Implant technology - unit 1 deck 2 Flashcards
What is the major role of all orthopaedic implants ?
To provide structural support
The ways in which different orthopaedic implants provide structural support differs but the actual mechanisms of load bearing are not that different.
T or F?
True
Fig 2A shows an internal fixation using plate and screws of a lower limb fracture. The load is shown as an axial force F (body weight), although there might also be bending and torsional loads.
In the case of the bone fixator, Figure 2A, we assume that the bones touch at the fracture site. We can identify three important regions, marked 1, 2 and 3 describe these regions in terms of what is happening to the load at each of them:
- Region 1 is where the screws fix the plate to the bone. Here, part of the applied load F1 in the bone above is transferred to the plate. This is a region of load transfer.
- Region 2 is at the fracture site, where the broken bones are supported by the plate. Here part of the load F is taken by the plate and part by the bone. This is a region of load sharing.
- Region 3 is where the screws fix the plate to the bone at the other side of the fracture. This is a region of load transfer, where the load transferred to the plate in Region 1 is transferred back to the bone.
Fig 2B . shows the intramedullary stem of a cemented joint replacement on the right The load is shown as an axial force F (body weight), although there might also be bending and torsional loads.
In the case of Figure 2B. We can identify three important regions, marked 1, 2 and 3 describe these regions in terms of what is happening to the load at each of them:
- Region 1 is one of load transfer, where part of the applied load F2 is transferred from the stem to the bone
- Region 2 is a middle region, where there is load sharing between the bone and the stem
- Region 3 is one of load transfer, where the remaining part of the load (i.e. that not transferred in Region 1) is transferred from the stem to the bone. The bone below this region, of course, takes all the load F2.
Note:
Some simplifications have been made e.g. the cement in the case of the prosthesis. The cement does share some of the load but not much because it is much less stiff than bone or metal. It was left out to simplify the models, but we will return to the subject in Unit 2 when we look at the influence of stiffness on load transfer and load sharing.
Appreciate this:
load transfer does not occur along the whole of the bone-implant interface - it only occurs at certain regions. For the bone plate example, the load is transferred at the bone screw regions, and for the intramedullary stem load transfer takes place at the end regions of the stem only.
Explain the terms load sharing and load transfer.
- For an implant attached to bone, the regions where the load is partly taken by the bone and partly taken by the implant are called regions of load sharing.
- The regions where load is transferred from an implant to a bone (or from a bone to an implant) are called regions of load transfer.
Give two examples where load transfer is combined with load sharing.
Two examples are a bone fixation plate and the stem of a joint replacement component
Recall shear stress during axial loading is caused by forces acting in opposite directions causing the surfaces/ planes to shear or slip within a material
Equation:
τ = V/ A
Shear stress = shearing force (V) / sheared area (A)
In composite materials (structures), how does load transfer occur and what is generated ?
- Load is passed from one material to another across the interfaces between them.
- This generates either interface stresses or relative movement at the interface
Appreciate this:
Interface stresses occur when the two materials are bonded together (composite materials/structures), relative movement occurs if these materials are not boneded or if they were bonded and the bond becomes loose. Loosening can be a serious complication in joint replacements.
How does the difference in the youngs modulus of the two materials bonded together creating a composite structure (material) affect the stresses which will be generated at the interfaces when a compresive load is applied?
The greater the difference in youngs modulus (E) of the two materials the greater the shear stresses generated
Consider the Fig A in the diagram:
The bottom material is more flexible than the top material, i.e. E2 < E1.
It will therefore compress more under loading and expand laterally more than the top material. If the two materials are bonded together then any lateral strain at the interface is the same for both materials, so a shear stress is generated at the interface, because one material is trying to expand more than the other one.
If the interface was not bonded and it is lubricated, sliding can of course occur freely so there are no shear stresses
Describe how the shear stresses will differ in Fig B if the top material is much more flexible (greater E) than the bottom compared to Fig A where the bottom material is more flexible than the top when a compressive load is applied
The shear stresses at the interface under the region of an applied load from above will be much greater than in Figure 3A.
An analogy for this example could be an office chair seat with a soft cushion and a hard base. The stiffer the cushion the more it can distribute the load over the support surface, hence a more flexible one will have a smaller area over which the shearing force is concentrated ==> greater shear stress (shear stress = V/ A)
This is why a very soft cushion resting on a hard base is not very comfortable to sit on.
The second example, shown in Figure 4 is one of load transfer under shear loading, the type which occurs between intramedullary stems and surrounding bone.
In the present example two rectangular bars with the same cross sectional shape are joined together and subjected to a compressive load (F). The transfer of the load from one bar to the other gives rise to a shear stress at the interface.
Describe how shear stress in these bars is not constant ?
Shear stress, t, is not constant across the whole length of the interface. (graph in pic shows this) Load transfer takes place at the end regions of the join between the bars, there is no shear stress in the central portion, which is a region of load sharing.
Note:
This is not perhaps what one would expect intuitively. If the load were transferred along the whole length of the interface, the shear stress would be equal to the applied force divided by the area being sheared, i.e. t =F/Ld (A), where d is the depth of the cross section. In reality, the shear stress reaches a much higher value than the average stress (shown by the dotted line). Only when the length of the area being sheared is small does the shear stress approach the average value (see Figure 4B). The high shear stresses that occur in practice can result in the interface bonds breaking which means that the components are no longer held together.
Why can stresses in reality be much higher than those calculated theoretically?
Due to non-uniformity of contact, non-uniformity of mechanical properties such as stiffness, and variations in surface contours.
Why are Sharp corners, notches and holes to be avoided in implants ?
- Because they induce high localised stress concentrations
In pic the closer the lines with arrows, the greater the stress concentrations.
Why is there a shear stress at a bone-implant interface?
A shear stress occurs at a bone-implant interface because the bone and implant each have a different material stiffness (Young’s modulus) so they try to deform by different amounts under the action of a load. If joined together they cannot deform separately so a shear stress develops between them - along the line of the interface.
What does the amount of load transferred from bone to implant or vice versa depends (i.e. at the load transfer region) on what?
How the loads are shared in the load sharing region
Youngs modulus (which represents material stiffness under axial and bending loads) and strength for implant materials commonly used in orthopaedic surgery are shown
Notice that the principal materials are metals, which are at least 10 times stiffer than cortical bone and many more times stiffer than cancellous bone
Note - Compressive strengths for the metals are not given because metals tend not to undergo compressive failure - they buckle or shear first. Also don’t memorise these fuck that
Under shear loading, including torsion, the material stiffness is measured by the shear modulus, rather than by Young‟s modulus.
State the equation used to calculate shear modulus
State the definition of stiffness
The force required to produce a unit deflection
State the equation for calculating the stiffness
A bar, which is made of a material with a Young‟s modulus E, and has a cross sectional area A and length L. It can be shown for axial (compressive or tensile) loading that the axial stiffness, which we will call SA, is expressed by the simple formula
When considering axial (compressive or tensile) loading what are the geometrical properties that affect the stiffness of the bar ?
Cross sectional area (A) and its length (L)
Note the structure becomes stiffer as the Young‟s modulus of the material increases but this is a constant unless you change the actual material
What is the approximate ratio of the material stiffness of stainless steel to that of cortical bone?
The ratio is approximately 200/20 = 10.
State the equation used to calculate axial rigidity (for compressive and tensile loading)
Appreciate this example:
Compare the axial rigidity of a long bone with that of a bone plate made of stainless steel. The figure also gives the values for E and A for the plate and the product EA. Although stainless steel material is ten times stiffer than cortical bone material, the plate has a cross section which is about ten times smaller so both structures have a similar axial rigidity (Rigidity = EA)
State the equation used to calculate bending rigidity
I = the second moment area
How does the second moment area (I) affect the bending rigidity
The further away a material is placed from the neutral axis the greater I is and therefore the more rigid it is when bent.
You can easily test this yourself by bending a ruler holding it in the two ways shown in Figure 7. You can see that there is more ruler material further away from the neutral axis in Figure 7B which makes it more difficult to bend
Describe how bending stiffness (rigidity) is affected by size
- The bigger the size the more stiff it is.
- The rod on the right, in Figure 8, has a higher value of I, so it will be stiffer when bent.
State the equation used to calculate torsional rigidity
- G is the Shear Modulus
- J is the polar second moment of area
State the simple equation relationship between polar second moment area and second moment area for circular cross-sections
What is the effect of size on the torsional rigidity on torsional rigidity ?
As size increases the torisonal rigidity increases
