Answer the following about a turning operation:
a) Estimate the machining time required in turning a single pass in a 1.25
m long annealed aluminum alloy round bar that is 75 mm in diameter,
using a high-speed steel (HSS) tool.
b) Estimate the time for a tungsten carbide (WC) tool. Take a constant
feed of 2 mm/rev and cutting speed for HSS tool of 4 m/s and for WC tool
of 7 m/s. Show all work.
c) Use a range of feed from 1 mm/rev up to 5 mm/rev and maximum
allowable cutting speeds for high-speed steel tool 3 m/s and 4 m/s, for
carbide tool 5 m/s and 7 m/s.
The solution
L=1.25m≈1250mm/D=75mm/v≈4m/s/f=2mm/rev
For (HSS)
𝑇=
𝐿
𝜋𝐷𝐿 𝜋 × 1250 × 75
=
=
≈ 36.825𝑚𝑖𝑛
𝑁𝑓
𝑣𝑓
4 × 103 × 2
For (WC)
𝜋𝐷𝐿 𝜋 × 75 × 1250
𝑇=
=
≈ 21.037𝑚𝑖𝑛
𝑣𝑓
7 × 103 × 2
2.A shaft is to be roughed in one cut from 95mm diameter to 91 mm and
finished to 90mm. The feed in roughing process = 4 finishing feed = 0.4
mm/rev. Calculate the total machining time if n = 345 r.p.m and the
workpiece length is 50 cm
The solution
N=344 r.m.p / L=500mm /f(rough)=.4mm/rev /f(finishing) =.1mm/rev
95 − 90
𝐴𝑡 =
= 2.5𝑚𝑚
2
95 − 91
91 − 90
𝑎𝑟 =
= 2𝑚𝑚 𝑎𝑓 =
= 0.5𝑚𝑚
2
2
𝐴𝑡 2.5
𝑖𝑟 =
=
≈ 1.
𝑎𝑟
2
𝐿 𝑖
𝐿𝑚 𝑖𝑓
𝑓𝑟 𝑁
𝑓𝑓 𝑁
𝑇𝑚 = 𝑚 𝑟 +
=
2.5 − 2
𝑖𝑓 =
=1
.5
500×1
.4×345
+
500×1
.1×345
≈ 18.12𝑚𝑖𝑛
2
3.Calculate the total machining time for a steel rod shown, which is
turned using a center lathe according to data and dimensions labeled in
the following figure.
The solution
Lm= 250+10=260mm.
90 − 74
𝐴𝑡 =
= 8𝑚𝑚
2
𝐴𝑡
8
𝑖=
=
≈ 3.2
𝑎𝑟 2.5
𝑖𝑟 = 3
𝐴𝑟 − 𝑖𝑟 𝑎𝑟 8 − 3 × 2.5
𝑖𝑓 =
=
=1
𝑎𝑓
.5
𝐿𝑚 𝑖𝑟 𝐿𝑚 𝑗𝑓 260 × 3
260 × 1
𝑇𝑚 =
+
=
+
= 11.123𝑚𝑖𝑛
𝑓𝑟 𝑁
𝑓𝑓 𝑁
250 × .5 250 × .15
3
4. A 120 mm long, 35 mm diameter steel rod is being reduced in
diameter to 25 mm by turning on a lathe. The workpiece length after
machining is 115 mm. The spindle speed is 425 r.p.m. and the tool is
traveling at an axial speed 0.17 m/min. If the depth of cut used in
roughing is 1.25 mm. Calculate:
a) The cutting speed, feed, the metal removal rate.
b) The total machining time,
where: n finish = 2 n rough, a rough = 2 a finish, and feed rate in roughing
= 2 feed rate in finishing
The solution
Facing
𝐷 35
𝐿𝑚 = =
+ 7 = 24.5𝑚𝑚
2
2
5
5 − 3 × 1.25
𝑖=
= 4 𝑖𝑟 = 4 − 1 = 3
𝑖𝑓 =
=2
1.25
. 625
𝐿𝑚 𝑖𝑟 𝐿𝑚 𝑗𝑓 24.5 × 3 24.5 × 2
𝑇𝑚 =
+
=
+
= 1.0088𝑚𝑖𝑛
𝑓𝑟 𝑁
𝑓𝑓 𝑁
170
85
Longitudinal
Lm=115+10=125mm
𝐿𝑚 𝑖𝑟 𝐿𝑚 𝑗𝑓 125 × 3 125 × 2
𝑇𝑚 =
+
=
+
= 5.15𝑚𝑖𝑛
𝑓𝑟 𝑁
𝑓𝑓 𝑁
170
85
Tm(total)=5.15+1.0088=6.1588min
𝑣 = 𝜋𝑁𝐷 = 𝜋 × 425 × 35 ≈ 46.73𝑚\𝑠
RMR=fvd=46.73×103×.4×1.25=23365mm3/min
4
5. A 150-mm-long, 12.5-mm-diameter 304 stainless steel rod is being
reduced in diameter to 12.0 mm by turning on a lathe. The spindle rotates
at N 400 rpm, and the tool is travelling at an axial speed of 200 mm/min.
Calculate the cutting speed, material-removal rate, cutting time.
The solution
a=.5mm. / N=400 r.p.m / fr=200mm/min. /D=12.5mm
𝑣 = 𝜋𝐷𝑁 = 𝜋 × 12.5 × 400 × 10−3 ≈ 15.7𝑚\𝑚𝑖𝑛
𝑓𝑟 200
𝑓= =
≈ .5𝑚𝑚/𝑟𝑒𝑣
𝑁 400
𝐷𝑜 − 𝐷𝑓 12.5 − 12
𝑑=
=
= .25𝑚𝑚
2
2
𝑀𝑅𝑅 = 𝑣𝑑𝑓𝑟 = 15.7 × 103 × 0.5 × .125 = 1963.5𝑚𝑚3 /𝑚𝑖𝑛
𝑇𝑚 =
𝐿
𝑓𝑟
=
150
200
= .75𝑚𝑖𝑛
6. A shaft is to be roughed in one cut from 95mm diameter to 90 mm and
finished to 88mm. The feed in roughing process = 4 finishing feed = 0.4
mm/rev. Calculate the total machining time if n = 450 r.p.m and the
workpiece length is 50 cm
The solution
95 − 88
𝐴𝑡 =
= 3.5𝑚𝑚
2
𝐴𝑡 3.5
𝑖=
=
1.4
𝑎𝑟 2.5
𝑖𝑟 = 1
𝑖𝑓 =
𝐴𝑡 − 𝑎𝑟 3.5 − 2.5
=
=1
𝑎𝑡
1
𝐿𝑚 𝑖𝑟 𝐿𝑚 𝑟𝑓
500 × 1
500 × 1
𝑇𝑚 =
+
=
+
= 13.89𝑚𝑖𝑛
𝑓𝑟 𝑁
𝑓𝑓 𝑁
. 4 × 450 . 1 × 450
7. A 140-mm-long, 12.5-mm-diameter 304 stainless steel rod is being
reduced in diameter to 12.0 mm by turning on a lathe. The spindle rotates
5
at N 600 rpm, and the tool is travelling at an axial speed of 250 mm/min.
Calculate the cutting speed, material-removal rate, cutting time
The solution
𝑣 = 𝜋𝐷𝑁 = 𝜋 × 12.5 × 600 × 10−3 = 23.56𝑚\𝑚𝑖𝑛
𝑓𝑟 250
𝑓= =
= .42𝑚𝑚\𝑟𝑒𝑣
𝑁 600
𝐷𝑜 − 𝐷𝑓 12.5 − 12
𝑑=
=
= .25𝑚𝑚
2
2
𝑀𝑅𝑅 = 𝑣𝑓𝑑 = 23.56 × .42 × .25 × 103 = 2473.8𝑚𝑚3 \𝑚𝑖𝑛
𝐿𝑚 140
𝑇𝑚 =
=
≈ .56𝑚𝑖𝑛
𝑓𝑟
250
6