The safe working pressures that pipes will sustain depends upon the materials of which they are made and the thickness of their walls. The ultimate stress that a material will sustain before rupture ensues, is known as the tensile strength of that material and is the resistance offered to its fibre being pulled apart.

Fig. 79

There are four notable stresses to which a pipe is subjected before rupture ensues; they are: Safe working pressure, elastic limit, absolute strength and bursting stress. The elastic limit of a seamless drawn pipe is generally about one-half its absolute strength; in the case of cast-iron and lead pipes, however, the elastic limit is much lower, owing to the low coefficients of elasticity for these metals. In water supply systems where the pressure is fairly constant and free from water hammer one-half the elastic limit of seamless pipes can be taken as their safe working strength. If the supply system is not properly fitted with air chambers and is equipped with quick-closing faucets, the static pressure should not exceed one-third of the elastic limit of the pipe. In the case of metals with a low coefficient of elasticity, the safe working pressure for dead loads can be taken as one-fourth, and for live loads as one-sixth of the elastic limit of the pipe. This allowance provides a suitable factor of safety for the excessive pressures inseparable from most water supply systems.

A dead load is one that is fairly constant in pressure; a live load is one that fluctuates in pressure or is seriously affected by water hammer.

If a pipe is subjected to a great internal pressure, but of not sufficient intensity to strain it beyond its elastic limit, the pipe will yield to the pressure, but will immediately return to its normal condition upon being released from the pressure. If, however, the elastic limit is exceeded, the pipe will yield to the pressure and assume a new shape which it will retain after the pressure is removed.

When the pressure in a pipe is sufficient to strain it beyond the elastic limit the pipe yields to the pressure and the alteration of form becomes greater and greater until the absolute strength of the pipe is reached. Any additional pressure will then cause a bursting strain and rupture the pipe.

The pressure at which lead pipe will burst can be found by the following rule:

Rule

Multiply the tensile strength of the metal in pounds per square inch by twice the thickness of the pipe in inches, and divide the product by the internal diameter of the pipe in inches. The result will be the pressure at which the pipe will burst.

Expressed as a formula: p=2000t2 d

In which p=bursting pressure in pounds per square inch

2000=tensile strength of the metal in pounds per square inch t=thickness of the metal in inches d=internal diameter of pipe in inches

Example

What is the bursting pressure of a lead pipe 3 inches in diameter and .5 inch thick?

2000X.5X2 2000 .__ a ,

Solution-----------------=---=666.6 pounds pressure.

3 3

The maximum thickness of a lead pipe that will burst under a given head of water can be found by the following rule:

Rule

Multiply the pressure of water in pounds per square inch by the internal diameter of the pipe in inches, and divide the product by twice the tensile strength of the metal in pounds per square inch. The result will be the thickness of the metal in inches.

Expressed as a formula: t= p d

4000 In which t=thickness of the metal in inches p=bursting pressure in pounds per square inch d=internal diameter of pipe in inches 4000=constant; two tensile strengths

Example

When the internal diameter of a pipe is 3 inches and the pressure to be sustained is 667 pounds per square inch, what is the minimum thickness of the metal that will withstand the pressure?

Solution

667X3 = 2001=.5inch

2000x2 4000

When the pressure of water is known the thickness of a lead pipe that will safely sustain that pressure can be found by the following rule:

Rule

Multiply the pressure in pounds per square inch by the coefficient of the factor of safety; multiply that product by the internal diameter of the pipe in inches, and divide the product by twice the tensile strength of the metal in pounds per square inch. The result will be the thickness of the metal in inches.

The coefficients of the factors of safety are: for a live load 6 and for a dead load 4.

Expressed as a formula: p c d t= -------.

In which t=thickness of the metal in inches p=pressure in pounds per square inch c=coefficient of factor of safety d=internal diameter of the pipe in inches

4000=constant; two tensile strengths

Example

Find the thickness of metal required for a 3-inch pipe to safely stand a pressure of 167 pounds per square inch, (a) when the system is equipped with self-closing faucets and no air chambers, (b) When compression cocks are used and a suitable air chamber provided.

Solution (a) 167X6X3=3006= .75 inch thick

2000x2 4000

(b) 167X4X3= 2004=.5 inch thick

2000x2 4000

Dimensions and weights of stock sizes of lead pipe can be found in the following table:

Table XXXI - Size And Weight Of Lead Pipes

Table XXXI - Size And Weight Of Lead Pipes - Continued