Respuesta :
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
The answer is "45.3 NM".
There at end of the movement, the forging force is given by
[tex]\to F = Y \times \pi \times r^2 \times [1 + (\frac{2 \mu r}{3h})][/tex]
h is the final height.
[tex]\to h = \frac{100}{2}= 50 \ mm[/tex]
The ultimate radius is determined by following a volume constancy law, which states that volumes before deformation measured amount following distortion.
[tex]\to \pi \times 75^2 \times 2 \times 100 = \pi \times r^2 \times 2 \times 50\\\\\to 75^2 \times 2 = r^2\\\\\to r^2 = 11250\\\\\to r = \sqrt{11250}\\\\\to r = 106 \ mm\\\\\to E = \In(\frac{100}{50})\\\\\to E = 0.69\\\\[/tex]
You may deduce from the graph flow that [tex]Y = 1000\ MPa[/tex], thus we use the formula.
[tex]= 1000 \times 3.14 \times 0.106^2 \times [1 + (\frac{ 2 \times 0.2 \times 0.106}{3 \times 0.05})]\\\\= 1000 \times 3.14 \times 0.011236 \times [1 + (\frac{ 0.0424}{0.15})]\\\\= 35.3 \times 1.2826\\\\ = 45.3 \ MN\\\\\\[/tex]
Therefore, the answer is "45.3 NM".
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