Intramedullary pin alone: which forces it does not control
2-year-old Labrador with a short oblique fracture of the mid-femoral shaft. Repair with a single smooth intramedullary pin filling 70% of the medullary canal is being considered.
Which statement about the forces acting on this fracture is correct?
chevron_right Show answer and explanation
A. A pin filling 70% of the canal neutralises all forces
Filling the canal improves resistance to bending, but a smooth round pin still does not stop the fragments rotating or sliding over each other.
B. The pin resists axial compression well; its only weak point is bending
It is the other way round: bending is precisely the force it controls best. In an oblique fracture, axial load makes the fragments slide and the fracture collapses.
C. The pin controls bending but not rotation or axial compression; it needs an adjunct or a different implant
Correct. To neutralise rotation and axial load it is combined with an external fixator (tie-in) or a plate (plate-rod), or a plate or interlocking nail is chosen instead.
D. Adding one cerclage wire around the short oblique fracture neutralises rotation
Cerclage wires only work in long oblique fractures (length at least twice the bone diameter), always two or more and tightly applied. On a short oblique fracture they slip and loosen.
lightbulb Key takeaway
Bending, rotation, axial compression and tension act on a fracture. The chosen implant must neutralise all of them. The typical failure of an intramedullary pin alone is rotational instability, leading to delayed union or non-union.
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Updated: October 31, 2026 · Educational content for veterinary professionals. It does not replace current clinical guidelines or the judgement of the attending veterinarian.