Chapter 26 Biomechanics
Introduction#
The stress–strain curve is a triple A-list subject. It always seems to be asked in viva examinations and is a definite top 10 core basic science question.
Structured oral examination question 1#
Stress–strain curve
Can you draw the stress–strain curve for stainless steel?
Drawing is a vital component of the FRCS (Tr & Orth) Exam. The stress–strain curve demonstrates how a material subjected to an increasing tensile load deforms until failure (Figure 26.1). Stress is force over area and has units of Newton per square metre. Strain is change in length over original length; It is usually expressed as a ratio or percentage.
Figure 26.1 Stress–strain curve for a typical metal.
Don’t wait to be asked ‘what are the units of stress?’ by the examiners. 1
How does force change the cross-sectional dimensions of the area it is acting on?
An object understress experiences strain in the transverse direction aswell as the horizontal direction. However , stress and strain are generally based on the original dimensions of the object.
An initial slightly off-the-wall question can derail a nervous candidate before they have time to setile and hast o be handled with skill. Strain is directly proportional to stress. This is known as Hooke’s law.
What happens to the molecular bonds when in this region of the graph?
This here is the yield point, which is the maximum stress up to which a material undergoes elastic deformation. The yield point is the point on the stress strain curve that indicates the limit of elastic behaviour and the beginning of plastic deformation. The deformation is permanent and if the stress is removed, strain is not completely recovered and the material does not return to its original state.
What happens to the bonds?
Molecular bonds are broken, and the molecules move too far apart to return to their original positions. The ultimate tensile strength is the maximum stress that the material can sustain before fracture. Stress at the fracture point can be slightly less than the ultimate tensile strength because the latter can cause the material to neck, which reduces its cross-sectional area and therefore the force required to fracture. The area under the stress–strain curve represents the energy absorbed per unit volume of the material.
You seem to be changing around and mixing up stiffness and strength. What exactly do you mean by these terms?
Stiffness is s tress–strain while strength is the load required to break a material and depends on plastic deformation. Stiffness and strength are terms often used interchangeably, but they are distinct mechanical properties.
Are you sure?
The steeper a stress–strain curve, the stiffer the material. Mechanical testing measures force on a construct that could consist of bone, ligament and possibly fixation device (plate) and the data obtained relates to properties of the construct as a whole. Force and displacement are normalized for an individual material into stress and strain. That is why we normalize force with its cross-section and displacement with its gauge length to arrive at stress and strain.
What is the difference between hardness and toughness?
Hardness describes a material’s resistance to localized surface plastic deformation, eg. Hardness determines the wear resistance of a material. Under the same loading conditions, a harder material has a greater wear resistance than a softer material.
So why is hardness important?
It has great relevance when.
Basic science applied to clinical relevance.
Structured oral examination question 2#
Stress–strain curve
Stress–strain curve – label all the points, axis and nomenclature and describe what the areas underneath signify. What is hardness?
The x-axis represents strain, which is change in length over original length. The y-axis is stress, which is force per unit area applied and has the units Newton per square metre (N/m2). The area under the stress–strain curve up to the elastic limit depicts the modulus of resilience (MR), which signifies the ability of material to store or absorb energy without permanent deformation. It has no association with the s tress–strain curve.
How does the Young’s modulus differ for isotropic and anisotropic materials?
The examiner has linked mechanical properties to Young’s modulus of elasticity to test whether a candidate is able to demonstrate a more advanced level of understanding. With an anisotropic material the Young’s modulus varies depending on the direction of loading Examples include cortical bone, ligaments. We have the yield point here, which is the start of plastic deformation, and the ultimate stress, which is the maximum stress the material can withstand.
Hold on; there are lots of different points on the graph that you haven’t mentioned ( Figure 26.2a).
Figure 26.2a Stress–strain curve. Various points on graph include: 1, proportional limit; 2, elastic limit; 3, yield point; 4, ultimate tensile strength and 5, failure.
The usual conventionist o define the yield strength, which is the intersection of the curve with a straight line parallel to the elastic deformation of 0.2% on the strain axis.
There are three points on the graph that in some materials are very close to each other and often difficult to differentiate. These are2: 1. Proportionality limit The stress at which Hooke’s law is no longer obeyed. 2. Elastic limit If the stress slightly exceeds the proportional limit, the stress–strain curve is no longer linear but the material may still respond elastically. The curve tends to bend and flatten out. This continues until the stress reaches the elastic limit. The elastic limit is the stress at which permanent deformation is seen and beyond this point the deformity will not completely recover if the force is removed. 3. Yield point The point in the stress–strain curve at which the curve levels off and plastic deformation begins to occur. What is the yield point and how does it differ from the elastic limit?
Some textbooks describe the yield point as very fractionally la ter in the curve than the elastic limit. So the elas tic limit is the point at which deformations tops being entirely reversible.
What is the difference between (offset) yield stress and yield point3?
As it is often difficult to pinpoint the exact stress at which plastic deformation begins in some materials, the (offset) yield stress is taken to be the stress needed to induce a specified amount of permanent strain, typically 0.2%.
What do we mean by the upper and lower yield point (Figure 26.2b)?
For certain materials, especially low-carbon steel alloys, the stress–strain curve produces both an upper yield point and a lower yield point. A fairly dramatic drop is then observed in the stress to the lower yield point, although the strain continues to increase. Eventually the material is strengthened by this deformation and the stress is increased with further straining (strain hardening).
Figure 26.2b Upper and lower yield points on stress–strain curve.
What is happening at the yielding, strain hardening and necking stage of plastic deformation?
I am not completely sure. This is not well explained in
The plastic region consists of different parts that
necking.
the material will undergo considerable elongation
(yielding) with litile or no increase in stress. This is indicated by the flatness of the stress–strain graph region in the plastic region.
Strain hardening is where the plastic deformation increases a material’s resistance to further deformation.
Latice defects occurring in the material become too much in number and they restrict each other’s movement.
An example is cold working of metal alloys. Strain hardening increases the yield point at the expense of lower ductility and toughness.
In the region after the ultimate strength point, stretching occurs with an actual reduction in the stress.
Although the stress the material can withstand is reduced after the ultimate strength point, this is not due to any loss of material strength but due to the reduction incr oss-sectional area of the bar.
Structured oral examination question 3#
Stress–strain curve
Draw me the stress–strain curves for materials used in THA (Figure 26.3a). What is the stress–strain curve for ceramic, UHMWPE and stainless steel?
Sharp intakes of breath if you haven’t read up beforehand, but relatively easy if you have worked through a pre-exam answer. Stainless steel 316 is moderately strong, tough and a highly ductile material (this allows bending before catastrophic failure).
Figure 26.3a Stress–strain curve for materials used in THA.
What is happening with this graph (Figure 26.3b). Can you interpret it for me?
Figure 26.3b Strain hardening of a material.
If at this stage, the specimen is unloaded, the strain does not recover along the original path AO, but moves along AB. If the specimen is reloaded immediately, the strain increases with stress from B to A but via another path (the slope of stress–strain line is steeper indicating that the material has got stiffer than before) and reaches the point C, after which it will follow the curvature if loading is continued. Ultimate strength increased from S1 to S2. Thus strain hardening reduces toughness.
What about UHMWPE used in joint arthroplasty. What are its properties?
UHMWPE has low friction and high impact strength, excellent toughness and low density, ease of fabrication, bio compatibility and bio stability. Its major drawback is wear.
That’s fine, we don’t need to go there.
Structured oral examination question 4#
Stress–strain curve
Stress–strain curve of different materials ceramicSS, plastic, bone, ligament), how are they different (Figure 26.4)?
Figure 26.4 Stress–strain curves of different materials.
Essentially this is describing and interpreting the s tress–strain curve for britile (ceramic), ductile (CoCr) and plas tic elas tic materials. Be careful with bone as the stress–strain curve is different for cortical and cancellous bone.
Describe how each material behaves when loaded.
This question is a different way of essentially testing the factual knowledge of the stress–strain curve.4
How is the mode of failure (fracture) different between britile and ductile materials ?
Materials fracture by a process of crack initiation and propagation. Crack progression in abri tile fracture is associated with litile plastic deformation whereas ductile fracture involves significant plastic deformation.
The appearance of abri tile fracture is characterized.
With a ductile fracture there is considerable deformation after.
Structured oral examination question 5#
Stress–strain curve
Can you drawout the stress-strain curve – label all the points, axis and areas of interest (Figure 26.5a)? What happens at the yielding, strain hardening and necking regions of the graph (see above)? Stress–strain curve of materials with three different moduli of elasticity . Pick three items from a display in front of me that match those stress–strain curves. Describe their material properties (ceramic/bone/ligament) (Figure 26.5b).
Figure 26.5a Stress–strain curve diagram for ductile material (stainless steel).
Figure 26.5b Stress–strain curve, different materials.
Abri tile material is one that exhibits a linear stress–strain relationship up to the point of failure. The classic example to mention is ceramic Other examples would include PMMA and glass (Figure 26.5c). Cortical bone displays anisotropic behaviour with its elastic modulus depending on the direction of loading. Initially , a large distance (strain) is travelled under minimal stress as the crimped fibres straighten out. The characteristic shape is produced by the increase in the number of collagen fibrils resisting the strain as the slack fibrils are straightened and stretched, reducing the crimp pattern.
Figure 26.5c Stress–strain curve for PMMA. Essentially britile with litile plastic deformation before failure.
What about the stress–strain curve of cortical vs cancellous bone (Figure 26.5d)?
The compressive stress–strain curve for cancellous bone shows an initially shorter elastic segment (lower yield point) and has a lower stiffness (< 10% that of cortical bone). This prolonged plastic deformation period explains why the total energy absorbed by cancellous bone under compression can exceed that of cortical bone.
Figure 26.5d Compressive stress–strain curve cortical vs. cancellous bone. There is a prolonged plateau in cancellous loading representing the collapse of trabeculae.
Structured oral examination question 6#
S-N curve
S-N curve (Figure 26.6a). What is this, what does it mean and label the axis? Relate this to both THA and TKA.
Figure 26.6a S-N curve. Specimens are tested in a series of decreasing stress levels until no failure occurs within a selected maximum number of cycles. The nearly horizontal portion of the curve defines the fatigue or endurance limit. If the applied stress is below the endurance limit of the material, the specimen is said to have an infinite life.
Stress on the y- axis is ploft ed against number (n) of cycles (millions) on the x-axis. The higher the peak stress produced in a given cycle of loading and unloading, the fewer cycles that can be sustained before failure. Fatigue or endurance limit is normally defined at 106 or 107 cycles.
So, what is the difference between fatigue strength and fatigue limit?
For some materials the S-N curve becomes horizontal at higher n values or there is a limiting stress level called the fatigue limit (some times called endurance limit) below which fatigue failure will not occur. For these materials, the fatigue response is specified as fatigue strength, which is defined as the stress level at which failure will occur for some specific number of cycles (e.g. 107, 108).
THA operate above the endurance limit, while TKA operate at the endurance limit.
It is important to avoid creating an y scratches or dents.
Fatigue failure in orthopaedic implants is.
Mechanical requirements of arthroplasty materials include a high yield point and endurance limit.
A concern is high and frequent loads on the hip joint. It is estimated around.
What do we mean by notch sensitivity?
Notch sensitivity is the extent to which the sensitivity of amate rialto fracture is increased by the presence of a surface inhomogeneity, e.g. Eventually the crack becomes sufficiently deep so that the stress concentration exceeds the fracture strength and sudden failure occurs. A notch causes nonuniform stress flow lines.
Stress–strain curve for a viscoelastic compound (Figure 26.6b). What is this? Explain viscoelasticity . Discuss creep/hysteresis/stress relaxation.
Figure 26.6b Viscoelastic materials exhibit a timed ela y in returning the material to original shape. Some energy is lost. Loading unloading curves are different.
Most biological tissues are viscoelastic (eg. tendon, ligament, bone, articular c artilag e).5 A viscoelastic material exhibits stress–strain behaviour that is time and rate dependent i.e. 1. Stress relaxation. 2. Creep. 3. Hysteresis. 4. Strain rate sensitivity . A viscoelastic material continues to deform when a constant stress is applied to it. This continued deformation is creep. The creep rate decreases with time. If a viscoelastic material is held at a constant strain, the stress will decrease with time. This is called stress relaxation Therefore, stress relaxation is the inverse of creep. Stress relaxation time decreases with time. A viscoelastic tissue does not follow the same path on a stress–strain graph. This is called hysteresis. This property allows viscoelastic materials to act as shock absorbers.
stronger and tougher when loaded at a higher strain rate.
This is called strain rate sensitivity and is duet o the.
There are various slightly differently worded definitions of cr eep/hysteresis/stress relaxation in the textbooks.
Can you give me some everyday examples of creep, stress relaxation?
The handle of a heavy shopping bag gets longer and thinner as you walk home. This is creep, a material stretching out over time when.
What about strain rate sensitivity?
Sorry, no. Blu tack.6
Structured oral examination question 7#
Draw stress–strain curve and mark the events. Draw the curve.
Usual standard question Remember ductility is.
Can you describe the biomaterial behaviour of material A, B and C (Figure 26.7a)?
Figure 26.7a Stress–strain curve for various materials.
A has high strength, low ductility and low toughness. B has high strength, high ductility and high toughness.
What do you mean by strength, ductility and toughness?
Strength is a somewhat imprecise term, but relates to the degree of resistance to deformation of amate rial. When comparing britile and ductile materials with the same ultimate strengths, the britile material isless tough as it has less area under the stress–strain curve.
What do we mean by creep, stress relaxation and hysteresis? Can you drawout the relevant graphs (Figure 26.7b–26.7d)?
Figure 26.7b, c and d Candidate drawings of creep, stress relaxation and hysteresis.
Figure 26.7b, c and d. Talk as you draw making sure your explanation is spot on for accuracy.
What about elastic materials?
Elastic materials do not exhibit energy dissipation or hysteresis as their loading and unloading curve is the same. Under fixed strain, elastic materials will reach a fixed stress and stay at that level with no relaxation.
Structured oral examination question 8#
Stress–strain curves for different materials
Stress–strain curves for a variety of different materials.
What is HA, how is it put on a Ti stem?
What is stainless steel?
Initials tress–strain curve questions can be a leadin (or prop) to go on and discuss biomaterials. The most common stainless steel in orthopaedics is 316 L. The number 316 refers to 3% molybdenum and 16% nickel added to a normal alloy of iron, carbon and chromium. The letter L indicates a low carbon content < 0.03%. Its use in arthroplasty surgery has been limited as Ti and CoCr alloys have better wear and corrosion resistance and lower stiffness. R elativ ely low biocompatibility and technical difficulties with MRI. Some newer SS contains a high nitrogen content that makes it stronger and more resistant to localized corrosion. It has poor wear characteristics and a high coefficient of friction, making it unsuitable for use as an articulating bearing surface in THA. The most commonly used orthopaedic titanium alloy is titanium 64. The numbers refer to alloying elements aluminium (6%) and vanadium (4%).
HA is used as an adjuvant surface coating on prosthetic.
Why is this?
When compared to cobalt chromium titanium demonstrates a 33% increase inbond strength. The modulus of elasticity of titanium is closer to bone, resulting inlesss tress shielding and bone resorption. As such titanium alloy is the preferred metallic substrate of choice.
What is the reason for this?
Porous ingrowth surfaces appear to have a greater inherent initial stability which encourages biological fixation. Fixation occurs by bony ongrowth on the implant surface, which requires a more extensive area of coating to secure the implant as this is a weaker method of fixation. As such HA is added to improve early stability, reduce micromotion in the immediate postoperative period and accelerate bone ongrowth onto the prosthesis surface.
Why do we use grit-blasted stems if ingrowth stems appear to have a better chance of initial stability and obtaining osseointegration?
The manufacturing process for porous ingrowth stems can result in diminished.
How is HA put on a Ti stem?
HA deposition is often achieved through the plasma spray technique, which is performed at high temperature (15,000°C) and under vacuum, by projecting HA particles on to the metallic material at a speed of 300 m/s.
H A is an osteoconductiv e agent that allows for more rapid closure of gaps. bone to prosthesis and prosthesis to bone), which clinically shortens the timet o biological fixation.
The optimal thickness of hydroxyapatite is 50–75 μm. Thicker.
Is there much difference in the clinical outcome or survivorship between HA-coated stems and uncoated stems.
From what I understand of the literature there isn’t much evidence to.
What radiological outcomes are you assessing?
Endosteal condensation is considered a sign of endosteal bone ingrowth on the surface of the femoral stem and suggests that femoral stem fixation is optimal. HA particles delaminated from the stem surface may induce osteolysis either by stimulating bone loss or by migration to the joint space producing third-body wear.
Can you please draw the stress–strain curve for steel?
The stress–strain curve for stainless steel is typical of that for a ductile material. The question may just require a simplified stress–strain diagram to be drawn.
Figure 26.8 Simplified stress–strain curve for steel.
The stress–strain curve for ligament vs. tendon?
Ligaments and tendons are predominantly made up of collagen. As such their stress–strain relationship is very similar to that of collagen. There are a few minor differences in the stress–strain curve between ligament and tendon. Because it is easier to stretch out the crimp of the collagen fibrils, this part of the stress–strain curve shows a relatively low stiffness. As individual fibrils within the ligament or tendon begin to fail, damage accumulates, stiffness is reduced and the ligament/tendon begins to fail.
Structured oral examination question 9#
Free body diagrams: elbow
Quite a large part of biomechanics involves drawing and explaining around free body diagrams (FB DIn the good old days of the past, if the examiners wanted topass.
Most candidates should breeze.
If examiners were unsure of a candidate they would ask them to draw a FBD of a hip and see how they got on. This was make or break for the candidate to redeem themselves, but if they messed up again then a couple more FBDs of a person holding a suitcase in one hand, or a suitcase in both hands would usually be enough to sink them.
If the examiners were keen to fail a candidate for whatever reason they would ask them straightup spinal biomechanics.
What do we mean by a free body diagram?
This is a method used to illustrate the various forces acting onas tructure such as a bone, and to illustrate how far from a joint or other pivot point these forces are acting. FBD show the locations and directions of all forces and moments acting on a body . They can not be used for dynamic equilibrium.
What are the assumptions made when drawing a free body diagram?
The assumptions made are that7: The weight of the body is concentrated at the exact centre of body mass. Muscles only act intension (no compressive forces).
The line of action of a muscle is along.
Joint reaction forces are assumed.
The joint acts only as a hinge (other.
What do we mean by a joint reaction force?
JRF is the force generated within a joint in response to external forces.
Can you draw a free body diagram of the elbow joint with an object in the hand (Figures 26.9a and 26.9b)?
Figure 26.9a Arm flexed at 90° at the elbow, with wrist and fingers rigid, holding a ball in palm of hand.
Figure 26.9b Free body diagram showing forearm holding a ball.
Candidates may be straight on asked to draw a free body diagram of a particular joint without the warmup preamble of general free body analysis assumptions (see above). It is reasonable to mention the general assumptions a t the beginning of the viva and then go on to joint-specific assumptions afterwards. My assumptions when drawing the FBD of the elbow are that: Arm is flexed 90°. Acting as a class III le ver.
Elbow fulcrum for the forearm lever.
It is a two-dimensional X–Y plane.
The axis about which an object rotates as the result of.
The entire mass of an object is considered to be concentrated at.
Centre of gravity (COG) is the point from which.
The centre of mass and the centre of gravity of an object are in the same position if the gravitational field in which the object exists is uniform. 1. G – weight of the forearm acting vertically downwards, 1.5 kg. 2. Wo – weight of object, 2 kg. 3. 4. Considering the rotational equilibrium of the forearm about IAR, summation of moments about O will be zero. Sum of clockwise (extension) = anticlockwise flexion) moments ∑ M = 0 Wf × 0.15 + Wo × 0.3 = 0.05 × Biceps 15N × 0.15 + 20N × 0.3 = B × 0.05 2.25 + 6 = 0.05B 165N = B (Brachialis force) As the forearm is in translational equilibrium the sum of the forces ∑ F = 0 acting on it is zero. There is no JR Fin the X-axis.
B – G – W – J (JRF) = 0
JRF(J) + 15 + 20 = 165 N
JRF = 165 – 35
JRF = 130 N
Learn a simplified FBD of the elbow that can be quickly drawn (Figure 26.9c). With the elbow flexed to 90° by the side of the body, brachialis is the main muscle that maintains this position.
If you are doing very well or very poorly then 26.9d).
Use the same methods as used for elbow flexion.
Figure 26.9c Candidate 20-second simplified FBD elbow.
Figure 26.9d Joint reaction force on the elbow joint during extension using the same method as that for elbow flexion.
∑ M = 0
(0.1 × W)–(0.03 × T) = 0
If W = 20N
T=(0.1 × 20N)/0.03
T = 67N
∑ F = 0
J–T–W =0
J = T + W
J = 67N + 20 N
J = 87N
For a brownie point, candidates may be asked what type of lever is occurring inflexion and extension. A class 1 lever with elbow extension whereas a class 3 lever with a flexed elbow. The elbow is held in 90° of flexion with the forearm positioned over the head and parallel to the ground.
Structured oral question 10#
Hip FBD
If a candidate can practise drawing out a standard FBD of a hip in around 20 seconds, then they will create more time to get further on forward with the question chain and sc ore some extra marks.
Can you draw a free body diagram of a hip joint when a person stands on one leg (Figure 26.10a and 26.10b)?
Figure 26.10a and 26.10b Candidate drawing of FBD hip.
Clockwise moment = Anticlockwise moment ∑ M = 0 Sum of moments is zero FAB × MFAB = FW × MW ∴ FAB = FW × MW MFAB If MFAB = 0.05 m If MW = 0.15 m If W = 600 N and 5/6 = 500 N FAB = 0.15 × 500 = 1500 N 0.05
Be careful with the line of action of the abductor muscles. The abductor force is three times closer to the fulcrum (0.05 m vs. 0.15 m). This is estimated by lengths of the limbs (calculate with scale drawings) or by trigonometry (Figure 26.10c).
Figure 26.10c A force triangle is drawn to calculate JRF -estimated by lengths of the limbs or by trigonometry JFR = FAB + FW
What class of lever is this?
This is a classI lever between the bodyweight and abductor force.
What happens to the joint reaction force in the hip if the patient has osteoarthritis of the hip and is given a walkingstick in the opposite hand (Figure 26.10d)? Can you draw this out for me?
- BW = 600 N
- A = 0.05 m
- B = 0.15 m
- C = 0.45 m
- FSTICK = 100 N
Sum of moments about the hip is zero
Clockwise = 500 × 0.15
Anticlockwise = F AB × 0.05 + 100 × 0.60
0.05 × FAB = 75 – 60
FAB = 15/0.05 = 300 N (1500 N without stick)
Figure 26.10d FBD hip with walkingstick.
What are the properties of a s tick?
Introducing a stick on the opposite side adds another anticlockwise moment. The joint reaction force is reduced by 80% with using a walkingstick in the contralateral hand.
How do you calculate the optimum length of shaft. How would a person walk if the stick is too short? And too long?
Apa tien t should stand upright inshoes they would normally use. The distance between the ground and end of the stick should r each to the distal wrist crease (or ulnostyloid joint). The method of measuring the distance from the greater trochanter to the ground is not accurate or effective. If it is too long, using it can cause shoulder pain and be uncomfortable.
What about carrying a suitcase on the opposite side as the weight-bearing leg. Can you draw this out for me (Figure 26.10e and 26.10f)?
Figure 26.10e and 26.10f FBD with patient carrying a suitcase opposite side of the weight-bearing limb.
Figure 26.10g and 26.10h FBD with patient carrying a suitcase on the same side as the weight-bearing limb.
Adding a suitcase on the opposite side introduces a.
BW = 600 N
Weight of suitcase = 250 N
A = 0.05 m
B = 0.15 m
C = 0.45 m
Sum of moments about the hip is zero
Clockwise moment = 500 × 0.15 + 250 × 0.6
Anticlockwise moment = FAB × 0.05
0.05FAB = 75+150
FAB = 225/0.05 = 4500 N
The abductor force that has to be generated.
Constructing a force triangle to calculate.
What about carrying a suitcase on the same side as the weight-bearing leg. Can you draw this out for me (Figure 26.10g and 26.10h)?
Adding a suitcase on the same side introduces an anticlockwise moment that aids the abductors. The amount of force needed to be generated by the hip abductors is reduced and
- Clockwise moment = 500 × 0.15
- Anticlockwise moment = FAB × 0.05 + 250 × 0.2
75 N = 0.05FAB + 50 N
25 N = 0.05FAB
FAB= 25/0.05 = 500 N
Despite the extra weight carried, the JRF is.
What about carrying a suitcase in both hands (Figure 26.10i and 26.10j)?
BW = 600 N
Weight of each suitcase = 250 N
A = 0.05 m
B = 0.15 m
C = 0.45 m
D = 0.20 m
The number of forces acting in this situation is 5
Force from both suitcases 500 N
JRF
Upper body force (5/6 BW) 500 N
Abductor muscle force FAB
Sum of moments about hip is zero
- Clockwise moment = 500 × 0.15 + 250 × 0.6
- Anticlockwise moment = FAB × 0.05 + 250 × 0.2
75 N + 150 N = 0.05FAB + 50 N
175 N = 0.05FAB
FAB= 175/0.05 = 3500 N
There is a large clockwise moment arm when carrying a suitcase in the non-weight-bearing side that is not equally balanced.
Figure 26.10i and 26.10j FBD with patient carrying a suitcase in both hands.
If a person is carrying two suitcases the bodyweight is more balanced, which shitis the C OM closer to the femoral head, thus decreasing the moment arm to the COM so the abductor muscles don’t have to work as hard (less force) to overcome the moment due to the weights leading to a lower reaction force.
What other methods are used to decrease the joint reaction force in the hip joint?
Augmenting the abductors or reducing the bodyweight moment achieves a reduction in the JR FActions that increase the abductor force include: Trendelenberg lurch – shitiing bodyweight nearer to the femoral head, thereby decreasing lever arm.
Medialization of THA cup (shitis centre of rotation.
What actions increase joint reaction?
Valgus neck–shaft angulation – decreases shear across joint.
Structured oral question 8#
Spine
The difficulty with a FBD of the spine is that for a candidate.
If a candidate is tight for time the temptation is to go straight to the crux of the question This.
For the loading conditions shown, calculate the erector spinae muscle force FM and the compressive and shear components of joint reaction force (FJC and FJS at the L5/S1 vertebrae red square) (Figure 26.11a).
Assume the person weighs 70 kg and litis a 20 kg weight. The spine is flexed approximately 35°. 1. 2. 3. For the body to be in moment equilibrium the sum of the moments acting on the lumbar spine must be zero. Clockwise and counterclockwise must balance. ∑ M = 0 W × 0.25 m + 200 N × 0.4 – FM × 0.05 = 0 450 N × 0.25 m + 200 N × 0.4 – E × 0.05 m = 0 E × 0.05 m = 112.5 Nm + 80 Nm E = 3850 N
Calculate the compressive force exerted on the disc.
C is the sum of the compressive forces acting over the disc which is inclined 35° to the transverse plane.
The force produced by the weight of the 35°.
P × cos 35°
which acts approximately at a right angle to the disc inclination.
The magnitude of C can be found through
∑ F = 0
W × cos 35° + P × cos 35° + E – C = 0
450 N × cos 35° + 200 N × cos 35° = 3850 N – C
C = 368.5 + 163.8 N = 3850 N
C = 4382 N
The shear component for the reaction force.
450 N × sin 35° + 200 × sin 35° – S
S = 373 N
Figure 26.11a FBD of spine.
Practise drawing out a FBD of the spine as it is definitely known to be asked in a viva and trying to drawout for the first time in the exam from first principles is generally going to be doomed to failure (Figure 26.11a).
Figure 26.11b Candidate’s diagram FBD spine. Influence of litiing technique on spinal forces.The upper bodyweight and force exerted by the weight act in front of the disc and create forward bending moments. LW:lever arm for bodyweight, LP-lever arm for weight carried in hand.
Notes
1. To try and catch a candidate out, to be clever.
2. As such, the examiners will move past a point if a candidate doesn’t know it.
3. These terms are often incorrectly interchanged in various internet PPP.
4. A different route whereby to get to.
5. As such this viva question can be asked.
6. Blu Tack is a reusable, putiy -like, pressure-sensitiv e adhesive produced by.
7. Practise beforehand being able to trot out these general assumptions while.