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
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.
9.2 BASIC ANALYSIS OF GEAR-TOOTH BENDING STRESS (LEWIS EQUATION)
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:
9.2 BASIC ANALYSIS OF GEAR-TOOTH BENDING STRESS (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)
9.2 BASIC ANALYSIS OF GEAR-TOOTH BENDING STRESS (LEWIS EQUATION)
• 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)
9.2 BASIC ANALYSIS OF GEAR-TOOTH BENDING STRESS (LEWIS EQUATION)
• 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.)
9.4 GEAR-TOOTH FATIGUE
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.)
9.4 GEAR-TOOTH FATIGUE
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
9.4 GEAR-TOOTH FATIGUE
BENDING ANALYSIS
9.4 GEAR-TOOTH FATIGUE
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
9.4 GEAR-TOOTH FATIGUE
BENDING ANALYSIS
9.4 GEAR-TOOTH FATIGUE
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.
9.4 GEAR-TOOTH FATIGUE
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.
9.4 GEAR-TOOTH FATIGUE
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)
9.4 GEAR-TOOTH FATIGUE
BENDING ANALYSIS
STRENGTH:
9.4 GEAR-TOOTH FATIGUE
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
9.4 GEAR-TOOTH FATIGUE
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.
9.4 GEAR-TOOTH FATIGUE
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:
9.5 GEAR-TOOTH SURFACE
FATIGUE ANALYSIS
9.5 GEAR-TOOTH SURFACE
FATIGUE ANALYSIS
9.5 GEAR-TOOTH SURFACE
FATIGUE ANALYSIS
9.5 GEAR-TOOTH SURFACE
FATIGUE ANALYSIS
9.5 GEAR-TOOTH SURFACE
FATIGUE ANALYSIS
9.5 GEAR-TOOTH SURFACE
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
9.5 GEAR-TOOTH SURFACE
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]
9.5 GEAR-TOOTH SURFACE
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