Introduction to Component Design and Stress Analysis

Component Design

Professor V N Achutha Naikan

IIT Kharagpur

Simple Stresses in Mechanical Components

• Components are subjected to various forces due to:

– Energy transmitted – Weight of machine – Frictional resistances

– Inertia of reciprocating parts – Change of temperature

– Lack of balance of moving parts

– Other forces

Load

• An external force acting on a component

• Types:

1. Dead or steady load: does not change magnitude and direction

2. Live or variable load: it changes continuously 3. Shock load: suddenly applied or removed

4. Impact load: it is applied with an initial velocity

Stress

• When external forces (P)act on a component, it develops internal forces (equal & opposite) to resist, at various sections of the component.

• The internal force per unit area (A) is known as stress. Typical unit is Pascal (Pa)

• 𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆, 𝜎𝜎 = 𝑃𝑃 𝐴𝐴

• 1 𝑃𝑃𝑃𝑃 = 1 𝑁𝑁/𝑚𝑚 2

• 1 𝑀𝑀𝑃𝑃𝑃𝑃 = 10 6 𝑁𝑁/𝑚𝑚 2

• 1 𝐺𝐺𝑃𝑃𝑃𝑃 = 10 9 𝑁𝑁/𝑚𝑚 2

Strain

• A component under external forces undergoes deformation ( 𝛿𝛿𝑙𝑙 ) from its original dimension ( 𝑙𝑙 )

• 𝑆𝑆𝑆𝑆𝑆𝑆𝑃𝑃𝑆𝑆𝑆𝑆, ∈= 𝛿𝛿𝛿𝛿 𝛿𝛿

Tension, Compression and Shear Stress

and Strain

Young’s Modulus

• Hook’s Law: When a material is loaded

within elastic limit, the stress is directly

proportional to strain

• 𝜎𝜎 ∝ 𝜖𝜖 𝑜𝑜𝑆𝑆, 𝜎𝜎 = 𝐸𝐸𝜖𝜖

• E : Young’s Modulus, or modulus of elasticity

• 𝐸𝐸 =

^𝜎𝜎_𝜖𝜖

=

_{𝐴𝐴𝛿𝛿𝛿𝛿}^{𝑃𝑃𝛿𝛿}

Young’s Modulus for Some Materials

Factor of Safety (FS)

• In engineering, a factor of safety (FS), also known as (and used

interchangeably with) safety factor (SF), expresses how much

stronger a system is than it needs to be for an intended load.

Design of Wire in a Elevator

• Elevator in a shopping complex.

• Function: Taking people and material up and down in a building

• Capacity: 10 people/800 Kg

• The elevator is operating on four steel ropes of circular CS

• Factor of Safety?

• Strength of rope material

against tension = 2500 Kg/sq

cm

Elevator Wire

• Elevator rope wire is made from bright phosphated and redrawn galvanized wires. They are

designed for use in traction- and deflection sheaves. Featuring high tensile strength and

excellent ductility, the wire helps increase the service life of your rope.

• Shape: Wire

• Bulk Material: C-steel

Note: Ductility is the ability of a material to be drawn or plastically deformed

without fracture.

Material strength of Steel Ropes

Design of Rope in an Elevator

Self weight of Lift = Kg

Max Applied Load Kg

No. of ropes=

Load on each rope= Kg

FS=

Load capacity on each rope = Kg

Strength of material Kg/sqcm

CS Area required= sqcm

dia of rope = cm

Dia of Standard rope available cm

Actual FS?

New FS =804.25/200

Actual strength of rope = Actual CSA * Strength

Design of Rope in an Elevator

Self weight of Lift = 10000 Kg

Max Applied Load 800 Kg

No. of ropes= 4

Load on each rope= 2700 Kg

FS= 2

Load capacity on each rope = 5400 Kg

Strength of material 2500 Kg/sqcm

CS Area required= 2.16 sqcm

dia of rope = 1.658372 cm

Dia of Standard rope available 1.8 cm

Actual FS?

Actual strength of rope = Actual CSA * Strength 6361.725

New FS = 2.356194

Standard Wires

Design of Rope in a Lift

• If redundancy is provided with two more

ropes what is the improved Factor of Safety?

• What are the uncertainties in this case?

• How do you take care these in the design?

A Simple Electrical Circuit

• Function: Lighting a bulb of 100 W.

• Design a circuit for this purpose

• Application: To light a newly constructed room (16 X 12 sq. ft.) in a house.

• What parts are required?

• What are the specifications of each part?

• Factor of Safety?

A Simple Electrical Circuit

Applied Voltage (V) V

Connected Load W

FS =Design Load W

Design Equation W = VI Cos(φ) Assume PF = 1

Current (I) = A

Available wires A

Available Switch Capacity A

Available Fuse Capacity A

Actual FS=?

Max Load Cap of Wire = W = VI Cos(φ) W Actual FS =

Design of a Simple Electrical Circuit

Applied Voltage (V) 120 V

Connected Load 100 W

FS = 2

Design Load 200 W

Design Equation W = VI Cos(φ)

Assume PF = 1 1

Current (I) = 1.666667 A

Available wires 2 A

Available Switch Capacity 5 A

Available Fuse Capacity 2 A

Actual FS=?

Max Load Cap of Wire = W = VI Cos(φ) 240 W

Actual FS = 2.4

Design of I-Section Beam

• Figure shows a simply supported beam carrying a uniformly distributed load ‘u’ in Kg/cm. The applied load follows a lognormal distribution with a mean of 3.5 Kg/cm. The mean yield strength properties of the beam material (steel) from material hand book

is1650 Kg/cm

²

.

Design the I-section beam. The required beam length is 5.32 m. Assume a factor of safety of 2.2.

Also estimate the dimensions of the section and the

requirement of extra material in case a higher FS = 3

is required.

Simply supported beam and its sectional

view

Solution

Solution…FS = 2.2

ISWB – Standard Wide Flange Beams

(IS 808: 1989)

Solution…

• From the Steel tables the required section of beam can be selected

• The selected beam is ISWB 175 with section modulus Z=172.5 cm³

The sectional dimensions are:

Depth of section h = 175 mm Width of flange, b = 125 mm Thickness of flange, t_f = 7.4 mm Thickness of web, t_w = 5.8 mm Section area, a = 28.11 cm² Radius at root, r₁ = 8 mm Radius at toe, r₂ = 4 mm Slope of flange, D = 96⁰

Solution…, FS = 3

Design of L section

Design

• Application – for supporting a passenger seat

• Load – 4 people to sit or to keep 6 packets of luggage - vibration load, other loads

• Type of load – tension, compression, torsion, bending…

- static/ dynamic / fatigue

• Factor of Safety – to be decided

• Reliability – target?

• Strength – to be calculated using failure theories

• Material – to be decided

• Dimensions – cross section, length, width, etc

• Tolerances – of each dimension – tolerance allocation

Design of Process

• Production / Manufacturing Technology - Casting

- Cutting from steel plates - From steel bars

- bending – cold/hot – direct/air/water/oil bath

- welding-fillet/butt/resistance/gas/hydrogen/electron beam / ultrasonic/friction welding, length, leg

- soldering – soft: tin-lead, hard: silver

- riveting – lap / butt / parallel / zigzag /no of rivets

with/without straps

- use screws - nuts & bolts

- adhesives – phenolic elastomers, vinyl-phenolics, epoxy-resin

• Reliability, cost, technical feasibility to be compared

Cost & its allocation

• Direct cost

- material - power

- consumables - labour

• Indirect cost

– Utility – Salaries

– Administrative – Tools

– Machinery – Building

– Transportation – Quality assurance – Maintenance

– Consultancy – Taxes - VAT

Safety & Risk

• Modes of failure

• Effects of each failure mode

• Risk of failures

• Maintainability and maintenance

Home Work

• Actual Electrical circuit will consists of Fans, Light bulbs, Television set, Music set, Geyser, Induction Cooker, Refrigerator, Air

Conditioners, MV Oven, etc.

• How will you proceed in designing such a

circuit(s)?

Electrical Amplifier

• A voltage amplifier is to be designed for a gain

of 10.

Water Pump

• A building design

engineer wants to pump water from the

underground sump to the 30

^th

floor of the building.

• What specifications are required?

• How this system can be designed, including

piping, storage tank, and

distribution network to

each flat?

Design of Simply Supported Beam

Cross sections of available beams

Development of a New UPS

• Is there a market gap available?

• What are the output specifications?

• New features?

• Cost?

• Design?

• Manufacture?

• Testing?

• Marketing policy?

• Service policy?

• Warranty policy?

End