3 Tensile stresses in the vessels

Cylindrical vessels consist of a cylindrical (boiler drum) part and two bottoms (ends), welded onto it. The ends may be circular flat, dished, spherical or conical. They may also be welded to the flange and then - in a form of screw cap - to the tank. Different types of containers have been presented in figure 3.1.

Fig. 3.1. Types of vessels: a) vertical, cylindrical, closed, b) cylindrical, with flat lid and conical bottom, c) cylindrical, horizontal, with two spherical lids, d) spherical with hatch.

The tensile stresses occurring in the tanks can be divided, depending on where they occur, into stress: in the walls, in flange bolts, in tank's fixture (suction and discharge piping, measuring connectors, valves, liquid level indicators, glass sighting tubes, lifting eye bolts etc.).

The theoretical basis concerning both tensile ?r and admissible kr stress together with described by the Hooke's law elongations ?l, presented in the previous chapter, are still valid, but only for linear loads and longitudinal elements, such as rods, bolts, or pipes.

Walls of the vessels, which are classified as flat and spatial elements, require individual theoretical introduction concerning stress and strength of thin-walled coatings.

A vessel is considered thin-walled if the actual thickness of the walls grz is less than the thickness limit ggr, which is linked to the diameter of the vessel D through the following relation:

(3.1)

3.1. Stresses in cylindrical walls - barrels

Cylindrical vessel consist of a cylindrical part called a barrel and two bottoms. If there is pressure p within the vessel, then it acts both on the surface of the barrel, as well as bottom surfaces. In case of barrel coating, shown in Fig. 3.2, there are two types of stress, circumferential ?ro and longitudinal ?rw. They stretch the coating as shown in an example of its rectangular element (Fig. 3.2).

Fig. 3.2. The tensile stresses in the cylindrical axisymmetric shell

In order to analyze the value of these stresses, let us imagine a vessel being cut by two planes perpendicular to each other: one perpendicular to the base - along the symmetry axis (Fig. 3.3) and second parallel (Fig. 3.4).

Fig. 3.3. Circumferential stress.

Fig. 3.4. Longitudinal breaking tensions

3.1.1. Tensile circumferential stresses

According to the definition, stress ?ro is the ratio of force F to the cross-sectional area A of the vessel, where the force is a product of pressure p and area of projected vessel surface parallel to the cross-section of that surface Ap.

(3.2)

thus, for D ? Dw and admissible tension stress kr: stresses ?ro and wall coating thickness g:

(3.3)

3.1.2. Longitudinal tensile stresses

Similar reasoning carried out for an imaginary longitudinal cross-section (Fig. 3.4) returns the dependence of longitudinal stress in cylindrical vessel walls.

After the imaginary cutting of the vessel with a plane, parallel to its base, through the action of a force F, the longitudinal tensile stress ?rw, arises at a plane A.

(3.4)

For thin-walled vessels (3.1), where D >> g, Dz = Dw + 2g ? Dw = D, circular ring area A equals approximately:

(3.5)

and then, after substituting (3.5) to (3.4) and shortening, we obtain:

(3.6)

3.2. Stresses and wall thickness of cylindrical vessels

A comparison of equations defining circumferential (3.3) and longitudinal (3.6) stresses shows that the former is, for identical conditions (p, D, kr), twice as high ?ro = 2?rw, therefore the wall thickness of the cylindrical vessel with prescribed pressure p and diameter D is equal to the thickness calculated from the circumferential stress gcyl = gro and hence - should be established from equation (3.3).

Exceptions include cases where the vessel contains a layer of liquid of height H and density ?, with hydrostatic pressure ph = H - ?, which is comparable to or greater than the gas pressure p inside the vessel. In such cases, longitudinal stresses ?rw might have equal, or even greater value than the circumferential ?ro. In this scenario, the thickness must be calculated from the higher one of these values, through one of the following relationships:

(3.7)

3.3. Stress and wall thickness of spherical vessels

The stress occurring in a tank wall or in the bottoms of spherical vessel, results from the symmetry condition for stresses in spherical coatings, where there is only one type of stress - the longitudinal tension stress ?rw.

The longitudinal stresses occurring in the cross-section of the vessel and their directions have been illustrated in Fig. 3.5 and relations (3.4) - (3.6), which show that wall thickness of a spherical vessel or bottom gd can be written as:

(3.8)

Fig. 3.5. Tensile stresses in the spherical shell

3.4. Calculation of bolts for flange connections in vessels

Example 3.1.

Calculate how many studs nsr with a diameter d = 36 mm (Figure 3.6) must be used to connect the cover with the vessels drum of diameter D = 2.2 m, if the pressure inside the vessel is p = 16 at, and kr = 130 MPa.

Fig. 3.6. Flange joint with a stud

SOLUTION:

Data: D = 2.2 m

p = 16 at ? 1.6 MPa

d = 36 mm = 3.6 -10-2 m

kr = 130 MPa

Searches: nsr

An upward force F acts on the cover of the vessel - it is derived from pressure, and equal to the product of the cross-section of the vessel and pressure. This force in studs cross-section, with a total area Asr generates tensile stress:

For technical reasons, the nearest value which is a multiple of 3 or 4 is chosen: nsr = 48.

Example. 3.2.

Fig. 3.7. The joint flange with hexagonal bolt screwed into the flange

Calculate how many steel bolts nsr, with kr = 120 MPa, and diameter d = 16 mm (M16) must twist the tank cap, of a diameter D = 1.4 m, wherein the pressure inside is p = 20 at.

SOLUTION:

Data: D = 1.4 m

p = 20 at = 2 MPa

kr = 120 MPa

d = 16 mm = 0.016 m

Searches: nsr

The number of bolts in the cover nsr can be calculated from tensile stress ?r, which on their surface Asr causes a force F, generated by a pressure pmaks, to act on the vessel lid Ap.