Complete Question
The complete question is shown on the first uploaded image
Answer:
a
The machining time is [tex]T_m = 0.866minutes[/tex]
b
The maximum metal removal rate is [tex]R_{mr} =4812.75mm^3 /sec[/tex]
Explanation:
From the question we are told that
The length of the work piece is [tex]L =300mm[/tex]
The width of the work piece is [tex]W = 125mm[/tex]
The cut Depth is [tex]D = 6mm[/tex]
The chip load is [tex]f = 0.27 mm/tooth[/tex]
The number of teeth is [tex]n = 4[/tex]
The cutting speed is [tex]v = 2.8m/s = 2800mm/s[/tex]
Now in order to obtain the actual machining time to make the pass across the surface
We first obtain the speed of rotation of the machine
The cutting velocity is mathematically represented as
[tex]v = \pi d N[/tex]
Where d is the diameter given as 150mm
N is the speed of rotation
Making N the subject of the formula
[tex]N =\frac{v}{\pi *d}[/tex]
[tex]= \frac{2800}{\pi *150 }[/tex]
[tex]= 5.941rev/sec[/tex]
Next we need to obtain the feed rate this is mathematically represented as
[tex]f_r = N nf[/tex]
Substituting the values accordingly
[tex]= 5.94 * 4 * 0.27[/tex]
[tex]= 6.417 mm/s[/tex]
The displacement of the milling operation is mathematically represented as
[tex]A = 0.5(d - \sqrt{d^2 - W^2} )[/tex]
[tex]= 0.5(150 - \sqrt{150^2 - 125^2} )[/tex]
[tex]= 0.5 * (150 - 82.9)[/tex]
[tex]= 33.5mm[/tex]
The machining time is mathematically represented as
[tex]T_m = \frac{(L + A)}{f_r}[/tex]
[tex]= \frac{(300 +33.5)}{6.417}[/tex]
[tex]=\frac{51.97}{60} = 0.866min[/tex]
The maximum metal removal rate is mathematically represented as
[tex]R_{mr} = W D f_r[/tex]
Substituting values accordingly
[tex]= 125 * 6* 6.417[/tex]
[tex]= 4812.75 mm^3/sec[/tex]
