AEE Mechanical Engineering Core · Chapter 3

Design of Machine Elements

What to remember

Design is a stress-and-strength check. Working stress = failure stress/factor of safety. Ductile parts use yield strength. Brittle parts use ultimate strength. Parts under varying load use fatigue criteria (Goodman, Soderberg).
Standard elements have standard formulas: shafts (T = πτd³/16), keys (shear and crushing), welds (throat 0.707s), rivets (tearing, shearing, crushing), bolts (core diameter), springs (Wahl factor), gears (Lewis equation) and bearings (L10 life).
Stress concentration raises local stress at holes, keyways, fillets and threads. It is vital under fatigue loading and is often ignored for static ductile loading.

1. Design basics

Factor of safety n = failure stress/allowable stress. Typical choice depends on certainty of load, material quality and consequence of failure.
Preferred numbers give standard series of sizes in geometric progression. R5 has ratio about 1.58, R10 about 1.26, R20 about 1.12 and R40 about 1.06. These are fifth, tenth, twentieth and fortieth roots of 10.
Stress concentration factor Kt = maximum stress/nominal stress. Notch sensitivity q = (Kf − 1)/(Kt − 1), so the fatigue factor Kf = 1 + q(Kt − 1). q lies between 0 and 1.
Ways to reduce stress concentration: generous fillet radii, relief grooves, gradual change of section, and removal of sharp corners.
Standardisation uses Indian Standards (IS), plus international codes such as ASME for shafts. Boiler pressure parts follow the Indian Boiler Regulations.

2. Fatigue

Endurance limit σe is the stress below which a part can take an infinite number of cycles. For steels, σe is about 0.5σu for lower strengths. Non-ferrous metals have no clear endurance limit.
Corrected endurance limit = σe′ × surface factor × size factor × load factor × reliability factor ÷ Kf.
Mean stress σm = (σmax + σmin)/2. Stress amplitude σa = (σmax − σmin)/2.
Criteria for combined mean and alternating stress:

Criterion	Line	Remark
Soderberg	σm/σy + σa/σe = 1/n	Most conservative
Modified Goodman	σm/σu + σa/σe = 1/n	Common practice
Gerber	Parabola through σe and σu	Fits ductile data well

The S-N curve plots stress against number of cycles. For steels it becomes flat near 10⁶ cycles.

3. Shafts, keys and couplings

Solid shaft in torsion: T = (π/16)τd³. Hollow shaft: T = (π/16)τ(D⁴ − d⁴)/D.
Bending: M = (π/32)σd³.
Combined bending M and twisting T:
Equivalent twisting moment Te = √(M² + T²) (maximum shear stress theory).
Equivalent bending moment Me = ½[M + √(M² + T²)] (maximum normal stress theory).
Shafts are also checked for rigidity (angle of twist) and for critical speed.
Power: P = 2πNT/60 (watts, T in N·m, N in rpm).
Keys transmit torque between shaft and hub. Types: sunk (rectangular or square), saddle, Woodruff, and round keys. Splines carry more torque than single keys.
Standard key proportions: width w = d/4 and thickness t = d/6 for a rectangular sunk key, where d is shaft diameter.
Key failure: force on key F = 2T/d. In shear: F = w·l·τ. In crushing: F = (t/2)·l·σc. A key is made weaker than the shaft so that it fails first and protects costly parts.
Rigid couplings: sleeve (muff), split-muff and flange couplings; they need exact shaft alignment. Flexible couplings: bushed-pin type; they tolerate small misalignment. Oldham coupling joins parallel shafts with a small offset. Universal (Hooke's) joint joins intersecting shafts.
Flange coupling bolts: T = n × (π/4)d_b² × τ × (D₁/2), with n bolts on pitch circle diameter D₁.

4. Fasteners: bolts, power screws and welds

Thread terms: pitch, lead (= pitch × number of starts), major diameter, minor (core) diameter. ISO metric threads have a 60° angle. The helix angle α is given by tanα = lead/(π d_mean).
Bolt under axial load P: core diameter dc = √[4P/(π σt)]. Bolts with initial tightening carry both preload and external load. Bolts under shear use the plain shank area.
Cylinder cover bolts: total load = (π/4)D²p. This is shared by all bolts.
Power screws (square, trapezoidal, ACME) convert rotary motion into linear motion. Torque to raise load W: T = W(d_m/2) tan(α + φ), with friction angle φ (tanφ = μ). Efficiency η = tanα/tan(α + φ). Maximum efficiency occurs at α = 45° − φ/2, and is (1 − sinφ)/(1 + sinφ).
Self-locking when φ ≥ α. Efficiency of a self-locking screw is below 50%.
Welded joints: throat of a fillet weld = 0.707 × leg size (s). Single fillet weld (shear): P = 0.707·s·l·τ. Double parallel fillet: P = 1.414·s·l·τ. Butt weld: P = t·l·σ.
Riveted joints: the plate or rivets fail in one of three ways, and strength is the least of these per pitch.

Failure mode	Resistance per pitch
Tearing of plate between holes	(p − d)·t·σt
Shearing of rivet (single shear)	(π/4)d²·τ per rivet
Crushing of rivet or plate	d·t·σc per rivet

Efficiency = least of the three/(p·t·σt). Unwin's formula: rivet hole diameter d = 6√t (mm, t in mm). Rivet pitch has a minimum of about 2d.

5. Springs and pressure vessels

Helical spring: mean coil diameter D, wire diameter d, spring index C = D/d (usually 4 to 12). Shear stress τ = K·8WD/(πd³). Wahl factor K = (4C − 1)/(4C − 4) + 0.615/C. It accounts for direct shear and curvature.
Deflection: δ = 8WD³n/(Gd⁴), with n active coils. Stiffness k = Gd⁴/(8D³n).
Leaf spring (semi-elliptic, n leaves, width b, thickness t, span L): σ = 3WL/(2nbt²), δ = 3WL³/(8Enbt³).
Surge is wave-like vibration along the coils in springs under fast loading, such as valve springs in engines.
Thin pressure vessel: t = pd/(2σt·η_j), with η_j the joint efficiency. Thick cylinders use Lame's equation. Boiler drums and steam pipes in thermal stations are designed with such rules.

6. Gears and bearings

Gear forces: tangential Wt = 2T/D. Radial Wr = Wt·tanφ. Helical gears also have axial thrust Wa = Wt·tanβ.
Lewis equation for beam strength of a tooth: Wt = σ·b·π·m·y = σ·b·m·Y, where Y = πy is the Lewis form factor. The factor depends on tooth number and pressure angle.
Velocity factor: Cv = 3/(3 + v) for ordinary cut gears, and 6/(6 + v) for accurately cut gears (v in m/s).
A gear can fail by tooth bending (Lewis), pitting (surface contact fatigue, Hertz), scoring or wear. Design is checked for both bending and wear strength.
Journal bearings: hydrodynamic lubrication builds an oil film. Sommerfeld number S = (ZN/p)(r/c)², where Z = viscosity, N = speed, p = bearing pressure, r/c = radius to clearance ratio. Bearing characteristic number = ZN/p. A low value means boundary lubrication and high friction. Petroff's equation gives friction for a lightly loaded bearing.
Rolling bearings: the life is the number of revolutions reached or exceeded by 90% of the bearings. Basic life L₁₀ = (C/P)^k million revolutions, where C is the dynamic load rating and P the equivalent load. k = 3 for ball bearings and 10/3 for roller bearings. Life in hours = 10⁶L/(60N).
Equivalent load P = XVFr + YFa (radial and axial parts).

7. Worked examples

Example 1 (shaft). Solid shaft d = 40 mm, τ = 50 MPa. T = (π/16)(50)(40³) = 628,318 N·mm ≈ 628 N·m. At 300 rpm: P = 2π × 300 × 628.3/60 ≈ 19.7 kW.

Example 2 (combined). M = 300 N·m, T = 400 N·m. Te = √(300² + 400²) = 500 N·m. Me = ½(300 + 500) = 400 N·m.

Example 3 (weld). Single fillet weld s = 8 mm, l = 100 mm, τ = 60 MPa. P = 0.707 × 8 × 100 × 60 ≈ 33.9 kN.

Example 4 (rivet). p = 50 mm, d = 20 mm, t = 10 mm, σt = 80, τ = 60, σc = 150 MPa. Tearing = 30 × 10 × 80 = 24,000 N. Shearing = 314.16 × 60 = 18,850 N. Crushing = 20 × 10 × 150 = 30,000 N. Least = 18,850 N. Plate strength = 50 × 10 × 80 = 40,000 N. Efficiency = 47.1%.

Example 5 (bolt). P = 7854 N, σt = 100 MPa. dc = √(4 × 7854/(π × 100)) = 10 mm.

Example 6 (bearing). C = 20 kN, P = 5 kN, ball bearing. L₁₀ = (20/5)³ = 64 million revolutions.

Example 7 (spring). C = 8: K = 31/28 + 0.615/8 = 1.107 + 0.077 = 1.184.

Exam traps

Soderberg uses yield strength, while Goodman uses ultimate strength. Soderberg is more conservative.
Equivalent torque Te = √(M² + T²) is not the same as equivalent bending moment Me = ½(M + Te).
Fillet weld throat is 0.707 s, not s.
A double parallel fillet weld carries twice the single weld load.
Key width is d/4 and thickness d/6, not the other way round.
Ball bearing exponent is 3. Roller bearing exponent is 10/3.
Efficiency of a self-locking screw is below 50%.
Wahl factor corrects for curvature and direct shear in springs. It does not change deflection.
A key is made weaker than the shaft and hub on purpose, so it fails first.

One-liners

1. Factor of safety = failure stress/allowable stress.
2. R10 preferred number series has ratio about 1.26.
3. Endurance limit of steel is about half the ultimate strength.
4. Soderberg line is the most conservative.
5. Te = √(M² + T²).
6. Fillet weld throat = 0.707 s.
7. Unwin's formula: d = 6√t.
8. Spring index C = D/d.
9. Lewis equation gives tooth bending strength.
10. Self-locking screw has efficiency below 50%.
11. Sommerfeld number is dimensionless.
12. L₁₀ life has 90% reliability.

Practice questions

The factor of safety of a machine part is defined as
1. failure stress divided by allowable (working) stress
2. yield strength multiplied by working stress
3. endurance limit divided by yield strength
4. working stress divided by failure stress
Answer
A. failure stress divided by allowable (working) stress
n = σfailure/σallowable.
The theoretical stress concentration factor Kt is defined as
1. ultimate stress divided by yield stress
2. maximum stress at the discontinuity divided by nominal stress
3. endurance limit divided by yield stress
4. nominal stress divided by maximum stress
Answer
B. maximum stress at the discontinuity divided by nominal stress
Kt depends only on the geometry of the notch.
Which design criterion for combined mean and alternating stress is the most conservative?
1. Soderberg line
2. Ultimate strength line
3. Modified Goodman line
4. Gerber parabola
Answer
A. Soderberg line
Soderberg uses yield strength on the mean-stress axis.
The throat thickness of a fillet weld of leg size s is
1. 0.5 s
2. 0.707 s
3. s
4. 1.414 s
Answer
B. 0.707 s
For an equal-leg 45° fillet, throat = s·cos45°.
For a standard rectangular sunk key on a shaft of diameter d, the width is
1. d/2
2. d/6
3. d/8
4. d/4
Answer
D. d/4
Width = d/4 and thickness = d/6.
The life exponent in the basic life formula L₁₀ = (C/P)^k for ball bearings is
1. 3
2. 10/3
3. 2
4. 4
Answer
A. 3
k = 3 for ball bearings and 10/3 for roller bearings.
The spring index of a helical spring is
1. outer diameter divided by free length
2. wire diameter divided by mean coil diameter
3. mean coil diameter divided by wire diameter
4. number of coils divided by pitch
Answer
C. mean coil diameter divided by wire diameter
C = D/d, usually between 4 and 12.
Unwin's formula for the diameter of a rivet hole is
1. d = 2t
2. d = 6√t
3. d = t/6
4. d = 1.6t
Answer
B. d = 6√t
With t in mm, d = 6√t mm.
The equivalent twisting moment for a shaft under bending moment M and torque T is
1. M + T
2. ½(M + T)
3. (M² + T²)
4. √(M² + T²)
Answer
D. √(M² + T²)
It is based on the maximum shear stress theory.
The efficiency of a self-locking power screw is
1. exactly 100%
2. more than 50%
3. less than 50%
4. always above 70%
Answer
C. less than 50%
Self-locking needs friction angle ≥ helix angle, which gives η < 50%.
A key is usually designed to be weaker than the shaft so that
1. it increases the shaft strength
2. it fails first and protects the shaft and hub
3. it reduces the shaft speed
4. it carries more torque
Answer
B. it fails first and protects the shaft and hub
A sacrificial key is cheap to replace.
The Wahl factor in the design of a helical spring accounts for
1. curvature of the wire and direct shear
2. fatigue of the wire
3. surge in the coils
4. variation of modulus with temperature
Answer
A. curvature of the wire and direct shear
K = (4C − 1)/(4C − 4) + 0.615/C.
An Oldham coupling is used to connect
1. two intersecting shafts
2. two collinear rigid shafts only
3. shafts at right angles only
4. two parallel shafts with a small lateral offset
Answer
D. two parallel shafts with a small lateral offset
It is a flexible coupling for parallel offset.
The Lewis equation is used to check a gear tooth for
1. surface pitting
2. scoring
3. bending strength
4. noise
Answer
C. bending strength
The tooth is treated as a cantilever beam.
The basic rated life L₁₀ of a rolling bearing is the life that
1. all bearings complete without failure
2. at least 90% of identical bearings complete or exceed
3. only 50% of bearings complete
4. 10% of bearings reach exactly
Answer
B. at least 90% of identical bearings complete or exceed
L₁₀ corresponds to 90% reliability.
The notch sensitivity factor q of a material lies between
1. 0 and 1
2. 1 and 2
3. 0 and 10
4. −1 and 0
Answer
A. 0 and 1
q = 0 means no effect of notch; q = 1 means full effect.
In the R10 series of preferred numbers, the ratio between successive terms is about
1. 1.58
2. 1.12
3. 1.26
4. 1.06
Answer
C. 1.26
R10 is the tenth root of 10.
A solid shaft of diameter 40 mm has an allowable shear stress of 50 MPa. The torque it can transmit is about
1. 1256 N·m
2. 314 N·m
3. 62.8 N·m
4. 628 N·m
Answer
D. 628 N·m
T = (π/16) × 50 × 40³ = 628,318 N·mm.
A torque of 100 N·m is transmitted at 600 rpm. The power is about
1. 628 W
2. 62.8 kW
3. 6.28 kW
4. 3.14 kW
Answer
C. 6.28 kW
P = 2πNT/60 = 2π × 600 × 100/60 = 6283 W.
A shaft carries a bending moment of 300 N·m and a torque of 400 N·m. The equivalent twisting moment is
1. 500 N·m
2. 350 N·m
3. 700 N·m
4. 100 N·m
Answer
A. 500 N·m
Te = √(300² + 400²) = 500 N·m.
A single transverse fillet weld has leg size 8 mm, length 100 mm and allowable shear stress 60 MPa. The load it can carry is about
1. 67.9 kN
2. 24 kN
3. 48 kN
4. 33.9 kN
Answer
D. 33.9 kN
P = 0.707 × 8 × 100 × 60 = 33,936 N.
In a single-riveted lap joint the pitch is 60 mm, rivet hole diameter 20 mm, plate thickness 10 mm and allowable tensile stress 80 MPa. The tearing resistance per pitch is
1. 64 kN
2. 32 kN
3. 48 kN
4. 16 kN
Answer
B. 32 kN
(p − d)·t·σt = 40 × 10 × 80 = 32,000 N.
A bolt carries a steady axial load of 7854 N with permissible tensile stress 100 MPa. The core diameter required is
1. 5 mm
2. 20 mm
3. 8 mm
4. 10 mm
Answer
D. 10 mm
dc = √(4P/(πσ)) = √(31416/314.16) = 10 mm.
A ball bearing has a dynamic load rating of 20 kN and carries an equivalent load of 5 kN. The L₁₀ life is
1. 64 million revolutions
2. 256 million revolutions
3. 16 million revolutions
4. 4 million revolutions
Answer
A. 64 million revolutions
L₁₀ = (20/5)³ = 64.
A bearing has L₁₀ life of 36 million revolutions and runs at 1000 rpm. Its life in hours is
1. 36 h
2. 6000 h
3. 600 h
4. 360 h
Answer
C. 600 h
Hours = 36 × 10⁶/(60 × 1000) = 600 h.
The Wahl stress factor for a helical spring of index 8 is about
1. 2.0
2. 1.18
3. 0.85
4. 1.50
Answer
B. 1.18
K = 31/28 + 0.615/8 = 1.107 + 0.077 = 1.184.
A helical spring has D = 50 mm, d = 5 mm, 10 active coils and G = 80 GPa. Under an axial load of 500 N the deflection is
1. 10 mm
2. 200 mm
3. 50 mm
4. 100 mm
Answer
D. 100 mm
δ = 8WD³n/(Gd⁴) = 8 × 500 × 125,000 × 10/(80,000 × 625) = 100 mm.
A gear transmits a torque of 100 N·m through a pitch circle diameter of 0.2 m. The tangential force on the tooth is
1. 2000 N
2. 500 N
3. 1000 N
4. 20 N
Answer
C. 1000 N
Wt = 2T/D = 200/0.2 = 1000 N.
A key on a 50 mm shaft transmits a torque of 250 N·m. Its width is 10 mm and length 40 mm. The shear stress in the key is
1. 25 MPa
2. 50 MPa
3. 100 MPa
4. 12.5 MPa
Answer
A. 25 MPa
F = 2T/d = 10,000 N; τ = 10,000/(10 × 40) = 25 MPa.
A power screw has tanα = 0.1 and coefficient of friction tanφ = 0.1. Its efficiency is about
1. 25%
2. 49.5%
3. 66.7%
4. 80%
Answer
B. 49.5%
tan(α + φ) = 0.2/0.99 = 0.202, so η = 0.1/0.202 = 0.495.
A part is subjected to a stress that varies between 120 MPa and 40 MPa. The stress amplitude is
1. 40 MPa
2. 20 MPa
3. 80 MPa
4. 160 MPa
Answer
A. 40 MPa
σa = (120 − 40)/2 = 40 MPa and σm = 80 MPa.
A part has σa = 40 MPa and σm = 80 MPa, with σe = 200 MPa and σu = 400 MPa. Using the modified Goodman line, the factor of safety is
1. 4.0
2. 1.5
3. 2.5
4. 2.0
Answer
C. 2.5
1/n = 40/200 + 80/400 = 0.4, so n = 2.5.
Consider the statements about fatigue criteria. 1. The Soderberg line uses the yield strength. 2. The Goodman line uses the ultimate strength. Which is/are correct?
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
C. Both 1 and 2
Both statements are correct.
Consider the statements about welds. 1. The throat of a fillet weld equals its leg size. 2. A double parallel fillet weld has load capacity 1.414 s·l·τ. Which is/are correct?
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
B. 2 only
Throat = 0.707 s, giving 2 × 0.707 = 1.414.
Consider the statements about rolling bearings. 1. The life exponent for ball bearings is 3. 2. The life exponent for roller bearings is also 3. Which is/are correct?
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
A. 1 only
Roller bearings use 10/3.
Consider the statements about rolling bearing life. 1. L₁₀ corresponds to 90% reliability. 2. Life in hours = 10⁶ L₁₀/(60 N), with N in rpm. Which is/are correct?
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
C. Both 1 and 2
Both statements are correct.
Consider the statements about keys. 1. A key is made stronger than the shaft. 2. A key can fail by shear or by crushing. Which is/are correct?
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
B. 2 only
A key is deliberately the weaker part.
Consider the statements about power screws. 1. A screw is self-locking if the friction angle is not less than the helix angle. 2. A self-locking screw has efficiency above 50%. Which is/are correct?
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
A. 1 only
A self-locking screw has efficiency below 50%.
Consider the statements about helical springs. 1. The Wahl factor reduces the deflection of the spring. 2. The spring index is defined as d/D. Which is/are correct?
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
D. Neither 1 nor 2
The Wahl factor corrects stress; the spring index is D/d.
Consider the statements about stress concentration in fatigue. 1. It should always be ignored in fatigue design. 2. The fatigue stress concentration factor is Kf = 1 + q(Kt − 1). Which is/are correct?
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
B. 2 only
Stress concentration is critical for fatigue.
Consider the statements about couplings. 1. A sleeve (muff) coupling is a rigid coupling. 2. A bushed-pin coupling is a flexible coupling. Which is/are correct?
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
C. Both 1 and 2
Both statements are correct.
Consider the statements about journal bearings. 1. A high value of ZN/p favours hydrodynamic lubrication. 2. A low value of ZN/p leads to boundary lubrication. Which is/are correct?
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
C. Both 1 and 2
Z = viscosity, N = speed, p = bearing pressure.
Consider the statements about gears. 1. The Lewis equation checks the wear strength of the tooth. 2. Hertz contact stress is used to check the bending of the tooth. Which is/are correct?
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
D. Neither 1 nor 2
Lewis checks bending; Hertz stress checks surface pitting.
Which pair of gear failure and design check is correctly matched?
1. Tooth breakage – Hertz contact stress
2. Pitting – Lewis bending equation
3. Scoring – Soderberg line
4. Tooth breakage – Lewis bending equation
Answer
D. Tooth breakage – Lewis bending equation
Breakage is a bending failure; pitting is a contact fatigue failure.
Which pair of rivet failure and resistance formula is correct?
1. Shearing of one rivet – (π/4)d²τ
2. Crushing – (π/4)d²σc
3. Tearing – (p + d)tσt
4. Shearing – dtσc
Answer
A. Shearing of one rivet – (π/4)d²τ
Tearing = (p − d)tσt; crushing = d·t·σc.

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