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Calculate the machining time required to reduce the shaft diameter by 6 mm if the maximum depth of cut is 2 mm, and also the parting off (cut off) time for the shaft, knowing that: the shaft diameter = 50 mm, shaft length = 500 mm, longitudinal feed = 0.2mm/rev, cross (traverse) feed = 0.1 mm/rev and cutting velocity (v) 16 m/min.

This is a problem about parting off then turning what I don't understand is the definition of depth of cut here I initially assumed that the depth of cut is 6mm but however when I read the max. depth of cut =2mm this confused me.

Also the turning part seems to be done on multiple stages longitudinal and transverse but exactly how do I calculate the number of stages?

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  • $\begingroup$ If this is supposed to help teach machining, it should specify tolerances. It helps to teach early on that if the maximum depth per pass is theoretically 2, and the desired result is a minimum of 8, that it must take 5 passes. However rather than 8, if it is 8+/-.1 and the machine can achieve +/-.01 then aiming for 7.95 allows it to become 4 passes. $\endgroup$
    – Abel
    Jun 3, 2022 at 12:23
  • $\begingroup$ @Abel Maybe yeah as here you can't achieve a 6mm shaft diameter ,so you'll have to make it 8mm as you mentioned . $\endgroup$ Jun 3, 2022 at 13:13
  • $\begingroup$ On a lathe DOC it would be measured along the x-axis when turning along the length. When facing it would be the z-axis. $\endgroup$
    – DKNguyen
    Jun 3, 2022 at 13:21
  • $\begingroup$ 8 was a purely hypothetical number for demonstration. Any relation to 8 elsewhere may be a coincidence. $\endgroup$
    – Abel
    Jun 3, 2022 at 13:43

2 Answers 2

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The maximum cutting depth is the deepest cut that can be made in one pass. Trying to cut too much will cause the cutter to dig in and achieve nothing usual. It can damage the machine. Consequently, each machine has a maximum depth it can cut safely and without doing any damage to itself or to the operator. If an item needs to be cut more deeply than the maximum depth of cut then multiple passes need to be made where some of the material can be removed.

In this situation, the minimum number of passes to achieve the required depth of cut be between 1.5 and 2.5. If you don't want to stress the machine too much a shallower depth of cut can be done, but that will require more passes of the cutting tool.

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    $\begingroup$ Might be worth mentioning in "each machine has..." that by machine, you mean a current state of a machine (including any loaded tools, any gearing settings, and even the part that it is to cut) since these all influence that max depth. $\endgroup$
    – Abel
    Jun 3, 2022 at 12:12
  • $\begingroup$ @Fred by your definition should it be $1.5 \approx 2$ passes as the machine cuts 2mm from the radius = 4mm from the diameter and not 3 $\endgroup$ Jun 3, 2022 at 13:06
  • $\begingroup$ @YoussefMohamed: You are correct. A cut a 2 mm will reduce the diameter by 4 mm. $\endgroup$
    – Fred
    Jun 3, 2022 at 18:52
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OK so you have a shaft of length 500 mm and diameter 50 mm. You need to reduce the diameter by 6 mm. For your material and setup, your cutting speed (that is, how fast the tool can travel over the shaft) is $\dot{c} = 16 m/min$, and your maximum depth of cut is 2 mm. Longitudinal feed is given as 0.2 mm/rev, transverse feed 0.1 mm/rev.

I am going to assume that machine set time, measurement time, tool change time, etc are all zero. I will also assume that there is no facing cut to be made, and that the original metal you would be using is greater than 500 mm, since you are parting it off. I am also going to assume that your machine can be set to ANY rpm.

Since your machine will reduce the radius, this means that your total longitudinal cuts should come out to $\Delta = 3mm$. This would be done most efficiently with two cuts, of $\delta_1 = 2mm$ and $\delta_2 = 1 mm$. You would then have a parting cut.

Pass 1: Longitudinal pass cutting $\delta_1 = 2mm$, across length $l=500mm$, original diameter $d_0 = 50 mm$. First, find the spindle RPM: $$\nu_1 = \frac{\dot{c}}{\pi d_0} = \frac{(16 m/min)(1000 mm/m)}{(3.1416/rev)(50 mm)}=102 rpm$$ Your travel speed is your longitudinal feed $f_z = 0.2 mm/rev)$ times your spindle speed: $$ \dot{z}_1=f_z \nu_1 = 20.4 mm/min$$ From there, you can get your tie taken: $$ l = \dot{z}_1t_1 \implies t_1=\frac{l}{\dot{z}_1}=\frac{500 mm}{20.4 mm/min} = 24.51 min$$ Pass 2: Your second pass should be identical to the first, except that the depth of cut is 1 mm. Normally, reducing your depth of cut means you can adjust your cutting speed by a factor $F_d$, but we will disregard that here since you don't have enough information to find it. Since your new diameter is $d_1=46mm$, we will recalculate the spindle RPM: $$\nu_2 = \frac{\dot{c}}{\pi d_1} = \frac{(16 m/min)(1000 mm/m)}{(3.1416/rev)(46 mm)}=111 rpm$$ $$\dot{z}_2=f_z\nu_2 = 22.2mm/min$$ $$t_2=\frac{l}{\dot{z}_2}=\frac{500 mm}{22.2 mm/min} = 22.52 min$$

Pass 3: This is the parting off. New diameter is $d_2=44 mm$, and transverse travel feed is $f_x=0.1 mm/rev$. Lets get our new spindle speed and travel speed: $$\nu_3 = \frac{\dot{c}}{\pi d_2} = \frac{(16 m/min)(1000 mm/m)}{(3.1416/rev)(44 mm)}=116 rpm$$ $$\dot{x}_3=f_x\nu_3=(0.1 mm/rev)(116 rpm)=11.6 mm/min$$ $$r = \frac{d}{2} = \dot{x}_3t_3 \implies t_3=\frac{d}{2 \dot{x}_3}=\frac{44 mm}{2(11.6 mm/min)}=1.90 min$$

Total time: Your total time is the sum of these three passes, since we are ignoring everything else: $$t=t_1+t_2+t_3 = 24.51 min + 22.52 min + 1.90 min = 48.93 min$$

Based on the assumptions made to solve the problem (and, barring a math error) this should be the machining time.

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