Role of uterine forces in intrauterine device embedment, perforation, and expulsion

Abstract

Background

The purpose of this study was to examine factors that could help reduce primary perforation during insertion of a framed intrauterine device (IUD) and to determine factors that contribute in generating enough uterine muscle force to cause embedment and secondary perforation of an IUD. The objective was also to evaluate the main underlying mechanism of IUD expulsion.

Methods

We compared known IUD insertion forces for “framed” devices with known perforation forces in vitro (hysterectomy specimens) and known IUD removal forces and calculated a range of possible intrauterine forces using pressure and surface area. These were compared with known perforation forces.

Results

IUD insertion forces range from 1.5 N to 6.5 N. Removal forces range from 1 N to 5.8 N and fracture forces from 8.7 N to 30 N depending upon device. Measured perforation forces are from 20 N to 54 N, and calculations show the uterus is capable of generating up to 50 N of myometrial force depending on internal pressure and surface area.

Conclusion

Primary perforation with conventional framed IUDs may occur if the insertion pressure exceeds the perforation resistance of the uterine fundus. This is more likely to occur if the front end of the inserter/IUD is narrow, the passage through the cervix is difficult, and the procedure is complex. IUD embedment and secondary perforation and IUD expulsion may be due to imbalance between the size of the IUD and that of the uterine cavity, causing production of asymmetrical uterine forces. The uterine muscle seems capable of generating enough force to cause an IUD to perforate the myometrium provided it is applied asymmetrically. A physical theory for IUD expulsion and secondary IUD perforation is given.

Keywords:

• IUD
• insertion forces
• removal forces
• fracture forces
• intrauterine pressure
• intrauterine surface area

Supplementary material

Mathematical glossary

a. The presenting surface area (psa) of an IUD inserter is related to the diameter by the formula π r², where r is the radius, ie, the diameter/2. $\text{Psa}={(\text{diameter}/2)}^2\times \pi$

b. Intrauterine force is a vector quantity, ie, it has both magnitude and direction. Intrauterine pressure is a scalar quantity, ie, it has magnitude only. $\text{Pressure}=\text{force}/\text{unit area},\text{ and therefore},\text{ Force}=\text{pressure}\times \text{unit area}$

The force in Newtons (N) of uterine contraction can be calculated from intrauterine pressure and intrauterine (endometrial cavity) surface area. $\text{Pressure is usually measured in Pascals and a Pascal}={\text{N}/\text{meters}}^2\text{ or N}=\text{Pascal}\times {\text{meters}}^2$

If intrauterine pressure is measured in mmHg and endometrial cavity surface area is measured in mm², then as $1\text{ kPascal}=7.5\text{ mmHg}$

and $\text{mmHg}=\frac{\text{kPascals}\times {10}^3}{7.5}=\text{Pascals}$

and $\begin{array}{c}{\text{mm}}^2={\text{meter}}^2\times {10}^{-6}\\ \text{Intrauterine force}(\mathrm{N})=\text{mmHg}/7.5\times {10}^3\times {\text{mm}}^2\times {10}^{-6}\\ =\text{mmHg/}7.5\times {10}^{-3}\times {\text{mm}}^2\end{array}$

Calculation of uterine force in Newtons is a summation of the overall forces related to a particular pressure and surface area. It does not provide the direction or directions of the component forces, and some of the forces may act in different directions and in opposition.

c. 1 lb force =4.4482 Newtons.

Disclosure

DW has been involved in the optimization of innovative drug delivery systems for use in the uterus. He is currently an advisor in devising new concepts in controlled release for contraception, gynecological treatment, and prevention of infectious diseases. NDG reports no conflict of interest in this work.