MECH 390 Mechanical Engineering Design Project …users.encs.concordia.ca/~nrskumar/Index_files/Mech390/Lecture...Engineering Design Project Lecture 3 Shaft Analysis and Fatigue Failure

Concordia University

Department of Mechanical, Industrial and

Aerospace Engineering

MECH 390 – Mechanical

Engineering Design Project

Lecture 3

Shaft Analysis and Fatigue Failure

Fall 2017

Supporting Axial Loads Axial loads must be supported through a bearing to the frame.

Generally best for only one bearing to carry axial load to shoulder

Allows greater tolerances and prevents binding

Providing for Torque Transmission

• Common means of transferring torque to shaft

– Keys

– Splines

– Setscrews

– Pins

– Press or shrink fits

– Tapered fits

• Keys are one of the most effective

– Slip fit of component onto shaft for easy assembly

– Positive angular orientation of component

– Can design key to be weakest link to fail in case of overload

Assembly and Disassembly

Shaft Design for Stress

• Stresses are only evaluated at critical locations

• Critical locations are usually

– On the outer surface

– Where the bending moment is large

– Where the torque is present

– Where stress concentrations exist

Shaft Design Procedure

1. Determine the rotational speed of the shaft.

2. Determine the power or the torque to be transmitted by the shaft.

3. Determine the design of the power-transmitting components or other devices that will be

mounted on the shaft, and specify the required location of each device.

4. Specify the location of bearings to support the shaft. Normally two and only two bearings

are used to support a shaft. The reactions on bearings supporting radial loads are

assumed to act at the midpoint of the bearings.

5. Propose the general form of the geometry for the shaft, considering how each element on

the shaft will be held in position axially and how power transmission from each element to

the shaft is to take place.

6. Determine the magnitude of torque that the shaft sees at all points.

7. Determine the forces that are exerted on the shaft, both radially and axially.

8. Resolve the radial forces into components in perpendicular directions, usually vertically

and horizontally.

9. Solve for the reactions on all support bearings in each plane.

10. Produce the complete shearing force and bending and torque moment diagrams to

determine the distribution of torque bending moments in the shaft.


11. Select the material from which the shaft will be made, and specify its condition: cold-

drawn, heat-treated, and so on. Suggested steel materials for shafts are plain carbon or

alloy steels

12. Determine an appropriate design stress, considering the manner of loading (smooth,

shock, repeated and reversed, or other)..

13. Analyze each critical point of the shaft to determine the minimum acceptable diameter of

the shaft at that point in order to ensure safety under the loading at that point. In general,

the critical points are several and include those where a change of diameter takes place,

where the higher values of torque and bending moment occur, and where stress

concentrations occur.

14. Specify the final dimensions, surface finishes, tolerances, geometric dimensioning details,

fillet radii, shoulder heights, keyseat dimensions, retaining ring groove geometry, and other

details for each part of the shaft, ensuring that the minimum diameter dimensions from

Step 13 are satisfied.

Because of simultaneous occurrence of torsional shear stresses and normal

stresses due to bending, stress analysis of shaft involves use of combined stress

approach.




• Typically the torque comes into the shaft at one gear and leaves the shaft at another gear.

A free body diagram of the shaft will allow the torque at any section to be determined. The

torque is often relatively constant at steady state operation. The shear stress due to the

torsion will be greatest on outer surfaces.

• The bending moments on a shaft can be determined by shear and bending moment

diagrams. Since most shaft problems incorporate gears or pulleys that introduce forces in

two planes, the shear and bending moment diagrams will generally be needed in two

planes. Resultant moments are obtained by summing moments as vectors at points of

interest along the shaft.

• Axial stresses on shafts due to the axial components transmitted through helical gears or

tapered roller bearings will almost always be negligibly small compared to the bending

moment stress. They are often also constant, so they contribute little to fatigue.

Consequently, it is usually acceptable to neglect the axial stresses induced by the gears

and bearings when bending is present in a shaft

Forces Exerted on Shafts by Machine Elements

Spur Gears

Torque:

T = 63 000 (P)/n

Tangential Force:

Wt = T/(D/2)

Radial Forces:

Wr = Wt tan ϕ

where P = power being transmitted in

hp

n = rotational speed in rpm

T = torque on the gear in lb.in

D = pitch diameter of the gear in inches


Spur Gears


Helical Gears

Torque: T = 63 000(P) / n [lb . In]

Tangential force:

Wt = (33 000)(P) / vt [lbf]

Axial force:

Wx = Wt tan ψ

Radial force:

Wr = Wt tan Φn

Φn is the normal pressure angle

ψ is the helix angle


Chain Sprockets

Force in Chain

Fc = T/(D/2)

Fcx = Fc cos

Fcy = Fc sin

D is Pitch dia of the sprocket

is the angle between tight side

and x-direction


V-Belt SheavesNet Driving Force

FN = F1 - F2

Net Driving Force

FN = T/(D/2)

Bending Force on the shaft carrying the

sheave is dependent on the sum,

FB= F1 + F2

For V-belt drives, the ratio is normally

taken to be

F1/F2 = 5

FB = C.FN

Bending Force on Shaft for V-Belt

Drive

FB = 1.5 FN = 1.5T/(D/2)

Bending Force on Shaft for Flat-Belt

Drive

FB = 2.0 FN = 2.0T/(D/2)

Stress Concentrations in Shafts

• Shafts typically contain:

Several diameters

Keyseats

Ring grooves

Other geometric discontinuities

• Stress concentration factors typically based on

diameter which is the objective of the design


• Stress analysis for shafts is highly dependent on stress

concentrations.

• Stress concentrations depend on size specifications, which

are not known the first time through a design process.

• Standard shaft elements such as shoulders and keys have

standard proportions, making it possible to estimate stress

concentrations factors before determining actual sizes.


Keyseats

Kt = 2.0 (profile)

Kt = 1.6 (sled runner)

A key.seat is a longitudinal groove cut into a

shaft for the mounting of a key

The profile keyseat is milled into the shaft,

using an end mill having a diameter equal to

the width of the key. The resulting groove is

flat-bottomed and has a sharp, square corner at

its end.

The sled runner keyseat is produced by a

circular milling cutter having a width equal to

the width of the key. As the cutter begins or

ends the keyseat, it produces a smooth radius.

For this reason, the stress concentration factor

for the sled runner keyseat is lower than that

for the profile keyseat.


Fillets

Kt = 2.5 (sharp fillet)

Kt = 1.5 (well-rounded fillet)

When a change in diameter occurs to

create a shoulder, a stress concentration

dependent on the ratio of the two

diameters and on the fillet radius occurs.

fillets classified in two categories:

sharp (Low fillet radii)

well-rounded (Large fillet radii)


Forces Exerted on Shafts

Forces and Moments Exerted on Shafts


Resulting Bending Moments in Shafts

Torque Diagram in Shafts

Torque Diagram in Shafts

Stresses in Shaft

Torsional shear stress

Bending stress

Axial tension/compression

Vertical shear stress

• Stresses are only evaluated at critical locations

• Critical locations are usually

– On the outer surface

– Where the bending moment is large

– Where the torque is present

– Where stress concentrations exist

Stresses in Shaft

• Standard stress equations can be customized for shafts

for convenience

• Axial loads are generally small and constant, so will be

ignored in this section

• Standard alternating and mean stresses

• Customized for round shafts

Stresses in Shaft

Combine stresses into von Mises stresses

Shafts in Bending and Torsion Only


• Substitute von Mises stresses into failure criteria equation.

For example, using modified Goodman line,

• Solving for d is convenient for design purposes


• For rotating shaft with steady bending and torsion

Bending stress is completely reversed, since a stress

element on the surface cycles from equal tension to

compression during each rotation

Torsional stress is steady

Previous equations simplify with Mm and Ta equal to 0

• Always necessary to consider static failure, even

in fatigue situation

Checking for Yielding in Shafts

• Use von Mises maximum stress to check for yielding,

Shafts Design for Fatigue Life

• Shaft Design Equation for Alternating Bending Moment (M) and Torque T

Design Procedure

• Find rotational speed and design power

• Find forces on shafts

components and reactions

• Draw torque, shear and moment diagrams

• Pick a material – ductile w/ %E ≥ 12%

Get Sy and compute Sn’ (Se)

• Determine initial design configuration

• Analyze each critical point

Change in diameter, large T, large M, Kt

Specify appropriate diameters

• Should check deflections at the end of the design

Design Procedure

• Stress analysis for shafts is highly dependent on stress concentrations.

• Stress concentrations depend on size specifications, which are not

known the first time through a design process.

• Standard shaft elements such as shoulders and keys have standard

proportions, making it possible to estimate stress concentrations

factors before determining actual sizes.

Example 7-2 Shigley’s Mechanical Engineering Design

Example 7-2Shigley’s Mechanical Engineering Design

Fig. 7-10



197 885

2” 5.75” 2.25”

R1R2

10*R2 = (197*2) + (885*7.75)

R2 = 725.3 and R1 = 356.7

540 2431

2” 5.75” 2.25”

R1R2

10*R2 = (2431*7.75) - (540*2))

R2 = 1776 and R1 = 115



Ka is surface factor based on

the type of surface machining;

Kb is size factor Kc is load; Kd

is temperature; and Ke

reliability


Fatigue Analysis

Fatigue Failure under Dynamic Loads

• A sequence of several, very complex phenomena encompassing several disciplines:

– motion of dislocations

– surface phenomena

– fracture mechanics

– stress analysis

– probability and statistics

• Begins as an consequence of reversed plastic deformation within a single crystallite but ultimately may cause the destruction of the entire component

• Influenced by a component’s environment

• Takes many forms:

– fatigue at notches

– rolling contact fatigue

– fretting fatigue

– corrosion fatigue

– creep-fatigue

Fatigue is not cause of failure per se but leads to the final fracture event.

Fatigue Failure Stages

1. Micro structural changes – nucleation of permanent

damage (mm)

2. Creation of microscopic cracks (mm)

These two steps = crack initiation = 99% of the total life!!!!!!!!!!!!!!!!!!!

Key: prevent cracks from forming at surface!!!!!!!!!!

Fatigue Crack Formation

Information Path for Strength and Fatigue Life Analysis

Component

Geometry

Loading

History

Stress-Strain

Analysis

Damage Analysis

Fatigue Life

Material

Properties

Characteristic parameters of the S - N curve are:

Se - fatigue limit corresponding to N = 1 or 2106 cycles for

steels and N = 108 cycles for aluminum alloys,

S103 - fully reversed stress amplitude corresponding to N = 103 cycles

m - slope of the high cycle regime curve

Fully reversed axial S-N curve for AISI 4130 steel. Note the break at the LCF/HCF transition and the endurance limit

10m A m

aS C N N

The S-N fatigue curve

Number of cycles, N

Str

ess a

mp

litu

de

, S

a(k

si)

Low cycle fatigue

Part 1

Infinite life

Part 3

100 101 102 103 104 105 106 107 10830

40

50

60

70

80

90

100

120

140

High cycle fatigue

Part 2

Sy

Se

Su

S103

Approximate endurance limit for various materials:

Magnesium alloys (at 108 cycles) Se = 0.35Su

Copper alloys (at 108 cycles) 0.25Su< Se <0.50Su

Nickel alloys (at 108 cycles) 0.35Su <Se < 0.50Su

Titanium alloys (at 107 cycles) 0.45Su <Se< 0.65Su

Al alloys (at 5x108 cycles) Se = 0.45Su (if Su ≤ 48 ksi) or Se = 19 ksi (if Su> 48 ksi)

Steels (at 106 cycles) Se = 0.5Su (if Su ≤ 200 ksi) or Se = 100 ksi (if Su>200 ksi)

Irons (at 106 cycles) Se = 0.4Su (if Su ≤ 60 ksi) or Se = 24 ksi (if Su> 60 ksi)

S – N curve

3 33

2 2

10 1010

1 1 1

10

1log log

3

A

m mmm mA m

e e e

aa aC

S SSm and C

or N C S

or AS S

S N N S

S

C

Endurance Limit

Most of available S - N fatigue data has been obtained from fully reversed rotational bending tests.

However, material behavior and the resultant S - N curves are different for different types of loading.

It concerns in particular the fatigue limit Se.

The stress endurance limit, Se, of steels (at 106 cycles) and the fatigue strength, S103 corresponding

to 103 cycles for three types of loading can be approximated as :

S103 = 0.90Su and Se = S106 = 0.5 Su - bending

S103 = 0.75Su and Se = S106 = 0.35 - 0.45Su - axial

S103 = 0.72Su and Se = S106 = 0.29 Su - torsion

103 104

105 106 107

0.1

0.3

0.5

1.0

S103

Se

Number of cycles, Log(N)

Re

lative

str

ess a

mp

litu

de

, S

a/S

u

Torsion

AxialBending

Mean Stress Effect

Fatigue Failure

Fatigue FailureWhile correcting for the mean stress and the notch effect use the fully corrected

S-N curve i.e. correct the fatigue curve for all effects and use the Goodman

equation below:

Where:

Sa- the nominal stress amplitude in the actual notched component

Sm –the nominal mean stress in the actual notched component

Su – the ultimate strength of the material

Se – the fully reversed (with Sm=0) fatigue limit (endurance limit) to be used with

the fully corrected S-N curve

1e

a m

u

S S

SS

Goodman Method

Good predictor of failure in ductile materials

experiencing fluctuating stress

Goodman Method

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MECH 390 Mechanical Engineering Design Project …users.encs.concordia.ca/~nrskumar/Index_files/Mech390/Lecture...Engineering Design Project Lecture 3 Shaft Analysis and Fatigue Failure