15ME328E
Process Planning and Cost Estimation
Handled by
Dr. U. Mohammed Iqbal & Dr. S. Murali
Associate Professor
Department of Mechanical Engineering, SRM IST
The contents in the presentation are from various sources including text books, reference books, internet sources.
For student study purpose.
Book:
1. Automation, Production Systems and Computer-Integrated Manufacturing by M. P. Groover, Pearson Publisher
2. Process planning and cost estimation by Dr. V. Jayakumar and J. Indu, Lakshmi Publications, ISBN: 978-9383103-03-4
3. Process planning and cost estimation by R. Kesavan, C. Elanchezhian, B. Vijaya Ramnath, New Age
International Publisher, 2nd edtion.
4. Process planning and cost estimation by M. Adithan, New Age International Publisher
2
3
4
Unit 1 – Process Planning
5
Manufacturing
 Manufacturing
 Latin: manus (hand) + factus (make)
 Manufacturing is the process of converting raw materials into useful products.
Processed
Material
Manufacturing
Process
Raw Material
Scarp and waste
6
Manufacturing vs Production
 When the raw material is used as input to produce goods with the use of machinery is known as
a Manufacturing. The process of transforming resources into finished products is known as
Production.
 In manufacturing, the use of machinery is a must whereas production can be done with or
without the use of machinery.
 In manufacturing, the output generated will be tangible in nature, i.e. goods only, but in the case
of production it produces both tangible and intangible outputs, i.e. goods as well as services.
 Men-machine setup should be there for manufacturing of goods, which is not in the case of
production, the only man is sufficient for producing output.
7
Production system
8
Production
 Production system is a collection of people, equipment, and procedures organized to perform
manufacturing
 Production system dived into two levels
Manufacturing
Facilities
systems
Factory and plant
layout
Production system
Business function
Manufacturing
support systems
Product design
Manufacturing
planning
Manufacturing
9
control
Facilities
 The Facilities in the production system include the factory, production machines and tooling,
material handling equipment, inspection equipment, and computer systems that control the
manufacturing operations.
Manual Work Systems
(Hand tools employed)
Manufacturing systems
Worker-Machine Systems
(Power tools employed)
Facilities
Automated Systems
Factory and plant layout
(without direct participation of
human)
10
Facilities
<Worker-Machine System>
<Manual Work Systems>
<Automated Systems>
Ref: Automation, Production Systems, and Computer-Integrated Manufacturing, Fouth Edition, by Mikell P. Groover
11
Manufacturing Support Systems
 Business functions - sales and marketing, order entry, cost accounting, customer billing
 Product design - research and development, design engineering, prototype shop
 Manufacturing planning - process planning, production planning, MRP, capacity planning
 Manufacturing control - shop floor control, inventory control, quality control
Ref: Automation, Production Systems, and Computer-Integrated Manufacturing, Fouth Edition, by Mikell P. Groover
12
Types of production
13
Types of production
Automation
Specialization
Skills
Ref: http://nptel.ac.in/courses/108105063
14
Types of production
15
Continuous/Mass production
 To produce standardized products with a standard set of process and operation sequence in
anticipation of demand. This ensures continuous production output.
 It is also termed as mass flow production or assembly line production.
 This system results in less work in process (WIP) inventory but involves high capital investment in
machinery and equipment.
 Large volume and small variety of output. E.g., oil refineries, cement manufacturing and sugar
factory, etc.
Ref: An overview of production systems and production planning and control, Dr. Hemant Sharma
16
Continuous/Mass production
 Characteristics
 Most of the machines are semi-automatic or automatic
 Semi-skilled workers are normally employed
 Cost of production per unit is very low owing to the high production rate
 Inventory of work in process is small
 Material handling also automatic (e.g. conveyors)
 Small varieties and large production volume
 Pre-defined line for product flow used
 Planning and control of the system is easier.
Ref: An overview of production systems and production planning and control, Dr. Hemant Sharma
17
Job Production / Unit Production
 It involves production as per customer’s specifications.
 This ensures the simultaneous production of large number of batches/orders. (Note: Each batch
or order comprises of a small lot of identical products and is different from other batches.)
 Characteristics
 Most suitable for heterogeneous products
 Man power should be skilled enough to deal with changing work conditions.
 Product cost is normally high because of high material and labor costs.
 Comparatively smaller investment in machines and equipment.
 Flexible in nature and it can be adapted to changes in product design/order size.
 Grouping of machines is done on functional basis.
Ref: An overview of production systems and production planning and control, Dr. Hemant Sharma
18
Intermittent Production / Batch Production
 Production of different types of products in small quantities (batches).
 Goods may be produced partly for inventory and partly for customer’s orders.
Example:components are made for inventory but they are combined differently for different
customers (automobile plants, printing presses, electrical goods plants)
 Characteristics
 Grouping of machines is done on functional basis.
 Semi-automatic and special purpose machines are used.
 Labor should be skilled enough to work upon different product batches.
 In process inventory usually high
Ref: An overview of production systems and production planning and control, Dr. Hemant Sharma
19
3S (Simplification, Standardization, Specialization)
 Simplification – is a process of product analysis through which unnecessary varieties and designs
are eliminated. Only a limited number of grades, types and sizes of the product are retained.
 Simplification is a point disagreement between production and marketing departments.
 Production department favors minimum variety as it reduces the number of set ups and
utilization of production facilities is increased and be able to produce the component at the
reduced cost compared to more variety.
 Marketing department favors higher variety helps them to sell more as it helps them to meet the
needs of the larger customer groups.
 Thus an optimum variety of types of products is to be manufactured.
20
3S (Simplification, Standardization, Specialization)
ADVANTAGES
1.
Reduce inventories of materials and components.
2.
Reduced investments in plant and machinery.
3.
Reduced space requirements of storage.
4.
Ease of planning and control.
5.
Reduction in selling price.
6.
Simplification of inspection and control.
DISADVANTAGES
1.
Not able to meet the needs of wide range of customer preferences.
2.
Possibility of loosing orders to competitors.
3.
Creates a constant source of conflict between marketing and production
21
3S (Simplification, Standardization, Specialization)
 Standardization – is the second step after simplification towards interchangeable manufacturing.
Having selected the varieties and grades of the products to be retained as much of its
manufacturing details are standardized as possible.
 Since manufacturing involves a large number of decisions from selection of raw material to the
process used for finishing, standardization of some of these items reduce unnecessary repetition
of work.
 Use of standard components reduces inventory costs, ensures interchangeability and makes
future maintenance easier.
 Standardization is the process of formulating and applying rules for an orderly approach to a
specific activity for the benefit of improving overall economy considering the safety requirements.
22
3S (Simplification, Standardization, Specialization)
23
3S (Simplification, Standardization, Specialization)
Classification of Standardization
1. Basic standardization- (scales and weights, voltages, limit and fits, surface texture etc.)
2. Dimensional standardization- (Engg. components such as nuts, bolts, screws, gears, keys)
3. Material standardization- (raw materials, lubricants, coolants, tools etc)
4. Equipment standardization- (specifications, location, layout etc.)
5. Process standardization- (turning, grinding, welding etc.)
6. Quantity standardization- (EOQ)
7. Safety measures standardization- (Rules and regulations)
8. Personnel standardization- (worker selection, training, wage rates and operating times)
9. Administrative standardization- (office methods and procedures for efficient working).
24
3S (Simplification, Standardization, Specialization)
 What is Interchangeability?
 The various components are manufactured in one or more batches by different persons on
different machines at different locations and are then assembled at one place.
 It is essential that the parts are manufactured in bulk to the desired accuracy and, at the
same time, adhere to the limits of accuracy specified. Manufacture of components under
such conditions is called interchangeable manufacture.
 Draw backs of Interchangeability
 Not economical to manufacture parts to a high degree of accuracy.
25
3S (Simplification, Standardization, Specialization)
 Interchangeability
 For example, assembly of a shaft and a part with a hole.
 The two mating parts are produced in bulk, say 1000 each. By interchangeable assembly any
shaft chosen randomly should assemble with any part with a hole selected at random,
providing the desired fit.
<hole>
<Shaft>
26
3S (Simplification, Standardization, Specialization)
 Specialization – In product specialization, a firm may produce and market only one or a limited
range of similar products. This leads to process and labor specialization, which increases
productivity and decreases costs.
 Specialization as applied to human activities on shop floor can be defined as ‘Division of Labour’.
This means that a worker performs one operation instead of completing the product and attains
proficiency in that and becomes a specialist in that.
 Specialization is applied to Products, Processes, Individuals, Companies, Jobs and Equipments.
 Specialization - In labour specialization,
 Specialization – In process specialization,
27
Product Design and Process Selection
 Product Design – the process of defining product characteristics
 Product design must support product manufacturability (the ease with which a product can
be made)
 Product
design
defines
a
product’s
characteristics
including
appearance,
materials,
dimensions, tolerances, performances, etc.
 Process selection - the development of the process necessary to produce the designed product.
28
Product Design and Process Selection
 Product Design Process

Step 1 - Idea Development - Someone thinks of a need and a product/service design to satisfy it:
customers, marketing, engineering, competitors, benchmarking, reverse engineering

Step 2 - Product Screening - Every business needs a formal/structured evaluation process: fit with
facility and labor skills, size of market, contribution margin, break-even analysis, return on sales

Step 3 – Preliminary Design and Testing - Technical specifications are developed, prototypes built,
testing starts

Step 4 – Final Design - Final design based on test results, facility, equipment, material, & labor skills
defined, suppliers identified
29
Product Design and Process Selection
 Product Design Process
30
Process planning, selection and analysis
 Product Design – it is the plan for the product and its components and subassemblies.
 Manufacturing Plan – is needed to convert the product design into a physical entity.
 Process Planning – the activity of developing such a manufacturing plan. It is the bridge
between product design and manufacturing.
 According to the American Society of Tool and Manufacturing Engineers, “Process planning is
the systematic determination of the methods by which a product is to be manufactured,
economically and competitively.”
Ref: Automation, Production Systems, and Computer-Integrated Manufacturing, Fouth Edition, by Mikell P. Groover
31
Purpose of Process planning
 It is to determine and describe the best process for each job so that,
 Specific requirements are established for which machines, tools and others equipment can be
designed or selected.
 The efforts of all engaged in manufacturing the product are coordinated.
 A guide is furnished to show the best way to use the existing or the providing facilties.
32
Overall Development of processing plan
33
Objective of Process planning
 The systematic determination of the engineering processes and systems to manufacture a
product competitively and economically is called operations planning. It is the stage between
design and production. The plan of manufacture considers functional requirements of the
product, quantity, tools and equipment, and eventually the costs for manufacture
34
Scope of Process planning
 Operation planning is a responsibility of the manufacturing organization. A
number of functional staff arrangements are possible. This process leads to the
same output despite organizational differences. The following are business
objectives for operations planning:
 New product manufacture
 Sales
 Quantity
 Effective use of facilities
 Cost reduction
35
Information Required for Process planning
 Quantity of work to be done along with product specifications.
 Quality of work to be completed.
 Availability of equipments, tools and personnels.
 Sequence in which operations will be performed on the raw material.
 Names of equipment on which the operations will be performed.
 Standard time for each operation.
 When the operations will be performed?
36
Process Planning (or) Operations Planning Activities
 Interpretation of design drawings  materials, dimensions, tolerances, surface finish, etc.
 Choice of process and sequence
 Choice of equipment  first prefer to use existing then go for purchasing new
 Choice of tools, dies, molds, fixtures, and gages
 Analysis of methods  layout, material handling, even workers poster or body movement
 Setting of work standards  time calculations for each operation
 Choice of cutting tools and cutting conditions  machining operations, often with reference to
standard handbook recommendations.
Ref: Automation, Production Systems, and Computer-Integrated Manufacturing, Fouth Edition, by Mikell P. Groover
37
Process Planning (or) Operations Planning Activities
Ref: Process planning and cost estimation, by Dr. V. Jayakumar
38
Process Planning Documentation / Operation Sheet / Route Sheet
 The route sheet lists the production operations and associated machine tools for each
component and subassembly of the product.
Ref: Process planning and cost estimation, by Dr. V. Jayakumar
39
Process Planning Documentation / Operation Sheet / Route Sheet
Ref: Process planning and cost estimation, by Dr. V. Jayakumar
40
Inputs and Outputs of Process Planning
Ref: Process planning and cost estimation, by Dr. V. Jayakumar
41
Process Planning vs Production Planning
 Process Planning  It is concerned with technical details such as the engineering and
technological issues of how to make the product, what types of equipment and tooling are
required to manufacture the product, etc.
 Production Planning  It is concerned with the logistics issues of making the product such as
ordering the materials and obtaining the resources required to make the produce.
Ref: Process planning and cost estimation, by Dr. V. Jayakumar
42
Process Planning Methods
 Methods
 Manual Process Planning
 Traditional approach
 Workbook approach
 Computer Aided Process Planning
 Retrieval (or variant) CAPP system
 Generative CAPP system
Ref: Process planning and cost estimation, by Dr. V. Jayakumar
43
Process Planning Methods
 Manual Process Planning
 Traditional approach
 Process
plan
prepared
manually
with
process
planner
knowledge/experience
 Workbook approach
 This is modified version of traditional approach. Once the drawing
interpretation is carried out, the suitable predetermined sequence of
operations are selected from the existing workbook.
Ref: Process planning and cost estimation, by Dr. V. Jayakumar
44
Process Planning Methods
 Manual Process Planning
 Advantages
 Very much suitable for small scale companies
 Highly flexible
 Low investment cost
 Disadvantages
 Requires skilled process planner
 Possibilities of error
 Increase paper work
 Inconsistent process plan result in reduced productivity
 It is not very responsive to chaning manufacturing environment, new tooling,
new processes, etc.
Ref: Process planning and cost estimation, by Dr. V. Jayakumar
45
Process Planning Methods
 Computer Aided Process Planning
 Benefits of CAPP
 Process rationalization and standardization
 Productivity improvement
 Product cost reduction
 Elimination of human error
 Reduction in time
 Reduced clerical effort and paper work
 Improve legibility
 Faster response to engineering changes
 Incorporation of other application programs
Ref: Process planning and cost estimation, by Dr. V. Jayakumar
46
Retrieval (Or Variant) CAPP
 Retrieval (Or Variant) CAPP
 Similar parts will have similar process plans
 A process plan for a new part is created by recalling, identifying and
retrieving an existing plan for a similar part, and making the necessary
modifications for the new part.
Ref: Process planning and cost estimation, by Dr. V. Jayakumar
47
Retrieval (Or Variant) CAPP
 Retrieval (Or Variant) CAPP
 Benefits: Once a standard plan has been
written, a variety of parts can be planned.
 Comparatively
simple
programming
and
installation with generative type.
 Planner has control of the final plan.
 Drawbacks: the components to be planned
are limited to similar components.
 Experienced planners required.
Commercial retrieval CAPP system:
MultiCapp and MIPLAN
Ref: Automation, Production Systems, and Computer-Integrated Manufacturing, Fouth Edition, by Mikell P. Groover
48
Generative CAPP
Generative CAPP
 Instead of retrieving and editing an existing plan contained in a computer
database, a generative system creates the process plan based on logical
procedures similar to those used by a human planner.
 Logical procedures: decision logic, formulae, algorithms and geometric
analysis.
 Part of the field of expert system, a branch of artificial intelligence.
Ref: Automation, Production Systems, and Computer-Integrated Manufacturing, Fouth Edition, by Mikell P. Groover
49
Generative CAPP
 Generative CAPP (continued…)
 An expert system is a computer program that is capable of solving complex
problems that normally can only be solved by a human with years of education and
experience.
Ref: Automation, Production Systems, and Computer-Integrated Manufacturing, Fouth Edition, by Mikell P. Groover
50
Generative CAPP
 Generative CAPP (continued…)
 First, the technical knowledge of manufacturing and the logic used by successful
process planners must be captured and coded into a computer program.
 In an expert system applied to process planning, the knowledge and logic of the
human process planners is incorporated into a so-called knowledge base.
 The generative CAPP system then uses that knowledge base to solve process
planning problems (i.e., create route sheets).
Ref: Automation, Production Systems, and Computer-Integrated Manufacturing, Fouth Edition, by Mikell P. Groover
51
Generative CAPP
 Generative CAPP (continued…)
 Second, a computer-compatible description of the part to be produced. This
description contains all of the pertinent data and information needed to
plan the process sequence.
 Two possible ways to provide description:
 The geometric model of the part that is developed on a CAD system
during product design
 A GT code number of the part that defines the part features in significant
detail.
Ref: Automation, Production Systems, and Computer-Integrated Manufacturing, Fouth Edition, by Mikell P. Groover
52
Generative CAPP
 Generative CAPP (continued…)
 Third, the capability to apply the process knowledge and planning logic contained
in the knowledge base to a given part description.
 i.e., the CAPP system uses its knowledge base to solve a specific problem –
planning the process for a new part. This problem solving procedure is referred to
as the inference engine in the terminology of expert system.
 By using its knowledge base and inference engine, the CAPP system synthesizes a
new process plan from scratch for each new part.
Ref: Automation, Production Systems, and Computer-Integrated Manufacturing, Fouth Edition, by Mikell P. Groover
53
Generative CAPP
 Generative CAPP (continued…)
 Advantages: It can generate consistent process plans rapidly
 New components can be planned as easily as existing components.
 Also has potential for integrating with an automated manufacturing facility to
provide detailed control information.
 Disadvantages: It is complex and very difficult to develop.
 Commercial Generative CAPP APPAS, CMPP, EXCAP and XPLAN
Ref: Automation, Production Systems, and Computer-Integrated Manufacturing, Fouth Edition, by Mikell P. Groover
54
Selection of Process Planning System
 Factors to be considered while selecting process planning system
 The general environment in which process planning is conducted
 The organizational structure of the company
 The technical expertise available to the company
 The needs and objectives of the company regarding the generation of
manufacturing information and process plans.
Ref: Automation, Production Systems, and Computer-Integrated Manufacturing, Fouth Edition, by Mikell P. Groover
55
Retrieval vs Generative
CAPPMethod
Variant
Generative
CodingSystem
Simple
Detailed
PartGeometry
Complicated
Simple
Similaritybetweenfamily
components
High
Low
Familysize
Large
Small
DataBase
Large
Small
Investment&Effort
High
Low
TimetoDevelop
Short
Long
Humanintervention
High
Small
Ref: Automation, Production Systems, and Computer-Integrated Manufacturing, Fouth Edition, by Mikell P. Groover
56
Process Analysis
 Process analysis also known as method study helps in finding better methods
of doing a job and this is achieved by eliminating unproductive and
unnecessary elements of the process.
 The process is analysed with the help of process charts and flow diagrams.
Objectives:
 To improve work methods and procedures.
 To determine the best sequence of doing work.
 To eliminate the waste and unproductive operations.
 To improve plant utilisation and material utilisation.
 To improve the working conditions and hence to improve labour efficiency.
 To eliminate unnecessary fatigue and therby effect economy in human effort.
57
PROCESS ANALYSIS
 Steps involved:
 Selection-Process or job keeping in view human, technical and economical factors.
 Recording- Record all facts regarding present and proposed work methods using
appropriate recording techniques(process charts and diagrams)
 Examining-Analyse the recorded facts to expose defects in the existing methods. The
purpose, place and sequence of operations should be critically examined.
 Developing new method- Evaluation, investigation and selection
 Defining the new method
 Installing the new method - Implementation of proposed method
 Maintaining the new method
58
PROCESS CHARTS
A process chart is a pictorial representation of the activities that occur in the work, method or
procedure, in which suitable symbols are used to represent various activities
Chart/Diagram
Application
Operation Process chart
(Outline process chart)
Gives bird’s eye view of process and records
Principal operations and inspecting
Flow process chart
(a) Man type
(b) Material type
(c) Equipment type
Sequence of activities performed by the worker
Sequence of activities performed on materials
Sequence of activities performed by equipment
Multiple activity chart
Chart activities of men and /or machines on a common
time scale
Two handed process chart
Activities performed by worker’s two hands
Travel chart
Movement of materials and / or men between departments
Flow and string diagrams
Path of movement of men and materials
59
Group Technology
 Group
technology is a manufacturing technique and philosophy to increase
production efficiency by exploiting the “underlying sameness” of component
shape, dimensions, process route, etc.
 Similar parts are arranged into part families. where each part family possesses
similar design and/or manufacturing characteristics.
Ref: Automation, Production Systems, and Computer-Integrated Manufacturing, Fouth Edition, by Mikell P. Groover
60
Group Technology Benefits
 Standardization of tooling, fixturing and setups
 Material handling is reduced because parts are moved within a machine cell
rather that within the entire factory.
 Process planning and production scheduling are simplified.
 Setup times are reduced, resulting in lower manufacturing lead times.
 Work-in-process is reduced
 Workers satisfaction usually improves when workers collaborate in a GT cell
 Higher quality work is accomplished using GT
Ref: Automation, Production Systems, and Computer-Integrated Manufacturing, Fouth Edition, by Mikell P. Groover
61
Traditional process layout
Ref: Automation, Production Systems, and Computer-Integrated Manufacturing, Fouth Edition, by Mikell P. Groover
62
Manufacturing with GT
Ref: Automation, Production Systems, and Computer-Integrated Manufacturing, Fouth Edition, by Mikell P. Groover
63
Part Family
 A collection of parts that possess similarities in geometric shape and size, or in
the processing steps used in their manufacture
 Two categories of part similarities can be distinguished
 design attributes, which are concerned with part characteristics such as
geometry, size, and material
 manufacturing attributes, which consider the sequence of processing
steps required to make a part.
 Both design and manufacturing attributes
Ref: Automation, Production Systems, and Computer-Integrated Manufacturing, Fouth Edition, by Mikell P. Groover
64
Ways to Identify Part Families
 Visual Inspection
 Using best judgment to group parts into appropriate families, based on the
parts or photos of the parts
 Parts classification and coding
 Identifying similarities and differences among parts and relating them by
means of a coding scheme
 Production flow analysis
 Using information contained on route sheets to classify parts
Ref: Automation, Production Systems, and Computer-Integrated Manufacturing, Fouth Edition, by Mikell P. Groover
65
Parts Classification and Coding
 Identification of similarities among parts and relating the similarities by means
of a numerical coding system
 Most time consuming of the three methods
 Must be customized for a given company or industry
 Reasons for using a coding scheme
 Design retrieval: access to a part that already exists
 Automated process planning: process plans for similar code parts
Ref: Automation, Production Systems, and Computer-Integrated Manufacturing, Fouth Edition, by Mikell P. Groover
66
Parts Classification and Coding
 The parts classification and coding systems are based on one of the following
 Part design attributes – major dimensions, basic internal and external shape,
length/diameter ratio, material type, tolerance, surface finish
 Part manufacturing attributes – major process, operation sequence, batch
size, annual production, machine tools, cutting tools, material type
 Both design and manufacturing attributes
Ref: Automation, Production Systems, and Computer-Integrated Manufacturing, Fouth Edition, by Mikell P. Groover
67
Coding Scheme Structures - Monocode
 Hierarchical structure or Monocode
 Interpretation of each successive digit depends on the value of the
preceding digit
Ref: Automation, Production Systems, and Computer-Integrated Manufacturing, Fouth Edition, by Mikell P. Groover
68
Coding Scheme Structures - Polycode
 Polycode or Attribute code or Chain-type structure
 Interpretation of each symbol is always the same
 No dependence on previous digits
Ref: Automation, Production Systems, and Computer-Integrated Manufacturing, Fouth Edition, by Mikell P. Groover
69
Make or Buy Decision
 Inevitably, the question arises whether a given part should be purch
ased from an outside vendor or made internally
Ref: Automation, Production Systems, and Computer-Integrated Manufacturing, Fouth Edition, by Mikell P. Groover
70
Make or Buy Decision
Ref: Automation, Production Systems, and Computer-Integrated Manufacturing, Fouth Edition, by Mikell P. Groover
71
Make or Buy Decision
Ref: Automation, Production Systems, and Computer-Integrated Manufacturing, Fouth Edition, by Mikell P. Groover
72
Break-Even Analysis
1.
Break-even analysis establishes the relationship among the factors affecting profit.
2.
It indicates at what level cost and revenue are in equilibrium.
3.
Presents to management the effect of changes in volume on profit.
4.
An economic analysis can be done by Break-even analysis.
Assumptions in Break-even Analysis: 1.
Selling prices will remain constant at all sales level.
2.
There is a linear relationship between sales volume and costs.
3.
Costs are divided into Fixed cost and Variable Cost.
4.
Production and sales quantities are equal (There is no inventory).
5.
No other factors will influence the cost expect the quantity.
73
Break-Even Point
 Break-even chart is a graphical representation of inter-relationship between
quantity produced, cost of producing and sales return.
 The total cost of production (fixed cost + variable cost) and total sales return
are plotted against quantity produced.
 The intersection of the total cost and total sales return lines gives the breakeven point.
74
Break-Even Analysis
1.
Break-even point refers to the level of sales (sales volume) at which the sales i
ncome (revenue) equal the total costs.
2.
It is a point at which the profit is zero.
3.
The quantities produced (sold) above the point results in profit and below the
point results in losses.
4.
The break-even point is reached when the fixed cost are completely recovered
75
Break-Even Chart
Break-even point
Cost
Fixed cost
Loss
Quantity
Break-Even Point
Let F = Fixed Cost.
Q = Quantity produced and Sold.
a = variable cost per unit.
b = sales price per unit.
Total Cost = Fixed cost + Variable cost.
= F + (variable cost per unit x Quantity produced and sold)
T.C = F + a.Q
Total income (T.I) = sales price per unit x Quantity produced and sold.
T.I = b.Q
Contribution = Sales – Variable cost = b – a.
77
Break-Even Point
At Break-even point Total cost equals Total income.
Therefore, Total Cost = Total Income.
F + a.Q = b.Q
Q = F / (b-a).
Break-even Quantity (Q) = Fixed Cost / Contribution.
MARGIN OF SAFETY
 Margin of safety is the difference between the existing level of output and the level of output
at B.E.P
Margin of safety = (Sales at BEP / Sales) x 100
78
Break-Even Point
ANGLE OF INCIDENCE
 The angle at which the sales line cuts the total cost line. The man
agement aims at large angle of incidence because large angle of i
ncidence indicates a high profit rate.
METHODS OF LOWERING BEP

Reduced Fixed Cost.

Reduce the Variable Cost.

Increase the slope of income line.
79
Break-Even Point
APPLICATIONS OF BEP: •
Safety Margin – refers to the extent to which an organization can afford to loose its sales before i
t starts incurring losses.
•
It helps to plan the profit.
•
It helps to compute up to what level the sales may be reduced or increased in competition to main
tain a particular level of profit.
•
It helps to make decisions about selection of equipment and processes among the alternatives ava
ilable.
•
It helps to take decisions on Make or Buy decisions.
•
It helps to decide the product mix and production mix.
80
Unit 2 – Costing and Estimation
Session Description of Topic
UNIT II:COSTING AND ESTIMATION
Objectives of costing and estimation : Functions and procedure
1
2
3
4
5
6
Contact hrs C-D-I-O
9
IOs
Reference
2
C
2
1
Introduction to costs, Computing material cost
1
C,D
2
1
Direct labor cost, Analysis of overhead costs
1
C,D
2
1
2
C,D
2
1
1
C,D
2
1,2
2
C,D
2
1
Factory expenses, Administrative expenses, Selling and distributing
expenses
Cost ladder ,Cost of product
Depreciation, Analysis of depreciation, Problems in depreciation
method
81
Costing or Cost Accounting
 Costing is the determination of an actual cost of a component after adding different expenses
increased incurred in various departments.
 Definition: Costing or cost accounting may be defined as a systematic procedure for recording
accurately every item of expenditure incurred on the manufacture of a product by different
sections of any manufacturing concern.
82
Costing or Cost Accounting
 Wheldon has defined costing as, “Costing is the classifying and recording the
appropriate allocation of expenditure for the determination of the costs of
products or services, and for presentation of the costs of products or services,
and for presentation of suitably arranged data for the purpose of control, and
guidance of management.
83
Objective of Costing or Cost Accounting
 Costing determination
 Fixing the discount
 For fixing selling price
 Pricing policy
 Cost control
 Budget preparation
 Comparison with estimate
 For preparing quotations / tenders
 Make or buy decisions
 Output targets
 Wastage reduction
 Legal provisions
 To suggest changes in design
 Purchasing new machines / plants
 Profit and loss
84
Method of Costing or Cost Accounting
 Job costing or order costing (eg: ship, airplane)
 Batch costing (eg. Group of product with similarities)
 Process costing (eg. Cement industry)
 Departmental costing (eg. automobile industry)
 Operating cost method or service cost (eg. Transport, water board, electricity)
 Unit cost method (eg. mines and quarries…..supply a uniform product)
 Multiple cost method (eg. forming industry….supply variety of products…)
85
Costing Estimating
 A cost estimate is an attempt to forecast the expenses that must be incurred to manufacture a
product.
 Definition: Cost estimating may be defined as the process of determining the probable cost of
the product before the start of its manufacture.
86
Knowledge required for Cost Estimating
 Cost estimating requires the knowledge of the following factors for calculating the probable cost
of the product:
 Design time
 Amount and cost of materials required
 Production time required
 Labour charges
 Cost of machinery, overheads and other expenses
 Use of previous estimates of similar parts
 Effect of volume of production on costing rates
 Effect of changes in facilities on costing rates
 Probable future changes in unit prices for materials, labour and expenses when the proposed
product is manufactured at a future date.
87
Importance of Cost Estimating
 The accurate estimation can enable the factory owner to make vital decisions such as
manufacturing and selling policies.
 What if job is Over-estimated??? The firm will not be able to compete with its competitors and
loses the order.
 What if job is Under-estimated??? The firm will face huge financial loss.
88
Importance of Cost Estimating
 Example
89
Objective of Cost Estimating
 To establish the selling price of a product for a quotation or contract, so as to ensure reasonable
profit to the company.
 To verify quotations submitted by vendors.
 To ascertain whether the proposed product can be manufactured and marketed profitably.
 To take make or buy decisions, i.e., determine whether the part or assembly can be make or buy
 To determine the most economical method process, or material for manufacturing a product.
 To establish the standard of performance that may be used to control costs.
 To prepare production budget.
 To evaluate alternate designs of product.
 To initiate means of cost reduction in existing production facilities
90
Functions of Cost Estimating
 Cost estimates are required to submit accurate tenders for getting the contracts.
 Cost estimates are required for the manufacturer to choose from various methods of production
the one which is likely to be most economical.
 Cost estimates are required for fixing the selling price of a product.
 Cost estimate gives detailed information of all the operations and their costs, thus setting a
standard to be achieved in actual practice.
 Cost estimate enables the management to plan for procurement of raw materials, tools, etc., and
to arrange the necessary capital, as it gives detailed requirement.
91
Types of Cost Estimates
 Types
 Preliminary cost estimates  this is based on incomplete data. These estimates are based on
assumptions and general information supplied by either the sales or engineering groups,
especially in areas of incomplete data.
 Final cost estimates  this is based on complete data for a product and hence it is the most
accurate estimate.
 Classification of cost estimates based upon design level
 Conceptual design phase (cost estimate accuracy ±30%)
 Preliminary design phase (accuracy ±20%)
 Detailed design phase (accuracy ±10%)
92
Method of Estimating
 Methods of estimating are:
 Conference Method (degree of accuracy depends on availability of data)
 Comparison Method (based on past experience and existing data)
These 2 methods
used when time
is main factor
 Detailed Analysis Method (estimate most reliable)
The selection of method to be determined by two factors: the information required and time
available.
93
Data Requirements for Cost Estimating
 General design specifications
 Total anticipated quantity and the rate of production
 Assembly or layout drawings
 List of the proposed subassemblies of the product
 Detail drawings and a bill of material of the product
 Test and inspection procedures and equipment
 Machine tool and equipment requirements
 Packaging and/or transportation requirements
 Manufacturing routings
 Detailed tool, machine tool, and equipment requirements
 Operation analysis and workplace studies
 Standard time data
 Material release data
 Subcontractor cost and delivery data
 Area and building requirements
94
Note: the greater the
number of items and
the completeness of
each item available to
the
estimator,
the
more accurate the
estimate.
Elements of Cost Estimation / Components of a Job Estimate / Constituents of Estimation
 The total estimated cost of a product consists of the following cost components.
 Design Cost
 Drafting Cost
 R&D Cost
 Materials Cost
 Labour Cost
 Inspection Cost
 Cost of tools, jigs, and fixture
 Overhead Cost
95
Elements of Cost Estimation / Components of a Job Estimate / Constituents of Estimation
 The total estimated cost of a product consists of the following cost components.
 Design Cost
𝐸𝑠𝑡𝑖𝑚𝑎𝑡𝑒𝑑 𝐷𝑒𝑠𝑖𝑔𝑛 𝐶𝑜𝑠𝑡 = 𝐸𝑠𝑡𝑖𝑚𝑎𝑡𝑒𝑑 𝑑𝑒𝑠𝑖𝑔𝑛 𝑡𝑖𝑚𝑒 ×
𝑆𝑎𝑙𝑎𝑟𝑦 𝑜𝑓 𝑑𝑒𝑠𝑖𝑔𝑛𝑒𝑟 𝑝𝑒𝑟 𝑢𝑛𝑖𝑡 𝑡𝑖𝑚𝑒
Where, the design time can be estimated on the basis of past data or experience. If we outsource
the design, then the money paid to be considered.
 Drafting Cost
𝐷𝑟𝑎𝑓𝑡𝑖𝑛𝑔 𝐶𝑜𝑠𝑡 =
𝐸𝑠𝑡𝑖𝑚𝑎𝑡𝑒𝑑 𝑡𝑖𝑚𝑒 𝑡𝑜 𝑏𝑒 𝑐𝑜𝑛𝑠𝑢𝑚𝑒𝑑 𝑏𝑦 𝑑𝑟𝑎𝑓𝑡𝑚𝑎𝑛 𝑖𝑛 𝑝𝑟𝑒𝑝𝑎𝑟𝑖𝑛𝑔 𝑑𝑟𝑎𝑤𝑖𝑛𝑔𝑠 ×
96
𝑆𝑎𝑙𝑎𝑟𝑦 𝑜𝑓 𝑑𝑟𝑎𝑓𝑡𝑠𝑚𝑎𝑛 𝑝𝑒𝑟 𝑢𝑛𝑖𝑡 𝑡𝑖𝑚𝑒
Elements of Cost Estimation / Components of a Job Estimate / Constituents of Estimation
 The total estimated cost of a product consists of the following cost components.
 Material Cost
 First list all materials required to manufacture the product.
 Estimate the weight of all the materials including wastage, allowance,
spoilage, and scrap
𝐸𝑠𝑡𝑖𝑚𝑎𝑡𝑒𝑑 𝑀𝑎𝑡𝑒𝑟𝑖𝑎𝑙𝑠 𝐶𝑜𝑠𝑡 = 𝐸𝑠𝑡𝑖𝑚𝑎𝑡𝑒𝑑 𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑒𝑎𝑐ℎ 𝑝𝑎𝑟𝑡 × 𝐸𝑠𝑡𝑖𝑚𝑎𝑡𝑒𝑑 𝑓𝑢𝑡𝑢𝑟𝑒 𝑝𝑟𝑖𝑐𝑒
 Labour Cost
𝐿𝑎𝑏𝑜𝑢𝑟 𝐶𝑜𝑠𝑡 = 𝐸𝑠𝑡𝑖𝑚𝑎𝑡𝑒𝑑 𝑙𝑎𝑏𝑜𝑢𝑟 𝑡𝑖𝑚𝑒 𝑛𝑒𝑒𝑑𝑒𝑑 𝑡𝑜 𝑝𝑟𝑜𝑑𝑢𝑐𝑒 𝑡ℎ𝑒 𝑝𝑟𝑜𝑑𝑢𝑐𝑡 × 𝐶𝑜𝑠𝑡 𝑜𝑓 𝑙𝑎𝑏𝑜𝑢𝑟 𝑝𝑒𝑟 ℎ𝑜𝑢𝑟
97
Elements of Cost Estimation / Components of a Job Estimate / Constituents of Estimation
 The total estimated cost of a product consists of the following cost components.
 Cost of tools, jigs, and fixtures
 Cost of a product includes the estimated cost and maintenance charges
for the tools, jigs, fixtures and dies required in the production.
 The cost of tools, jigs, fixtures, etc., are estimated considering their present
prices, market trends and the number of times a particular tool can be
used during its lift time.
𝑇𝑜𝑜𝑙 𝑐𝑜𝑠𝑡 𝑝𝑒𝑟 𝑢𝑛𝑖𝑡 𝑝𝑟𝑜𝑑𝑢𝑐𝑒𝑑 =
98
𝐸𝑠𝑡𝑖𝑚𝑎𝑡𝑒𝑑 𝑐𝑜𝑠𝑡
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑗𝑜𝑏𝑠
Elements of Cost Estimation / Components of a Job Estimate / Constituents of Estimation
 The total estimated cost of a product consists of the following cost components.
 Cost of R&D
 Considerable time and money has to be spent on R & D work.
 The research may be theoretical, experimental, or developmental research.
 Inspection Cost
 Should consider the cost of inspection equipment's, gauges and
consumables, and wages to the inspectors
 Overhead Cost
 Administrative expenses, selling and distribution expenses are added to
the overhead cost. i.e., expenses other than the director material cost,
director labour cost, direct expenses.
99
Difference between Cost Estimation and Cost Accounting
S. No.
Particular
1.
Nature of
cost
2.
3.
4.
5.
Estimating
Costing
It gives the probable cost of the product before the st It gives the actual cost of the product after
art of its manufacture.
adding different expenses incurred in vario
us departments.
Quality of Estimation requires a highly technical knowledge he Costing requires the knowledge of account
personnel nce an estimator is basically an engineer.
s and therefore costing is done by account
required
ants.
Duration of Estimating is carried out before the actual production Costing usually starts with the issue of ord
process
of a product.
er for production of a product and ends aft
er the product is dispatched on sale.
Main
i) To establish the selling price of a product for a quo i) To determine the actual cost of the final
objectives tation or contract.
product.
ii) To take make or buy decisions of parts.
ii) To form a basis for fixing the selling pri
iii) To ascertain whether the proposed product can be ce.
manufactured and marketed profitably.
iii) To check the accuracy of estimates.
iv) To help in detecting the undesirable wa
stages and expenses.
Organising Estimating work is done under the planning departm Costing work is done under the accounting
100
department ent.
department.
Cost Estimation Procedure
 Step 1: Study the cost estimation request thoroughly and understand it completely.
 Step 2: Analyse the product and decide the requirements and specifications of the
product.
 Step 3: Prepare the list of all the parts of the product and their bill of materials
 Step 4: Take make or buy decisions and prepare separate lists of parts to be
manufactured within the plant and parts to be purchased outside the plant.
 Step 5: Estimate the materials cost for the parts to be manufactured in the plant.
𝑴𝒂𝒕𝒆𝒓𝒊𝒂𝒍 𝒄𝒐𝒔𝒕 = 𝑾𝒆𝒊𝒈𝒉𝒕 𝒐𝒇 𝒕𝒉𝒆 𝒎𝒂𝒕𝒆𝒓𝒊𝒂𝒍 × 𝑴𝒂𝒕𝒆𝒓𝒊𝒂𝒍 𝒄𝒐𝒔𝒕 𝒑𝒆𝒓 𝒖𝒏𝒊𝒕 𝒘𝒆𝒊𝒈𝒉𝒕
101
Cost Estimation Procedure
 Step 6: Determine the cost of parts to be purchased from outside
 Step 7: Make a manufacturing process plan for the parts to be manufactured in the plant
 Step 8: Estimate the machining time for each operations listed in manufacturing
process plan
 Step 9: Determine the direct labour cost
𝑫𝒊𝒓𝒆𝒄𝒕 𝑳𝒂𝒃𝒐𝒖𝒓 𝒄𝒐𝒔𝒕 = 𝑻𝒐𝒕𝒂𝒍 𝒐𝒑𝒆𝒓𝒂𝒕𝒊𝒐𝒏 𝒕𝒊𝒎𝒆 × 𝑳𝒂𝒃𝒐𝒖𝒓 𝒘𝒂𝒈𝒆 𝒓𝒂𝒕𝒆
102
Cost Estimation Procedure
 Step 10: Determine the prime cost by adding direct expenses, direct material cost, and
director labour cost.
𝑷𝒓𝒊𝒎𝒆 𝒄𝒐𝒔𝒕 = 𝑫𝒊𝒓𝒆𝒄𝒕 𝒍𝒂𝒃𝒐𝒖𝒓 𝑪𝒐𝒔𝒕 + 𝑫𝒊𝒓𝒆𝒄𝒕 𝒎𝒂𝒕𝒆𝒊𝒓𝒂𝒍 𝒄𝒐𝒔𝒕 + 𝑫𝒊𝒓𝒆𝒄𝒕 𝒆𝒙𝒑𝒆𝒏𝒔𝒆𝒔
 Step 11: Estimate the factory overheads, which include all indirect expenditure
incurred during production such as indirect material cost, indirect labour cost,
depreciation and expenditure on maintenance of the plant, machinery, power, etc.
 Step 12: Estimate the administrative expenses
 Step 13: Estimate the selling and distribution expenses, which include packing and
delivery charges, advertisement charges, etc.
103
Cost Estimation Procedure
 Step 14: Now calculate the total cost of the product.
𝑻𝒐𝒕𝒂𝒍 𝒄𝒐𝒔𝒕
= 𝑷𝒓𝒊𝒎𝒆 𝑪𝒐𝒔𝒕 + 𝑭𝒂𝒄𝒕𝒐𝒓𝒚 𝑶𝒗𝒆𝒓𝒉𝒆𝒂𝒅𝒔 + 𝑨𝒅𝒎𝒊𝒏𝒊𝒔𝒕𝒓𝒂𝒕𝒊𝒗𝒆 𝒆𝒙𝒑𝒆𝒏𝒔𝒆𝒔 + 𝑺𝒆𝒍𝒍𝒊𝒏𝒈 𝒂𝒏𝒅 𝒅𝒊𝒔𝒕𝒓𝒊𝒃𝒖𝒕𝒊𝒐𝒏 𝒆𝒙𝒑𝒆𝒏𝒔𝒆𝒔
 Step 15: Decide the profit and add the profit to the total cost to fix the selling price of
the part.
𝑺𝒆𝒍𝒍𝒊𝒏𝒈 𝑷𝒓𝒊𝒄𝒆 = 𝑻𝒐𝒕𝒂𝒍 𝒄𝒐𝒔𝒕 + 𝑷𝒓𝒐𝒇𝒊𝒕
 Step 16: Finally estimate the time of delivery in consultation with the production and
sales department.
104
Cost Estimation Procedure
105
Cost Estimation
 Labour Cost  It is the cost spent to the workers who are directly or indirectly
involved in manufacturing operations.
 Direct labour cost  Direct labourer is one who actually works and processes
the materials to convert it into the final shape. The cost associated with direct
labour is called direct labour cost.
 Indirect labour cost  Indirect labourer is one who is not directly employed
in the manufacturing of the product but his services are used in some indirect
manner. (ex: supervisor, inspectors, foreman, storekeeper, gatekeeper, crane driver, etc)
106
Determination of Direct Labour Cost
 In order to calculate the labour cost, an estimator must have the knowledge of
(i) all the operations involved
(ii) the tools used for production
(iii) the machines used for production
107
Determination of Direct Labour Cost
 Factors while calculating the time required for a particular job.
(i) Set up time
(ii) Operation time (floor to floor time)
(a) Handling time (b) Machining time
(iii) Tear down time
(iv) Miscellaneous allowances (Standard time = Basic Time + Allowances)
a) Personal allowances (5 % for male and 7% for female out of total work time)
b) Fatigue allowances (5 % of total time)
c) Contingency allowances (less than 5% of total time)
d) Process allowances
e) Interference allowances
f) Special allowances
108
Determination of Direct Labour Cost
 Standard time = Basic Time + Allowances
 Normal time (or) Basic time = Observed time x Rating factor
 Rating factor = (Observed time/Normal time) x 100
Here, rating factor also known as performance rating, is a leveling factor to convert
observed timings into normal timings.
109
Material Cost
 Material cost consists of the cost of materials which are used in the manufacture
of product.
 Direct material cost
 Indirect material cost
110
Direct Material Cost
 It is the cost of those materials which are directly used for the manufacture of
the product and become a part of the finished product.
 This expenditure can be directly allocated and charged to the manufacture of a
specific product or job and includes the scrap and waste that has been cut away
from original bar or casting.
111
Direct Material Cost
112
Indirect Material Cost
 In addition to direct materials a number of other materials are necessary to help
in the conversion of direct materials into final shape. Though these materials are
consumed in the production, they don’t become part of the finished product and
their cost cannot be directly booked to the manufacture of a specific product.
 The indirect materials include oils, general tools, greases, sand papers, coolants,
cotton waste etc. The cost associated with indirect materials is called indirect
material cost.
113
Calculate Material Cost (Example)
 Calculate the material cost of the slide block shown in fig. The weight of material
is 7.2gm/cc and one kg of material cost Rs. 6.25.
Density (ρ) = 7.2 gm/cc
= 7.2 x 10-6 kg/mm3
Unit cost
114
= Rs. 6.25 per kg.
Calculate Material Cost (Example)
 Solution
Volume of “A” = 175x150x185 = 4856250 mm3
Volume of “B” = 300x25x185 = 1387500 mm3
Volume of “C” = Volume of “B” = 1387500 mm3
Volume of “D” = 300x25x25 = 187500 mm3
Volume of “E” = Volume of “D” = 187500 mm3
Volume of “F” = (π/4)d2 l = (π/4)x652x185 = 613886.84 mm3
115
Calculate Material Cost (Example)
 Total Volume = Vol. of A + Vol. of B + Vol. of C + Vol. of D + Vol. of E - Vol. of F
= 7392363.16 mm3
Weight of the block = Total volume x Density = 53.225 kg
Material cost = Weight x Cost per unit weight = Rs. 332.66
116
Expenses
 Apart from material and labour cost in each factory there are several other
expenditures such as cost of special layouts, designs, etc. hire of special tools
and equipments; depreciation charges of plants and factory building; building
rent; cost of transportation, salaries and commissions to salesman etc. All these
expenditures are known as overheads or expenses.
 Except direct material and direct labour cost, all other expenditures
 The expenses include indirect material cost and indirect cost and such other
expenses.
117
Direct Expenses
 Direct expenses also known as chargeable expenses include any expenditure
other than direct material or direct labour incurred on a specific cost unit.
 These are the expenses which can be charged directly to a particular job and are
done for that specific job only.
 For example, hire of special tools and equipment, cost of special jigs and fixtures
or some special patterns and its maintenance cost, costs of layouts, designs and
drawings or experimental work on a particular job etc.
118
Indirect Expenses
 These are known as overhead charges, burden or on cost.
 All the expenses over and above prime cost are indirect expenses.
 Overhead is the sum of indirect labour cost, indirect material cost and other
expenses including service which cannot be conveniently charged to specific cost
unit.
119
Cost Ladder
120
Material densities
121
Area of plane figures
122
Area of plane figures
123
Volume Calculation
124
Volume Calculation
125
Calculate Material Cost (Example)
Problem # An isometric view of a work piece is shown in figure. What will be the
weight of the mild steel material required to produce it. The density of material is
2.681 gm/cc. Find also the material cost if its rate is Rs. 100 per kg. All dimensions
are in mm.
126
Calculate Material Cost (Example)
Solution #
Density (ρ) = 2.681gm/cc
1. Weight of the material?
= 2671 kg/m3
2. Material Cost.
Unit cost
Volume of “A” =
𝜋
4
𝑑 2 𝑙= 13304.64 mm3
Volume of “B” =
𝜋
4
𝑑2 𝑙= 29486.99 mm3
Volume of “C” = 32 x 32 x15 = 15360 mm3
Volume of “D”
𝜋
=
4
𝑑 2 𝑙 = 15707.96 mm3
127
= Rs. 100 per kg.
Calculate Material Cost (Example)
Solution #
1. Weight of the material?
2. Material Cost.
Total Volume = Vol. of A + Vol. of B + Vol. of C + Vol. of D
= 73859.59 mm3 or 73859.59 x 10-9 m3
Weight of the material = Total volume x Density = 73859.59 x 10-9 m3 x 2671 kg/m3
Weight of the material
Material Cost
= 0.198 kg
= Weight x cost per unit = Rs. 19.80
128
Calculate Material Cost (Example)
Problem # Estimate the weight of material required for manufacturing 220 pieces
of shaft as shown in figure. The shafts are made of mild steel which weight 7.87
gm/cm3 and cost Rs. 100 per kg. Also calculate the material cost for 220 such
shafts.
129
Calculate Material Cost (Example)
Solution #
n = 220 pcs; ρ = 7.87 gm/cm3 = 7.87 x 10-6 kg/mm3
𝜋
4
Volume of “A” =
ℎ
3
𝑑 2 𝑙= 2309.07 mm3
Volume of “B” = (𝑎1 + 𝑎1 +
Volume of “C” =
𝜋
4
𝑎1 𝑎2 )= 10775.675 mm3
𝑑 2 𝑙= 12507.46 mm3
𝜋
4
Volume of “D” = 𝑑 2 𝑙 = 7793.11 mm3
𝜋
4
Volume of “E” = (𝐷 2 −𝑑 2 ) 𝑙 = 2261.95 mm3
130
Calculate Material Cost (Example)
Solution #
Total Volume = Vol. of A + Vol. of B + Vol. of C + Vol. of D + Vol. of E
= 35647.265 mm3
Weight of the shaft = 35647.265 mm3 x 7.87 x 10-6 kg/mm3 = 0.2805 kg
Then, weight of 220 shafts = 0.2805 x 220 = 61.72 kg
Cost of materials = 61.72 x 100 = Rs. 6172
131
Calculate Material Cost (Example)
Problem # Calculate the material cost of 20 gun metal bushes as per the diagram.
Assume the density of gun metal as 8.3 gm per cc and its cost is Rs. 350 per kg.
Consider 10 % material loss during process. All dimension are in mm.
132
Calculate Material Cost (Example)
Solution #
ρ = 8.3 gm/cm3
𝜋
4
Volume of “A” =
𝑑 2 𝑙= 32π cm3
𝜋
4
Volume of “B” = 𝑑 2 𝑙= 9π cm3
Volume of “C” =
𝜋
4
𝑑 2 𝑙= 16π cm3
𝜋
4
Volume of “D” = 𝑑 2 𝑙 = 22.5π cm3
Volume of bush = 2A + 2B + C – D = 237.3 cm3
133
Calculate Material Cost (Example)
Solution #
ρ = 8.3 gm/cm3
Weight of bush = = 237.3 cm3 x 8.3 gm/cm3 = 1.97 kg
Hence, weight of material required,
20 bushes = 1.97 x 20 x 1.1
(considering 10 % loss)
= 43.34 kg
Material Cost = 43.34 x 350 = Rs. 15169
134
Calculate Material Cost (Example)
Problem # Find the cost of material for the machine part shown. Density of
material may be taken as 8.2 gm/cu. cm. The cost of material is Rs. 30 per kg.
Assume 20 % wastage of material of the finished component and 6 holes for bolt.
135
Calculate Material Cost (Example)
Solution #
ρ = 8.2 gm/cm3 = 8.2 x 10-6 kg/mm3
Total volume = Vol. of A – [(Vol. of B + Vol. of C + Vol. of F + 6(Vol. of D + Vol. of E)]
= 400452.95 mm3
Wastage of material i.e., material cost in machining processes = 20 %
= 400452.95 x 0.02 = 80090.59 mm3
Gross volume of the machine part = 400452.95 + 80090.59 = 480543.54 mm3
Then, total weight = 3.94 kg
Hence, material cost = Rs. 118.20
136
Calculate Standard time (Example)
Problem # In a manual operation, observed time for a cycle of operation is 0.5
minute and the rating factor as observed by the time study engineer is 125 %. All
allowance put together is 15 % of normal time. Estimate the standard time.
Standard time = Basic Time or normal time + Allowances
Basic Time or normal time = Observed time x Rating factor = 0.5 x (125/100) = 0.625 min
Allowance = 15 % of normal time = 0.15 x 0.625 = 0.09375 min
Therefore, Standard time = 0.0975 + 0.625 = 0.718 min.
137
Calculate Standard time (Example)
Problem # In a manufacturing process, the observed time for one cycle of
operation is 0.75 minute. The rating factor is 110 %. The following are the various
allowances as percentage of normal time. Personal allowance = 3%, Relaxation
allowance = 10%; Delay allowance = 2%. Estimate the standard time.
138
Calculate Standard time (Example)
Solution #
Normal time = Observed time x Rating factor = 0.75 x (110/100) = 0.825 min
Total allowance = (Personal allowance + relaxation allowance + delay allowance) of
normal time
= (3% + 10% + 2%) x 0.825 min = 0.12375 min
Therefore, standard time = 0.825 + 0.12375 = 0.948 min
139
Example problem,
Problem # Calculate prime cost, factory cost, production cost, total cost and selling
price per item form the data given below for the year 2003-04.
140
Example problem,
Solution #
(i) Director material used = Stock of raw material on 1-04-2003 + raw material
purchased – stock of raw material on 31-03-2004
= 25000 + 40000 – 15000 = Rs. 50000.
(ii) Direct labour = Rs. 14000
(iii) Direct expenses = Rs. 1000
Prime cost = Direct material + Direct labour + Direct expenses
= 50000 + 14000 +1000 = Rs. 65000
141
Example problem,
Solution #
Factory cost = Prime cost + Factory expenses
= Rs. 65000 + Rs. 9750 = Rs. 74750
Production cost = Factory cost + Administrative expenses
= Rs. 74750 + 6,500 = Rs. 81250
Total cost = Production cost + Selling expenses
= 81250 + 3250 = Rs. 84500
Selling price = total cost + profit = 84500 + (10 % x 84500) = Rs. 92950.
142
Example problem,
Solution #
Factory cost/item = 74750/650 = Rs. 115
Production cost/item = 81250/650 = Rs. 125
Prime cost/item = 65000/650 = Rs. 100
Total cost/item = 84500/650 = Rs. 130
Selling price/item = 92950/650 = Rs. 143
143
Example problem,
Problem # Calculate the selling price per unit from the following data:
Direct material cost = Rs. 8,000
Direct labour cost = 60 percent of direct material cost
Direct expenses = 5 percent of direct labour cost
Factory expenses = 120 percent of direct labour cost
Administrative expenses = 80 percent direct labour cost
Sales and distribution expenses = 10 percent of direct labour cost
Profit = 8 percent of total cost
No. of pieces produced = 200
144
Example problem,
Solution # Calculate the selling price per unit from the following data:
Direct material cost = Rs. 8,000
Direct labour cost = 60 % x 8000 = Rs. 4800
Direct expenses = 5 % 4800 = Rs. 240
Prime cost = 8000 + 4800 + 240 = Rs. 13040
Factory expenses = 120 % x 4800 = Rs. 5760
Administrative expenses = 80 % x 4800 = Rs. 3840
Sales and distribution expenses = 10 % x 4800 = Rs. 480
Total cost = Prime cost + Factory expenses + Office expenses + Sales and distribution expenses
= 13040 + 5760 + 3840 +480 = Rs. 23120
145
Example problem,
Solution # Calculate the selling price per unit from the following data:
Profit = 8% x 23120 = Rs. 1849.60
Selling price = 23120 + 1849.60 = Rs. 24970
Selling price / unit = 24970 / 200 = Rs. 125.
146
Example problem,
Problem # A factory is producing 1000 high tensile fasteners per hour on a
machine. The material cost is Rs. 375, labour cost is Rs. 245 and direct expense is
Rs. 80. The factory oncost is 150 percent of the total labour cost and office oncost
is 30 percent of the factory cost. If the selling price of each fastener is Rs. 1.30,
calculate whether there is loss or gain and by what amount ?
147
Example problem,
Solution #
For 1000 fasteners,
Material cost = Rs. 375.00
Labour cost = Rs. 245.00
Direct expenses = Rs. 80.00
Factory on-cost = 150 % x 245 = Rs. 367.50
Factory cost = 375 + 245 + 80 + 367.50 = Rs. 1067.50
Office on-cost = 30 % x 1067.50 = Rs. 320.25
Total cost for 1000 fasteners = Rs. 1067.50 + 320.25 = Rs. 1387.75
Cost per fastener = 1387.75/1000 = Rs. 1.387
Selling price = Rs. 1.30
As selling price is lower than total cost per fastener, the management will suffer a loss.
Loss per fastener = (1.39 – 1.30) = Rs. 0.09
Loss per 1000 fastener = 0.09 × 1000 = Rs. 90
148
Depreciation
 The reduction in the value and efficiency of the plant, equipment or any fixed
asset because of wear and tear, due to passage of time, use and climatic
conditions is known as depreciation.
 Depreciation may be defined as a method for spreading the cost of a fixed asset
over the life, or expected years of use, of the asset.
149
Reasons for Depreciation
Wear and tear
Depreciation
Depreciation due to
physical conditions
Physical decay
Accident
Poor maintenance and
neglect
Depreciation due to
functional conditions
150
Inadequacy
Obsolescence
Method of Depreciation
 Various methods of calculating the depreciation funds are:
 Straight line method
 Diminishing balance method
 Sinking fund method
 Annuity method
 Sum of years digits method
 Insurance policy method
 Machine-hour method
 Production-unit method
 Revaluation method
 Retirement method
151
Straight line method (Depreciation)
 Straight line method  the amount of depreciation is distributed over the useful
life of the machine in equal periodic instalments.
𝐶 −𝑆
𝐷=
𝑅𝑢𝑝𝑒𝑒𝑠
𝑛
o Simple method,
however, this method
does not consider the
maintenance and
repair charges.
C = Initial cost of the machine in rupees,
S = Scrap (or salvage) value in rupees,
n = Estimated life of the machine in years, and
D = Depreciation amount per year.
152
Straight line method (Depreciation) – Example
Problem: A CNC machine was purchased for Rs. 1,25,000 on 15th June 1995, the erection and
installation cost was Rs. 10,000. The CNC machine is to be replaced by a new one on 14th June
2010. If the estimated scrap value is Rs. 25,000, what should be the rate of depreciation and
depreciation fund on June 14th 2002.
If after 9 years of running, some machine parts are replaced and the estimated replacements cost is
Rs. 4000. What will be the new rate of depreciation?
Solution:
Total cost (C) = Machine cost = Rs. 1,25,000 + Rs. 10,000 = Rs. 1,35,000
Life of CNC machine (n) = 15th June 1995 to 14th June 2010 = 15 years.
Scrap value = Rs. 25,000
Then, Depreciation D = Rs. 7333.33
153
𝐷=
𝐶 −𝑆
𝑅𝑢𝑝𝑒𝑒𝑠
𝑛
Straight line method (Depreciation) – Example
Life of CNC machine (n) = 15th June 1995 to 14th June 2002 = 7 years.
Therefore, Rate of depreciation = Rs. 7333.33 x 7 = Rs. 51,333.33
ii. Replacement cost = Rs. 4000
After 9th year, the book value or current value of the machine.
= Initial cost (C) – Depreciation cost for 9 years = 1,35,000 –(7333.33 x 9) = Rs. 69,000.3
= Rs. 69,000.3 + Replacement cost = Rs. 73,000.03 i.e., new value of the machine (C’)
Then D’ = (73,000.3 – 25,000)/6 = Rs. 8000
154
Diminishing balance method (Depreciation) – Example
𝑆
𝑌𝑒𝑎𝑟𝑙𝑦 𝑑𝑒𝑝𝑟𝑒𝑐𝑖𝑎𝑡𝑖𝑜𝑛 𝑓𝑎𝑐𝑡𝑜𝑟, 𝑝 = 1 −
𝐶
155
1 𝑛
Diminishing balance method (Depreciation) – Example
Problem: A certain machine was purchased for Rs. 25000 and it was presumed that it will last for 20 years. It
was also considered that by selling the scrap of the machine, the residual value will be Rs. 4000. If the
depreciation is charged by reducing balance method, find out the depreciation fund after and depreciation
amount for the 3rd year. Also find out the percentage by which value of the machine is reduced every year.
Given: C = Rs. 25000; S = Rs. 4000
Ans: Percentage by which value of the machine is reduced every year = 𝑝 = 1 −
𝑆 1 𝑛
𝐶
Value of machine after 1st year = C (1 – p) = 25000 (1-0.08755) = Rs. 22811.09
Then, depreciation fund after 1st year = 25000 – 22811.09 = Rs 2188.91
Value of machine after 2nd year = C (1- p) = 22811.009 (1-0.08755) = Rs. 20813.98
Then, depreciation fund after 2nd year = 22811.09-20813.98 = Rs. 1997.11
Value of machine after 3rd year = C (1 – p) = 20813.98 (1-0.08755) = Rs. 18990.82
Then, depreciation fund after 3rd year = 20813.98 – 18990.82 = Rs. 1823.16
Depreciation fund after 3rd year = 2188.91 + 1997.11 + 1823.16 = Rs. 6009.18
156
= 0.08755
Sinking fund method (Depreciation) – Example
𝐷=
𝐶−𝑆 𝑟
(1+ 𝑟)𝑛 −1
r = rate of interest on depreciated fund in percentage,
157
Sinking fund method (Depreciation) – Example
Problem: A power hacksaw machine was purchased for Rs. 25000. After 5 years the
machine was values at Rs. 10000. Find out the depreciation amount according to
the sinking fund method, the rate of interest being 5%.
Solution:
Given: C = 25000; n = 5 years; S = Rs. 10000; r = 5%
Then D= ?
𝐷=
𝐶−𝑆 𝑟
(1+ 𝑟)𝑛 −1
= Rs. 2714.62
158
Annuity method (Depreciation) – Example
𝐶 (1 + 𝑟)𝑛 −𝑆] [1 − (1 + 𝑟)
𝐷=
[1 − 1 + 𝑟 𝑛 ]
r = rate of interest on depreciated fund in percentage,
159
Annuity method (Depreciation) – Example
Problem: Find out the depreciation annuity by the annuity charging method after 4
years, when the cost of machine is Rs. 15000 and the scrap value is Rs. 3000 only.
Take rate of interest as 5%
Solution:
Given: C = 15000; n = 4 years; S = Rs. 3000; r = 5%
Then D= ?
𝐶 (1 + 𝑟)𝑛 −𝑆] [1 − (1 + 𝑟)
𝐷=
= 𝑅𝑠. 3534.14
𝑛
[1 − 1 + 𝑟 ]
160
Machine hour method (Depreciation) – Example
𝐶 −𝑆
𝐷𝑒𝑝𝑟𝑒𝑐𝑖𝑎𝑡𝑖𝑜𝑛 𝑟𝑎𝑡𝑒 𝑝𝑒𝑟 ℎ𝑜𝑢𝑟 =
𝑇𝑜𝑡𝑎𝑙 𝑚𝑎𝑐ℎ𝑖𝑛𝑒 ℎ𝑜𝑢𝑟𝑠 𝑜𝑣𝑒𝑟 𝑢𝑠𝑒𝑓𝑢𝑙 𝑙𝑖𝑓𝑒 𝑜𝑓 𝑚𝑎𝑐ℎ𝑖𝑛𝑒
In this method, the rate of depreciation is calculated by a fixed rate per hour of
production. Here the life of a machine is estimated in terms of its working /
production hours.
161
Machine hour method (Depreciation) – Example
Problem: The estimated life of a shaper is 8 years and it works 12 hours a day. The
initial cost of the shaper is Rs. 85000 and scrap value after 8 years is Rs. 8500. If
the machine works for 3330 hours in a year, calculate the rate of depreciation
charged per hour by the machine hour basis method. Also calculate the rate of
depreciation charged annually.
162
Machine hour method (Depreciation) – Example
Solution:
Given: C = Rs. 85000; n = 8 years; S = Rs. 8500;
First, life of machine in hours = 8 years x 12 hours/day x 365 days/year
= 35040 hours
𝐶 −𝑆
(a) 𝐷𝑒𝑝𝑟𝑒𝑐𝑖𝑎𝑡𝑖𝑜𝑛 𝑟𝑎𝑡𝑒 𝑝𝑒𝑟 ℎ𝑜𝑢𝑟 = 𝑇𝑜𝑡𝑎𝑙 𝑚𝑎𝑐ℎ𝑖𝑛𝑒 ℎ𝑜𝑢𝑟𝑠 𝑜𝑣𝑒𝑟 𝑢𝑠𝑒𝑓𝑢𝑙 𝑙𝑖𝑓𝑒 𝑜𝑓 𝑚𝑎𝑐ℎ𝑖𝑛𝑒 = Rs. 2.183 per hour
(b) Rate of depreciated charged annually = Depreciation rate per hour x Number of hours in a year
= Rs. 2.183 x (12 x 365) = Rs. 9561.54 per year
163
Production unit method (Depreciation) – Example
𝐶 −𝑆
𝐷𝑒𝑝𝑟𝑒𝑐𝑖𝑎𝑡𝑖𝑜𝑛 𝑐𝑜𝑠𝑡 𝑝𝑒𝑟 𝑢𝑛𝑖𝑡 =
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑢𝑛𝑖𝑡𝑠 𝑖𝑡 𝑖𝑠 𝑒𝑥𝑝𝑒𝑐𝑡𝑒𝑑 𝑡𝑜 𝑝𝑟𝑜𝑑𝑢𝑐𝑒
In this method, the rate of depreciation is calculated by a fixed rate per hour of
production. Here the life of a machine is expressed in terms of the number of
units that a machine is expected to produce over its estimated life.
164
Production unit method (Depreciation) – Example
Problem: Calculate the depreciation rate per unit of a machine whose initial cost is
Rs. 2,25,000 and the estimated residual cost after a useful life of the 10 years is Rs.
20,000. It is estimated that the machine will work 50 weeks a year of 46 hours a
week. The rate of production is estimated to be 10 units per hour.
165
Production unit method (Depreciation) – Example
Solution:
C = Rs. 2,25,000; n = 10 years ; S = Rs. 20,000
Hence, 𝐷𝑒𝑝𝑟𝑒𝑐𝑖𝑎𝑡𝑖𝑜𝑛 𝑐𝑜𝑠𝑡 𝑝𝑒𝑟 𝑢𝑛𝑖𝑡 =
𝐶 −𝑆
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑢𝑛𝑖𝑡𝑠 𝑖𝑡 𝑖𝑠 𝑒𝑥𝑝𝑒𝑐𝑡𝑒𝑑 𝑡𝑜 𝑝𝑟𝑜𝑑𝑢𝑐𝑒
Number of units the machine is expected to produce
= Life of the machine in years x No. of working hours in a
year x Production units per hour = 10 x (50 x 46) x 10 = 2,30,000 units.
Hence, 𝐷𝑒𝑝𝑟𝑒𝑐𝑖𝑎𝑡𝑖𝑜𝑛 𝑐𝑜𝑠𝑡 𝑝𝑒𝑟 𝑢𝑛𝑖𝑡 =
𝐶 −𝑆
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑢𝑛𝑖𝑡𝑠 𝑖𝑡 𝑖𝑠 𝑒𝑥𝑝𝑒𝑐𝑡𝑒𝑑 𝑡𝑜 𝑝𝑟𝑜𝑑𝑢𝑐𝑒
166
= Rs. 0.89
Unit 3 – ESTIMATION OF COSTS IN DIFFERENT SHOPS
Session Description of Topic
1
2
3
4
5
Contact hrs C-D-I-O
Estimation in foundry shop: Pattern cost, Casting
cost
Cost estimation in Foundry shop
Forging: Types, Operations, Estimation of Losses a
nd time in forging
Estimation of Forging cost
Cost estimation in Forging shop: Tutorials
167
IOs
Reference
2
C
3
1
2
C,D
3
1,2
2
C
3
1
1
2
C
C,D
3
3
1,2
1,2
Casting
 Casting process is one of the earliest metal shaping techniques known to human
being.
 Casting is a manufacturing process by which a liquid material is usually poured
into a mould, which contains a hollow cavity of the desired shape, and then
allowed to solidify.
 The solidified part is also known as a casting, which is ejected or broken out of
the mould to complete the process.
168
Casting
169
Casting
Adapted from: Automation, Production Systems and Computer-Integrated Manufacturing by M. P. Groover, Pearson Publisher
170
Casting
Adapted from: Automation, Production Systems and Computer-Integrated Manufacturing by M. P. Groover, Pearson Publisher
171
Casting
172
Pattern
 In casting, a pattern is a replica of the object to be cast, used to prepare the
cavity into which molten material will be poured during the casting process.
 The quality of the casting produced depends upon the material of the pattern,
its design, and construction.
173
Pattern
 Single piece pattern
 Split piece pattern
 Loose piece pattern
 Match plate pattern
 Sweep pattern
 Gated pattern
 Skeleton pattern
 Cope and Drag pattern
174
Pattern Material
 Patterns may be constructed from the following materials: wood, metals and
alloys, plastic, polystyrene, rubber, wax, and resins.
 Each material has its own advantages, limitations, and field of application.
175
Pattern Allowance
 Why are allowance necessary?
 Final dimensions of casting are difference from pattern because of various
reasons….??????????
 Types of allowances
 Shrinkage Allowance
 Machining Allowance
 Draft (or) Taper Allowance
 Distortion Allowance
 Rapping (or) Shake Allowance
176
Shrinkage Allowance
 Liquid shrinkage refers to reduction in volume when metal changes from liquid
to solid state. Riser are used to compensate this.
177
Shrinkage Allowance
 Solid shrinkage refers to reduction in volume when metal loses temperature in
solid state. To compensate this shrinkage allowance is used.
178
Shrinkage Allowance
 How to overcome shrinkage allowance issue?
 Pattern is made slightly bigger. This difference in size of the pattern is called
shrinkage allowance.
 Amount of allowance depends upon type of material, its composition,
pouring temperature etc.
179
Shrinkage Allowance
 Shrinkage Allowance
180
Types of Pattern Allowance
181
Machining Allowance
 It’s given to get better surface finish.
 Provided to compensate for machining on casting.
 Pattern is made slightly bigger is size.
 Amount of allowance depends upon size and shape of casting, type of material,
machining process to be used, degree of accuracy and surface finish required
etc.
 A layer of 1.5–2.5 mm thick material has to be provided all round the casting
182
Machining Allowance
183
Draft or Taper Allowance
 Provided to facilitate easy withdrawal of the pattern.
 Typically it ranges from 1 degree to 3 degree for wooden patterns.
184
Draft or Taper Allowance
185
Distortion Allowance
 A U-shaped casting will be distorted during cooling with the legs diverging,
instead of parallel.
 To compensate, the pattern is made with legs converged
cools, the legs straighten and remain parallel.
186
but, as the casting
Pattern Cost
 Total pattern cost consists of following elements
1. Direct material cost
2. Director labour cost
3. Overheads
𝐷𝑖𝑟𝑒𝑐𝑡 𝑀𝑎𝑡𝑒𝑟𝑖𝑎𝑙 𝑐𝑜𝑠𝑡 =
𝐺𝑟𝑜𝑠𝑠 𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑎𝑡𝑡𝑒𝑟𝑛
× 𝐶𝑜𝑠𝑡 𝑝𝑒𝑟 𝑢𝑛𝑖𝑡 𝑤𝑒𝑖𝑔ℎ𝑡
𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙 𝑖𝑛𝑐𝑙𝑢𝑑𝑖𝑛𝑔 𝑝𝑎𝑡𝑡𝑒𝑟𝑛 𝑎𝑙𝑙𝑜𝑤𝑎𝑛𝑐𝑒𝑠
𝐷𝑖𝑟𝑒𝑐𝑡 𝐿𝑎𝑏𝑜𝑢𝑟 𝑐𝑜𝑠𝑡 = 𝐸𝑠𝑡𝑖𝑚𝑎𝑡𝑒𝑑 𝑡𝑖𝑚𝑒 𝑡𝑜 𝑚𝑎𝑛𝑢𝑓𝑎𝑐𝑡𝑢𝑟𝑒 𝑡ℎ𝑒 𝑝𝑎𝑡𝑡𝑒𝑟𝑛 × 𝐿𝑎𝑏𝑜𝑢𝑟 𝑟𝑎𝑡𝑒
𝑇𝑜𝑡𝑎𝑙 𝑃𝑎𝑡𝑡𝑒𝑟𝑛 𝑐𝑜𝑠𝑡 = 𝐷𝑖𝑟𝑒𝑐𝑡 𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙 𝑐𝑜𝑠𝑡 + 𝐷𝑖𝑟𝑒𝑐𝑡 𝑙𝑎𝑏𝑜𝑢𝑡 𝑐𝑜𝑠𝑡 + 𝑂𝑣𝑒𝑟ℎ𝑒𝑎𝑑𝑠
187
COST ESTIMATION IN FOUNDRY SHOP
 Pattern making section
 Sand-mixing section
 Core-making section
 Mould making section
 Melting section
 Fettling section
 Inspection section
<Fettling>
188
COST ESTIMATION IN FOUNDRY SHOP
 The total cost of manufacturing a component consists of following elements :
1. Material cost.
2. Labour cost.
3. Direct other expenses.
4. Overhead expenses.
189
COST ESTIMATION IN FOUNDRY SHOP
1. Material cost = Direct and indirect material cost
(a) Direct material cost
Step 1: To calculate the net weight of the casting
Net weight of casting = Volume of material required for casting x density of material
Step 2: Add the weight of process scarp i.e., weight of runners, gates, risers, etc,
This is generally taken as 15 to 20% of the net weight of casting.
Step 3: Add the weight of metal lost in oxidation in furnace and in cutting gates,
spills, and over arm etc., which is not recoverable. This is usually 8 to 10% of the
net weight of casting.
Step 4: Find out the cost of material
Cost of material = Gross weight of casting x Cost per unit weight
190
COST ESTIMATION IN FOUNDRY SHOP
Step 5: Subtract the return value of scrap, if any, from the cost of material to get
the direct material cost
Direct material cost = Cost of material – Return value of scrap
(b) Indirect material cost
(i) Materials required in melting the metal such as coal, limestone, other fluxes,
etc.,
(ii) Materials used in core shop for making the cores i.e., oils, binders, and
refractories, etc.
The expenditure made on these materials is generally expressed as per kg of
casting weight and is covered under overhead
costs.
191
COST ESTIMATION IN FOUNDRY SHOP
2. Labour cost
(i) The cost of labour involved in making the cores, baking of cores and moulds
is based on the time taken for making various moulds and cores.
(ii) The cost of labour involved in firing the furnace, melting and pouring of the
metal. Cleaning of castings, fettling, painting of castings etc., is generally
calculated on the basis of per kg of cast weight.
192
COST ESTIMATION IN FOUNDRY SHOP
3. Direct Other Expenses
Direct expenses include the expenditure incurred on patterns, core boxes, cost of
using machines and other items which can be directly identified with a particular
product. The cost of patterns, core boxes etc., is distributed on per item basis.
4. Overhead Expenses
The overheads consist of the salary and wages of supervisory staff, pattern shop
staff and inspection staff, administrative expenses, water and electricity charges etc.
The overheads are generally expressed as percentage of labour charges.
The cost of a cast component is calculated by adding the above constituents.
193
Problems Cost Estimation in Foundry Shop
Problem # Estimate the total cost of 20 C.I. flanged pipe casting shown. Assuming
the following data: Cost of C.I. = Rs. 30/kg; Cost of process scrap = Rs. 7/kg;
Process scrap = 2% of net weight of casting; Moulding and pouring charges = Rs.
15/piece; Casting removal and cleaning = Rs. 5/piece; Administrative overheads =
5% factory cost; Selling overheads = 70% administrative overheads.
194
Problems Cost Estimation in Foundry Shop
Solution #
Net volume of cast component = Volume (A+B+C-D) = 1,903,805.15 mm3
Assume density of CI = 7.2 gg/cc
Hence, net weight of cast component = 1,903,805.15 x 7.2 x 10-6 = 13.71 kg
(i) To find material cost:
Process scrap = 2 % of net weight of casting = 2% x 13.71 = 0.274 kg
Hence, gross material required = Net weight + process scrap = 13.984 kg
Cost of C.I. = Rs. 30 /kg
Therefore, cost of gross material = 13.984 x 30 = Rs. 419.52
Cost of process scrap = Rs. 7/kg
Then, cost of process scrap = 0.274 x 7 195
= Rs. 1.918
Problems Cost Estimation in Foundry Shop
Solution #
Then, Material cost per piece = Cost of gross material – Cost of process scrap
= Rs. 419.52 – Rs. 1.918 = Rs. 417.60
(ii) To find labour cost:
Moulding and pouring charges = Rs. 15/piece
Casting removal and cleaning = Rs. /piece
Hence, total labour cost per piece = 15 + 5 = Rs. 20
196
Problems Cost Estimation in Foundry Shop
Solution #
(iii) To find overheads
Factory cost = Material cost + labour cost + Director expenses, if any.
= Rs. 417.60 + Rs. 20 + 0 = Rs. 437.60
Administrative overheads = 5% of factory cost = 5% x Rs. 437.60 = Rs. 21.88
Selling overheads = 70% of administrative cost = 7% x Rs. 21.88 = Rs. 15.31
(iv) To find total cost of 20 C.I. flanged pipe:
Total costing cost/piece = Factory cost + administrative cost + Selling overheads
= Rs. 437.60 + Rs. 21.88 +Rs.15.31 = Rs. 474.79 = Rs. 474.79 x 20 = Rs. 9495.80
197
Problems Cost Estimation in Foundry Shop
Problem # 20 number of gun metal bevel gear blank shown are to be cast in the
factory from the pattern supplied by the customer. Estimate the selling price of
each piece from the following data: cost of molten gun metal = Rs. 350 per kg;
Scrap return value = Rs. 150 per kg; Process scrap = 10% net weight of casting;
Administrative overheads = Rs. 5 per kg; Profit = 15% of manufacturing cost;
density of gun metal = 8.73 gg/cc.
198
Problems Cost Estimation in Foundry Shop
Solution #
Net volume = Volume (A + B +C)
𝐻
Volume A = Volume of frustum of cone =
𝑎1 + 𝑎2 + 𝑎1 𝑎2
3
Volume B = Volume of frustum of cone = 116.35 cm3
Volume C = 17.18 cm3
= 57.06 cm3
Net volume of a bevel gear blank = Volume (A + B +C) = 190.59 cm3
Net weight = net volume x density = 1.644 kg
Process scrap = 10 % of net weight = 10 % x 1.664 = 0.1664 kg
Weight of gun metal required = 1.664 + 0.1664 = 1.8304 kg
199
Problems Cost Estimation in Foundry Shop
Solution #
(i) To find material cost:
Cost of molten gun metal = 1.8304 x 350 = Rs. 640.64
Scrap return value = 0.1664 x 150 = Rs. 24.96
Material cost = 640.64 – 24.963 = Rs. 615.68
(ii) To find labour cost:
Moulding and pouring charges per piece = Rs. 20
Casting, removal and gate cutting charges per piece = Rs. 7
Fettling charges per piece = Rs. 2
Hence, total labour cost = 20 + 7 + 2 = Rs. 29
200
Problems Cost Estimation in Foundry Shop
Solution #
(iii) To find overheads:
Moulding and pouring factory overhead =
25
60
× 60 = 𝑅𝑠. 25
Casting, removal and gate cutting factory overhead =
Fettling factory overhead =
3
60
10
60
× 40 = 𝑅𝑠. 6.67
× 40 = 𝑅𝑠. 2
Hence, Total factory overhead = 25 + 6.67 + 2 = Rs. 33.67
Administrative overheads = Rs. 5 per kg.
Hence, administrative overheads for 1.8304 kg = 5 x 1.8304 = Rs. 9.15
201
Problems Cost Estimation in Foundry Shop
Solution #
(iii) To find selling price:
Manufacturing cost = Material cost + labour cost + factory overheads +
administrative overheads
= 615.68 + 29 + 33.67 + 9.15 = Rs. 687.50
Profit = 15 % of manufacturing cost = 15% x 687.50 = Rs. 103.12
Hence, selling price per piece = Manufacturing cost + Profit
= 687.50 + 103.12 = Rs. 790.62
202
Problems Cost Estimation in Foundry Shop
Problem # Find out the cost of wood required for making the pattern of a M.S. Bell
crank lever. The finished drawing of M.S. Bell crank lever is given. Wood is available
at the rate of Rs. 12000/m3. Holes are to be bored on machines after casting.
Consider the metal shrinkage and pattern finishing allowances.
Also estimate the cost of pattern, if
(i) Pattern maker is available at the rate of Rs. 320 per 8 hours working day and
takes 4 hours in preparing the pattern.
(ii) Overhead charges are 15% of labour cost.
To find: (i) Cost of wood required for making the pattern
(ii) Cost of pattern.
203
Problems Cost Estimation in Foundry Shop
204
Problems Cost Estimation in Foundry Shop
Solution # (i) Cost of wood required for making the pattern
After considering the shrinkage allowance of 10 mm / metre, the sketch of pattern
is shown.
205
Problems Cost Estimation in Foundry Shop
Solution # (i) Cost of wood required for making the pattern
Split the pattern drawing into four simple parts A, B, C, and D, While estimating the
amount of rough wood required for each part, wood finishing allowance of 3 mm
on each side should be considered.
206
Problems Cost Estimation in Foundry Shop
Solution # (i) Cost of wood required for making the pattern
Also, to make the pattern economical, each pattern is prepared separately and then
fixed together to get the complete pattern instead of preparing the whole pattern
from one piece of wood.
Volume of rough wood required:
207
Problems Cost Estimation in Foundry Shop
Solution # (i) Cost of wood required for making the pattern
208
Problems Cost Estimation in Foundry Shop
Solution # (i) Cost of wood required for making the pattern
209
Problems Cost Estimation in Foundry Shop
Solution # (ii) Cost of pattern
210
Problems Cost Estimation in Foundry Shop
Problem # Find out the cost of wood required for making the pattern of a ‘lathe
centre’ . Wood is available at the rate of Rs. 12000/m3. Consider the pattern
allowance. To find: Cost of wood required for making the pattern.
211
Problems Cost Estimation in Foundry Shop
Solution #
To find the amount of wood required for making the pattern, metal shrinkage
allowance and pattern machining allowance should be considered.
Shrinkage allowance= 10 mm/metre
Pattern making allowance = 2 mm on each side on casting
212
Problems Cost Estimation in Foundry Shop
Solution #
Pattern machining allowance = 2 mm on each side on casting
213
Problems Cost Estimation in Foundry Shop
Solution #
Now we can estimate volume of wood required, by considering A, B, C, D by
considering wood finishing allowance of 3 mm on each side.
214
Problems Cost Estimation in Foundry Shop
Solution #
Now we can estimate volume of wood required, by considering A, B, C, D
215
Problems Cost Estimation in Foundry Shop
Solution #
Now we can estimate volume of wood required, by considering A, B, C, D
216
Problems Cost Estimation in Foundry Shop
Solution #
Now we can estimate volume of wood required, by considering A, B, C, D
217
Problems Cost Estimation in Foundry Shop
Solution #
Total amount of wood required. And Cost of wood.
218
Problems Cost Estimation in Foundry Shop
Solution #
With same problem. Estimate the cost of pattern, if
i. Labour rate for pattern maker is Rs. 40 per hour and pattern required 3 hours
for completion
ii. Overhead charges are 50% of material cost
To find: Cost of pattern.
219
Bulk Deformation
 Metal forming operations which cause significant shape change by deforming
metal parts whose initial form is bulk rather than sheet.
 Starting forms:
 Cylindrical bars and billets
 Rectangular billets and slabs and similar shapes
 These processes stress metal sufficiently to cause plastic flow into the desired
shape.
 Performed as cold, warm, and hot working operations.
220
Importance of Bulk Deformation
 In hot working, significant shape change can be accomplished.
 In cold working, strength is increased during shape change.
 Little or no waste – some operations are near net shape or net shape processes
 The parts require little or no subsequent machining
221
Four basic Bulk Deformation Processes
 Rolling – slab or plate is squeezed between opposing rolls
 Forging – work is squeezed and shaped between opposing dies
 Extrusion – work is squeezed through a die opening, thereby taking the shape
the shape of the opening
 Wire and bar drawing – diameter of wire or bar is reduced by pulling it through
a die opening.
222
Forging
 Forging is the working of metal into a useful shape by hammering or pressing.
 Forging machines are now capable of making parts ranging in size of a bolt to a
turbine rotor.
 Most forging operations are carried out hot. (cold and hot forging)
223
Hot Forging
 During hot forging, the temperature reaches above the recrystallization point of
the metal.
 This kind of extreme heat is necessary in avoiding strain hardening of the metal
during deformation.
 Isothermal forging is used to prevent the oxidation of certain metals, like super
alloys.
 Up to 1150°C for Steel
 360 to 520 °C for Al-Alloys
 700 to 800°C for Cu-Alloys
224
Hot Forging
 Advantages
 Forged parts possess high ductility and offers great resistance to impact and fatigue loads.
 Forging refines the structure of the metal.
 It results in considerable saving in time, labor and material as compared to the production of
similar item by cutting from a solid stock and then shaping it.
 Forging distorts the previously created unidirectional fiber as created by rolling and increases
the strength by setting the direction of grains.
225
Hot Forging
 Disadvantages
 Rapid oxidation in forging of metal surface at high temperature results in scaling which wears
the dies.
 The close tolerances in forging operations are difficult to maintain.
 Forging is limited to simple shapes and has limitation for parts having undercuts
 Some materials are not readily worked by forging.
 The initial cost of forging dies and the cost of their maintenance is high.
226
Cold Forging
 Cold forging deforms metal while it is below its recrystallization point.
 Cold forging is generally preferred when the metal is already a soft metal, like aluminium.
 This process is usually less expensive than hot forging and the end product requires little, if any,
finishing work.
227
Cold Forging
 Advantages
Produces net shape or near-net shape parts
Cold forging is also less susceptible to contamination problems
Final component features a better overall surface finish.
Minimizes the cost
Easier to impart directional properties
 Disadvantages
 The metal surfaces must be clean and free of scale before forging occurs
 The metal is less ductile
 Residual stress may occur
 Heavier and more powerful equipment is needed
 Stronger tooling is required
228
Application of Forging
 Forging is generally carried out on carbon alloy steels, wrought iron, copper-base alloys,
aluminium alloys, and magnesium alloys.
 Stainless steels, nickel based super-alloys, and titanium are forged especially for aerospace.
 Forged automobile components include connecting rods, crankshafts, wheel spindles, axle beams,
pistons, gears, and steering arms.
229
Types of Forging
 Smith forging - hand forging, the component is made by hammering the
heated material on an anvil. The hammering may be done by hand or machine.
 Drop forging - The forging is done by using the impressions machined on a pair
of die blocks.
 Press forging - In this method the metal is squeezed into desired shape in dies
using presses. Instead of rapid impact blows of hammer, pressure is applied
slowly. This method is used for producing accurate forgings.
230
Types of Forging
 Machine forging or Upset forging - In machine forging or upset forging the
metal is shaped by making it to flow at right angles to the normal axis. The
heated bar stock is held between two dies and the protruding end is hammered
using another die. In upset forging the cross-section of the metal is increased
with a corresponding reduction in its length.
 Roll forging - Roll forging is used to draw out sections of bar stock, i.e.,
reducing the cross-section and increasing the length. Special roll forging
machines, with dies of decreasing cross-section are used for roll forging.
231
Types of Forging
 Open-die forging – work is compressed between two flat dies, allowing metal to
flow laterally with minimum constraint.
 Impression – die forging – die contains cavity or impression that is imparted to
workpart
 Metal flow is constrained so that flash is created.
 Flashless forging – workpart is completely constrained in die
 No excess flash is created
232
Types of Forging
233
Types of Forging
234
Types of Forging
235
Material Losses in Forging
 Shear loss – the blank required for forging a component is cut from billets or
long bars. The material equal to the product of thickness of sawing blade and
cross-section of bar is lost for each cut.
 The material consumed in the form of saw-dust or pieces of smaller dimensions
left as defective pieces is called shear loss. This is usually taken as 5% of the net
weight of forging.
236
Material Losses in Forging
 Tonghold loss – Drop forging operations are performed by holding the stock at
one end with the help of tongs.
 A small length, about 2.0 – 2.5 cm and equal to diameter of stock is added to
the stock for holding.
Tonghold loss = Area of X-section of bar × Length of tonghold
237
Material Losses in Forging
 Tonghold loss –The tonghold loss is equal to the volume of the protections.
 For example, the tonghold loss for a bar of 2 cm diameter will be
238
Material Losses in Forging
 Scale loss –As the forging process is performed at very high temperature, the
Oxygen from air forms iron oxide by reacting with hot surface.
 This iron oxide forms a thin film called scale, and falls off from surface at each
stroke of hammer.
 Scale loss is taken as 6% of net weight.
239
Material Losses in Forging
 Flash loss – When dies are used for forging, some metal comes out of the die at
the parting line of the top and bottom halves of the die. This extra metal is
called flash.
 Flash is generally taken as 20 mm wide and 3 mm thick.
 Flash loss = Volume of flash x Density of the material
 Volume of flash = Circumference of component at parting line x Cross-sectional
area of flash.
 Cross sectional area of flash = Flash thickness x Flash width
240
Material Losses in Forging
 Sprue loss – When the component is forged by holding the stock with tongs,
the tonghold and metal in the die are connected by a portion of metal called
the sprue or runner. This is cut off when product is completed.
 Sprue loss is taken as 7 % of net weight
241
Estimation of Cost of Forgings
Total forging cost = Labour cost + Material cost + direct expenses
+ Overhead cost
242
Estimation of Cost of Forgings
1. Cost of direct materials
Step 1: Net weight = volume of forging x density of material
Step 2: Gross weight = Net weight + material loss in process
In case of smith or hand forging, only scale loss and shear loss are to be added to
net weight. But in case of die forging, all the losses are taken into account.
Step 3: Direct material cost = Gross weight x Price of raw material per kg
Step 4: To select the diameter: The greatest section of the forging gives the
diameter of the stock to be used. and length of stock
Step 5: To select the length of stock =
𝐺𝑟𝑜𝑠𝑠 𝑤𝑒𝑖𝑔ℎ𝑡
𝐷𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙 𝑋 𝐶𝑟𝑜𝑠𝑠−𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑠𝑡𝑜𝑐𝑘
243
Estimation of Cost of Forgings
2. Cost of direct labour
Direct labour cost = time for forging per piece (in hours) x labour rate per hour
244
Estimation of Cost of Forgings
3. Overheads
 Supervisory charges, depreciation of plant and machinery, consumables, power
and lighting charges, office expenses etc.
 Generally expressed as % of direct labour cost.
245
Estimation of Cost of Forgings
Problem: A bar stock of 3 cm diameter and 2 m long is to be converted into (i)
square bar 3 cm side; (ii) hexagonal bar 3 cm side; (iii) rectangular bar of sides 3
cm x 2 cm. Calculate the length of bar made in each case. Assume hand forging
and neglect losses.
Solution:
Stock diameter, D = 3 cm
Stock length, L = 2 m = 200 cm
Hence, volume of the bar =
𝜋 2
𝐷 𝐿
4
= 1413.72 𝑐𝑚3
246
Estimation of Cost of Forgings
Solution:
i) Length of square 3 cm side:
Let a = side of square = 3 cm
Volume of square bar = Area of square x length = a2 x L = (3)2 x L = 9L cm3
If neglecting losses means: Volume of square bar = Volume of stock bar
Then, 9L =1413.72 𝑐𝑚3
L = 157.08 cm
247
Estimation of Cost of Forgings
Solution:
ii) Length of hexagonal bar of side 3 cm:
Volume of hexagonal bar = Area of hexagonal base x length
=[
3 3
2
× 𝑎2 ] x L = 23.38 L cm3
If neglecting losses means: Volume of hexagonal bar = Volume of stock bar
Then, 23.38L =1413.72 𝑐𝑚3
L = 60.46 cm
248
Estimation of Cost of Forgings
Solution:
iii) Length of rectangular bar of sides 3 cm x 2 cm:
Volume of rectangular bar = Area of rectangle x length
= [ 3 × 2] x L = 6 L cm3
If neglecting losses means: Volume of hexagonal bar = Volume of stock bar
Then, 6L =1413.72 𝑐𝑚3
L = 235.62 cm
249
Estimation of Cost of Forgings
Problem: 200 pieces of bolt are to be made by upsetting from 25 mm diameter.
What is the length of each bolt before upsetting? What is the length of the rod is
required if 3.5 % of the length goes as scrap?
250
Estimation of Cost of Forgings
Solution:
Volume of the bolt =
𝜋
4
40
2
× 22 +
Area of cross-section of the stock bar
𝜋
4
𝜋
=
4
25
2
25
× 112.5 = 82,869.3 𝑚𝑚3
2
= 490.87 𝑚𝑚3
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑏𝑜𝑙𝑡
𝐴𝑟𝑒𝑎 𝑜𝑓 𝑐𝑟𝑜𝑠𝑠−𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑡𝑜𝑐𝑘 𝑏𝑎𝑟
Length of each bolt before upsetting =
= 168.82 𝑚𝑚
Length for 200 pieces = 168.82 x 200 = 33,764.25 mm
Scrap = 3.5 % of the length = 3.5 % x 33,764.25 = 1,181.75 mm
Hence, net length required after scrap for 200 bolts = 33,764.25 + 1,181.75
= 34946 mm
251
Estimation of Cost of Forgings
Problem: Calculate the net weight and gross weight for the component shown.
Density of material used is 7.86 gm/cc. Also calculate: i. length of 14 mm diameter
bar required to forge one component. ii. Cost of forging/piece. If: Material cost =
Rs. 80/ kg. Labour cost = Rs. 5/ piece. Overheads = 150 % of labour cost.
252
Estimation of Cost of Forgings
Solution:
Net volume of forged component=
56.76 𝑐𝑐
𝜋
4
42 × 2 + 32 × 2.5 + 22 × 2 + 1.42 × 5 =
Net weight = Net volume x density = 446 gms
Losses:
Shear loss = 5 % of net weight = 22.30 gms
Scale loss = 6 % of net weight = 26.76 gms
Sprue loss = 7 % of net weight = 31.22 gms
Tonghold loss = volume of protrusion = 2 x Area of cross-section of bar x 7.86
= 24.22 gms
253
Estimation of Cost of Forgings
Solution:
 Flash loss = Volume of flash x Density of the material
 Volume of flash = Circumference of component at parting line x Cross-sectional
area of flash.
 Cross sectional area of flash = Flash thickness x Flash width
Generally flash taken as 20 mm wide and 3 mm thick.
Hence, Cross sectional area of flash = 2 x 0.3 = 6 cm
Periphery of parting line (or) Circumference of component at parting line =
= [2(2+2.5+2+5)+1.4+(2-1.4)+(3-2)+(4-3)+4] = 31.0
Flash loss = Volume of flash x Density of the material = 31.0 x 6 x 7.86 = 146 gms
254
Estimation of Cost of Forgings
Solution:
Total material loss = 22.3 + 26.8 + 146 + 24.22 +31.22 = 250 gms
Gross weight = net weight + losses = 446 + 250 = 696 gms
i) New length of 14 mm of bar required per piece
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑓𝑜𝑟𝑔𝑖𝑛𝑔
56.76
=
= 𝜋 2 = 36.86 cm
𝐴𝑟𝑒𝑎 𝑜𝑓 𝑐𝑟𝑜𝑠𝑠−𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑏𝑎𝑟
4
1.4
Direct material cost = 696 x 10-3 x 80 = Rs. 55.68
Direct labour cost = Rs. 5 per piece
Over heads = 150 % of labour cost = 1.5 x 5 = Rs. 7.5
ii). Cost per piece = 55.68 + 5 +7.5 = Rs. 68.18
255
Estimation of Cost of Forgings
Problem: Calculate the net weight and gross weight for the manufacture of 500
levers shown in figure. The material weighs 7.8 gm/cc and the total losses account
for 25 % of net weight of the lever.
Also calculate: i) Length of 3 cm diameter required / component,
ii) The cost of forging 500 pieces if the material costs Rs. 80/ kg, labour cost is Rs.
12 per piece and overheads are 25 % of material cost.
256
Estimation of Cost of Forgings
Solution:
n = 500, Density = 7.8 x 10-6 kg/mm3
To find: (i) Net weight and gross weight of 500 levers.
(ii) Length of 3 cm diameter required /component
(iii) The cost of forging 500 pieces.
257
Estimation of Cost of Forgings
Solution:
n = 500, Density = 7.8 x 10-6 kg/mm3
(i) Net weight and gross weight of 500 levers.
Net weight = Net volume x Density
Hence, Net volume =
𝜋
4
× 25
2
× 50 +
= 58016.81 mm3
Hence, net weight = 0.452 kg / piece.
𝜋
4
× 15
2
× 30 +[(60x60)x8]-
𝜋
4
× 10
2
×8
For 500 levers, = 0.452 x 500 = 226 kg
Total losses = 25 % of net weight = 25 % x 0.452 = 0.113 kg
Hence, Gross weight = net weight + total losses = 0.452 + 0.113 = 0.565 kg/piece.
258
Estimation of Cost of Forgings
Solution:
Hence, Gross weight = net weight + total losses = 0.452 + 0.113 = 0.565 kg/piece.
Therefore, gross weight of 500 levers = 500 x 0.565 = 282.5 kg.
(ii) Length of 30 mm diameter bar required /component
Area of cross-section of 30 mm dia bar =
Gross volume of one lever =
Then, Length =
𝐺𝑟𝑜𝑠𝑠 𝑤𝑒𝑖𝑔ℎ𝑡
𝐷𝑒𝑛𝑠𝑖𝑡𝑦
=
𝜋
4
× 30
0.565
𝐷𝑒𝑛𝑠𝑖𝑡𝑦
𝐺𝑟𝑜𝑠𝑠 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑜𝑛𝑒 𝑙𝑒𝑣𝑒𝑟
𝐴𝑟𝑒𝑎 𝑜𝑓 𝑐𝑟𝑜𝑠𝑠 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 30 𝑚𝑚 𝑑𝑖𝑎 𝑏𝑎𝑟
259
2
= 706.86 mm2
= 72435.9 mm3
= 102.47 𝑚𝑚
Estimation of Cost of Forgings
Solution:
(iii) Cost of forging 500 pieces:
Material cost = Rs. 80/kg; labour cost = Rs. 12/piece; overhead = 25% of material
cost
Total material cost = Gross weight of 500 levers x material cost per kg.
= 282.5 x 80 = Rs. 22600
Total labour cost = No. of pieces x labour cost per piece = 500 x 12 = Rs. 6000
Overhead cost = 25 % x Rs. 22600 = Rs. 5650.
Total cost of forging = Rs. 22600 + Rs. 6000 + Rs. 5650 = Rs. 34250
260
Estimation of Cost of Forgings
Problem: 500 pieces of a component as shown are to be drop forged from 80 mm
diameter stock bar. Calculate the cost of manufacturing, if material cost is Rs. 2750
per meter. Forging charges Rs. 1.50 per cm2 of surface area to be forged. Overhead
expenses to be 12% of the cost of the material cost. Consider all possible losses
during operations.
261
Estimation of Cost of Forgings
Solution:
i) To find material cost:
𝜋
4
Net volume of the finished material = 2 [(40x40)40] + [
80 2 120] = 731186 𝑚𝑚3
Shear loss = 5% of net volume = 36559 𝑚𝑚3
Scale loss = 6% of net volume = 43871 𝑚𝑚3
Sprue loss = 7% of net volume = 51183 𝑚𝑚3
Tonghold loss = taking 20 mm extra length required to be held in tong.
= Area of cross-section of bar x length of the tonghold
=
𝜋
4
80
2
x 20 = 0.6283 𝑚𝑚3
262
Estimation of Cost of Forgings
Flash loss = Taking flash width = 20 mm and flash thickness = 3 mm,
Volume of flash = periphery of parting line of dies x flash width x flash thickness
= [2(50+120+40)+40+(80-40)+(80-40)+40]x20x3 = 33600 𝑚𝑚3
Total material loss = 171496 𝑚𝑚3
Gross volume of material required = Net volume + Material loss = 902682 𝑚𝑚3
Area of cross-section of 40 mm square bar stock = 40 x 40 = 1600 𝑚𝑚2
263
Estimation of Cost of Forgings
Hence, length of bar stock required =
𝐺𝑟𝑜𝑠𝑠 𝑣𝑜𝑙𝑢𝑚𝑒
𝐴𝑟𝑒𝑎 𝑜𝑓 𝑐𝑟𝑜𝑠𝑠𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑏𝑎𝑟
= 564.2 𝑚𝑚
Given: Material cost = Rs. 2750 per metre.
Therefore, total material cost = 564.2 x10-3 x 2750 = Rs. 1551.55
2. Cost of forging:
Since, Forging charges Rs. 1.50 per cm2 of surface area given. We have to calculate
surface area of the shaft.
264
Estimation of Cost of Forgings
2. Cost of forging:
Since, Forging charges Rs. 1.50 per cm2 of surface area given. We have to calculate
surface area of the shaft.
Surface area of the shaft = (40 x 40)+(4 x 40 x 40) + (40 x 40)+(4 x 40 x 40) +
𝜋
4
× 802 − (40 × 40) +
𝜋
4
× 802 − (40 × 40) + 𝜋 × 80 × 120 = 530.13 cm2
Total cost of forging = 530.13 x 1.5 = Rs. 795.2
265
Estimation of Cost of Forgings
3. Overhead expenses:
Overheads = 12% of cost of the materials = 15% x 1551.55 = Rs. 186.19
4. To find the total forging cost of 500 pieces of component
Total cost/forged shaft = material cost + forging cost + overheads = Rs. 2532.94
For 500 piece, the total cost = 500 x Rs. 2532.94 = Rs. 12,66,470
266
Estimation of Cost of Forgings
Problem: Calculate the cost of forging a crank shaft as shown in Figure. The
forging is to be made out of a bar stock of 50 mm and following data is available:
i) Material price = Rs. 80/kg
ii) Direct labour charges= Rs. 23/piece
iii) Overhead charges = 150% of material cost
iv) Density of material = 7.5 gm/cc
v) Losses = 28% of net weight
All dimensions in mm.
267
Estimation of Cost of Forgings
268
Estimation of Cost of Forgings
269
Estimation of Cost of Forgings
Problem: Estimate the cost of manufacturing a high carbon steel spanner as shown
in figure to be made by die forging. The following data are available:
i) Batch size = 500 pieces
ii) Die cost per batch = Rs. 500
iii) Stock cutting charges = Rs. 100 per batch
iv) Set up and machine operation cost = Rs. 200 per batch
v) Labour charges = Rs. 100 per batch
vi) Density of steel = 8.5 gm/cc
vii) Cost of high carbon steel = Rs. 70 per kg
viii) Losses = 24% of net weight
120
270
Estimation of Cost of Forgings
Solution:
120
Net volume of spanner = Volume (A + B + C – D – E ) = ?
Volume
Volume
Volume
Volume
Volume
(A) =(15 x 6) x (120-20-16) = 7560 mm3
(B) = 𝜋𝑟 2 × 𝑇ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠 = 12566.37 mm3
(C) = 𝜋𝑟 2 × 𝑇ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠 = 8042.48mm3
(D) = (12x20) 10 = 2400 mm3
(E ) = (10 x 16) 10 = 1600 mm3
Net volume of spanner = Volume (A + B + C – D – E ) = 24168.85 mm3
271
Estimation of Cost of Forgings
Solution:
Net weight = net volume of spanner x density of steel = 0.2054 kg
Gross weight = Net weight + 24 % of net weight = 0.2547 kg
Material cost = Rs. 70 per kg
Total material cost = Rs. 17.83
272
Estimation of Cost of Forgings
Solution:
Labour cost = Rs. 100 per batch
Labour cost per piece = Rs.100/500 pieces = Rs. 0.2
Die cost per batch = Rs. 500
Die cost per piece = Rs. 500 / 500 pieces = Rs. 1
273
Estimation of Cost of Forgings
Solution:
Stock cutting charges = Rs. 100 per batch
Stock cutting charges per piece = Rs.100/500 pieces = Rs. 0.2
Set-up and machine operation cost = Rs. 200
Set-up and machine operation cost per piece = Rs. 200 / 500 pieces = Rs. 0.4
Hence, total cost of manufacturing a spanner = material cost + labour cost + die
cost + stock cutting charges + set=up and machine operation cost
Hence, total cost of manufacturing a spanner = 17.83 + 0.2 + 1 + 0.2 + 0.4 = Rs. 19.63
274
Unit 4 – ESTIMATION OF COSTS IN FABRICATION SHOPS
Session Description of Topic
1
2
3
4
5
6
Contact hrs C-D-I-O
Welding, Types of weld joints, Gas welding
Estimation of Gas welding cost, Gas cutting
Arc welding: Equipment's, Cost Estimation
Cost estimation in Welding shop: Tutorials
Estimation in sheet metal shop, Shearing and forming
Cost estimation in Sheet metal shop
275
1
1
1
2
2
2
C
C
C
C,D
C
C,D
IOs
Reference
3
3
3
3
3
3
1
1
1
1,2
1
1,2
Welding
Welding is the process of joining similar or dissimilar metals by the
application of heat.
Welding can be done with or without application of pressure and
with or without the addition of filler metal.
While welding, the edges of metal pieces are either melted or
brought to plastic condition.
Welding used for Permanent Joint276
Welding
The filler material has a similar composition and melting point as the
base metal.
A flux is required in some welding processes, so as to remove the
oxide layers, in the form of fusible slag which floats on the molten
metal. It also provides a shield preventing the re-formation of the
oxide layer.
277
Types of Welding
Plastic or pressure welding or solid state welding
 The piece of metal to be joined are heated to a plastic state and
forced together by external pressure
 Ex: Resistance welding
Fusion or non-pressure welding
 The material at the joint is heated to a molten state and allowed to solidify
 Ex: Gas welding, Arc welding
278
Welding
279
Classification of Welding Processes
Gas welding (Oxy-Acetylene)
Arc welding (Metal arc)
Resistance welding
Solid state welding
Thermo-chemical welding
Low temperature welding
280
281
Gas welding
The most commonly used gas welding is oxy-acetylene welding. The
high temperature required for welding is obtained by the application
of a flame from mixture of oxygen and acetylene gas.
The filler material is used to fill the gap between the parts to be
welded.
The welding technique used may be leftward welding or rightward
welding.
282
Gas welding (oxy-acetylene welding)
283
284
Gas welding (oxy-acetylene welding)
285
Method of welding
Leftward or forward or forehand welding
 Direction of weld is RHS to LHS
 Blow pipe angle 60°to 70° from surface of W/P
 Filler road angle 30° to 40° from surface of W/P
 Vertical joint are welded
 This method is used for welding plates upto 5 mm thick.
 No edge preparation is required in case of the plates of thickness
upto 3 mm.
286
Method of welding
Rightward or Backward or Backhand welding
 Direction of welds is LHS to RHS
 Blow pipe angle 40° to 50° from surface of W/P
 Filler rod angle 40° to 50° from surface of W/P
 Horizontal and Overhead joint are welded
 This method is adopted for welding thicker plates.
 Faster by 20 to 25% compared to leftward
287
Method of welding
288
Method of welding
289
Method of welding
290
Gas cutting
 Gas cutting or flame cutting is a process of cutting metals into pieces using oxyacetylene flame.
 It is used for cutting plates of thickness upto 150 mm.
 The equipment used for gas cutting is similar to that of gas welding equipment
except the tip of the torch.
 The torch tip has a central hole for oxygen jet with surrounding holes for
preheating flames.
 The oxy-acetylene flame is used for preheating purpose and the high pressure
oxygen jet does the cutting work.
291
Gas cutting
292
Method of welding
293
Arc welding
294
Method of welding
295
Method of welding
296
Method of welding
297
Types of Weld Joints
298
Estimation of Cost of Welding
Estimation of welding cost
 The cost of welding depends on the welding process used, the
type of joint, materials, and labour employed in making and
inspecting the joint.
299
Estimation of Cost of Welding
Estimation of welding cost
 Direct material cost
 Direct labour cost
 Direct expenses
 Overhead expenses
300
Estimation of Cost of Welding
 Direct material cost
 Cost of base materials to be welded i.e., sheet, plate, rolled
section, casting or forging.
 Cost of consumables such as electrodes, flux, O2, C2H2, etc
 Direct labour cost
 Preparation or pre-welding labour cost
 Welding cost
 Post welding or finishing cost
301
Estimation of Cost of Welding
 Direct expenses
 Cost of power consumed
 Cost of welding fixtures
302
Estimation of Cost of Welding
 Direct expenses
 Cost of power consumed
 Cost of welding fixtures
 Overhead expenses
 The overheads consists of all other charges which include the
salaries of supervisors, indirect labour charges, depreciations of
welding tools and auxiliaries, administrative expenses, water and
electricity charges, etc.
303
Estimation of Cost of Welding
 Problem: A butt joint between two square metal plates of 250 cm x
250 cm is made using electric arc welding. If the rate of welding is 5
meter/hour, calculate the time required to complete ten such
welding operations.
304
Estimation of Cost of Welding
 Solution:
Plate size = 250 cm x 250 cm
Rate of welding is 5 meter/hour
Calculate the time required to complete ten such welding operations.
Time required/weld =
1
𝑅𝑎𝑡𝑒 𝑜𝑓 𝑤𝑒𝑙𝑑𝑖𝑛𝑔
× 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑤𝑒𝑙𝑑 =
Hence, for 10 such weld = 10 x 0.5 = 5 hours
305
1
5
× 2.5 = 0.5 ℎ𝑜𝑢𝑟𝑠/weld
Estimation of Cost of Welding
 Problem: A lap welded joint is to be made as shown in figure
Length of joint = 1.5 metre
306
Estimation of Cost of Welding
 Solution:
Time per meter run of weld =
1
10
ℎ𝑟𝑠 = 6 𝑚𝑖𝑛𝑢𝑡𝑒𝑠
Cost of power consumed per meter run of weld
=
30×250
1000
×
6
60
×
1
0.5
𝑉×𝐴
=
1000
×
1
0.4
×
1
60%
1
𝐸
1
𝑟
× × ×𝐶
× 3 = Rs. 11.25
𝐶𝑜𝑠𝑡 𝑜𝑓 𝑙𝑎𝑏𝑜𝑢𝑟/ℎ𝑜𝑢𝑟
Cost of labour per meter of weld length =
𝑊𝑒𝑙𝑑𝑖𝑛𝑔 𝑠𝑝𝑒𝑒𝑑 𝑖𝑛 𝑚/ℎ𝑟
40
=
10
×
𝑡
60
×
1
𝐿𝑎𝑏𝑜𝑢𝑟 𝑎𝑐𝑐𝑜𝑚𝑝𝑙𝑖𝑠ℎ𝑚𝑒𝑛𝑡 𝑓𝑎𝑐𝑡𝑜𝑟
= Rs. 6.66 / meter of weld length
307
Estimation of Cost of Welding
 Solution:
Cost of electrodes per meter of weld = 0.350 x 8 = Rs. 2.80
Total direct cost per meter of weld = 11.5 + 6.66 + 2.80 = Rs. 20.71
Overhead charges per meter of weld = 80% direct charges
= 80% x 20.71 = Rs. 16.60
Total charges for welding one meter length of joint = 20.71 + 16.60 = Rs. 37.31
As this is a double fillet weld, lap joint length of weld = 1.5 x 2 = 3 meters
Hence, total charges of making the welded joint = Rs. 37.31 x 3 = Rs. 112
308
Estimation of Cost of Welding
 Problem: What is the material cost of welding two plates of size 300
mm length and 150 mm width and 8 mm thickness to make a piece
of 300 x 300 mm approximately.
Use rightward technique with no edge preparation cost. Take overall
cost of oxygen as Rs. 15 per cu. metre., cost of acetylene at Rs. 60 per
cu. Metre, cost of filler metal Rs. 50 per kg and 1 cu.cm of filler metal
weighs 11.28 gms. Assume dia of filler rod = 4 mm. filler rod used per
metre of weld = 3.5 metre. Rate of welding = 2.1 metre/hour.
Consumption of O2 = consumption of C2H2 = 7.1 m3/hr
309
Estimation of Cost of Welding
 Solution: What is the material cost of welding two plates?
Length of weld = 0.3 m
i) Find the Cost of filler material?
Filler rod used per metre of weld = 3.4 m
Hence, rod used per 0.3 m = 3.4 x 0.3 = 1.02 m
Volume of filler rod used =
𝜋 2
𝑑 𝑙
4
=
𝜋
4
310
× (0.04)2 × 1.02 = 1.282 x 10-5 m3
Estimation of Cost of Welding
Weight of filler rod = volume x density =
1.282 x 10-5 x 11.28 x 103 = 0.1446
Actual cost of filler material = 0.1446 x Rs. 50 = Rs. 7.23
(ii). Find cost of O2 and C2H2 consumed:
Given: Rate of welding (or) Speed = 2.1 metre /hour
Hence, time to weld 300 mm length =
1
2.1
× 0.3 = 0.143 ℎ𝑜𝑢𝑟𝑠
Given: Consumption of O2 = consumption of C2H2 = 7.1 m3/hr
Hence, actual consumption of O2 = 7.1 x 0.143 = 1.0153 m3
311
Estimation of Cost of Welding
Hence, actual consumption of O2 = 7.1 x 0.143 = 1.0153 m3
Cost of O2 consumed = 1.0153 x Rs. 15 = Rs. 15.23
Cost of C2 H2 consumed = 1.0153 x Rs. 60 = Rs. 60.92
(iii) To find total material cost of welding:
Total material cost = Cost of filler material + Costs of O2 & C2 H2 consumed
= 7.23 + 15.23 + 60.92 = Rs. 83.38
312
Estimation of Cost of Welding
 Problem: Estimate the material cost for welding 2 flat pieces of Mild Steel 15 x
16 x 1 cm size at an angle of 90 degree by gas welding. Neglect edge
preparation cost and assume: Cost of O2 = Rs. 15/m3; Cost of C2H2 = Rs. 60/m3;
Density of filler metal = 7 gm/cc; cost of filler metal = Rs. 50/kg; filler rod dia =
5 mm; filler rod required 4.5 m/m of welding. Assume O2 consumption = 0.7
cu.m/hr; C2H2 consumption = 0.5 cu.m/hr; Welding time = 30 min/m of welding.
313
Estimation of Cost of Welding
 Solution: Estimate the material cost for welding 2 flat pieces?
(i) Cost of filler material
(ii) Cost of O2 and C2H2 consumed
(iii) Total material cost
314
Estimation of Cost of Welding
 Solution: Estimate the material cost for welding 2 flat pieces?
The total length of weld = 160 mm = 0.16 m
(i) Cost of filler material
Filler rod required per metre of weld = 4.5 m
Actual filler rod required = 4.5 x 0.16 = 0.72 m = 72 cm
Weight of filler rod = Volume x density =
𝜋
4
0.5
2
× 72 × 7 = 98.98 gms
Hence, cost of filler material = 98.98 x 10-3 x Rs. 50 = Rs. 4.95
315
Estimation of Cost of Welding
 Solution: Estimate the material cost for welding 2 flat pieces?
The total length of weld = 160 mm = 0.16 m
(ii) Cost of O2 and C2H2 consumed
Welding time = 30 min/m
Hence, welding time for total length of weld = 0.16 x 30 = 4.8 min (or) 0.08 hours
Cost of O2 = Consumption x Cost = (0.7 x 0.08) x Rs.15 = Rs. 0.84
Cost of C2H2= Consumption x cost = (0.5 x 0.08) x Rs. 60 = Rs. 2.40
(iii) Total material cost = Cost of filler material + Cost of O2 & C2H2
= 4.95 + 0.84 + 2.40 = Rs. 8.19
316
Estimation of Cost of Welding
 Problem: A closed water tank of dimensions 50 x 50 x 50 cm is to be welded
from a metallic sheet of size 55 x 50 x 1 cm. What is the cost of material
involved if the rates of oxygen, acetylene and filler materials are Rs. 15 per cu.
metre, Rs. 60 per cu. Metre, and Rs. 50 per kg, respectively.
Find also the labour cost, overhead charges, prime cost and factory cost of making
50 such tanks, if worker gets Rs. 40 per hour. Take density of filler metal as 11.28
gm/cc.
317
Estimation of Cost of Welding
 Solution:
Given: Rate of O2 = Rs. 15/m3; Rate of C2H2 = Rs. 60/m3;
Rate of filler material = Rs. 50 per kg; n = 50; labour rate = Rs. 40 / hour;
Density = 11.28 gm/cc; n = 50 tanks; Labour rate = Rs. 40 / hour
To find:
(1) Material cost
(2) Labour cost
(3) Overheads
(4) Factory cost.
318
Estimation of Cost of Welding
 Solution:
Total length of weld = AB + BC + CD + DA + A’B’ + B’C’ + C’D’ + D’A’ + AA’ + BB’
+ CC’ + DD’ = 12 x 50 = 600 cm or 0.6 m
319
Estimation of Cost of Welding
 Solution:
Here, there is no data on consumption of O2 and C2H2 values given.
Thickness of sheet is 1 cm > 5 mm, then rightward welding technique is employed.
For 10 mm plate = 1.00 – 1.30
So assume O2 and C2H2 = 1.15 m3/hr
Dia of rod = 5 mm
Rate of welding = 1.7 – 2.0, assume = 1.85 m/hr
filler rod used per metre of weld = 4.5 metres.
320
Estimation of Cost of Welding
(i) Material cost:
a) Cost of filler material = ?
Filler rod used per 0.6 m of weld = 4.5 x 0.6 = 2.7 m
Weight of filler rod = volume of filler rod x density = 0.6 kg.
Then, actual cost of filler material = 0.6 x Rs. 50 = Rs. 30
b) Cost of O2 and C2H2 consumed
Rate of welding = 1.85 m/hr
Hence, time to weld 0.6 m length -
1
1.85
× 0.6 = 0.324 ℎ𝑟 𝑜𝑟 19.44 𝑚𝑖𝑛
Actual consumption of O2 = 1.15 m3/hr x 0.324 = 0.3726 m3
Cost of O2 = 15 x 0.3726 = Rs. 5.59
321
Estimation of Cost of Welding
Cost of C2H2 = 60 x 0.3726 = Rs. 22.36
Total material cost = 30 + 5.59 + 22.36 = Rs. 57.95
ii) Labour cost
Time to weld = 19.44 min
Assuming, edge preparation time = 80% of time to weld. = 15.55 min
Total labour time = 19.44 + 15.55 = 34.99 min.
Labour cost = Rs. 40 /hour
Hence, labour cost = Rs. 40/60 x 34.99 = Rs. 23.33
322
Estimation of Cost of Welding
iii) Overheads
Assuming, overheads = 100% of labour cost = Rs. 23.33
iv) Prime cost
Prime cost = material cost + labour cost = 57.95 + 23.33 = Rs. 81.28
v) Factory cost to make 50 tanks
Factory cost = Prime cost + Overheads = 81.28 + 23.33 = Rs. 104.61/tank
Hence, for 50 tanks = 50 x Rs. 104.61 = Rs. 5230.50
323
Estimation of Cost of Welding
Problem: Two plates each 1.2 m long and 8 mm thick are to be welded. A 60 degree Vee is
prepared by means of gas cutting before welding is to be commenced. The cost of O2 is Rs. 15/m3
and of C2H2 Rs. 60/m3. The labour charges are Rs. 40/hr. The filer material costs Rs. 50 per kg. Using
rightward technique, find the cost of cutting and welding. Take density of filer material as 10 gm/cc.
The following data is also available.
For cutting (for 10 mm thick plate)
Cutting speed = 20 m/hr
Consumption of O2 = 2 m3/hr
Consumption of C2H2 = 0.2 m3/hr
Data for rightward welding (for 8 mm thick plate)
Consumption of O2 = 0.78 m3/hr
Consumption of C2H2 = 0.8 m3/hr
Diameter of filler rod used = 4 mm
Filler rod used per metre of weld = 3.4 m
Rate of wedling = 2.25 m/hr
Fine. Cost of gas cutting and Cost of gas welding.
324
Estimation of Cost of Welding
Solution:
i) Cost of gas cutting
The length of cut = AB = BC =
8 𝑚𝑚
𝐶𝑜𝑠 30°
= 9.24 𝑚𝑚 … . 𝑠𝑎𝑦 𝑎𝑏𝑜𝑢𝑡 10 𝑚𝑚
Given: For 10 mm thick plate cutting speed = 20 m/hr
Time taken to cut 2 plates of 1.2 m length each for 60 degree Vee preparation =
1.2
20
Consumption of O2 for cutting = 2 m3/hr x 0.12 hr = 0.24 m3
Cost of O2 for cutting = 0.24 m3 x Rs. 15 = Rs. 3.6
Consumption of C2H2 for cutting = 0.2 m3/hr x 0.12 hr = 0.024m3
Cost of C2H2 for cutting = 0.024m3 x Rs. 60 = Rs. 1.44
Total material cost = 3.6 + 1.44 = Rs. 5.04
Labour cost of cutting at the rate of Rs. 40 per hour, Hence, 40 x 0.12 hr = Rs. 4.8
Therefore, total cost for cutting = 5.04 + 4.8 = Rs. 9.84
325
× 2 = 0.12 ℎ𝑟
Estimation of Cost of Welding
ii) Cost of gas welding
a) Cost of filler material
b) Cost of O2 and C2H2
c) Cost of labour
a) Cost of filler material
The length of weld = 1.2 m
Length of filler rod used = 3.4 x 1.2 = 4.76 m
𝜋
Weight of filler rod = volume of filler rod used x density= [ 4 × 10−3 2 × 4.76] × 10 × 103 =0.598 kg
4
Cost of filler rod used = 50 x 0.598 = Rs. 29.90
b) Cost of O2 and C2H2
1
1
Time taken for welding =
× 𝐿𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑤𝑒𝑙𝑑 =
× 1.2 = 0.533 ℎ𝑟
𝑅𝑎𝑡𝑒 𝑜𝑓 𝑤𝑒𝑑𝑙𝑖𝑛𝑔
2.25
Consumption of O2 for welding = 0.78 m3/hr x 0.533 hr = 0.416 m3
Cost of O2 for welding = 0.416 m3 x Rs. 15 = Rs. 6.24
Consumption of C2H2 for welding = 0.8 m3/hr x 0.533 hr = 0.4264m3
Cost of C2H2 for welding = 0.4264m3 x Rs. 60 = Rs. 25.58
Total material cost = 29.90 + 6.24 + 25.58 = Rs. 61.72
326
Estimation of Cost of Welding
c) Cost of labour
Labour cost of welding at the rate of Rs. 40/hr
Time taken for welding = 0.533 hr
Hence, labour cost = 40 x 0.533 = Rs. 21.32
Hence, Total cost of gas welding = Rs. 61.72 + Rs. 21.32 = Rs. 83.04
327
Estimation of Cost of Welding
Problem: Estimate the electric arc welding cost for a cylindrical boiler drum 3 m x 1.2 m diameter
which is to be made from 15 mm thick mild steel plates. Both the ends are closed by welding
circular plates to the drum. Cylindrical portion is welded along the longitudinal seam and welding is
done both in inner and outer sides. Assume the following data:
1. Rate of welding = 2m/hr on inner side and 2.5 m/hr on outer side.
2. Length of electrode required = 1.5 m/metre of welding
3. Cost of electrode = Rs. 20/metre
4. Power consumption = 4 kWhr/metre of weld
5. Power charges = Rs. 8/kWhr
6. Labour charges = Rs. 40/hr
7. Overhead charges = 90% of prime cost
8. Discarded electrodes = 6%
9. Fatigue and setting up time = 5% of welding time
328
Estimation of Cost of Welding
Solution:
Length of boiler = 3 m
Welding is done both in inner and outer
Diameter of boiler = 1.2 m
Hence, length of weld = 2 × 𝜋 × 𝑑𝑖𝑎. 𝑜𝑓 𝑏𝑜𝑖𝑙𝑒𝑟 + 2 𝑡𝑖𝑚𝑒𝑠 × 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑏𝑜𝑖𝑙𝑒𝑟
= (2 x 𝜋 x 1.2) + (2 x 3) = 13.54 m
i)
Electrode cost
given, Length of electrode required = 1.5 m/metre of weld
Hence, for 13.54 m weld = 1.5 x 13.54 = 20.31 m
Discarded electrodes = 6% = 6% x 20.31 = 1.22 m
Therefore, total length of electrodes required = 20.31 +1.22 = 21.53 m
Then, cost of electrodes at Rs. 20/m = 20 x 21.53 = Rs. 430.60
329
Estimation of Cost of Welding
ii) Labour cost
Given: the side plates are welded on single side and longitudinal seam is welded on both sides
Given: Rate of welding = 2m/hr on inner side and 2.5 m/hr on outer side.
Length of weld on inner side = 3 m
Therefore, time taken for inside weld =
3
2
= 1.5 ℎ𝑟𝑠
Length of weld on outside of boiler = 2 × 𝜋 × 𝑑𝑖𝑎. 𝑜𝑓 𝑏𝑜𝑖𝑙𝑒𝑟 + 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑏𝑜𝑖𝑙𝑒𝑟 = 10.54 m
Therefore, time taken for outside weld =
10.54
2.5
= 4.216 ℎ𝑟𝑠
Net time required for welding = 1.5 + 4.216 = 5.716 hrs
Given: Fatigue and setting up allowance = 5% of welding time = 5% x 5.716 = 0.286 hrs
Total welding time required = 5.716 + 0.286 = 6 hrs
Then, labour cost at Rs.40/hr = 40 x 6 = Rs. 240
330
Estimation of Cost of Welding
iii) Power cost
Given: Power consumption = 4 kWhr/metre of weld
Hence, power consumption for total length of weld = 13.54 x 4 = 54.16 kWhr
Cost of power at Rs. 8/kWhr = 8 x 54.16 = Rs. 433.28
iv) Overhead charges
Given: overhead charges = 90% of prime cost
Then, Prime cost = Cost of electrodes + Cost of labour + Cost of Power = Rs. 1103.88
Therefore, overhead charges = 90% x Rs. 1103.88 = Rs. 993.49
v) Total welding cost
Total welding cost = Prime cost + overhead charges = Rs. 2097.37
331
Sheet Metal forming
 Sheet metal forming is a grouping of many complementary processes that are
used to form sheet metal parts.
332
Shearing Operations
 Sheet metal cutting operation along a straight line between two cutting edges
 Typically used to cut large sheets
Front View
Side View
333
Shearing Operations
 Shearing – mechanical cutting of material without the formation of chips
 Curved blades may be used to produce different shapes
 Blanking
 Piercing
 Notching
 Trimming
334
Shearing Operations
 Shearing – mechanical cutting of material without the formation of chips
 The workpiece is stressed beyond its ultimate strength.
 The stresses caused in the metal by the applied forces will be shearing stresses.
335
Shearing Operations
 Shearing operation includes
 Piercing
 Blanking
 Notching
 Slitting
 Parting
 Shaving
 Trimming
336
Shearing Operations
 Punching
 It is a cutting operation by which various shaped holes are made in sheet metal.
 Hole is desired product.
 Blanking
 It is the operation of cutting a flat shape sheet metal.
 The article punched out is called the blank.
337
Shearing Operations
 Punching
 1/3 of material is cut and 2/3 of material fractures
338
Shearing Operations
 Blanking
 For thicker and softer materials generally higher angular clearance is given. In most cases, 2°
of angular clearance is sufficient.
339
Shearing Operations
 Clearance
 To small  less than optimal fracture and excessive forces
 To large  oversized burr
340
Trimming
 When parts are produced by die casting or drop forging, a small amount of extra
metal gets spread out at the parting plane.
 This extra metal, called flash, is cut off before the part is used, by an operation
called trimming.
341
Coining
 Coining is a closed die forging process, in which pressure is applied on the
surface of the forging in order to obtain closer tolerances, smoother surfaces and
eliminate draft.
 Closed die forging is a process in which forging is done by placing the work
piece between two shaped dies.
The pressure involved in coining process is about 1600Mpa.
342
Embossing
 Embossing  Similar like coining, however, embossing dies possess matching
cavity contours.
 The punch containing the positive contour
 The dies containing the negative contour
 Whereas coining dies may have quite different cavities in the two die halves
 The operation is also sometimes used for making decoration items like number plates or
name plates, jewelry, etc.
343
Coining vs Embossing
<Embossing>
<Coining>
344
Stretch Forming
 Tensile force is applied on the metal which is placed over the die
 Large deformation for ductile metal can be achieved only by this process
 Sheet is first wrapped around the block and the tensile load is increased through
jaws until sheet is plastically deformed to final shape.
345
Bending
 It is an operation by which straight length is converted to curved like drums,
channels. (Straining of the metal around a straight axis)
(a) Bending of sheet metal; (b) both compression and tensile elongation of the metal occur in bending.
346
Bending
 During the bending: the metal on the inside of the neutral plane is compressed,
while the metal on the outside is stretched.
 The metal is plastically deformed so that the bend takes a permanent set upon
removal of the stresses that caused it.
 Bending produces little or no change in the thickness of the sheet metal.
347
Bending
 Bending operations are performed using punch and die tooling.
 Types:
 V-bending, performed with a V-die
 Edge-bending, performed with a wiping die
Bending methods: (a) V-bending and (b) edge-bending; (1) before and (2) after bending. v = motion,
F = applied bending force, Fh = blank.
348
Bending Force
 Bending Force  Force required to perform bending operation
 Factors
 Geometry of the punch and die
 Strength, thickness, and length of the sheet metal
F
K bf (TS ) wt 2
D
where F = bending force, N
(TS) = tensile strength of the sheet metal, MPa
w = width of part in the direction of the bend axis, mm
t = stock thickness, mm
D = die opening dimension, mm
(a) V-die, (b) wiping die.
V-bending, Kbf = 1.33;
Edge bending, Kbf = 0.33.
349
Springback effect in Bending
350
Springback effect in Bending
 Springback effect  In bending, after plastic deformation there is an elastic
recovery this recovery is called springback.
 Low carbon steels spring back is 1– 2°, while for medium carbon steel it is 3–4°
 Compensation for spring back
 Over bending of part
 Bottoming and ironing
 Allowances in die and punch
351
Tube Forming or Bending
 Tube forming require special tooling to avoid buckling and folding.
 The oldest method of bending a tube or pipe is to pack the inside with loose
particles, commonly used sand and bend the part in a suitable fixture.
 This technique prevents the tube from buckling. After the tube has been bent,
the sand is shaken out. Tubes can also be plugged with various flexible internal
mandrels.
<Buckling>
352
Tube Forming or Bending
353
Unit 5 – ESTIMATION OF MACHINING TIMES AND COSTS
Session Description of Topic
1
2
3
4
5
6
7
Contact hrs C-D-I-O
Machine shop operations, Estimation of Machining time
Estimation of machining time for turning, knurling and facing
operations : Tutorials
Estimation of machining time for reaming, threading and tapp
ing operations : Tutorials
Estimation of machining time for drilling, boring : Tutorials
Estimation of machining time for shaping, planning : Tutorials
Estimation of machining time for milling and grinding operati
ons : Tutorials
Case studies: Estimation of cost for a product
354
IOs
Reference
1
C
4
1,4
1
C,D
4
1,2
1
C,D
4
1,2
2
2
C,D
C,D
4
4
1,2
1,2
2
C,D
4
1,2
1
C,D
4
6
Machine Shop Operations
What is unique in this process?
355
Mechanics of Metal Cutting
 A cutting tool exerts compressive force on the workpiece which stresses the work
material beyond the yield point and therefore metal deform plastically and shears
off.
356
Mechanics of Metal Cutting
 Plastic flow takes place in a localized region called the shear plane.
 Sheared material begins to flow along the cutting tool face in the form of chips.
 Applied compressive force is cutting force
357
Mechanics of Metal Cutting
What is required for machining?
 Depth of cut  pre-set interference between tool and work piece
 Feed  motion to bring in additional material for machining
 Speed  what generates the basic wedge and cuts
358
Metal Cutting
Turning
359
Metal Cutting
Drilling
360
Metal Cutting
Milling
361
Cutting Tool
 Single point cutting edge tool
 Multiple point cutting edge tool
 Left-handed or right-handed
Example: Right handed / Left handed
362
Machining Processes
363
Machining Time
 To estimate the total cost of any product involving machining operations, the
machining cost is to be estimated primarily.
1. To estimate machining cost, first need to estimate machining time required.
2. After knowing machining time and machining rate, the total machining cost can
be calculated.
364
Machining Time
 What is Machining Time?
 It is the time for which the machine works on the component, i.e., from the
time when the tool touches the work to when the tool leaves the component
after completion of operation.
 It depends on the type and extent of machining required, material being
machined, speed, feed, depth of cut and number of cuts required.
 Also, set-up time, handling time, tear down time, down time and allowances
to the workers to be considered.
365
Terms used in Machining Time
 Length of cut
𝐿𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑐𝑢𝑡 = 𝐴𝑝𝑝𝑟𝑜𝑎𝑐ℎ 𝑙𝑒𝑛𝑔𝑡ℎ + 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑤𝑜𝑟𝑘𝑝𝑖𝑒𝑐𝑒 𝑡𝑜 𝑏𝑒 𝑚𝑎𝑐ℎ𝑖𝑛𝑒𝑑 + 𝑜𝑣𝑒𝑟 𝑡𝑟𝑎𝑣𝑒𝑙
366
Terms used in Machining Time
 Feed  It is the distance, through which the tool advances into the workpiece during one
revolution of the workpiece or the cutter.
Unit: millimetres per revolution (mm/rev) or millimetres per stroke (mm/str)
 Depth of cut  It is the thickness of the layer of metal removed in one cut or pass, measured in a
direction perpendicular to the machined surface.
The depth of cut is always perpendicular to the direction of feed motion.
Unit: millimetres
367
Terms used in Machining Time
 Cutting speed  The relative speed between the tool and the job.
The relative term, since either the tool or the job or both may be moving during cutting.
Unit: metres per minute
𝐶𝑢𝑡𝑡𝑖𝑛𝑔 𝑠𝑝𝑒𝑒𝑑 = 𝑆 =
1000 𝑋 𝑆
𝑁=
𝜋𝐷
368
𝜋𝐷𝑁
1000
𝑟𝑝𝑚
𝑚/𝑚𝑖𝑛
Calculation of machining time for lathe operations
 Turning is the process of removing the excess material from the workpiece by means of a single
pointed cutting tool
𝑁=
1000 ×𝑆
𝜋𝐷
Where, S = Cutting speed in meters/min
D = Diameter of the job to be turned in mm
T = Time required for turning in minutes
f = feed per revolution in mm
L = Length of job (or stock) to be turned in mm
N = Revolutions of the job per minute in rpm
𝐿𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑡ℎ𝑒 𝑗𝑜𝑏 𝑡𝑜 𝑏𝑒 𝑡𝑢𝑟𝑛𝑒𝑑
𝑇𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛 𝑡𝑜 𝑡𝑢𝑟𝑛 𝐿 𝑚𝑚 𝑙𝑒𝑛𝑔𝑡ℎ = 𝑇 =
𝐹𝑒𝑒𝑑 𝑟𝑒𝑣 × 𝑟𝑝𝑚
𝐿
𝑇=
𝑓×𝑁
 Considering number of cuts
𝐿
𝑇=
× 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑐𝑢𝑡𝑠 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑
𝑓×𝑁
369
Calculation of machining time for lathe operations
 Here, N – revolutions of the job per minute in “rpm” to be calculated if based on mean diameter
of the job.
𝑁=
1000 ×𝑆
𝜋𝐷𝑎𝑣𝑔
𝐷𝑎𝑣𝑔 =
Where, D = Diameter of the job before turning in mm
d = Diameter of the job after turning in mm
𝐷+𝑑
2
 Considering over travel and approach
𝐴+𝐿+𝑂
𝑇=
× 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑐𝑢𝑡𝑠 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑
𝑓×𝑁
Where, O = Over travel in mm
A = Approach length in mm
370
Calculation of machining time for lathe operations
 The depth of cut should not exceed 3 mm in roughing operation and 0.75 mm in finishing.
 Number of cuts =
𝐷𝑒𝑝𝑡ℎ 𝑜𝑓 𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙 𝑡𝑜 𝑏𝑒 𝑟𝑒𝑚𝑜𝑣𝑒𝑑
𝐷𝑒𝑝𝑡ℎ 𝑜𝑓 𝑐𝑢𝑡
=
𝐷 −𝑑
2 × 𝐷𝑒𝑝𝑡ℎ 𝑜𝑓 𝑐𝑢𝑡
371
Problems
Problem#1 A 80 mm diameter shaft is to be reduced to 60 mm diameter. If the depth of cut is 2.5
mm per pass, calculate the minimum number of passes required.
Solution:
Diameter of the job before turning = D = 80 mm
Diameter of the job after turning = d = 60 mm
Depth of cut = 2.5 mm
Number of passes =
𝐷𝑒𝑝𝑡ℎ 𝑜𝑓 𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙 𝑡𝑜 𝑏𝑒 𝑟𝑒𝑚𝑜𝑣𝑒𝑑
𝐷𝑒𝑝𝑡ℎ 𝑜𝑓 𝑐𝑢𝑡
=
𝐷−𝑑
2 ×𝐷𝑒𝑝𝑡ℎ 𝑜𝑓 𝑐𝑢𝑡
372
=
90−60
2 ×2.5
= 4 𝑝𝑎𝑠𝑠𝑒𝑠
Problems
Problem#2 Estimate the machining time to turn a 3 cm diameter mild steel bar 10 cm long, down to
2.5 cm diameter in a single cut, using high speed steel tool. Assume the cutting speed of the tool to
be 30 m/min and a feed of 0.4 mm per revolutions.
Solution:
Diameter of the job before turning = D = 3 cm (30 mm)
Diameter of the job after turning = d = 2.5 cm (25 mm)
Length of job = L = 10 cm (100 mm)
Single cut = S = 30 m/min
Feed = f = 0.4 mm/rev.
𝑁=
1000 ×𝑆
𝜋𝐷
= 318.31 rpm
Machining time required = T =
𝐿
𝑓×𝑁
= 0.785 min
373
Problems
Problem#3 Estimate the total time taken to turn a 15 cm long, 2.5 cm diameter mild steel rod to a
diameter of 2.3 cm in a single cut. Take cutting speed as 30 m/min, feed 0.1 mm/rev and the
mounting time in a self-centering 3-jaw chuck as 45 sec. Neglect time taken setting up tools, etc.
Solution:
Diameter of the job before turning = D = 2.5 cm (25 mm)
Diameter of the job after turning = d = 2.3 cm (23 mm)
Length of job = L = 15 cm (150 mm)
Cutting speed = Single cut = S = 30 m/min
Feed = f = 0.1 mm/rev.
Mounting time = 45 sec
374
Problems
Solution:
N
1000 ×𝑆
=
𝜋𝐷
= 381.97 rpm
Time for turning, T =
𝐿
𝑓 ×𝑁
=
150
0.1 ×381.97
= 3.926 𝑚𝑖𝑛
Given, The job mounting time = 45 sec (or) 0.75 min
Hence, total machining time = time for turning + job mounting time = 4.676 min
375
Problems
Problem#4 A mild steel bar is turned on a lathe, the cut being taken 0.6 cm deep
and feed/rev as 0.15 cm. Assume the cutting speed 25 m/min, find the time
required to reduce the bar from 25 to 20 cm diameter. Length is taken as 40 cm.
Solution:
Depth of cut = 0.6 cm (6 mm); f = 0.15 cm/rev or 1.5 mm/rev; S = 25 m/min;
D = 25 cm or 250 mm; d = 20 cm or 200 mm; L = 40 cm or 400 mm.
Time required to turn?
376
Problems
Solution:
Depth of cut = 0.6 cm (6 mm); f = 0.15 cm/rev or 1.5 mm/rev; S = 25 m/min;
D = 25 cm or 250 mm; d = 20 cm or 200 mm; L = 40 cm or 400 mm.
Time required to turn?
Number of passes =
𝑁=
1000 ×𝑆
𝜋𝐷
=
𝐷𝑒𝑝𝑡ℎ 𝑜𝑓 𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙 𝑡𝑜 𝑏𝑒 𝑟𝑒𝑚𝑜𝑣𝑒𝑑
𝐷𝑒𝑝𝑡ℎ 𝑜𝑓 𝑐𝑢𝑡
1000 ×𝑆
𝜋𝐷
1000 ×25
𝜋×250
=
Time required for turning = T =
𝐿
𝑓×𝑁
=
𝐷 −𝑑
2 ×𝐷𝑒𝑝𝑡ℎ 𝑜𝑓 𝑐𝑢𝑡
=
250 −200
2×6
= 4.166 𝑜𝑟 5 𝑝𝑎𝑠𝑠𝑒𝑠
= 31.83 𝑟𝑝𝑚
× 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑐𝑢𝑡𝑠 =
377
400
1.5 ×31.83
× 5 = 41.89 𝑚𝑖𝑛
Problems
Problem#5 What is the machining time to turn the dimensions given in Fig. The
material is brass, the cutting speed with HSS tool being 60 m/min, and the feed is
7.5 m/rev., depth of cut is 3 mm per pass.
378
Problems
Solution:
S = 60 m/min; f = 7.5 mm/rev; Depth of cut = 3 mm per pass
Machining time required to turn?
Step 1: For turning: dia 60 mm to dia 40 mm with L = 95 mm
=
60 −40
2 ×3
Time required for turning = T1 =
𝐿1
𝑋
𝑓 𝑋 𝑁1
Number of pass =
1000 𝑆
𝜋𝐷
N1=
=
𝐷 −𝑑
2 ×𝐷𝑒𝑝𝑡ℎ 𝑜𝑓 𝑐𝑢𝑡
1000 𝑋 60
𝜋𝑋60
= 3.33 𝑜𝑟 4 𝑝𝑎𝑠𝑠𝑒𝑠
= 318.3 𝑟𝑝𝑚
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑝𝑎𝑠𝑠𝑒𝑠 =
379
95
𝑋
7.5 𝑋 318.3
4 = 0.159 𝑚𝑖𝑛
Problems
Solution:
S = 60 m/min; f = 7.5 mm/rev; Depth of cut = 3 mm per pass
Machining time required to turn?
Step 2: For turning: dia 40 mm to dia 30 mm with L = 55 mm
Number of pass =
1000 𝑆
𝜋𝐷
N2=
=
𝐷 −𝑑
2 ×𝐷𝑒𝑝𝑡ℎ 𝑜𝑓 𝑐𝑢𝑡
1000 𝑋 60
𝜋𝑋40
=
40 −30
2 ×3
= 1.66 𝑜𝑟 2 𝑝𝑎𝑠𝑠𝑒𝑠
= 477.46 𝑟𝑝𝑚
Time required for turning = T2 =
𝐿2
𝑋
𝑓 𝑋 𝑁2
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑝𝑎𝑠𝑠𝑒𝑠 =
Hence, total turning time T = T1 + T2 = 0.1897 min
380
55
𝑋
7.5 𝑋 477.46
2 = 0.0307 𝑚𝑖𝑛
Problems
Problem#6 Calculate the machining time of a centrifugal pump shaft shown in Fig.
The bar stock diameter is 75 mm and finished dimensions are shown in sketch and
all are in mm. Assume cutting speed as 350 rpm for a depth of cut 2 mm and
feed/revolution as 0.2 mm.
381
Problems
Solution: N = 350 rpm; f = 0.2 mm/rev; Depth of cut = 2 mm
Machining time required to turn?
Step 1: For turning: dia 75 mm to dia 70 mm with L = 170 mm
Number of cuts =
75−70
2𝑋2
Time required = T1 =
= 1.25 𝑜𝑟 2 𝑝𝑎𝑠𝑠𝑒𝑠
𝐿1
𝑋
𝑓𝑋𝑁
𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑝𝑎𝑠𝑠𝑒𝑠 =
170
𝑋
0.2 𝑋 350
2 = 4.857 𝑚𝑖𝑛
Step 2: For turning: dia 70 mm to dia 35 mm with L = 110 mm
Number of cuts =
70−35
2𝑋2
Time required = T2 =
= 8.75 𝑜𝑟 9 𝑝𝑎𝑠𝑠𝑒𝑠
𝐿2
𝑋
𝑓𝑋𝑁
𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑝𝑎𝑠𝑠𝑒𝑠 =
110
𝑋
0.2 𝑋 350
9 = 14.14 𝑚𝑖𝑛
Hence, total turning time T = T1 + T2 = 18.997 min
382
Problems
Problem#7 A 25 cm diameter bar is revolving at 65 rpm. How much time will take to
machine 45 degree by 3.125 mm chamfer as shown in fig. And feed is 0.3 mm.
Solution: D = 25 cm or 250 mm; N = 65 rpm; L = 3.125 mm; f = 0.3 mm
Time for chamfering (T)
𝐿
=
𝑓𝑋𝑁
=
3.125
0.3 𝑋 65
= 0.16 𝑚𝑖𝑛
383
Problems
Problem#8 A mild steel bolt 24 mm diameter is rotating at 370 rpm. How much
machining time is required to make 45 deg chamfer by 8 mm. Take feed as 0.25
mm/rev.
Solution: D = 24 mm; N = 370 rpm; f = 0.25 mm/rev
L = ? L = AB
Sin 45 =
8
𝐴𝐵
; therefore AB = 11.314 mm
𝐿
𝑓𝑋𝑁
Time for chamfering (T) =
=
11.314
0.25 𝑋 370
= 0.153 𝑚𝑖𝑛
384
Problems
385
Problems
Problem#9 Estimate the machining time to face both ends of a workpiece of 30 mm
steel rod in one cut, if the cutting speed is 30 m/min and cross feed as 0.2 mm/rev.
Solution: D = 30 mm; S = 30 m/min; f = 0.2 mm/rev
L = D/2 = 30/2 = 15 mm
N=
1000 𝑆
𝜋𝐷
=
1000 ×30
𝜋×30
= 318.31 𝑟𝑝𝑚
Time for facing on one end, T =
𝐿
𝑓𝑋𝑁
= 0.2356 𝑚𝑖𝑛.
Hence, total time for facing both ends = 2 x 0.2356 = 0.471 min
386
Problems
Problem#10 Calculate the machining time to face on a cast iron flange shown in Fig.
Assume speed of rotation of the job as 80 rpm and cross feed as 0.6 mm/rev.
Solution: N = 80 rpm; f = 0.6 mm/rev
1
2
1
2
L = (𝐷 − 𝑑) = (150 − 80) = 35 mm
Time for facing on one end, T =
𝐿
𝑓𝑋𝑁
= 0.729 𝑚𝑖𝑛.
387
Problems
Problem#11 A taper gauge as shown in Fig. is to be knurled by a high speed
knurling tool. Assuming cutting speed of 25 m/min and a feed of 0.3 mm/rev., find
the time required for knurling.
Solution: S = 25 m/min; f = 0.3 mm/rev.
N=
1000 𝑆
𝜋𝐷
=
1000 ×25
𝜋×22
= 361.71 𝑟𝑝𝑚
Time for knurling, T =
𝐿
𝑓𝑋𝑁
=
105
0.3 𝑋 361.71
= 0.9676 𝑚𝑖𝑛.
388
Problems (Drilling)
Problem#12 Find the time required to drill 6 holes in a casted flange each of 1 cm
depth, if the hole diameter is 1.5 cm. Assume cutting speed as 20 m/min and feed
as 0.02 cm/rev.
Solution: S =20 m/min; f = 0.2 mm/rev; D = 15 mm ; L = 10 mm
𝑁=
1000 𝑆
𝜋𝐷
= 424.41 rpm
𝐿
Time for drilling = T =
= 0.118 min ie., for 6 holes = 0.118 X 6 = 0.708 min
𝑓𝑋𝑁
389
Problems (drilling & boring)
Problem#13 Calculate the time required for drilling a component as shown in fig.
assume the cutting speed as 30 m/min and feed as 0.25 mm/rev.
Solution: S =30 m/min; f = 0.25 mm/rev;
Step 1: Time for drilling of 100 mm for dia 15 mm
Ans: N = 636.62 rpm & Time = 0.628 min
Step 2: Time for boring of 40 mm length for dia 25 mm
Ans: N = 381.97 rpm & Time = 0.419min
Total time = 1.047 min
390
Tapping
 It is the operation of cutting internal threads with the help of a tool called tap.
 Formula used: The tap has to cut its own lead before it actually starts tapping. For practical
purposes, the lead is usually taken as half the nominal diameter of the tap used.
Actual tapped length = L + (D/2)
Time taken for tapping =
𝐿𝑒𝑛𝑔𝑡ℎ 𝑡𝑟𝑎𝑣𝑒𝑙𝑙𝑒𝑑 𝑏𝑦 𝑡𝑎𝑝
𝐹𝑒𝑒𝑑
𝑟𝑒𝑣
𝑋 (𝑟𝑝𝑚)
=
𝐿+
𝐷
2
𝑃𝑖𝑡𝑐ℎ 𝑋 (𝑟𝑝𝑚)
Time for returning the tap is taken as ½ of time required to push.
Hence, total time for tapping =
3
(𝐿
2
𝐷
+ 2)
𝑃𝑖𝑡𝑐ℎ 𝑋 (𝑟𝑝𝑚)
If, Thread Per Centrimeter (T.P.C) given. Then Pitch =
391
1
𝑃.𝑇.𝐶
Problems (tapping)
Problem#14 Estimate the time required for tapping a hole with 25 mm tap (3 mm
pitch tap) to a length of 50 mm. For return stroke the speed is 1.5 times the cutting
speed. Take the cutting speed as 8 m/min.
Solution: S =8 m/min; D =25 mm; Pitch = 3 mm; L = 50 mm.
𝑁=
1000 𝑆
𝜋𝐷
= 101.86 rpm
Time taken for tapping =
𝐿+
𝐷
2
25
50 + 2
=
𝑃𝑖𝑡𝑐ℎ 𝑋 (𝑟𝑝𝑚) 3 𝑋 101.86
= 0.2045 𝑚𝑖𝑛
Return stroke speed = 1.5 times the cutting speed.
Hence, return time =
1
𝑋
𝑟𝑒𝑡𝑢𝑟𝑛 𝑠𝑡𝑜𝑟𝑘𝑒 𝑠𝑝𝑒𝑒𝑑
𝑇𝑎𝑝𝑝𝑖𝑛𝑔 𝑡𝑖𝑚𝑒 = 0.1363 min
392
So total time for tapping = 0.2045 + 0.1363
= 0.3408 min
Thread cutting or Threading or Screw Cutting
 It is the process of removing material to produce helix on external or internal circular surface for
fastening purposes.
 Time for threading =
𝐿𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑐𝑢𝑡
𝑃𝑖𝑡𝑐ℎ 𝑜𝑟 𝑙𝑒𝑎𝑑 𝑋 𝑁
𝑋 𝑁𝑜. 𝑜𝑓 𝑐𝑢𝑡𝑠
Pitch for a single start thread = Pitch of the thread
Pitch for a multi-start thread = Pitch x Number of starts.
 Number of cuts
=
25
𝑇ℎ𝑟𝑒𝑎𝑑𝑠 𝑝𝑒𝑟 𝑐𝑚
𝑓𝑜𝑟 𝑒𝑥𝑡𝑒𝑟𝑛𝑎𝑙 𝑡ℎ𝑟𝑒𝑎𝑑𝑠
=
32
𝑇ℎ𝑟𝑒𝑎𝑑𝑠 𝑝𝑒𝑟 𝑐𝑚
𝑓𝑜𝑟 𝑖𝑛𝑡𝑒𝑟𝑛𝑎𝑙 𝑡ℎ𝑟𝑒𝑎𝑑𝑠
393
Problems (threading)
Problem#15 Estimate the time required for cutting 3 mm pitch threads on a mild
steel bar of 2.8 cm dia and 8 cm long. Assume the cutting speed for threading as
15 m/min.
Solution: S = 15 m/min; Pitch = 3 mm or 0.3 cm; D = 28 mm; L = 80 cm.
Ans: N =
1000 𝑆
𝜋𝐷
=170.52 rpm
Threads per cm =
1
𝑃𝑖𝑡𝑐ℎ
Number of cuts =
25
𝑇ℎ𝑟𝑒𝑎𝑑𝑠 𝑝𝑒𝑟 𝑐𝑚
Time for threading =
= 3.333
𝑓𝑜𝑟 𝑒𝑥𝑡𝑒𝑟𝑛𝑎𝑙 𝑡ℎ𝑟𝑒𝑎𝑑𝑠 = 7.5 say 8.
𝐿𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑐𝑢𝑡
𝑃𝑖𝑡𝑐ℎ 𝑜𝑟 𝑙𝑒𝑎𝑑 𝑋 𝑁
𝑋 𝑁𝑜. 𝑜𝑓 𝑐𝑢𝑡𝑠= 1.25 min
394
Problems (Threading)
Problem#16 Estimate the threading time on a lathe to cut square thread with the following assumptions:
Length of thread = 260 mm;
Number of thread = 3 per cm;
Cutting speed (S) = 25 m/min;
Diameter of thread = 43.75 mm;
Depth of thread = 1.64 mm;
Depth of cut = 0.125 mm/pass.
Solution: N =
1000 𝑆
𝜋𝐷
=181.89 rpm
𝐷𝑒𝑝𝑡ℎ 𝑜𝑓 𝑡ℎ𝑟𝑒𝑎𝑑
𝐷𝑒𝑝𝑡ℎ 𝑜𝑓 𝑐𝑢𝑡
Number of cuts =
Pitch =
1
Threads per cm
=
1.64
0.125
= 13.12 𝑜𝑟 14..
= 0.333 cm
Hence, threading time =
𝐿𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑐𝑢𝑡
𝑃𝑖𝑡𝑐ℎ 𝑜𝑟 𝑙𝑒𝑎𝑑 𝑋 𝑁
𝑋 𝑁𝑜. 𝑜𝑓 𝑐𝑢𝑡𝑠 = 6 min.
395
Problems
Problem#17 A mild streel bar 120 mm long and 40 mm in diameter is turned to 38
mm diameter and was again turned to a diameter of 35 mm over a length of 50
mm as shown in fig. The bar was chamfered at both the ends to give a chamfer of
45deg x 4 mm after facing. Calculate the machining time. Assume cutting speed of
50 m/min and feed 0.3 mm/rev. The depth of cut is not to exceed 3 mm in any
operations.
396
Problems 17
Solution: S = 50 m/min; f = 0.3 mm/rev; Depth of cut = 3 mm; D = 40 mm;
Step 1: Turning
N1 = 1000 𝑆 𝜋𝐷 = 397.89 rpm
T1 = L f N = 1.005 min
Step 2: Turning
N2 = 1000 𝑆 𝜋𝐷 = 418.83 rpm
T2 = L f N = 0.4 min
Step 3: Facing
Length of cut for facing = 38/2 = 19 mm; D = 38 mm; S = 50 m/min
N3 = 418.83 rpm
T3 = L f N = 0.1512 min per one end, then
397 for both ends = 0.302 min
Problems 17
Step 4: Chamfering
Chamfering 45 deg x 4 mm on both ends
L = 4 mm; D = 38 mm; S = 50 m/min
N4 = 1000 𝑆 𝜋𝐷 = 418.83 rpm
T4 = L f N = 0.032 min per one side.
Hence for both sides = 0.032 x 2 = 0.064 min
Total machining time = 1.771 min
398
Problem 18
Calculate the machining time required to produce one piece of the component
shown in figure starting from 25 mm bar. The following data is available.
For turning: Cutting speed = 40 m/min; Feed = 0.4 mm/rev; Depth of cut = 2.5
mm/per pass. For thread cutting: Cutting speed = 8 m/min
399
Problem 18
Solution:
Step 1: Turing 25 mm dia to 15 mm dia for the length of 50 mm.
N1 = 1000 𝑆 𝜋𝐷 = 509.3 rpm
Number of cuts = (D-d)/(2 times depth of cut) = 2
T1 = (L f N) 𝑋 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑐𝑢𝑡𝑠 = 0.491 min
Step 2: Turing 15 mm dia to 10 mm dia for the length of 30 mm.
N2 = 1000 𝑆 𝜋𝐷 = 848.82 rpm
No. of cuts = (D-d)/(2 times depth of cut) = 1
T2 = (L f N) 𝑋 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑐𝑢𝑡𝑠 = 0.088 min
400
Problem 18
Solution:
Step 3: Threading M10 x 1.5; L = 20 mm; Pitch = 1.5 mm; D = 10 mm; S =8 m/min
N3 = 1000 𝑆 𝜋𝐷 = 254.65 rpm
Thread per cm = 1/Pitch = 1/1.5 = 6.666
Number of cuts = 25/Threads per cm =25/6.666 =3.75 or 4.
T3 = (L Pitch X N) 𝑋 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑐𝑢𝑡𝑠 = 0.209 min
Hence, total machining time = 0.788 min
401
Problem 19
A 60 mm rod of aluminium to be machined on a lathe, the finished size is shown in
figure. The length of rod is 175 mm. Determine the total machining time and
material cost, if material is purchased at the rate of R. 110 per kg. Assuming cutting
speed 30 m/min and feed 0.2 mm/rev. Take density of aluminium as 2.7 gm/cc.
Depth of cut not to exceed 2.5 mm.
402
Problem 19
Solution: S = 30 m/min; f = 0.2 mm/rev; density = 2.7 gm/cc. Estimate total
machining time and material cost?
Step 1: Facing both ends to reduce from 175 mm to 165 mm.
Length of cut = L/2 = 60/2 = 30 mm; D = 60
N1 = 1000 𝑆 𝜋𝐷 = 159.15 rpm
T1 = (L f N) = 0.942 min for one end, 1.884 min for both ends.
Step 2: Turning from 60 mm dia to 55 mm dia for 165 mm.
N2 = 1000 𝑆 𝜋𝐷 = 159.15 rpm
Number of cuts = (D-d)/(2 times depth of cut) =1
T2 = (L f N) 𝑋 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑐𝑢𝑡𝑠 = 5.184 min
403
Problem 19
Step 3: Turning from 55 mm dia to 20 mm dia for 20 mm length.
N3 = 1000 𝑆 𝜋𝐷 = 173.62 rpm
Number of cuts = (D-d)/(2 times depth of cut) =7
T3 = (L f N) 𝑋 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑐𝑢𝑡𝑠 = 4.032 min
Step 4: Grooving for 5 mm length. Here, N = 159.15 rpm
T4 = (L f N) = 0.157 min
Step 5: Taper turning from 55 mm dia to 20 mm dia for 30 mm length.
N5 = 1000 𝑆 𝜋𝐷 = 173.62 rpm
Number of cuts = (D-d)/(2 times depth of cut) =7
T5 = (L f N) 𝑋 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑐𝑢𝑡𝑠 = 6.047 min
404
Problem 19
Total time for machining = 17.304 min
Material cost: Total weight x cost of material per kg
Total weight = Total volume x Density
Total volume = Volume of cylinders + Volume of frustum
𝜋
4
𝜋
4
𝜋𝑋30
(202
3
=[ (552 ) . 155]+ [ (202 ) . 20]+[
+ 552 − 20𝑋55] = 281828.57 mm3
Total weight = 281828.57 X 2.7 X 10-6 = 0.761 kg
Hence, material cost = 0.61 x Rs. 110 = Rs. 83.71
405
Problem 20
A mild steel shaft, shown in figure is to be turned from a 36 mm diameter bar. The
complete machining consists of the following steps:
i) Facing 36 mm dia on both sides.
ii) Turning to 30 mm dia
iii) Drilling 10 mm hole
iv) Knurling.
With H.S.S. tool the cutting speed is 60 m/min. The feed for facing 0.2 mm/rev, feed
for knurling 0.3 mm/rev and feed for drilling is 0.08 mm/rev. Depth of cut should
not exceed 2.5 mm in any operation. Find the machining time to finish the job.
406
Problem 20
Solution:
Step 1.Facing 36 mm dia on both sides.
S = 60 m/min; f = 0.2 mm/rev; D= 36 mm; L = 36/2 = 18 mm
N1 = 1000 𝑆 𝜋𝐷 = 530.52 rpm
T1 = (L f N) = 0.17 min for one end, 0.34 min for both ends
Step 2. Turning from 36 mm dia to 30 mm dia.
S = 60 m/min; f = 0.3 mm/rev; D= 36 mm; d = 30 mm; L = 58 mm
Number of cuts = (D-d)/(2 times depth of cut) =1.2 or 2.
N2 = 1000 𝑆 𝜋𝐷 = 530.52 rpm
T2 = (L f N) = 0.364 min
407
Problem 20
Step 3.Drilling 10 mm dia for 35 mm length.
S = 60 m/min; f = 0.08 mm/rev; D= 10 mm; L = 35 mm
N3 = 1000 𝑆 𝜋𝐷 = 1909.86 rpm
T3 = (L f N) = 0.23 min
Step 4. Knurling for 12 mm length
S = 60 m/min; f = 0.3 mm/rev; D= 36 mm; L = 12 mm
N4 = 1000 𝑆 𝜋𝐷 = 530.52 rpm
T4 = (L f N) = 0.075 min
Total machining time = 1.009 min
408
Estimation of Machining Time for Shaping, Planning
 Shaping and Planning are carried out on reciprocating machines with a single
point cutting tool.
 Cutting action takes places only in the forward stroke and the return stroke is idle
stroke. So to reduce the total time, the time required for idle stroke is reduced by
increasing the speed of idle stroke.
409
Estimation of Machining Time for Shaping, Planning
 Formula used:
1)
Effective cutting speed (C) =
𝐿
1000
𝑋 𝑁 m/min
Where, L = Length of forward stroke in mm (including clearance on each ends)
N = Number of forward strokes/min.
2) Cutting speed (S) =
𝐾=
𝐿 (1+𝐾)
1000
𝑋 𝑁 m/min
𝑇𝑖𝑚𝑒 𝑓𝑜𝑟 𝑟𝑒𝑡𝑢𝑟𝑛 𝑠𝑡𝑟𝑜𝑘𝑒
𝑇𝑖𝑚𝑒 𝑓𝑜𝑟 𝑓𝑜𝑟𝑤𝑎𝑟𝑑 𝑠𝑡𝑟𝑜𝑘𝑒
3) Total time for one cut (T) =
𝐿
𝑆 𝑋 1000
+
𝐿𝑋𝐾
𝑆 𝑋 1000
=
𝐿 (1+𝐾)
𝑆 𝑋 1000
Now, if w = width of the job in mm, and f = feed per stroke.
Then, the number of double strokes required to complete one cut on full length = w/f
𝐿 (1+𝐾)
Hence, total time for completing one cut = T =𝑆 𝑋 1000 𝑋
𝑤
𝑓
4) Total time required, If p = Number of cuts (or passes)410required, then T =
𝐿 (1+𝐾)
𝑆 𝑋 1000
𝑋
𝑤
𝑓
𝑋𝑝
Problem 21 (Shaping operations)
Find the time required on the shaper to complete one cut on a plate 600 x 900 mm
(or 900 x 600 mm). If the cutting speed is 6 m/min. The return time to cutting time
ratio is 1:4 and the feed is 2 mm/stroke. The clearance at each end along the length
is 75 mm.
Solution: Plate size 600 x 900 mm; S = 6 m/min; f = 2 mm/stroke; clearance at each
side = 75 mm; K = 1:4 = 0.25
Length of storke = Length of plate + clearances = 900 + 2 x 75 = 1050 mm
T𝑖𝑚𝑒 𝑡𝑜 𝑐𝑜𝑚𝑝𝑙𝑒𝑡 𝑜𝑛𝑒 𝑐𝑢𝑡 𝑜𝑛 𝑓𝑢𝑙𝑙 𝑙𝑒𝑛𝑔𝑡ℎ =
𝐿 (1+𝐾) 𝑤
X
𝑆 𝑋 1000 𝑓
Note: Either 600 X 900 or 900 X 600, maximum value should be taken411
as Length
= 65.625 min
Problem 22 (Shaping operations)
Determine the time required to shape a block 400 x 150 mm on a shaper working with a cutting
speed of 12 m/min and cross feed of 0.85 mm/stroke. Ratio of return stroke to cutting stroke speed
is 3:2. Take allowance as 25 mm on each side on length and 5 mm on each side on width.
Solution: Plate size 400 x 150 mm; S = 12 m/min; f = 0.85 mm/stroke; clearance at each side = 25
mm for length & 5 mm for width; K = 3:2 (speed) so convert into time = 2:3 (time) = 0.666
Length of stroke = Length of plate + clearances = 400 + 2 x 25 = 550 mm
Width = width of the block + clearances = 150 + 2 x 15 = 180 mm
T𝑖𝑚𝑒 𝑡𝑜 𝑐𝑜𝑚𝑝𝑙𝑒𝑡 𝑜𝑛𝑒 𝑐𝑢𝑡 𝑜𝑛 𝑓𝑢𝑙𝑙 𝑙𝑒𝑛𝑔𝑡ℎ =
412
𝐿 (1+𝐾) 𝑤
X
𝑆 𝑋 1000 𝑓
= 13.23 min
Problem 23 (Shaping operations)
Find the time required on a shaper to machine a plate 1100 X 500 mm, if the cutting speed is 16
m/min. The ratio of return stroke time is 2:3. The clearance at each end is 20 mm along the length
and 15 mm on width. Two cuts are required, one roughing cut with cross feed of 2 mm per stroke
and one finishing cut with feed of 1.25 mm per stroke.
Solution: S = 16 m/min; K (time = 2: 3 = 2/3 = 0.666; Feed for rough = 2 mm/stroke; Feed for finish
= 1.25 mm/stroke.
Length of stroke = Length of plate + Clearances = 1100 + 2 x 20 = 1140 mm
Cross travel of table = Width of plate + Clearances = 500 + 2 x 15 = 530 mm
Since, rough and finishing cut required, Time for one complete stroke =
413
𝐿 (1+𝐾)
𝑆 𝑋 1000
= 0.1187 min
Problem 23 (Shaping operations)
Number of strokes for rough cut =
𝐶𝑟𝑜𝑠𝑠 𝑡𝑟𝑎𝑣𝑒𝑙 𝑜𝑓 𝑡𝑎𝑏𝑙𝑒
Number of strokes for Finish cut =
𝐶𝑟𝑜𝑠𝑠 𝑡𝑟𝑎𝑣𝑒𝑙 𝑜𝑓 𝑡𝑎𝑏𝑙𝑒
𝐹𝑒𝑒𝑑
𝑆𝑡𝑟𝑜𝑘𝑒
𝐹𝑒𝑒𝑑
𝑆𝑡𝑟𝑜𝑘𝑒
= 530/2 = 265 no’s
= 530/1.25 = 424 no’s
Hence, total number of stroke required = 265 + 424 = 689 no’s
Total machining time = 689 X 0.1187 = 81.79 min.
414
Problem 24 (Planning operations)
Estimate the planning time for a casting 1.25 m long and 0.5 m wide which is machined on a planner
having cutting speed of 12 m/min and a return speed of 30 m/min. Two cuts are required: one
roughing with a depth of 3.125 mm and a feed of 0.1 mm/rev and other finishing with a depth of
0.125 mm and using a feed of 0.125 mm.
Solution:
Length of stroke = L = Length of casting + Approach + over-run.
= 1250 mm + 25 mm + 25 mm = 1300 mm….(approach & over-run assume based on length of the
workpiece.)
Cross feed = w = Width of casting + approach + over-run
= 500 + 15 + 15 = 530 mm….. ….(approach & over-run assume based on width of the workpiece.)
𝑅𝑒𝑡𝑢𝑟𝑛 𝑠𝑡𝑟𝑜𝑘𝑒 𝑡𝑖𝑚𝑒
𝐶𝑢𝑡𝑡𝑖𝑛𝑔 𝑠𝑡𝑟𝑜𝑘𝑒 𝑡𝑖𝑚𝑒
K=
=
𝐶𝑢𝑡𝑡𝑖𝑛𝑔 𝑠𝑡𝑟𝑜𝑘𝑒 𝑠𝑝𝑒𝑒𝑑
𝑅𝑒𝑡𝑢𝑟𝑛 𝑠𝑡𝑟𝑜𝑘𝑒 𝑠𝑝𝑒𝑒𝑑
=
12
30
= 0.4
415
Problem 24 (Planning operations)
𝑅𝑒𝑡𝑢𝑟𝑛 𝑠𝑡𝑟𝑜𝑘𝑒 𝑡𝑖𝑚𝑒
𝐶𝑢𝑡𝑡𝑖𝑛𝑔 𝑠𝑡𝑟𝑜𝑘𝑒 𝑡𝑖𝑚𝑒
K=
=
𝐶𝑢𝑡𝑡𝑖𝑛𝑔 𝑠𝑡𝑟𝑜𝑘𝑒 𝑠𝑝𝑒𝑒𝑑
𝑅𝑒𝑡𝑢𝑟𝑛 𝑠𝑡𝑟𝑜𝑘𝑒 𝑠𝑝𝑒𝑒𝑑
=
Hence, time for one complete stroke =
12
30
= 0.4
𝐿 (1+𝐾)
𝑆 𝑋 1000
= 0.152 min
Number of strokes for rough cut =
𝐶𝑟𝑜𝑠𝑠 𝑡𝑟𝑎𝑣𝑒𝑙 𝑜𝑓 𝑡𝑎𝑏𝑙𝑒
Number of strokes for Finish cut =
𝐶𝑟𝑜𝑠𝑠 𝑡𝑟𝑎𝑣𝑒𝑙 𝑜𝑓 𝑡𝑎𝑏𝑙𝑒
𝐹𝑒𝑒𝑑
𝑆𝑡𝑟𝑜𝑘𝑒
𝐹𝑒𝑒𝑑
𝑆𝑡𝑟𝑜𝑘𝑒
= 530/0.1 = 5300 no’s
= 530/0.125 = 4240 no’s
Hence, total number of stroke required = 5300 + 4240 = 9540 no’s
Total machining time = 9540 X 0.152 = 1450 min.
416
Estimation of Time for Milling Operations
 Milling is the operation of removing material from a surface by using a rotatory multipoint tool
called cutter.
 Various operations can be done including profile milling, slotting, key way cutting, indexing
operations, etc.
Formula used:
𝐿
1. Milling Time (T) =
𝑓𝑁
N=
1000 𝑋 𝑆
𝜋𝐷
in sec
in rpm
Where, S – cutting speed in m/min, and D = cutter diameter in mm.
417
Estimation of Time for Milling Operations
Formula used:
2. To find feed/rev: Since the milling cutter is a multi-point cutter, the feed will be
given by
Feed per revolution = Feed per tooth X Number of cutter teeth
3. To find time taken per cut:
Time taken per cut =
𝐿𝑒𝑛𝑔ℎ𝑡 𝑜𝑓 𝑐𝑢𝑡 (𝑇𝑜𝑡𝑎𝑙 𝑡𝑎𝑏𝑙𝑒 𝑡𝑟𝑎𝑣𝑒𝑙)
𝐹𝑒𝑒𝑑
𝑟𝑒𝑣
𝑋(𝑟𝑝𝑚 𝑜𝑓 𝑐𝑢𝑡𝑡𝑒𝑟)
Total table travel = Length of job + Cutter approach + Over-travel
So time take per cut = =
𝐿𝑒𝑛𝑔ℎ𝑡 𝑜𝑓 𝑗𝑜𝑏+𝑐𝑢𝑡𝑡𝑒𝑟 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ+𝑜𝑣𝑒𝑟𝑡𝑟𝑎𝑣𝑒𝑙
𝐹𝑒𝑒𝑑
𝑟𝑒𝑣
𝑋(𝑟𝑝𝑚 𝑜𝑓 𝑐𝑢𝑡𝑡𝑒𝑟)
418
Estimation of Time for Milling Operations
 Determination of Added Table Travel (i.e., Cutter approach + Over travel)
419
Estimation of Time for Milling Operations
 Determination of Added Table Travel (i.e., Cutter approach + Over travel)
420
Estimation of Time for Milling Operations
 Determination of Added Table Travel (i.e., Cutter approach + Over travel)
421
Estimation of Time for Milling Operations
 Determination of Added Table Travel (i.e., Cutter approach + Over travel)
422
Problem 25 (Face Milling operations)
A face milling cutter of 150 mm diameter is used to give a cut on a block 500 mm x 250 mm. The
cutting speed is 50 m/min and feed 0.2 mm/revolution. Calculate the time required to complete one
cut.
Solution: D = 150 mm; S = 50 m/min; F = 0.2 mm/rev.
N
=
1000 𝑆
= 106.1 rpm.
𝜋𝐷
Length of the job = 500 mm; Width of the job = 250 mm.
Here, the cutter diameter (D) < width (w) of the job. i.e.,150 mm < 250 mm
Hence, Approach = 0.5 D; and Over-travel = 7 mm  Assumptions
So, added table travel = 0.5 D + 7 mm = 0.5 X 150 + 7 = 82 mm
𝐿𝑒𝑛𝑔ℎ𝑡 𝑜𝑓 𝑗𝑜𝑏+𝑐𝑢𝑡𝑡𝑒𝑟 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ+𝑜𝑣𝑒𝑟𝑡𝑟𝑎𝑣𝑒𝑙
Therefore, milling time/cut =
𝐹𝑒𝑒𝑑
𝑟𝑒𝑣
𝑋(𝑟𝑝𝑚 𝑜𝑓 𝑐𝑢𝑡𝑡𝑒𝑟)
423
=
500+82
0.2 𝑋 106.1
= 27.43 𝑚𝑖𝑛
Problem 26 (Face Milling operations)
A 20cm x 5 cm C. I. surface is to be faced on milling machine with a cutter having a dia of 10 cm
and 16 teeth. If the cutting speed and feed are 50 m/min and 5 cm/min respectively, determine the
milling time, rpm of the cutter and feed per tooth.
Solution: L = 200 mm; w = 50 mm; S = 50 m/min; f = 50 mm/min; D = 100 mm; number of cutter
teeth = 16
1.
To find cutter rpm: N =
2.
To find feed per tooth:
1000 𝑆
𝜋𝐷
= 159.15 rpm
𝐹𝑒𝑒𝑑 𝑝𝑒𝑟 𝑚𝑖𝑛
𝑟𝑝𝑚 𝑋 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑡𝑒𝑒𝑡ℎ
=
50
159.15 𝑋 16
= 0.0196 𝑚𝑚/𝑡𝑜𝑜𝑡ℎ
1
2
Here, D > w, ie., 100 mm > 50 mm, So, the added table travel = [𝐷 − 𝐷 2 − 𝑤 2 = 6.698 mm
1.
Milling time (T) =
𝐿𝑒𝑛𝑔ℎ𝑡 𝑜𝑓 𝑗𝑜𝑏+𝐴𝑑𝑑𝑒𝑑 𝑡𝑎𝑏𝑙𝑒 𝑡𝑟𝑎𝑣𝑒𝑙
𝐹𝑒𝑒𝑑 𝑝𝑒𝑟 𝑚𝑖𝑛
= 4.134 min..(here, Feed per min used because D >
w, so only one cut is sufficient)
424
Problem 27 (Slot Milling operations)
A 3 cm deep slot is to be milled with a 8 cm diameter cutter. The length of the slot is 30 cm. What
will be the total table travel to complete the cut? If the cutting speed is 20 metres/min and feed per
tooth is 0.2 mm, estimate the milling time. The cutter has 24 teeth and one cut is sufficient for the
slot.
Solution: Depth of cut, d = 3 cm or 30 mm; Length of the slot = 30 cm or 300 mm; Cutter diameter
D= 8 cm or 80 mm; Number of cutter teeth = 24; S =20 m/min; f = 0.2 per tooth.
1. To find the total table travel to complete the cut:
Total table travel = Length of work piece + added table travel
For slot milling, the added table travel is =
𝐷𝑑 − 𝑑 2 = 38.73 𝑚𝑚
Hence, total table travel = 300 + 38.73 = 338.73 mm
425
Problem 27 (Slot Milling operations)
Solution (continue): Depth of cut, d = 3 cm or 30 mm; Length of the slot = 30 cm or 300 mm; Cutter
diameter D= 8 cm or 80 mm; Number of cutter teeth = 24; S =20 m/min; f = 0.2 mm per tooth.
2. To estimate the milling time:
Milling time =
N=
𝐿𝑒𝑛𝑔ℎ𝑡 𝑜𝑓 𝑗𝑜𝑏+𝑐𝑢𝑡𝑡𝑒𝑟 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ+𝑜𝑣𝑒𝑟𝑡𝑟𝑎𝑣𝑒𝑙
𝐹𝑒𝑒𝑑
𝑟𝑒𝑣
𝑋(𝑟𝑝𝑚 𝑜𝑓 𝑐𝑢𝑡𝑡𝑒𝑟)
1000 𝑆
= 79.58 rpm
𝜋𝐷
Feed/rev = Feed per tooth x number of cutter teeth = 0.2 X 24 = 4.8 mm/rev
338.73
So Milling time, T =
4.8 𝑋 79.58
= 0.886 𝑚𝑖𝑛.
426
Problem 28 (Slot Milling operations)
A T-Slot is to be cut in a C.I. slab as shown in Figure. Estimate the machining time. Take cutting
speed 25 m/min, feed is 0.25 mm/rev. Diameter of cutter for channel milling is 80 mm.
Solution: S = 25 m/min; f = 0.25 mm/rev; Dia of cutter D = 80 mm.
427
Problem 28 (Slot Milling operations)
Solution: S = 25 m/min; f = 0.25 mm/rev; Dia of cutter D = 80 mm.
Step 1: Cutting a channel of 20 mm wide and 35 mm deep (d=35 mm) along the 260 mm length
 For slot milling, Added table travel =
𝐷𝑑 − 𝑑 2 = 39.68 mm
 Total table travel or tool travel = Length of job + added table travel = 260 + 39.68 = 299.68 mm
 N = (1000 x S)/ (πD) = 100 rpm
 Time for cutting slot, T1 = L/fN = 299.68/0.25 X 100 = 11.987 min.
Step 2: Cutting T-Slot of dimensions 60 X 20 mm with a T-slot cutter.
Dia of cutter = 60 mm
 N = (1000 x S)/ (πD) = 132.63 rpm
Since, diameter of cutter = width of slab, therefore
So, T = T1 + T2 = 20.733 mm
 Over-travel of tool = 0.5 X diameter of cutter = 30 mm
Hence, total tool or table travel (L)= 260 + 30 = 290 mm.
 Time taken for cut the T-slot = T2 = L/fN = 299.68/0.25
X 100 = 8.746 min
428
Estimation of Time for Grinding Operations
 Grinding is the process of metal removal by abrasion.
 It employs a multipoint cutting tool called grinding wheel.
 It is generally a finishing operation, which removes comparatively very small
amount of material.
 Methods: Surface grinding, Cylindrical grinding.
429
Estimation of Time for Grinding Operations
 Cylindrical grinding
Formula used:
 Grinding time/cut =
𝐿𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑐𝑢𝑡
𝐹𝑒𝑒𝑑
𝑋 𝑟𝑝𝑚
𝑟𝑒𝑣
=
L0 – w + 5
𝑤
𝑤
( 2 𝑜𝑟 4 )𝑋 𝑁
 Length of cut, L = Length of workpiece – width of grinding wheel (w) + approach
L = L0 – w + 5
 The longitudinal feed of the grinding wheel:
Feed/rev =
𝑤
2
for rough grinding
=
𝑤
4
for finish grinding
 Total Grinding time =
L0 – w + 5
𝑤
𝑤
( 2 𝑜𝑟 4 )𝑋 𝑁
X Number of cuts required
 Number of cuts or number of pass required = n =
𝐷𝑒𝑝𝑡ℎ 𝑜𝑓 𝑠𝑡𝑜𝑐𝑘 𝑡𝑜 𝑏𝑒 𝑟𝑒𝑚𝑜𝑣𝑒𝑑 (𝑑)
𝐷𝑒𝑝𝑡ℎ 𝑝𝑒𝑟 𝑐𝑢𝑡 (𝑡)
430
=
𝑑
𝑡
Problem 29 (Cylindrical Grinding operations)
Find the time required for doing rough grinding of a 15 cm long steel shaft to reduce its diameter
from 4 to 3.8 cm with the grinding wheel of 2 cm face width. Assume cutting speed as 15 m/min and
depth of cut as 0.25 min.
Solution: S = 15 m/min; Depth of cut (t) = 0.25 mm per pass; L0 = 150 mm; D = 40 mm; cutter face
width (w) = 2 cm or 20 mm.
Feed for rough cut =
𝑤
2
=
20
2
= 10𝑚𝑚/𝑟𝑒𝑣
N = (1000 x S)/ (πD) = 119.37 rpm
Number of pass required = n =
𝑑
𝑡
=
(40−38)/2
0.25
= 4 𝑝𝑎𝑠𝑠𝑒𝑠
Length of cut = L = L0 – w + 5 = 135 mm
Total grinding time =
L0 – w + 5
𝑤
𝑤
( 2 𝑜𝑟 4 )𝑋 𝑁
135
𝑋4=
10 𝑋 119.37
X Number of cuts required =
431
0.452 min
Problem 30 (Cylindrical Grinding operations)
Estimate the grinding time to finish a shaft from 38.5 mm to 30 mm diameter. Length of shaft as 300
mm. Assume the following dat. Width of grinding wheel = 50 mm; depth of cut in roughening
operation = 0.785 mm; depth of cut in finishing operation = 0.05 mm; cutting speed 12 m/min.
Assume 1 mm on diameter to be finished grounded and remaining rough ground.
Solution: S = 12 m/min; L0 = 300 mm; D = 38.5 mm; cutter face width (w) = 50 mm; Depth of cut (t)
= 0.05 mm for finishing; Depth of cut (t) = 0.785 mm for roughing
Step 1: Grinding time required for roughing operation from 38.5 mm dia to 31 mm dia
Length of cut L= L0 – w + 5 = 255 mm
Feed for rough cut =
𝑤
2
=
50
2
= 25𝑚𝑚/𝑟𝑒𝑣
N = (1000 x S)/ (πD) = 99.21 rpm
Number of pass required = n =
Total grinding time =
L0 – w + 5
𝑤
𝑤
( 2 𝑜𝑟 4 )𝑋 𝑁
𝑑
𝑡
=
(38.5−31)/2
0.785
= 4. . 77 𝑜𝑟 5 𝑝𝑎𝑠𝑠𝑒𝑠
255
𝑋5=
25 𝑋 99.21
X Number of cuts required =
432
0.514 min
Problem 30 (Cylindrical Grinding operations)
Step 2: Grinding time required for finishing operation from 31 mm dia to 30 mm dia
Length of cut L= 255 mm
Feed for rough cut =
𝑤
4
=
50
4
= 12.5𝑚𝑚/𝑟𝑒𝑣
N = (1000 x S)/ (πD) = 123.22 rpm
Number of pass required = n =
Total grinding time =
L0 – w + 5
𝑤
𝑤
( 2 𝑜𝑟 4 )𝑋 𝑁
𝑑
𝑡
=
(31−30)/2
0.05
= 10 𝑝𝑎𝑠𝑠𝑒𝑠
255
𝑋10=
12.5 𝑋 123.22
X Number of cuts required =
Hence, total time = T1 + T2 = 2.169 min
433
1.655 min
Thank You
&
Best Wishes
434