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Question 41

Topic: Biomechanics & Biomaterials

A 55-year-old active man undergoes a total hip arthroplasty. The surgeon opts for a highly cross-linked polyethylene (HXLPE) liner. Which of the following best describes the mechanical trade-off associated with the increased cross-linking of the polyethylene?

. Increased wear resistance with decreased fatigue strength
. Decreased wear resistance with increased fracture toughness
. Increased oxidation resistance with increased yield strength
. Decreased oxidation resistance with decreased wear resistance
. Increased wear resistance with increased fracture toughness

Correct Answer & Explanation

. Increased wear resistance with decreased fatigue strength


Explanation

Highly cross-linked polyethylene undergoes irradiation to increase wear resistance by forming cross-links between polymer chains. However, this process decreases bulk mechanical properties such as fatigue strength, fracture toughness, and yield strength.

Question 42

Topic: Biomechanics & Biomaterials

A 58-year-old woman with a long history of erosive osteoarthritis of the index finger PIP joint presents with severe, unremitting pain and a progressive 'gull-wing' deformity. Radiographs confirm advanced joint space narrowing, central subchondral collapse, and significant osteophyte formation. She has failed all conservative treatments. The biomechanical changes in EOA, particularly the central collapse, disrupt normal joint congruity and alter load distribution. Which of the following best describes the primary consequence of this central collapse and altered load distribution in the index finger PIP joint?

. Increased stability due to compensatory osteophyte formation.
. Enhanced range of motion due to reduced articular surface contact.
. Exacerbated pain, stiffness, and functional disability due to disrupted mechanics and inflammation.
. Improved intrinsic muscle balance leading to better pinch strength.
. Reduced susceptibility to further deformity due to bone remodeling.

Correct Answer & Explanation

. Exacerbated pain, stiffness, and functional disability due to disrupted mechanics and inflammation.


Explanation

Correct Answer: CThe case content explains that in EOA, the inflammatory process leads to active chondrolysis and subchondral bone erosion, often centrally, resulting in the characteristic 'gull-wing' deformity. This central collapse disrupts the normal joint congruity and alters load distribution. The biomechanics section states: 'The index finger PIP joint's role in forceful pinch and grip exacerbates these changes, leading to increased pain, stiffness, deformity (e.g., flexion contracture, lateral deviation), and functional disability.' Therefore, exacerbated pain, stiffness, and functional disability are the primary consequences.Incorrect Options:A:While osteophyte formation is a compensatory response, it restricts motion and contributes to pain through impingement, rather than increasing overall joint stability in a functional sense, especially when central collapse is present. The overall effect is instability and dysfunction.B:Disrupted articular surface contact due to central collapse and osteophytes typically leads to restricted range of motion and stiffness, not enhanced motion.D:The case states that 'The intrinsic muscles can become imbalanced, further contributing to progressive deformity,' which would negatively impact pinch strength and function, not improve it.E:The central collapse and altered load distribution, coupled with the inflammatory process, make the joint more susceptible to further deformity and destruction, not less.

Question 43

Topic: Biomechanics & Biomaterials

Which factor primarily determines the bending stiffness of an intramedullary nail construct?

. The material's yield strength.
. The number of interlocking screws.
. The cross-sectional area moment of inertia of the nail.
. The friction coefficient between the bone and nail.
. The length of the nail.

Correct Answer & Explanation

. The cross-sectional area moment of inertia of the nail.


Explanation

Correct Answer: CThe bending stiffness of a structural element, like an IM nail, is primarily determined by its Young's modulus (material stiffness) and its area moment of inertia (I). The area moment of inertia is highly dependent on the nail's diameter and cross-sectional geometry. A larger diameter nail, even with the same material, will have a significantly higher area moment of inertia and thus greater bending stiffness (Stiffness is proportional to E*I). The number of interlocking screws contributes to rotational and translational stability but does not directly dictate intrinsic bending stiffness of the nail itself. Yield strength relates to plastic deformation, and length influences deflection but not intrinsic stiffness.

Question 44

Topic: Biomechanics & Biomaterials

A titanium intramedullary nail is selected over a stainless steel nail of identical dimensions for a tibia fracture. Based on material properties, how does the titanium nail influence the biomechanical environment?

. It provides higher bending stiffness and absolute stability.
. It has a higher modulus of elasticity, increasing stress shielding.
. It has a lower modulus of elasticity, reducing bending stiffness and decreasing stress shielding.
. It prevents endochondral ossification by minimizing interfragmentary strain.
. It drastically increases the torsional rigidity compared to stainless steel.

Correct Answer & Explanation

. It has a lower modulus of elasticity, reducing bending stiffness and decreasing stress shielding.


Explanation

Titanium has a lower modulus of elasticity than stainless steel, making it less stiff and biomechanically closer to cortical bone. This allows more load-sharing, reduces stress shielding, and promotes callus formation.

Question 45

Topic: Biomechanics & Biomaterials

Compared to stainless steel, titanium alloy intramedullary nails exhibit which of the following biomechanical characteristics?

. Higher modulus of elasticity
. Increased stiffness in bending
. Closer modulus of elasticity to cortical bone
. Greater risk of galvanic corrosion
. Higher notch sensitivity resulting in sudden catastrophic failure

Correct Answer & Explanation

. Closer modulus of elasticity to cortical bone


Explanation

Titanium alloy has a lower modulus of elasticity than stainless steel, making it biomechanically closer to that of cortical bone. This relative flexibility decreases stress shielding and allows for more favorable load sharing during fracture healing.

Question 46

Topic: Biomechanics & Biomaterials

According to the polar area moment of inertia, the torsional rigidity of a solid cylindrical intramedullary nail is proportional to its radius raised to which power?

. First
. Second
. Third
. Fourth
. Fifth

Correct Answer & Explanation

. Fourth


Explanation

The torsional rigidity of a solid cylinder is determined by its polar moment of inertia, which is proportional to the fourth power of its radius (r^4). Therefore, even small increases in nail diameter exponentially increase its torsional stability.

Question 47

Topic: Biomechanics & Biomaterials

The primary biomechanical consequence of utilizing a slotted (open-section) intramedullary nail compared to a solid (closed-section) nail of identical diameter and material is:

. Increased torsional rigidity
. Significantly decreased torsional rigidity
. Increased bending stiffness in the coronal plane
. Enhanced endosteal revascularization
. Promotion of primary rather than secondary bone healing

Correct Answer & Explanation

. Significantly decreased torsional rigidity


Explanation

Slotted nails have a significantly lower polar moment of inertia compared to solid nails, resulting in substantially decreased torsional rigidity. This makes them more prone to torsional deformation and failure under rotational loads.

Question 48

Topic: Biomechanics & Biomaterials

An initially statically locked intramedullary nail used to treat a 4-cm segmental tibial defect presents at 6 months with broken distal interlocking screws. What is the primary biomechanical cause of this failure?

. Galvanic corrosion between a titanium nail and stainless steel screws
. Cyclic loading over a large working length without bony support leading to fatigue failure
. Stress shielding caused by an oversized nail diameter
. Premature dynamization of the proximal segment
. Excessive torsional rigidity of the intramedullary nail

Correct Answer & Explanation

. Cyclic loading over a large working length without bony support leading to fatigue failure


Explanation

In cases of segmental defects or severe comminution, the nail and screws bear all physiological loads because there is no cortical bone contact to share the load. This large working length concentrates cyclic stresses on the screws, eventually causing fatigue failure prior to union.

Question 49

Topic: Biomechanics & Biomaterials

A solid intramedullary nail is upgraded from a 10 mm diameter to a 12 mm diameter. Assuming identical material properties and working length, by what approximate factor does the torsional rigidity of the nail increase?

. 1.2
. 1.4
. 2.1
. 3.0
. 4.0

Correct Answer & Explanation

. 2.1


Explanation

The torsional rigidity of a solid cylindrical device is proportional to the polar moment of inertia, which scales with the radius to the fourth power (r^4). Increasing the diameter from 10 mm to 12 mm increases rigidity by a factor of (1.2)^4, which is approximately 2.1.

Question 50

Topic: Biomechanics & Biomaterials

When comparing a titanium alloy intramedullary nail to a stainless steel nail of identical dimensions and design, the titanium nail biomechanically exhibits:

. Higher modulus of elasticity and increased bending stiffness
. Higher modulus of elasticity and decreased bending stiffness
. Lower modulus of elasticity and decreased bending stiffness
. Lower modulus of elasticity and increased bending stiffness
. Identical modulus of elasticity and decreased fatigue strength

Correct Answer & Explanation

. Lower modulus of elasticity and decreased bending stiffness


Explanation

Titanium alloy has a lower modulus of elasticity compared to stainless steel, making it less stiff and more flexible. This lower modulus allows the nail to more closely approximate the stiffness of cortical bone, potentially improving load sharing and reducing stress shielding.

Question 51

Topic: Biomechanics & Biomaterials

An intramedullary nail is manufactured from a titanium alloy rather than 316L stainless steel. What is the primary biomechanical difference regarding the material properties of the titanium nail?

. Higher modulus of elasticity leading to increased stiffness
. Lower modulus of elasticity resulting in stiffness closer to cortical bone
. Higher ultimate tensile strength preventing fatigue failure
. Lower notch sensitivity compared to stainless steel
. Higher galvanic corrosion potential in vivo

Correct Answer & Explanation

. Lower modulus of elasticity resulting in stiffness closer to cortical bone


Explanation

Titanium alloys possess a lower modulus of elasticity compared to stainless steel, making them less rigid and closer to the natural stiffness of cortical bone. However, they tend to be more notch sensitive.

Question 52

Topic: Biomechanics & Biomaterials

Regarding the biomechanical footprint of intramedullary implants, which of the following characteristics best minimizes the 'stress shielding' effect on the surrounding diaphyseal bone?

. High area moment of inertia
. Use of 316L stainless steel rather than titanium
. Matching the modulus of elasticity of the implant to that of cortical bone
. Maximal interference fit at the isthmus without locking screws
. Increasing the wall thickness of a hollow nail

Correct Answer & Explanation

. Matching the modulus of elasticity of the implant to that of cortical bone


Explanation

Stress shielding occurs when a rigid implant bears too large a share of physiological loads, causing disuse osteopenia in adjacent bone. Using materials with a lower modulus of elasticity closer to cortical bone, like titanium, promotes healthier load sharing.

Question 53

Topic: Biomechanics & Biomaterials

What is the primary effect of over-reaming the medullary canal by 2 mm beyond the native isthmus diameter on the inherent torsional strength of the diaphyseal bone itself?

. It increases torsional strength due to the resulting thermal hyperemic response
. It has no measurable effect on torsional strength as long as the periosteum is intact
. It decreases torsional strength by reducing the bone mass at the outer radius
. It significantly decreases torsional strength by removing bone from the inner radius
. It shifts the neutral axis of the bone, increasing its overall bending strength

Correct Answer & Explanation

. It decreases torsional strength by reducing the bone mass at the outer radius


Explanation

The torsional strength of a tubular bone depends heavily on its geometry, specifically the polar moment of inertia. Reaming removes cortical bone from the inner radius, which measurably reduces the remaining diaphyseal bone's inherent resistance to torsion.

Question 54

Topic: Biomechanics & Biomaterials

Modern conventional antegrade tibial intramedullary nails are manufactured with a distinct proximal bend, known as the Herzog curve. What is the primary biomechanical purpose of this design feature?

. To increase the ultimate tensile strength of the titanium alloy
. To accommodate the extra-articular anterior starting point while aligning the nail with the straight diaphyseal canal
. To prevent rotation of the proximal fragment without the need for locking screws
. To avoid injury to the deep peroneal nerve during insertion
. To facilitate a completely intra-articular entry portal

Correct Answer & Explanation

. To accommodate the extra-articular anterior starting point while aligning the nail with the straight diaphyseal canal


Explanation

The Herzog curve allows the nail to enter via the proximal, slightly anterior extra-articular starting portal and then smoothly transition into the straight mechanical axis of the tibial diaphysis.

Question 55

Topic: Biomechanics & Biomaterials

Galvanic corrosion in orthopedic implants occurs under which of the following conditions?

. Motion between two modular components of the same metal
. Implantation of two dissimilar metals in electrolytic contact
. Crevice formation in a single metal implant
. Exposure of the implant to a high-pH environment
. Fatigue loading of an intramedullary nail

Correct Answer & Explanation

. Implantation of two dissimilar metals in electrolytic contact


Explanation

Galvanic corrosion is an electrochemical process that occurs when two dissimilar metals are placed in electrical contact within an electrolytic solution like body fluids. The less noble metal acts as an anode and corrodes preferentially.

Question 56

Topic: Biomechanics & Biomaterials

A 50-year-old male with a history of chronic alcoholism and poor nutrition presents with a nonunion of a mid-shaft femoral fracture treated with an intramedullary nail. Radiographs show a persistent fracture gap and sclerotic bone ends. The nail itself appears intact. What is the most critical biomechanical factor that contributes to the risk of fatigue failure of the intramedullary nail in this nonunion scenario?

. The patient's poor nutritional status and comorbidities.
. The use of a titanium alloy nail instead of stainless steel.
. The material's ultimate tensile strength being too low.
. The prolonged, cyclical loading of the implant due to the absence of bone healing.
. The presence of interlocking screws at both ends of the nail.

Correct Answer & Explanation

. The prolonged, cyclical loading of the implant due to the absence of bone healing.


Explanation

Correct Answer: DThe correct answer is D. Fatigue failure of an intramedullary nail occurs when the implant is subjected to repeated stresses below its ultimate strength over a prolonged period. In a nonunion, the bone is not healing, meaning the implant continues to bear the majority of the physiological load indefinitely. This prolonged, cyclical loading, often for months or years beyond the expected healing time, eventually exhausts the implant's fatigue life, leading to fracture or failure of the nail. The implant is designed to be a temporary load-sharing device, not a permanent load-bearing one in the absence of bone healing.Option A (patient's comorbidities) contributes to the nonunion itself, but the direct biomechanical cause of nail fatigue failure is the implant's inability to offload to healed bone.Option B (titanium vs. stainless steel) relates to Young's Modulus and stress shielding, but while material choice affects fatigue life, it's not themost criticalfactor in a nonunion whereanyimplant will eventually fail if not offloaded.Option C (ultimate tensile strength) is a material property, but fatigue failure occurs below this limit. The issue is thedurationandrepetitionof loading, not necessarily that the ultimate strength was too low for a single load.Option E (interlocking screws) are essential for stability, but their presence does not prevent fatigue failure if the bone does not heal and continues to load the nail cyclically.

Question 57

Topic: Biomechanics & Biomaterials

A 42-year-old male sustains a mid-shaft femoral fracture. The surgeon is debating between using a stainless steel nail or a titanium alloy nail. Biomechanically, what property of titanium alloys contributes to their perceived advantage in reducing stress shielding compared to stainless steel nails?

. Higher density, providing more mass for stability.
. Increased hardness, making it more resistant to wear.
. Lower Young's Modulus, making the implant less stiff and closer to bone's elasticity.
. Superior fatigue strength, allowing for longer implant life.
. Greater coefficient of friction with bone, enhancing stability.

Correct Answer & Explanation

. Lower Young's Modulus, making the implant less stiff and closer to bone's elasticity.


Explanation

Correct Answer: CThe correct answer is C. Young's Modulus (or modulus of elasticity) is a measure of a material's stiffness or resistance to elastic deformation under stress. Titanium alloys (e.g., Ti-6Al-4V) generally have a lower Young's Modulus (approximately 110 GPa) compared to stainless steel (approximately 200 GPa) or cobalt-chromium alloys (approximately 230 GPa). Cortical bone has a Young's Modulus of approximately 17-20 GPa. Biomechanically, an implant with a Young's Modulus significantly higher than bone will bear a disproportionate amount of the load, leading to stress shielding of the adjacent bone. A lower Young's Modulus, like that of titanium, brings the implant's stiffness closer to that of bone, thereby reducing the magnitude of stress shielding. Less stress shielding means the bone carries more physiological load, which is thought to be beneficial for bone remodeling and strength, potentially promoting fracture healing.Option A (higher density) is incorrect; titanium is actually less dense than stainless steel, which is a benefit for weight but not directly related to stress shielding.Option B (increased hardness) is not the primary factor for stress shielding; hardness relates to resistance to indentation.Option D (superior fatigue strength) is often debated and depends on specific alloy and design, but it's not the primary reason for reduced stress shielding.Option E (greater coefficient of friction) is not a primary biomechanical advantage for reducing stress shielding; stress shielding is about load transfer through the material's stiffness.

Question 58

Topic: Biomechanics & Biomaterials

A 30-year-old male sustains a transverse midshaft femoral fracture. During open reduction and internal fixation with a conventional plate, the surgeon performs 'pre-bending' of the plate. What is the primary purpose of this pre-bending technique when applied to a transverse or short oblique diaphyseal fracture?

. To ensure uniform screw purchase along the plate length.
. To facilitate primary bone healing by promoting absolute stability.
. To prevent gapping on the opposite cortex when achieving compression.
. To reduce the modulus of elasticity of the plate, thereby decreasing stress shielding.
. To provide dynamic compression across the fracture site.

Correct Answer & Explanation

. To prevent gapping on the opposite cortex when achieving compression.


Explanation

Correct Answer: CPre-bending a conventional plate is a critical step for transverse or short oblique fractures. When the plate is applied and screws are tightened, the plate attempts to straighten out against the bone. This straightening action drives the fracture fragments together, creating compression on the far cortex (the cortex opposite the plate) and preventing gapping on that side. This enhances interfragmentary compression across the entire fracture plane, which is essential for achieving absolute stability and promoting primary bone healing. Without pre-bending, compression of the near cortex (under the plate) can lead to distraction and gapping of the far cortex, compromising stability. Option A is incorrect as pre-bending does not primarily affect screw purchase uniformity. Option B is a consequence of successful compression, not the direct purpose of pre-bending itself. Option D is incorrect; pre-bending does not alter the plate's modulus of elasticity. Option E is incorrect; dynamic compression is achieved through eccentric drilling, not pre-bending.

Question 59

Topic: Biomechanics & Biomaterials

A 55-year-old patient is 2 years status post-ORIF of a tibia fracture with a rigid locking plate. Radiographs show complete bone healing, but also significant cortical thinning beneath the plate. The patient is considering hardware removal. What is the primary concern when considering 'stress shielding' in this context?

. The potential for the implant to corrode over time due to bodily fluids.
. The concentration of stress at the screw-bone interface leading to screw loosening.
. The bone adjacent to the implant experiencing reduced physiological loading, leading to disuse osteopenia.
. The stress on the plate itself, causing fatigue failure of the implant.
. The uneven distribution of compressive forces across the fracture site.

Correct Answer & Explanation

. The bone adjacent to the implant experiencing reduced physiological loading, leading to disuse osteopenia.


Explanation

Correct Answer: CStress shielding occurs when a rigid implant (like a plate) carries a disproportionate amount of the physiological load, thereby 'shielding' the underlying bone from mechanical stress. According to Wolff's Law, bone adapts to the loads placed upon it. If the bone is shielded from stress, it can lead to disuse osteopenia, weakening of the bone, and potentially refracture after implant removal. This is a significant long-term concern with highly rigid plate constructs, particularly locking plates, as the bone becomes accustomed to not bearing its full load. Options A, B, D, and E describe other potential issues or general biomechanical concepts, but do not directly define the phenomenon of stress shielding and its impact on bone health.

Question 60

Topic: Biomechanics & Biomaterials

A surgeon uses a cannulated screw system for a femoral neck fracture. What is the primary advantage of cannulation?

. To allow for easier removal of the screw in the future.
. To reduce the overall weight of the implant, minimizing stress shielding.
. To permit precise screw placement over a guide wire.
. To allow for simultaneous injection of bone cement.
. To enhance screw-bone interface for better purchase.

Correct Answer & Explanation

. To permit precise screw placement over a guide wire.


Explanation

Correct Answer: CCannulated screws have a hollow central channel that allows them to be inserted over a pre-placed K-wire or guide wire. This is a significant advantage, particularly in fractures where precise screw placement is critical (e.g., femoral neck, scaphoid, malleoli). The K-wire is first inserted under fluoroscopic guidance to ensure optimal position, and then the cannulated drill and screw are advanced over it, ensuring accurate screw trajectory without repeated attempts that can compromise bone quality.