# (20)1.Consider the Papers Presented in Class to Arrive at the T/F Answer for the Following ## ME 482 - Exam #2

Nov. 6, 2006

Name______

#### Problems

(20)1.Consider the papers presented in class to arrive at the T/F answer for the following (circle answer):

a) In the “Prediction of tool life in end milling by response surface methodology” paper tool life
is a function of 5 model parameters.TF X

b)In the paper of question a) above response surface is a term related to the tool-life
contour curves shown in the paper.T XF

c)In the “Approach to optimization of cutting conditions by using artificial neural networks” paper
labor costs and overhead costs are separate parameters (differs from Groover’s presentation).

T XF

d)In the paper of question c) above the artificial neural network algorithms are the same as the
optimization algorithm.TF X

(40) 2.In a turning operation on plain carbon steel whose Brinell hardness = 275 HB, the cutting speed is set at 200 m/min and depth of cut = 6.0 mm. The lathe motor is rated at 25 kW, and its mechanical efficiency = 90%. Using the appropriate specific energy value from Table 21.2, determine the maximum feed that can be set for this operation.

Solution:

From Table 21.2, U = 2.8 N-m/mm3 = 2.8 J/mm3

MRR = vfd = (200 m/min)(103 mm/m)(6 mm)f = 1200(103)f mm3/min = 20(103)f mm3/s

Available power Pc = Pg E = 25(103)(0.90) = 22.5 (103) = 22,500W

Required power Pc = (2.8 N-m/mm3)( 20 x 103) f = 56,000 fEqn (21.21)

Setting available power = required power,

22,500 = 56,000 f

f = 22,500/56,000 = 0.402 mm (this should be interpreted as mm/rev for a turning operation)

However, for this feed (feed = to = depth of cut), correction factor in Figure 21.14 = 0.9; thus,

U = 2.8(0.90) = 2.52 N-m/mm3

and an iterative calculation procedure is required to match the unit power value with the feed, taking the correction factor into account.

Required Pc = (2.52)(20 x 103) f = 50,400 f

Again setting available power = required power,

22,500 = 50,400 f

f = 22,500/50,400 = 0.446 mm/rev

One more iteration using the correction factor yields a value around f = 0.45 mm/rev.

(40)3.A NC lathe cuts two passes across a cylindrical workpiece under automatic cycle. The operator loads and unloads the machine. The starting diameter of the work is 3.0 in. and its length = 10 in. The work cycle consists of the following steps (with element times given in parentheses where applicable): 1 – Operator loads part into machine, starts cycle (1.0 min); 2 – NC lathe positions tool for first pass (0.1 min); 3 – NC lathe turns first pass (time depends on cutting speed); 4 – NC lathe repositions tool for second pass (0.4 min); 5 – NC lathe turns second pass (time depends on cutting speed); and 6 – Operator unloads part and places it in tote pan (1.0 min). In addition, the cutting tool must be periodically changed. This tool change time takes 1.0 min. The feedrate = 0.007 in/rev and the depth of cut for each pass = 0.1 in. The cost of the operator and the machine = \$39/hr and the tool cost = \$2/cutting edge. The applicable Taylor tool life equation has parameters: n = 0.26 and C = 900 ft/min. Determine: a) the cutting speed for minimum cost per piece, b) the average time required to complete one production cycle, c) cost of production cycle, and d) if the setup time for this job is 3.0 hours and the batch size = 300 parts, how long will it take to complete the batch?

Solution

(a)Co = \$39/hr = \$0.65/min.

vmin = 900[0.65/((1/.26 - 1)(0.65 x 1.0 + 2.00))]0.26 = 900[0.65/(2.846 x 2.65)]0.26 = 476 ft/min. (Eq 24.15)

(b)Tmin = (1/.26 - 1)(0.65 x 1 + 2.0)/0.65 = 2.846(2.65/0.65) = 11.6 min. (Eq 24.16)

Tm = (3)(10)/(476 x 12 x 0.007) = 2.36 min/pc. Assume both passes have equal Tm. (Eq 24.5)

np = 11.6/2.36 = 4.9 passes/tool life

Since there are two passes/workpiece, np = 2.45 pc/tool life

Tc = 2.5 + 2 x 2.36 + 1.0/2.45 = 7.63 min/pc. (Eq 24.4)

(c) Cc = 0.65(2.5 + 2 x 2.36) + (0.65 x 1 + 2.00)/2.45 = \$5.77/pc (Eq 24.13)

(d) Time to complete batch = 3.0(60) + 300(7.63) = 2469 min = 41.15 hr.