Question

Three tool materials (high-speed steel, cemented carbide, and ceramic) are to be compared for the same turning operation on a batch of 50 steel parts. For the high-speed steel tool, the Taylor equation parameters are n = 0.130 and C = 80 (m/min). The price of the HSS tool is $20.00, and it is estimated that it can be ground and reground 15 times at a cost of $2.00 per grind. Tool change time is 3 min. Both carbide and ceramic tools are inserts and can be held in the same mechanical toolholder. The Taylor equation parameters for the cemented carbide are n = 0.30 and C = 650 (m/min), and for the ceramic: n = 0.6 and C = 3500 (m/min). The cost per insert for the carbide is $8.00, and for the ceramic is $10.00. There are six cutting edges per insert in both cases. Tool change time = 1.0 min for both tools. The time to change a part = 2.5 min. Feed = 0.30 mm/rev, and depth of cut = 3.5 mm. Cost of operator and machine time = $40/hr. Part diameter = 73 mm, and length = 250 mm. Setup time for the batch = 2.0 hr. For the three tooling cases, compare (a) cutting speeds for minimum cost, (b) tool lives, (c) cycle time, (d) cost per production unit, and (e) total time to complete the batch. (f) What is the proportion of time spent actually cutting metal for each tool material?

Answer 1

Among all three tools, the ceramic tool is taking the least time for the production of a batch, however, machining from the HSS tool is taking the highest time.

Step-by-step explanation:

The optimum cutting speed for the minimum cost

`V_(opt)= (C)/(\left[\left(T_c+(C_e)/(C_m)\right)\left((1)/(n)-1\right)\right]^n)\;\cdots(i)`

Where,

C,n = Taylor equation parameters

`T_h` =Tool changing time in minutes

`C_e`=Cost per grinding per edge

`C_m`= Machine and operator cost per minute

On comparing with the Taylor equation
`VT^n=C`,

Tool life,

`T= \left[ \left(T_t+(C_e)/(C_m)\right)\left((1)/(n)-1\right)\right]}\;\cdots(ii)`

Given that,

Cost of operator and machine time
`=\$40/hr=\$0.667/min`

Batch setting time = 2 hr

Part handling time:
`T_h=2.5` min

Part diameter:
`D=73 mm`
`=73* 10^(-3) m`

Part length:
`l=250 mm=250* 10^(-3) m`

Feed:
`f=0.30 mm/rev= 0.3* 10^(-3) m/rev`

Depth of cut:
`d=3.5 mm`

For the HSS tool:

Tool cost is $20 and it can be ground and reground 15 times and the grinding= $2/grind.

So,
`C_e=`
`\$20/15+2=\$3.33/edge`

Tool changing time,
`T_t=3` min.

`C= 80 m/min`

`n=0.130`

(a) From equation (i), cutting speed for the minimum cost:

`V_(opt)= \frac {80}{\left[ \left(3+(3.33)/(0.667)\right)\left((1)/(0.13)-1\right)\right]^(0.13)}`

`\Rightarrow 47.7` m/min

(b) From equation (ii), the tool life,

`T=\left(3+(3.33)/(0.667)\right)\left((1)/(0.13)-1\right)\right]}`

`\Rightarrow T=53.4` min

(c) Cycle time:
`T_c=T_h+T_m+(T_t)/(n_p)`

where,

`T_m=` Machining time for one part

`n_p=` Number of pieces cut in one tool life

`T_m= (l)/(fN)` min, where
`N=(V_(opt))/(\pi D)` is the rpm of the spindle.

`\Rightarrow T_m= (\pi D l)/(fV_(opt))`

`\Rightarrow T_m=(\pi * 73 * 250* 10^(-6))/(0.3* 10^(-3)* 47.7)=4.01 min/pc`

So, the number of parts produced in one tool life

`n_p=\frac {T}{T_m}`

`\Rightarrow n_p=\frac {53.4}{4.01}=13.3`

Round it to the lower integer

`\Rightarrow n_p=13`

So, the cycle time

`T_c=2.5+4.01+(3)/(13)=6.74` min/pc

(d) Cost per production unit:

`C_c= C_mT_c+(C_e)/(n_p)`

`\Rightarrow C_c=0.667*6.74+(3.33)/(13)=\$4.75/pc`

(e) Total time to complete the batch= Sum of setup time and production time for one batch

`=2*60+ {50* 6.74}{50}=457 min=7.62 hr`.

(f) The proportion of time spent actually cutting metal

`=(50*4.01)/(457)=0.4387=43.87\%`

Now, for the cemented carbide tool:

Cost per edge,

`C_e=`
`\$8/6=\$1.33/edge`

Tool changing time,
`T_t=1min`

`C= 650 m/min`

`n=0.30`

(a) Cutting speed for the minimum cost:

`V_(opt)= \frac {650}{\left[ \left(1+(1.33)/(0.667)\right)\left((1)/(0.3)-1\right)\right]^(0.3)}=363m/min` [from(i)]

(b) Tool life,

`T=\left[ \left(1+(1.33)/(0.667)\right)\left((1)/(0.3)-1\right)\right]=7min` [from(ii)]

(c) Cycle time:

`T_c=T_h+T_m+(T_t)/(n_p)`

`T_m= (\pi D l)/(fV_(opt))`

`\Rightarrow T_m=(\pi * 73 * 250* 10^(-6))/(0.3* 10^(-3)* 363)=0.53min/pc`

`n_p=\frac {7}{0.53}=13.2`

`\Rightarrow n_p=13` [ nearest lower integer]

So, the cycle time

`T_c=2.5+0.53+(1)/(13)=3.11 min/pc`

(d) Cost per production unit:

`C_c= C_mT_c+(C_e)/(n_p)`

`\Rightarrow C_c=0.667*3.11+(1.33)/(13)=\$2.18/pc`

(e) Total time to complete the batch
`=2*60+ {50* 3.11}{50}=275.5 min=4.59 hr`.

(f) The proportion of time spent actually cutting metal

`=(50*0.53)/(275.5)=0.0962=9.62\%`

Similarly, for the ceramic tool:

`C_e=`
`\$10/6=\$1.67/edge`

`T_t-1min`

`C= 3500 m/min`

`n=0.6`

(a) Cutting speed:

`V_(opt)= \frac {3500}{\left[ \left(1+(1.67)/(0.667)\right)\left((1)/(0.6)-1\right)\right]^(0.6)}`

`\Rightarrow V_(opt)=2105 m/min`

(b) Tool life,

`T=\left[ \left(1+(1.67)/(0.667)\right)\left((1)/(0.6)-1\right)\right]=2.33 min`

(c) Cycle time:

`T_c=T_h+T_m+(T_t)/(n_p)`

`\Rightarrow T_m=(\pi * 73 * 250* 10^(-6))/(0.3* 10^(-3)* 2105)=0.091 min/pc`

`n_p=\frac {2.33}{0.091}=25.6`

`\Rightarrow n_p=25 pc/tool\; life`

So,

`T_c=2.5+0.091+(1)/(25)=2.63 min/pc`

(d) Cost per production unit:

`C_c= C_mT_c+(C_e)/(n_p)`

`\Rightarrow C_c=0.667*2.63+(1.67)/(25)=$1.82/pc`

(e) Total time to complete the batch

`=2*60+ {50* 2.63}=251.5 min=4.19 hr`.

(f) The proportion of time spent actually cutting metal

`=(50*0.091)/(251.5)=0.0181=1.81\%`