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3 years ago

The strength and stiffness of a properly constructed composite buildup depends primarily on?

Engineering

1 answer:

Flauer [41]3 years ago

4 0

Answer:orientation of piles

Explanation:

The stiffness and strength of a properly constructed composite buildup depend upon the orientation of piles to the load direction.

Structural composite materials are designed and manufactured to withstand certain specific stress loads. The ability to withstand these stress loads depends, to a large extent, on how the fibers are arranged. The fiber orientation of the parts must be the same as that of the original form.

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Nec ________ covers selection of time-delay fuses for motor- overload protection.

Murljashka [212]

Nec Article 430 covers selection of time-delay fuses for motor- overload protection.

<h3>What article in the NEC covers motor overloads?</h3>

Article 430 that is found in National Electrical Code (NEC) is known to be state as “Motors, Motor Circuits and Controllers.” .

Note that the article tells that it covers areas such as motors, motor branch-circuit as well as feeder conductors, motor branch-circuit and others.

Therefore, Nec Article 430 covers selection of time-delay fuses for motor- overload protection.

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3 years ago

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3 years ago

Why is it important to review your plan when you write an algorithm?

user100 [1]

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3 years ago

A pin must be inserted into a collar of the same steel using an expansion fit. The coefficient of thermal expansion of the metal

nirvana33 [79]

Answer:

a) the temperature to which the pin must be cooled for assembly is $T_2 = -101.89^ \ ^0}C$

b) the radial pressure at room temperature after assembly is $P_f = 62.8 \ MPa$

c) the safety factor in the resulting assembly = 6.4

Explanation:

Coefficient of thermal expansion $\alpha = 12.3*10^{-6} \ ^0 C$

Yield strength $\sigma_y$ = 400 MPa

Modulus of elasticity (E) = 209 GPa

Room Temperature $T_1$ = 20°C

outer diameter of the collar $D_o = 95 \ mm$

inner diameter of the collar $D_i = 60 \ mm$

pin diameter $D_p$ = $60.03 \ mm$

Clearance c = 0.06 mm

The temperature to which the pin must be cooled for assembly can be calculated by using the formula:

$(D_i - c )-D_p = \alpha * D_p(T_2-T_1)$

$(60-0.06)-60.03=12.3*10^{-6}*60.03(T_{2}-20^{0}C)$

-0.09 = $7.38369*10^{-4}$ $(T_{2}-20^{0}C)$

-0.09 = $7.38369*10^{-4}T_2$ $\ \ - \ \ 0.01476738$

$-0.09 + 0.01476738$ = $7.38369*10^{-4}T_2$

−0.07523262 = $7.38369*10^{-4}T_2$

$T_2 = \frac{-0.07523262}{7.38369*10^{-4}}$

$T_2 = -101.89^ \ ^0}C$

To determine the radial pressure at room temperature after assembly ;we have:

$P_f = \frac{E * (D_p-D_i)(D_o^2-D_1^2)}{D_i*D_o} \\ \\ \\ P_f = \frac{209*10^9* 0.03(95^2-60^2)}{60*95^2} \\ \\ P_f = 62815789.47 \ Pa \\ \\ P_f = 62.8 \ MPa$

c) the safety factor of the resulting assembly is calculated as:

safety factor = $\frac{Yield \ strength }{walking \ stress}$

safety factor = $\frac{400}{62.8}$

safety factor = 6.4

Thus, the safety factor in the resulting assembly = 6.4

4 0

4 years ago