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lidiya [134]
2 years ago
5

The mechanical properties of a metal may be improved by incorporating fine particles of its oxide. Given that the moduli of elas

ticity of the metal and oxide are, respectively, 52 GPa and 380 GPa, what is the (a) upper-bound, and (b) lower-bound modulus of elasticity values (in GPa) for a composite that has a composition of 90 vol% of oxide particles.
Engineering
1 answer:
Goshia [24]2 years ago
7 0

Answer:

a) 347.2 GPa

b) 233.02 GPa

Explanation:

a) To find the upper bound modulus of elasticity, we use the formula:

E_c(u) = E_mV_m + E_pV_p

Where,

Volume fraction= V

E = modulus

Em=52GPa

Ep=380GPa

Vp=90%=0.90

Vm= 100%-90%=10%=0.10

We now have:

E_c(u) = (52*0.1)+(380*0.90)

= 5.2+342

= 347.2 GPa

b) For the lower bound modulus of elasticity, we use:

E_c(l) = \frac{E_mE_p}{E_pV_m+E_m+V_p}

=\frac{52*380}{(52*0.90)+(380*0.10)}

=233.02 GPa

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An air compressor of mass 120 kg is mounted on an elastic foundation. It has been observed that, when a harmonic force of amplit
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Answer:

equivalent stiffness is 136906.78 N/m

damping constant is 718.96 N.s/m

Explanation:

given data

mass = 120 kg

amplitude = 120 N

frequency = 320 r/min

displacement = 5 mm

to find out

equivalent stiffness and damping

solution

we will apply here frequency formula that is

frequency ω = ω(n) √(1-∈ ²)      ......................1

here  ω(n) is natural frequency i.e = √(k/m)

so from equation 1

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k × ( 1 - 2∈²) = 33.51² ×120

k × ( 1 - 2∈²) = 134752.99    .....................2

and here amplitude ( max ) of displacement is express as

displacement = force / k  ×  (  \frac{1}{2\varepsilon \sqrt{1-\varepsilon ^2}})

put here value

0.005 = 120/k   ×  (  \frac{1}{2\varepsilon \sqrt{1-\varepsilon ^2}})  

k ×∈ × √(1-2∈²) = 1200       ......................3

so by equation 3 and 2

\frac{k\varepsilon \sqrt{1-\varepsilon^2})}{k(1-2\varepsilon^2)} = \frac{12000}{134752.99}

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solve it and we get

∈ = 1.00396

and

∈ = 0.08869

here small value we will consider so

by equation 2 we get

k × ( 1 - 2(0.08869)²) = 134752.99

k  = 136906.78 N/m

so equivalent stiffness is 136906.78 N/m

and

damping is express as

damping = 2∈ √(mk)

put here all value

damping = 2(0.08869) √(120×136906.78)

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