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defon
3 years ago
14

A series of end-milling cuts is currently used to produce an aluminum part that is an aircraft component. The purpose of the mac

hining operation is to remove 95% of the part weight to create a structural frame. A total of 4.0 min is lost during the milling cycle due to tool repositioning. The part has a length = 1.6 m, width=0 m, and mm. The operation uses a four-tooth indexable-insert end mill with mm at a cutting speed 600 m/min, chip load 0.15 mm/tooth, and average cross-sectional area of cut-240 mm. High-speed machining has to replace the conventional milling process. The same chip load and average area of cut will be used, but the cutting speed will be increased to 3600 m/min , and the time lost for tool repositioning will be reduced to 2.0 min.
Determine :
(a) The cycle time of the current milling operation and
(b) The cycle time of the proposed HSM operation.
(c) Is this part a good candidate for high-speed machining?
Engineering
1 answer:
laila [671]3 years ago
3 0

Answer:

what-

Explanation:

I dont understand

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Explanation:

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A structural component in the form of a wide plate is to be fabricated from a steel alloy that has a plane-strain fracture tough
jeyben [28]

Complete question:

A structural component in the form of a wide plate is to be fabricated from a steel alloy that has a plane strain fracture toughness of 98.9 MPa root m (90 ksi root in.) and a yield strength of 860 MPa (125,000 psi). The flaw size resolution limit of the flaw detection apparatus is 3.0 mm (0.12 in.). If the design stress is one-half of the yield strength and the value of Y is 1.0, determine whether or not a critical flaw for this plate is subject to detection.

Answer:

Since the flaw 17mm is greater than 3 mm the critical flaw for this plate is subject to detection

so that critical flow is subject to detection  

Explanation:

We are given:

Plane strain fracture toughness K = 98.9 MPa \sqrt{m}

Yield strength Y = 860 MPa

Flaw detection apparatus = 3.0mm (12in)

y = 1.0

Let's use the expression:

oc = \frac{K}{Y \sqrt{pi * a}}

We already know

K= design

a = length of surface creak

Since we are to find the length of surface creak, we will make "a" subject of the formula in the expression above.

Therefore

a= \frac{1}{pi} * [\frac{k}{y*a}]^2

Substituting figures in the expression above, we have:

= \frac{1}{pi} * [\frac{98.9 MPa \sqrt{m}} {10 * \frac{860MPa}{2}}]^2

= 0.0168 m

= 17mm

Therefore, since the flaw 17mm > 3 mm the critical flow is subject to detection  

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