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2 years ago

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Consider a cylindrical nickel wire 2.1 mm in diameter and 3.2 × 104 mm long. Calculate its elongation when a load of 280 N is ap

plied. Assume that the deformation is totally elastic and that the elastic modulus for nickel is 207 GPa (or 207 × 109 N/m2).

1 answer:

pshichka [43]2 years ago

4 0

Answer:

Total elongation will be 0.012 m

Explanation:

We have given diameter of the cylinder = 2.1 mm

Length of wire $L=3.2\times 10^4mm$

So radius $r=\frac{d}{2}=\frac{2.1}{2}=1.05mm=1.05\times 10^{-3}m$

Load F = 280 N

Elastic modulus = 207 Gpa

Area of cross section $A=\pi r^2=3.14\times (1.05\times 10^{-3})^2=3.461\times 10^{-6}m^2$

We know that elongation in wire is given by $\delta =\frac{FL}{AE}$ , here F is load, L is length, A is area and E is elastic modulus

So $\delta =\frac{FL}{AE}=\frac{280\times 32}{3.461\times 10^{-6}\times 207\times 10^9}=0.012m$

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A crystalline grain of aluminum in a metal plate is situated so that a tensile load is oriented along the [1 1 1] direction. Wha

Ivenika [448]

Answer: required tensile stress is 0.889 MPa

Explanation:

Given that;

tensile load is oriented along the [1 1 1] direction

shear stress is 0.242 MPa along [1 0 1] in the (1 1 -1) plane

first we determine

λ which is Angle between [1 1 1] and [1 0 1]

so

cosλ = [ 1(1) + 1(0) + 1(1) ] / [ √(1² + 1² + 1²) √(1² + 0² + 1²)]

= 2 / √3√2 = 2/√6

Next, we determine ∅ which is angle between [1 1 1] and [1 1 -1]

so,

cos∅ = [ 1(1) + 1(1) + 1(-1) ] / [ √(1² + 1² + 1²) √(1² + 1² + (-1)²)]

cos∅ = [ 2-1] / [√3√3 ]

cos∅ = 1/3

Now, we know that;

σ = T_stress/cosλcosθ

so we substitute

σ = 0.242 / ( 2/√6 × 1/3 )

σ = 0.242 / 0.2721

σ = 0.889 MPa

Therefore, required tensile stress is 0.889 MPa

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2 years ago

How do scientists and engineers use math to help them?

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2 years ago

Compute the volume percent of graphite, VGr, in a 3.2 wt% C cast iron, assuming that all the carbon exists as the graphite phase

Yanka [14]

Answer:

The volume percentage of graphite is 10.197 per cent.

Explanation:

The volume percent of graphite is the ratio of the volume occupied by the graphite phase to the volume occupied by the graphite and ferrite phases. The weight percent in the cast iron is 3.2 wt% (graphite) and 96.8 wt% (ferrite). The volume percentage of graphite is:

$\%V_{gr} = \frac{V_{gr}}{V_{gr}+V_{fe}} \times 100\,\%$

Where:

$V_{gr}$ - Volume occupied by the graphite phase, measured in cubic centimeters.

$V_{fe}$ - Volume occupied by the graphite phase, measured in cubic centimeters.

The expression is expanded by using the definition of density and subsequently simplified:

$\%V_{gr} = \frac{\frac{m_{gr}}{\rho_{gr}} }{\frac{m_{gr}}{\rho_{gr}}+\frac{m_{fe}}{\rho_{fe}}}\times 100\,\%$

Where:

$m_{fe}$ , $m_{gr}$ - Masses of the ferrite and graphite phases, measured in grams.

$\rho_{fe}, \rho_{gr}$ - Densities of the ferrite and graphite phases, measured in grams per cubic centimeter.

$\%V_{gr} = \frac{1}{1+\frac{\frac{m_{fe}}{\rho_{fe}} }{\frac{m_{gr}}{\rho_{gr}} } }\times 100\,\%$

$\%V_{gr} = \frac{1}{1 + \left(\frac{\rho_{gr}}{\rho_{fe}} \right)\cdot\left(\frac{m_{fe}}{m_{gr}} \right)} \times 100\,\%$

If $\rho_{gr} = 2.3\,\frac{g}{cm^{3}}$ , $\rho_{fe} = 7.9\,\frac{g}{cm^{3}}$ , $m_{gr} = 3.2\,g$ and $m_{fe} = 96.8\,g$ , the volume percentage of graphite is:

$\%V_{gr} = \frac{1}{1+\left(\frac{2.3\,\frac{g}{cm^{3}} }{7.9\,\frac{g}{cm^{3}} } \right)\cdot \left(\frac{96.8\,g}{3.2\,g} \right)} \times 100\,\%$

$\%V_{gr} = 10.197\,\%V$

The volume percentage of graphite is 10.197 per cent.

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3 years ago

Consider fully developed laminar flow in a circular pipe. If the viscosity of the fluid is reduced by half by heating while the

gladu [14]

Answer:

The pressure drop across the pipe also reduces by half of its initial value if the viscosity of the fluid reduces by half of its original value.

Explanation:

For a fully developed laminar flow in a circular pipe, the flowrate (volumetric) is given by the Hagen-Poiseulle's equation.

Q = π(ΔPR⁴/8μL)

where Q = volumetric flowrate

ΔP = Pressure drop across the pipe

μ = fluid viscosity

L = pipe length

If all the other parameters are kept constant, the pressure drop across the circular pipe is directly proportional to the viscosity of the fluid flowing in the pipe

ΔP = μ(8QL/πR⁴)

ΔP = Kμ

K = (8QL/πR⁴) = constant (for this question)

ΔP = Kμ

K = (ΔP/μ)

So, if the viscosity is halved, the new viscosity (μ₁) will be half of the original viscosity (μ).

μ₁ = (μ/2)

The new pressure drop (ΔP₁) is then

ΔP₁ = Kμ₁ = K(μ/2)

Recall,

K = (ΔP/μ)

ΔP₁ = K(μ/2) = (ΔP/μ) × (μ/2) = (ΔP/2)

Hence, the pressure drop across the pipe also reduces by half of its initial value if the viscosity of the fluid reduces by half of its value.

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