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Solved Manufacturing Process Problems: Lot Cost, Milling and Turning Examples

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Manufacturing Processes Exam: Problems and Solutions

Course: Manufacturing Processes I
Authors: Ing. Sánchez V., Victoriano and Ing. Emiliano Loayza H.
Institution: Universidad Nacional del Callao, Faculty of Mechanical-Energy Engineering

Note: In the notebook, write the course name, code, surnames and first names. The development of each problem must be literal, without abbreviations of any kind, and must be done only with black or blue pen. Pencil or red pen is prohibited.

Problem No. 1: Production Cost of a Batch

The production cost of a batch of 600 units of a mechanical device for serial work is calculated with the following expenses: raw material is US$1200, tooling expenses are US$60. Machine setup time is 10 hours for the construction of the machine. The manufacturing cycle is 50 minutes per piece, and the fatigue time plus service time is 9 min. The initial cost of the machine to be used is US$18000, with a useful life of 10 years under normal conditions. Maintenance costs US$1500 monthly and the operator costs US$400 per month. General workshop expenses amount to US$300 per month. The investment rate is 6% monthly. The machine operation in one annual shift is 2000 hours.

Under the proposed conditions, determine:

  1. The time required to manufacture the entire batch.
  2. The cost of the machine to produce.
  3. The hourly machine operation cost.
  4. The manufacturing cost per unit produced.
  5. The unit cost to produce.

Solution

Data:

  • Number of production units: n = 600 units
  • Raw material: US$1200 (material cost)
  • Tooling cost: CHta = US$60
  • Machine setup time: TP = 10 hours
  • Manufacturing cycle time: TC = 50 min/piece
  • Fatigue plus service time: Tf + Ts = 9 min
  • Initial cost of the assembly: C0 = US$18000
  • Useful life of the assembly: N = 10 years
  • Maintenance cost: M = US$1500 per month
  • Operator cost: US$400 per month
  • General workshop expenses: B = US$300 per month
  • Investment rate: i = 6% monthly

1. Time required to manufacture the entire batch in hours.

Standard time: Te = Tc + Tf + Ts = (50 + 9) = 59 min.

Calculated time: The total manufacturing time for the whole batch in hours will be TT = Tcal × n = 60 min/piece × 600 = 600 hours.

2. Cost of the machine to produce.

Where: Co = US$18000, H = 2000 hours, N = 10 years.

Maintenance cost, average investment interest, and the total cost of the machine to produce are calculated from these values.

3. Hourly machine cost.

Operator salary cost: US$400 × 12 = US$4800 per year.

Operator labor cost per hour is obtained by dividing this by the annual operating hours.

General workshop expenses per hour: B = US$800 per month (US$800 × 12 = US$9600 per year).

Hourly machine cost is then obtained by adding these components.

4. Manufacturing cost per unit produced.

Material cost, tooling cost per piece, and machine cost per piece are added.

5. Unit cost per piece.

Problem No. 2: Milling Steel Parts

A series of steel parts measuring 88 mm wide, 150 mm long, with a depth of cut of 5 mm, is to be milled. A milling machine with a 2 kW motor is available, and the transmission efficiency of the machine may be considered as 75%. The available speeds and feeds are:

  • n = 34, 53, 87, 137, 210, 340, 860, 955 rpm
  • Va = 14, 23, 42, 80, 120, 195, 300, 490 mm/min

For the indicated work, a face cutter of 95 mm diameter with 10 teeth is used, whose edge position angle is 60°. Cutting speed should not exceed 24 m/min, nor 0.09 mm/tooth feed per tooth. Considering that the working range of the specific mean cutting pressure is 0.1 Kw·min/cm³, and considering the most advantageous conditions, determine:

  1. The rotational speed of the main spindle.
  2. The power required in one cutting pass.
  3. The number of passes of equal depth.
  4. Execution time in one pass.
  5. The maximum undeformed chip thickness.

Solution

Data:

  • Feed per tooth: Z = 0.09 mm/tooth
  • Width = 88 mm
  • Length = 150 mm
  • Depth = 5 mm
  • Power: Pm = 2 kW
  • Efficiency: η = 75%
  • Cutter diameter: D = 95 mm
  • Number of teeth: Z = 10
  • Edge angle: Kr = 60°
  • Cutting speed: Vc = 24 m/min

1. Rotational speed of the main spindle.

We select from the table n = 53 rpm.

The feed is: Va = fz × Z × n, where Z = number of teeth.

Specific coefficient of the material is determined, the feed speed is calculated, and the cutting area is found.

The removal rate is then calculated.

2. Power required in one cutting pass.

Verifying the effective power: Pe < Pc. If not satisfied, modify the depth of cut.

3. Number of passes of equal depth.

Two passes will be made with a depth of 2.5 mm, and a cutting power of 0.924 kW.

4. Machining time in one pass.

5. Maximum deformed chip thickness.

Undeformed chip thickness is calculated from the feed and the edge geometry.

Problem No. 3: Orthogonal Cutting Experiment

An orthogonal cutting experiment is being carried out by turning a tube of 140 mm diameter and 8 mm wall thickness. The tube rotates at 100 rpm. A rake angle of 15° is used, and the tool feed is proposed as 0.3 mm/rev. The cutting force is 300 kgf, the thrust force is 100 kgf, both acting on the tool. The uncut chip thickness is 0.3 mm. It is also known that one meter of chip length obtained weighs 30 grf, and the specific weight of the material is 7.8 grf/cm³.

Determine:

  1. The shear angle.
  2. The apparent friction coefficient between the chip and the material.
  3. The shear force.
  4. The mean cutting stress acting on the shear plane.
  5. The specific cutting energy (Kw·min/cm³).

Solution

Data:

  • Diameter = 140 mm
  • Thickness = 8 mm
  • Rake angle = 15°
  • Feed = 0.3 mm/rev
  • n = 100 rpm
  • Cutting force = 300 kgf
  • Thrust force = 100 kgf

1. Shear angle.

Calculation of the deformed chip thickness and the cutting ratio.

2. Mean friction coefficient between the chip and the material.

3. Shear force.

4. Mean cutting stress acting on the shear plane.

5. Specific cutting energy.

Problem No. 4: Universal Lathe Transmission

On a universal lathe, motion is transmitted to the main spindle through trapezoidal pulleys and equal opposed stepped cones (100, 150 and 200 mm), as shown in the attached figure. The rotational speed of the motor is 2500 rpm, and its minimum speed with reducer for the main spindle is 80 rpm. Determine:

  1. The diameter of trapezoidal pulley A.
  2. The direct speeds of the main spindle.
  3. The reduced speeds of the main spindle.
  4. The standardized module of the system.
  5. The pitch diameters of the gears.

Solution

1. Diameter of trapezoidal pulley A.

Total transmission ratio. Minimum rotational speed: n_min = 80 rpm.

Pulley diameter calculation based on the stepped cones.

2. Direct speeds of the main spindle.

2500 × 80 = 175 × n1, from which n1 = 100 and n2 = 200 are obtained.

3. Reduced speeds of the main spindle.

4. Standardized module of the system.

5. Pitch diameters of the gears:

  • d1 = m × Z1 = 5 × 18 = 90 mm
  • d2 = m × Z2 = 5 × 48 = 240 mm
  • d3 = m × Z3 = 5 × 18 = 90 mm
  • d4 = m × Z4 = 5 × 48 = 240 mm

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