Design of Mechanical Element 1: Gear-Tooth Strength...Gear Horsepower Capacity for Tooth-Bending...

Chapter 9: Design of

Mechanical Element 1: Gear-Tooth Strength

DR. AMIR PUTRA BIN MD SAAD

C24-322

[email protected] | [email protected]

mech.utm.my/amirputra

GEAR

9.1 INTRODUCTION

BACK TO NATURE

9.1 INTRODUCTION

Having dealt with gear geometry and force analysis, we now turn to the question

of how much power or torque a given pair of gears will transmit without tooth

failure.

9.1 INTRODUCTION

The two primary failure modes for gears are

i. Tooth breakage – from excessive bending stress

ii. Surface pitting / wear – from excessive contact stress

Flank pitting –

surface contact

Root cracking –

bending stress

9.2 MODE OF TOOTH FAILURE

9.2 BASIC ANALYSIS OF GEAR-TOOTH BENDING STRESS (LEWIS EQUATION)

An equation for estimating the bending stress in gear teeth in which the toothform entered into the formulation was presented by Wilfred Lewis to PhiladelphiaEngineers Club in 1892.


1. The load is applied to the tip of a single tooth.

2. The radial component of the load, 𝐹𝑟, is negligible.

3. The load is distributed uniformly across the full face width.

4. Stress concentration in the tooth fillet is negligible. Stress concentration factors were unknown in Mr. Lewis’s time but are now known to be important. This will be taken into account later.

5. Force due to tooth sliding friction are negligible.

Assumptions made in deriving Lewis’ equation:


• The section modulus I/c is Ft2/6, and therefore the bending stress is

𝜎 =𝑀

Τ𝐼 𝑐=

6𝐹𝑡ℎ

𝑏𝑡2

• The maximum stress in a gear tooth occurs at point a as shown in figure 14-1b.

• Using the similarity of triangles, we can write:

𝑥

Τ𝑡 2=

Τ𝑡 2

ℎ→ 𝑥 =

𝑡2

4ℎl

x

t/2

l x

t/2

t/2

(a)

(b)


• From equation (a): we can rewrite it as the following:

𝜎 =6𝐹𝑡ℎ

𝑏𝑡2=

𝐹𝑡𝑏

1

Τ𝑡2 6ℎ=

𝐹𝑡𝑏

1

Τ𝑡2 4 ℎ

1

Τ4 6

• Substitute (b) into (c) and multiply the numerator and denominator by the circular pitch p, we find:

𝜎 =𝐹𝑡𝑝

𝑏( Τ2 3)𝑥𝑝

• Letting y = 2x/3p, we have

𝜎 =𝐹𝑡𝐹𝑦𝑝

• The factor y is called the Lewis form factor.

(c)


• Most engineers prefer to employ the diametral pitch in determining the stresses. This is done by substituting 𝑝 = 𝜋/𝑃 and 𝑦 = 𝑌/𝜋 in previous equation. This gives

𝜎 =𝐹𝑡𝑃

𝑏𝑌- US Customary

𝜎 =𝐹𝑡𝑏𝑚𝑌

- SI unit

𝑌 =2𝑥𝑃

3

where,

• Above equation considers only the bending of the tooth. And the effect of the radial load 𝐹𝑟 is neglected.

9.3 LEWIS FORM FACTOR

Y = 0.334

9.4 GEAR-TOOTH FATIGUE

BENDING ANALYSIS

𝜎 =𝐹𝑡𝑃

𝑏𝐽𝐾𝑣𝐾𝑜𝐾𝑚

where,

i. 𝑃 = Diametral Pitch

ii. 𝐹𝑡 = Tangential Force

iii. 𝑏 = Face width

iv. 𝐽 = Spur Gear Geometry Factor [Refer Figure 15.23]

v. 𝐾𝑣 = Velocity or Dynamic Factor [Refer Figure 15.24]

vi. 𝐾𝑜 = Overload Factor [Refer Table 15.1]

vii. 𝐾𝑚 = Mounting Factor [Refer Table 15.2]

Geometry factor J for standard spur gears (based on tooth fillet radius of0.35/P).(From AGMA Information Sheet 225.01; also see AGMA 908-B89.)


BENDING ANALYSIS

𝐽𝑝𝑖𝑛𝑖𝑜𝑛 = 0.235 (N=18) 𝐽𝑔𝑒𝑎𝑟 = 0.28 (N=36)

Geometry factor J for standard spur gears (based on tooth fillet radius of0.35/P).(From AGMA Information Sheet 225.01; also see AGMA 908-B89.)


BENDING ANALYSIS

𝐷: 𝐾𝑣 =1200 + 𝑉

1200

𝐸: 𝐾𝑣 =600 + 𝑉

600

𝐵: 𝐾𝑣 =78 + 𝑉

78

𝐴: 𝐾𝑣 =78 + 𝑉

78

𝐶: 𝐾𝑣 =50 + 𝑉

50


BENDING ANALYSIS


BENDING ANALYSIS

where,

i. 𝐶𝐿 = Load Factor [𝐶𝐿= 1.0 for bending]

ii. 𝐶𝐺 = Size or Gradient Factor [𝐶𝐺 = 1.0 for P > 5 or 𝐶𝐺 = 0.85 for P ≤ 5]

iii. 𝐶𝑆 = Surface Condition Factor

iv. 𝑘𝑟 = Reliability Factor [Use Table 15.3]

v. 𝑘𝑡 = Temperature Factor [For steel gear, 𝑘𝑡 = 1.0 < 160°F or 𝑘𝑡= 620/(460 + T) for T > 160°F]

vi. 𝑘𝑚𝑠 = Mean Stress Factor [𝑘𝑚𝑠 = 1.0 for idler gear and 𝑘𝑚𝑠 = 1.4 for one-way bending]

vii. 𝑆𝑛′ = Standard R.R Moore endurance limit

𝑆𝑛 = 𝑆𝑛′ 𝐶𝐿𝐶𝐺𝐶𝑆𝑘𝑟𝑘𝑡𝑘𝑚𝑠


BENDING ANALYSIS


BENDING ANALYSIS

SAFETY FACTOR

The safety factor for bending fatigue can be taken as the ratio of fatigue strength

to fatigue stress:

𝑛 =𝑆𝑛𝜎

Since factors 𝐾𝑜, 𝐾𝑚, and 𝑘𝑟 have been taken into account separately, the “safety

factor” need not be as large as would otherwise be necessary. Typically, a safety

factor of 1.5 might be selected, together with a reliability factor corresponding to

99.9 percent reliability.


BENDING ANALYSIS

SAMPLE PROBLEM

Gear Horsepower Capacity for Tooth-Bending Fatigue Failure

Figure above shows a specific application of a pair of spur gears, each withface width, b = 1.25 in. Estimate the maximum horsepower that the gearscan transmit continuously with only a 1 percent chance of encounteringtooth-bending fatigue failure.


BENDING ANALYSIS

where,

i. 𝐶𝐿 = 1 (bending load)

ii. 𝐶𝐺 = 1 (since P > 5) *[𝐶𝐺 = 0.85 for P ≤ 5]

iii. 𝐶𝑆 = 0.68 (Pinion) and = 0.70 (Gear) *machined surface

iv. 𝑘𝑟 = 0.814

v. 𝑘𝑡 = 1 (Temperature should be < 160 0F)

vi. 𝑘𝑚𝑠 = 1.4 (One-way bending)

vii. 𝑆𝑛′ = 290/4 = 72.5 ksi (Gear) 𝑆𝑛 = 57.8 ksi (Gear)

𝑆𝑛′ = 330/4 = 82.5 ksi (Pinion) 𝑆𝑛 = 63.9 ksi (Pinion)

𝑆𝑛 = 𝑆𝑛′ 𝐶𝐿𝐶𝐺𝐶𝑆𝑘𝑟𝑘𝑡𝑘𝑚𝑠

modification factors (Empirical Data)


BENDING ANALYSIS

STRENGTH:


BENDING ANALYSIS

𝜎 =𝐹𝑡𝑃

𝑏𝐽𝐾𝑣𝐾𝑜𝐾𝑚

The bending fatigue stress is estimated as follow

STRESS:

𝑃 = 10 𝑏 = 1.25 𝐽𝑝𝑖𝑛𝑖𝑜𝑛 = 0.235 (N=18) 𝐽𝑔𝑒𝑎𝑟 = 0.28 (N=36)

𝑉 =𝜋𝑑𝑝𝑛𝑝

12

=𝜋

1810

1720

12

= 811 𝑓𝑝𝑚

= 1.68

𝐾𝑜 = 1.25

𝐾𝑚 = 1.6

𝜎𝑝 = 114𝐹𝑡 𝜎𝑔 = 96𝐹𝑡

Therefore,

and


BENDING ANALYSIS

𝐾𝑣 =1200 + 811

1200

𝑛 = 1

63,900 = 114𝐹𝑡 , 𝐹𝑡 = 561 (pinion)

57,800 = 96𝐹𝑡 , 𝐹𝑡 = 602 (gear)

The transmitted power, ሶ𝑊

Hence, the pinion is the weaker member.


BENDING ANALYSIS

ሶ𝑊 =𝐹𝑡𝑉

33000=

561 811

33000= 13.8 ℎ𝑝

9.5 GEAR-TOOTH SURFACE

FATIGUE ANALYSIS

𝐶𝑃 = 0.5641

𝜋1 − 𝑣𝑃

2

𝐸𝑃+

1 − 𝑣𝐺2

𝐸𝐺

𝐼 =𝑠𝑖𝑛𝜙 𝑐𝑜𝑠𝜙

2

𝑅

𝑅 + 1𝑅 =

𝑑𝑔

𝑑𝑝

𝜎𝐻 = 𝐶𝑝𝐹𝑡

𝑏𝑑𝑝𝐼𝐾𝑣𝐾𝑜𝐾𝑚

𝑆𝐻 = 𝑆𝑓𝑒𝐶𝐿𝑖𝐶𝑅

STRESS:

STRENGTH:


FATIGUE ANALYSIS


FATIGUE ANALYSIS

For the gears in above problem, estimate the maximum horsepower thatthe gears can transmit with only a 1 percent chance of a surface fatiguefailure during 5 years of 40 hours/week, 50 weeks/year operation.

Gear Horsepower Capacity for Tooth Surface Fatigue Failure


FATIGUE ANALYSIS

𝑆𝐻 = 𝑆𝑓𝑒𝐶𝐿𝑖𝐶𝑅

STRENGTH:

𝑆𝑓𝑒 = 122 ksi

𝐶𝐿𝑖 = 0.8 𝑙𝑖𝑓𝑒 = 1720 60 40 50 5 = 1.03 × 109 𝑐𝑦𝑐𝑙𝑒𝑠

𝐶𝑅 = 1 [ 99 % Reliability ]

𝑆𝐻 = 122 0.8 1 = 97.6 ksi

𝜎𝐻 = 𝐶𝑝𝐹𝑡

𝑏𝑑𝑝𝐼𝐾𝑣𝐾𝑜𝐾𝑚

STRESS:

𝐾𝑣 = 1.68

𝐾𝑜 = 1.25

𝐾𝑚 = 1.6

𝑏 = 1.25 𝑖𝑛

𝑑𝑝 = 1.8 𝑖𝑛

𝐼 = 0.107

𝐶𝑝 = 2300 𝑝𝑠𝑖

[𝑆𝑓𝑒 = 0.4 𝐵ℎ𝑛 − 10 = 0.4 330 − 10 = 122 𝑘𝑠𝑖]


FATIGUE ANALYSIS

𝜎𝐻 = 2300𝐹𝑡

1.25 1.8 0.1071.68 1.25 1.6 = 8592 𝐹𝑡

STRESS:

8592 𝐹𝑡 = 97600 psi 𝐹𝑡 = 129 lb

The transmitted power, ሶ𝑊

ሶ𝑊 =𝐹𝑡𝑉

33000=

129 811

33000= 3.2 ℎ𝑝

SF: 𝒏 = 𝟏

Design of Mechanical Element 1: Gear-Tooth Strength...Gear Horsepower Capacity for Tooth-Bending...

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