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ArbitrLikvidat [17]
3 years ago
6

An object is supported by a crane through a steel cable of 0.02m diameter. If the natural swinging of the equivalent pendulum is

0.95 rad/s and the natural time period of the axial vibration is found to be 0.35 sec. What is the mass of the object.
Engineering
1 answer:
devlian [24]3 years ago
4 0

Answer:

22.90 × 10⁸ kg

Explanation:

Given:

Diameter, d = 0.02 m

ωₙ = 0.95 rad/sec

Time period, T = 0.35 sec

Now, we know

T= 2\pi\sqrt{\frac{L}{g}}

where, L is the length of the steel cable

g is the acceleration due to gravity

0.35= 2\pi\sqrt{\frac{L}{9.81}}

or

L = 0.0304 m

Now,

The stiffness, K is given as:

K = \frac{\textup{AE}}{\textup{L}}

Where, A is the area

E is the elastic modulus of the steel = 2 × 10¹¹ N/m²

or

K = \frac{\frac{\pi}{4}d^2\times2\times10^11}{0.0304}

or

K = 20.66 × 10⁸ N

Also,

Natural frequency, ωₙ = \sqrt{\frac{K}{m}}

or

mass, m = \sqrt{\frac{K}{\omega_n^2}}

or

mass, m = \sqrt{\frac{20.66\times10^8}{0.95^2}}

mass, m = 22.90 × 10⁸ kg

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masya89 [10]

Answer:

The tube surface temperature immediately after installation is 120.4°C and after prolonged service is 110.8°C

Explanation:

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pV = 0.596 kg/m³

ΔHL = 2257 kJ/kg

CpL = 4.217 kJ/kg K

uL = 279x10⁻⁶Ns/m²

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σ = 58.9x10³N/m

When the water boils on the surface its heat flux is:

q=0.149h_{fg} \rho _{v} (\frac{\sigma (\rho _{L}-\rho _{v})}{\rho _{v}^{2} }  )^{1/4} =0.149*2257*0.596*(\frac{58.9x10^{-3}*(957.9-0.596) }{0.596^{2} } )^{1/4} =18703.42W/m^{2}

For copper-water, the properties are:

Cfg = 0.0128

The heat flux is:

qn = 0.9 * 18703.42 = 16833.078 W/m²

q_{n} =uK(\frac{g(\rho_{L}-\rho _{v})     }{\sigma })^{1/2} (\frac{c_{pL}*deltaT }{c_{fg}h_{fg}Pr  } \\16833.078=279x10^{-6} *2257x10^{3} (\frac{9.8*(957.9-0.596)}{0.596} )^{1/2} *(\frac{4.127x10^{3}*delta-T }{0.0128*2257x10^{3}*1.76 } )^{3} \\delta-T=20.4

The tube surface temperature immediately after installation is:

Tinst = 100 + 20.4 = 120.4°C

For rough surfaces, Cfg = 0.0068. Using the same equation:

ΔT = 10.8°C

The tube surface temperature after prolonged service is:

Tprolo = 100 + 10.8 = 110.8°C

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A eutectic alloy is which of the following (two correct answers): (a) the composition in an alloy system at which the two elemen
NeTakaya

Answer:

Option (c) and option (d)

Explanation:

Eutectic system is one in which a solid and homogeneous mixture of two or more substances resulting in the formation of super lattice is formed which can melt or solidify at a temperature lower than the melting point of any individual metal.

Eutectic alloys are those which have its components mixed in a specific ratio.

It is the composition in an alloy system for which both the  liquidus and solidus temperatures are equal.

Eutectic alloys have the composition in which the melting point of the metal is lower than the other alloy composition.

3 0
3 years ago
1. Fatigue equations are based solely on theoretical assumptions. Experimental data is only used to verify the theory. a. True.b
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Answer:

1.  b. False

2. b. False

3.  b. False

4.  b. False

5. a. True

6. a. True

7.  b. False

8.  b. False

9. a. True

Explanation:

1. The fatigue properties of a material  are determined by series of test.

2. For most steels there is a level of fatigue limit below which a component will survive an infinite number of cycles, for aluminum and titanium a fatigue limit can not be defined, as failure will eventually occur after enough experienced cycles.

3. Although there is a cyclic stress, there are also stresses complex circumstances involving tensile to compresive and constant stress, where the solution is given into the mean stress and the stress amplitude or stress range, which is double the stress amplitude.

4. Low‐cycle fatigue is defined as few thousand cycles and high cycle fatigue is around more than 10,000 cycles.

5. The number of cycles for failure on brittle materials are less and determined compared with the ductile materials.

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7.  A SN-Curve for a given material, is a plot displayed on logarithmic scales of the magnitude of an alternating stress in relation to the number of cycles to failure

8. The strain life method measures the strain resistance of local stresses and strains around stress concentration that controls the fatigue life of the material. It is more accurate than determining fatigue performance as the stress-life method is for long life millions of cycles in elastic stresses, but an it gets an effective stress concentration in fatigue loading.

9. Linear Elastic Fracture Mechanics (LEFM) states that the material is isotropic and linear elastic so, when the stresses near the crack surpasses the material fracture toughness, the crack grows.

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3 years ago
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