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Biomechanics & Biomaterials MCQs

Practice Set 6 of 88

This practice set contains high-yield board review questions covering key concepts in Biomechanics & Biomaterials. Each clinical scenario is designed to test your diagnostic and management skills relevant to this subspecialty.

Question 101

Topic: Biomechanics & Biomaterials

A 30-year-old patient with a simple transverse midshaft femoral fracture is treated with a stainless steel plate. If a titanium plate of identical geometry were used instead, what would be the primary biomechanical difference relevant to bone healing?

. The titanium plate would provide greater absolute stiffness, leading to faster healing.
. The titanium plate would have a higher fatigue life, reducing implant failure.
. The titanium plate would cause less stress shielding due to a lower modulus of elasticity.
. The titanium plate would be more resistant to corrosion in the biological environment.
. The titanium plate would allow for greater interfragmentary compression at the fracture site.

Correct Answer & Explanation

. The titanium plate would cause less stress shielding due to a lower modulus of elasticity.

Explanation

Correct Answer: CThe primary biomechanical difference between stainless steel and titanium plates of identical geometry, relevant to bone healing, lies in their modulus of elasticity. Titanium and its alloys have a modulus of elasticity closer to that of cortical bone (approximately 110 GPa for titanium vs. 200 GPa for stainless steel vs. 17-20 GPa for cortical bone). A lower modulus of elasticity means the implant is less stiff. When a less stiff implant is used, it shares more load with the bone, allowing more physiological stress to be transmitted to the healing fracture. This reduces the phenomenon of 'stress shielding,' where a very stiff implant carries too much load, leading to disuse osteoporosis in the adjacent bone and potentially delayed healing. While titanium is indeed more corrosion-resistant (Option D) and has excellent biocompatibility, its lower modulus of elasticity is the key biomechanical advantage in terms of load sharing and stress shielding. Fatigue life (Option B) can be complex and depends on many factors, but titanium generally has good fatigue properties. Neither material inherently provides greater stiffness (Option A) or compression (Option E) for the same geometry; rather, it's the relative stiffness to bone that is critical.

Question 102

Topic: Biomechanics & Biomaterials

During a femoral nailing, a surgeon elects to use a 12 mm solid intramedullary nail instead of a 10 mm solid nail of the exact same material. The torsional rigidity of the 12 mm nail is increased by approximately what factor compared to the 10 mm nail?

. 1.2
. 1.4
. 2.1
. 2.5
. 4.0

Correct Answer & Explanation

. 2.1

Explanation

The torsional rigidity of a solid cylinder is proportional to its polar moment of inertia, which scales with the radius to the fourth power (r^4). Calculating (6^4 / 5^4) yields 1296 / 625, which is an increase by a factor of approximately 2.07.

Question 103

Topic: Biomechanics & Biomaterials

A 25-year-old male sustains a highly comminuted tibia fracture from a high-speed motorcycle crash. Bone fractures with significantly more comminution at high loading rates compared to low-speed falls due to which biomechanical property of bone?

. Anisotropy
. Viscoelasticity
. Osteoconduction
. Creep
. Stress relaxation

Correct Answer & Explanation

. Viscoelasticity

Explanation

Bone is a viscoelastic material, meaning its mechanical properties change based on the rate of loading. At higher strain rates, bone becomes stiffer, absorbs more energy before failure, and ultimately releases that energy explosively, causing high comminution.

Question 104

Topic: Biomechanics & Biomaterials

During open reduction of a clavicle fracture, the surgeon extensively bends a reconstruction plate back and forth to match the bone's S-shape. This repetitive plastic deformation increases the risk of implant failure primarily by inducing which material phenomenon?

. Stress shielding
. Creep
. Cold working (strain hardening)
. Galvanic corrosion
. Viscoelastic relaxation

Correct Answer & Explanation

. Cold working (strain hardening)

Explanation

Repeated bending of a metal plate beyond its yield point causes cold working (strain hardening), which increases the metal's brittleness and decreases its fatigue life. This makes the implant highly susceptible to early fatigue failure before the fracture heals.

Question 105

Topic: Biomechanics & Biomaterials

A patient undergoes ORIF of a bimalleolar ankle fracture. A stainless steel plate is applied to the fibula, but titanium screws are mistakenly used to secure it. This mixing of dissimilar metals places the construct at highest risk for which complication?

. Fretting wear
. Galvanic corrosion
. Stress shielding
. Early fatigue failure
. Decreased screw pullout strength

Correct Answer & Explanation

. Galvanic corrosion

Explanation

Galvanic corrosion occurs when two dissimilar metals with different anodic indices are placed in electrical contact within an electrolytic environment (like body fluid). This accelerates corrosion of the less noble metal, potentially leading to implant failure or localized tissue toxicity.

Question 106

Topic: Biomechanics & Biomaterials

A retrieved broken femoral plate from a nonunion is subjected to mechanical testing. Analysis shows the plate failed due to repetitive sub-maximal loading at stress levels well below its ultimate tensile strength. This failure mode is defined as:

. Yield failure
. Fatigue failure
. Creep
. Brittle failure
. Ductile failure

Correct Answer & Explanation

. Fatigue failure

Explanation

Fatigue failure occurs when a material undergoes repetitive cyclic loading at stress levels below its yield point or ultimate tensile strength, eventually leading to microscopic crack propagation and macroscopic failure. It is a common cause of implant failure in nonunions.

Question 107

Topic: Biomechanics & Biomaterials

Which of the following modifications to a cortical screw will most significantly increase its resistance to bending forces (bending stiffness)?

. Increasing the outer thread diameter
. Increasing the inner core diameter
. Increasing the thread pitch
. Decreasing the screw length
. Decreasing the inner core diameter

Correct Answer & Explanation

. Increasing the inner core diameter

Explanation

The bending stiffness of a screw is proportional to the radius of its core (inner) diameter to the fourth power. Therefore, increasing the core diameter drastically increases its bending strength and resistance to fatigue failure.

Question 108

Topic: Biomechanics & Biomaterials

Cortical bone is a viscoelastic material. How does its biomechanical behavior change when it is loaded at a very high rate, such as during high-energy trauma?

. It becomes more ductile and yields at lower forces
. It exhibits decreased ultimate tensile strength
. It becomes stiffer and can absorb more energy before failure
. Its modulus of elasticity decreases significantly
. It undergoes plastic deformation earlier

Correct Answer & Explanation

. It becomes stiffer and can absorb more energy before failure

Explanation

As a viscoelastic material, cortical bone becomes stiffer and stronger when loaded rapidly. This allows it to absorb more energy before fracturing during high-velocity impacts, which typically leads to more comminuted fracture patterns upon failure.

Question 109

Topic: Biomechanics & Biomaterials

A 45-year-old patient undergoes ORIF using a stainless steel plate and titanium screws. What specific biomechanical or biomaterial complication is most likely to occur at the screw-plate interface over time?

. Crevice corrosion
. Galvanic corrosion
. Fretting corrosion
. Stress shielding
. Aseptic loosening secondary to high modulus mismatch

Correct Answer & Explanation

. Galvanic corrosion

Explanation

Galvanic corrosion occurs when two dissimilar metals (such as titanium and stainless steel) are placed in direct physical contact within an electrolytic environment like body fluids. This electrochemical reaction leads to the rapid corrosion of the less noble metal.

Question 110

Topic: Biomechanics & Biomaterials

A rigid, thick plate is applied to the anterior femur. Years later, dual-energy x-ray absorptiometry reveals severe osteopenia of the underlying anterior femoral cortex. What biomechanical principle explains this phenomenon?

. Wolff's law driven stress shielding
. Galvanic corrosion
. Perren's strain theory
. Viscoelastic creep
. Fatigue failure

Correct Answer & Explanation

. Wolff's law driven stress shielding

Explanation

Stress shielding occurs when a highly rigid implant offloads the normal physiological stresses from the underlying bone. According to Wolff's law, bone remodels in response to mechanical stress; therefore, the lack of stress leads to localized osteopenia.

Question 111

Topic: Biomechanics & Biomaterials

Cortical screws typically have a smaller pitch compared to cancellous screws. What does the term "pitch" refer to in screw biomechanics?

. The angle of the screw thread relative to the core
. The distance between adjacent screw threads
. The difference between the outer diameter and core diameter
. The length of the non-threaded shaft
. The ratio of bone volume to metal volume between threads

Correct Answer & Explanation

. The distance between adjacent screw threads

Explanation

Screw pitch is defined as the linear distance traveled by the screw with one full 360-degree turn, which corresponds to the distance between adjacent threads. Cancellous screws have a larger pitch to capture more bone per turn in less dense trabecular bone.

Question 112

Topic: Biomechanics & Biomaterials

Which of the following accurately describes the typical failure mechanism of a locked plating construct under excessive axial load compared to a non-locked construct?

. Screws fail sequentially starting from the fracture site
. The plate undergoes friction-induced loosening
. Screws pull out sequentially from the far cortex
. The construct typically fails as a single unit via simultaneous screw pullout or plate breakage
. Failure occurs primarily through galvanic corrosion at the screw head

Correct Answer & Explanation

. The construct typically fails as a single unit via simultaneous screw pullout or plate breakage

Explanation

Because a locked plate acts as a single fixed-angle beam, the screws do not fail sequentially. Under excessive load, the entire construct typically fails simultaneously, often through catastrophic en masse screw pullout or plate breakage.

Question 113

Topic: Biomechanics & Biomaterials

Cortical bone exhibits different mechanical properties depending on the direction of the applied load. It is strongest in compression and weakest in shear. What is the biomechanical term for this property?

. Viscoelasticity
. Isotropy
. Anisotropy
. Hysteresis
. Plasticity

Correct Answer & Explanation

. Anisotropy

Explanation

Anisotropy refers to a material having different mechanical properties when loaded in different directions. Cortical bone is highly anisotropic, being strongest under longitudinal compression and weakest under transverse shear forces.

Question 114

Topic: Biomechanics & Biomaterials

A surgeon decides to use a titanium intramedullary nail instead of a stainless steel nail for a tibial shaft fracture. Which of the following accurately describes a key biomechanical difference between titanium alloy (Ti-6Al-4V) and 316L stainless steel?

. Titanium has a higher modulus of elasticity than stainless steel
. Titanium is substantially more rigid than stainless steel of the same dimensions
. Titanium has a modulus of elasticity closer to that of cortical bone
. Stainless steel exhibits greater notch sensitivity than titanium
. Titanium relies exclusively on its carbon content for fatigue resistance

Correct Answer & Explanation

. Titanium has a modulus of elasticity closer to that of cortical bone

Explanation

Titanium alloy has a lower modulus of elasticity than stainless steel, making it less rigid and closer to the modulus of elasticity of cortical bone. This property significantly reduces stress shielding at the fracture site compared to stiffer stainless steel implants.

Question 115

Topic: Biomechanics & Biomaterials
A 30-year-old male sustains a high-energy transverse femoral shaft fracture. The orthopedic surgeon is considering an intramedullary nail for fixation. To maximize the nail's resistance to bending and torsional forces, which design parameter, if increased, would yield the most significant biomechanical advantage?
. The nail's material Young's Modulus
. The nail's overall length
. The nail's outer diameter
. The nail's surface roughness
. The nail's ultimate tensile strength

Correct Answer & Explanation

. The nail's outer diameter

Explanation

The resistance of an intramedullary nail to bending and torsional forces is primarily determined by its Area Moment of Inertia (I) and Polar Moment of Inertia (J), respectively. For a circular cross-section, both I and J are proportional to the fourth power of the diameter. This means that even a small increase in the outer diameter of the nail leads to a disproportionately large increase in its bending and torsional stiffness.

Question 116

Topic: Biomechanics & Biomaterials

A biomechanical engineer is tasked with designing a new, lightweight, yet stiff, intramedullary implant for a long bone. Given the choice between a solid cylindrical design and a hollow cylindrical design, both made of the same material and having the same overall mass, which design would offer superior resistance to bending and torsion?

. The solid cylindrical design, due to its continuous material
. The hollow cylindrical design, by distributing material further from the neutral axis
. Both designs would offer equal resistance if their cross-sectional areas are identical
. The solid cylindrical design, if its length is minimized
. The hollow cylindrical design, only if its inner diameter is very small

Correct Answer & Explanation

. The hollow cylindrical design, by distributing material further from the neutral axis

Explanation

Correct Answer: BRationale:For a given amount of material (and thus mass), a hollow cylindrical design offers superior resistance to bending and torsion compared to a solid cylindrical design. This is because the Area Moment of Inertia (I) and Polar Moment of Inertia (J) are maximized when the material is distributed as far as possible from the neutral axis of bending or the central axis of torsion. A hollow cylinder achieves this by concentrating its mass at the periphery, leading to a significantly higher I and J for the same cross-sectional area or mass. This principle is why long bones are tubular.A) The solid cylindrical design, due to its continuous material:While continuous, the material near the neutral axis contributes very little to the MOI, making it less efficient for bending/torsion resistance compared to a hollow design of the same mass.C) Both designs would offer equal resistance if their cross-sectional areas are identical:If cross-sectional areas are identical, a hollow cylinder would have a larger outer diameter and thus a much higher MOI than a solid cylinder of the same area, making this statement incorrect.D) The solid cylindrical design, if its length is minimized:Minimizing length increases stiffness, but this is independent of the cross-sectional geometry's inherent MOI. The comparison is about the efficiency of the cross-section itself.E) The hollow cylindrical design, only if its inner diameter is very small:A hollow design is efficient even with a larger inner diameter, as long as the material is distributed peripherally. The key is thedistributionof material, not just a small inner diameter.

Question 117

Topic: Biomechanics & Biomaterials

A 55-year-old patient with a history of metastatic breast cancer develops a lytic lesion in the femoral diaphysis. This lesion significantly increases the risk of a pathological fracture. The primary biomechanical reason for this increased risk, related to the bone's geometry, is a reduction in the bone's:

. Young's Modulus
. Ultimate compressive strength
. Area Moment of Inertia
. Bone mineral density
. Trabecular bone volume

Correct Answer & Explanation

. Area Moment of Inertia

Explanation

Correct Answer: CRationale:A lytic lesion in the femoral diaphysis involves the destruction and removal of bone tissue. This directly reduces the effective cross-sectional area of the bone, particularly the cortical bone, and redistributes the remaining material closer to the neutral axis or eliminates it entirely. This geometric change leads to a dramatic reduction in the Area Moment of Inertia (I) at the lesion site. Since the bone's resistance to bending and torsional forces is directly proportional to its MOI, a reduced MOI makes the bone significantly weaker and highly susceptible to pathological fractures under normal physiological loads.A) Young's Modulus:Young's Modulus is a material property. While the quality of the remaining bone might be affected, the primary and most dramatic impact of a lytic lesion on structural integrity is geometric (MOI).B) Ultimate compressive strength:This is a material property. While the bone's material strength might be compromised, the geometric weakening (MOI reduction) is the predominant factor for overall structural failure.D) Bone mineral density:BMD is a measure of bone mass per unit volume. While a lytic lesion reduces BMD locally, the biomechanical consequence of this reduction, in terms of resistance to bending, is best captured by the Area Moment of Inertia.E) Trabecular bone volume:The femoral diaphysis is primarily cortical bone. While trabecular bone is present in metaphyses, a diaphyseal lytic lesion primarily affects cortical bone and its MOI.

Question 118

Topic: Biomechanics & Biomaterials

For a comminuted open tibia fracture, an external fixator is applied. To maximize the bending and torsional stiffness of the frame, which adjustment would be most effective?

. Increasing the number of pins per fragment
. Using smaller diameter pins
. Increasing the distance between the connecting rods and the bone axis
. Decreasing the length of the connecting rods
. Using a more flexible connecting rod material

Correct Answer & Explanation

. Increasing the distance between the connecting rods and the bone axis

Explanation

Correct Answer: CRationale:The stiffness of an external fixator frame is highly dependent on its geometric configuration, particularly the Area Moment of Inertia (I) of the overall construct. Increasing the distance between the connecting rods and the bone axis (i.e., making the frame larger) significantly increases the effective Area Moment of Inertia of the frame. This is because the resistance to bending and torsion is maximized when the structural elements are distributed further from the neutral axis. This leverage effect dramatically enhances the frame's bending and torsional stiffness, providing greater stability to the fracture.A) Increasing the number of pins per fragment:More pins can improve load sharing and pin-bone interface stability, but the geometric arrangement of the frame's main load-bearing elements (rods) relative to the bone has a more profound effect on overall frame stiffness.B) Using smaller diameter pins:Smaller diameter pins would decrease their individual Area Moment of Inertia, making them less stiff and potentially increasing pin bending and failure.D) Decreasing the length of the connecting rods:While shorter rods can increase stiffness (stiffness is inversely proportional to length cubed), this option refers to the length of the individual rods, not the distance from the bone axis, which is a more powerful determinant of overall frame MOI.E) Using a more flexible connecting rod material:A more flexible material (lower Young's Modulus) would decrease the stiffness of the connecting rods and thus the overall frame.

Question 119

Topic: Biomechanics & Biomaterials

A researcher is studying the biomechanics of a long bone during a twisting injury. Which specific moment of inertia is most relevant for quantifying the bone's resistance to this torsional (twisting) force?

. Mass Moment of Inertia
. Area Moment of Inertia
. Polar Moment of Inertia
. First Moment of Area
. Centroidal Moment of Inertia

Correct Answer & Explanation

. Polar Moment of Inertia

Explanation

Correct Answer: CRationale:ThePolar Moment of Inertia (J)is the geometric property that quantifies a cross-section's resistance to torsional (twisting) deformation. It is analogous to the Area Moment of Inertia (I) for bending. For a circular cross-section, J is proportional to the diameter to the fourth power (J ~ d4). Understanding J is crucial for analyzing how bones and implants resist twisting forces.A) Mass Moment of Inertia:This describes a body's resistance to changes in its rotational motion (angular acceleration), not its resistance to torsional deformation under a static or quasi-static twist. It involves the mass distribution of the entire body.B) Area Moment of Inertia:This (also known as the second moment of area) quantifies a cross-section's resistance to bending deformation, not torsional deformation.D) First Moment of Area:This is used to locate the centroid (neutral axis) of a cross-section and is relevant for shear stress calculations, but not directly for resistance to bending or torsion.E) Centroidal Moment of Inertia:This is a specific type of Area Moment of Inertia calculated about the centroidal axis. While related to bending, it is not the specific term for torsional resistance.

Question 120

Topic: Biomechanics & Biomaterials

A surgeon chooses a 'dynamic' plating technique for a comminuted humeral shaft fracture, aiming to promote secondary bone healing. This approach typically involves a construct that allows for controlled micro-motion at the fracture site. How is the Area Moment of Inertia (I) of the plate typically managed in such a strategy?

. The plate is designed with a maximal I to ensure absolute rigidity.
. The construct is designed to achieve a lower effective Area Moment of Inertia or a longer working length to decrease overall construct stiffness.
. Area Moment of Inertia is irrelevant, as only the material's Young's Modulus matters for dynamic healing.
. The plate's I is increased to compensate for a smaller number of screws.
. The plate's I is kept constant, but the screw design is altered for flexibility.

Correct Answer & Explanation

. The construct is designed to achieve a lower effective Area Moment of Inertia or a longer working length to decrease overall construct stiffness.

Explanation

Correct Answer: BRationale:Dynamic plating strategies aim to promote secondary bone healing by allowing controlled micro-motion at the fracture site. This requires a construct with relatively lower overall stiffness compared to rigid fixation. This lower stiffness is achieved by either using a plate with an intrinsically lower Area Moment of Inertia (e.g., a thinner or narrower plate) or, more commonly, by increasing the plate's working length (the unsupported segment of the plate bridging the fracture). Increasing the working length significantly reduces the construct's bending stiffness (stiffness is inversely proportional to the cube of the working length for a given plate MOI), thereby allowing the desired micro-motion.A) The plate is designed with a maximal I to ensure absolute rigidity:This describes a rigid fixation strategy, which aims for primary bone healing, not dynamic plating for secondary healing.C) Area Moment of Inertia is irrelevant, as only the material's Young's Modulus matters for dynamic healing:Both the material's Young's Modulus (E) and the plate's Area Moment of Inertia (I) contribute to bending stiffness (EI). MOI is highly relevant for controlling stiffness.D) The plate's I is increased to compensate for a smaller number of screws:Increasing MOI would increase stiffness, which is contrary to the goal of dynamic plating.E) The plate's I is kept constant, but the screw design is altered for flexibility:While screw design can influence construct flexibility, the primary method to achieve controlled micro-motion in plating is by adjusting the plate's MOI or working length.

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