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Flexor tendon injury is frequently encountered by the hand surgeon. Surgical repair of these tendons is a technical challenge, and is often accompanied by temporary immobilization of the finger to avoid tendon rupture. Different surgical techniques have been developed for flexor tendon repair using sutures[
Different conventional knotted repair technique
In 1967 McKenzie developed a 3-0 nylon multi strand suture with protruding barbs to anchor the tendon
The two types of Mckenzie's suture with protruding barbs
This new knotless technique exhibited multiple advantages. First, the load is not concentrated on the knot but instead dispersed over an extended area of the tendon[
Recently, the application of barbed sutures for tenorraphy has been of great interest. Previous research focused mainly on the mechanical properties of sutures[
Cut depth and cut angle of a barbed suture
In addition to cut angle and depth, we will discuss a third geometrical factor: the cross sectional configuration of the sutures and their effect on suture strength.
The finite element analysis of the barbed suture was implemented on a commercially available package: ABAQUS 6.14. The geometric model consists of a three-dimensional deformable cylinder with an elliptical cross-sectional area to represent the suture, whereas the barb is introduced as a cell partition. Simulations were performed while varying three main factors: aspect ratio of the ellipse (ρ), cut angle and cut depth.
Initial studies relied on geometric dimensions of commercially available barbed sutures in order to gain insight regarding their performance. Existing sutures have a circular cross-sectional area[
The main motivation behind varying the aspect ratio of the suture arises from the elliptical nature of the flexor tendons cross-section. It is anticipated that better conformity between the suture and the tendon is achieved when both possess similar aspect ratios. Widths and heights of Flexor Digitorum Superficialis and Profundus tendons (FDS and FDP respectively) for the different fingers were obtained from the literature[
Dimensions of flexor digitorum superficialis and profondus tendons
Index | Middle | Ring | Little | ||
---|---|---|---|---|---|
FDST | Major axis ( |
2.5 | 3.25 | 2.75 | 1.5 |
Cross-sectional area (Area) (mm^{2}) | 8.36 | 11.54 | 10.46 | 4.04 | |
Minor axis ( |
1.064 | 1.13 | 1.21 | 0.86 | |
Aspect ratio (ρ = a/b) | 2.35 | 2.88 | 2.27 | 1.75 | |
FDPT | Major axis ( |
2.5 | 3 | 2.5 | 2.25 |
Cross-sectional area (Area) (mm^{2}) | 11.4 | 14.44 | 13.42 | 8.84 | |
Minor axis ( |
1.452 | 1.532 | 1.708 | 1.25 | |
Aspect ratio ( |
1.722 | 1.96 | 1.46 | 1.8 |
The dimensions of the smallest tendon should be taken as the reference design such that the developed suture is suitable for use on all flexor tendons. As
Based on these findings, the values of “ρ” were varied between 1/3 and 3. The effect of the aspect ratio on the strength of the suture was investigated for constant cut angle and cut depth. Also, three different cut depths and cut angles: 0.07-mm, 0.12-mm, and 0.19-mm as depths and 150 154 and 160 as angles were studied for a constant “ρ”.
Furthermore, the radius of the suture (or the equivalent ellipse axes) was increased from 0.3 mm to a larger diameter that is suitable with both the dimensions of the smallest tendon (FDS of little finger) and the optimum aspect ratio. It is desired that the optimized suture accommodates about 30% of the tendon’s cross-sectional area. The diagram in
Flowchart clarifying the plan followed throughout the analysis
The values for diameter, cut angle, and cut depth were set for a circular cross-sectional area (CSA). When considering elliptical profiles, the major and minor axes were determined such that the area of the ellipse is equal to that of the circle. Furthermore, while the cut angles remain the same, the cut depths should be adapted for the ellipses in order to retain the same amount of intact area after inducing a cut in the sutures. The performed calculations are detailed below.
Finding major and minor axes:
and for an ellipse,
therefore,
To find equivalent cut depth, the remaining area of the ellipse
Diagram of an ellipse with a cut on its major axis
(
Where “a” and “b” are the major and minor axes respectively, and “d” is the distance from the centre of the cut (d = a - cut depth).
Since the remaining area of the ellipse should be equal to that of the circle, Equation (1) was used with a = b = r (radius of the circle) along with the desired cut distance “d” in order to determine the remaining area of the circle. Once the later was known, it was set as the remaining area of the ellipse and the equation was solved for the equivalent “d”. The corresponding cut depth is simply found by subtracting the major axis from the cut distance; that is, cut depth = a - d
Ellipse equivalent dimensions and cut depth
Ellipse equivalence | |||||
---|---|---|---|---|---|
Aspect ratio (ρ = a/b) | 1/3 | 1/2 | 1 | 2 | 3 |
Major axis a (mm) | 0.1732 | 0.2121 | 0.3 | 0.4243 | 0.5196 |
Minor axis b (mm) | 0.5196 | 0.4243 | 0.3 | 0.2121 | 0.1732 |
Cross-sectional area (Area) (mm^{2}) | 0.2827 | 0.2827 | 0.2827 | 0.2827 | 0.2827 |
Remaining area (RA) (mm^{2}) | 0.2059 | 0.2059 | 0.2059 | 0.2059 | 0.2059 |
Cut distance d (mm) | 0.0635 | 0.0778 | 0.11 | 0.1556 | 0.1905 |
Cut depth (mm) | 0.1097 | 0.1344 | 0.19 | 0.2687 | 0.3291 |
For simplification purposes, a single barb was included in the model with five different geometries used for this experimental analysis
Different geometries used in modeling
The material properties of polypropylene which correspond to the material usually used for flexor tendon repair were obtained after performing tensile tests to failure on 12 different suture specimens. This allowed us to determine polypropylene’s short-term linear elastic behavior, which requires the definition of an instantaneous Young’s Modulus and a Poisson’s ratio
Elastic properties of propylene
Mean | Standard Deviation | |
---|---|---|
Young’s Modulus (MPa) | 1668.67 | 573.37 |
Poisson’s ratio | 0.36878 | 0.06347 |
Tensile strength (MPa) | 353.15 | 48.81 |
The time-dependent behavior, on the other hand, was specified as previously described, via inputting into ABAQUS normalized shear relaxation data with respect to the initial stress along with the respective time instants as obtained from the relaxation tests. ABAQUS then performs a curve fitting analysis to determine the prony series coefficients. Computed values for the normalized shear
Shear relaxation curve from experimental data (red) and ABAQUS curve fitting (blue)
Prony series coefficients for polypropylene obtained from ABAQUS
I | g_{i} | k_{i} | t_{i} |
---|---|---|---|
1 | 0.31542 | 0 | 0.63594 |
2 | 0.32027 | 0 | 10.243 |
It is noteworthy to point out that the relaxation times gotten from experimentation were scaled by 100 when inputted into ABAQUS such that the total time becomes 36 s instead of 3600 s (60 min). This was done in order to avoid lengthy simulations and to be able to capture polypropylene’s relaxation behavior within the model.
As for damage initiation, the maximum principal stress was chosen[
Applied to polypropylene
For a pre-existing crack, damage evolution is much more important for the analysis. Based on the principles of LEFM, a power law mixed mode energy-based criterion was selected with the power set as 1. The critical Mode I, Mode II, and Mode III energy release rates (G_{Ic}, G_{IIc}, and G_{IIIc} respectively) should be specified. These values were obtained as critical stress intensity factors K_{Ic} and K_{IIc} form the literature[
Where E is the Young’s Modulus and
For
and
Or alternatively,
For Mode III, it was assumed that the critical energy release rate G_{IIIc} was equal to that of mode II due to the lack of fracture toughness data for this out-of-plane shear. As such, G_{Ic}, G_{IIc}, and G_{IIIc} were set to be equal to 15.663
For damage stabilization, viscous regularization was used to specify viscosity coefficients[
ALLSE and ALLVD for the whole model with respect to time
For the definition of the crack in ABAQUS[
Defining the crack using XFEM
As aforementioned, XFEM based on VCCT was implemented as the fracture criterion to model crack propagation. A contact interaction property was created with energy-based mixed mode power law. Maximum tangential shear stress was chosen to be the direction of crack growth. A tolerance of 0.05 and viscosity of 1E-005 were specified. Tolerance for unstable crack propagation was toggled on and kept at its default value of 0.2.
Mode I, Mode II, and Mode III critical energy release rates were given the same values as those previously specified for the damage evolution criterion. As for the powers
Input for mixed-mode power law
Mode I | Mode II | Mode III | a_{m} | a_{n} | a_{o} |
---|---|---|---|---|---|
15.663 | 3.0323 | 3.0323 | 1 | 2 | 1 |
In order to visualize a crack when using XFEM analysis, PHILSM and PSILSM should be requested as output. PSILSM is required to view the initial crack front, while PHILSM is necessary to observe the location of the crack. Also, the output STATUSXFEM should be chosen in order to perceive the status of the enriched elements. A value of 0 for this output means that the element is undamaged, a value of 1 means that the element has been completely cut by the crack, and a value between 1 and 0 signifies that the element is damaged but some traction forces remain.
The performance of the barbed suture was investigated under two loading conditions. The first case studies the strength of the suture and simulates a tensile test whereby one end is held fixed while the other is subjected to a finite displacement. The second case investigates the strength of a single barb. The boundary conditions are similar to those of the first load case, but in this analysis, the surface of the barb is constrained in the direction of application of the displacement[
Three loading situations were performed. In loading case 1, the right extremity was fixed by an ENCASTRE boundary condition applied in the initial step and propagated to the following Static and Visco steps. As for the left edge, a displacement/rotation boundary condition was specified at the surface. The displacement in z-direction, U3 was set to 0 initially. However, it was modified in the first Static step to a ramped displacement of magnitude 0.5-mm and was propagated to the subsequent Visco step
Boundary conditions for load case 1
Average stresses and strains in the body of the suture below the crack tip were of interest at the onset of failure. For this reason, an element set, Elem1, was created and associated with a Field Output Request to report strains and stresses in the z-direction, NE33 and S33 respectively
The position of Elem1 set highlighted in red
In loading case 2, the left extremity was fixed by an ENCASTRE boundary condition applied in the initial step and propagated to the following Static and Visco steps. As for the right edge, a displacement/rotation boundary condition was specified at the surface. The displacements in x, y, and z-directions, U1, U2, and U3 respectively as well as the rotations in these directions, UR1, UR2, and UR3 respectively were set to 0 initially. However, U3 was modified in the first Static step to a ramped displacement of magnitude 0.3-mm and was propagated to the subsequent Visco step. Furthermore, a third boundary condition was applied at the crack which was constrained in the direction of application of the displacement; roller in z-direction. Since the created partition to represent the barb was defined as a crack, loads and boundary conditions could not be defined exactly at the surface. For this reason, a new partition was created by offsetting the plane of the crack by 0.005-mm. The roller boundary condition was applied at that surface
Boundary conditions for load case 2
Since the main concern from applying this test case is to investigate the strength of a single barb, average stresses at the surface of the crack were monitored at the onset of failure. For this reason, an element set, BabrElem, was created and associated with a Field Output Request to report Von-mises stresses
The position of BarbElem set highlighted in red
Finally, loading case 3 was set in order to investigate the behavior of the suture when loads are applied to both, the body of the suture and the surface of the barb, so a third test case which combines the former test conditions was applied. The significance of this study lies in its resemblance to the real-life suture-tendon interaction. These loading conditions were considered for the final optimized geometry only.
After assessing the forces exerted in the different repaired FDS & FDP tendons during the five different protocols of rehabilitation
Peak forces recorded in FDP and FDS tendons during five rehabilitation exercises for two different wrist positions
Forces recorded | Place and hold | Active finger flexion | Isolated FDP flexion | Isolated FDS flexion | Tenodesis | ||||
---|---|---|---|---|---|---|---|---|---|
Wrist neutral | Wrist flexed | Wrist neutral | Wrist flexed | Wrist neutral | Wrist flexed | Wrist neutral | Wrist flexed | ||
Peak FDP forces (N) | |||||||||
Mean | 3.6 | 3.1 | 6.5 | 5.9 | 25.5 | 23.8 | 3.1 | 2.9 | 2.8 |
SD | 3.1 | 2.8 | 5.1 | 4.7 | 20.4 | 19.6 | 5.2 | 7.5 | 4.8 |
Maximum | 10.6 | 10.1 | 17.3 | 17.8 | 73.8 | 74.7 | 16.0 | 22.7 | 15.8 |
Peak FDS forces (N) | |||||||||
Mean | 4.9 | 7.7 | 2.9 | 3.5 | 4.3 | 4.2 | 12.9 | 14.1 | 2.7 |
SD | 2.6 | 5.6 | 6.7 | 12.9 | 3.1 | 5.1 | 6.4 | 8.0 | 1.0 |
Maximum | 10.9 | 23.7 | 25.6 | 47.5 | 12.9 | 20.0 | 24.2 | 32.9 | 4.6 |
A first factor of safety of 18% was added in order to withstand the tendon softening and the decrease in the repair strength during the early days after tendon repair and a second factor of safety of 30% was considered to compensate the risk of gap formation in the zone of repair. So the force of 25 N becomes 37 N and this was comforted by the literature which advices an ultimate repair strength of at least 30 to 55 N with a mean force of 40 N when an active rehabilitation maneuvers are involved[
So in loading case 3, the left extremity was subjected to a displacement/rotation boundary condition. The displacements in x, y, and z-directions, U1, U2, and U3 respectively as well as the rotations in these directions, UR1, UR2, and UR3 respectively were set to 0 initially. However, U3 was modified in the first Static step to a ramped displacement of magnitude 0.096-mm equivalent to 40N, as obtained from the load-displacement curve generated for the model on ABAQUS
Force-Displacement curve as obtained from ABAQUS
Boundary conditions for load case 3
The impact of the quality of the mesh on the correctness of the generated solution can never be overemphasized. A high-quality mesh is one that serves the objective of the model; obtain a solution with the needed accuracy by using only as many degrees of freedom as necessary.
The mesh is refined until a critical result, such as the maximum stress in a specific location, converges; that is, it does not change significantly with further refinement (h-method).
A mesh convergence analysis was performed for this study for the two loading cases. The investigation was carried out for the circular suture (ρ = 1) having a cut angle of 154, and a cut depth 0.19-mm.
In loading case 1, meshing was done using linear tetrahedral elements (C3D4). A mesh sensitivity analysis was carried to determine the optimum mesh size among the tested values: 0.015, 0.02, 0.025, 0.027, 0.03, 0.031, 0.035 and 0.037. The corresponding average strains at onset of failure for the set Elem1 for each mesh size are summarized in
Average strains at failure (%) detected for different mesh sizes of linear tetrahedral elements
Mesh convergence C3D4 | ||
---|---|---|
Mesh size | Number of elements | Failure strain (%) |
0.015 | 376,896 | 19.2 |
0.02 | 181,649 | 17.76 |
0.025 | 98,739 | 16.7 |
0.027 | 85,062 | 24 |
0.03 | 60,375 | 24.2 |
0.031 | 53,559 | 18.74 |
0.035 | 40,507 | 16.8 |
0.037 | 35,679 | 16.82 |
No significant trend was observed and convergence was obviously not attained. An error of 31% between the maximum and minimum values of the strain was noted. To resolve this problem, a local structured mesh of brick elements with reduced integration (C3D8R) was generated in the critical region near the crack tip while the global mesh type was kept as tetrahedral elements C3D4
Mesh for load case 1 consisting of structured brick elements near the crack
Again, a mesh sensitivity analysis was performed for the mesh sizes: 0.015, 0.022, 0.024, 0.025, 0.027, 0.03, 0.035, 0.04 and 0.05. For each mesh size, the corresponding average strain at onset of failure was recorded with the results summarized in
Average strains at failure (%) detected for different mesh sizes of mixed brick and tetrahedral elements
Mesh convergence C3D8R | ||
---|---|---|
Mesh size | Number of elements | Failure strain (%) |
0.015 | 469,222 | 19.45 |
0.022 | 174,020 | 18.65 |
0.024 | 134,521 | 17.8 |
0.025 | 124,261 | 18 |
0.027 | 98,441 | 18 |
0.03 | 80,790 | 19.12 |
0.035 | 46,826 | 20.3 |
0.04 | 32,946 | 19.5 |
0.05 | 16,527 | 18.8 |
Finally, a converged mesh has been obtained with a maximum error of 12% between the maximum and minimum values of the strain. A mesh size of 0.025 (124,261 elements) was chosen, taking into consideration both accuracy of the results and the required simulation time.
In loading case 2, in order to avoid the difficulties faced in loading case 1 particularly the possibility of nodes connections forming new elements instead of being deleted at the crack site, linear tetrahedral elements (C3D4) were used to mesh the whole model. For the mesh sizing, a mesh convergence study was carried out considering the following sizes; 0.02, 0.025, 0.03 and 0.035.
For each mesh size, the corresponding average maximum principal stress at the surface of the barb when a crack started to propagate (BarbElem) was recorded
Average maximum principal stress (MPa) at failure detected for different mesh sizes of linear tetrahedral elements
Mesh convergence C3D4 | ||
---|---|---|
Mesh size | Number of elements | Maximum principal stress (MPa) |
0.02 | 183,967 | 32.5 |
0.025 | 108,708 | 29.9 |
0.03 | 66,309 | 31.59 |
0.035 | 41,139 | 32.5 |
A converged mesh has been attained with a maximum error of 8% between the maximum and minimum values of the maximum principal stress and a mesh size of 0.025 (108,708 elements) was chosen for the same reasons as in loading case 1
Mesh for load case 3 consisting of linear tetrahedral elements
The results of the finite element of the models detailed above subjected to load cases 1 and 2 are reported. The attained optimized suture geometry is then tested under the loading case 3.
At first, the aspect ratios were varied for a constant cross-sectional area equivalent to that of a circle having a radius of 0.3-mm, a constant cut angle (CA) of 154 and cut depth (CD) of 0.19-mm for both loading conditions. The second step was to investigate the effect of varying CA and CD for a constant.
With constant cut angle and cut depth, in the loading case 1, an ENCASTRE and a displacement/rotation boundary conditions were applied to the suture geometry as thoroughly described in the boundary conditions and loads section. Stresses and strains in the z-direction, S33 and NE33 respectively, in the body of the suture (Elem1) were reported at the onset of crack propagation and compared among the different cross-sections
Average stresses (MPa) and strains (%) at failure for load case 1, constant CSA, CA = 154°, CD = 0.19 mm
Results for load case 1 | ||
---|---|---|
Elem 1 | ||
Aspect ratio (ρ) | NE33 (%) | S33 (MPa) |
1/3 | 19.23 | 280 |
1/2 | 18.65 | 275.92 |
1 | 17.98 | 262 |
2 | 18.05 | 264 |
3 | 18.1 | 265 |
So, changing the aspect ratio can affect the strength of the suture. The highest strains and stresses are observed for aspect ratio 1/3. As such, ρ = 1/3 is the strongest, followed by 1/2, 3, 2, and finally 1. A circular suture is found to be the weakest among all configurations. However, since the differences in the magnitudes of NE33 and S33 are not significant, the results cannot be considered conclusive. No clear trend is detected.
In this circular cross-sectional model, the direction of propagation of the crack is almost perpendicular to the applied displacement and the rupture failure is spotted. Similar behaviors are observed for all other aspect ratios
Initial situation of the suture under load case 1
A propagating crack in a circular suture subjected to load case 1
In loading case 2, the results reveal an evident trend
Average stresses (MPa) and their corresponding forces at failure for load case , constant CSA, CA = 154°, CD = 0.19 mm
Results for load case 2 | |||||
---|---|---|---|---|---|
Barb area (mm^{2}) | BarbElem | ||||
Aspect ratio (ρ) | a (mm) | b (mm) | Area (mm^{2}) | Von-mises (MPa) | Force (N) |
1/3 | 0.48343 | 0.250245 | 0.19 | 49 | 9.31 |
1/2 | 0.39474 | 0.306407 | 0.19 | 45.09 | 8.57 |
1 | 0.27911 | 0.433423 | 0.19 | 39.25 | 7.46 |
2 | 0.19732 | 0.612951 | 0.19 | 36 | 6.84 |
3 | 0.16115 | 0.750711 | 0.19 | 34.5 | 6.56 |
The direction of crack propagation for a circular cross-section model is being almost parallel to the applied displacement. Under such boundary conditions, peeling rather than rupture failure is detected. All other aspect ratios behave in a similar manner
A propagating crack in a circular suture subjected to load case 2
The results obtained for constant cross-sectional area, constant cut angle and cut depth for both loading cases 1 & 2, cannot offer a decisive conclusion regarding the best cross-sectional configuration. A mere 7% increase in %NE33 is observed when comparing the weakest suture (ρ = 1) with the strongest one (ρ = 1/3). Also, it is not clear which condition leads to an increased suture strength, decreasing the aspect ratio below one or increasing it. On the other hand, a 30% rise in the barb force is detected for LC2 when changing the aspect ratio from 3 to 1/3.
So since the aspect ratio ρ of the tendon is 2, a suture with an aspect ratio of 1/2, given the findings up till this point, appears to be most suitable. Furthermore, since the smallest major and minor axes among the tendons are a = 1.5-mm and b = 0.858-mm respectively, the radius (or the equivalent ellipse axes) of the suture can be increased from 0.3 mm to a larger diameter that is suitable with both the dimensions of the smallest tendon and the optimum aspect ratio. It is desired that the optimized suture accommodates about 30% of the tendon’s cross-sectional area. Consequently, a suggested geometry has the following characteristics: ρ = 1/2, a = 0.25-mm and b = 0.5-mm.
The results of a constant suture aspect ratio , with different cut angles and cut depths are reported below. A circular cross-section was used with a radius of 0.3536-mm equivalent to an ellipse having a = 0.25-mm and b = 0.5-mm. While the cut angles remain the same, the cut depths were adapted for the ellipse
Equivalent dimensions and cut depths for ρ = 1/2
Ellipse equivalence | |||
---|---|---|---|
Aspect ratio | |||
1/2 | 1 | ||
Major axis a (mm) | 0.25 | 0.3536 | |
Minor axis b (mm) | 0.5 | 0.3536 | |
Cross-sectional Area (CSA) (mm^{2}) | 0.3928 | 0.3928 | |
CD = 0.07 mm | Remaining Area (RA) (mm^{2}) | 0.3726 | 0.3726 |
Cut distance d (mm) | 0.2005 | 0.2836 | |
Cut depth (mm) | 0.0495 | 0.07 | |
CD = 0.12 mm | Remaining Area (RA) (mm^{2}) | 0.3485 | 0.3485 |
Cut distance d (mm) | 0.16515 | 0.2336 | |
Cut depth (mm) | 0.0849 | 0.12 | |
CD = 0.19 mm | Remaining Area (RA) (mm^{2}) | 0.3078 | 0.3078 |
Cut distance d (mm) | 0.11565 | 0.1636 | |
Cut depth (mm) | 0.1344 | 0.19 |
In loading case 1, the boundary conditions were applied to a cylinder with a circular cross-section of r = 0.3536-mm for the cut angles 150, 154, and 160, and cut depths 0.07-mm, 0.12-mm and 0.19-mm. The outputs of interest are the same as before for the same load case
Average stresses (MPa) and strains (%) at failure for load case 1, ρ = 1 for varying cut angles and cut depths
Results for load case 1, ρ = 1, r = 0.3536 mm | |||
---|---|---|---|
Elem 1 | |||
Cut angle (o) | Cut depth (mm) | NE33 (%) | S33 (MPa) |
160 | 0.19 | 21.55 | 306 |
0.12 | 24 | 325 | |
0.07 | 29.63 | 384 | |
154 | 0.19 | 18.1 | 264 |
0.12 | 21.18 | 297 | |
0.07 | 26.25 | 350 | |
150 | 0.19 | 17 | 250 |
0.12 | 20.05 | 285 | |
0.07 | 25.06 | 336 |
For a constant cut angle, decreasing the cut depth leads to stronger sutures that fracture at higher average strains. This result is expected since a smaller cut depth is synonymous with a smaller crack. On the other hand, for a constant cut depth, increasing the cut angle results in stronger geometries. From this analysis, the best configuration is CA = 160 and CD = 0.07-mm.
In loading case 2, the analysis was performed for a circular suture of r = 0.3536-mm for the cut angles and cut depths 150, 154, and 160, and 0.07-mm, 0.12-mm and 0.19-mm respectively. The outputs of interest are the same as before for the same load case
Average stresses (MPa) and their corresponding forces at failure for load case 2, ρ = 1 for varying cut angles and cut depths
Results for load Case 2, ρ = 1, r = 0.3536 mm | ||||||
---|---|---|---|---|---|---|
Barb area (mm^{2}) | BarbElem | |||||
Cut angle (o) | Cut depth (mm) | a (mm) | b (mm) | Area (mm^{2}) | Von-mises (MPa) | Force(N) |
160 | 0.19 | 0.313477 | 0.555523 | 0.2735 | 37.76 | 10.33 |
0.12 | 0.265451 | 0.350857 | 0.1463 | 47.39 | 6.93 | |
0.07 | 0.211197 | 0.204666 | 0.0679 | 54.9 | 3.73 | |
154 | 0.19 | 0.313477 | 0.433423 | 0.2134 | 39.08 | 8.34 |
0.12 | 0.265451 | 0.273741 | 0.11414 | 41.22 | 4.7 | |
0.07 | 0.211197 | 0.159682 | 0.053 | 56.9 | 3.02 | |
150 | 0.19 | 0.313477 | 0.38 | 0.18712 | 36.035 | 6.74 |
0.12 | 0.265451 | 0.24 | 0.1 | 43.68 | 4.368 | |
0.07 | 0.211197 | 0.14 | 0.0464 | 58.42 | 2.71 |
For a constant cut angle, decreasing the cut depth results in a weaker barb that can withstand lower forces. On the other hand, for a constant cut depth, increasing the cut angle results in stronger barbs. This is expected, for when both CA and CD increase, the barb area increases as well, allowing it to endure higher forces. From this analysis, the best configuration is CA = 160 and CD = 0.19-mm
With respect to aspect ratios, it was concluded that ρ = 1/3 presents the strongest design as compared to other aspect ratios. Taking into consideration the tendon’s cross-section, it was decided that ρ = 1/2 is most suitable.
Concerning cut angles and cut depths, it was found from the load case 1, that the best configuration corresponds to CA = 160 and CD = 0.07-mm. For the second test case, CA = 160 and CD = 0.19-mm were found to offer the best results. As such, a cut angle of 160 is evidently superior. However, a decision on the cut angle is not direct. A middle value between 0.19-mm and 0.07-mm was chosen; that is, 0.12-mm.
Therefore, the optimum geometry has an aspect ratio ρ = 1/2, CA = 160 and CD = 0.12-mm. Load cases 1 and 2 were performed again for this optimized geometry
Results for the optimized geometry for load cases 1 and 2
Results for ρ = 1/2, CA = 160, CD = 0.12 mm | |||
---|---|---|---|
Loading case 1 | Loading case 2 | ||
Elem 1 | BarbElem | ||
NE33 (%) | S33 (MPa) | Von-mises (MPa) | Force (N) |
26.88 | 362 | 52.69 | 7.71 |
Comparing the results with those obtained for a circular cross-section
Results for ρ = 1, CA = 160°, CD = 0.12 mm subjected to both load cases 1 and 2
Results for ρ = 1, CA = 160, CD = 0.12 mm | |||
---|---|---|---|
Loading case 1 | Loading case 2 | ||
Elem 1 | BarbElem | ||
NE33 (%) | S33 (MPa) | Von-mises (MPa) | Force (N) |
24 | 325 | 47.39 | 6.93 |
Since the values for both cut angle and cut depth were varied for a circular cross-section, the observed trend was to be validated for ρ = 1/2. The test was performed once for a constant CA of 160 and two different values for CD: 0.19-mm and 0.12-mm, and another time for a constant CD of 0.19-mm while changing the CA from 160 to 154.
For a constant cut depth, decreasing the cut angle resulted in a weaker suture and a barb that can tolerate lower forces (decrease of 14% in %NE33 and 33% in the force), as observed for ρ = 1
Summary of results for ρ =1 and ρ =1/2, for varying cut angles and cut depths under both load cases
Loading case 1 | Loading case 2 | ||||
---|---|---|---|---|---|
Results for ρ = 1/2, a = 0.25 mm, b = 0.5 mm | |||||
Elem 1 | BarbElem | ||||
Cut angle (o) | Cut depth (mm) | NE33 (%) | S33 (MPa) | Von-mises (MPa) | Force (N) |
160 | 0.19 | 21.71 | 311 | 61. 9 | 16.93 |
0.12 | 26.88 | 362 | 52.69 | 7.71 | |
154 | 0.19 | 18.7 | 275 | 52.8 | 11.27 |
Results for ρ = 1, a = 0.3536 mm | |||||
160 | 0.19 | 21.55 | 306 | 37.76 | 10.33 |
0.12 | 24 | 325 | 47.39 | 6.93 | |
154 | 0.19 | 18.1 | 264 | 39.08 | 8.34 |
A loading case 3 was performed for the optimized cut angle of 160and cut depth of 0.12-mm, but for ρ = 1, since generating a working mesh for ρ = 1/2 was problematic under the prescribed boundary conditions.
Reporting the findings for this test case, a small difference was noticed in comparison with the second loading condition. A force of 6.26 N was obtained as opposed to 6.93 N (10% difference). The same trend is expected for ρ = 1/2 but most likely, the percent variation differs. However, this decrease in barb strength would be accounted for in safety factors when determining the number of barbs required to be anchored within the tendon
Results for optimized cut angle and cut depth for ρ = 1 for load cases 2 and 3
Results for ρ = 1, CA = 160, CD = 0.12 mm | |
---|---|
BarbElem | |
Von-mises (MPa) | Force (N) |
Loading case 3 | |
42.8 | 6.26 |
Loading case 2 | |
47.39 | 6.93 |
As aforementioned, an estimation of the number of barbs required to hold a tendon force equivalent to 40 N can be simply found by dividing this force by the one withstood by a single barb.
For ρ = 1/2, CA = 160, CD = 0.12-mm, the force at the barb surface was recorded to be 7.71 N. This value represents the force at which crack propagation initiates. Certainly, the barb should be loaded at forces lower than the threshold. And since for the third loading case, a value lower than 7.71 is anticipated, the force at the barb surface is assumed to be 5 N, 35% less. Taking into consideration a factor of safety of 3, the required number of barbs to be anchored in the tendon to withstand 40 N becomes:
Hence, 24 barbs are required. Note that this number is a mere estimate.
The mechanisms that led to the obtained results are discussed and the outcomes of the finite element analysis are not accepted at face value.
For a constant cut angle and cut depth, it was noticed that decreasing the aspect ratio below ρ = 1 resulted in stronger sutures, or more accurately, barbs. For this case, since the remaining areas of the ellipses are the same, the only variables are the cut length
With this in mind, we suspect that the main reason behind the observed results is the stress intensity factor. For Modes I and II,
Stresses intensity factors for Modes I, II and III
Stress intensity factors | |||
---|---|---|---|
Aspect ratio | KI [MPa (m)^{1/2}] | KII [MPa (m)^{1/2}] | KIII [MPa (m)^{1/2}] |
1/3 | 105 | -122 | -4.11 |
1/2 | 106 | -119 | -1.6 |
1 | 57.7 | -79.5 | -1.14 |
2 | 79.4 | -120 | 2.34 |
3 | 66.3 | -124 | -0.484 |
It is evident that mixed-mode crack growth takes place with a slight Mode II dominance. The observed trend for varying the aspect ratio is a result of the combination of the three Modes, governed by the mixed-mode power law. The resultant stress intensity factor, or more specifically, energy release rate is suspected to be highest for ρ = 3 followed by 2, 1, 1/2 and finally 1/3. This explains the decrease in suture/barb strength as the aspect ratio is varied from 1/3 to 3.
For a constant aspect ratio, two opposing behaviors were noted. A stronger suture is obtained for a decreasing cut depth but an increasing cut angle. However, with respect to barb strength, an increase in both cut angle and cut depth enhanced the performance of the barb.
Considering the first case, for a constant cut angle, decreasing the cut depth means that the crack size has been reduced (smaller barb), leading to an increased strength of the body of the suture. As for a constant cut depth, the increase in cut angle results in a longer cut length, which can be observed as the moment arm that translates the forces at the barb surface into compressive stresses at the crack tip
Plot of shear stresses (S23) for ρ = 1, CA = 154°, CD = 0.19 mm for load case 1. Notice the initiation of crack propagation mid crack length
For the second case, the logic is more straightforward. Increasing both the cut angle and cut depth leads to larger barb areas that can tolerate higher forces.
In conclusion, the extended finite element method has been successfully implemented on ABAQUS to predict crack growth in viscoelastic material. It was successfully used to assess the mechanical performance of barbed sutures for the repair of flexor tendon injuries and ultimately suggest a new optimized suture geometry for a novel single-stranded repair technique. Without doubt, the outcome of this work presents a solid foundation towards the use of a new barbed suture design for a novel single-stranded repair technique, but for a more comprehensive assessment, more investigations are still needed. For future studies, the use of material with superior mechanical properties, as polyglyconate instead of polypropylene, should be investigated in order to increase the suture strength. The different investigations for mechanical properties evaluation should be performed at 37 instead of doing them at room temperature, as the material is going to be implemented into the human body. The Numbers of barbs and their distribution on the suture should be assessed more accurately taking into consideration that not all barbs will carry the same load. Finally, the loading investigations should be performed on a suture with multiple barbs simultaneously instead of doing them on only one barb in order to assess the interaction between barbs.
Medical and surgical part of this work: Bakhach J, Oneissi A, Karameh R
Literature review, photos design and tables: Bakhach D
Biomechanical investigations and studies: Hiba K, Elie S
Materials and data come from within the organization; they follow the privacy policy and can be repeated again.
None.
All authors declared that there are no conflicts of interest.
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© The Author(s) 2018.