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77julia77 [94]
2 years ago
9

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, 67 GPa and 390 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 44 vol% of oxide particles.
Engineering
1 answer:
Lilit [14]2 years ago
4 0

Answer:

A) 209.12 GPa

B) 105.41 GPa

Explanation:

We are given;

Modulus of elasticity of the metal; E_m = 67 GPa

Modulus of elasticity of the oxide; E_f = 390 GPa

Composition of oxide particles; V_f = 44% = 0.44

A) Formula for upper bound modulus of elasticity is given as;

E = E_m(1 - V_f) + (E_f × V_f)

Plugging in the relevant values gives;

E = (67(1 - 0.44)) + (390 × 0.44)

E = 209.12 GPa

B) Formula for upper bound modulus of elasticity is given as;

E = 1/[(V_f/E_f) + (1 - V_f)/E_m]

Plugging in the relevant values;

E = 1/((0.44/390) + ((1 - 0.44)/67))

E = 105.41 GPa

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Let Stiffness be denoted by 'K' for each mounting, then for 4 mountings it is 4K

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Using the given formula:

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\frac{A}{F} will give the tranfer function

Therefore,

\frac{A}{F} = \frac{1}{\sqrt{(4K - 120\ ^{2})}}

0.1\times 10^{-3} =  \frac{1}{\sqrt{(4K - 120\ ^{2})}}

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2 years ago
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ycow [4]

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The length= 64 inches

Ends are fixed Le= 64/2 = 32 inches

Factor Of Safety (FOS) = 3. 0

E= 10.6× 10^6 ps

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The square cross-section= ia^4/12

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6500= 3.142^2 × 10^6 × a^4/12×32^2

a^4= 0.81 => a=0.81 inches => a=0.95 inches

Given σy= 4000ps

σallowable= σy/3= 40000/3= 13333. 33psi

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σactual=6500/0.9025

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3 years ago
The electron concentration in silicon at T = 300 K is given by
puteri [66]

Answer:

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Explanation:

From the question we are told that:

Temperature of silicon T=300k

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Electron diffusion coefficient is Dn = 25cm^2/s \approx 2.5*10^{-3}

Electron mobility is \mu n = 960 cm^2/V-s \approx0.096m/V

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Generally the equation for the semiconductor is mathematically given by

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