Free-Body Diagram. Supports & Connections II. Supports & Connections I. Free-Body Diagram. Active & Reactive units.

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1 Free-Body Diagram Supports & Connections I Supports & Connections II Free-Body Diagram Active & Reactive units 1

2 Consistency & Inconsistency Dependent on the desirability of the force system developed, when a straight wire is tied into the bracket of the active and the reactive unit, force systems can be classified as consistent or inconsistent. Consistent force systems The force system developed when a wire is inserted in a bracket comprises forces and moments. If both the rotation generated by the moment and the translation resulting from the force are leading to the planned tooth movement, the force system is said to be consistent. Inconsistent force systems An inconsistent force system is unable to produce the desired combination of forces and moments. When the moment produced is desired, the force is undesirable and vice versa. Statically determinate systems A force system is called statically determinate when all unknown forces and moments in the force system can be calculated from the force and moment equilibrium formulas. The cantilever is the most important example of a statically determinate orthodontic appliance. Statically indeterminate systems A force system is called statically indeterminate when some of the unknown forces and moments in the force system cannot be calculated from the force and moment equilibrium formulas. Additional information should then be employed to determine these unknown quantities. The force system developed in a two-tooth segment when connected by a continuous straight wire is determined by the angles of the brackets with respect to the wire passing through the centers of the brackets and the interbracket distance. In the V-bend the forces and moments at the brackets cannot be calculated straight away. However using symmetry considerations, the forces must be zero and the moments opposite to one another. Burstone & Konig, AJO,

3 Geometry I θ A /θ B = 1.0 M A /M B = 1.0 Geometry II Geometry III θ A /θ B = 0.5 θ A /θ B = 0.0 M A /M B = 0.8 M A /M B = 0.5 Geometry IV Geometry V θ A /θ B = -0.5 θ A /θ B = M A /M B = 0.0 M A /M B =

4 Geometry VI θ A /θ B = -1.0 F A = F B = 0.0 M A /M B = -1.0 M B = (K * θ B ) / L, F = (M A + M B ) / L, with: K = (3.853 θ A /θ B ) * W s W s = M s * C s L - interbracket distance M s - material stiffness C s - cross-sectional stiffness Step-bend force systems Step-bends: forces Step-bends: moments 4

5 Step-bends: moments V-bends: forces V-bends: moments V-bends: moments a / L = 0.5 F A = F B = 0.0 M A = - M B 1 / 3 < a / L < 2 / 3 M A & M B opposite 5

6 a / L = 2 / 3 M B = 0; a / L = 1 / 3 M A = 0 a / L > 2 / 3 M A & M B same direction a / L < 1 / 3 M A & M B same direction V-bends in the occlusal plane Special loops 1. vertical loop 2. L-loop 3. rectangular loop 4. T-loop Force system L-loop influence of the vertical discrepancy Force system L-loop influence of the horizontal length G 6

7 Force system rectangular loop influence of the vertical discrepancy Force system rectangular loop influence of the horizontal length G Force system T-loop influence of the vertical discrepancy Force system T-loop influence of the horizontal length G 7

Shear Force and Moment Diagrams

C h a p t e r 9 Shear Force and Moment Diagrams In this chapter, you will learn the following to World Class standards: Making a Shear Force Diagram Simple Shear Force Diagram Practice Problems More Complex