A Note on Bursting Pressure

| November/December 1995

735 Riddle Road Cincinnati, Ohio 45220

The internal combustion generation hardly conceives that, not so long also, the topic of bursting pressure inspired raging debate. In the steam era, when boilers proliferated, tempers, like water, simmered on that subject. Knowledge of bursting pressure and the closely related concepts of working pressure and factors of safety mattered not only to designers of boilers and engines but also to investigators at explosion sites, insurance companies, implement dealers, and engineers. Equations for finding bursting pressure and working pressure abounded. Most formulas did not differ so much in substance as in degree of refinement; however, misunderstandings provoked controversy between one school of thought and another. In fact, to comprehend bursting pressure did require uncommon concentration. Descriptions in reputable mechanical engineering manuals and science textbooks from the late nineteenth and early twentieth centuries helped to unravel the mysteries of bursting pressure.

One such handbook, The Power Catechism, published by McGraw Hill in 1897, asked engineering students to imagine a ring like that depicted in Figure 1. If the ring were twelve inches in diameter and one inch wide (or, to be more exact, slong), and if it could contain an internal pressure of no more than one-hundred pounds to the square inch, the force tending to burst the ring would be 12 x 1 x 100, or 1,200 pounds. That force, thus, consists of the product found by multiplying the diameter, the length, and the pounds of pressure per square inch.

Now, a reasonable human being might think that the circumference, not the diameter, had better be used in the equation. After all, is not the internal pressure pushing against the whole ring, not merely an invisible diameter? Figure 2, however, depicts the physical law that pressure in any direction is effective on only those surfaces at right angles to that direction. Symbolized by the arrows, the pressure pushes against the ring and is resolved into the horizontal lines shown in the diagram. When placed end to end, these horizontal lines add up to a diameter.

A reasonable person might also consider it logical to assume that, if the pressure is, in effect, pulling a diameter in one direction, then an equal pressure must be pulling a diameter in the opposite direction. Figure 3 depicts such thinking. If this diagram represents the ring discussed earlier, then one arrow signifies a pressure of 1,200 pounds in one direction, and the other arrow indicates a pressure of 1,200 pounds in the opposite direction. A person might conclude that the bursting pressure is twice 1,200 pounds, or 2,400 pounds. The twin stresses of 1,200 pounds, though, were exerted against each other. As The Power Catechism points out, if two equally-strong people pull to the extent of their strength on opposite ends of a rope, the effect on the rope is the same as when one of the people pulls on the rope with its opposite end attached to a solid post. In the same way, one side of the ring pulls against the other side, and the tension in the ring is the force with which each side pulls not double the force. The bursting pressure for the ring, therefore, remains 1,200 pounds.

So far, the ring in these examples has been considered to have length (one inch) and diameter (twelve inches) but no thickness. Suppose it were one quarter of an inch thick. Further, assume it were made of boiler steel having a tensile strength of sixty thousand pounds per square inch. A bar of such steel one square inch in cross section breaks when a pull of sixty thousand pounds is applied in the direction of the bar's length. To tear the steel ring into two halves requires that it be torn apart at two points. Assume that, after the break, the area of each of these points will measure exactly one quarter of an inch by one inch. The area exposed by one of the breaks, then, will equal one-quarter of a square inch. The sum of the two areas will be one half of a square inch. To burst these two points apart will need a force equal to half of sixty thousand pounds per square inch, or thirty-thousand pounds. This force must be exerted on twelve square inches (the diameter of the ring multiplied by its length of one inch). The bursting pressure will be thirty thousand divided by twelve, or 2,500 pounds per square inch. The formula for this operation follows: