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igor_vitrenko [27]
3 years ago
11

A copper wire of original diameter .80 m exhibits a maximum tensile load/ strength at an engineering stress= 248.2 mpa. its duct

ility is measured as 75 reduction of area. Determine the true strain at failure.
Engineering
1 answer:
ddd [48]3 years ago
4 0

Answer:

True Strain at failure = 1.386

Explanation:

For the question, ductility is given in reduction In the area to be 0.75

Let the initial Cross sectional Area of wire be Ao

And the final cross sectional Area of wire of cross sectional Area of wire at fracture be A

(Ao - A)/Ao = ductility = 0.75

Ao - A = 0.75Ao

A = Ao - 0.75Ao

A = 0.25 Ao

True Strain = In (Lf/Lo)

To obtain the ratio of the lengths of wire,

The volume of the wire stays constant, that is, Vo = Vf; Vo = Ao × Lo and Vf = A × Lf

AoLo = ALf

Lf/Lo = Ao/A

In (Lf/Lo) = In (Ao/A) = In (Ao/0.25Ao) = In 4 = 1.386

True Strain = 1.386

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Three bars each made of different materials are connected together and placed between two walls when the temperature is 12 oC. D
slega [8]

Answer:

F = 9.11 x 10³ N = 9.11 KN

Explanation:

The areas, lengths, young's modulus, and coefficient of linear thermal expansion are given in the diagram. First we find the equivalent change in length due to temperature change:

ΔL = (ΔL)steel + (ΔL)brass + (ΔL)Copper

ΔL = (∝s)(Ls)(ΔT) + (∝b)(Lb)(ΔT) + (∝c)(Lc)(ΔT)

where,

ΔL = Equivalent Change in Length = ?

ΔT = Change in Temperature = 25°C - 12°C = 13°C

Ls = Length of Steel Segment = 300 mm = 0.3 m

Lb = Length of Brass Segment = 200 mm = 0.2 m

Lc = Length of Copper Segment = 100 mm = 0.1 m

Therefore,

ΔL = (12 x 10⁻⁶ °C⁻¹)(0.3 m)(13 °C) + (21 x 10⁻⁶ °C⁻¹)(0.2 m)(13 °C) + (17 x 10⁻⁶ °C⁻¹)(0.1 m)(13 °C)

ΔL = 46.8 x 10⁻⁶ m + 54.6 x 10⁻⁶ m + 22.1 x 10⁻⁶ m

ΔL = 123.5 x 10⁻⁶ m   ----------------------- equation (1)

Now, we calculate this deflection in terms of an applied force (F):

ΔL = (F)(Ls)/(Es)(As) + (F)(Lb)/(Eb)(Ab) + (F)(Lc)/(Ec)(Ac)

ΔL = (F)(0.3 m)/(200 x 10⁹ Pa)(200 x 10⁻⁶ m²) + (F)(0.2 m)/(100 x 10⁹ Pa)(450 x 10⁻⁶ m²) + (F)(0.1 m)/(120 x 10⁹ Pa)(515 x 10⁻⁶ m²)

ΔL = F(7.5 x 10⁻⁹ m/N + 4.44 x 10⁻⁹ m/N + 1.61 x 10⁻⁹ m/N)

ΔL = F(13.55 x 10⁻⁹ m/N)   --------------------- equation (1)

Comparing equation (1) and equation (2):

123.5 x 10⁻⁶ m = F(13.55 x 10⁻⁹ m/N)

F = (123.5 x 10⁻⁶ m)/(13.55 x 10⁻⁹ m/N)

<u>F = 9.11 x 10³ N = 9.11 KN</u>

6 0
3 years ago
The top surface of an L = 5­mm­thick anodized aluminum plate is irradiated with G = 1000 W/m2 while being simultaneously exposed
puteri [66]

Answer:

J=1963W/m^2

q_{rad}=963w/m^2

\triangle T= -0.378k/s

Explanation:

From the question we are told that:

L=5mm => 5*10^{-3}\\Irradiation G=1000W/m^2\\h=50W/m^2\\T_{infinity} = 25C.\\Plate\ temperature\ T_p= 400 K\\\alpha=0.14\\E=0.76

at Temp=400K

E=2702kg/m^2,c=949J/kgk

Generally the equation for Radiosity is mathematically given by

J=eG+\in E_p

J=(1-\alpha)G+\in \sigma T^4

J=(1-0.14)1000+0.76 (5.67*10^_{8}) (400)^4

J=1963W/m^2

Generally the equation for net radiation heat flux q_{rad} is mathematically given by

q_{rad}=J-G\\q_{rad}=1963-1000

q_{rad}=963w/m^2

Generally the equation for and the rate of plate temp \triangleT is mathematically given by

\triangle T= -\frac{q_{con} +q_{rad}}{Ecl}

\triangle T= \frac{45(400)-(30+273+963)}{(2702*949*0.005)}

\triangle T= -0.378k/s

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3 years ago
The purification of hydrogen gas is possible by diffusion through a thin palladium sheet. Calculate the number of kilograms of h
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Man I don't have a clue, but this stuff sounds interesting
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3 years ago
What is included in the environmental impact assessment process, such as the use of geographic information systems?
IgorLugansk [536]

Answer:

The combination of GISs with associated data sources, such as remote sensing imagery, is now common in environmental monitoring and assessment. ... Using a GIS as an environmental modeling tool allows modelers to incorporate database capabilities, data visualization, and analytical tools in a single integrated environment.

Explanation:

I hope it helps...

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4 0
3 years ago
Consider a mild steel specimen with yield strength of 43.5 ksi and Young's modulus of 29,000 ksi. It is stretched up to a point
mezya [45]

Answer:

0.05%

Explanation:

From the question, we have;

The yield strength of the mild steel, \sigma _c = 43.5 ksi

Young's modulus of elasticity, ∈ = 29,000 ksi

The total strain, \epsilon _c = 0.2% = 0.002

The inelatic strain \epsilon_c^{in} is given as follows;

\epsilon_c^{in} = \epsilon _c - \sigma _c/∈

Therefore, we have;

\epsilon_c^{in} = 0.002 - 43.5/(29,000) = 0.0005

Therefore, the inelastic strain, \epsilon_c^{in} = 0.0005 = 0.05%

Taking the inelastic strain as the residual strain, we have;

The residual strain = 0.05%

7 0
4 years ago
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