The basic rating life L₁₀ of a rolling bearing is the life:

that 90% of a large group of identical bearings complete or exceed before fatigue failure

at which 10% of the bearings fail immediately

that the average bearing achieves

of the bearing at its maximum speed

The load on a ball bearing is doubled at the same speed. Its L₁₀ life in revolutions becomes:

one-eighth of its earlier value

one-quarter of its earlier value

one-sixteenth of its earlier value

For hydrodynamic lubrication of a journal bearing, which of the following is essential?

relative motion of the surfaces dragging a viscous fluid into a converging wedge

an external pump supplying oil at high pressure

dry contact of the surfaces at a low speed

a perfectly concentric shaft and bush

The STL file format used to send a model to a 3D printer represents the part surface as:

a Boolean tree of solid primitives

Which of the following statements about additive manufacturing are true?

Fused deposition modelling builds a part by extruding a molten thermoplastic filament layer by layer

In selective laser sintering the unfused powder supports the part during the build

Internal channels and lattices can be made without special tooling

Fused deposition modelling builds a part by extruding a molten thermoplastic filament layer by layer

In selective laser sintering the unfused powder supports the part during the build

Internal channels and lattices can be made without special tooling

The stair-step effect on sloping surfaces does not occur

Which of the following statements about CAD and CAM are true?

A cubic Bezier curve passes through its first and last control points

STEP and IGES are neutral formats for exchanging CAD data between systems

A boundary representation stores the faces, edges and vertices that enclose a solid

A cubic Bezier curve passes through its first and last control points

STEP and IGES are neutral formats for exchanging CAD data between systems

A boundary representation stores the faces, edges and vertices that enclose a solid

A post-processor is unnecessary because every CNC controller reads the same G-code

The life equation L₁₀ = (C/P)ᵖ uses the exponent p equal to:

10/3 for ball bearings and 3 for roller bearings

1 for ball bearings and 3 for roller bearings

Machine Design II: Bearings, Brakes and Clutches, Springs, CAD and CAM and Their Integration, and Additive Manufacturing

The second half of Section B2.3 completes the machine elements and then turns to the computer. Rolling and sliding contact bearings, brakes and clutches and springs finish the list of machine elements the section names. Then come the basic concepts of Computer Aided Design (CAD) and Computer Aided Manufacturing (CAM) and the tools that integrate them, and additive manufacturing. GATE asks a bearing life in hours, a friction coefficient, a clutch or brake torque, a spring rate, a point on a Bezier curve, and what a print process does.

1. Rolling contact bearings

A rolling bearing puts balls or rollers between two races, replacing sliding by rolling. Deep-groove ball bearings take radial and modest axial load at high speed; angular-contact ball bearings take combined loads, and are paired to take thrust both ways; cylindrical roller bearings carry heavy radial loads; tapered roller bearings carry combined heavy radial and thrust loads; thrust bearings carry axial load only; self-aligning types tolerate misalignment. Their fatigue is by subsurface cracking and spalling of the races.

The basic dynamic load rating C is the constant radial load a bearing can carry for one million revolutions with 90% reliability, and L₁₀ is the life that 90% of a group survive: L₁₀ = (C/P)ᵖ million revolutions, p = 3 for ball bearings and 10/3 for roller bearings, with P the equivalent load P = XVF_r + YF_a. In hours, L₁₀h = 10⁶(C/P)ᵖ/(60n). For C = 30 kN, P = 10 kN and n = 1000 rpm, L₁₀ = 27 million revolutions and L₁₀h = 27 × 10⁶/(60 000) = 450 h. Because of the cube, doubling the load cuts the life to one-eighth, and doubling the speed halves the hours.

2. Sliding contact bearings

A journal bearing supports a shaft on a film of oil. In hydrodynamic lubrication the shaft’s rotation drags oil into a converging wedge and builds a pressure that carries the load with no metal contact; it needs relative motion, a wedge, and a viscous fluid. Below a certain speed or above a certain load the film thins to mixed and then boundary lubrication, with high friction and wear. The governing group is the bearing characteristic number μN/P (viscosity times speed over the bearing pressure), or the dimensionless Sommerfeld number S = (μN/P)(r/c)², with r/c the ratio of journal radius to radial clearance, near 1000. Friction is high at small μN/P (boundary), falls to a minimum, and rises again as the film thickens, so a design point is chosen safely to the right of the minimum.

For a lightly loaded, concentric journal, Petroff’s equation gives the coefficient of friction f = 2π²(μN/P)(r/c), with N in revolutions per second, P the projected bearing pressure and μ in Pa·s. For μ = 0.02 Pa·s, N = 20 rev/s, P = 1 MPa and r/c = 1000, f = 2π² × (0.02 × 20/10⁶) × 1000 = 19.74 × 4 × 10⁻⁴ = 7.9 × 10⁻³. The friction is more than an order of magnitude below that of dry sliding, which is the point of the film. A bearing is also chosen for its material: soft, embeddable, conformable liners such as white metal (babbitt) on a steel backing protect the shaft.

3. Brakes and clutches

A clutch transmits torque by friction between surfaces engaged by an axial force W, a brake absorbs energy by friction. For a friction pair of mean radius R_m, coefficient μ, axial force W and n friction surfaces the torque is T = μ W R_m n. For a disc clutch of inner radius R₂ and outer R₁ the mean radius depends on the assumption. Uniform wear (the usual design basis, once the plates have worn in) gives R_m = (R₁ + R₂)/2; uniform pressure (a new clutch) gives R_m = (2/3)(R₁³ − R₂³)/(R₁² − R₂²), slightly larger, so a new clutch transmits more. For a single-plate clutch with both faces working (n = 2), μ = 0.3, W = 2000 N, R₁ = 100 mm and R₂ = 60 mm, uniform wear gives R_m = 80 mm and T = 0.3 × 2000 × 0.08 × 2 = 96 N·m; uniform pressure gives R_m = 81.7 mm.

A band brake wraps a flexible band round a drum through an angle θ, and the tensions on the tight and slack sides are related by the capstan equation T₁/T₂ = e^μθ; the braking torque is (T₁ − T₂)r. For μ = 0.25, θ = 180° and T₂ = 200 N: T₁ = 200 e^0.785 = 438.7 N, the braking force 238.7 N, and on a drum of radius 0.25 m the torque is 59.7 N·m. A shoe (block) brake presses a shoe on the drum, and a brake is self-locking or self-energising when the friction moment on the lever helps the applied force, and self-locking when it holds the brake on with no force at all. Brakes turn kinetic energy into heat, and their temperature rise, ΔT = E/(mc), often limits them.

4. Springs

A helical compression spring of wire diameter d, mean coil diameter D and N active coils has, from torsion of the wire, the rate k = Gd⁴/(8D³N) and the deflection δ = 8FD³N/(Gd⁴). The shear stress is τ = K_w · 8FD/(πd³), where the Wahl factor K_w = (4C − 1)/(4C − 4) + 0.615/C corrects for direct shear and coil curvature and depends on the spring index C = D/d. For d = 4 mm, D = 32 mm, N = 10 and G = 80 GPa, k = 80 × 10⁹ × (0.004)⁴/(8 × 0.032³ × 10) = 7812 N/m = 7.81 N/mm; with C = 8, K_w = 31/28 + 0.0769 = 1.18. A small index means a stiff, highly stressed spring, and springs are usually kept between C = 4 and 12.

Springs in series add their compliances, 1/k = 1/k₁ + 1/k₂, and in parallel their rates, k = k₁ + k₂. A leaf spring is a beam of graduated leaves, stressed uniformly in bending, and it absorbs energy with the friction between leaves as damping; a torsion spring stores energy in twisting a wire or bar, and a Belleville washer is a coned disc with a non-linear rate. Springs under fluctuating load fail by fatigue, so shot peening and set removal (presetting) are applied, and buckling limits a slender compression spring whose free length exceeds about four times its mean diameter unless it is guided.

5. Computer aided design, computer aided manufacturing and their integration

CAD uses a computer to create, modify, analyse and document a design. Its geometric models are wireframe (edges only, ambiguous), surface (faces, no volume) and solid. Solids are built by constructive solid geometry (CSG), Boolean union, intersection and difference of primitives such as blocks and cylinders, or stored as a boundary representation (B-rep), the faces, edges and vertices that enclose the volume. Modern systems are parametric and feature-based: a dimension or a relation is a driver that regenerates the model when changed. Points move by homogeneous 3 × 3 (2D) or 4 × 4 (3D) matrices that compose translation, rotation and scaling by multiplication, in an order that matters.

Free-form curves use Bezier and B-spline forms. A cubic Bezier curve with control points P₀…P₃ is P(t) = (1−t)³P₀ + 3t(1−t)²P₁ + 3t²(1−t)P₂ + t³P₃, for 0 ≤ t ≤ 1. It passes through the first and last control points, is tangent to the first and last segments of the control polygon, and lies within its convex hull. For P₀ = (0, 0), P₁ = (1, 2), P₂ = (3, 2) and P₃ = (4, 0) at t = 0.5, P = (P₀ + 3P₁ + 3P₂ + P₃)/8 = (16, 12)/8 = (2, 1.5). B-splines give local control of a curve, since moving a control point changes only a nearby segment, and NURBS add weights that represent circles and conics exactly.

CAM uses the computer to plan, manage and control production. From the CAD model it generates tool paths, the cutter-location data, and a post-processor converts them to the G-code of a particular machine’s controller; simulation checks for collisions before metal is cut. The link between the two is the CAD/CAM integration: a common database, neutral data-exchange formats such as IGES and STEP, a shared feature model that lets computer-aided process planning read a part’s features, and links to product-data management. The gain is one model, no re-drawing, fewer transcription errors, and a shorter time from design to part; the cost is that changes in the model must flow through to the tooling and the programs.

6. Additive manufacturing

Additive manufacturing builds a part by adding material layer by layer from a CAD model, unlike subtractive machining. The model is exported as an STL file, a mesh of triangles that approximates its surface, then sliced into layers of a chosen thickness, and each layer is deposited or fused. The main processes are: fused deposition modelling (FDM), extruding a molten thermoplastic filament through a nozzle; stereolithography (SLA), curing a liquid photopolymer with ultraviolet light; selective laser sintering (SLS), fusing a bed of polymer powder with a laser, where the unfused powder supports the part; selective laser melting and direct metal laser sintering, the same for metal powders; laminated object manufacturing, bonding sheets; and binder jetting, printing a binder onto a powder bed.

The strengths are geometric freedom (internal channels, lattices and consolidated assemblies that cannot be machined), no tooling, fast turnaround and economy at low volume, and near-net-shape parts with little waste. The limits are the stair-step effect of the finite layer thickness, which leaves a rough surface on slopes and calls for finishing, anisotropy (a part is weaker between layers), slow build speed at volume, residual stresses and porosity in metal, and the need for supports. The build time is the number of layers times the time per layer: a 60 mm part in 0.2 mm layers of 30 s each takes 300 × 30 = 9000 s, that is 2.5 h, and halving the layer thickness doubles the time while improving the finish.

ℹ️ Why STL, and what it loses

STL stores only triangles, with no units, colour, or exact curves, so a smooth cylinder becomes a faceted one and a coarse export shows as facets on the part. A finer mesh gives a smoother part and a larger file. Newer formats (AMF, 3MF) carry units, colour and materials.

Key takeaways

L₁₀ = (C/P)ᵖ million revolutions, p = 3 for balls; L₁₀h = 10⁶(C/P)ᵖ/60n; doubling the load cuts the life eightfold.
Hydrodynamic lubrication needs a wedge, motion and viscosity; Petroff f = 2π²(μN/P)(r/c).
Clutch T = μ W R_m n with R_m = (R₁ + R₂)/2 for uniform wear; band brake T₁/T₂ = e^μθ.
Spring rate k = Gd⁴/8D³N; Wahl factor (4C − 1)/(4C − 4) + 0.615/C; series adds compliance, parallel adds rate.
A Bezier curve interpolates its end control points; CAD/CAM meet through neutral formats such as STEP; additive manufacturing slices an STL into layers and shows stair-stepping.

Practice questions (17)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

The basic rating life L₁₀ of a rolling bearing is the life:
1. that 90% of a large group of identical bearings complete or exceed before fatigue failure
2. at which 10% of the bearings fail immediately
3. that the average bearing achieves
4. of the bearing at its maximum speed
Show answer
Answer: A — that 90% of a large group of identical bearings complete or exceed before fatigue failure
L₁₀ is the 10th-percentile life, with 90% reliability. The average (median) life of a group is about five times L₁₀, so quoting L₁₀ as an average would be far too pessimistic.
A deep-groove ball bearing with a basic dynamic load rating of 30 kN carries an equivalent radial load of 10 kN at 1000 rpm. Its L₁₀ life is ____ hours.
Numerical answer — type the value.
Show answer
Answer: 450
L₁₀ = (C/P)³ = 27 million revolutions, and L₁₀h = 27 × 10⁶/(60 × 1000) = 450 h. Using the exponent 10/3, which belongs to roller bearings, would give 38.9 million revolutions and 649 h.
The load on a ball bearing is doubled at the same speed. Its L₁₀ life in revolutions becomes:
1. one-eighth of its earlier value
2. half of its earlier value
3. one-quarter of its earlier value
4. one-sixteenth of its earlier value
Show answer
Answer: A — one-eighth of its earlier value
L₁₀ ∝ P⁻³ for a ball bearing, so doubling P divides the life by 2³ = 8. For a roller bearing the exponent 10/3 would give a factor of about 10.
For hydrodynamic lubrication of a journal bearing, which of the following is essential?
1. relative motion of the surfaces dragging a viscous fluid into a converging wedge
2. an external pump supplying oil at high pressure
3. dry contact of the surfaces at a low speed
4. a perfectly concentric shaft and bush
Show answer
Answer: A — relative motion of the surfaces dragging a viscous fluid into a converging wedge
The film pressure is generated by the motion itself. A pump-fed film is hydrostatic, not hydrodynamic. A perfectly concentric journal would carry no load, because it is the eccentricity that forms the wedge.
A lightly loaded journal bearing has lubricant viscosity 0.02 Pa·s, journal speed 20 rev/s, bearing pressure 1 MPa and radius-to-clearance ratio 1000. By Petroff’s equation the coefficient of friction is ____ × 10⁻³, correct to one decimal place.
Numerical answer — type the value.
Show answer
Answer: 7.9
f = 2π²(μN/P)(r/c) = 19.74 × (0.02 × 20/10⁶) × 1000 = 19.74 × 4 × 10⁻⁴ = 7.9 × 10⁻³. The speed must be in revolutions per second and the pressure in pascals.
A single-plate clutch with both faces effective has inner and outer radii of 60 mm and 100 mm, a coefficient of friction of 0.3 and an axial force of 2000 N. Assuming uniform wear, the torque it can transmit is ____ N·m.
Numerical answer — type the value.
Show answer
Answer: 96
R_m = (100 + 60)/2 = 80 mm, and T = μ W R_m n = 0.3 × 2000 × 0.08 × 2 = 96 N·m. Counting one friction face gives 48 N·m, and the outer radius alone, 120 N·m, would overstate it.
For a clutch with outer radius 100 mm and inner radius 60 mm, the mean friction radius on the assumption of uniform pressure is ____ mm, correct to one decimal place.
Numerical answer — type the value.
Show answer
Answer: 81.7
R_m = (2/3)(R₁³ − R₂³)/(R₁² − R₂²) = (2/3)(10⁶ − 216 000)/(10 000 − 3600) = (2/3)(122.5) = 81.67 mm, a little more than the 80 mm of uniform wear, which is why a new clutch grips harder.
A band brake wraps 180° round a drum of radius 0.25 m, with a coefficient of friction of 0.25. The tension on the slack side is 200 N. Taking e^0.25π = 2.1933, the braking torque is ____ N·m, correct to one decimal place.
Numerical answer — type the value.
Show answer
Answer: 59.7
T₁ = T₂ e^μθ = 200 × 2.1933 = 438.7 N, so T₁ − T₂ = 238.7 N and the torque is 238.7 × 0.25 = 59.7 N·m. Using the tight-side tension alone, 438.7 × 0.25 = 109.7 N·m, forgets the slack side’s pull in the opposite sense.
A helical compression spring has wire diameter 4 mm, mean coil diameter 32 mm and 10 active coils, and the shear modulus of the wire is 80 GPa. The spring rate is ____ N/mm, correct to two decimal places.
Numerical answer — type the value.
Show answer
Answer: 7.81
k = Gd⁴/(8D³N) = 80 000 × 4⁴/(8 × 32³ × 10) = 20.48 × 10⁶/(2.62 × 10⁶) = 7.81 N/mm (G in N/mm²). Using the outer diameter in place of the mean coil diameter would understate the rate.
A helical spring has a mean coil diameter of 32 mm and a wire diameter of 4 mm. The Wahl stress-correction factor, using K_w = (4C − 1)/(4C − 4) + 0.615/C, is ____, correct to two decimal places.
Numerical answer — type the value.
Show answer
Answer: 1.18
C = D/d = 8, so K_w = (32 − 1)/(32 − 4) + 0.615/8 = 31/28 + 0.0769 = 1.1071 + 0.0769 = 1.184 ≈ 1.18. The factor is above 1 because the inside of the coil carries a higher stress than the straight-bar formula gives.
Two springs of rates k₁ and k₂ are connected in series. The equivalent rate is:
1. k₁k₂/(k₁ + k₂)
2. k₁ + k₂
3. (k₁ + k₂)/2
4. √(k₁k₂)
Show answer
Answer: A — k₁k₂/(k₁ + k₂)
In series both springs carry the same force and the deflections add, so the compliances add: 1/k = 1/k₁ + 1/k₂, which is always softer than either. k₁ + k₂ is the parallel combination.
The STL file format used to send a model to a 3D printer represents the part surface as:
1. a mesh of triangular facets
2. a set of exact NURBS surfaces
3. a stack of pixel layers
4. a Boolean tree of solid primitives
Show answer
Answer: A — a mesh of triangular facets
STL (stereolithography) stores triangles only, an approximation that is finer for a finer mesh. Exact surfaces belong to STEP and IGES, and a Boolean tree of primitives is CSG.
A part 60 mm tall is printed by a layer-by-layer process in layers 0.2 mm thick, and each layer takes 30 s to print. The build time is ____ hours.
Numerical answer — type the value.
Show answer
Answer: 2.5
Layers = 60/0.2 = 300, and the time is 300 × 30 = 9000 s = 2.5 h. Using 0.2 mm layers without converting the height, 60 × 30 s, would give 0.5 h.
A cubic Bezier curve has control points P₀ = (0, 0), P₁ = (1, 2), P₂ = (3, 2) and P₃ = (4, 0). The y-coordinate of the point on the curve at the parameter t = 0.5 is ____.
Numerical answer — type the value.
Show answer
Answer: 1.5
At t = 0.5 the weights are 1/8, 3/8, 3/8, 1/8, so y = (0 + 3 × 2 + 3 × 2 + 0)/8 = 12/8 = 1.5, and x = (0 + 3 + 9 + 4)/8 = 2. The curve stays below the control points’ y = 2, inside their convex hull.
Which of the following statements about additive manufacturing are true?
1. Fused deposition modelling builds a part by extruding a molten thermoplastic filament layer by layer
2. In selective laser sintering the unfused powder supports the part during the build
3. Internal channels and lattices can be made without special tooling
4. The stair-step effect on sloping surfaces does not occur
Show answer
Answer: A — Fused deposition modelling builds a part by extruding a molten thermoplastic filament layer by layer; B — In selective laser sintering the unfused powder supports the part during the build; C — Internal channels and lattices can be made without special tooling
(A) The FDM process. (B) The powder bed acts as the support. (C) The geometric freedom that makes the process valuable. (D) False: the finite layer thickness leaves steps on any surface that is neither vertical nor horizontal, which is why finishing is often needed.
Which of the following statements about CAD and CAM are true?
1. A cubic Bezier curve passes through its first and last control points
2. STEP and IGES are neutral formats for exchanging CAD data between systems
3. A boundary representation stores the faces, edges and vertices that enclose a solid
4. A post-processor is unnecessary because every CNC controller reads the same G-code
Show answer
Answer: A — A cubic Bezier curve passes through its first and last control points; B — STEP and IGES are neutral formats for exchanging CAD data between systems; C — A boundary representation stores the faces, edges and vertices that enclose a solid
(A) The curve interpolates the end points only and merely approximates the inner ones. (B) The standard neutral exchange formats. (C) The B-rep definition. (D) False: controllers differ in their codes and cycles, and the post-processor translates the neutral cutter-location data into the dialect of the machine in use.
The life equation L₁₀ = (C/P)ᵖ uses the exponent p equal to:
1. 3 for ball bearings and 10/3 for roller bearings
2. 10/3 for ball bearings and 3 for roller bearings
3. 2 for both types
4. 1 for ball bearings and 3 for roller bearings
Show answer
Answer: A — 3 for ball bearings and 10/3 for roller bearings
Point contact in a ball bearing gives p = 3 and the line contact of a roller bearing 10/3, because a line contact spreads the load and is less sensitive to it. The two values are often swapped in recall.