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3 years ago

13

A 4140 steel shaft, heat-treated to a minimum yield strength of 100 ksi, has a diameter of 1 7/16 in. The shaft rotates at 600 r

pm and transmits 40 hp through a gear. The hub upon which the gear is seated is 1.5 in wide. You have in stock 1137 OQT 1300 carbon steel that you are considering using as a key to seat the gear hub. What dimensions will the key need to be? Is this material appropriate for the application? If not, select a more appropriate material.

2 answers:

MrMuchimi3 years ago

8 0

Answer:

The key dimensions; (4.54mm× 6.75mm×4.5mm)

Explanation

Check attachment

velikii [3]3 years ago

4 0

Answer:

Explanation:

4140-40 I’d pick wood

I hope this helps! :)

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Turn the motor around in the circuit. What happens?

kenny6666 [7]

Answer:

It will create a massive drag and pretty much stop the motor.

Explanation:

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3 years ago

If the bolt head and the supporting bracket are made of the same material having a failure shear stress of 'Tra;i = 120 MPa, det

Nina [5.8K]

Answer:

P=361.91 KN

Explanation:

given data:

brackets and head of the screw are made of material with T_fail=120 Mpa

safety factor is F.S=2.5

maximum value of force P=??

<em>solution:</em>

to find the shear stress

T_allow=T_fail/F.S

=120 Mpa/2.5

=48 Mpa

we know that,

V=P

<u>Area for shear head:</u>

A(head)=π×d×t

=π×0.04×0.075

=0.003×πm^2

<u>Area for plate:</u>

A(plate)=π×d×t

=π×0.08×0.03

=0.0024×πm^2

now we have to find shear stress for both head and plate

<u>For head:</u>

T_allow=V/A(head)

48 Mpa=P/0.003×π ..(V=P)

P =48 Mpa×0.003×π

=452.16 KN

<u>For plate:</u>

T_allow=V/A(plate)

48 Mpa=P/0.0024×π ..(V=P)

P =48 Mpa×0.0024×π

=361.91 KN

the boundary load is obtained as the minimum value of force P for all three cases. so the solution is

P=361.91 KN

note:

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3 years ago

A flywheel made of Grade 30 cast iron (UTS = 217 MPa, UCS = 763 MPa, E = 100 GPa, density = 7100 Kg/m, Poisson's ratio = 0.26) h

hram777 [196]

Answer:

N = 38546.82 rpm

Explanation:

$D_{1}$ = 150 mm

$A_{1}= \frac{\pi }{4}\times 150^{2}$

= 17671.45 $mm^{2}$

$D_{2}$ = 250 mm

$A_{2}= \frac{\pi }{4}\times 250^{2}$

= 49087.78 $mm^{2}$

The centrifugal force acting on the flywheel is fiven by

F = M ( $R_{2}$ - $R_{1}$ ) x $w^{2}$ ------------(1)

Here F = ( -UTS x $A_{1}$ + UCS x $A_{2}$ )

Since density, $\rho = \frac{M}{V}$

$\rho = \frac{M}{A\times t}$

$M = \rho \times A\times t$ $M = 7100 \times \frac{\pi }{4}\left ( D_{2}^{2}-D_{1}^{2} \right )\times t$

$M = 7100 \times \frac{\pi }{4}\left ( 250^{2}-150^{2} \right )\times 37$

$M = 8252963901$

∴ $R_{2}$ - $R_{1}$ = 50 mm

∴ F = $763\times \frac{\pi }{4}\times 250^{2}-217\times \frac{\pi }{4}\times 150^{2}$

F = 33618968.38 N --------(2)

Now comparing (1) and (2)

$33618968.38 = 8252963901\times 50\times \omega ^{2}$

∴ ω = 4036.61

We know

$\omega = \frac{2\pi N}{60}$

$4036.61 = \frac{2\pi N}{60}$

∴ N = 38546.82 rpm

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zavuch27 [327]

Explanation:

i think option 4 is correct answer because itsrelated to animal not plants.

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Arlecino [84]

Personally, I would go to the space station. The space station has extreme different levels of technology and abilities, plus who doesn’t want to go to space.

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