Tube and drum stresses will be explained by considering a seamless cylinder enclosed with heads in each end, as shown in Fig. 3.2.
Pressure applied to the interior of this cylinder is distributed equally to all areas. This pressure causes the cylinder to be subjected to stress in three places.
First, the total force on each head is equal to the applied pressure (in pounds per square inch) multiplied by the area of the head in square inches. These forces A and B (Fig. 3.2a) produce tensile stress in the circumferential section (Fig. 3.2b).
Second, these same forces on the heads have a tendency to make them bulge (Fig. 3.3).
Third, stress is created along a longitudinal section MN of the cylinder (Fig. 3.4a). Assume that the cylinder is cut in half at section ABCD (Fig. 3.5). Further assume that the lower half of the cylinder is
Figure 3.2 Tension in circumferential section of a drum: (a) seamless cylinder; (b) cross section of cylinder showing metal in tension.
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Figure 3.3 Bulging tendency of flat heads.
Figure 3.4 Tension in a longitudinal section of a drum: (a) internal forces tending to cause longitudinal failure; (b) radi-al forces on inside of drum.
Figure 3.5 Diagrammatic illustration of transverse forces on the longitudinal section of a cylinder. The forces act on projected areas A, B, C, and D and produce tension stresses in sections AD and BC.
replaced by a heavy plate. The remaining vessel consists of half a cylinder and a flat plate. The force on the plate equals the applied pressure multiplied by the area ABCD. (The rectangular surface ABCD is known as the projected area of the curved surface.) This force is resisted by two strips of metal AB and CD. Each of these strips has a cross-sectional area equal to the length of the cylinder multiplied by the thickness of metal in the cylinder.
An example will show the application of this reasoning to the calcu-lations of the stresses in an actual cylindrical shell.
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Design and Construction of Boilers
Example A cylindrical shell, as in Fig. 3.2a and b, has an inside diameter of 6 ft and is 20 ft long; it is made of 1⁄2-in plate and is subjected to a pressure of 100 psi. (1) Find the stress in pounds per square inch on the metal in the circumferential section of the shell. (2) What is the stress in pounds per square inch in the cylindrical sides due to the transverse pressure?
Solution (1) Total pressure on the head is what tends to cause failure; for this reason the head must be held on the sides by additional metal.
Total pressure on the head:
A 4 D2
(see App. B)
A 62 144 in2/ft2 4071.51 in2
P 4071.51 100 407,151 lb total pressure The area in the metal ring (Fig. 3.2b) that must support this load:
C D (see App. B) A C thickness
A 6 12 in/ft 3.1416 1⁄2 113.10 in2of metal The force in each square inch of metal is
4
(2) Projected area (refer to Fig. 3.5):
A 6 20 144 in2/ft2 17,280 in2 Total pressure exerted on this area:
P 17,280 100 1,728,000 lb
But since this pressure is held by both sides of the shell, one side must hold half, or 1,728,000 2 864,000 lb. The area of the strip of metal that sup-ports this pressure is 12 in/ft 20 1⁄2 120 in2. The force on this area equals 864,000 120 7200 psi.
From these calculations the force on each square inch of metal tending to pull the drum apart in a transverse section is 3600 psi.
The force on the longitudinal section is 7200 psi. The force on the lon-gitudinal section is two times as much as the force on the transverse section. This fact must be considered in the determination of plate thickness requirements.
In determining the internal design pressure of a boiler drum consid-eration must be given to the tensile strength of the steel used in the construction, the thickness of the plate, the dimensions of the drum,
3.1416
4
146 Chapter Three
the permissible factor of safety, and the efficiency of the longitudinal joint. The tensile strength of steel and its ability to withstand temper-ature have been increased by the use of alloys and improved methods of manufacture. Steel varies in tensile strength from 50,000 to 100,000 psi. When the tensile strength of a boilerplate is not known, it may be assumed to be 45,000 psi for wrought iron and 55,000 psi for steel when calculating the internal design pressure. The metal in a boiler must never be subjected to a pressure greater than or even approaching the elastic limit. This is the reason there is a definite limit to the pressure used in applying a hydrostatic test. For the same reason, safety valves are used to limit the pressure when a boiler is in operation.
The required thickness of boilerplates is reduced proportionally by the use of high-tensile-strength steel. Sometimes drums are designed with plates of a greater thickness in that section which is drilled to receive the tubes. This procedure, together with welded construction, results in a decrease in drum weight. The thickness of the plate must always be sufficient for the specified design pressure; in addition, there is also a minimum-thickness requirement based on drum diameter.
For a given pressure, the drum metal thickness must be increased when the drum diameter is increased. Solve the previous problem using a shell 4 ft in diameter in place of the 6-ft-diameter shell.
The maximum stress to which a boilerplate may be subjected varies from one-fourth to one-seventh the tensile strength of the material. This comparison of tensile strength to actual working stress is known as the factor of safety (FS). If the tensile strength of a boilerplate is 55,000 psi, the stress to which it may be subjected varies from 55,000/7 7857 to 50,000/4 12,500 psi, depending on its age, type of construction, and condition. A minimum factor of safety of 4 may be used in determining the plate thickness of new boilers. In no case may the internal design pressure be increased above that allowable for new boilers. Secondhand boilers shall have a factor of safety of at least 51⁄2 unless constructed according to American Society of Mechanical Engineers (ASME) rules, when the factor shall be at least 5.
Note that the actual design values are per the latest ASME code requirements that must be adhered to by the boiler designer.
The equation for setting the minimum required wall thickness is per the ASME code (Sec. VIII) and is as follows:
t
SE PR
0.6P