Duplex Stainless Steel – Density – Strength – Hardness – Melting Point

About Duplex Stainless Steel

Duplex stainless steels, as their name indicates, are a combination of two of the main alloy types. They have a mixed microstructure of austenite and ferrite, the aim usually being to produce a 50/50 mix, although in commercial alloys the ratio may be 40/60.  Their corrosion resistance is similar to their austenitic counterparts, but their stress-corrosion resistance (especially to chloride stress corrosion cracking), tensile strength, and yield strengths (roughly twice the yield strength of austenitic stainless steels) are generally superior to that of the austenitic grades. In duplex stainless steels, carbon is kept to very low levels (C<0.03%). Chromium content ranges from 21.00 to 26.00%, nickel content ranges from 3.50 to 8.00% and these alloys may contain molybdenum (up to 4.50%). Toughness and ductility generally fall between those of the austenitic and ferritic grades. Duplex grades usually divided into three sub-groups based on their corrosion resistance: lean duplex, standard duplex and superduplex. Superduplex steels have enhanced strength and resistance to all forms of corrosion compared to standard austenitic steels. Common uses are in marine applications, petrochemical plant, desalination plant, heat exchangers and papermaking industry. Today, the oil and gas industry is the largest user and has pushed for more corrosion resistant grades, leading to the development of superduplex steels.

The resistance of stainless steels to the chemical effects of corrosive  agents is based on passivation. For passivation to occur and remain stable, the Fe-Cr alloy must have a minimum chromium content of about 10.5% by weight, above which passivity can occur and below which it is impossible. Chromium can be used as a hardening element and is frequently used with a toughening element such as nickel to produce superior mechanical properties.

Duplex Stainless Steels – SAF 2205 – 1.4462

A common duplex stainless steel is SAF 2205 (a Sandvik-owned trademark for a 22Cr duplex (ferritic-austenitic) stainless steel), which contains typically 22% chromium and 5% nickel. It has excellent corrosion resistance and high strength, 2205 is the most widely used duplex stainless steel. Applications of SAF 2205 are in the following industries:

• Transport, storage and chemical processing
• Processing equipment
• High chloride and marine environments
• Oil and gas exploration
• Paper machines

Summary

Density of Duplex Stainless Steel

Typical densities of various substances are at atmospheric pressure. Density is defined as the mass per unit volume. It is an intensive property, which is mathematically defined as mass divided by volume:  ρ = m/V

In words, the density (ρ) of a substance is the total mass (m) of that substance divided by the total volume (V) occupied by that substance. The standard SI unit is kilograms per cubic meter (kg/m³). The Standard English unit is pounds mass per cubic foot (lbm/ft³).

Density of Duplex Stainless Steel is 7800 kg/m³.

Example: Density

Calculate the height of a cube made of Duplex Stainless Steel, which weighs one metric ton.

Solution:

Density is defined as the mass per unit volume. It is mathematically defined as mass divided by volume: ρ = m/V

As the volume of a cube is the third power of its sides (V = a³), the height of this cube can be calculated:

The height of this cube is then a = 0.504 m.

Properties of Duplex Stainless Steel

Material properties are intensive properties, that means they are independent of the amount of mass and may vary from place to place within the system at any moment. The basis of materials science involves studying the structure of materials, and relating them to their properties (mechanical, electrical etc.). Once a materials scientist knows about this structure-property correlation, they can then go on to study the relative performance of a material in a given application. The major determinants of the structure of a material and thus of its properties are its constituent chemical elements and the way in which it has been processed into its final form.

Mechanical Properties of Duplex Stainless Steel

Materials are frequently chosen for various applications because they have desirable combinations of mechanical characteristics. For structural applications, material properties are crucial and engineers must take them into account.

Strength of Duplex Stainless Steel

In mechanics of materials, the strength of a material is its ability to withstand an applied load without failure or plastic deformation. Strength of materials basically considers the relationship between the external loads applied to a material and the resulting deformation or change in material dimensions. Strength of a material is its ability to withstand this applied load without failure or plastic deformation.

Ultimate Tensile Strength

Ultimate tensile strength of duplex stainless steels – SAF 2205 is 620 MPa.

The ultimate tensile strength is the maximum on the engineering stress-strain curve. This corresponds to the maximum stress that can be sustained by a structure in tension. Ultimate tensile strength is often shortened to “tensile strength” or even to “the ultimate.”  If this stress is applied and maintained, fracture will result. Often, this value is significantly more than the yield stress (as much as 50 to 60 percent more than the yield for some types of metals). When a ductile material reaches its ultimate strength, it experiences necking where the cross-sectional area reduces locally. The stress-strain curve contains no higher stress than the ultimate strength. Even though deformations can continue to increase, the stress usually decreases after the ultimate strength has been achieved. It is an intensive property; therefore its value does not depend on the size of the test specimen. However, it is dependent on other factors, such as the preparation of the specimen, the presence or otherwise of surface defects, and the temperature of the test environment and material. Ultimate tensile strengths vary from 50 MPa for an aluminum to as high as 3000 MPa for very high-strength steels.

Yield Strength

Yield strength of duplex stainless steels – SAF 2205 is 440 MPa.

The yield point is the point on a stress-strain curve that indicates the limit of elastic behavior and the beginning plastic behavior. Yield strength or yield stress is the material property defined as the stress at which a material begins to deform plastically whereas yield point is the point where nonlinear (elastic + plastic) deformation begins. Prior to the yield point, the material will deform elastically and will return to its original shape when the applied stress is removed. Once the yield point is passed, some fraction of the deformation will be permanent and non-reversible. Some steels and other materials exhibit a behaviour termed a yield point phenomenon. Yield strengths vary from 35 MPa for a low-strength aluminum to greater than 1400 MPa for very high-strength steels.

Young’s Modulus of Elasticity

Young’s modulus of elasticity of duplex stainless steels – SAF 2205 is 200 GPa.

The Young’s modulus of elasticity is the elastic modulus for tensile and compressive stress in the linear elasticity regime of a uniaxial deformation and is usually assessed by tensile tests. Up to a limiting stress, a body will be able to recover its dimensions on removal of the load. The applied stresses cause the atoms in a crystal to move from their equilibrium position. All the atoms are displaced the same amount and still maintain their relative geometry. When the stresses are removed, all the atoms return to their original positions and no permanent deformation occurs. According to the Hooke’s law, the stress is proportional to the strain (in the elastic region), and the slope is Young’s modulus. Young’s modulus is equal to the longitudinal stress divided by the strain.

Hardness of Duplex Stainless Steel

Brinell hardness of duplex stainless steels – SAF 2205 is approximately 217 MPa.

In materials science, hardness is the ability to withstand surface indentation (localized plastic deformation) and scratching. Hardness is probably the most poorly defined material property because it may indicate resistance to scratching, resistance to abrasion, resistance to indentation or even resistance to shaping or localized plastic deformation. Hardness is important from an engineering standpoint because resistance to wear by either friction or erosion by steam, oil, and water generally increases with hardness.

Brinell hardness test is one of indentation hardness tests, that has been developed for hardness testing. In Brinell tests, a hard, spherical indenter is forced under a specific load into the surface of the metal to be tested. The typical test uses a 10 mm (0.39 in) diameter  hardened steel ball as an indenter with a 3,000 kgf (29.42 kN; 6,614 lbf) force. The load is maintained constant for a specified time (between 10 and 30 s). For softer materials, a smaller force is used; for harder materials, a tungsten carbide ball is substituted for the steel ball.

The test provides numerical results to quantify the hardness of a material, which is expressed by the Brinell hardness number – HB. The Brinell hardness number is designated by the most commonly used test standards (ASTM E10-14[2] and ISO 6506–1:2005) as HBW (H from hardness, B from brinell and W from the material of the indenter, tungsten (wolfram) carbide). In former standards HB or HBS were used to refer to measurements made with steel indenters.

The Brinell hardness number (HB) is the load divided by the surface area of the indentation. The diameter of the impression is measured with a microscope with a superimposed scale. The Brinell hardness number is computed from the equation:

There are a variety of  test methods in common use (e.g. Brinell, Knoop, Vickers and Rockwell). There are tables that are available correlating the hardness numbers from the different test methods where correlation is applicable. In all scales, a high hardness number represents a hard metal.

Example: Strength

Assume a plastic rod, which is made of Duplex Stainless Steel. This plastic rod has a cross-sectional area of 1 cm². Calculate the tensile force needed to achieve the ultimate tensile strength for this material, which is: UTS = 620 MPa.

Solution:

Stress (σ) can be equated to the load per unit area or the force (F) applied per cross-sectional area (A) perpendicular to the force as:

therefore, the tensile force needed to achieve the ultimate tensile strength is:

F = UTS x A = 620 x 10⁶ x 0.0001 = 62 000 N

Thermal Properties of Duplex Stainless Steel

Thermal properties of materials refer to the response of materials to changes in their thermodynamics/thermodynamic-properties/what-is-temperature-physics/”>temperature and to the application of heat. As a solid absorbs thermodynamics/what-is-energy-physics/”>energy in the form of heat, its temperature rises and its dimensions increase. But different materials react to the application of heat differently.

Heat capacity, thermal expansion, and thermal conductivity are properties that are often critical in the practical use of solids.

Melting Point of Duplex Stainless Steel

Melting point of duplex stainless steels – SAF 2205 steel is around 1450°C.

In general, melting is a phase change of a substance from the solid to the liquid phase. The melting point of a substance is the temperature at which this phase change occurs. The melting point also defines a condition in which the solid and liquid can exist in equilibrium.

Thermal Conductivity of Duplex Stainless Steel

The thermal conductivity of duplex stainless steels – SAF 2205 is 19 W/(m.K).

The heat transfer characteristics of a solid material are measured by a property called the thermal conductivity, k (or λ), measured in W/m.K. It is a measure of a substance’s ability to transfer heat through a material by conduction. Note that Fourier’s law applies for all matter, regardless of its state (solid, liquid, or gas), therefore, it is also defined for liquids and gases.

The thermal conductivity of most liquids and solids varies with temperature. For vapors, it also depends upon pressure. In general:

Most materials are very nearly homogeneous, therefore we can usually write k = k (T). Similar definitions are associated with thermal conductivities in the y- and z-directions (ky, kz), but for an isotropic material the thermal conductivity is independent of the direction of transfer, kx = ky = kz = k.

Example: Heat transfer calculation

Thermal conductivity is defined as the amount of heat (in watts) transferred through a square area of material of given thickness (in metres) due to a difference in temperature. The lower the thermal conductivity of the material the greater the material’s ability to resist heat transfer.

Calculate the rate of heat flux through a wall 3 m x 10 m in area (A = 30 m²). The wall is 15 cm thick (L₁) and it is made of Duplex Stainless Steel with the thermal conductivity of k₁ = 19 W/m.K (poor thermal insulator). Assume that, the indoor and the outdoor temperatures are 22°C and -8°C, and the convection heat transfer coefficients on the inner and the outer sides are h₁ = 10 W/m²K and h₂ = 30 W/m²K, respectively. Note that, these convection coefficients strongly depend especially on ambient and interior conditions (wind, humidity, etc.).

Calculate the heat flux (heat loss) through this wall.

Solution:

As was written, many of the heat transfer processes involve composite systems and even involve a combination of both conduction and convection. With these composite systems, it is often convenient to work with an overall heat transfer coefficient, known as a U-factor. The U-factor is defined by an expression analogous to Newton’s law of cooling:

The overall heat transfer coefficient is related to the total thermal resistance and depends on the geometry of the problem.

Assuming one-dimensional heat transfer through the plane wall and disregarding radiation, the overall heat transfer coefficient can be calculated as:

The overall heat transfer coefficient is then: U = 1 / (1/10 + 0.15/19 + 1/30) = 7.08 W/m²K

The heat flux can be then calculated simply as: q = 7.08 [W/m²K] x 30 [K] = 212.42 W/m²

The total heat loss through this wall will be: q_loss = q . A = 212.42 [W/m²] x 30 [m²] = 6372.67 W