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Tamiku [17]
3 years ago
11

Calculate the wire pressure for a round copper bar with an original cross-sectional area of 12.56 mm2 to a 30% reduction of area

, given that the included die angle is 300 with a coefficient of friction (μ) of 0.08 and the yield stress for copper is 350 MPa.
Engineering
1 answer:
dybincka [34]3 years ago
8 0

Answer:153.76 MPa

Explanation:

Initial Area\left ( A_0\right )=12.56 mm^2

Final Area\left ( A_f\right )=0.7\times 12.56 mm^2=8.792 mm^2

Die angle=30^{\circ}

\alpha =\frac{30}{2}=15^{\circ}

\mu =0.08

Yield stress\left ( \sigma _y \right )=350 MPa

B=\mu cot\left ( \aplha\right )=0.2985

\sigma _{pressure}=\sigma _y\left [\frac{1+B}{B}\right ]\left [ 1-\frac{A_f}{A_0}\right ]^B

\sigma _{pressure}=350\left [\frac{1+0.2985}{0.2985}\right ]\left [ 1-\frac{8.792}{12.56}\right ]^{0.2985}

\sigma _{pressure}=153.76 MPa

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A 03-series cylindrical roller bearing with inner ring rotating is required for an application in which the life requirement is
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\mathbf{C_{10} = 137.611 \ kN}

Explanation:

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x_D = \text{bearing life in million revolution} \\  \\ x_D = \dfrac{60 \times L_h \times N}{10^6} \\ \\ x_D = \dfrac{60 \times 40 \times 10^3 \times 520}{10^6}\\ \\ x_D = 1248 \text{ million revolutions}

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C_{10}=1.4\times 2600 \times \pmatrix \dfrac{1248}{0.02+(4.439) \bigg(In(\dfrac{1}{0.9}) \bigg)^{\dfrac{1}{1.483}}} \end {pmatrix} ^{^{^{\dfrac{1}{\dfrac{10}{3}}}

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C_{10} = 3640 \times \bigg[\dfrac{1248}{0.9933481582}\bigg]^{\dfrac{3}{10}}

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