reference, and then kinetic and potential energy  are taken as zero.

change in mechanical energy is givrn as

e_{in} - e_{out} = \frac{P}{\rho} - 0 = gh = 9.81 \times 50( \frac{1 kJ/kg}{1000 m^2/s^2}

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\Delta \dot E_{mec} = \dot m (e_{in} - e_{out}) = 5000 \times 0.491 = 2455 kW

\eta_{OVERALL}  = \frac{\dot W}{\Delta \dot E_{mec}} = \frac{1862}{2455} = 0.76

B) \eta -{gen} = \frac{\eta_{overall}}{\eta_{gen}} = \frac{0.76}{0.95} = 0.80

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amid [387]

Answer:

rectifier

Explanation:

7 0
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