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2 years ago

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A rate of 0.42 minute per piece is set for a forging operation. The operator works on the job for a full eight-hour day and prod

uces 1,500 pieces. Use a standard hour plan.
Required:
a. How many standard hours does the operator earn?
b. What is the operator's efficiency for the day?
c. If the base rate is 9.80 per hour, compute the earnings for the day.
d. What is the direct labor cost per piece at this efficiency?
e. What would be the proper piece rate (rate expressed in money) for this job, assuming that the above time standard is correct?

1 answer:

MrRa [10]2 years ago

6 0

Answer:

b. What is the operator's efficiency for the day?

AND

e. What would be the proper piece rate (rate expressed in money) for this job, assuming that the above time standard is correct?

Explanation:

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The solution for the given problem is done below.

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A milling operation was used to remove a portion of a solid bar of square cross section. Forces of magnitude P = 18 kN are appli

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The smallest allowable depth is $d=16.04 \mathrm{mm}$ for the milled portion of bar.

<u>Explanation:</u>

Given,

Magnitude of force, $\mathbf{p}=18 \mathrm{kN}$

$a=30 \mathrm{mm}$

$=0.03 \mathrm{m}$

Allowable stress, $\sigma_{a l l}=135 \mathrm{MPa}$

cross sectional area of bar,

$A=a \times d$

$A=a d$

e - eccentricity

$e=\frac{a}{2}-\frac{d}{2}$

The internal forces in the cross section are equivalent to a centric force P and a bending couple M.

$M=P e$

$=P\left(\frac{a}{2}-\frac{d}{2}\right)$

$=\frac{P(a-d)}{2}$

Allowable stress

$\sigma=\frac{P}{A}+\frac{M c}{I}$

$c=\frac{d}{2}$

Moment of Inertia,

$I=\frac{b d^{3}}{12}$

$=\frac{a d^{3}}{12}$

$\therefore \sigma=\frac{P}{a d}+\frac{\frac{P(a-d)}{2} \times \frac{d}{2}}{\frac{a d^{3}}{12}}$

$\sigma=\frac{P}{a d}+\frac{3 P(a-d)}{a d^{2}}\\$

$\sqrt{x} \sigma\left(a d^{2}\right)=P d+3 P(a-d)$

$\sigma\left(a d^{2}\right)=P d+3 P a-3 P d$

$\sigma\left(a d^{2}\right)=(P-3 P) d+3 P a$

$\left(\sigma a d^{2}\right)=-2 P d+3 P a$

$\sigma d^{2}=-\frac{2 P}{a} d+3 P$

By substituting values we get,

$\left(135 \times 10^{6}\right) d^{2}+\frac{2 \times 18 \times 10^{3}}{0.03} d-3\left(18 \times 10^{3}\right)=0$

$\left(135 \times 10^{6}\right) d^{2}+\left(12 \times 10^{5}\right) d-54 \times 10^{3}=0$

On solving above equation we get, $d=0.01604 \mathrm{m}\\$

$d=16.04 \mathrm{mm}$

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