ME2303-NOL

Notes on Lesson

Faculty Name : V. KIRAN KUMAR Code : AU28

Subject Name : DESIGN OF MACHINE

ELEMENTS

Code : ME2303

Year : III Semester : V

Degree & Branch : B.E. – AUTOMOBILE Section : -

Failure Theories

Stress in machine components should be accurately computed.

Designer must understand material limits to ensure a safe design.

Design Factor Factor of Safety (N)

Suitable values depend on inherent danger, certainty of calculations, certainty of material properties, etc.

FailureComponent at Stress

Stress ExpectedN

Static Stresses - Brittle Materials Percent elongation < 5%

• for parts in tension

• for parts in compression

• for parts with general stress

tus

N

cus

N

ct uu ssN211

Example

The Gray Cast Iron (Grade 40) cylinder carries an axial compressive load of 75,000 lbs and a torque of 20,000 in lbs. Compute the resulting design factor.

Ø4.00”

Ø5.00”

R0.25”

R0.25”

Static Stresses - Ductile Materials

Percent elongation > 5% Distortion Energy Theory

Define von Mises Stress

• For nominal stress

• For localized stress

'ys

N

'us

N

2122

21'

Static Stresses - Ductile Materials

Percent elongation > 5% Maximum Shear Stress Theory

• For nominal stress

• For localized stress

maxmax 2yys ss

N

maxuss

N

Example

Specify a diameter for the middle portion of the rod, if it is to be made from AISI 1040-hot rolled steel.

5000 lbs

450

Example

For the seat support shown, specify a standard structural tube to resist static loads shown. The tube has properties similar to AISI 1020 hot-rolled steel. Use a design factor of 3.

200 lb

400 lb

20”14”

Repeated Loads

Time

Stress alt

mean

Example

The notched bar is machined from AISI 1020 steel. This bar is subjected to a load that varies from 2000 lb to 3000 lb. Determine the mean and alternating nominal stresses.

0.1” R

1”

1.25”

.75”

Fatigue Strength

R.R. Moore Test

103 104 105 106 107 108

Endurance Strength, sn

Cycles to Failure, N (log)

Alternating Stress, a

Motor

Endurance Strength

sn = Endurance strength Listed in tables

If no information is available, use

• sn 0.5 su (Steel)

• sn 0.4 su (Aluminum)

Adjusted Endurance Strength

The data from the standard R.R. Moore test is adjusted for a particular application.

sn’ = Adjusted endurance strength

= (Cs) (Cm) (Cst) (CR) (sn)

Size and Stress Type Factors

– Cs = Size Factor

• D< 0.4 in Cs = 1.0

• 0.4 < D 2.0 in Cs = (D/0.3)-0.068

• 2.0 < D 10.0 in Cs = D-0.19

For rectangular sections, D=.808(h b)1/2

– Cst = Stress Type Factor

• = 1.0 for bending• = 0.80 for axial tension• = 0.50 for torsion

Material and Reliability Factor– Cm = Material Factor

• = 1.0 for wrought steel• = 0.80 for cast steel• = 0.70 for cast iron

– CR = Reliability Factor

• 50% CR = 1.0

• 90% CR = 0.90

• 99% CR = 0.81

• 99.9% CR = 0.75

Example

The notched bar is machined from AISI 1020 steel. This bar is subjected to a load that varies from 2000 lb to 3000 lb. Determine the endurance limit of the material.

0.1” R

1”

1.25”

.75”

Repeated Stresses - Ductile Materials

Distortion Energy Theory

Define repeated von Mises Stress

• Solderberg criterion

n

at

y

m

s

K

sN '

''1

mmmmm 2122

21'

aaaaa 2122

21'

Repeated Stresses - Ductile Materials

Maximum Shear Stress Theory

sn

at

sy

m

s

K

sN '

)()(1 maxmax

• ssy = 0.5 sy

• s’sn = 0.5 sn

Example

The notched bar is machined from AISI 1020 steel. This bar is subjected to a load that varies from 2000 lb to 3000 lb. Comment on the robustness of the design.

0.1” R

1”

1.25”

.75”

Example

Comment on the robustness of a 1-1/4” round bar made from AISI 1213 C-D steel. It carries a constant tensile load of 1500 lbs, a bending load that varies from 0 to 800 lbs at the senter of the 48” length and a constant torque of 1200 in lbs.

48”

Connect power transmission components.

Inherently subjected to transverse loads and torsion.

Shafts

Shaft Forces Gears

As before

Wr

Wt

T

Shaft Forces

Chains

Fslack = 0

Ftight

D

TFtight

2

D

T

Shaft Forces

V-belts

Fslack

Ftight

D

TFtight

5.2

D

T

D

TFslack 2

Shaft Forces

Flat belts

Fslack

Ftight

D

TFtight

3

D

T

D

TFslack

Material Properties For steady load (torsion)

sys=.5sy

For fatique load ( bending)

sn’=cs cR sn

cT = 1 (bending)

cm = 1 (wrought steel)

Stress Concentrations

Keyseats– Sled Runner Kt = 1.6

– Profile Kt = 2.0

– Woodruff Kt = 1.5

Stress Concentrations

Shoulders– Sharp, Bearing (r/d .03) Kt = 2.5

– Round, Gear Bore (r/d .17) Kt = 1.5

Grooves– Retaining Rings Kt = 1.5

Try not to let Kt’s overlap. Leave .10 - .15” between

Strength Analysis

Bending stress

Torsion stress

S

MK

I

cMK tt

S

T

J

rT

2

32

3DS

c

I

r

J2For round sections

For round sections

Strength Analysis

Mohr’s circle and Solderberg

S

sTsMK

N

ynt22 /

43

'/1

Suggested Design Factors:

• N=2 smooth operation

• N=3 typical industrial operation

• N=4 shock or impact loading

Minimum Acceptable Diameter

The designer must size the shaft.– Solve for appropriate diameters

22 /

4

3'/

32ynt sTsMK

ND

Example

Determine a suitable diameter for a shaft made from AISI 1144 OQT 1000. It is subjected to a reversing bending moment of 3000 ft lbs and a steady torque of 1800 ft lbs. The shaft has a profile keyway.

Example

The shaft shown is part of a grain drying system At A, a 34 lb. propeller-type fan requires 12 hp

when rotating at 475 rpm. A flat belt pulley at D delivers 3.5 hp to a screw

conveyor handling the grain. All power comes to the shaft through the v-belt at

C.

Using AISI 1144 cold drawn steel, determine the minimum acceptable diameter at C.

Example

12” 10” 10” 4”

A B C D E

Sheave C150

Sheave D

Components used to securely mount power transmitting elements on a shaft.

Shafts Accessories

Axial

Rotational

Keys Allow torque to be transferred from a shaft

to a power transmitting element (gear, sprocket, sheave, etc.)

Key Design Use a soft, low strength material

(ie, low carbon steel)

Standard size H=W=1/4 D Design length

based on strength

H

L W

Standard Key Sizes

Shaft Dia. (in) W (in)

2

22 WDHDS

T S

HW

.005.2

22

inWDHD

T

Key Design

Key Shear

Failure Theory

Length

yDWs

TNL

4

DLW

T

A

F 2

TD

LWssN yy

42

D

T

D

TF

2

2/

Example

Specify a key for a gear (grade 40, gray cast iron) to be mounted on a shaft (AISI 1144, hot rolled) with a 2.00 in. diameter. The gear transmits 21000 lb-in of torque and has a hub length of 4 in.

Retaining Rings Also known as snap rings Provides a removable shoulder to lock

components on shafts or in bores. Made of spring steel, with a high shear

strength. Stamped, bent-wire, and spiral-wound.

Retaining Ring Selection

Based on shaft diameter & thrust force

Set Screws Setscrews are fasteners that hold collars, pulleys, or gears on shafts. They are categorized by drive type and point style.

Standard Set Screw Sizes

Set Screw Holding

Pins

A pin is placed in double shear Holds torsion and axial loads

ysD

NTd

8

D

d

Hole is made slightly smaller than the pin (FN1 fit)

Example

Specify a pin for a gear (grade 40, gray cast iron) to be mounted on a shaft (AISI 1144, hot rolled) with a 2.00 in. diameter. The gear transmits 21000 lb-in of torque and has a hub length of 4 in.

Roll Pins

Easier disassembly

Collars

Creates a shoulder on shaft without increasing stock size.

Held with either set screw or friction (clamped)

Mechanical Couplings Couplings are used to join two shafts Rigid couplings are simple and low cost.

But they demand almost perfect alignment of the mating shafts.

Misalignment causes undue forces and accelerated wear on the shafts, coupling, shaft bearings, or machine housing.

Mechanical Couplings

In connecting two shafts, misalignment is the rule rather than the exception. It comes from such sources as bearing wear, structural deflection, thermal expansion, or settling machine foundations.

When misalignment is expected, a flexible coupling must be used.

Mechanical Couplings Selection factors include:

Amount of torque (or power & speed)

Shaft Size

Misalignment tolerance

Fasteners, Powers Screws, Connections

Helical thread screw was an important invention.

Power Screw, transmit angular motion to liner motion

Transmit large or produce large axial force

It is always desired to reduce number of screws

Definition of important Terminologies

Major diameter d, Minor diameter dr Mean dia or pitch diameter dp

Lead l, distance the nut moves for one turn rotation

Single and Double threaded screws

Double threaded screws are stronger and moves faster

Screw Designations

United National Standard UNS

International Standard Organization

Roots and crest can be either flat or round

Pitch diameter produce same width in the thread and space,

Coarse thread Designated by UNC

Fine Thread UNF, is more resistance to loosening, because of its small helix angle.

They are used when Vibration is present Class of screw, defines its fit, Class 1 fits have

widest tolerances, Class 2 is the most commonly used

Class three for very precision application Example:1in-12 UNRF-2A-LH, A for Ext. Thread

and B for Internal, R root radius Metric M10x1.5 10 diameter mm major

diameter,1.5 pitch

Some important Data for UNC, UNF and M threads Lets Look at the Table 8-1 on Page 398

Square and Acme Threads are used for the power screw

Preferred pitch for Acme Thread

d, in 1/4 5/16 3/8 1/2 5/8 3/4 7/8 1 1 1/4

p,in 1/16 1/14 1/12 1/10 1/8 1/6 1/6 1/5 1/5

Mechanics of Power Screws

Used in design to change the angular motion to linear motion, Could you recall recent failure of power screw leading to significant causalities

What is the relationship between the applied torque on power screw and lifting force F

Torque for single flat thread

)sec

sec(

2

fld

fdlFdT

m

mmR

)(2

)(2

fld

lfdFdT

fld

fdlFdT

m

mmL

m

mmR

If the thread as an angle α, the torque will be

Wedging action, it increases friction

Stresses in the power Screw

pnd

F

A

V

pnd

F

pnd

Fd

T

tr

trb

tmB

3

2

3

6

2/

163

Shear stress in the base of the screw

Bearing stress

Bending stress at the root of the screw

Shear stress in the thread

nt number of engaged thread

Loading to the fasteners and their Failure considerations

Bolts are used to clamp two or more parts

It causes pre tension in the bolt Grip length is the total thickness of parts and washersl

ld

lt

t2

ldh

L’ effective grip= h+t2 if t2<d

=h=d/2 for t2 d

lt=L’- ld

Failure of bolted or riveted joints

Type of Joints Lap Joint (single Joint) But Joint

Example 1

Example 2

Example 2

Example 3

Weld

Weld under Bending

Springs

Used to: Exert force Store energy

Flexible machine elements

Spring Rate

Effective springs have a linear deflection curve. Slope of the spring deflection curve is the rate

Force

Deflection

k1

L

Fk

Example

A compression spring with a rate of 20 lb/in is loaded with 6 lbs and has a length of 1.5 in. Determine the unloaded spring length (free length)

Geometry

Wire diameter, Dw (Standard gages)

Mean Diameter, Dm

Dm = Do - Dw

Do

Di

Dw

L

Dw

Spring Parameters

Spring index

C > 5 (manufacturing limits) Active coils, Na

= N for plain ends

= N-1 for ground ends

= N-2 for closed ends

w

m

D

DC

Deflection

Deflection for helical springs

w

a

w

am

GD

NFC

GD

NFD 3

4

3 88

Spring rate for helical springs

a

w

NC

GDk

38

G = Shear modulus

Example

A helical compression spring is formed from 35 gage music wire with 10-1/4 turns and an O.D. if 0.850 in. It’s ends are squared. The free length is 2 inches. Determine the force to press the spring solid.

Stress Analysis

Spring wire is in torsion

F

V

T2

8

wD

CFK

J

rT

Wahl factor, KAccounts for the curvature of the

wire

CC

CK

615.

44

14

Example

A helical compression spring is formed from 35 gage music wire with 10-1/4 turns and an O.D. if 0.850 in. It’s made from A228 and the ends are squared. The free length is 2 inches.

If the spring is repeatedly compressed to 1.3 in, do you expect problems?

Design Procedure

Select a material Compute required spring rate Estimate Dm based on size constraints

Determine required Dw (use K=1.2)

Select standard wire Verify actual stress is satisfactory. Compute number of coils required.

Example

Design a helical compression spring to exert a force of 22 lbs when compressed to a length of 1.75 in. When its length is 3.0 in, it must exert 5 lb. The spring will be cycled rapidly. Use ASTM A401 steel wire.

Provides support for machine elements, while allowing smooth motion.

=0.001 - 0.005

Rolling Element Bearings

Types

Single-rowRadial Ball

Angular Contact Ball

Radial Roller

Angular Roller

Types

Spherical Roller

Needle

Thrust

Tapered Roller

Ball Bearings

Stress Analysis

Contact Stress

c=300,000 is not unusual

Balls, rollers and races are made from extremely high strength steel

ex. AISI 52100

sy = 260,000 psi

su=322,000 psi

Bearing Load/Life

Test (fatigue) data

L10 Life (cycles)

Radial Load (lbs)k

P

P

L

L

2

1

1

2

Empirical relationship:

•k=3.0 (ball)

•k=3.33 (roller)

Example

A bearing is mounted on a shaft rotating at 1200 rpm. The bearing has been tested to have a L10 life of 300 hrs, when loaded with 500 lbs. Determine the expected L10 life, if the load is increased to 700 lbs.

Manufacturer’s Data Vendors publish the

Basic Dynamic Load rating (C) of a bearing at an L10 life of 1 million cycles.

Bearing Selection

Determine the design life (in cycles) Determine the design load

Pd = V R

Calculate the required basic dynamic load

Select a bearing with (C > Creq’d) and a bore that closely matches the shaft diameter.

kd

ddreq

LPC

1

6' 10

•V=1 for inner race rotation•V=1.2 for outer race rotation

Example

Specify suitable bearings for a shaft used in an grain dryer. The shaft rotates at 1700 rpm. The required supporting loads at the bearing are

and the minimum acceptable diameter is 2.16”.

RBx=589 lbRBy=164 lb

Mounting of Bearings

Shaft/bearing bore has a light interference fit. Housing/outer race has a slight clearance fit.

Check manufacturers catalog

Match maximum permissible fillet radius. Shaft or housing shoulders not to exceed

20% of diameter.

Mounted Bearings

Pillow block

Bearing is inserted into a cast housing, with base or flange slots, which can be readily attached to a machine base.

Bearings with Varying Loads Compute a weighted average load based on

duty cycle.

p

i

ip

im N

NFF

1

Fm=equivalent load

Fi= load level for condition i

Ni= cycles for condition i

p = exponent for load/life

Example

Bearing 6211 is carrying the following load cycle, while rotating at 1700 rpm.

Stage Load (lbs) Time (min)

1 600 480

2 200 115

3 100 45

Compute the bearing L10 life in minutes.

Radial & Thrust Loads

Calculate an equivalent load

P=VXR +YT

T=thrust loadX factors dependingY on bearing

=

Thrust factors, Y

Deep -groove, ball bearings

X = 0.56 for all values of Y

Example

A bearing is to carry a radial load of 650 lb and a thrust load of 270 lb. Specify a suitable single-row, deep-groove ball bearing if the shaft rotates at 1150 rpm and the design life is 20,000 hrs.