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Orthopaedic Biomechanics > Ch. 11 - Articulating Surfaces > Flashcards

Ch. 11 - Articulating Surfaces Flashcards

1
Q

Why is it important to minimise friction in total joint replacements?

A

Friction leads to large shear forces which:

  • increase the risk of loosening
  • increase the release of wear debris
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2
Q

What are 5 common joint material combinations?

A
  1. Metal/PTFE
  2. Metal/PE
  3. Metal/Metal
  4. Ceramic/Ceramic
  5. Ceramic/PE
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3
Q

Why was the Metal/PTFE combination for joint replacements initially popular but then discontinued?

A

The use of PTFE provided a smooth articulation but the material degraded quickly

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4
Q

Which material combination is preferred for joint replacement components?

A

Metal/PE are preferred. Given PE’s low coefficient of friction, they provide smooth articulation with low wear.

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5
Q

In a Metal/PE joint replacement, where does most of the wear occur?

A

In the PE component

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6
Q

Name 3 types of surface damage in a joint replacement. Which one leads to the small accumulation of small debris particles?

A
  1. Abrasive wear*
  2. Pitting
  3. Delamination
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7
Q

What does biocompatibility mean?

A

It means:

  1. The material must not harm the body
  2. The body must not harm the material
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8
Q

A PE component passes a biocompatibility test as a bulk material. Can we safely use it for a joint replacement based on this evidence?

A

Not necessarily. We must consider that the response of the body to very small debris particles may be different than that to the bulk material.

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9
Q

What is the role of bearing surfaces in a total joint replacement?

A

They must:

  1. Transmit normal joint forces
  2. Allow for normal motion
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10
Q

How does the body’s biological response vary for small vs. large amounts of particle debris from a joint replacement?

A
  • Small amounts are eliminated through the perivascular lymph spaces.
  • Larger amounts that cannot be eliminated this way lead to the body releasing an agent to attack the bone at its weakest point: the bone-implant interface. This eventually may cause osteolysis.
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11
Q

Why is osteolysis more common in the hip than in the knee?

A

This is possibly due to the larger size of the hip and a higher rate of deposition.

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12
Q

How can osteolysis be visualized in a radiographic image?

A

Increased transparency of the bone

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13
Q

What are potential sources of debris in a joint replacement?

A
  1. Articulating surfaces
  2. PE/metal interface in metal backed components
  3. Screw head contacting PE
  4. Screw holes in PE
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14
Q

Why would a prosthetic articular component have more holes in it than necessary? What are the possible complications of this?

A

More holes than necessary provide surgical flexibility. However, unused holes provide channels for debris to migrate to the bone.

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15
Q

Can initial conformity btw PE liner and metal backing eliminate the chance of wear? Why or why not?

A

No, because in changes in geometry and material properties may occur in vivo over time (due to creep for example).

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16
Q

Describe 2 mechanisms by which cracking/fracture of PE components in an acetabular cup might occur.

A
  1. Interlock grooves in cemented acetabular cups compromise the structural integrity.
  2. Rim failure may occur if the rim is too thin to support the load. Rim failure could then lead to disassociation of the polyethylene liner in an uncemented acetabular cup.
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17
Q

Describe the deformation that occurs between metal and PE components in a joint replacement.

A
  • Metal acts as a rigid indenter, such that its deformation is negligible
  • PE is compressed btw the metal component and the underlying material (bone, cement, metal-backing)
  • PE component conforms to shape of the metal surface
  • PE is subject to tensile, compressive and shear stresses
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18
Q

The magnitude of the stress caused by contact between 2 components in a joint replacement depends on the magnitude of __________.

A

The joint load.

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19
Q

The level of damage in the components of a joint replacement is a function of ________.

A
  1. Patient weight/mass
  2. Relative joint motion
  3. Time since implantation + Activity level (# of cycles)
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20
Q

Is surface damage a fatigue problem?

A

There is strong circumstantial evidence that this is the case.

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21
Q

Where can damage associated with stress occur in a joint replacement component?

A
  • At the PE component surface

- Within the PE component

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22
Q

What 2 types of stresses occur at the PE component surface of a joint replacement?

A
  1. Normal compressive stress (contact stress)

2. Tangential stress due to friction

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23
Q

Under which circumstances is it appropriate to ignore (tangential) stresses due to friction?

A

The bearing surface between a metal and PE component is of low friction, such that tangential stresses can thus be ignored and only contact stresses need be considered.
However, friction cannot be ignored in the following 2 cases:
1. If one of the surface is damaged or improperly manufactured
2. If we want information on damage mechanisms even in undamaged PE components

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24
Q

How do contact stresses acting on the PE surface vary with angle?

A

At the surface, contact stresses acting perpendicular to the surface are the largest compressive stresses. They diminish rapidly (non-linearly) with a greater angle from the perpendicular and are mostly gone by the time the angle = 25 degrees.

25
Q

How to compressive stresses vary through the thickness of the PE?

A

They decrease non-linearly with depth

26
Q

What type of stresses exist within the PE component due to joint contact?

A
  1. Tangential compressive stresses

2. Tangential tensile stresses

27
Q

What causes tangential compressive stresses within the PE component upon joint contact?

A

The PE at the centre of the contact area expands radially during compression, but is resisted by the surrounding material.

28
Q

What causes tangential tensile stresses within the PE component upon joint contact?

A

The surface needs to stretch/change shape to conform to the metal part when loaded. Since stretching occurs near the edge of contact, tensile stresses are largest near the surface.

29
Q

How are contact stresses in the knee different from those in the hip? Why?

A

In the knee we see higher contact stresses because it is a less conforming joint than the hip and therefore has a smaller contact area.

30
Q

Where does maximum shear stress occur for nonconforming joints vs. in conforming joints?

A
  • Conforming: at the surface

- Non-conforming: just below the surface (~1mm)

31
Q

Surface damage is due to a combination of which 3 stress components?

A
  1. Maximum principal stress
  2. Minimum principal stress
  3. Maximum shear stress
32
Q

What does the maximum principal stress represent?

A

The maximum tangential stress

33
Q

What does the minimum principal stress represent?

A

The largest magnitude compressive stress

34
Q

How does the size distribution of debris particles vary with conformity of the joint replacement?

A

It varies with the conformity of the joint:

  • Conforming joints = smaller particles in larger amounts
  • Non-conforming joints = larger particles in smaller amounts
35
Q

Describe 2 mechanisms by which crack propagation results in pitting in non-conforming joints.

A
  1. Cracks propagate from the surface into the PE (maximum principal stress)
  2. Subsurface crack propagate toward the surface, releasing debris into the joint (maximum shear stress)
36
Q

Explain how contact stresses in a knee joint could result in fatigue.

A

Fatigue could result due to the large range of stresses the surface is subjected to during normal use. The contact area changes with flexion/extension and the tension and compression are exerted on the joint at the edge and centre of contact, respectively.

37
Q

Describe the mechanism by which crack propagation results in delamination.

A

Subsurface cracks continue to propagate parallel to the surface, resulting in the formation of a separate sheet of material.

38
Q

What causes the crack propagation direction to turn?

A
  1. Tangential stresses at the edge of the contact area decrease rapidly with depth
  2. For frictionless contact, shear stresses reach a maximum about 1mm below the surface
  3. Maximum principal stress direction change which depth

This results in a changing stress field, which causes the crack propagation direction to change accordingly.

39
Q

Which parameters should be reduced in order to reduce stresses associated with fatigue and crack propagation?

A
  1. Range of maximum principal stresses

2. Magnitude of maximum shear stresses

40
Q

How can surface damage be reduced?

A

By reducing increasing PE strength or reducing contact stresses. (Reducing contact stresses also reduces principal and shear stresses).

41
Q

What do contact stresses depend on?

A
  1. Changes in loading (magnitude + direction)
  2. Conformity of articulating surfaces
  3. Thickness of PE
  4. Material stiffness (mostly of PE, but also of metal)
42
Q

How can contact stresses be determined? What different information is provided by each of these techniques?

A
  1. Computationally - provides internal PE stresses + contact stresses
  2. Experimentally - provides real contact stresses for real geometries (hard to approximate computationally)
43
Q

How are contact stresses distributed? Where is it the greatest? Elaborate by considering the case of 2 spherical components.

A

Generally, contact stresses are not uniform over the contact area, as contact surfaces are never entirely conforming. Contact stresses are greatest where the PE surface displacement is greatest.

For 2 spherical components:
- Contact shape is circular
- Maximum contact stress occurs at center of contact
- Minimum contact stress occurs at edge of contact
(contact stress decreases radially)

44
Q

What could be the reason behind surface waviness?

A
  1. Manufacturing process
  2. Design specifications
  3. Surface deformation
45
Q

What is the effect of changing the value of Rt (radius of tibial component) or Rf (radius of femoral component) such that they are further apart?

A

If Rt and Rf are further apart, then this means joint conformity will decrease, ultimately causing maximum contact stress to increase.

46
Q

What is the effect of increasing PE thickness on contact stress?

A

Increasing PE thickness would lead to reduced contact stresses. (think: stiffness = EA/t)

47
Q

What is the effect of increasing the modulus of PE on contact stress?

A

Increasing E_t leads to increased material stiffness which results in increased maximum contact stress.

48
Q

Describe the rod analogy to understand the effect of varying PE thickness and stiffness on contact stresses.

A

Let us consider that the PE supports the indenter as a collection of parallel rods aligned along the direction of loading. Then the stiffness under the axial load:
K = EA/L
This analogy ignores shear stress due to the independence of the rods.

49
Q

Describe 2 methods to calculate contact stresses in PE components.

A
  1. Hertz Theory
    - Simplest method but with limited interpretation.
    - Assumes small contact area, similar material properties, and planar surface of contact.
  2. Finite Element Analysis
    - Works for varying levels of complexity
    - Is compatible with non-linear material behaviour
50
Q

According to Hertz theory, what happens to contact area and maximum pressure if we increase material stiffness?

A

Contact area is reduced and maximum pressure is increased. (Not linearly though)

51
Q

According to Hertz theory, what happens to contact area and maximum pressure if we increase the radius of the cup?

A

Increasing the radius of the cup causes the term C_G to decrease, such that the radius of the contact area also decreases and maximum pressure increases. (Non-linear relationship)

52
Q

How do you predict the estimate for maximum contact stress obtained using Hertz Theory will compare the that obtained using FEA?

A

Hertz Theory alwas overestimates the maximum contact stress.

53
Q

Describe the trade-off that exists between pressure and sliding distance as parameters to reduce wear.

A

Wear is proportional to both pressure and sliding distance. However, they are both a function of diameter. As diameter increased, pressure decreases and sliding distance increases.

54
Q

Where do maximum shear stresses occur in an acetabular component and how does surface damage compare to that normally seen in a tibial component?

A
  • Maximum shear stress occurs close to the articulating surface.
  • Fewer pits than in a tibial component
  • Delamination is very rare
  • Wear is most likely due to abrasion caused by large shear stresses at the surface
55
Q

Are efforts to increase conformity in tibial components focused on the AP or ML direction? Why?

A

Efforts to increase conformity are focused on the ML direction.
Most knee replacement designs mimic the native AP geometry of the femoral condyles. The min. AP radius of the tibial component is defined, as it must be larger than that of the femoral component. This radius is then further increased to provide joint laxity in extension and allow cruciate ligament retaining designs to function properly.
Finally, contact stresses are more dramatically affected by changes in the ML radius.

56
Q

When conformity maximised on the ML direction of a tibial component?

A

When the condyle and plateau are flat in the ML direction.

57
Q

What are the possible complications of a flat design tibial component?

A

Medially directed loads during normal gate will allow the lateral condyle to lift-off, which will in turn:

  1. increase contact stresses
  2. move contact stresses twd the edge of the PE component due to tilting
  3. increase fracture stresses
  4. possibly crush the trabecular bone due to edge loading
58
Q

Discuss the validity of the following statement:

A tibial component that is curved on the ML direction is preferable over a flat design.

A

While in a curved component, the contact area will not shift to the edge as we have better joint conformity, this increase in radial conformity will eliminate the laxity necessary for soft tissue loading. If all the load is then transferred through the prosthesis, the risk of bone-implant interface failure is high.
A better rule of thumb is to choose the component radii such that contact stresses are minimised while still providing the appropriate joint laxity.

Orthopaedic Biomechanics (10 decks)

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