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RideAnS [48]
3 years ago
12

A solid cylindrical workpiece made of 304 stainless steel is 150 mm in diameter and 100 mm is high. It is reduced in height by 5

0%, at room temperature, by open-die forging with flat dies. Assume that the coefficient of friction is 0.2. Calculate the forging force at the end of the stroke.
Engineering
1 answer:
goblinko [34]3 years ago
7 0

Answer:

45.3 MN

Explanation:

The forging force at the end of the stroke is given by

F = Y.π.r².[1 + (2μr/3h)]

The final height, h is given as h = 100/2

h = 50 mm

Next, we find the final radius by applying the volume constancy law

volumes before deformation = volumes after deformation

π * 75² * 2 * 100 = π * r² * 2 * 50

75² * 2 = r²

r² = 11250

r = √11250

r = 106 mm

E = In(100/50)

E = 0.69

From the graph flow, we find that Y = 1000 MPa, and thus, we apply the formula

F = Y.π.r².[1 + (2μr/3h)]

F = 1000 * 3.142 * 0.106² * [1 + (2 * 0.2 * 0.106/ 3 * 0.05)]

F = 35.3 * [1 + 0.2826]

F = 35.3 * 1.2826

F = 45.3 MN

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Ymorist [56]
Torch body if I’m wrong I’m really sorry that’s what I got
3 0
3 years ago
A long, circular aluminum rod is attached at one end to a heated wall and transfers heat by convection to a cold fluid.
Trava [24]

Answer:

a. Heat removal rate will increase

b. Heat removal rate will decrease

Explanation:

Given that

One end of rod is connected to the furnace and rod is long.So this rod can be treated as infinite long fin.

We know that heat transfer in fin given as follows

Q_{fin}=\sqrt{hPKA}\ \Delta T

We know that area

A=\dfrac{\pi}{4}d^2

Now when diameter will triples then :

A_f=\dfrac{\pi}{4}{\left (3d \right )}^2

A_f=9A

Q'_{fin}=\sqrt{9hPKA}\ \Delta T

Q'_{fin}=3\sqrt{hPKA}\ \Delta T

Q'_{fin}=3Q

So the new heat transfer will increase by 3 times.

Now when copper rod will replace by aluminium rod :

As we know that thermal conductivity(K) of Aluminium is low as compare to Copper .It means that heat transfer will decreases.

3 0
3 years ago
Check Your Understanding: True Stress and Stress A cylindrical specimen of a metal alloy 47.7 mm long and 9.72 mm in diameter is
VARVARA [1.3K]

Answer:

The answer is "583.042533 MPa".

Explanation:

Solve the following for the real state strain 1:

\varepsilon_{T}=\In \frac{I_{il}}{I_{01}}

Solve the following for the real stress and pressure for the stable.\sigma_{r1}=K(\varepsilon_{r1})^{n}

K=\frac{\sigma_{r1}}{[\In \frac{I_{il}}{I_{01}}]^n}

Solve the following for the true state stress and stress2.

\sigma_{r2}=K(\varepsilon_{r2})^n

     =\frac{\sigma_{r1}}{[\In \frac{I_{il}}{I_{01}}]^n} \times [\In \frac{I_{i2}}{I_{02}}]^n\\\\=\frac{399 \ MPa}{[In \frac{54.4}{47.7}]^{0.2}} \times [In \frac{57.8}{47.7}]^{0.2}\\\\ =\frac{399 \ MPa}{[ In (1.14046122)]^{0.2}} \times [In (1.21174004)]^{0.2}\\\\ =\frac{399 \ MPa}{[ In (1.02663509)]} \times [In 1.03915873]\\\\=\frac{399 \ MPa}{0.0114161042} \times 0.0166818905\\\\= 399 \ MPa \times 1.46125948\\\\=583.042533\ \ MPa

4 0
3 years ago
The contact angle between the mercury surface and capillary tube wall is______ A) Less than 90 B) Equal to 90 C) Greater than 90
MakcuM [25]

Answer:

The Answer to the question is :

Explanation:

The contact angle between the mercury surface and capillary tube wall is Greater than 90.

If the surface of the solid is hydrophobic, the contact angle will be greater than 90 °. On very hydrophobic surfaces the angle can be greater than 150º and even close to 180º.

8 0
3 years ago
Pls paki answers naman Ito pls sa inyo​
netineya [11]

:nanjan lang din yan sa binasa mo ne........

7 0
2 years ago
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