Centroids and area moments of inertia

[M16] Hibbeler table "Geometric properties of line and area elements" + ZigZag summary

Mass moments of inertia

[M16] Hibbeler table "Center of gravity and mass moment of inertia of homogeneous solids" + ZigZag summary

Work and energy

[D1-8] "Planar kinetics of rigid body: work and energy"

• No work is done by pin reactions, the normal reaction, the weight in horizontal motion, or friction in rolling without slipping. Internal forces cancel.
• For a system add the T of every member according to its own motion (links rotating, a bar in curvilinear translation, a disc in GPM). At an extreme position a rolling disc stops and its centre becomes the IC of the rod attached to it.

About these notes

Every line comes from the source tagged at its end, [label p.n], where n is the PDF page. Nothing is added from memory. ⚠ marks a suspected error in the source; see "⚠ Flags" near the end. The lectures are by Prof. Dr. Mohamed Agwa, 1444/2022-2023. The slides follow Shigley's Mechanical Engineering Design, so equation and table numbers are Shigley's. The sections are by Eng. Ehsan Khaled. Units: the formulas are written as printed. Where the source gives a US form and an SI form, both are listed.

Equation and table numbers are Shigley's Mechanical Engineering Design, as in the lecturer's slides. Worked examples were recomputed; the results are in the "Worked-example checks" section.

Course description and introduction

[MD-1]

1.1 Course data

• Lecture #1 is titled "Mechanical Systems Design (DPE 301): Course Description and Introduction to Mechanical Systems Design", by Professor Dr. Mohamed Agwa [MD-1 p.1].
• Design (1) syllabus: code DPE202, "Machine elements Design", prerequisites DPE102 & DPE103 [MD-1 p.2].
  • Contact hours: Cr 3, Lec 2, Tutorial 3, Lab –, Total 5. Marks: CW 50 (50%), O/P –, FE 50 (50%), total 100 [MD-1 p.2].
  • Category: Compulsory (MR) [MD-1 p.2].
  • Objective: knowledge about different machine elements, their application and methods of design under different static and fatigue loads [MD-1 p.2].
  • Topics: fundamentals of machine design; fits and tolerance, geometrical tolerances and machine marks; design considerations (design criteria, selection of materials, dimensions and shapes, production, assembly and maintenance, etc.); design for static loads; design for dynamic loads and fatigue strength; design of riveted joints; design of power screws and threaded joints; design of welded joints; design of helical and leaf springs; design of shrink-fitted joints; design of axles and shafts for strength, rigidity and critical speed [MD-1 p.2].
  • Textbook for Design (1): Chapters 1–9 of Richard Budynas and Keith Nisbett, Shigley's Mechanical Engineering Design, McGraw-Hill Education (2019) [MD-1 p.2].
• Design (2) syllabus (this course): code DPE 301, "Mechanical Systems Design", prerequisite DPE202 [MD-1 p.3].
  • Contact hours: Cr 3, Lec 2, Tutorial 2, Lab –, Total 4. Marks: CW 50 (50%), O/P –, FE 50 (50%), total 100 [MD-1 p.3].
  • Category: Compulsory (MR) [MD-1 p.3].
  • Objective: knowledge about different mechanical systems, its components, application and methods of design under different static and fatigue loads [MD-1 p.3].
  • Topics: fundamentals of power transmission elements design; design of rolling bearing; design of sliding bearing; design of rigid and flexible couplings; design of brakes and clutches; design of flat and V belts; design of sprockets and chains; design of spur, helical, bevel and worm gears (printed "warm gears") [MD-1 p.3].
• Shigley chapters shown as the course material (Shigley lecture-slide covers) [MD-1 p.4–10]:
  • Chapter 11 Rolling-Contact Bearings (10th ed.) [p.4]; Chapter 12 Lubrication and Journal Bearings (9th ed.) [p.5];
  • Chapter 13 Gears – General [p.6]; Chapter 14 Spur and Helical Gears [p.7]; Chapter 15 Bevel and Worm Gears [p.8] (10th ed. in SI units);
  • Chapter 16 Clutches, Brakes, Couplings, and Flywheels (9th ed.) [p.9]; Chapter 17 Flexible Mechanical Elements (10th ed. in SI units) [p.10].

1.2 Projects: Table 1-2 "Problem Numbers for Linked End-of-Chapter Problems" [MD-1 p.11]

Each row is one chain of linked Shigley problems (⚠ F13 on the duplicated 13-48).

1.3 Example mechanical systems (pictures only, no text beyond labels) [MD-1 p.12–19]

• Hoist: motor (θ, damping Bm) → gearbox (ratio i) → drum (radius rd, inertia Jd) → rope tension F over a pulley → payload m moving up z [MD-1 p.12].
• "Typical Gear box": input power P1, input torque M1, input speed n1 on the input shaft; output power P2, output torque M2, output speed n2 on the output shaft; power loss Pv; housing or casing; bearings [MD-1 p.13].
• Compound reverted gear train, gears 2, 3, 4, 5 in a housing of height Y [MD-1 p.14].
• Two-stage reducer layout: shaft 1 (IN) with pinion A on bearings a, b; shaft 2 with gear B and pinion C on bearings c, d; shaft 3 (OUT) with gear D on bearings e, f; housing [MD-1 p.15].
• Test-rig photo: motor, drive-shaft support, coupling, gearbox, magnetic powder brake [MD-1 p.16].
• Conveyor drive: drive → coupling → gearbox (bevel stage then spur) → drive pulley between two bearings carrying the load [MD-1 p.17].
• Winch: belt drive, reducer, motor, coupling, drum, rope arrangement with guide rod and reciprocating screw, chain drive; with a 3-D model [MD-1 p.18].
• Hoist project layout: electric motor with clutch → flat belt (pulleys D1, D2, center distance X) → shaft 1 with bevel pinion Z2 → shaft 2 (bevel Z3, pinion Z4) → shaft 3 (Z5, Z6) → shaft 4 (Z7; bearings E, F; dimensions 250 and 200 mm; center distance C67) → flange coupling → shaft 5 with brake → drum Dd → load. Bearings A–F [MD-1 p.19].
• Lecture #1 has no formulas or worked examples [MD-1 p.1–19].

1.4 Student advice note "ملف مهم جدا جدا" (Machine Design, تصميم الماكينات) [MD-IMP]

A handwritten Arabic note (6 pp) by a former student, signed but unnamed, titled "تصميم الماكينات" (Machine Design). It contains advice and course organisation only, no technical content. The year it describes is not stated. The points, translated:

• The design course this year is easier than the second-year one [MD-IMP p.1].
• The course is out of 150 marks: 50 coursework (أعمال سنة), split among the midterm, the project, and attendance at lectures and sections; 100 for the final (⚠ F21: [MD-1 p.3] gives CW 50 + FE 50 = 100) [MD-IMP p.1].
• "Of course no need to say that the midterm and the final are رش" (student slang; the note does not explain it), followed by a sad face [MD-IMP p.1].
• The course is taught by Dr. Mohamed Agwa (د. محمد عجوة) and Eng. Mohamed Ashry (م. محمد عشري) [MD-IMP p.2]. (The section handouts used in these notes are by Eng. Ehsan Khaled.)
• The writer's opinion: Dr. Agwa is, without exaggeration, the best in the faculty for marking, explanation and the way he treats students. Eng. Ashry explains well, but the writer advises relying on the doctor first and last [MD-IMP p.2].
• Lecture rules the writer reports: nobody enters the lecture late, nobody brings food or drink, and nobody uses a mobile phone during the lecture [MD-IMP p.2].
• Midterm and final: the doctor offers two options [MD-IMP p.3]:
  1. Enter the exam with whatever you want (including the problems you solved), but the exam will contain problems you have never seen.
  2. The students collect money and photocopy the lectures, without the problems in them; the exam problems then come from the lectures or from the reference (those he asked the students to solve), 3 to 5 problems per chapter.
  • The writer strongly recommends the second option.
• The doctor asks for any papers or tables students want added to be sent to him to collect. Files named "ورق الميدترم" (midterm papers) and "ورق الفاينال" (final papers) hold all the additions; students should go through them well and talk to the doctor if they want to add something [MD-IMP p.4].
• For the teaching assistant (المعيد): the syllabus may not be finished in the sections; this is no problem because the doctor's explanation is sufficient [MD-IMP p.4].
• While studying, print the doctor's lectures, because these are the papers taken into the exam. Studying from a laptop leaves you confused in the exam; printing is better so you are used to the paper [MD-IMP p.4–5].
• A file named "محاضرات صوتية" (audio lectures) holds recordings of the doctor's lectures; listen to them when something is unclear, with the lecture paper in front of you [MD-IMP p.5].
• Most important: beware of studying from any material of the Power department (قسم باور). Its requirements and problem data are completely different, as is the order of explanation; the syllabus is the same, but each doctor has his own requirements in the problems [MD-IMP p.5].
• The note closes with good wishes and a signature, and a joke picture captioned "لما تصمم Brake وتطلع Non safe" (when you design a brake and it comes out non-safe): a sofa built from car seats and wheels [MD-IMP p.6].

Belts: general and flat belts

[MD-2]

2.1 Types and arrangements

• Belt types (Table 17-1): flat, round, V and timing [MD-2 p.4].
• Flat belt: joint yes; size t = 0.03–0.20 in (0.75–5 mm); no upper limit on center distance [MD-2 p.4].
• Round belt: joint yes; d = 1/8 to 3/4 in; no upper limit on center distance [MD-2 p.4].
• V belt: no joint; b = 0.31–0.91 in (8–19 mm); center distance limited [MD-2 p.4].
• Timing belt: no joint; p = 2 mm and up; center distance limited [MD-2 p.4].
• Fig 17-2 shows (a) a nonreversing open belt, (b) a reversing crossed belt and (c) a reversing open-belt drive. Crossed belts must be separated to prevent rubbing if high-friction materials are used [MD-2 p.5].
• Open belt compared with crossed belt [MD-2 p.6]:
  • open: shafts in the same plane; crossed: two planes.
  • open: pulleys turn in the same direction; crossed: opposite directions.
  • open: wrap angle < 180°; crossed: > 180°.
  • open: lower power capacity; crossed: higher.
  • open: shorter belt; crossed: longer.
  • open: no rubbing, so longer life; crossed: rubs, so shorter life.
  • open: vibrates (whips) at larger center distance; crossed: free from such vibration.
• Quarter-twist drive (Fig 17-3): an idler guide pulley must be used if motion is to be in both directions [MD-2 p.7].
• Loose-pulley drive (Fig 17-4): it removes the need for a clutch; a fork shifts the belt. Fig 17-5 shows variable-speed belt drives [MD-2 p.8].

2.2 Geometry

Figure: md-belt-open-geometry.jpg (Fig 17-1a) [MD-2 p.9].

• Open belt: θd = π − 2 sin⁻¹[(D − d)/(2C)]; θD = π + 2 sin⁻¹[(D − d)/(2C)] (17-1). D = large pulley diameter, d = small pulley diameter, C = center distance, θ = angle of contact [MD-2 p.9].
• Open belt length: L = [4C² − (D − d)²]^½ + ½(D θD + d θd) (17-2) [MD-2 p.10].
• Crossed belt: θ = π + 2 sin⁻¹[(D + d)/(2C)] (17-3); L = [4C² − (D + d)²]^½ + ½(D + d)θ (17-4) [MD-2 p.11].

2.3 Tensions (belting equation)

• Belt element (Fig 17-6): F = (F2 − m r² ω²) exp(fθ) + m r² ω² (17-5). F1 = tight-side tension, F2 = loose-side tension, b = width, t = thickness, m = mass per unit length, ω = angular velocity. θd = φ [MD-2 p.12].
• F1 = (F2 − m r² ω²) exp(fφ) + m r² ω² (17-6) [MD-2 p.13].
• (F1 − Fc)/(F2 − Fc) = exp(fφ) (17-7), where Fc = m r² ω² [MD-2 p.13].
• F1 − F2 = (F1 − Fc)[exp(fφ) − 1]/exp(fφ) (17-8) [MD-2 p.13].
• Centrifugal (hoop) tension: Fc = m r² ω², with ω = V/r [MD-2 p.14].
• US form: V = π d n/12 ft/min; w = 12 γ b t lbf/ft (γ in lbf/in³; b, t in in); Fc = (w/g)(V/60)² = (w/32.17)(V/60)² [MD-2 p.14].
• SI form: w = γ b t; Fc = (w/g)V² [MD-2 p.14].
• Torque: ΔF = F1 − F2 = 2T/d [MD-2 p.15].
• F1 = Fi + Fc + ΔF/2 = Fi + Fc + T/d (f); F2 = Fi + Fc − ΔF/2 = Fi + Fc − T/d (g) [MD-2 p.16].
  • Fi = initial tension, Fc = hoop tension, ΔF/2 = tension due to torque T.
• F1 − F2 = 2T/d (h); F1 + F2 = 2Fi + 2Fc [MD-2 p.16].
• Fi = (F1 + F2)/2 − Fc (i) [MD-2 p.17].
• Fi = (T/d)[exp(fφ) + 1]/[exp(fφ) − 1] (17-9) [MD-2 p.17].
• F1 = Fc + Fi · 2exp(fφ)/[exp(fφ) + 1] (17-10) [MD-2 p.17].
• F2 = Fc + Fi · 2/[exp(fφ) + 1] (17-11) [MD-2 p.18].
• Power: H = (F1 − F2)V/33 000 (j), in hp with lbf and ft/min [MD-2 p.18].

2.4 Allowable tension and correction factors (flat belts)

• Manufacturers give an allowable tension Fa (or σall) per unit width. Belt life is several years [MD-2 p.19].
• The severity of flexing gives the pulley correction factor Cp. Speeds above 600 ft/min need the velocity correction factor Cv [MD-2 p.19].
• Cv = 1 for polyamide and urethane belts. For leather, read Cv from Fig 17-9 [MD-2 p.19–20].
• Ks is the service factor for excursions of load; nd is the design factor for exigencies [MD-2 p.19].
• Design power: Hd = Hnom Ks nd. The highlighted box on the slide prints "Ha = Hnom Ks nd" (⚠ F1) [MD-2 p.19].
• Allowable largest tension: (F1)a = b Fa Cp Cv (17-12) [MD-2 p.20].
  • (F1)a in lbf; b = belt width, in; Fa = manufacturer's allowed tension, lbf/in; Cp from Table 17-4; Cv from Fig 17-9 (leather) or 1 (polyamide, urethane).
• Fig 17-9 (Cv for leather belts): Cv against V (0–30 m/s) for 4.5 mm, 5 mm, 7 & 8 mm and 9 mm thicknesses. Cv falls from 1.0 to about 0.7 at 30 m/s. Data: Machinery's Handbook, 20th ed., 1976, p. 1047 [MD-2 p.21].
• The same chart in ft/min (0–6000) is on p.22, for thicknesses 11/64, 13/64, 18/64 & 20/64 and 25/64 in [MD-2 p.22].

Table 17-4. Pulley correction factor Cp for flat belts (columns: small-pulley diameter, in; values are averages approximated from Habasit Engineering Manual curves) [MD-2 p.23]

Table 17-2 (SI). Properties of some flat and round belt materials [MD-2 p.25]

Columns: size (mm); minimum pulley diameter (mm); allowable tension per unit width at 3 m/s (10³ N/m); specific weight (kN/m³); coefficient of friction f.

• Footnotes [MD-2 p.27]:
  • a: add 2 in to the pulley size for belts 8 in wide or more;
  • b: source Habasit;
  • c: friction cover of acrylonitrile-butadiene rubber on both sides;
  • d: source Eagle Belting;
  • e: value at 6% elongation, 12% maximum allowable.
• The US version of Table 17-2 is on p.26 (e.g., leather 1-ply t = 11/64 in: 3 in, 30 lbf/in, 0.035–0.045 lbf/in³, f = 0.4) [MD-2 p.26].

2.5 Flat-belt analysis steps [MD-2 p.24]

1. Find exp(fφ) from the belt-drive geometry and friction.
2. Find Fc from the geometry and speed.
3. Find the torque: T = 63 025 Hnom Ks nd/n (US). The box on the slide gives T = Hnom Ks nd/(2πn).
4. Find (F1)a − F2 = 2T/d.
5. Find (F1)a from Tables 17-2 and 17-4 and Eq (17-12).
6. F2 = (F1)a − [(F1)a − F2].
7. Find Fi from Eq (i).
8. Check the friction development: f′ < f, where f′ = (1/φ) ln{[(F1)a − Fc]/(F2 − Fc)}.
9. Factor of safety: nfs = Ha/(Hnom Ks).

2.6 Worked example 17-1 (flat belt) [MD-2 p.28–31]

• Data [p.28]:
  • Polyamide A-3 belt, 150 mm wide (belt 150 × 3.3 mm, γ = 11 kN/m³), 11 kW.
  • Light shock, Ks = 1.25; nfs ≥ 1.1.
  • Horizontal drive, shafts 2.4 m apart, loose side on top.
  • Driving pulley 150 mm at 1750 rpm; driven pulley 450 mm.
• Solution chain [p.29–31]:
  • φ = 3.0165 rad; exp(0.8 × 3.0165) = 11.17.
  • V = 13.7 m/s.
  • w = 11 000(0.15)(0.0033) = 5.4 N/m; Fc = (5.4/9.81)(13.7)² = 103.3 N.
  • T = 1.25(1.1)(11 000)/(2π·1750/60) = 82.5 N·m.
  • (F1)a − F2 = 2(82)/0.15 = 1093.3 N (⚠ F2).
  • Fa = 18 kN/m, Cv = 1, Cp = 0.70 → (F1)a = 0.15(18 000)(0.70)(1) = 1890 N.
  • F2 = 796.7 N; Fi = 1240.4 N.
  • Design power = 11(1.25)(1.1) = 15.125 kW.
  • f′ = 0.314 < 0.8, so no slip.
  • nfs = 15.125/(11 × 1.25) = 1.1.
• Sheet problems: Shigley 17-2 and 17-4 [MD-2 p.31]. Both are solved in [MD-S1] (2.7 below).

2.7 Section 1: flat-belt formula sheet and solved problems [MD-S1]

The section handout "Machine Design, Sec. (1) Flat Belt" is by Eng. Ehsan Khaled (11 pp; pp. 1–4 typed, pp. 5–11 handwritten, CamScanner) [MD-S1 p.1].

• Table 17-2 (US units), properties of some flat- and round-belt materials (diameter d, thickness t, width w); columns: size, in; minimum pulley diameter, in; allowable tension per unit width at 600 ft/min, lbf/in; specific weight, lbf/in³; coefficient of friction [MD-S1 p.1]:

• The footnote letters are those of [MD-2 p.27] (section 2.4). Table 17-4 (Cp) is repeated with the same values as section 2.4 [MD-S1 p.1].
• Fig 17-9 (Cv for leather belts, in ft/min: 11/64, 13/64, 18/64 and 20/64, 25/64 in) and the open- and crossed-belt geometry sketches with Eqs (17-1)–(17-4) are repeated [MD-S1 p.2].
• Formula sheet (US units) [MD-S1 p.2–3]:
  • Allowable largest tension: F1a = b Fa Cp Cv; Cp from Table 17-4; Cv = 1 for polyamide and urethane, Fig 17-9 for leather [p.2].
  • Linear speed of belt: v (ft/min) = π d(in) n(rpm)/12 [p.3].
  • Centrifugal force: Fc = m r² ω² = (w/g)(V/60)² = (w/32.17)(V/60)², with w = 12γbt lbf/ft [p.3].
  • Transmitted torque: H = Hnom Ks nd; T (lbf·in) = 63 025H/n [p.3].
  • Transmitted horsepower: H = (F1 − F2)V/33 000 [p.3].
  • Loose-side tension: F1a − F2 = 2T/d [p.3].
  • Initial tension: Fi = (F1a + F2)/2 − Fc [p.3].
  • Coefficient of friction: φ = φd = π − 2 sin⁻¹[(D − d)/2C]; f′ = (1/φ) ln[(F1a − Fc)/(F2 − Fc)] [p.3].
  • Factor of safety: nfs = Ha/(Hnom Ks) [p.3].
• Dip: when flat belts were used for long center distances, the weight of the belt itself can provide the initial tension. The static belt deflects to a curve (a word is missing in the handout: "deflects to an curve"), and the dip from a straight belt can be measured against a stretched music wire [MD-S1 p.4]:
  • dip = C² w/(96 Fi); dip in in, C = center-to-center distance in in, w = weight per foot of the belt in lbf/ft, Fi = initial tension in lbf. Sketch: two pulleys at distance C, tension Fi at each end, dip at mid-span.
• Sheet problem P17-2 [MD-S1 p.5–8]:
  • Statement: a 6 in wide polyamide F-1 flat belt connects a 2 in diameter pulley with an angular velocity ratio of 0.5; center distance 9 ft; the smaller pulley turns at 1750 rpm delivering 2 hp; Ks = 1.25. (a) Find Fc, Fi, (F1)a and F2 at the maximum tension limit. (b) Find Ha, nfs and the belt length. (c) Find the dip [p.5].
  • Given: b = 6 in; d = 2 in; d/D = 1/2 → D = 4 in; C = 9 ft; n = 1750 rpm; Hnom = 2 hp; Ks = 1.25. Table 17-2: t = 0.05 in, γ = 0.035 lbf/in³ [p.5].
  • w = 12(0.035)(6)(0.05) = 0.126 lbf/ft; V = π(2)(1750)/12 = 916.3 ft/min; Fc = (0.126/32.17)(916.3/60)² = 0.9134 lbf [p.6].
  • Fa = 35 lbf/in (Table 17-2); Cp = 0.7 (Table 17-4); Cv = 1 → (F1)a = 6(35)(0.7)(1) = 147 lbf [p.6].
  • H = Hnom Ks nd = 2(1.25)(1) = 2.5 hp; T = 63 025(2.5)/1750 = 90 lbf·in; 147 − F2 = 2(90)/2 → F2 = 147 − 90 = 57 lbf [p.6].
  • Fi = (147 + 57)/2 − 0.9134 = 101.08 lbf [p.7].
  • Ha = (147 − 57)(916.3)/33 000 = 2.5 hp; nfs = 2.5/(2 × 1.25) = 1 ("as expected") [p.7].
  • Arabic note: the friction coefficient is checked to make sure the friction the design relies on is below the material's friction coefficient; if not, the belt slips, F1 and F2 change, and Ha (and so nfs) changes [p.7].
  • θd = π − 2 sin⁻¹[(4 − 2)/(2 × 9 × 12)] = 3.123 rad; f′ = (1/3.123) ln[(147 − 0.9134)/(57 − 0.9134)] = 0.307; Table 17-2 gives f = 0.5, so f′ < f as required [p.7].
  • θD = π + 2 sin⁻¹[(4 − 2)/(2 × 9 × 12)] = 3.16 rad; L = [4(9 × 12)² − (4 − 2)²]^½ + ½(4 × 3.16 + 2 × 3.12) = 225.4 in [p.8].
  • dip = (9 × 12)²(0.126)/(96 × 101.08) = 0.1515 in [p.8].
• Sheet problem P17-4 [MD-S1 p.9–11]:
  • Statement: a flat-belt drive of two 4 ft diameter cast-iron pulleys spaced 16 ft apart; select a belt type to transmit 60 hp at a pulley speed of 380 rpm; service factor 1.1, design factor 1 [p.9].
  • Function: Hnom = 60 hp, n = 380 rpm, Ks = 1.1, C = 16 ft; nd = 1; belt material polyamide; d = D = 4 × 12 = 48 in; unknowns t and b [p.9].
  • Table 17-2: d = 48 in is available with a wide range of materials; choose one and calculate the other variables. Choosing polyamide A-3: t = 0.13 in, γ = 0.042 lbf/in³, Fa = 100 lbf/in, f = 0.8. Solve for the unknown width b [p.9].
  • Cp = 1 (Table 17-4); Cv = 1 (polyamide) → (F1)a = b(100)(1)(1) = 100b lbf [p.9–10].
  • V = π d n = π(4)(380) = 4775 ft/min (d in ft); w = 12(0.042)b(0.13) = 0.0655b lbf/ft; Fc = (0.0655b/32.17)(4775/60)² = 12.89b lbf [p.10].
  • T = 63 025 Hnom Ks nd/n = 63 025(60)(1.1)(1)/380 = 10 946.44 (unit written "lbf/in") [p.10].
  • 100b − F2 = 2(10 946.44)/48 → F2 = 100b − 456.1 lbf [p.10].
  • φ = φd = π − 2 sin⁻¹(0) = π. Assuming full friction development: 0.8 = (1/π) ln[(100b − 12.89b)/(100b − 456.1 − 12.89b)], written as "π × 0.8 = (1/π) ln(…)"; e^(0.8π) = 87.11b/(87.11b − 456.1); 87.11b = 12.34(87.11b − 456.1) → b = 5.6 in (⚠ F19) [p.11].
  • A belt width greater than 5.6 in will develop friction less than 0.8, so b = 6 in is taken [p.11].

2.8 Sheet 3: flat-belt problems (Shigley 17-1 to 17-9) [MD-SH3]

The file is a phone screenshot of the problem list at the end of Shigley Chapter 17 ("Richard Budynas, Keith…", book pp. 927–928); it gives statements only, no answers [MD-SH3 p.1]. Short statements:

• 17-1: Example 17-2 resulted in a 10 in wide A-3 polyamide flat belt. Show that the value of F1 restoring f to 0.80 is F1 = [(ΔF + Fc) exp fφ − Fc]/(exp fφ − 1), and compare the initial tensions.
• 17-2: 6 in wide polyamide F-1 belt, 2 in pulley, velocity ratio 0.5, C = 9 ft, 1750 rev/min, 2 hp, Ks = 1.25: (a) Fc, Fi, (F1)a, F2 at the maximum tension limit; (b) Ha, nfs, belt length; (c) the dip. Solved in [MD-S1 p.5–8] (1.7): Fc = 0.9134 lbf, (F1)a = 147 lbf, F2 = 57 lbf, Fi = 101.08 lbf, Ha = 2.5 hp, nfs = 1, L = 225.4 in, dip = 0.1515 in.
• 17-3: double all geometric dimensions of the drive of Problem 17-2 and compare the effect on the problem parameters.
• 17-4: two 4 ft cast-iron pulleys 16 ft apart; select a belt type to transmit 60 hp at 380 rev/min; Ks = 1.1, nd = 1.0. Solved in [MD-S1 p.9–11] (1.7): polyamide A-3, b = 5.6 in computed (⚠ F19), b = 6 in chosen.
• 17-5: using the recurring forms w = 12γbt = a1 b, (F1)a = Fa b Cp Cv = a0 b, Fc = wV²/g = (a1 b/32.174)(V/60)² = a2 b, (F1)a − F2 = 2T/d = 33 000 Hd/V = 33 000 Hnom Ks nd/V, F2 = a0 b − 2T/d and fφ = ln[((F1)a − Fc)/(F2 − Fc)] = ln[(a0 − a2)b/((a0 − a2)b − 2T/d)], show that b = [1/(a0 − a2)](33 000 Hd/V)·exp(fφ)/(exp(fφ) − 1).
• 17-6: return to Example 17-1: (a) the minimum initial tension Fi that would put the drive as built at the point of slip; (b) with that tension, the belt width b giving nfs = nd = 1.1; (c) for that b, the corresponding (F1)a, Fc, Fi, F2, power and nfs at the tension limit; (d) what have you learned?
• 17-7: take the drive of Example 17-1 and double the belt width; compare Fc, Fi, F2, Ha, nfs and dip at the tension limit.
• 17-8: belted pulleys place loads on shafts (bending) and on bearings; using Fig 17-7 develop an expression for the load the belt places on the pulley, then apply it to Example 17-2.
• 17-9: "The line shaft illustrated in the figure is used to transmit power from an elec…" (the rest of the statement is cut off in the screenshot).

V-belts, timing belts and chains

[MD-3]

3.1 Sections and lengths

Table 17-9 (SI). Standard V-belt sections; belt angle 40° [MD-3 p.2]

• US version (in; hp) [MD-3 p.3]:
  • A: 1/2, 11/32, 3.0, 1/4–10
  • B: 21/32, 7/16, 5.4, 1–25
  • C: 7/8, 17/32, 9.0, 15–100
  • D: 1 1/4, 3/4, 13.0, 50–250
  • E: 1 1/2, 1, 21.6, 100 and up
• Table 17-10, inside circumferences of standard V-belts, mm [MD-3 p.4]:
  • A: 650, 775, 825, 875, 950, 1050, 1150, 1200, 1275, 1325, 1375, 1425, 1500, 1550, 1600, 1650, 1700, 1775, 1875, 1950, 2000, 2125, 2250, 2400, 2625, 2800, 3000, 3200.
  • B: 875 … 2800, 3000, 3200, 3275, 3400, 3450, 3950, 4325, 4500, 4875, 5250, 6000, 6750, 7500.
  • C: 1275 … 10 500. D: 3000 … 16 500. E: 4500 … 16 500.
  • The inch version is on p.5.
• Table 17-11, length conversion: add to the inside circumference to get the pitch length. A +32, B +45, C +72, D +82, E +112 mm [MD-3 p.6]. In inches: 1.3, 1.8, 2.9, 3.3, 4.5 [MD-3 p.7].
• Pitch length: Lp = 2C + π(D + d)/2 + (D − d)²/(4C) (17-16a). D and d are pitch diameters [MD-3 p.8].
• Center distance: C = 0.25{[Lp − (π/2)(D + d)] + √([Lp − (π/2)(D + d)]² − 2(D − d)²)} (17-16b) [MD-3 p.8].

3.2 Power rating

• Ratings are based on, for example, 24 000 h or 10⁸–10⁹ belt passes. An undersized belt set augmented by one belt can end up substantially oversized [MD-3 p.9].
• Allowable power per belt: Ha = K1 K2 Htab (17-17). K1 = angle-of-wrap factor (Table 17-13); K2 = belt-length factor (Table 17-14) [MD-3 p.9].

Table 17-12 (SI). "Horsepower Ratings of Standard V-Belts": power per belt against sheave pitch diameter (mm) and belt speed (m/s). The power unit is not printed; the worked example uses the values as kW [MD-3 p.10, p.25]. (⚠ F3 on A-85 at 20 m/s.)

• The US version of Table 17-12 (hp against 1000–5000 ft/min) is on p.11; it is not transcribed here [MD-3 p.11].

Table 17-13. Angle-of-wrap correction factor K1 (VV = V sheave on both pulleys; V-flat = V sheave driving a flat pulley) [MD-3 p.12]

Table 17-14 (SI). Belt-length correction factor K2: multiply the rated power per belt by this factor [MD-3 p.13]

• The inch version of Table 17-14 is on p.14 [MD-3 p.14].

3.3 Tensions, number of belts, factor of safety

• The Gates Rubber Co. gives an effective friction coefficient of 0.5123 for grooves: (F1 − Fc)/(F2 − Fc) = exp(0.5123 φ) (17-18) [MD-3 p.15].
• Design power: Hd = Hnom Ks nd (17-19), with Ks from Table 17-15 [MD-3 p.16].
• Number of belts: Nb ≥ Hd/Ha, Nb = 1, 2, 3, … (17-20) [MD-3 p.16].

Table 17-15. Suggested service factors Ks for V-belt drives [MD-3 p.17]

• Centrifugal tension: Fc = Kc(V/2.4)² with V in m/s; Fc = Kc(V/1000)² with V in ft/min (17-21). Kc is from Table 17-16 [MD-3 p.18].
• Tension difference: ΔF = 63 025(Hd/Nb)/[n(d/2)] (17-22). ΔF in lbf, Hd in hp, n in rpm, d in in [MD-3 p.18].
• F1 = Fc + ΔF exp(fφ)/[exp(fφ) − 1] (17-23) [MD-3 p.18].
• F2 = F1 − ΔF (17-24); Fi = (F1 + F2)/2 − Fc (17-25) [MD-3 p.19].
• Factor of safety: nfs = Ha Nb/(Hnom Ks) (17-26) [MD-3 p.21].

Table 17-16. Some V-belt parameters (Gates Rubber Co.) [MD-3 p.20]

• The sketch on the same slide gives the narrow-section widths: 3V = 3/8 in, 5V = 5/8 in, 8V = 1 in [MD-3 p.20].
• The worked example treats Kb as lbf·in and converts 576 lbf·in ≈ 65 N·m [MD-3 p.28].

3.4 Belt life (passes)

• Bending induces flexural stresses in the belt [MD-3 p.22–23].
• Number of passes: Np = [(K/T1)^−b + (K/T2)^−b]^−1 (17-27) [MD-3 p.22–23].
• Peak tensions: T1 = F1 + (Fb)1 = F1 + Kb/d; T2 = F1 + (Fb)2 = F1 + Kb/D [MD-3 p.22–23].
• Life in hours: t = Np Lp/(720 V) (17-28), with Lp in in and V in ft/min [MD-3 p.23].

Table 17-17. Durability parameters for some V-belt sections (K units are not printed; minimum sheave diameter in mm) [MD-3 p.24]

3.5 V-belt design procedure (the chain used in Example 17-4) [MD-3 p.9, p.15–21, p.25–29]

1. Find Lp: inside circumference plus the Table 17-11 addition.
2. Find C from Eq (17-16b).
3. Find φ = θd from Eq (17-1).
4. Find Htab from Table 17-12, interpolating in speed.
5. Find K1 (Table 17-13) and K2 (Table 17-14); Ha = K1 K2 Htab.
6. Hd = Hnom Ks nd; Nb ≥ Hd/Ha.
7. Tensions: Fc (Kc), ΔF, F1, F2, Fi.
8. nfs from Eq (17-26).
9. Life: T1, T2, Np, t.

3.6 Worked example 17-4 [MD-3 p.25–29]

• Data [p.25]:
  • 7.46 kW split-phase motor at 1750 rpm driving a rotary pump, 24 h/day.
  • d = 188 mm, D = 280 mm; three B2800 belts.
  • Ks = 1.2, plus 0.1 for continuous duty.
• Geometry and power rating [p.26–27]:
  • V = 17 m/s; Lp = 2800 + 45 = 2845 mm; C = 1054 mm; φ = 3.054 rad; exp(0.5123 × 3.054) = 4.781.
  • Htab: interpolating 3.35 kW (15 m/s) and 3.74 kW (20 m/s) at 17 m/s gives 3.506 kW per belt (3.5 kW).
  • Wrap angle 175°, so K1 = 0.99; K2 = 1.05; Ha = 3.64 kW.
  • Hd = 7.46(1.2 + 0.1)(1) = 9.7 kW; Nb ≥ 2.67, so 3 belts.
• Tensions and factor of safety [p.27–28]:
  • Kc = 0.965 → Fc = 0.965(17/2.4)² = 48.4 N.
  • ΔF = (9700/3)/[π(1750/60)(0.188)] = 188 N.
  • F1 = 286 N; F2 = 286 − 188 = 98 N; Fi = (286 + 98)/2 − 48.4 = 144 N.
  • nfs = 3.64(3)/[7.46(1.3)] = 1.13.
• Life [p.28–29]:
  • Kb = 576 lbf·in ≈ 65 N·m → Fb1 = 65/0.188 = 346 N; Fb2 = 65/0.28 = 232 N.
  • T1 = 286 + 346 = 632 N; T2 = 286 + 232 = 518 N.
  • K = 5309, b = 10.926 → Np = 11(10⁹) passes.
  • This is outside the validity range of Eq (17-27), so life is reported as > 10⁹ passes: t > 10⁹(2.845)/[3600(17)] = 46 487 h.

3.7 Timing belts and roller chain (descriptive only)

• Timing belts are made of rubberized fabric and steel wire. Their teeth fit into grooves on the sprockets (Fig 17-15) [MD-3 p.30].
• A timing belt does not stretch or slip, so it transmits power at a constant angular-velocity ratio. It needs no initial tension, so fixed-center drives can be used [MD-3 p.30].
• Timing belts remove the restriction on speed [MD-3 p.31].
• Disadvantages: first cost, the need to groove the sprockets, and dynamic fluctuations at the belt-tooth meshing frequency [MD-3 p.31].
• Roller chain (Fig 17-16: roller diameter, width, pitch p, strand spacing) and chain–sprocket engagement (Fig 17-17) are shown as pictures only [MD-3 p.32–33].
• Sheet problems: Shigley 17-18, 17-19, 17-20 [MD-3 p.33]. 17-18 and 17-19 are solved in [MD-S2] (3.8 below).

3.8 Section 2: V-belt formula sheet, US tables and solved problems [MD-S2, MD-S2T]

The section handout "Machine Design, Sec. (2) V Belt" is by Eng. Ehsan Khaled (13 pp; pp. 1–6 typed, pp. 7–13 handwritten) [MD-S2 p.1]. The separate "Sec2 tables" file (الجداول, 6 pp) is page-for-page identical to MD-S2 pp. 1–6 [MD-S2T p.1–6]. Everything below is in US units; the SI versions are in 3.1–3.4.

• Table 17-9 (US) is repeated as in 3.1 (A 1/2, 11/32 in, 3.0 in, 1/4–10 hp … E 1½, 1 in, 21.6 in, 100 hp and up); the sketch shows the belt angle αb = 40° [MD-S2 p.1].
• Table 17-10, inside circumferences of standard V belts, in [MD-S2 p.1]:
  • A: 26, 31, 33, 35, 38, 42, 46, 48, 51, 53, 55, 57, 60, 62, 64, 66, 68, 71, 75, 78, 80, 85, 90, 96, 105, 112, 120, 128.
  • B: 35, 38, 42, 46, 48, 51, 53, 55, 57, 60, 62, 64, 65, 66, 68, 71, 75, 78, 79, 81, 83, 85, 90, 93, 97, 100, 103, 105, 112, 120, 128, 131, 136, 144, 158, 173, 180, 195, 210, 240, 270, 300.
  • C: 51, 60, 68, 75, 81, 85, 90, 96, 105, 112, 120, 128, 136, 144, 158, 162, 173, 180, 195, 210, 240, 270, 300, 330, 360, 390, 420.
  • D: 120, 128, 144, 158, 162, 173, 180, 195, 210, 240, 270, 300, 330, 360, 390, 420, 480, 540, 600, 660.
  • E: 180, 195, 210, 240, 270, 300, 330, 360, 390, 420, 480, 540, 600, 660.
• Table 17-11 (in): add A 1.3, B 1.8, C 2.9, D 3.3, E 4.5 to the inside circumference to obtain the pitch length [MD-S2 p.1].
• Lp and C as in 3.1 (D, d = pitch diameters of the large and small sheaves); Ha = K1 K2 Htab, with Ha = allowable power per belt, K1 = angle-of-wrap (φ) correction factor (Table 17-13), K2 = belt-length correction factor (Table 17-14), Htab = tabulated value (Table 17-12) [MD-S2 p.2].
• Table 17-13 is repeated (values as in 3.2) with two footnotes [MD-S2 p.2]:
  • A curve fit for the VV column in terms of φ: K1 = 0.143 543 + 0.007 468φ − 0.000 015 052φ² for 90° ≤ φ ≤ 180°.
  • VV drives: driver and driven pulleys are both V pulleys. A V-flat drive uses a V pulley, typically the smaller on the driver shaft, and a flat pulley on the driven shaft.

Table 17-14 (US). Belt-length correction factor K2, nominal belt length in inches [MD-S2 p.3]

Table 17-12 (US). Horsepower ratings of standard V belts (hp per belt; belt speed in ft/min) [MD-S2 p.3]

• Design power Hd = Hnom Ks nd (Hnom = nominal power, Ks = service factor from Table 17-15, nd = design factor). The number of belts Nb is usually the next higher integer to Hd/Ha: Nb ≥ Hd/Ha, Nb = 1, 2, 3, … Table 17-15 is repeated as in 2.3 [MD-S2 p.4].
• (F1 − Fc)/(F2 − Fc) = exp(0.5123φ) (Gates); Fc = Kc(V/1000)² with V in ft/min, or Kc(V/2.4)² with V in m/s; Kc from Table 17-16 [MD-S2 p.4].
• Table 17-16 (Kb, Kc) is repeated as in 3.3 [MD-S2 p.5].
• ΔF = 63 025(Hd/Nb)/[n(d/2)], ΔF in lbf, Hd in hp, n in rpm, d in in; F1 = Fc + ΔF exp(f′φ)/[exp(f′φ) − 1]; F2 = F1 − ΔF; Fi = (F1 + F2)/2 − Fc; nfs = Ha Nb/(Hnom Ks) [MD-S2 p.5].
• Durability: bending induces flexural stresses in the belt; the belt tension that induces the same maximum tensile stress is (Fb)1 at the driving sheave and (Fb)2 at the driven pulley. T1 = F1 + (Fb)1 = F1 + Kb/d; T2 = F1 + (Fb)2 = F1 + Kb/D, Kb from Table 17-16 [MD-S2 p.6].
• Life in hours t = Np Lp/(720V), Lp in in and V in ft/min; number of passes 1/Np = (K/T1)^−b + (K/T2)^−b; K and b from Table 17-17. If Np > 10⁹, report Np = 10⁹ [MD-S2 p.6].

Table 17-17 (US). Durability parameters for some V-belt sections [MD-S2 p.6]

• Sheet problem P17-18 [MD-S2 p.7–9]:
  • Statement: two B85 V belts drive a 5.4 in driving sheave at 1200 rpm and a 16 in driven sheave. Find the power capacity based on a service factor of 1.25, and the center distance [p.7].
  • Given: Nb = 2, B85, d = 5.4 in, D = 16 in, n = 1200 rpm, Ks = 1.25; required Hnom and C [p.7].
  • Table 17-11: Lp = 85 + 1.8 = 86.8 in. C = 0.25[A + √(A² − B)] with A = Lp − (π/2)(D + d) = 86.8 − (π/2)(16 + 5.4) = 53.185 and B = 2(D − d)² = 2(16 − 5.4)² = 224.72 → C = 26.05 in [p.7].
  • θd = 180° − 2 sin⁻¹[(16 − 5.4)/(2 × 26.05)] = 156.5°; (D − d)/C = 0.407 → K1 = 0.94 (Table 17-13); K2 = 1 (Table 17-14) [p.8].
  • V = π(1200)(5.4)/12 = 1696.46 ft/min; interpolating between 1000 ft/min (1.59) and 2000 ft/min (2.62) in Table 17-12: Htab = 2.31 hp/belt; Ha = 0.94(1)(2.31) = 2.17 hp/belt [p.8].
  • Assuming nfs = 1: 1 = 2.17 × 2/(Hnom × 1.25) → Hnom = 3.49 hp [p.9].
• Sheet problem P17-19 [MD-S2 p.10–13]:
  • Statement: a 60 hp four-cylinder internal-combustion engine drives a medium-shock brick-making machine on two shifts per day. Two 26 in sheaves about 12 ft apart, sheave speed 400 rpm. Select a Gates Rubber V-belt arrangement, find the factor of safety, and estimate the life in passes and hours [p.10].
  • Given: Hnom = 60 hp, medium shock, D = d = 26 in, C = 12 ft, n = 400 rpm [p.10].
  • C = 144 in: Lp = 2(144) + π(26 + 26)/2 + 0 = 369.68 in. Choosing a D360 belt: Lp,new = 360 + 3.3 = 363.3 in (Table 17-11) [p.10].
  • A = 363.3 − (π/2)(2 × 26) = 281.62 in; B = 2(D − d)² = 0 → C = 0.25[281.62 × 2] = 140.8 in [p.10–11].
  • θd = 180°, (D − d)/C = 0 → K1 = 1; K2 = 1.1 (Table 17-14) [p.11].
  • V = π(400)(26)/12 = 2722.7 ft/min; interpolating D, 17.0 in and up, between 2000 (13.9) and 3000 ft/min (18.1): Htab = 16.94 hp/belt; Ha = 1(1.1)(16.94) = 18.63 hp/belt [p.11].
  • Table 17-15: Ks = 1.4; Hd = 60(1.4)(1) = 84 hp; Nb = 84/18.63 = 4.51 → 5 belts [p.11–12].
  • nfs = 18.63(5)/(60 × 1.4) = 1.109 [p.12].
  • ΔF = 63 025(84/5)/(400 × 13) = 203.62 lbf/belt; Kc = 3.498 → Fc = 3.498(2722.7/1000)² = 25.9 lbf/belt [p.12].
  • exp(0.5123π) = 5; F1 = 25.9 + 203.62(5)/(5 − 1) = 280.425 lbf/belt; F2 = 280.425 − 203.62 = 76.805 lbf/belt; Fi = 152.715 lbf/belt [p.12].
  • T1 = T2 = F1 + Kb/d (as d = D); Kb = 5680 → T1 = T2 = 280.425 + 5680/26 = 498.9 lbf [p.13].
  • K = 4208, b = 11.105 (Table 17-17): Np = [2(4208/498.9)^−11.105]^−1 = 9.6(10⁹) passes [p.13].
  • t = 10⁹(363.3)/(720 × 2722.7) = 185 324.6 h (Np capped at 10⁹) [p.13].
• Problem 17-20 (listed in [MD-3 p.33]) is not solved in the section handout.

3.9 Sheet 2: V-belt and flat-belt problems (Shigley 17-15 to 17-24) [MD-SH2]

The file is a screenshot of Shigley problems 17-15 to 17-24 (the top line is the end of 17-14, "… stainless steel belt."); statements only [MD-SH2 p.1]. Short statements:

• 17-15: beryllium-copper metal flat belt, Sf = 56.67 kpsi, transmits 5 hp at 1125 rev/min with a life goal of 10⁶ belt passes; shafts 20 in apart, centerlines horizontal; f = 0.32; Ks = 1.25, design factor 1.1; driven shaft at one-third of the motor-pulley speed. Specify the belt, pulley sizes and initial tension at installation.
• 17-16: for the conditions of 17-15 use a 1095 plain carbon-steel heat-treated belt; the driving pulley hub requires a pulley outside diameter of 3 in or more. Specify belt, pulley sizes and initial tension.
• 17-17: a single Gates Rubber V belt delivers engine power to the wheel-drive transmission of a riding tractor; 5 hp single-cylinder engine, at most 60% of this power transmitted to the belt; driving sheave 6.2 in, driven 12.0 in; belt as close to a 92 in pitch length as possible; engine governed to a maximum of 3100 rev/min. Select a belt and assess the factor of safety and the belt life in passes.
• 17-18: two B85 V belts, 5.4 in driving sheave at 1200 rev/min, 16 in driven sheave; power capacity for Ks = 1.25, and the center distance. Solved in [MD-S2 p.7–9] (2.8): C = 26.05 in, Ha = 2.17 hp/belt, Hnom = 3.49 hp.
• 17-19: 60 hp four-cylinder IC engine, medium-shock brick-making machine, two shifts per day; two 26 in sheaves about 12 ft apart at 400 rev/min. Select a Gates Rubber V-belt arrangement, find the factor of safety and the life in passes and hours. Solved in [MD-S2 p.10–13] (2.8): five D360 belts, nfs = 1.109, Np = 9.6(10⁹) passes (report 10⁹), t = 185 324.6 h.
• 17-20: reciprocating air compressor with a 5 ft diameter flywheel 14 in wide at 170 rev/min; eight-pole squirrel-cage induction motor, nameplate 50 bhp at 875 rev/min. (a) Design a Gates Rubber V-belt drive from the motor to the compressor flywheel. (b) Can cutting V-belt grooves in the flywheel be avoided by using a V-flat drive?
• 17-21: geometry of a V-flat drive: (a) if the earth's equator were an inextensible string snug to a spherical earth and 6 ft of string were spliced in and arranged concentric to the equator, how far off the ground is the string? (b) Using this, formulate the modifications to mG, φd, φD, Lp and C for a V-flat drive. (c) How would you revise part (b) of Problem 17-20?
• 17-22: 2 hp electric motor at 1720 rev/min drives a blower at 240 rev/min. Select a V-belt drive, specify standard V belts, sheave sizes and the center distance; the motor size limits the center distance to at least 22 in.
• 17-23: the standard roller-chain number gives pitch, proportions, series and strands, e.g., 100H-2: pitch 10/8 in, standard proportions, heavy series, two strands. In Example 17-5, ascertain the pitch from the selected chain number and confirm from Table 17-19.
• 17-24: equate Eqs (17-32) and (17-33) to find the rotating speed n1 at which the power equates and marks the division between the premaximum and postmaximum power domains; (a) show that n1 = [0.25(10⁶)Kr N1^0.42/p^(2.2−0.07p)]^(1/2.4) (the rest is cut off).
• Chains (Example 17-5, Table 17-19, Eqs 17-32/17-33) are not covered by the lecture or the sections; see "Not covered by the sources".

Gears: types, nomenclature and geometry

[MD-4]

4.1 Types and nomenclature

• Types shown: spur, helical, rack and pinion, bevel, miter, worm and worm gear, screw gears, internal gears (Figs 13-1 to 13-4) [MD-4 p.2–4].
• Fig 13-5 nomenclature: top land, face width, addendum circle, face, flank, circular pitch, tooth thickness, width of space, pitch circle, addendum, dedendum, clearance, fillet radius, dedendum circle, clearance circle, bottom land [MD-4 p.5].
• Diametral pitch P = N/d (13-1): teeth per inch of pitch diameter [MD-4 p.6].
• Module m = d/N (13-2), mm [MD-4 p.6].
• Circular pitch p = πd/N = πm (13-3); pP = π (13-4) [MD-4 p.6].

Table 13-1. Standard tooth systems for spur gears (Pd = diametral pitch of the pinion) [MD-4 p.7]

• Table 13-2, tooth sizes in general use. Cutters are generally available for these sizes [MD-4 p.8]:
  • Diametral pitch, coarse: 2, 2¼, 2½, 3, 4, 6, 8, 10, 12, 16.
  • Diametral pitch, fine: 20, 24, 32, 40, 48, 64, 80, 96, 120, 150, 200.
  • Module, preferred: 1, 1.25, 1.5, 2, 2.5, 3, 4, 5, 6, 8, 10, 12, 16, 20, 25, 32, 40, 50.
  • Module, next choice: 1.125, 1.375, 1.75, 2.25, 2.75, 3.5, 4.5, 5.5, 7, 9, 11, 14, 18, 22, 28, 36, 45.
• Table 13-4, standard helical-gear dimensions (in, deg) [MD-4 p.9]:
  • addendum 1.00/Pn; dedendum 1.25/Pn;
  • pinion pitch diameter NP/(Pn cos ψ); gear pitch diameter NG/(Pn cos ψ);
  • normal arc tooth thickness π/Pn − Bn/2 (Bn = normal backlash) (⚠ F20);
  • base diameters: pinion d cos φt, gear D cos φt;
  • base helix angle tan⁻¹(tan ψ cos φt);
  • external gears: center distance (D + d)/2; outside diameters D + 2a and d + 2a; root diameters D − 2b and d − 2b;
  • internal gears: center distance (D − d)/2; inside diameter D − 2a; root diameter D + 2b.
• Table 13-3, 20° straight bevel gears [MD-4 p.10]:
  • working depth hk = 2.0/P; clearance c = (0.188/P) + 0.002 in;
  • gear addendum aG = 0.54/P + 0.460/[P(m90)²]; gear ratio mG = NG/NP;
  • equivalent 90° ratio: m90 = mG when Γ = 90°; m90 = √(mG cos γ/cos Γ) when Γ ≠ 90°;
  • face width F = 0.3A0 or F = 10/P, whichever is smaller (A0 = cone distance);
  • minimum numbers of teeth (pinion/gear): 16/16, 15/17, 14/20, 13/30.
• Common pressure angles are 20° and 25°; the old standard was 14.5° [MD-4 p.11].
• Face width: 3p < F < 5p, i.e., 3π/P < F < 5π/P [MD-4 p.11].

4.2 Action, contact ratio, interference

• Line cd is the pressure line (generating line, line of action). It gives the direction of the resultant force between the gears [MD-4 p.12].
• Layout: draw the pitch circles in contact, then the pressure line at the desired pressure angle, then the base circles tangent to the pressure line [MD-4 p.13].
• Base radius: rb = r cos φ (13-6) [MD-4 p.14].
• First contact (point a) is where the pinion flank touches the gear tip. Last contact (point b) is where the pinion tip touches the gear flank. ab is the line of action; the angle of action is the angle of approach plus the angle of recess [MD-4 p.15].
• A rack is a spur gear of infinite pitch diameter; its sides are straight lines at the pressure angle to the line of centers. Base pitch pb = pc cos φ (13-7) [MD-4 p.16].
• Arc of action qt = qa + qr [MD-4 p.20].
• Contact ratio mc = qt/p (13-8): the average number of pairs of teeth in contact [MD-4 p.20].
• mc = Lab/(p cos φ) (13-9). It should be at least 1.2 (mounting inaccuracies, impact, noise) [MD-4 p.21].
• Interference is contact of non-conjugate portions; it occurs when contact falls below the base circle. The generating process removes the interfering portion (undercutting) [MD-4 p.22].
• Smallest pinion for a one-to-one ratio: NP = [2k/(3 sin²φ)](1 + √(1 + 3 sin²φ)) (13-10). k = 1 for full depth, 0.8 for stub [MD-4 p.23].
• Smallest pinion for ratio mG = NG/NP = m: NP = {2k/[(1 + 2m) sin²φ]}(m + √(m² + (1 + 2m) sin²φ)) (13-11) [MD-4 p.23].
• Largest gear with a given pinion: NG = (NP² sin²φ − 4k²)/(4k − 2NP sin²φ) (13-12) [MD-4 p.24].
• Smallest pinion that runs with a rack: NP = 2k/sin²φ (13-13) [MD-4 p.24].

Interference limits (full depth) [MD-4 p.25–26]

4.3 Bevel and helical geometry

• Straight bevel gears transmit motion between intersecting shafts (Fig 13-3) [MD-4 p.27].
• Pitch angles: tan γ = NP/NG, tan Γ = NG/NP (13-14). Fig 13-20 shows the cone distance A0, face F, back cone, back-cone radius rb and pitch diameter DG [MD-4 p.28].
• Helical teeth are cut at the helix angle to the axis. They add an axial force on the shaft and bearings, and engage gradually, giving a smoother transition [MD-4 p.29].
• ψ = helix angle; the normal pressure angle φn differs from the transverse pressure angle φt [MD-4 p.30].
• pt = transverse circular pitch (plane of rotation) [MD-4 p.31].
• pn = pt cos ψ (13-16); px = pt/tan ψ (13-17); Pn = Pt/cos ψ (13-18); pn Pn = π [MD-4 p.31].
• cos ψ = tan φn/tan φt (13-19) [MD-4 p.32].
• Helical, smallest pinion for 1:1: NP = [2k cos ψ/(3 sin²φt)](1 + √(1 + 3 sin²φt)) (13-21) [MD-4 p.36].
• Helical, ratio m: NP = {2k cos ψ/[(1 + 2m) sin²φt]}[m + √(m² + (1 + 2m) sin²φt)] (13-22) [MD-4 p.36].
• Helical, largest gear: NG = (NP² sin²φt − 4k² cos²ψ)/(4k cos ψ − 2NP sin²φt) (13-23) [MD-4 p.37].
• Helical, smallest pinion with a rack: NP = 2k cos ψ/sin²φt (13-24) [MD-4 p.37].

4.4 Gear trains

• n3 = |N2/N3| n2 = |d2/d3| n2 (13-29), n in rev or rpm [MD-4 p.38].
• For the train of Fig 13-27: n6 = −(N2 N3 N5)/(N3 N4 N6) n2 [MD-4 p.39].
• Train value e = (product of driving tooth numbers)/(product of driven tooth numbers) (13-30); nL = e nF (13-31) [MD-4 p.39].
• Rough guide: one pair gives up to 10:1. Beyond that, compound two gears on the same shaft (Fig 13-28) [MD-4 p.40].

4.5 Worked examples [MD-4]

• Ex 13-1 [p.18–19]:
  • 16T pinion, 40T gear, P = 2, a = 1/P, b = 1.25/P, φ = 20°.
  • p = π/2 = 1.571 in; dP = 8 in, dG = 20 in; C = 14 in.
  • Base radii: pinion 3.759 in, gear 9.397 in.
  • With C increased by ¼ in: dP′ = 8.143 in, dG′ = 20.357 in; φ′ = cos⁻¹[3.759/(8.143/2)] = 22.59°.
• Ex 13-2 [p.33]:
  • Stock helical gear: φn = 20°, ψ = 25°, Pt = 6 teeth/in, 18T.
  • d = 3 in; pt = 0.5236 in; pn = 0.4745 in; px = 1.123 in; Pn = 6.620 teeth/in; φt = 21.88°.
• Helical pinion 17T driving 34T [p.34–35] (⚠ F11 on the example labels):
  • RH, ψ = 30°, φn = 20°, Pn = 5.
  • pn = 0.6283 in; pt = 0.7255 in; px = 1.25 in; pnb = 0.590 in; Pt = 4.33.
  • φt = 22.8°; a = 0.200 in; b = 0.250 in; dP = 3.926 in; dG = 7.852 in.
• Two-stage 30:1 (±1%) train [p.41–42]:
  • Each stage √30 = 5.4772; minimum pinion 16T by Eq (13-11) at 20°; 16√30 = 87.64 ≈ 88; e = (88/16)² = 30.25, within 1%.
  • Exact 30:1: e = (6)(5); N3 = N5 = 16 → N2 = 96, N4 = 80; e = (96/16)(80/16) = 30.
• Sheet problems: Example 13-5 and Problems 13-4, 13-5, 13-7, 13-8, 13-20, 13-23 [MD-4 p.42].

4.6 Section 3: introduction to gears [MD-S3]

The section handout "Machine Design, Sec. (3) Introduction to Gears" is by Eng. Ehsan Khaled (10 pp, typed). It restates the lecture material of 4.1–4.4 with Shigley figures. It has no worked examples. The items below are those it words differently or adds.

• Types of gears [MD-S3 p.1–2]:
  • Spur gears have teeth parallel to the axis of rotation and transmit power between parallel shafts.
  • Helical gears have teeth inclined to the axis of rotation. They can be used for the same applications as spur gears and are then not as noisy, because of the more gradual engagement of the teeth during meshing. Sometimes they transmit motion between nonparallel shafts.
  • Bevel gears have teeth formed on conical surfaces and are used mostly for transmitting motion between intersecting shafts.
  • Worm gearsets transmit rotary motion between nonparallel and nonintersecting shafts.
• Nomenclature (مصطلحات) [MD-S3 p.3]:
  • Pitch circle: a theoretical circle on which all calculations are usually based; the pitch circles of mating gears are tangent. Pitch diameter: the diameter of the pitch circle.
  • Pinion: the smaller of two mating gears; the larger is often called the gear.
  • Circular pitch p: the distance on the pitch circle from a point on one tooth to the corresponding point on the adjacent tooth, i.e., tooth thickness plus width of space.
  • Module m: pitch diameter divided by the number of teeth, customarily in mm; the index of tooth size in SI.
  • Diametral pitch P: number of teeth divided by the pitch diameter, the reciprocal of the module; used only with US units, in teeth per inch.
  • Addendum a: radial distance between the top land and the pitch circle. Dedendum b: radial distance from the bottom land to the pitch circle. Whole depth ht = addendum + dedendum.
  • Clearance circle: a circle tangent to the addendum circle of the mating gear. Clearance c: the amount by which the dedendum of a gear exceeds the addendum of its mate.
  • Face width: when F < 3p a larger gear is needed to carry the load, which needs more space and costs more; when F > 5p the load is likely to be distributed nonuniformly. So 3p ≤ F ≤ 5p, i.e., 3πm ≤ F ≤ 5πm.
• P = N/d; m = d/N; p = πd/N = πm; pP = π. P in teeth per inch, N = number of teeth, d = pitch diameter (in or mm), m = module (mm), p = circular pitch (in or mm) [MD-S3 p.4].
• The line ab has three names: pressure line, generating line, line of action. The angle of action is the sum of the angles of approach and recess. The pressure angle φ is the angle between the pressure line and the perpendicular to the line of centers O1O2; usually 20° or 25°. The pressure line is tangent to both base circles: rbi = ri cos φ, i = 1, 2 [MD-S3 p.4].
• A rack is a spur gear with an infinite pitch diameter; its base pitch is pb = pc cos φ [MD-S3 p.4].
• Pictures and relations repeated from the lecture: internal gear and pinion; straight bevel gears tan γ = NP/NG, tan Γ = NG/NP (γ, Γ = pitch angles of pinion and gear); parallel helical gears pn = pt cos ψ, px = pt/tan ψ, Pn = Pt/cos ψ (from pn Pn = π), cos ψ = tan φn/tan φt [MD-S3 p.5].
• Worm gears: as in 5.0 (Worm gearing): λ and ψG, px and pt, dG = NG pt/π, C^0.875/3.0 ≤ dW ≤ C^0.875/1.7, L = px NW, tan λ = L/(π dW); FG = length of a tangent to the worm pitch circle between its intersections with the addendum circle [MD-S3 p.6].
• A tooth system is a standard that specifies the relationships among addendum, dedendum, working depth, tooth thickness and pressure angle. The standards were originally planned to attain interchangeability of gears of all tooth numbers but of the same pressure angle and pitch [MD-S3 p.7].
• Tables 13-1, 13-2, 13-3 are repeated as in 4.1 [MD-S3 p.7]. Table 13-4 is repeated, but its normal arc tooth thickness is printed π/(2Pn) − Bn/2 (⚠ F20) [MD-S3 p.8]. Table 13-5 (worm pressure angles and tooth depths) is repeated as in 5.0 [MD-S3 p.8].
• Number of teeth [MD-S3 p.8–9]:
  • The pinion is the smallest driving gear. Based on the tooth system and φ, the pinion must have more teeth than the minimum NP, to avoid interference between the gears and so avoid undercut: NP = [2k/(3 sin²φ)](1 + √(1 + 3 sin²φ)), k = 1 for full depth, 0.8 for stub.
  • If mG = NG/NP = m > 1: NP = {2k/[(1 + 2m) sin²φ]}(m + √(m² + (1 + 2m) sin²φ)).
  • Largest gear: NG = (NP² sin²φ − 4k²)/(4k − 2NP sin²φ).
• Contact ratio: the average number of pairs of teeth in contact; mc = qt/p, qt = qa + qr; mc = Lab/(p cos φ) (figure: arc of approach qa, arc of recess qr, pressure line, addendum and pitch circles, Lab) [MD-S3 p.9].
• Gear trains: n3 = |N2/N3| n2 = |d2/d3| n2 (n = revolutions or rev/min, N = number of teeth, d = pitch diameter); for gears 2–6, n6 = −(N2 N3 N5)/(N3 N4 N6) n2; e = ±(product of driving tooth numbers)/(product of driven tooth numbers); nL = e nF [MD-S3 p.9–10].
• e is positive if the last gear rotates in the same sense as the first and negative if opposite. A train value of up to 10 to 1 can be obtained with one pair; to obtain more, compound two gears onto the same shaft [MD-S3 p.10].

4.7 Sheet 4: gear problems (Shigley 13-1 to 13-25) [MD-SH4]

The file (شيت4, 3 pp) is a scan of the Shigley Chapter 13 problem list, book pp. 724–728; problems marked * are linked to other chapters (Table 1-2). Statements only [MD-SH4 p.1–3]. Short statements:

• 13-1: 17T spur pinion, P = 8 teeth/in, 1120 rev/min, drives a gear at 544 rev/min; find NG and the theoretical center distance [p.1].
• 13-2: 15T spur pinion, m = 3 mm, 1600 rev/min; driven gear 60T; find the gear speed, circular pitch and theoretical center distance [p.1].
• 13-3: spur gearset, m = 6 mm, velocity ratio 4, pinion 16T; find NG, the pitch diameters and the theoretical center distance [p.1].
• 13-4: 21T spur pinion with a 28T gear, P = 3 teeth/in, φ = 20°; draw the gears with one tooth on each; tabulate addendum, dedendum, clearance, circular pitch, tooth thickness, base-circle diameters, arcs of approach, recess and action, base pitch and contact ratio [p.1].
• 13-5: 20° straight bevel pinion, 14T, P = 6 teeth/in, drives a 32T gear, shafts at right angles in one plane; find (a) cone distance, (b) pitch angles, (c) pitch diameters, (d) face width [p.1].
• 13-6: parallel helical gearset, 20T pinion driving 36T; RH ψ = 30°, φn = 25°, Pn = 4 teeth/in; find (a) normal, transverse and axial circular pitches, (b) normal base circular pitch, (c) Pt and φt, (d) addendum, dedendum and pitch diameter of each gear [p.1].
• 13-7: parallel helical gearset, 19T pinion driving 57T; LH ψ = 30°, φn = 20°, mn = 2.5 mm; find (a) pn, pt, px, (b) Pt and φt, (c) addendum, dedendum and pitch diameters [p.1].
• 13-8: minimum pinion teeth to avoid interference, 20° spur, for ratios (a) 2:1, (b) 3:1, (c) 4:1, (d) 5:1 [p.1].
• 13-9: repeat 13-8 with 25° [p.1].
• 13-10: spur φ = 20°, avoiding interference: (a) smallest pinion that runs with itself; (b) smallest pinion at mG = 2.5 and the largest gear possible with it; (c) smallest pinion that runs with a rack [p.1].
• 13-11: repeat 13-10 for a helical gearset with φn = 20° and ψ = 30° [p.1].
• 13-12: φn = 20°, Pt = 6 teeth/in, ψ = 30°, 2:1 reduction; choose the smallest acceptable full-depth pinion and gear tooth counts to avoid interference [p.1].
• 13-13: repeat 13-12 with ψ = 45° [p.1].
• 13-14: full-depth spur gears; the smallest pressure angle φ obtainable without undercutting for a 9-tooth pinion meshing with a rack [p.1].
• 13-15: 18T helical pinion driving 32T; LH ψ = 25°, φn = 20°, mn = 3 mm; find (a) pn, pt, px, (b) transverse module and φt, (c) the pitch diameters [p.1].
• 13-16: double-reduction helical gearset driven through shaft a at 700 rev/min; gears 2, 3: Pn = 12 teeth/in, ψ = 30°, φn = 20°; gears 4, 5: Pn = 8, ψ = 25°, φn = 20°; N2 = 12, N3 = 48, N4 = 16, N5 = 36 (figure with bearings A–F, dimensions in inches). Find (a) the thrust direction of each gear on its shaft, (b) speed and direction of shaft c, (c) center distance between shafts [p.2].
• 13-17: shaft a at 600 rev/min in the direction shown (gears 20T ψ = 30° RH, 40T, 8T ψ = 60° RH, 17T ψ = 30° RH, 20T, 60T); find the speed and direction of shaft d [p.2].
• 13-18: gears and pulleys driving gear 9; pulley 2 (6 in dia) at 1200 rev/min drives a 10 in pulley; gears 18T, 38T, 48T, 20T, a 3T RH worm and a 36T worm gear 9. Determine the speed and direction of gear 9 [p.2–3].
• 13-19: a pair of helical gears (17½° normal pressure angle; 8 normal DP, 12T, 23° ψ, 40T; 540 rev/min) and a pair of miter gears (4P, 32T, 32T); find (a) the speed of shaft c, (b) the distance between shafts a and b, (c) the pitch diameter of the miter gears [p.3].
• 13-20: compound reverted gear train as a speed increaser giving exactly 45:1; φ = 20°; choose tooth numbers to minimize size while avoiding interference; same diametral pitch for all gears [p.3].
• 13-21: repeat 13-20 with 25°. 13-22: repeat 13-20 for exactly 30:1. 13-23: repeat 13-20 for approximately 45:1 [p.3].
• 13-24: compound reverted gearbox transmitting 25 hp, input 2500 rev/min, output 280–300 rev/min, 20° spur gears; choose tooth numbers minimizing size, output within range, avoiding interference [p.3].
• 13-25: automotive differential, N2 = 16, N3 = 48, N4 = 14, N5 = N6 = 20; drive shaft at 900 rev/min. (a) Wheel speeds in a straight line on a good road; (b) right wheel jacked up, left on a good road: right-wheel speed; (c) rear-wheel drive parked with the right wheel on wet ice: does (b) hint at what happens when starting off? [p.3].
• The lecture assigns Example 13-5 and Problems 13-4, 13-5, 13-7, 13-8, 13-20, 13-23 from this list [MD-4 p.42].

Gear force analysis

[MD-41, MD-S4]

Figure: md-helical-force-components.jpg (W resolved into Wt, Wr, Wa for a helical tooth) [MD-S4 p.4].

5.0 Lecture treatment of gear forces [MD-41]

Lecture file "Lecture_4-1" is titled on its first slide "Lecture #4, Design of Mechanical Elements: Gears: Spur, Bevel and Helical" [MD-41 p.1]. Its pages 1–42 repeat the Lecture #4 deck page for page (section 4 above; the MD-4 page numbers equal the MD-41 page numbers). Pages 43–88 add force analysis for spur, bevel, helical and worm gearing; page 89 is "Thank you / Questions" [MD-41 p.89]. It does not contain the Lewis or AGMA stress equations.

• Extra items in pages 1–42 not listed in section 4 [MD-41]:
  • Photos of gearboxes and a bevel-gear shaft [p.2]; a sketch labelling line of contact, pitch point, line of action, base circle, the tooth force W, pressure angle φ and line of centers; pictures of a full-depth tooth and a stub tooth (pitch circle and base circle marked) [p.7].
  • Footnote to Table 13-2: tooth sizes are standardized by the American Gear Manufacturers Association (AGMA); www.agma.org [p.8].
  • Fig 13-14, internal gear: internal gear 3 (ω3) meshing with pinion 2 (ω2), with pitch, base, addendum and dedendum circles and the pressure line [p.17].

Spur gearing [MD-41 p.43–49]

• Fig 13-32: (a) pinion 2 on shaft a driving gear 3 on shaft b, pressure angle φ; (b) FBD of the pinion: F32 at φ, shaft force Fa2, torque Ta2; (c) FBD of the gear: F23, Fb3, Tb3 [p.43].
• Transmitted load Wt is the tangential load, Wt = F^t_32. It is the useful component of the force, the one that transmits the torque: T = (d/2)Wt (Fig 13-33: F32 resolved into F^r_32 and F^t_32; Fa2 into F^r_a2 and F^t_a2) [p.44].
• Transmitted power: H = Tω = (Wt d/2)ω (13-33) [p.45].
• Pitch-line velocity (the linear velocity of a point on the gear at the pitch-circle radius; a common term in tabulating gear data): V = πdn/12 (13-34), V in ft/min, d in in, n in rev/min [p.45].
• Wt = 33 000 H/V (13-35): Wt in lbf, H in hp, V in ft/min [p.46].
• SI: Wt = 60 000 H/(πdn) (13-36): Wt in kN, H in kW, d in mm, n in rev/min [p.46].
• Example 13-7 (idler) [p.47–49]:
  • Data: pinion 2 (20T) at 1750 rev/min transmits 2.5 kW to idler gear 3 (50T), which drives gear 4 (30T); 20° full depth, m = 2.5 mm. Draw the FBD of gear 3 (Fig 13-34).
  • d2 = 20(2.5) = 50 mm; d3 = 50(2.5) = 125 mm.
  • Wt = 60 000(2.5)/[π(50)(1750)] = 0.546 kN = F^t_23; F^r_23 = 0.546 tan 20° = 0.199 kN; F23 = 0.546/cos 20° = 0.581 kN.
  • The idler transmits no torque, so F^t_43 = 0.546 kN, F^r_43 = 0.199 kN, F43 = 0.581 kN.
  • F^x_b3 = −(F^t_23 + F^r_43) = −(−0.546 + 0.199) = 0.347 kN; F^y_b3 = −(F^r_23 + F^t_43) = −(0.199 − 0.546) = 0.347 kN.
  • Fb3 = √(0.347² + 0.347²) = 0.491 kN.

Bevel gearing [MD-41 p.50–56]

• Wt = T/r_av (13-37), where r_av is the pitch radius at the midpoint of the tooth for the gear under consideration [p.50].
• Wr = Wt tan φ cos γ; Wa = Wt tan φ sin γ (13-38) [p.50].
• The forces act at the center of the tooth (Fig 13-35); the resultant W has a tangential component Wt, a radial component Wr and an axial component Wa [p.50].
• Direction sketch: pinion as drive gear, gear as driven gear, showing Ft1, Fr1, Fx1 on the pinion and Ft2, Fr2, Fx2 on the gear for both rotation senses [p.51].
• Example 13-8 [p.52–56]:
  • Data: bevel pinion (15T, module 5) at 600 rev/min transmits 3.75 kW to a 45T gear (Fig 13-36a). Average pitch radii 32 mm (pinion) and 90 mm (gear); dimensions 165, 90, 75, 60, 32, 90, 225 mm. Bearings A and C take the thrust. Find the bearing forces on the gearshaft.
  • γ = tan⁻¹(75/225) = 18.4°; Γ = tan⁻¹(225/75) = 71.6° [p.53].
  • V = 2π(32)(600)/60 = 2011 mm/s; Wt = H/V = 3750/2.011 = 1865 N [p.53].
  • With Γ replacing γ: Wr = 1865 tan 20° cos 71.6° = 214 N; Wa = "Wt tan φ cos Γ" = 1865 tan 20° sin 71.6° = 644 N (⚠ F14) [p.53].
  • Wt acts in +x, Wa in −y (Fig 13-36b) [p.55].
  • RG (D to G) = 90i − (60 + 32)j = 90i − 92j; RC (D to C) = −(60 + 90)j = −150j [p.55].
  • ΣM_D: RG × W + RC × FC + T = 0 (1), i.e., (90i − 92j) × (−214i − 644j + 1865k) + (−150j) × (F^x_C i + F^y_C j + F^z_C k) + Tj = 0 (2) [p.55].
  • After the cross products: (−171 580i − 167 850j − 77 712k) + (−150F^z_C i + 150F^x_C k) + Tj = 0, giving T = 168j N·m, F^x_C = 518 N, F^z_C = −1144 N (3) [p.56].
  • ΣF: FD + FC + W = 0 (4); (F^x_D i + F^z_D k) + (518i + F^y_C j − 1144k) + (−214i − 644j + 1865k) = 0 (5) [p.56].
  • F^y_C = 644 N, so FC = 518i + 644j − 1144k N; FD = −303i − 721k N [p.56].
  • The analysis of the pinion shaft is similar [p.56].

Helical gearing [MD-41 p.57–64]

• Wr = W sin φn; Wt = W cos φn cos ψ; Wa = W cos φn sin ψ (13-39), with cos ψ = tan φn/tan φt (13-19) [p.57].
• Wr = Wt tan φt; Wa = Wt tan ψ; W = Wt/(cos φn cos ψ) (13-40) [p.57].
• W = total force; Wr = radial component; Wt = tangential component, also called the transmitted load; Wa = axial component, also called the thrust load (Fig 13-37) [p.57].
• Direction sketches: (I) right-hand pinion as drive gear with left-hand gear as driven gear; (II) left-hand pinion as drive gear with right-hand gear as driven gear; plus a photo of hands showing spur, left-hand and right-hand helical gears [p.58].
• Example 13-9 [p.59–64]:
  • Data: 750 W motor at 1800 rev/min, clockwise viewed from +x; 18T helical pinion, φn = 20°, ψ = 30°, normal module 3.0 mm, meshing with a 36T gear; bearings A and B 250 mm apart, pinion 75 mm beyond B (Fig 13-38). The thrust is taken out at A.
  • φt = tan⁻¹(tan 20°/cos 30°) = 22.8° [p.60].
  • mt = mn/cos ψ = 3/cos 30° = 3.46 mm; dp = 18(3.46) = 62.3 mm; V = π(62.3)(1800)/60 = 5871.6 mm/s = 5.87 m/s ("From Eq. 13-1, 13-2, 13-18") [p.60].
  • Wt = H/V = 750/5.87 = 128 N; Wr = 128 tan 22.8° = 54 N; Wa = 128 tan 30° = 74 N; W = 128/(cos 20° cos 30°) = 157 N [p.62].
  • Scalar solution: Wr in −y, Wa in −x, Wt in +z at C (Fig 13-39). F^x_A = Wa = 74 N. ΣM_z: −(54)(325) + (74)(62.3/2) + 250F^y_B = 0 → F^y_B = 61 N; ΣF_y → F^y_A = 7 N. ΣM_y: 250F^z_B − (128)(325) = 0 → F^z_B = 166 N; ΣF_z → F^z_A = 38 N. T = Wt dp/2 = 128(62.3/2) = 3982 N·mm (⚠ F15) [p.63].
  • Vector solution: W = −74i − 54j + 128k N; RB = 250i; RC = 325i + 31.15j. ΣM_A: RB × FB + T + RC × W = 0 → (250F^y_B k + 250F^z_B j) − Ti + (3987i − 41 600j − 15 245k) = 0, so T = 4 kN·mm, F^y_B = 61 N, F^z_B = 166 N. FA = −FB − W = 74i − 7j + 38k N [p.64].

Worm gearing [MD-41 p.65–88]

• Pictures: worm and worm gear; worm types (a) straight face, (b) hobbed straight face, (c) concave face; worm nomenclature: axial pitch, normal pitch, lead angle, pitch diameter, bore diameter, hub diameter, hub widths (L) and (R) [p.65–66].
• It is common to specify the lead angle λ for the worm and the helix angle ψG for the gear; the two are equal for a 90° shaft angle [p.67].
• It is common to specify the axial pitch px for the worm and the transverse circular pitch pt for the gear; these are equal if the shaft angle is 90° [p.67].
• The pitch diameter of the gear is measured on the plane containing the worm axis: dG = NG pt/π (13-25) (Fig 13-24: worm pitch diameter dw, root diameter, pitch cylinder, helix, helix angle ψW, lead L, lead angle λ) [p.67].
• The worm pitch diameter is not related to the number of teeth, so the worm may have any pitch diameter; it should be the same as the pitch diameter of the hob used to cut the worm-gear teeth [p.68].
• Recommended worm pitch diameter against center distance C: C^0.875/3.0 ≤ dW ≤ C^0.875/1.7 (13-26) [p.68].
• Lead and lead angle: L = px NW (13-27); tan λ = L/(π dW) (13-28). Sketches: single start L = p, double start L = 2p, triple start L = 3p [p.68].

Table 13-5. Recommended pressure angles and tooth depths for worm gearing [MD-41 p.69]

• Face width FG of a worm gear should equal the length of a tangent to the worm pitch circle between its points of intersection with the addendum circle (Fig 13-25) [p.70].
• Schematics of right engagement (worm torque τw, gear torque τg) and force diagrams with Fn, Ff, λ [p.71].
• Friction neglected: the only force exerted by the gear is W (Fig 13-40), with Wˣ = W cos φn sin λ; Wʸ = W sin φn; Wᶻ = W cos φn cos λ (13-41) [p.72].
• In spur gears the relative tooth motion is primarily rolling (pure rolling at the pitch point). Between worm and worm-gear teeth it is pure sliding, so friction plays an important role [p.73].
• Direction sketches: (I) right-hand worm gear pair, (II) left-hand worm gear pair, each with the worm as drive gear and the worm wheel as driven gear [p.74].
• The gear forces are opposite to the worm forces: W_Wt = −W_Ga = Wˣ; W_Wr = −W_Gr = Wʸ; W_Wa = −W_Gt = Wᶻ (13-42) [p.75].
• Relative motion in worm gearing is sliding, so friction is much more significant than in other types of gears. The normal force W produces a friction force Wf = fW, with fW cos λ in the −x direction and fW sin λ in the +z direction [p.76].
• With friction (13-43): Wˣ = W(cos φn sin λ + f cos λ); Wʸ = W sin φn; Wᶻ = W(cos φn cos λ − f sin λ) [p.76].
• Wf = fW = f W_Gt/(f sin λ − cos φn cos λ) (13-44); W_Wt = W_Gt (cos φn sin λ + f cos λ)/(f sin λ − cos φn cos λ) (13-45) [p.76].
• Efficiency η = W_Wt(without friction)/W_Wt(with friction) (a); substituting (13-45) with f = 0 in the numerator gives η = (cos φn − f tan λ)/(cos φn + f cot λ) (13-46) [p.77].
• Table 13-6 (f = 0.05, φn = 20°): same values as in 5.3 below [p.78].
• VG = pitch-line velocity of the gear; VW = pitch-line velocity of the worm; f depends on the sliding velocity VS; vectorially VW = VG + VS, so VS = VW/cos λ (13-47) (Fig 13-41) [p.79].
• Fig 13-42: representative f for worm gearing, based on good lubrication, against VS = 0–2000 ft/min (f axis 0–0.10). Curve B for high-quality materials (case-hardened steel worm with a phosphor-bronze gear); curve A for more friction (cast-iron worm with a cast-iron gear) [p.80].
• Example 13-10 [p.81–88]:
  • Data: 2-tooth right-hand worm, 1 hp at 1200 rev/min, 30T worm gear; gear Pt = 6 teeth/in, face 1 in; worm pitch diameter 2 in, face 2½ in; φn = 14½°; curve B of Fig 13-42. Gear shaft on bearings A and B (Fig 13-43; dimensions 1½ in and 2½ in).
  • (a) px = pt = π/6 = 0.5236 in; dG = 30/6 = 5 in; C = (2 + 5)/2 = 3.5 in; L = 0.5236(2) = 1.0472 in; λ = tan⁻¹[1.0472/(π·2)] = 9.46° [p.82].
  • (b) Directions: by the right-hand rule the worm thrust is +z; with the bolt-and-nut analogy the gear surface in contact moves in −z, so the gear rotates clockwise about x (right thumb in −x) [p.83].
  • VW = π(2)(1200)/12 = 628 ft/min; nG = (2/30)(1200) = 80 rev/min; VG = π(5)(80)/12 = 105 ft/min [p.83].
  • VS = 628/cos 9.46° = 637 ft/min; W_Wt = 33 000(1)/628 = 52.5 lbf (in −x, as in Fig 13-40); f = 0.03 from Fig 13-42 [p.84].
  • W = Wˣ/(cos φn sin λ + f cos λ) = 52.5/(cos 14.5° sin 9.46° + 0.03 cos 9.46°) = 278 lbf [p.84].
  • Wʸ = 278 sin 14.5° = 69.6 lbf; Wᶻ = 278(cos 14.5° cos 9.46° − 0.03 sin 9.46°) = 264 lbf [p.85].
  • On the gear: W_Ga = −Wˣ = 52.5 lbf; W_Gr = −Wʸ = −69.6 lbf; W_Gt = −Wᶻ = −264 lbf (FBD Fig 13-44) [p.85–86].
  • B is made the thrust bearing to put the gearshaft in compression: F^x_B = −52.5 lbf [p.87].
  • ΣM_z: −(52.5)(2.5) − (69.6)(1.5) + 4F^y_B = 0 → F^y_B = 58.9 lbf. ΣM_y: (264)(1.5) − 4F^z_B = 0 → F^z_B = 99 lbf [p.87].
  • ΣF_y: −69.6 + 58.9 + F^y_A = 0 → F^y_A = 10.7 lbf. ΣF_z: −264 + 99 + F^z_A = 0 → F^z_A = 165 lbf. ΣM_x: −(264)(2.5) + T = 0 → T = 660 lbf·in [p.88].
  • Because of the frictional loss, this output torque is less than the product of the gear ratio and the input torque [p.88].
• Sheet problems: Problems 13-32, 13-39, 13-49, 13-55, 13-57 [MD-41 p.88]. These are the problems solved in [MD-S4] (5.4 below).

5.1 Notation and procedure

• The frame is numbered 1, the input gear 2, then the gears 3, 4, … in order; shafts are a, b, c, … [MD-S4 p.1].
• F23 = force exerted by gear 2 on gear 3; F2a = force of gear 2 on shaft a; Fa2 = force of shaft a on gear 2. Superscripts r and t mark the radial and tangential components (e.g., F^t_43) [MD-S4 p.1].
• Steps: (1) get the gearing parameters (dimensions, angles); (2) mark the directions of rotation to find the force directions; (3) draw the free-body diagram; (4) calculate the forces and power, then apply the equilibrium equations [MD-S4 p.2].
• Direction rules (the notes are in Arabic) [MD-S4 p.8]:
  • Spur: the tangential force on the driving gear opposes its rotation; on the driven gear it acts with the rotation. The radial force points toward the gear center (it tries to separate the two gears).
  • Bevel: same tangential and radial rules; the axial force acts along the shaft axis, tending to separate the gears.
  • Helical and worm: the axial direction comes from the right- or left-hand rule (fingers curled with the rotation, thumb gives the axial force on the driver). The driven gear gets the opposite axial force.

5.2 Transmitted load

• Wt = F^t_32 [MD-S4 p.2].
• US: Wt = 33 000 H/V, with Wt in lbf, H in hp, V = pitch-line velocity in ft/min [MD-S4 p.2].
• SI: Wt = 60 000 H/(π d n), with Wt in kN, H in kW, d in mm, n in rev/min [MD-S4 p.2].
• Torque T = (d/2)Wt; power H = Tω = (Wt d/2)ω [MD-S4 p.2].

5.3 Components by gear type

• Spur: Wr = Wt tan φ (as used in the examples) [MD-S4 p.9, p.11].
• Bevel [MD-S4 p.3]:
  • Wt = T/r_av, where r_av = pitch radius at the mid-face (or from H and V).
  • Wr = Wt tan φ cos γ.
  • Wa = Wt tan φ sin γ.
• Helical [MD-S4 p.4]:
  • Wt = W cos φn cos ψ (or from H and V).
  • Wr = W sin φn.
  • Wa = W cos φn sin ψ.
  • Also cos ψ = tan φn/tan φt.
• Worm, friction neglected [MD-S4 p.5]:
  • Wˣ = W cos φn sin λ; Wʸ = W sin φn; Wᶻ = W cos φn cos λ.
  • W_Wt = −W_Ga = Wˣ; W_Wr = −W_Gr = Wʸ; W_Wa = −W_Gt = Wᶻ.
• Worm with friction: the friction force Wf = fW has a component fW cos λ in −x and fW sin λ in +z [MD-S4 p.6]. Then:
  • Wˣ = W(cos φn sin λ + f cos λ).
  • Wʸ = W sin φn.
  • Wᶻ = W(cos φn cos λ − f sin λ).
  • Wf = fW = f W_Gt/(f sin λ − cos φn cos λ).
  • W_Wt = W_Gt (cos φn sin λ + f cos λ)/(f sin λ − cos φn cos λ).
• Worm efficiency: η = W_Wt(without friction)/W_Wt(with friction) = (cos φn − f tan λ)/(cos φn + f cot λ) [MD-S4 p.6].
• Table 13-6, efficiency of worm gearsets for f = 0.05 [MD-S4 p.6]:

• The worm friction coefficient depends on the sliding velocity [MD-S4 p.7]:
  • VW = VG + VS (vector sum); VG = gear pitch-line velocity, VW = worm pitch-line velocity; VS = VW/cos λ.
  • Fig 13-38 plots f against VS (0–2000 ft/min).
  • Curve B is for high-quality materials (case-hardened steel worm with a phosphor-bronze gear). Curve A is for more friction (cast-iron worm with a cast-iron gear).
• Worm steps [MD-S4 p.7–8]:
  • The worm and gear teeth have the same hand.
  • Get the worm and gear pitch-line velocities and the sliding velocity.
  • Read f from Fig 13-38, then find the forces from the laws and the FBD.

5.4 Worked problems [MD-S4]

• P13-32, spur with an idler [p.9–10]:
  • Data: 75 kW at 1000 rpm, m = 5 mm, 20°; N2 = 17, N3 = 34 (idler), N4 = 51 → d = 85, 170, 255 mm.
  • F^t_23 = 60 000(75)/(π·85·1000) = 16.85 kN; F^r_23 = 16.85 tan 20° = 6.1335 kN.
  • The idler transmits no torque, so F^t_43 = 16.85 kN and F^r_43 = 6.1335 kN.
  • F^x_3b = 2(16.85) = 33.7 kN, F^y_3b = 0, so F3b = 33.7 kN.
  • T2 = 16.85(85/2) = 716.125 N·m; T4c = 716.125(51/17) = 2148.375 N·m.
• P13-39, spur with an idler pair [p.11–12]:
  • Data: m = 12 mm, 20°, 1800 rpm, 150 kW; N = 18, 32, 18, 48 → d = 216, 384, 216, 576 mm.
  • F^t_23 = 7.36 kN; F^r_23 = 2.68 kN.
  • Idler: F^t_54(216/2) = 7.36(384/2) → F^t_54 = 13.08 kN; F^r_54 = 4.76 kN.
  • F^x_b = 4.76 − 7.36 = −2.61 kN; F^y_b = 13.08 − 2.68 = 10.42 kN; Fb = 10.742 kN.
• P13-49, straight bevel [p.13–16]:
  • Data: 16T 20° pinion driving 32T, 2.5 hp at 240 rpm; bearing A takes radial and thrust.
  • DG = 8 in, DP = 4 in; γ = 26.56°, Γ = 63.43°; face 1.5 in.
  • Mean diameters: dp = 2(2 − 1.5 sin 26.56°/2) = 3.33 in; dG = 6.66 in.
  • V = 209.23 ft/min; Wt = 394.3 lbf; Wr = 128.37 lbf; Wa = 64.17 lbf.
  • ΣM_A (vector) gives FBz = −815.4, FBx = −222.732 and T = −656.5 lbf·in; FB = 845.27 lbf (radial).
  • ΣF gives FA = (94.362, 64.17, 421.1) lbf; |FA| = 436.3 lbf; radial 431.54 lbf; thrust 64.17 lbf.
• P13-55, helical with an idler [p.17–19]:
  • Data: 16T, φt = 20°, ψ = 15°, m = 4 mm, idler 36T, driven 28T, 1600 rpm, 6 kW; d = 64, 144, 112 mm.
  • φn = tan⁻¹(cos 15° tan 20°) = 19.37°.
  • Wt = 1.119 kN; W = Wt/(cos φn cos ψ) = 1.228 kN; Wr = 0.407 kN; Wa = 0.3 kN.
  • T2 = 35.81 N·m.
  • Shaft forces: F2a = (1.119i + 0.407j + 0.3k) kN; Fb = (−0.712i + 0.712j) kN; Fc = (−0.407i − 1.119j − 0.3k) kN.
• P13-57, worm [p.20–23] (⚠ F9):
  • Data: RH single-tooth hardened-steel worm, 2000 W at 600 rpm, 48T cast-iron gear, px = 25 mm, φn = 14.5°, dw = 100 mm.
  • dG = 48(25)/π = 381.97 mm; L = 25 mm; λ = tan⁻¹[25/(π·100)] = 4.55°.
  • Vw = 3.14 m/s; Vs = 3.15 m/s ≈ 620 ft/min; curve B gives f = 0.03.
  • W_Wt = 2000/3.14 = 637 N; W = 5969.58 N; Wy = 1494.66 N; Wz = 5747 N.
  • Printed bearing results: FB = 2207.18 N (radial); FA radial 3692.5 N, thrust 5747 N; T = 31.85 N·m. Bearing A takes the thrust so that the shaft is in compression.

Gear tooth stresses: Lewis equation and AGMA method

[MD-5]

Lecture file "Lecture_5.pdf" (92 pp) is titled "Machine Design (DPE 301), Lecture #5, Design of Mechanical Elements: Gears: Spur, Helical", Prof. Dr. Mohamed Agwa, 1444/2022-2023 [MD-5 p.1]. It is Shigley Chapter 14 (not a fatigue lecture), although it uses the Marin factors of [MD-F] (section 10) in Example 14-2.

6.1 Introduction

• The chapter is devoted primarily to analysis and design of spur and helical gears to resist bending failure of the teeth and pitting failure of the tooth surfaces [MD-5 p.2].
• Failure by bending occurs when the significant tooth stress equals or exceeds either the yield strength or the bending endurance strength [MD-5 p.2].
• A surface failure occurs when the significant contact stress equals or exceeds the surface endurance strength [MD-5 p.2].
• The American Gear Manufacturers Association (AGMA) has for many years been the responsible authority for the dissemination of knowledge on the design and analysis of gearing [MD-5 p.2].
• The general AGMA approach requires a great many charts and graphs, too many for a single chapter; many are omitted by choosing a single pressure angle and using only full-depth teeth. This reduces complexity but still allows a basic understanding of the approach [MD-5 p.2].

6.2 Lewis bending equation

• Wilfred Lewis introduced (1892) an equation for estimating the bending stress in gear teeth in which the tooth form enters the formulation; it still remains the basis for most gear design today [MD-5 p.3].
• Fig 14-1: (a) cantilever of depth t, width F, length l with load Wt at the end; (b) gear tooth with W resolved into Wr and Wt at the tip, inscribed parabola touching the fillet (radius rf) at point a, distance x, tooth thickness t at the root section [MD-5 p.3–4]. A tooth-nomenclature picture (face width, top land, addendum circle, circular pitch, tooth thickness, width of space, pitch circle, addendum, dedendum, working depth, whole depth, clearance, fillet radius, dedendum circle, clearance circle, flank, bottom land) is repeated [MD-5 p.3].
• Cantilever model: σ = M/(I/c) = 6Wt l/(F t²) [MD-5 p.4].
• Similar triangles: (t/2)/x = l/(t/2), i.e., x = t²/(4l) or l = t²/(4x) [MD-5 p.4].
• σ = 6Wt l/(F t²) = (Wt/F)·1/(t²/6l) = (Wt/F)·1/(t²/4l)·1/(4/6); σ = Wt p/[F(2/3)x p] [MD-5 p.4].
• y = 2x/(3p), where p is the circular pitch; y is called the Lewis form factor [MD-5 p.4].
• σ = Wt/(F p y) (14-1) [MD-5 p.4–5].
• With p = π/P and Y = πy: Lewis equation σ = Wt P/(F Y) (14-2); Lewis form factor Y = 2xP/3 (14-3) [MD-5 p.5].

Table 14-2. Values of the Lewis form factor Y (normal pressure angle 20°, full-depth teeth, diametral pitch of unity in the plane of rotation) [MD-5 p.6]

6.3 Dynamic effects: velocity factor (Barth)

• When a pair of gears is driven at moderate or high speed and noise is generated, dynamic effects are certainly present. Carl G. Barth first expressed the velocity factor in the nineteenth century; the velocity factor Kv accounts for this [MD-5 p.7].
• V in ft/min [MD-5 p.7]:
  • Kv = (600 + V)/600, cast iron, cast profile (14-4a)
  • Kv = (1200 + V)/1200, cut or milled profile (14-4b)
  • The Barth equation is often modified for cut or milled teeth; AGMA later added: Kv = (50 + √V)/50, hobbed or shaped profile (14-5a); Kv = √[(78 + √V)/78], shaved or ground profile (14-5b)
• V in m/s [MD-5 p.8]:
  • Kv = (3.05 + V)/3.05, cast iron, cast profile (14-6a)
  • Kv = (6.1 + V)/6.1, cut or milled profile (14-6b)
  • Kv = (3.56 + √V)/3.56, hobbed or shaped profile (14-6c)
  • Kv = √[(5.56 + √V)/5.56], shaved or ground profile (14-6d)
• Lewis equation with velocity factor: US σ = Kv Wt P/(F Y) (14-7); metric σ = Kv Wt/(F m Y) (14-8) [MD-5 p.9].
• These are acceptable for general estimation of stresses in gear teeth and form the basis of the AGMA method, which is the preferred approach. They are in general use for estimating the capacity of gear drives when life and reliability are not important considerations, and are useful for a preliminary estimate of gear sizes [MD-5 p.9].

6.4 Examples 14-1 and 14-2 (Lewis)

• Example 14-1 [MD-5 p.10–11]:
  • Data: stock spur gear, P = 8 teeth/in, face 1½ in, 16 teeth, 20° full depth; AISI 1020 as-rolled; nd = 3; rate the horsepower at 1200 rev/min for moderate applications.
  • "Moderate applications" implies rating by the yield strength. Table A-20: Sut = 55 kpsi, Sy = 30 kpsi; σall = 30/3 = 10 kpsi.
  • d = N/P = 16/8 = 2 in; V = π(2)1200/12 = 628 ft/min; Kv (14-4b) = (1200 + 628)/1200 = 1.52.
  • Y = 0.296 (16 teeth); Wt = F Y σall/(Kv P) = 1.5(0.296)10 000/[1.52(8)] = 365 lbf.
  • hp = Wt V/33 000 = 365(628)/33 000 = 6.95 hp.
• Example 14-2 (same gear, infinite life in bending) [MD-5 p.12–]:
  • S′e = 0.5Sut = 0.5(55) = 27.5 kpsi (Eq 6-8, p. 290).
  • Machined surface (Table 6-3, p. 298): a = 2.70, b = −0.265; ka = 2.70(55)^−0.265 = 0.934 (Eq 6-19).
  • Size factor: from Table 13-1, l = 1/P + 1.25/P = 1/8 + 1.25/8 = 0.281 in.
  • Tooth thickness (Sec 14-1, Eq (b)): t = (4lx)^½ with x = 3Y/(2P) from Eq (14-3): x = 3(0.296)/[2(8)] = 0.0555 in; t = [4(0.281)0.0555]^½ = 0.250 in [MD-5 p.13].
  • The tooth is a cantilever of rectangular section; a rectangular section h × b has equivalent rotating-beam diameter de = 0.808(hb)^½ (6-24): de = 0.808(Ft)^½ = 0.808[1.5(0.250)]^½ = 0.495 in [MD-5 p.13].
  • kb = (de/0.30)^−0.107 = (0.495/0.30)^−0.107 = 0.948 (Eq 6-20) [MD-5 p.13].
  • kc = 1 (Eq 6-25); no temperature or reliability data, so kd = ke = 1 [MD-5 p.14].
  • In general a gear tooth is subjected only to one-way bending; exceptions include idler gears and gears in reversing mechanisms. One-way bending is accounted for by a miscellaneous-effects Marin factor kf [MD-5 p.14].
  • One-way bending: σa = σm = σ/2, σ = largest repeatedly applied bending stress from Eq (14-7). σm = mean (midrange) stress; σa = alternating stress (stress amplitude) [MD-5 p.14].
  • Goodman locus σa/S′e + σm/Sut = 1 with σa = σm gives σa = S′e Sut/(S′e + Sut) [MD-5 p.14].
  • Replace σa with σ/2 and S′e with 0.5Sut in the denominator: σ = 2S′e Sut/(0.5Sut + Sut) = 2S′e/(0.5 + 1) = 1.33S′e, so kf = σ/S′e = 1.33 [MD-5 p.15].
  • Gerber locus σa/S′e + (σm/Sut)² = 1 with σa = σm: σa = (Sut²/2S′e)[−1 + √(1 + 4S′e²/Sut²)]; with σa = σ/2 and Sut = S′e/0.5: σ = (S′e/0.5²)[−1 + √(1 + 4(0.5)²)] = 1.66S′e, so kf = 1.66 [MD-5 p.15].
  • A Gerber locus runs in and among fatigue data and Goodman does not, so kf = 1.66 is used [MD-5 p.15].
  • Marin: Se = ka kb kc kd ke kf S′e = 0.838(0.948)(1)(1)(1)1.66(27.5) = 36.3 kpsi (⚠ F16: ka = 0.934 on p.12) [MD-5 p.15].
  • Fatigue stress-concentration factor: for a 20° full-depth tooth the root-fillet radius is typically rf = 0.300/P = 0.300/8 = 0.0375 in. Fig A-15-6: r/d = rf/t = 0.0375/0.250 = 0.15. Since D/d = ∞, approximate with D/d = 3: Kt = 1.68. Fig 6-26: q = 0.62. Eq (6-32): Kf = 1 + (0.62)(1.68 − 1) = 1.42 [MD-5 p.16].
  • Design factor nd = 3 applied to load or strength; Eq (6-30): σmax = Kf σall = Se/nd, so σall = Se/(Kf nd) = 36.3/[1.42(3)] = 8.52 kpsi [MD-5 p.16].
  • Wt = F Y σall/(Kv P) = 1.5(0.296)8520/[1.52(8)] = 311 lbf; hp = 311(628)/33 000 = 5.9 hp [MD-5 p.16].
  • These results should be accepted only as preliminary estimates, to alert you to the nature of bending in gear teeth [MD-5 p.16].

6.5 Fatigue stress-concentration factor for gear teeth

• In Ex 14-2 the resources (Fig A-15-6, rectangular filleted bar in bending, σ0 = Mc/I, c = d/2, I = td³/12) did not directly address stress concentration in gear teeth [MD-5 p.17].
• A photoelastic investigation by Dolan and Broghamer reported in 1942 is a primary source of information on stress concentration; Mitchiner and Mabie interpret the results in terms of the fatigue stress-concentration factor Kf [MD-5 p.17].
• Kf = H + (t/r)^L (t/l)^M (14-9), where H = 0.34 − 0.458 366 2φ; L = 0.316 − 0.458 366 2φ; M = 0.290 + 0.458 366 2φ; r = (b − rf)²/[(d/2) + b − rf] [MD-5 p.18].
  • l and t are from the layout of Fig 14-1; φ = pressure angle; rf = fillet radius; b = dedendum; d = pitch diameter. Comparing Kf from (14-9) with the Fig A-15-6 approximation of Ex 14-2 is left as an exercise [MD-5 p.18].
  • The slide shows photoelastic fringe pictures of a tooth and of a contact [MD-5 p.18].

6.6 Surface durability (contact stress)

• The failure of the surfaces of gear teeth is generally called wear [MD-5 p.19].
• Pitting is a surface fatigue failure due to many repetitions of high contact stresses [MD-5 p.19].
• Other surface failures: scoring (scuffing), a lubrication failure; abrasion, wear due to the presence of foreign material. Photos: slight pitting, moderate pitting, scuffed tooth surfaces against the original surface [MD-5 p.19].
• The gear mesh is modelled as contact between two cylinders, Eq (3-74): pmax = 2F/(πbl); pmax = largest surface pressure, F = force pressing the two cylinders together, l = length of cylinders [MD-5 p.20].
• Half-width of the contact band: b = {(2F/πl)[(1 − ν1²)/E1 + (1 − ν2²)/E2]/(1/d1 + 1/d2)}^½ (14-10); ν1, ν2, E1, E2 = elastic constants; d1, d2 = cylinder diameters [MD-5 p.20].
• For gearing, replace F by Wt/cos φ, d by 2r and l by the face width F; substitute b into Eq (a) and replace pmax by σC, the surface compressive stress (Hertzian stress) [MD-5 p.21]:
  • σC² = [Wt/(πF cos φ)]·(1/r1 + 1/r2)/[(1 − ν1²)/E1 + (1 − ν2²)/E2] (14-11); r1, r2 = instantaneous radii of curvature of the pinion and gear profiles at the point of contact.
• The critical location is usually at the pitch line: r1 = dP sin φ/2, r2 = dG sin φ/2 (14-12) [MD-5 p.21].
• Elastic coefficient (from the denominator of 14-11): Cp = {1/[π((1 − νP²)/EP + (1 − νG²)/EG)]}^½ (14-13) [MD-5 p.22].
• With Cp and the velocity factor: σC = −Cp[(Kv Wt/(F cos φ))(1/r1 + 1/r2)]^½ (14-14). Useful for estimating, and the basis of the preferred AGMA approach [MD-5 p.22].

6.7 Example 14-3 (contact stress) [MD-5 p.23–24]

• Data: the pinion of Ex 14-1/14-2 mated with a 50-tooth gear of ASTM No. 50 cast iron; tangential load 382 lbf; estimate the factor of safety against surface fatigue. The surface endurance strength of cast iron is estimated from Sc = 0.32 HB kpsi for 10⁸ cycles [p.23].
• Table A-5: EP = 30 Mpsi, νP = 0.292; EG = 14.5 Mpsi, νG = 0.211 → Cp = {1/(π[(1 − 0.292²)/30(10⁶) + (1 − 0.211²)/14.5(10⁶)])}^½ = 1817 [p.23].
• dP = 2 in; dG = 50/8 = 6.25 in; r1 = 2 sin 20°/2 = 0.342 in; r2 = 6.25 sin 20°/2 = 1.069 in [p.23].
• F = 1.5 in, Kv = 1.52: σC = −1817[(1.52(380)/(1.5 cos 20°))(1/0.342 + 1/1.069)]^½ = −72 400 psi (⚠ F17: 380 used, 382 given) [p.24].
• Table A-24: HB = 262 for ASTM No. 50 cast iron → Sc = 0.32(262) = 83.8 kpsi [p.24].
• Contact stress is not linear with the transmitted load. If the factor of safety is the loss-of-function load divided by the imposed load, the ratio of loads is the ratio of stresses squared: n = Sc²/σC² = (83.8/72.4)² = 1.34 [p.24].
• One is free to define the factor of safety as Sc/σC. Suppose this gear has a bending-fatigue factor of safety of 1.20 and a surface-fatigue factor of 1.34: since both are load ratios and 1.34 > 1.20, the threat is bending fatigue. If the surface factor is defined as Sc/σC = √1.34 = 1.16, then 1.20 > 1.16 but the threat is still not from surface fatigue. Either definition may be used; one requires squaring before numbers can be compared [MD-5 p.25].
• Sheet problems: Problems 14-4, 14-11, 14-15 [MD-5 p.25].

6.8 AGMA method: stress equations

• AGMA provides a recommended method for gear design. It includes bending stress and contact stress as two failure modes, incorporates modifying factors for various situations, and imbeds much of the detail in tables and figures. Two fundamental stress equations are used: one for bending stress and one for pitting resistance (contact stress) [MD-5 p.26].
• Bending stress (14-15) [MD-5 p.27]:
  • US: σ = Wt Ko Kv Ks (Pd/F)(Km KB/J)
  • SI: σ = Wt Ko Kv Ks (1/(b mt))(KH KB/YJ)
  • Wt = tangential transmitted load, lbf (N); Ko = overload factor; Kv = dynamic factor; Ks = size factor; Pd = transverse diametral pitch; F (b) = face width of the narrower member, in (mm); Km (KH) = load-distribution factor; KB = rim-thickness factor; J (YJ) = geometry factor for bending strength, which includes the root-fillet stress-concentration factor Kf; mt = transverse metric module.
• Contact stress (14-16) [MD-5 p.28]:
  • US: σc = Cp √[Wt Ko Kv Ks (Km/(dP F))(Cf/I)]
  • SI: σc = ZE √[Wt Ko Kv Ks (KH/(dw1 b))(ZR/ZI)]
  • Cp (ZE) = elastic coefficient, √(lbf/in²) (√(N/mm²)); Cf (ZR) = surface-condition factor; dP (dw1) = pitch diameter of the pinion, in (mm); I (ZI) = geometry factor for pitting resistance. The other terms are as in (14-15).
• View the terms of (14-15) as advice on items the designer should consider whether or not the voluntary standard is followed: transmitted load magnitude; overload; dynamic augmentation of transmitted load; size; geometry (pitch and face width); distribution of load across the teeth; rim support of the tooth; Lewis form factor and root-fillet stress concentration [MD-5 p.29].

6.9 AGMA strengths

• AGMA uses allowable stress numbers rather than strengths; they are called strengths here for consistency with the textbook. They are only for use with the AGMA stress values and should not be compared with other true material strengths [MD-5 p.30].
• Representative bending strengths: Table 14-3 (steel gears) and Table 14-4 (iron and bronze gears); Figs 14-2, 14-3 and 14-4 are used as indicated in the tables. The tables assume repeatedly applied loads at 10⁷ cycles and 0.99 reliability [MD-5 p.30].
• Allowable bending stress (14-17) [MD-5 p.31]:
  • US: σall = (St/SF)·YN/(KT KR); SI: σall = (St/SF)·YN/(Yθ YZ)
  • St = allowable bending stress, lbf/in² (N/mm²); YN = stress-cycle factor for bending stress; KT (Yθ) = temperature factors; KR (YZ) = reliability factors; SF = AGMA factor of safety, a stress ratio.

Table 14-3. Repeatedly applied bending strength St at 10⁷ cycles and 0.99 reliability for steel gears (Source: ANSI/AGMA 2001-D04) [MD-5 p.32]

• The table footnotes (numbered 1–7) are referenced but their text is not reproduced on the slide [MD-5 p.32].
• Fig 14-2, through-hardened steels, St against Brinell hardness HB (150–450); "metallurgical and quality control procedure required" for grade 2 [MD-5 p.33]:
  • Grade 1: St = 77.3HB + 12 800 psi (SI: St = 0.533HB + 88.3 MPa)
  • Grade 2: St = 102HB + 16 400 psi (SI: St = 0.703HB + 113 MPa)
• Fig 14-3, nitrided through-hardened steels (i.e., AISI 4140, 4340), St against core hardness HB (250–350) [MD-5 p.34]:
  • Grade 1: St = 82.3HB + 12 150 psi (SI: 0.568HB + 83.8 MPa)
  • Grade 2: St = 108.6HB + 15 890 psi (SI: 0.749HB + 110 MPa)
• Fig 14-4, nitriding steels, St against core hardness HB (250–350) [MD-5 p.35]:
  • Grade 1 Nitralloy: St = 86.2HB + 12 730 psi (SI: 0.594HB + 87.76 MPa)
  • Grade 2 Nitralloy: St = 113.8HB + 16 650 psi (SI: 0.784HB + 114.81 MPa)
  • Grade 1 2.5% chrome: St = 105.2HB + 9280 psi (SI: 0.7255HB + 63.89 MPa)
  • Grade 2 2.5% chrome: St = 105.2HB + 22 280 psi (SI: 0.7255HB + 153.63 MPa)
  • Grade 3 2.5% chrome: St = 105.2HB + 29 280 psi (SI: 0.7255HB + 201.91 MPa)

Table 14-4. Repeatedly applied bending strength St for iron and bronze gears at 10⁷ cycles and 0.99 reliability (Source: ANSI/AGMA 2001-D04) [MD-5 p.36]

6.10 Allowable contact stress

• σc,all (14-18) [MD-5 p.37]:
  • US: σc,all = (Sc/SH)·ZN CH/(KT KR); SI: σc,all = (Sc/SH)·ZN ZW/(Yθ YZ)
  • Sc = allowable contact stress, lbf/in² (N/mm²); ZN = stress-cycle factor; CH (ZW) = hardness-ratio factors for pitting resistance; KT (Yθ) = temperature factors; KR (YZ) = reliability factors; SH = AGMA factor of safety, a stress ratio.
  • Values of Sc are found in Fig 14-5 and Tables 14-5, 14-6 and 14-7 [MD-5 p.37]. (Table 14-5 itself is not shown in the deck.)

Table 14-6. Repeatedly applied contact strength Sc at 10⁷ cycles and 0.99 reliability for steel gears (Source: ANSI/AGMA 2001-D04) [MD-5 p.38]

• Fig 14-5, contact strength Sc at 10⁷ cycles and 0.99 reliability for through-hardened steel gears, against HB (150–450) [MD-5 p.39]:
  • Grade 1: Sc = 322HB + 29 100 psi (SI: 2.22HB + 200 MPa)
  • Grade 2: Sc = 349HB + 34 300 psi (SI: 2.41HB + 237 MPa)

Table 14-7. Repeatedly applied contact strength Sc at 10⁷ cycles and 0.99 reliability for iron and bronze gears (Source: ANSI/AGMA 2001-D04) [MD-5 p.40]

6.11 Geometry factors J and I

• J (YJ in metric) accounts for the shape of the tooth in the bending stress equation. It includes a modification of the Lewis form factor Y, the fatigue stress-concentration factor Kf and the tooth load-sharing ratio mN [MD-5 p.41].
• The load-sharing ratio mN equals the face width divided by the minimum total length of the lines of contact [MD-5 p.41].
• J = Y/(Kf mN) (14-20); mN = pN/(0.95Z) (14-21), where pN = normal base pitch and Z = length of the line of action in the transverse plane (distance Lab in Fig 13-11; the sketch shows arc of approach qa, arc of recess qr, pressure line, addendum circles, pitch circle) [MD-5 p.41].
• Values of Y and Z are found in the AGMA standards. Z = [(rP + a)² − rbP²]^½ + [(rG + a)² − rbG²]^½ − (rP + rG) sin φt (14-25); rP, rG = pitch radii; rbP, rbG = base-circle radii. From Eq (13-6): rb = r cos φt (14-26) [MD-5 p.42].
• For the common spur gear with 20° pressure angle, J is read directly from Fig 14-6. For helical gears with 20° normal pressure angle, use Figs 14-7 and 14-8 [MD-5 p.42].
• Fig 14-6, spur-gear geometry factor J: J (0.20–0.60) against the number of teeth for which J is desired (12 to ∞); one curve per number of teeth in the mating gear (17, 25, 35, 50, 85, 170, 1000) for load applied at the highest point of single-tooth contact, plus a lower curve for load applied at the tip of the tooth. Inset: generating rack, 1 pitch, addendum 1.000, whole depth 2.400, 20°, tip radius 0.35 rT; pinion and gear addendum 1.000. Source: AGMA 218.01, consistent with tabular data of the current AGMA 908-B89; convenient for design purposes [MD-5 p.43].
• Helical J: get J′ from Fig 14-7, which assumes the mating gear has 75 teeth; get the multiplier from Fig 14-8 for a mating gear with other than 75 teeth; J = multiplier × J′ [MD-5 p.44].
  • Fig 14-7: J′ (0.30–0.70) against helix angle ψ (0–35°), curves for 20, 30, 60, 150, 500 teeth; the curves peak around 15° and fall toward 35° [MD-5 p.44].
  • Fig 14-8: modifying factor (0.85–1.05) against ψ (0–35°), curves for 20, 30, 50, 75 (= 1.00), 150, 500 teeth in the mating element; about 0.93 for 20 teeth and about 1.04 for 500 teeth at small ψ [MD-5 p.45].
• Surface-strength geometry factor I (ZI in metric), called the pitting resistance geometry factor by AGMA (14-23) [MD-5 p.46]:
  • External gears: I = [cos φt sin φt/(2mN)]·mG/(mG + 1)
  • Internal gears: I = [cos φt sin φt/(2mN)]·mG/(mG − 1)
  • φt = transverse pressure angle; speed ratio mG = NG/NP = dG/dP (14-22); mN = pN/(0.95Z) (14-21), with mN = 1 for spur gears; normal base pitch pN = pn cos φn (14-24), pn = normal circular pitch.

6.12 Elastic coefficient Cp (ZE)

• Obtained from Eq (14-13) or from Table 14-8 [MD-5 p.47].

Table 14-8. Elastic coefficient Cp (ZE), √psi (√MPa) (Source: AGMA 218.01). Columns: gear material and EG, psi (MPa) [MD-5 p.48]

6.13 Dynamic factor Kv (AGMA)

• Kv accounts for increased forces with increased speed and is affected by the manufacturing quality of the gears [MD-5 p.49].
• A set of quality numbers Qv defines tolerances for gears manufactured to a specified accuracy. Qv 3 to 7 include most commercial-quality gears; Qv 8 to 12 are of precision quality [MD-5 p.49].
• Kv = [(A + √V)/A]^B with V in ft/min; Kv = [(A + √(200V))/A]^B with V in m/s (14-27); A = 50 + 56(1 − B), B = 0.25(12 − Qv)^(2/3) (14-28) [MD-5 p.50].
• Kv can also be read from Fig 14-9 [MD-5 p.50].
• Maximum recommended velocity for a given quality number (the end point of each Qv curve): (Vt)max = [A + (Qv − 3)]² ft/min, or [A + (Qv − 3)]²/200 m/s (14-29) [MD-5 p.50].
• Fig 14-9: Kv (1.0–1.8) against pitch-line velocity Vt (0–10 000 ft/min) for Qv = 5 to 11; a shaded band at Kv ≈ 1.02–1.11 is labelled "very accurate gearing". The curves follow (14-27) and end at (14-29) (ANSI/AGMA 2001-D04, Annex A) [MD-5 p.51].

6.14 Overload factor Ko

• Ko accounts for the probability of an increase in the nominal tangential load due to the particular application. Recommended values (driven machine: uniform / moderate shock / heavy shock): power source uniform 1.00 / 1.25 / 1.75; light shock 1.25 / 1.50 / 2.00; medium shock 1.50 / 1.75 / 2.25 [MD-5 p.52]. This is the same table as in [MD-S5 p.1–2] (section 7.2).

6.15 Surface-condition factor Cf (ZR)

• Accounts for detrimental surface finish. No values are currently given by AGMA; use 1 for normal commercial gears [MD-5 p.53].
• Cf (ZR) is used only in the pitting resistance equation (14-16). It depends on surface finish as affected by (but not limited to) cutting, shaving, lapping, grinding, shotpeening; residual stress; plastic effects (work hardening) [MD-5 p.53].
• Standard surface conditions for gear teeth have not yet been established. When a detrimental surface-finish effect is known to exist, AGMA specifies Cf greater than unity [MD-5 p.53].

6.16 Size factor Ks

• Reflects nonuniformity of material properties due to size. It depends on tooth size; diameter of part; ratio of tooth size to diameter of part; face width; area of stress pattern; ratio of case depth to tooth size; hardenability and heat treatment [MD-5 p.54].
• It accounts for the fatigue size effect and the non-uniformity of material properties for large sizes. AGMA has not established size factors; use 1 for normal gear sizes [MD-5 p.54].
• One could apply the fatigue size-factor method of Ch. 6, where this size factor is the reciprocal of the Marin size factor kb. With the known gear-tooth geometry: Ks = 1/kb = 1.192(F√Y/P)^0.0535 [MD-5 p.54].

6.17 Load-distribution factor Km (KH)

• Accounts for non-uniform distribution of load across the line of contact; depends on mounting and face width [MD-5 p.55].
• Currently only defined for: F/dP ≤ 2; gears mounted between bearings; face widths up to 40 in; contact across the full width of the narrowest member [MD-5 p.55].
• Face load-distribution factor: Km = Cmf = 1 + Cmc(Cpf Cpm + Cma Ce) (14-30) [MD-5 p.56].
• Cmc = 1 for uncrowned teeth, 0.8 for crowned teeth (14-31) (sketch of a crowned tooth) [MD-5 p.56].
• Cpf (14-32) [MD-5 p.56]:
  • F/(10d) − 0.025 for F ≤ 1 in
  • F/(10d) − 0.0375 + 0.0125F for 1 < F ≤ 17 in
  • F/(10d) − 0.1109 + 0.0207F − 0.000 228F² for 17 < F ≤ 40 in
  • b/(10d) − 0.025 for b ≤ 25 mm
  • b/(10d) − 0.0375 + 4.92(10⁻⁴)b for 25 < b ≤ 425 mm
  • b/(10d) − 0.1109 + 8.15(10⁻⁴)b − 3.53(10⁻⁷)b² for 425 < b ≤ 1000 mm
• Ce = 0.8 for gearing adjusted at assembly, or compatibility improved by lapping, or both; 1 for all other conditions (14-35) [MD-5 p.57].
• Cpm = 1 for a straddle-mounted pinion with S1/S < 0.175; 1.1 with S1/S ≥ 0.175 (14-33). Fig 14-10: S = bearing span (centerline to centerline); S1 = offset of the gear-face centerline from the mid-span [MD-5 p.57].
• Gear mounting (Schmid, Hamrock and Jacobson, Fundamentals of Machine Elements, 3rd ed., Fig 15.7, CRC Press 2014): (a) straddle mounting, gear between bearings; (b) overhang mounting. Deep-groove bearings are shown, but often a bearing better suited for thrust is required in at least one location [MD-5 p.58].
• Cma = A + BF + CF² (14-34), constants from Table 14-9, or read Cma directly from Fig 14-11 [MD-5 p.59].

Table 14-9. Empirical constants A, B, C for Eq (14-34), face width F in inches (Source: ANSI/AGMA 2001-D04; see ANSI/AGMA 2101-D04, pp. 20–22, for the SI formulation) [MD-5 p.59]

• Fig 14-11, mesh alignment factor Cma (0–0.90) against face width F (0–40 in): curve 1 open gearing, curve 2 commercial enclosed, curve 3 precision enclosed, curve 4 extra-precision enclosed gear units; curve-fit equations in Table 14-9 (ANSI/AGMA 2001-D04) [MD-5 p.60].

6.18 Hardness-ratio factor CH (ZW)

• The pinion is subjected to more cycles than the gear, so it is often hardened more than the gear. CH accounts for the difference in hardness of pinion and gear. CH is applied only to the gear: CH = 1 for the pinion [MD-5 p.61].
• For the gear: CH = 1.0 + A′(mG − 1.0) (14-36), mG = NG/NP = dG/dP (14-22) [MD-5 p.61]:
  • A′ = 8.98(10⁻³)(HBP/HBG) − 8.29(10⁻³) for 1.2 ≤ HBP/HBG ≤ 1.7
  • A′ = 0 for HBP/HBG < 1.2
  • A′ = 0.006 98 for HBP/HBG > 1.7
• Fig 14-12 (graph form of 14-36): CH (1.00–1.14) against single-reduction gear ratio mG (0–20), lines for HBP/HBG = 1.2 to 1.7; "when HBP/HBG < 1.2, use CH = 1" (through-hardened steel; ANSI/AGMA 2001-D04) [MD-5 p.62].
• If the pinion is surface-hardened to 48 Rockwell C or greater, the softer gear can work-harden during operation. Then CH = 1 + B′(450 − HBG) (14-37), B′ = 0.000 75 exp[−0.0112fP], where fP = surface finish of the pinion as root-mean-square roughness Ra in μin [MD-5 p.63].
• Fig 14-13 (surface-hardened steel pinion): CH (1.00–1.16) against gear Brinell hardness HBG (180–400) for fP = 16, 32, 64 μin; when fP > 64 use CH = 1.0 (ANSI/AGMA 2001-D04) [MD-5 p.63].

6.19 Stress-cycle factors YN and ZN

• AGMA strengths are for 10⁷ cycles; the stress-cycle factors account for other design cycles. Fig 14-14 gives YN for bending; Fig 14-15 gives ZN for contact stress [MD-5 p.64].
• Fig 14-14, repeatedly applied bending strength stress-cycle factor YN (ANSI/AGMA 2001-D04), N = 10² to 10¹⁰, all lines meeting at YN = 1 near 3(10⁶) cycles [MD-5 p.65]:
  • Flat below 10³ cycles: 400 HB ≈ 3.4, case carburized, 250 HB, nitrided ≈ 2.0, 160 HB ≈ 1.6 (read from the graph).
  • Sloping lines: 400 HB YN = 9.4518N^−0.148; case carb. YN = 6.1514N^−0.1192; 250 HB YN = 4.9404N^−0.1045; nitrided YN = 3.517N^−0.0817; 160 HB YN = 2.3194N^−0.0538.
  • Above about 3(10⁶) cycles, a shaded band between YN = 1.3558N^−0.0178 (upper) and YN = 1.6831N^−0.0323 (lower).
  • Note: the choice of YN in the shaded area is influenced by pitchline velocity, gear material cleanliness, residual stress, material ductility and fracture toughness.
• Fig 14-15, pitting resistance stress-cycle factor ZN (ANSI/AGMA 2001-D04) [MD-5 p.66]:
  • ZN = 2.466N^−0.056 (flat at about 1.5 below 10⁴ cycles); nitrided ZN = 1.249N^−0.0138 (flat at about 1.1 below 10⁴ cycles); the lines meet at ZN = 1 at 10⁷ cycles.
  • Above 10⁷ cycles, a shaded zone below ZN = 1.4488N^−0.023.
  • Note: the choice of ZN in the shaded zone is influenced by lubrication regime, failure criteria, smoothness of operation required, pitchline velocity, gear material cleanliness, material ductility and fracture toughness, residual stress.

6.20 Reliability factor KR (YZ) and temperature factor KT (Yθ)

• KR accounts for statistical distributions of material fatigue failures; it does not account for load variation. Use Table 14-10. Since reliability is highly nonlinear, interpolate with the least-squares fit (14-38) [MD-5 p.67]:
  • KR = 0.658 − 0.0759 ln(1 − R) for 0.5 < R < 0.99
  • KR = 0.50 − 0.109 ln(1 − R) for 0.99 ≤ R ≤ 0.9999
• Table 14-10: R = 0.9999 → KR = 1.50; 0.999 → 1.25; 0.99 → 1.00; 0.90 → 0.85; 0.50 → 0.70 [MD-5 p.67].
• KT: AGMA has not established values. For temperatures up to 250°F (120°C), KT = 1 is acceptable; for higher temperatures the factor should be greater than unity [MD-5 p.68].

6.21 Rim-thickness factor KB

• When the rim thickness is not sufficient to provide full support for the tooth root, bending fatigue failure may occur through the gear rim rather than at the tooth fillet; then a stress-modifying factor KB is recommended. KB adjusts the estimated bending stress for the thin-rimmed gear and is a function of the backup ratio mB. (Picture "The Involute Form": width of face, tooth thickness, tooth space, circular pitch, face, flank, thickness of rim, clearance) [MD-5 p.69].
• KB = 1.6 ln(2.242/mB) for mB < 1.2; KB = 1 for mB ≥ 1.2 (14-40); backup ratio mB = tR/ht (14-39), tR = rim thickness below the tooth root, ht = tooth height (Fig 14-16: KB from 2.4 at mB = 0.5 down to 1.0 at mB = 1.2, then constant up to mB = 10) [MD-5 p.70].

6.22 Safety factors SF and SH; comparison

• SF and SH are included as design factors in the strength equations and can be solved for and used as factors of safety [MD-5 p.71]:
  • SF = [St YN/(KT KR)]/σ = fully corrected bending strength/bending stress (14-41)
  • SH = [Sc ZN CH/(KT KR)]/σc = fully corrected contact strength/contact stress (14-42)
• Or set SF (SH) equal to unity in (14-17)/(14-18) and solve for the traditional factor of safety n = σall/σ [MD-5 p.71].
• Bending stress is linear with transmitted load; contact stress is not. To compare factors of safety between the failure modes and find which is critical: compare SF with SH² when deciding whether bending or wear is the threat to function; for crowned gears compare SF with SH³ [MD-5 p.72].

6.23 Summary roadmaps (Figs 14-17 and 14-18)

• Fig 14-17, summary for bending of gear teeth [MD-5 p.73]: dP = NP/Pd; V = πdn/12; Wt = 33 000H/V; σ = Wt Ko Kv Ks (Pd/F)(Km KB/J) (14-15), with Ko from the overload table, Kv Eq (14-27) p. 748, Ks 1 [or Eq (a), Sec 14-10] p. 751, Km Eq (14-30) p. 751, KB Eq (14-40) p. 756, J Fig 14-6 p. 745; σall = (St/SF)YN/(KT KR) (14-17), with 0.99(St)10⁷ from Tables 14-3, 14-4 pp. 740–741, YN Fig 14-14 p. 755, KR Table 14-10/Eq (14-38) pp. 756, 755, KT = 1 if T < 250°F; SF = [St YN/(KT KR)]/σ (14-41). These are the same roadmaps reproduced in [MD-S5 p.1–2] (section 7.1).
• Fig 14-18, summary for surface wear of gear teeth [MD-5 p.74]: σc = Cp(Wt Ko Kv Ks Km Cf/(dP F I))^½ (14-16), with Cp Eq (14-13)/Table 14-8 pp. 736, 749, Cf = 1, I Eq (14-23) p. 747; σc,all = (Sc/SH)ZN CH/(KT KR) (14-18), with 0.99(Sc)10⁷ Tables 14-6, 14-7 pp. 743–744, ZN Fig 14-15 p. 755, CH Section 14-12 (gear only) pp. 753–754; SH = [Sc ZN CH/(KT KR)]/σc (14-42), CH for the gear only.

6.24 Example 14-4 (AGMA spur analysis) [MD-5 p.75–]

• Data [p.75]: 17-tooth 20° spur pinion at 1800 rev/min transmits 4 hp to a 52-tooth disk gear; Pd = 10 teeth/in; F = 1.5 in; quality No. 6; straddle-mounted with bearings immediately adjacent. Pinion: grade 1 steel, 240 HB surface and through-hardened core. Gear: through-hardened grade 1 steel, 200 HB surface and core. ν = 0.30; JP = 0.30, JG = 0.40; E = 30(10⁶) psi. Smooth loading (motor and load). Pinion life 10⁸ cycles, R = 0.90; YN = 1.3558N^−0.0178, ZN = 1.4488N^−0.023. Uncrowned; commercial enclosed gear unit. Find (a) SF in bending, (b) SH in wear, (c) the threat to each gear and to the mesh.
• Figs 14-17 and 14-18 are used as guides [p.76].
• dP = 17/10 = 1.7 in; dG = 52/10 = 5.2 in; V = π(1.7)1800/12 = 801.1 ft/min; Wt = 33 000(4)/801.1 = 164.8 lbf [p.76].
• Uniform loading: Ko = 1. Qv = 6: B = 0.25(12 − 6)^(2/3) = 0.8255; A = 50 + 56(1 − 0.8255) = 59.77; Kv = [(59.77 + √801.1)/59.77]^0.8255 = 1.377 [p.77].
• Table 14-2: YP = 0.303 (17 teeth); interpolation for NG = 52 gives YG = 0.412. Eq (a) of Sec 14-10 with F = 1.5 in: (Ks)P = 1.192(1.5√0.303/10)^0.0535 = 1.043; (Ks)G = 1.192(1.5√0.412/10)^0.0535 = 1.052 [p.77].
• Km (14-30): uncrowned Cmc = 1; Cpf = 1.5/[10(1.7)] − 0.0375 + 0.0125(1.5) = 0.0695; bearings immediately adjacent Cpm = 1; commercial enclosed (Fig 14-11) Cma = 0.15; Ce = 1 → Km = 1 + (1)[0.0695(1) + 0.15(1)] = 1.22 [p.78].
• Constant-thickness gears: KB = 1. mG = 52/17 = 3.059. N(pinion) = 10⁸, N(gear) = 10⁸/3.059: (YN)P = 1.3558(10⁸)^−0.0178 = 0.977; (YN)G = 1.3558(10⁸/3.059)^−0.0178 = 0.996 [p.78].
• Table 14-10, R = 0.9: KR = 0.85. Fig 14-18: KT = 1, Cf = 1. Eq (14-23), mN = 1: I = (cos 20° sin 20°/2)(3.059/(3.059 + 1)) = 0.121. Table 14-8: Cp = 2300 √psi [p.79].
• Table 14-3 → Fig 14-2 (grade 1): (St)P = 77.3(240) + 12 800 = 31 350 psi; (St)G = 77.3(200) + 12 800 = 28 260 psi. Table 14-6 → Fig 14-5: (Sc)P = 322(240) + 29 100 = 106 400 psi; (Sc)G = 322(200) + 29 100 = 93 500 psi [p.79].
• Fig 14-15: (ZN)P = 1.4488(10⁸)^−0.023 = 0.948; (ZN)G = 1.4488(10⁸/3.059)^−0.023 = 0.973 [p.80].
• HBP/HBG = 240/200 = 1.2 → A′ = 8.98(10⁻³)(1.2) − 8.29(10⁻³) = 0.002 49; CH = 1 + 0.002 49(3.059 − 1) = 1.005 [p.80].
• (a) Pinion bending: (σ)P = 164.8(1)1.377(1.043)(10/1.5)(1.22(1)/0.30) = 6417 psi; (SF)P = 31 350(0.977)/[1(0.85)]/6417 = 5.62 [p.81].
• Gear bending: (σ)G = 164.8(1)1.377(1.052)(10/1.5)(1.22(1)/0.40) = 4854 psi; (SF)G = 28 260(0.996)/[1(0.85)]/4854 = 6.82 [p.81].
• (b) Pinion wear: (σc)P = 2300[164.8(1)1.377(1.043)(1.22/(1.7(1.5)))(1/0.121)]^½ = 70 360 psi; (SH)P = 106 400(0.948)/[1(0.85)]/70 360 = 1.69 [p.82].
• Gear wear: only Ks changes, so (σc)G = [(Ks)G/(Ks)P]^½(σc)P = (1.052/1.043)^½ 70 360 = 70 660 psi; with CH = 1.005, (SH)G = 93 500(0.973)1.005/[1(0.85)]/70 660 = 1.52 [p.83].
• (c) Pinion: compare (SF)P = 5.62 with (SH)P² = 1.69² = 2.86, so the threat in the pinion is wear. Gear: (SF)G = 6.82 against 1.52² = 2.31, so the threat in the gear is also wear [p.83].

6.25 Example 14-5 (AGMA helical analysis) [MD-5 p.84–]

• Data [p.84]: 17-tooth 20° normal pressure-angle helical pinion ("normal pitch-angle" as printed), right-hand helix angle 30°, 1800 rev/min, 4 hp to a 52-tooth helical gear; normal diametral pitch 10 teeth/in; F = 1.5 in; quality number 6; straddle-mounted, bearings immediately adjacent. Through-hardened steel: 240 HB (pinion) and 200 HB (gear), surface and core. Smooth transmission (electric motor and centrifugal pump). Pinion life 10⁸ cycles, R = 0.9, use the upper curves of Figs 14-14 and 14-15. Find (a) SF, (b) SH, (c) the threat to each gear and to the mesh.
• Unchanged from Ex 14-4 [p.85]: Ko = 1, YP = 0.303, YG = 0.412, mG = 3.059, (Ks)P = 1.043, (Ks)G = 1.052, (YN)P = 0.977, (YN)G = 0.996, KR = 0.85, KT = 1, Cf = 1, Cp = 2300 √psi, (St)P = 31 350 psi, (St)G = 28 260 psi, (Sc)P = 106 380 psi (106 400 in Ex 14-4), (Sc)G = 93 500 psi, (ZN)P = 0.948, (ZN)G = 0.973, CH = 1.005.
• Eq (13-18): Pt = Pn cos ψ = 10 cos 30° = 8.660 teeth/in; dP = 17/8.660 = 1.963 in; dG = 52/8.660 = 6.005 in [p.85].
• V = π(1.963)1800/12 = 925 ft/min; Wt = 33 000(4)/925 = 142.7 lbf [p.85].
• B = 0.8255, A = 59.77: Kv = [(59.77 + √925)/59.77]^0.8255 = 1.404 [p.86].
• Eq (13-19): φt = tan⁻¹(tan 20°/cos 30°) = 22.80° [p.86].
• rP = 1.963/2 = 0.9815 in; rG = 6.004/2 = 3.002 in; a = 1/Pn = 1/10 = 0.1; (rb)P = 0.9815 cos 22.80° = 0.9048 in; (rb)G = 3.002 cos 22.80° = 2.767 in [p.86].
• Eq (14-25): Z = √[(0.9815 + 0.1)² − 0.9048²] + √[(3.004 + 0.1)² − 2.769²] − (0.9815 + 3.004) sin 22.80° = 0.5924 + 1.4027 − 1.5444 = 0.4507 in (⚠ F18 on 3.002/3.004, 2.767/2.769) [p.87].
• Since the first two terms are less than 1.5444, the equation for Z stands [p.87].
• Eq (14-24): pN = pn cos φn = (π/Pn) cos 20° = (π/10) cos 20° = 0.2952 in (the slide calls pN the "normal circular pitch") [p.87].
• Eq (14-21): mN = 0.2952/[0.95(0.4507)] = 0.6895; Eq (14-23): I = [sin 22.80° cos 22.80°/(2(0.6895))][3.06/(3.06 + 1)] = 0.195 [p.87].
• Fig 14-7: J′P = 0.45, J′G = 0.54; Fig 14-8 multipliers 0.94 and 0.98 → JP = 0.45(0.94) = 0.423, JG = 0.54(0.98) = 0.529 [p.88].
• Cpf = 1.5/[10(1.963)] − 0.0375 + 0.0125(1.5) = 0.0577; Cmc = 1, Cpm = 1, Cma = 0.15 (Fig 14-11), Ce = 1 → Km = 1 + (1)[0.0577(1) + 0.15(1)] = 1.208 [p.88].
• (a) Pinion bending, Eq (14-15) using Pt: (σ)P = 142.7(1)1.404(1.043)(8.66/1.5)(1.208(1)/0.423) = 3445 psi; (SF)P = 31 350(0.977)/[1(0.85)]/3445 = 10.5 [p.89].
• Gear bending: (σ)G = 142.7(1)1.404(1.052)(8.66/1.5)(1.208(1)/0.529) = 2779 psi; (SF)G = 28 260(0.996)/[1(0.85)]/2779 = 11.9 [p.89].
• (b) Pinion wear: (σc)P = 2300[142.7(1)1.404(1.043)(1.208/(1.963(1.5)))(1/0.195)]^½ = 48 230 psi; (SH)P = 106 400(0.948)/[1(0.85)]/48 230 = 2.46 [p.90].
• Gear wear: (σc)G = (1.052/1.043)^½ 48 230 = 48 440 psi; (SH)G = 93 500(0.973)1.005/[1(0.85)]/48 440 = 2.22 [p.91].
• (c) Pinion: 10.5 against 2.46² = 6.05, so the threat is wear. Gear: 11.9 against 2.22² = 4.93, so the threat is also wear. For the meshing gearset, wear controls [p.91].
• Sheet problems: Example 14-5, Problems 14-20 and 14-22 [MD-5 p.91]. (Problem 14-20 is solved in [MD-S5], sections 7.4–7.5.)
• Page 92 is "Thank you / Questions" [MD-5 p.92].

Spur and helical gear design, AGMA (as used in the section)

[MD-S5]

The lecture derivation of these factors (Lecture #4-1) is not available. The equation and figure numbers below are the ones printed in the Shigley roadmaps [MD-S5 p.1–2].

7.1 Bending (ANSI/AGMA 2001-D04, US units) [MD-S5 p.1]

• dP = NP/Pd; V = π d n/12; Wt = 33 000 H/V.
• Bending stress: σ = Wt Ko Kv Ks (Pd/F)(Km KB/J) (14-15). Where each factor comes from:
  • Ko: overload table (below).
  • Kv: Eq (14-27).
  • Ks: 1, or Eq (a) of Sec 14-10.
  • Km: Eq (14-30).
  • KB: Eq (14-40).
  • J: Fig 14-6.
• Allowable bending stress: σall = (St/SF)[YN/(KT KR)] (14-17). Where each factor comes from:
  • St: the 0.99 reliability, 10⁷-cycle value from Tables 14-3 and 14-4.
  • YN: Fig 14-14.
  • KT: 1 if T < 250°F.
  • KR: Table 14-10 or Eq (14-38).
• Bending factor of safety: SF = [St YN/(KT KR)]/σ (14-41).
• To decide whether bending or wear is the threat, compare SF with SH². For crowned gears, compare SF with SH³.

7.2 Wear (contact) [MD-S5 p.2]

• Contact stress: σc = Cp[Wt Ko Kv Ks (Km/(dP F))(Cf/I)]^½ (14-16). Where each factor comes from:
  • Cp: Eq (14-13), Table 14-8.
  • Cf: 1.
  • I: Eq (14-23).
• Allowable contact stress: σc,all = Sc ZN CH/(SH KT KR) (14-18). Where each factor comes from:
  • Sc: the 0.99 reliability, 10⁷-cycle value from Tables 14-6 and 14-7.
  • ZN: Fig 14-15.
  • CH: Section 14-12, gear only.
• Wear factor of safety: SH = [Sc ZN CH/(KT KR)]/σc (14-42).

Overload factor Ko [MD-S5 p.1–2]

• The analysis finds σ, σall, σc and σc,all for both pinion and gear, then SF = σall/σ and SH = σc,all/σc [MD-S5 p.3].
• Pinion and gear may differ in Ks, J, YN, ZN, St, Sc and CH [MD-S5 p.3].
• A helical gear differs from a spur gear in only four factors: Km, Kv, I and J [MD-S5 p.3].

7.3 Factor formulas as used in the worked problem (P14-20)

• Kv = [(A + √V)/A]^B, with B = 0.25(12 − Qv)^(2/3) and A = 50 + 56(1 − B) [MD-S5 p.5].
• Ks = 1.192(F√Y/P)^0.0535, with the Lewis form factor Y from Table 14-2 [MD-S5 p.5].
• Km = 1 + Cmc(Cpf Cpm + Cma Ce) [MD-S5 p.6]:
  • Cmc = 1 for uncrowned teeth;
  • Cpf = F/(10d) − 0.0375 + 0.0125F for 1 < F ≤ 17 in;
  • Cpm = 1 for gears centered between the bearings;
  • Cma = A + BF + CF², Table 14-9; for commercial enclosed units A = 0.127, B = 0.0158, C = −0.930(10⁻⁴);
  • Ce = 1.
• KB = 1 was assumed because no data were given for tR and ht [MD-S5 p.6].
• J (spur) from Fig 14-6 [MD-S5 p.7].
• J (helical) = J′ from Fig 14-7 × the modifying factor from Fig 14-8 [MD-S5 p.15].
• St (grade 1, through-hardened, Fig 14-2): St = 77.3 HB + 12 800 psi [MD-S5 p.7].
• YN (Fig 14-14): YN = 1.3558 N^−0.0178; the gear cycles are N_G = N_P/mG [MD-S5 p.7].
• KT = 1; KR = 0.85 from Table 14-10 (90% reliability in this problem) [MD-S5 p.4, p.8].
• Cp = 2300 √psi (Table 14-8, steel) [MD-S5 p.8].
• Cf = 1 assumed [MD-S5 p.8].
• I = [cos φt sin φt/(2 mN)][mG/(mG + 1)], with mN = 1 for spur gears [MD-S5 p.8].
• Helical [MD-S5 p.15]:
  • mN = pN/(0.95 Z), with pN = pn cos φn.
  • Z = [(rP + a)² − rbP²]^½ + [(rG + a)² − rbG²]^½ − (rP + rG) sin φt.
  • a = 1/Pn.
• Sc (grade 1, through-hardened, Fig 14-5): Sc = 322 HB + 29 100 psi [MD-S5 p.8].
• ZN (Fig 14-15): ZN = 1.4488 N^−0.023 [MD-S5 p.9].
• CH = 1 for the pinion. For the gear, CH = 1 + A′(mG − 1), where A′ = 0 when HBP/HBG = 1 [MD-S5 p.9].
• Gear stresses can be scaled from the pinion values [MD-S5 p.10–11]:
  • σG = (KsG/KsP)(JP/JG) σP;
  • σall,G = (YNG/YNP) σall,P;
  • σc,G = (KsG/KsP)^0.5 σc,P.

7.4 Worked problem P14-20 (spur) [MD-S5 p.4–13]

• Data [p.4]:
  • Commercial enclosed drive, 5 hp, pinion at 300 rev/min, 20°, NP = 16, NG = 48.
  • Grade 1 steel, through-hardened, 200 HB; quality No. 6; uncrowned; centered between bearings.
  • Pinion life 10⁸ cycles, R = 0.90; trial Pd = 6 teeth/in, F = 2 in.
• Geometry and factors:
  • dp = 2.667 in, dG = 8 in; V = 209.4 ft/min; Wt = 787.8 lbf [p.4].
  • Ko = 1; B = 0.8255; A = 59.77; Kv = 1.196 [p.5].
  • YP = 0.296; YG interpolated between 43T (0.397) and 50T (0.409) = 0.4056 [p.5].
  • Ks,P = 1.088; Ks,G = 1.097; Cpf = 0.0625; Cma = 0.158; Km = 1.2205 [p.6].
  • JP = 0.27, JG = 0.38; St = 28 260 psi; YN,P = 0.977, YN,G = 0.996 [p.7].
  • I = 0.1205; Sc = 93 500 psi [p.8–9]; ZN,P = 0.948, ZN,G = 0.973 [p.9].
• Bending [p.9–10]:
  • σP = 13 901.8 psi; σall,P = 32 482.37 psi; SF,P = 2.34.
  • σG = 9959.3 psi; σall,G = 33 114 psi; SF,G = 3.32.
• Wear [p.10–11]:
  • σc,P = 101 476.3 psi; σc,all,P = 104 280 psi; SH,P = 1.0276.
  • σc,G = 101 895.14 psi; σc,all,G = 107 030 psi; SH,G = 1.05.
• Conclusion [p.11]: pinion SF 2.34 > SH² = 1.056, and gear SF 3.32 > SH² = 1.1025, so wear is the threat for both. The hardness should be increased.
• Part (b) [p.12–13] (⚠ F10):
  • Set SH,P = 2 with 3p ≤ F ≤ 5p. This gives (104 280)²/[(2300)²·648.94·(Pd/F)] = 4, so Pd/F = 0.8.
  • Then 3π/0.8 ≤ F² ≤ 5π/0.8, so 3.43 ≤ F ≤ 4.43. Result: F = 4 in, Pd = 3.2 teeth/in.

7.5 Worked problem P14-20 repeated with helical gears [MD-S5 p.14–18]

• Data: φn = 20°, ψ = 30°, Pn = 6.
• Geometry:
  • Pt = Pn cos ψ = 5.196 teeth/in; dp = 3.079 in; dG = 9.238 in [p.14].
  • V = 241.8 ft/min; Wt = 682.3 lbf [p.14].
• Factors that differ from the spur case:
  • Kv = 1.21; Cpf = 0.0525 (⚠ F12) [p.14].
  • Km = 1.2105 [p.15].
  • J′P = 0.45, J′G = 0.54; modifying factors 0.98 and 0.94 → JP = 0.441, JG = 0.5076 [p.15].
  • φt = tan⁻¹(tan 20°/cos 30°) = 22.8°; pN = (π/6) cos 20° = 0.492 in [p.15].
  • rbP = 1.419 in; rbG = 4.258 in; a = 0.167 in [p.15].
  • Z = 0.7466; mN = 0.6937; I = 0.193 [p.16].
• Bending [p.16–17]:
  • σP = 6405.53 psi; σall,P = 32 482.37 psi; SF,P = 5.07.
  • σG = 5611.12 psi; σall,G = 33 114.1 psi; SF,G = 5.9.
• Wear [p.17–18]:
  • σc,P = 69 567.53 psi; σc,all,P = 104 280 psi; SH,P = 1.5.
  • σc,G = 69 854.67 psi; σc,all,G = 107 030 psi; SH,G = 1.53.

Clutches and brakes

[MD-6]

8.1 Definitions and classification

• A clutch connects and disconnects two collinear shafts. It is similar to a coupling, but involves friction and heat dissipation [MD-6 p.2].
• A brake stops the rotation of a shaft. Braking is by friction as a stationary part bears on a moving part [MD-6 p.2].
• Heat dissipation is a problem. Brake fade occurs during continuous application because of heat [MD-6 p.2].
• Square-jaw and spiral-jaw clutches are shown [MD-6 p.2].
• Clutch requirements: let the vehicle stop while the transmission is in gear; give a smooth take-off from dead stop; give smooth gear changes; not slip under heavy loads or full power [MD-6 p.3].
• A brake applies artificial frictional resistance to a moving body to retard or stop it. It absorbs kinetic energy, potential energy, or both [MD-6 p.3].
• Brakes and clutches are essentially the same devices, associated with rotation. Brakes absorb KE and convert it to heat; clutches transmit power between two shafts [MD-6 p.4].
• Model (Fig 16-1a): two inertias I1 and I2 at ω1 and ω2 (one may be zero for a brake) are brought to the same speed. Slippage dissipates energy and raises the temperature [MD-6 p.5].
• Quantities of interest: actuating force, torque transmitted, energy loss, temperature rise [MD-6 p.5].
• Torque depends on the actuating force, f and geometry (statics, per configuration). Temperature rise depends on the energy loss and is independent of the type; it depends on the geometry of the heat-dissipating surfaces [MD-6 p.6].
• Classification: rim type with internal expanding shoes; rim type with external contracting shoes; band type; disk or axial type; cone type; miscellaneous [MD-6 p.6].
• Internal-shoe rim clutches have three elements: the mating frictional surface, the means of transmitting torque and the actuating mechanism. They are classified as expanding-ring, centrifugal, magnetic, hydraulic or pneumatic [MD-6 p.9].

8.2 Internal expanding shoes (drum)

• The pressure p at angle θ from the hinge pin is proportional to sin θ, with maximum pa at θa: p/sin θ = pa/sin θa [MD-6 p.18].
• p = (pa/sin θa) sin θ (16-1): a sinusoidal distribution [MD-6 p.19].
• Short shoe (Fig 16-6a): the largest pressure pa is at the end of the shoe (θa = θ2) [MD-6 p.19].
• Long shoe (Fig 16-6b): the largest pressure pa is at θa = 90° [MD-6 p.20].
• Normal force on an element: dN = p b r dθ = pa b r sin θ dθ/sin θa, where b = face width. Fig 16-7 shows the hinge-pin reactions Rx, Ry and the force F (Fx, Fy) at distance c from the hinge pin [MD-6 p.21].
• Moment of the friction forces: Mf = ∫ f dN (r − a cos θ) = (f pa b r/sin θa) ∫θ1→θ2 sin θ (r − a cos θ) dθ (16-2) [MD-6 p.23].
  • Integrated form: (f pa b r/sin θa)[(−r cos θ) − a(½ sin²θ)], evaluated from θ1 to θ2.
• Moment of the normal forces: MN = ∫ dN (a sin θ) = (pa b r a/sin θa) ∫θ1→θ2 sin²θ dθ (16-3) [MD-6 p.24].
  • Integrated form: (pa b r a/sin θa)(θ/2 − ¼ sin 2θ), evaluated from θ1 to θ2.
• Equilibrium: F c − MN + Mf = 0, so F = (MN − Mf)/c (16-4) [MD-6 p.25].
• If MN = Mf the shoe is self-locking and no actuating force is needed. To avoid self-locking: MN > Mf (16-5) [MD-6 p.25].
• Torque: T = ∫ f r dN = (f pa b r²/sin θa) ∫ sin θ dθ = f pa b r²(cos θ1 − cos θ2)/sin θa (16-6) [MD-6 p.26].
• If rotation reverses, the friction forces reverse. For counterclockwise rotation: F = (MN + Mf)/c (16-7) [MD-6 p.27].
• A = ∫ sin θ cos θ dθ = (½ sin²θ), from θ1 to θ2; B = ∫ sin²θ dθ = (θ/2 − ¼ sin 2θ), from θ1 to θ2 (16-8) [MD-6 p.28].
• Hinge-pin reactions, clockwise rotation (Fig 16-7) (16-9) [MD-6 p.28]:
  • Rx = (pa b r/sin θa)(A − fB) − Fx
  • Ry = (pa b r/sin θa)(B + fA) − Fy
• Hinge-pin reactions, counterclockwise rotation (16-10) [MD-6 p.28]:
  • Rx = (pa b r/sin θa)(A + fB) − Fx
  • Ry = (pa b r/sin θa)(B − fA) − Fy

8.3 External contracting shoes

• Mf and MN are the same as Eqs (16-2) and (16-3) (Fig 16-11) [MD-6 p.39–40].
• Clockwise [MD-6 p.40]:
  • F = (MN + Mf)/c (16-11)
  • Rx = (pa b r/sin θa)(A + fB) − Fx (16-12)
  • Ry = (pa b r/sin θa)(fA − B) + Fy (16-12)
• Counterclockwise [MD-6 p.41]:
  • F = (MN − Mf)/c (16-13)
  • Rx = (pa b r/sin θa)(A − fB) − Fx (16-14)
  • Ry = (pa b r/sin θa)(−fA − B) + Fy (16-14)

8.4 Symmetrical pivoted shoe

• A special case: the pivot is placed symmetrically, and so that the moment of the friction forces about it is zero (Fig 16-12a) [MD-6 p.43].
• Symmetry means θ1 = θ2. Setting Mf = 2∫0→θ2 (f dN)(a cos θ − r) = 0 gives a = 4r sin θ2/(2θ2 + sin 2θ2) (16-15) [MD-6 p.44].
• Rx = 2∫0→θ2 dN cos θ = (pa b r/2)(2θ2 + sin 2θ2) (16-16) [MD-6 p.45].
• Ry = 2∫0→θ2 f dN cos θ = (pa b r f/2)(2θ2 + sin 2θ2) (16-17) [MD-6 p.45].
• Rx = N and Ry = fN; the torque is T = a f N (16-18) [MD-6 p.45].

8.5 Band brakes

• Flexible clutch and brake bands are used in power excavators, hoisting and other machinery [MD-6 p.46].
• Because of friction and drum rotation, the actuating force P2 is less than the pin reaction P1 (Fig 16-13) [MD-6 p.47].
• P1/P2 = e^(fφ) (16-19); T = (P1 − P2)D/2 (16-20) [MD-6 p.48].
• Pressure: p = P/(br) = 2P/(bD) (16-21). The pressure is proportional to the band tension [MD-6 p.48].
• The maximum pressure occurs at the toe: pa = 2P1/(bD) (16-22), where b = band width [MD-6 p.48].

8.6 Axial (disk) clutches

• Fig 16-14: single-plate clutch; A = driver, B = driven plate (keyed to the driven shaft), C = actuator [MD-6 p.50].
• Fig 16-16: friction disk with outside diameter D and inside diameter d. We want the axial force F that produces torque T at pressure p. Two methods are used, depending on the construction [MD-6 p.51].
• Uniform wear: pa occurs at r = d/2, so p r = pa d/2 [MD-6 p.52].
  • F = ∫d/2→D/2 2πpr dr = (π pa d/2)(D − d) (16-23)
  • T = ∫ 2π f p r² dr = (π f pa d/8)(D² − d²) (16-24)
  • T = (Ff/4)(D + d) (16-25)
• Uniform pressure [MD-6 p.53]:
  • F = (π pa/4)(D² − d²) (16-26)
  • T = 2π f pa ∫d/2→D/2 r² dr = (π f pa/12)(D³ − d³) (16-27)
  • T = (Ff/3)(D³ − d³)/(D² − d²) (16-28)

8.7 Disk brakes: annular pad and circular (button) pad

• Annular pad (Fig 16-19): r̄ locates the line of action of F. The effective radius re is the radius of an equivalent shoe of infinitesimal radial thickness [MD-6 p.58].
• Annular pad, uniform wear [MD-6 p.59]:
  • F = (θ2 − θ1) pa ri (ro − ri) (16-33)
  • T = ½(θ2 − θ1) f pa ri (ro² − ri²) (16-34)
  • re from f F re = T: re = (ro + ri)/2 (16-35)
  • r̄ = [(cos θ1 − cos θ2)/(θ2 − θ1)](ro + ri)/2 (16-36)
• Annular pad, uniform pressure [MD-6 p.60]:
  • F = ½(θ2 − θ1) pa (ro² − ri²) (16-37)
  • T = ⅓(θ2 − θ1) f pa (ro³ − ri³) (16-38)
  • re = (2/3)(ro³ − ri³)/(ro² − ri²) (16-39)
  • r̄ = (cos θ1 − cos θ2)/(θ2 − θ1)(ro³ − ri³)/(ro² − ri²) (16-40)
• θ1 and θ2 are in radians [MD-6 p.62].
• Circular pad caliper (Fig 16-20, pad radius R, eccentricity e): numerical integration is needed because the boundaries are hard to handle in closed form. Table 16-1 (by Fazekas) gives the parameters [MD-6 p.65].
• Circular pad: re = δe (16-41); F = πR² pav (16-42); T = f F re (16-43) [MD-6 p.66].

Table 16-1. Parameters for a circular-pad caliper brake (Source: G. A. Fazekas, "On Circular Spot Brakes," Trans. ASME, J. Engineering for Industry, vol. 94, Series B, no. 3, Aug 1972, pp. 859–863) [MD-6 p.66]

8.8 Cone clutch

• Parts (Fig 16-21): cup, cone, spring, shifting groove, cone angle α. The contact element is dr/sin α (Fig 16-22) [MD-6 p.69–71].
• Uniform wear [MD-6 p.72]:
  • F = ∫ p dA sin α = ∫ (pa d/2r)(2πr dr/sin α)(sin α) = (π pa d/2)(D − d) (16-44)
  • T = ∫ r f p dA = [π f pa d/(8 sin α)](D² − d²) (16-45)
  • T = [Ff/(4 sin α)](D + d) (16-46)
• Uniform pressure [MD-6 p.73]:
  • F = (π pa/4)(D² − d²) (16-47)
  • T = [π f pa/(12 sin α)](D³ − d³) (16-48)
  • T = [Ff/(3 sin α)](D³ − d³)/(D² − d²) (16-49)

8.9 Energy and temperature rise

• US: ΔT = H/(Cp W) (16-54). ΔT in °F; Cp = specific heat, Btu/(lbm·°F) (use 0.12 for steel or cast iron); W = mass of the clutch or brake parts, lbm [MD-6 p.74].
• SI: ΔT = E/(Cp m) (16-55). ΔT in °C; Cp in J/(kg·°C) (use 500 for steel or cast iron); m in kg [MD-6 p.74].
• Heat in Btu: H = E/9336 (16-53) [MD-6 p.74].
• Rate of heat loss: Hloss = ħCR A(T − T∞) = (hr + fv hc)A(T − T∞) (16-57) [MD-6 p.75]. Terms:
  • Hloss, Btu/s; ħCR = overall heat-transfer coefficient, Btu/(in²·s·°F).
  • hr = radiation component, hc = convective component (both Fig 16-24a); fv = ventilation factor (Fig 16-24b).
  • T = disk temperature, T∞ = ambient temperature (°F); A = lateral surface area, in².
• Fig 16-24a: hr and hc (10⁻⁶ Btu/(s·in²·°F)) in still air against temperature rise T − T∞ = 0–700 °F [MD-6 p.76].
• Fig 16-24b: ventilation factor fv against forced ventilation velocity, 0–80 ft/s (Tolo-o-matic) [MD-6 p.77].
• Energy to stop a rotary inertia I: E = ½(I/9336)(ωo² − ωf²) (16-58). Temperature rise per stop: ΔT = E/(W Cp) (16-59) [MD-6 p.78].
• Steady state: (Tmin − T∞)/(Tmax − T∞) = exp(−β t1), with β = ħCR A/(W Cp) [MD-6 p.78].
• Tmax = T∞ + ΔT/[1 − exp(−β t1)] (16-60). t1 is the average time in seconds between two applications of the brake [MD-6 p.78].

Table 16-3. Characteristics of friction materials for brakes and clutches [MD-6 p.82]

• Sources printed under the table: Ferodo Ltd.; Scan-pac; Raybestos; Gatke Corp.; General Metals Powder Co.; D. A. B. Industries; Friction Products Co. [MD-6 p.82].
• Sheet problems: 16-1, 16-9, 16-13, 16-16, 16-19. Projects: Table 1-2 problem groups; Example 18-1, p. 936 in Shigley [MD-6 p.82–83].
• The single-plate clutch sketch labels the friction surface "μs = 1.6" (⚠ F8) [MD-6 p.49].

8.10 Worked examples [MD-6]

• Ex 16-2, internal two-shoe brake (Fig 16-8) [p.29–36]:
  • Data: drum 300 mm diameter; the same force F on each shoe; b = 32 mm; molded asbestos, f = 0.32; pa ≤ 1000 kPa.
  • The right-hand shoe is self-energizing: θ1 = 0, θ2 = 126°, θa = 90°; a = √(112² + 50²) = 122.7 mm; c = 100 + 112 = 212 mm.
  • Mf = 304 N·m; MN = 788 N·m; F = (788 − 304)/(100 + 112) = 2.28 kN.
  • TR = 366 N·m.
  • Left-hand shoe from Eq (16-7): pa = 443 kPa; TL = 162 N·m; total T = 528 N·m.
  • A = 0.3273; B = 1.3373; D = pa b r/sin θa = 4.8 kN.
  • Right hinge pin: Rx = −1.410 kN; Ry = 4.839 kN; R = 5.04 kN.
  • Left hinge pin: Rx = 0.678 kN; Ry = 0.538 kN; R = 0.866 kN.
• Ex 16-3, annular pads, uniform wear [p.61–64]:
  • Data: ri = 3.875 in, ro = 5.50 in, 108°, f = 0.37; two hydraulic cylinders 1.5 in in diameter; T = 13 000 lbf·in.
  • θ1 = 36°, θ2 = 144°; T per pad = 6500 lbf·in.
  • pa = 315.8 psi; F = 3748 lbf; re = 4.688 in; r̄ = 4.024 in; p_hydraulic = 2121 psi.
• Ex 16-4, button pad [p.67–68]:
  • Data: dry sintered metal, R = ½ in, e = 2 in, pmax/2 = 175 psi, f = 0.31.
  • R/e = 0.25 → δ = 0.963, pmax/pav = 1.290.
  • re = 1.926 in; pav = 135.7 psi; F = 106.6 lbf; T = 63.65 lbf·in (one side).
• Ex 16-5, caliper brake temperature [p.79–81] (⚠ F7):
  • Data: 24 stops/h from 250 rev/min; air speed 25 ft/s; I = 289 (unit printed as lbm·in·s); steel disk γ = 0.282 lbm/in³, Cp = 0.108 Btu/(lbm·°F), 6 in diameter, ¼ in thick; A = 50 in²; T∞ = 70°F.
  • t1 = 3600/24 = 150 s (printed as 60²/24).
  • Assuming a 200°F rise: hr = 3.0(10⁻⁶), hc = 2.0(10⁻⁶), fv = 4.8 → ħCR = 12.6(10⁻⁶).
  • W = 1.99 lbm; E = 10.6 Btu; β = 2.93(10⁻³) s⁻¹; ΔT = 49.3°F.
  • Tmax = 209°F; Tmin = 160°F. Iterating on hr and hc converges to Tmax = 220°F, Tmin = 171°F.
  • Table 16-3 gives 570–660°F continuous for dry sintered metal, so there is no danger of overheating.

Rolling-contact bearings

[MD-7]

Figure: md-bearing-nomenclature.jpg (Fig 11-1) [MD-7 p.2].

9.1 Nomenclature and types

• Fig 11-1 parts: width, corner radius, outer ring, inner ring, shoulders, bore, outside diameter, inner-ring ball race, separator (retainer), outer-ring ball race, face [MD-7 p.2].
• Ball bearings (Fig 11-2): deep groove, filling notch, angular contact, shielded, sealed, external self-aligning, double row, self-aligning, thrust, self-aligning thrust [MD-7 p.4].
• Roller bearings (Fig 11-3, Timken): straight cylindrical, spherical roller thrust, tapered roller thrust, needle, tapered roller, steep-angle tapered roller [MD-7 p.5].

9.2 Life and load-rating definitions

• Bearing failure: spalling or pitting of an area of 0.01 in² [MD-7 p.6].
• Life: the number of revolutions (or hours at a given speed) to failure, for one bearing [MD-7 p.6].
• Rating life: the life that 10% of a group of bearings fail to reach. It is also called the minimum life or L10 life [MD-7 p.6].
• Median life: the life at which 50% of the sample has failed, for many groups. It is also called the average life or average median life. The median life is typically 4 or 5 times the L10 life [MD-7 p.6].
• Each manufacturer chooses its own rating life. The most common is 10⁶ rev. Timken is a well-known exception: 3000 h at 500 rev/min = 90(10⁶) rev [MD-7 p.7].
• These rating-life levels are low for today's bearings, but they are kept because the rating life is an arbitrary reference point [MD-7 p.7].
• Catalog load rating C10: the constant radial load that causes 10% of a group to fail at the manufacturer's rating life [MD-7 p.8].
  • It depends on bearing type, geometry, accuracy of fabrication and material.
  • It is called the basic dynamic load rating (or basic load rating) if the rating life is 10⁶ rev.
  • It is a reference value, not an actual load.
• Static load rating C0: the static radial load that produces a permanent deformation of 0.0001d at the most heavily stressed contact of rolling element and race (d = roller diameter) [MD-7 p.9].
  • It is used to check for permanent deformation, and is tabulated with C10 (Table 11-2).
• Equivalent radial load Fe: the load that does the same damage as the combined radial and thrust loads [MD-7 p.9].

9.3 Load–life relation

• Load against life on a log–log plot is approximately linear (Fig 11-4) [MD-7 p.10].
• F L^(1/a) = constant (11-1); a = 3 for ball bearings; a = 10/3 for roller bearings (cylindrical and tapered) [MD-7 p.10].
• F1 L1^(1/a) = F2 L2^(1/a) (11-2) [MD-7 p.11].
• With R = catalog rating conditions and D = desired design conditions: FR LR^(1/a) = FD LD^(1/a) (a). If life 𝓛 is in hours at n rev/min: L = 60 𝓛 n (b) [MD-7 p.12].
• FR(𝓛R nR 60)^(1/a) = FD(𝓛D nD 60)^(1/a) (c). Terms: FR = catalog rating (lbf or kN); 𝓛R = rating life, h; nR = rating speed, rev/min; FD = desired radial load; 𝓛D = desired life, h; nD = desired speed [MD-7 p.13].
• C10 = FR = FD(LD/LR)^(1/a) = FD[𝓛D nD 60/(𝓛R nR 60)]^(1/a) (11-3) [MD-7 p.13–14].
• FD and LD come from the problem statement; LR is stated by the manufacturer (many catalogs rate at LR = 10⁶ rev). C10 is used to find a suitable bearing in the catalog [MD-7 p.14].
• xD = LD/LR is a dimensionless multiple of the rating life [MD-7 p.14].

9.4 Reliability (Weibull)

• At constant load, the life distribution is right-skewed; the Weibull distribution is a good candidate [MD-7 p.16–17].
• R = exp{−[(x − x0)/(θ − x0)]^b} (11-4) [MD-7 p.17]. Terms:
  • x = L/L10 (dimensionless life); x0 = guaranteed ("minimum") value of x;
  • θ = characteristic parameter (the 63.2121 percentile of x);
  • b = shape parameter, ≈ 1.5 for rolling-contact bearings.
• Mean: μx = x0 + (θ − x0)Γ(1 + 1/b) (11-6) [MD-7 p.18].
• Standard deviation: σ̂x = (θ − x0)√[Γ(1 + 2/b) − Γ²(1 + 1/b)] (11-7) [MD-7 p.18].
• Γ is the gamma function, tabulated in Table A-34 for n = 1.00–2.00. For n > 2 use Γ(n) = (n − 1)Γ(n − 1); e.g., Γ(5.42) = 32.4245 [MD-7 p.18–19].
• Life at reliability R: x = x0 + (θ − x0)[ln(1/R)]^(1/b) (11-8) [MD-7 p.18].
• Constant-reliability contours (Fig 11-5) [MD-7 p.22–25]:
  • The catalog point A is at (C10, x10 = 1) on the R = 0.90 contour; the design point D is at (FD, xD) on R = RD. The designer moves from D to A via B.
  • Along BD: FB = FD(xD/xB)^(1/a).
  • Along AB: xB = x0 + (θ − x0)[ln(1/RD)]^(1/b).
• C10 = af FD {xD/[x0 + (θ − x0)(ln(1/RD))^(1/b)]}^(1/a) (11-9) [MD-7 p.25].
  • af = application factor: a factor of safety for overload, dynamic loading and uncertainty (Table 11-5).
  • When RD = 0.90, the denominator equals 1 and the equation reduces to (11-3).
• Approximation: ln(1/RD) = ln[1/(1 − pf)] ≈ pf = 1 − RD, where pf = probability of failure [MD-7 p.27]. This gives:
  • C10 ≈ af FD {xD/[x0 + (θ − x0)(1 − RD)^(1/b)]}^(1/a), valid for R ≥ 0.90 (11-10).
• Typical Weibull parameters (usually given by the catalog) [MD-7 p.26]:

• Combined reliability of several bearings is the product of the individual reliabilities. For two bearings on a shaft, R = RA RB. Identical bearings should each have the square root of the total desired reliability. Bearings that are not identical need not have equal reliabilities, as long as the total is met [MD-7 p.29].

9.5 Dimension series and catalog tables

• The ABMA dimension-series code gives the relative size of the boundary dimensions of the cross-section for metric bearings. It has two digits [MD-7 p.30–31]:
  • First digit = width series 0, 1, 2, 3, 4, 5, 6 (width increasing).
  • Second digit = diameter series 8, 9, 0, 1, 2, 3, 4 (outside diameter increasing).
• Shaft and housing shoulder diameters dS and dH should be adequate to ensure good bearing support (Fig 11-8) [MD-7 p.33].

Table 11-2. Dimensions and load ratings for single-row 02-series deep-groove and angular-contact ball bearings [MD-7 p.32] (also partly on p.15)

Table 11-3. Dimensions and basic load ratings for cylindrical roller bearings [MD-7 p.34]

9.6 Combined radial and thrust load

• When a ball bearing carries both an axial thrust Fa and a radial load Fr, an equivalent radial load Fe that does the same damage is used. The plot of Fe/(VFr) against Fa/(VFr) is obtained experimentally (Fig 11-6) [MD-7 p.35].
• V is the rotation factor, for the difference in ball rotation between outer-ring and inner-ring rotation: V = 1 for inner-ring rotation, V = 1.2 for outer-ring rotation [MD-7 p.35].
• The data are approximated by two straight lines [MD-7 p.36]:
  • Fe/(VFr) = 1 when Fa/(VFr) ≤ e (11-11a)
  • Fe/(VFr) = X + Y Fa/(VFr) when Fa/(VFr) > e (11-11b)
  • X is the ordinate intercept and Y the slope.
• Single equation: Fe = Xi V Fr + Yi Fa (11-12), with i = 1 when Fa/(VFr) ≤ e and i = 2 when Fa/(VFr) > e. X and Y depend on the geometry and construction of the bearing [MD-7 p.37].
• X and Y for a specific bearing come from its catalog; Table 11-1 gives representative values. X and Y are functions of e, which is a function of Fa/C0 (C0 = basic static load rating) [MD-7 p.38].