Implant technology - unit 2 deck 3 Flashcards

1
Q

State what rigidity and stiffness is and their relationship

A
  • Rigidity of a bar is an indication of its ability to resist axial deformation. Rigidity = EA
  • Stiffness of a bar is the force required to produce a unit deflection. k = F/ Δl

Relationship is k = EA/ l (EA = rigidity) so stiffness is equal to rigidity per unit length

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2
Q

Describe the load transfer between the stem component of a hip arthroplasty and the femur, with reference to Huiskes explanation of load transfer in a cemented femoral stem

A

The stem transfers some of the load proximally and the rest distally to the femur/cement with a central load sharing region

Huiskes:

At the proximal end of the femur, the stem takes all the load, F, but some ofthis is transferred to the bone in the proximal load transfer region. Distally, the rest of the load is transferred to the bone

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3
Q

How does the rigidity (or stiffness as referring to their relationship the more rigid the stem the more stiff) of a femoral stem affect load sharing with the bone/cement?

A

The more rigid the stem, the more load it takes proximally. This means that a more rigid stem results in more proximal stress shielding of the femur.

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4
Q

In order to reduce stress shielding what can be done ?

A

Use a less stiff stem

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5
Q

what are the problems with using a less stiff stem component and therefore why a compromise between keeping stress-sheilding low and keeping the proximal interface shear stresses low must be made in prosthesis design?

A
  • Using a lower stiffness stem decreases stress sheilding but increases shear stresses generated at the proximal bone-stem or bone-cement-stem interfaces and in the cement itself.
  • These shear stresses can be large enough to shear apart the interlocking grip at the bone-cement and stem-cement interfaces resulting in loosening ==> a compromise must be made.
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6
Q

State the equation used to calculate the load (Lb) taken by the bone in the central load sharing region when a femoral stem is used

A
Rb = rigidity of bone 
Rt = rigidity of combined bone-stem composite structure respectively
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7
Q

State the equation used to calculate the load (Ls) taken by the stem in the central load sharing region when a femoral stem is used

A
Rs = rigidity of stem 
Rt = rigidity of combined bone-stem composite structure respectively
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8
Q

Consdering the equations used to calculate the load taken by the bone and the stem in the load sharing region, why can the rigidity of the bone cement be excluded ?

A

. The rigidity of the cement is so low so takes a very small proportion of the load in the load sharing region

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9
Q

Knowing the proportion of load taken by both the bone and the stem in the load sharing region (using the equations), how can we determine how much load is transferred from the stem to the bone at the proximal and distal ends of the stem?

A

e.g. its calculated that 40% of the total load was taken by the bone in the load sharing region

this means that only 40% of the total load was transferred from the stem to the femur at the proximal end, so the other 60% was transferred distally.

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10
Q

How is the load transferred from the stem to the bone (via the cement if present) and where is this highest and what does its magnitude depend on?

A

Shear stress - highest at the ends of the stem

Its magnitude depends on the magnitude of the shear force, which in turn depends on how much load is transferred to the bone proximally and distally, e.g. the greater the proximal load transfer from the stem to the bone, the greater the proximal shear stress at the stem-bone interface.

Using a stiffer material to increase the rigidity of the stem results in a lower proximal shear stress (because less load is transferred to the bone proximally) but increases stress shielding of the proximal femur. It also increases the shear stress in the distal load transfer region because more load is transferred distally.

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11
Q

what are isoelastic stems and why are they not viable

A

stems with the same stiffness as bone

reduce stress shielding but cause high shear stress at bone-stem or bone-cement-stem interface causing implant failure

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12
Q

We cannot easily design a femoral stem which gives both low stress shielding and low interface stresses unless it has what?

A

a collar to rest on the top of the femur

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13
Q

What is the effect of tapers on shear stresses in the cement layer of cemented stems?

A

They act to reduce the proximal and distal cement stresses considerably as in a non-tapered stem load is transfered soley as shear stresses proximally and distally, whereas in tapered stems it allows some of the load to be transfered by compression, the greater the taper the greater the compression

[The stresses in the stem are higher than those in most commercial designs because there is less material where it is tapered, but it is still strong enough to take the loads placed on it without fracturing. because stress = F/A]

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14
Q

Explain why the relative rigidities of the stem and femur influence the proximal load transfer stresses.

A

The relative rigidities of the stem and the femur determine how much load is taken by each in the load sharing region.

A stiff stem takes a large proportion of the total applied load in the load sharing region, so less is transferred to the bone proximally. This means that the proximal shear forces, and hence the shear stresses, associated with load transfer are low.

A low modulus stem takes less load in the load sharing region, so more load is transferred proximally, which means higher shear forces and interface shear stresses.

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15
Q

What are the main disadvantages of using an isoelastic stem?

A

The proximal shear stresses are too high for cement and hydroxyapatite to withstand so interface bonds fail

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16
Q

what does cement allow for in prostheses

A

allows good contact between bone and stem, avoiding high stress concentrations. Should allow good contact along the full length of the prosthesis

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17
Q

What does the magnitude of stresses within cement depend on ?

A

Its thickness and stiffness

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18
Q

What happens if cement is too thin or too thick?

A
  • too thin = very high cement stress and bone reabsorption at proximal end of femur
  • too thick = high cement stress

==> cement failure and loosening

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19
Q

what is the optimum cement thickness

A

layer of 3mm to 7mm proximally

minimum 2mm distally

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20
Q

what is the effect of collar on load transfer

A

2 sides to the debate; one side thinks they work, the other don’t

However, both collared and collarless stems are used succesfully

For using collar:
- allows compressive transfers load from stem to the bone proximally, helping to reduce stress shielding and proximal interface shear stress

Against using collar:
- collar acts as a pivot, causing fretting wear at the pivot and high stem stresses distally

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21
Q

what is the potential mechanism for cement to fail

A
  • cement has low shear strength and its interface bonds are weak, particularly with the stem
  • stem-cement bonds loosens, stresses in cement increases, leads to cracks and eventually cement failure
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22
Q

How can the stem-cement bond be strengthened ?

A
  • by coating stem with PMMA
  • by roughening the stem surface
  • by applying a porous coating to stem surface
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23
Q

How do cementless stems work and how do they reduce stress shielding ?

A

They rely on a press-fit which promotes hoop stresses in the bone which reduces stress shielding

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24
Q

Why are cementless stems are now surface coated usually with hydroxyapatite?

A
  • It helps bone ingrowth and potentially eliminates metal debris from bone-metal abrasion.
  • Also helps bone to bond to a larger area of the stem which lessens the chance of failure of the bond under subsequent loading.
25
Q

What is the drawback of coating cementless stems fully with hydroxyapatite?

A

fully coated stems promote stress shielding of the bone

26
Q

what are disadvantages of cement-less stems

A

Lack of distal contact causes thigh pain - can be prevented by distal anchoring of the stem

bone ingrowth might not occur:

  • caused excessive movement at bone-stem interface after surgery.
  • fibrous tissue occurs instead preventing any bony ingrowth

After few years, bone-hydroxyapatite bone breaks down

27
Q

What is the main advantage of completely coating the surface of a stem with hydroxyapatite as opposed to just partial coating?

A

Gives the best chance for adequate bone ingrowth to hold the prosthesis in place

28
Q

What is a general rule of thumb for all stem shapes?

A
  • all are tapered to prevent subsidence
  • many have a proximal wedge so that the stem can rest of the bone, allowing transmission of compressive forces as well as shear forces
29
Q

why is the shape of the stem important in cement-less femoral implants

A

stem needs to be in contact with a large proportion of the femur

If the outer dimensions of the stem at any point along its length are smaller than the corresponding inner dimensions of the medullary canal, there will be a gap that cannot be filled. (seen in fig.)

Careful stem selection is required to overcome potential shape problems

30
Q

why are different stem shapes needed for the elderly compared to the young?

A

cortical bone thins and the canal gradually becomes wider as you age
[particularly in women after the age of 50 to 60]

31
Q

What does the bending stress in a femoral stem to a large extent depend on ?

A

The bending moment

32
Q

how can the bending moment be reduced in hip protheses to reduce bending stress in the stem

A

reduce the offset distance from the head to the neutral axis of the stem

can be done by 2 ways:

  • reducing the length of the neck of the stem (pic A)
  • increasing the angle between the long axis and the axis of the neck i.e. make it more vertical (pic B)
33
Q

What is the con of utilising either of the 2 ways of reducing bending moments in hip prostheses and explain how they do this?

A

They increase the joint reaction force, giving rise to greater wear and acetabular bone-implant stresses

Taking moments about point A on the pelvis;

ΣMA = 0 ==> W.d - J.x = 0

so J = W.d/ x

Moving the line of action of the joint force laterally in order to reduce its lever arm, as shown in Fig. A and B, decreases x so J increases.

34
Q

why is reducing the bending moment in hip prosthesis not as big an issue anymore?

A
  • modern stem materials are much stronger than those of 20 years ago
    • therefore, fracture of the stem due to bending stresses less of an issue

thus, there is now a trend towards creating a more anatomically accurate stem (because the 2 ways of reducing bending moment increase joint force and wear), which increases the offset distance, reduces the magnitude of J and therefore reduces joint wear.

35
Q

How does a tapered wedge help proximal load transfer in hip prosthesis ?

A

Because it transfers significant proportion of the load in compression, rather than shear

36
Q

State 2 essential requirements of replacement hip joints

A
  • must have low contact friction between the bearing surfaces
  • the bearing materials used must wear as slowly as possible.
37
Q

What is the equation used to calculate friction force ?

A

F = μN

F = the friction force (parallel to the surface) required to start the block moving

μ = coefficient of friction 
N = normal contact force
38
Q

what is friction force dependant on

A
  1. surface properties of the two materials in contact
  2. magnitude of the load pressing them together [N] i.e. magnitude of joint reaction force
39
Q

how does a synovial joint coefficient of friction compared to HDP on metal beating [i.e. artificial joint]

A

synovial joint has between 3 and 100 times less friction than artificial joint

40
Q

why is HDP used instead of alternative materials with lower coefficient of friction

A

alternative materials are either toxic or have poorer wear properties

41
Q

With HDP as one of the bearing surfaces what is usually the other bearing surface made of in hip prostheses ?

A
  • Cobalt chrome, stainless steel is less common now due to having worse corrosion resistance
  • Ceramics are starting to be used as they have excellent corrosion resistance, low friction and cause less wear on HDP
42
Q

what is one of the consequences of replacement joints having a higher contact friction force than normal joints

A

much higher shear force transmitted to the acetabulum
- can be high enough to cause loosening at the fixation interface

43
Q

why do most hip replacements use small heads

A

interface force for the small head is much less than it would be in a bigger head (refer to pic)

  • The friction force (F) is the same in both cases as it is independent of area
  • you will see that r1/R1 is much less than r2/R2, so the interface force F1 for the small head is also much less than the force F2.

[It is also important to bear in mind from the formula F = μN above that the frictional force at the joint surface depends on the magnitude of the contact force pressing on the joint, and that this is dependent on the magnitude of the joint reaction force.

44
Q

Why do many authorities favour a return to a anatomical offset on the stem of hip prosthesis ?

A
  1. the formula F = mN above that the frictional force at the joint surface depends on the magnitude of the contact force pressing on the joint, and that this is dependent on the magnitude of the joint reaction force.
  2. many hip replacements, including the Charnley, are designed with a small offset (lever arm) to reduce the bending moment on the hip.
  3. The consequence of this is a higher joint reaction force. This means that the joint friction force, F, increases and so too does the interface shear force F1. A higher joint reaction force also increases wear in the joint,
45
Q

State and describe the 2 main types of wear that occur between bearing surfaces

A

Adhesive and abrasive wear

  • Adhesive wear occurs because the two bearing surfaces stick to each other when they are pressed together and one, usually the softer is torn off by the harder one.
  • Abrasive wear occurs because surfaces are not perfectly smooth.
46
Q

How is adhesive wear minimised ?

A
  • bearing surfaces are made up of materials which have a low level of adhesion
  • Lubricants provide a layer between the two materials which reduces wear
47
Q

How is abrasive wear minimised ?

A
  • hip joints must have highly polished surfaces
  • Good circulation of lubricant so that wear particles can be removed and not rub against the bearing surfaces
48
Q

what is another cause of abrasive wear in hip joint replacements

A

Entry of particles from outside the bearing (third bodies). In joint prostheses, cement particles often find their way into replacement joints and accelerate the wear process.

49
Q

3 factors affect the volume of adhesive wear - what is the formula for calculating abrasive wear which incorportates them?

A

This implies that the softer material will wear away first (refer to P) and that it will wear away more, the softer it is, the more it is loaded and the more it moves.

50
Q

HDP wear fragments and particles (from the acetabulum bearing surface as HDP makes this bit of a hip joint) can migrate considerable distances within an implant - what effect do they have on the implant?

A
  • They cause intense inflammatory reactions over a no. of years, can cause bone reabsorption and prosthesis loosening
  • They are one of the major problems associated with aseptic (non-infected) loosening of prosthetic components
51
Q

what are the methods of reducing wear based on the equation for calculating it ?

A

reduce loading on the joint (effects of stem geometry on joint reaction force previously disscused about ofset etc)
- i.e. reduce N

keep the sliding distance as small as practically possible
- i.e. reduce s

find alternative materials for the replacement head that reduce wear in the HDP
- i.e. reduce c

52
Q

how is the sliding distance kept as small as possible?

A

use a femoral head with a small radius
- wears less than one with a large radius as the bearing surface moves less as slides

Pic shows by rotating an angle (θ), the distance, D, moved by the bearing surface of the larger head is much greater than the distance, d, moved by the smaller head.

53
Q

what ceramic material is being looked at as an alternative material to reduce wear in the HDP and why?

A

Zirconia

  • Because it is more stratch resistant than cobalt chrome so they are less prone to cause abrasive wear of HDP.
  • It is also tougher, more fracture resistant, harder and more scratch resistant than other ceramics
    • It produces less that half as much HDP as other ceramics
    • some manufacturers now use material for femoral head
54
Q

What are the 2 disadvantages on wear when using a smaller femoral head ?

A
  1. The rate of depth of wear is greater than it would be for a larger head because its contact area is less ==> wear through the bearing surface of the cup quicker. Resulting in lose of the joints ROM as the material wears, because the neck of the stem of the femoral component contacts the cup
  2. An increased likelihood of dislocation in the post-op period. This is because of the increased likelihood of neck impingement on the edge of the cup.
55
Q

Which of the following affect the frictional force on a bearing:

  • contact load
  • contact area
  • material properties of the surface to which the load is applied
  • material properties of both surfaces
A

contact load and material properties of both surfaces

56
Q

Why is HDP used as a bearing surface in replacement joints and what is its main disadvantage ?

A
  • Because it has a low coefficent of friction with metals
  • Dis - fragments of HDP produce adverse tissue reaction that leads to bone reabsoption
57
Q

State two reasons why small diameter heads are used in hip replacements?

A
  • Reduce cup-bone interface shear stresses, reducing risk of loosening
  • Produce less volume of wear of HDP than large diameter heads
58
Q

What are the two main disadvantages of using a small diameter head?

A
  • high contact stresses
  • rate of depth of wear is greater due to reduced contact area