Concordia University

INDU 411

Computer Integrated Manufacturing

Assignment # 1:

Due February 1st, 2011 before the class

Onur KuzgunkayaWinter 2011

Question 1:

Compare the differences of mass production and flexible production system paradigm. Explain their business model and technological enablers.

Mass production business model is based on maximizing the production capacity to reduce the unit production cost which in turn will generate more demand that will create the need for further increase in capacity (also called economies of scale). The objective is to minimize the cost through the decomposition of manufacturing steps to simple tasks to create high efficiency. This could be accomplished by the interchangeable part concept (enabler).

On the other hand flexible production paradigm relies on generating revenue by increasing the product variety. Each product variant might have small demand but the total demand received by this system should be at a level to justify production cost. This business model is based on economies of scope, i.e by increasing the product scope the company is able to capture a variety of markets so that the total production quantity is at higher levels. This can be achieved by two main enablers. The first enabler is that the products should have a certain degree of similarity so that we can identify product families. The second enabler is to use the flexible production technology that is capable of processing part families without the need for changeovers between different model. This will allow a mixed model production line without the cost of setups.

Question 2:

A milling operation is performed on a NC machining center. Total travel distance = 300 mm in a direction parallel to one of the axes of the worktable. Cutting speed = 1.25 m/s and chip load = 0.05 mm. The end milling cutter has four teeth and its diameter = 20.0 mm. The axis uses a dc servomotor whose output shaft is coupled to a lead screw with pitch = 6.0 mm. The feedback sensing device is an optical encoder which emits 250 pulses per revolution. Determine (a) feed rate and time to complete the cut, (b) rotational speed of the motor and (c) pulse rate of the encoder at the feed rate indicated.

Solution: (a) N = (1.25 x 60 m/min)/(2010-3 m/rev) = 1193 rpm.

fr = N f nt = 1193(0.05)(4) = 238 mm/min.Tm = 300/238 = 1.26 min

(b) N = fr /p = (238 mm/min)/(6 mm/rev) = 39.67rpm.fp = ns N = 250(39.67/60) = 165.3 Hz

Question 3:

A DC servomotor is used to drive the x-axis of an NC milling machine table. The motor is coupled directly to the table leadscrew, which has 1.6 threads/cm. An optical encoder is used to provide the feedback measurement. It is connected to the leadscrew using a 1:5 gear ratio (one turn of the leadscrew converts to 5 turns of the encoder disk). The optical encoder emits 125 pulses per revolution. To execute a certain programmed instruction, the table must be moved on x-y plane from point (3.5, 1.5) to point (1.0, 7.2) in a straight line trajectory at a feed rate of 7.5mm/min (a) What is the minimum travel distance that can be controlled? (b) Determine the rotational speed of the motor and (c) the frequency of the pulse train emitted by the optical encoder when the desired feed rate is achieved.

p=1/1.6=6.25mm ns=125 =360/125=2.88°

a) Due to the gear between leadscrew and optical encoder we have to include the gear ratio ==>

b) x-y plane move from (3.5,1.5) to (1, 7.2) x=2.5 y=5.7

We have to find vx to define the rotational speed of the motor that actuates the x-axis displacement

arctan(5.7/2.5)=66.318° so vx=7.5*cos(66.318)=3mm/min

N=0.48rpm

c) Rotational speed of the encoder 0.48x5=2.4rpm


Question 4:

The outline of the following part is to be profile milled using a 30 mm diameter end mill with four teeth. The part is 15 mm thick. Cutting speed = 150 m/min and feed = 0.085 mm/tooth. Use the lower left corner of the part as the origin in the x-y axis system. Two of the holes in the part have already been drilled and will be used for clamping the part during profile milling.Use absolute positioning.

Solution: The part will be fixtured so that its top surface is 40 mm above the surface of the machine tool table, and the x-y plane of the axis system will be defined 40 mm above the table surface. As given, a 30 mm diameter end mill with four teeth will be used.

Cutting speed is specified as 150 m/min. Rotational speed of the cutter is calculated as N = 150/(30 x 10-3) = 1592 rev/min. Given a feed = 0.085 mm/tooth, feed rate is calculated as 1592(4)(0.085) = 541 mm/min.

Cutter diameter data has been manually entered into offset register 05. At the beginning of the job, the cutter will be positioned so that its center tip is at a target point located at x = -50, y = -50, and z = + 10. The program begins with the tool positioned at this location.

NC part program code
N001 G71 G90 G92 X-050.0 Y-050.0 Z010.0;
N002 G00 Z-015.0 S1592 M03;
N003 G01 G42 Y0 D05 F541;
N004 G01 X200.0;
N005 G01 Y050.0;
N006 G01 X150.0;
N007 G02 X125.0 Y075.0 R025.0;
N008 G01 Y100.0;
N009 G01 X025.0;
N010 G01 X0 Y050.0;
N011 G01 Y0;
N012 G40 G00 X-050.0 Y-050.0 Z010.0 M05;
N013 M30; / Comments
Define origin of axes.
Rapid to cutter depth, turn spindle on.
Bring tool to starting y-value, start cutter offset.
Mill lower part edge.
Mill right straight edge.
Mill horizontal step above two 8 mm holes
Circular interpolation around arc.
Mill vertical step above arc.
Mill top part edge.
Mill angled edge at left of part.
Mill vertical edge at left of part.
Rapid move to target point, cancel offset, spindle stop.
End of program, stop machine.