About Duralumin
Aluminium alloys of 2000 series are alloyed with copper, they can be precipitation hardened to strengths comparable to steel. Formerly referred to as duralumin, they were once the most common aerospace alloys, but were susceptible to stress corrosion cracking and are increasingly replaced by 7000 series in new designs. In addition to aluminium, the main materials in duralumin are copper, manganese and magnesium.
Duralumin (also called duraluminum, duraluminium, duralum, dural(l)ium, or dural) is a strong, lightweight alloy of aluminium discovered in 1910 by Alfred Wilm, a German metallurgist. He discovered that after quenching, an aluminium alloy containing 4% copper would slowly harden when left at room temperature for several days. This process is now known as natural aging. He also designed an alloy (Duralumin) suitable for strengthening by this process in what is now known as preciptitation hardening. Although an explanation for the phenomenon was not provided until 1919, duralumin was one of the first “age hardening” alloys used.
Summary
| Name | Duralumin |
| Phase at STP | solid |
| Density | 2780 kg/m3 |
| Ultimate Tensile Strength | 450 MPa |
| Yield Strength | 300 MPa |
| Young’s Modulus of Elasticity | 76 GPa |
| Brinell Hardness | 120 BHN |
| Melting Point | 570 °C |
| Thermal Conductivity | 140 W/mK |
| Heat Capacity | 900 J/g K |
| Price | 6 $/kg |
In terms of age hardening, solution annealed aluminium-copper alloys can be aged naturally at room temperature for four days or more to obtain maximum properties such as hardness and strength. This process is known as natural aging. At room temperature, the solubility of copper in aluminium drops to a small fraction of 1%. At this point, the copper solute is locked inside the aluminium lattice (matrix), but must “precipitate” out of the supersaturated aluminium lattice. The aging process also can be accelerated to a matter of hours after solution treatment and quenching by heating the supersaturated alloy to a specific temperature and holding at that temperature for a specified time. This process is called artificial aging.
Duralumin is relatively soft, ductile and easily workable under normal temperature. The alloy can be rolled, forged and extruded into various forms and products. The light weight and high strength of duralumin when compared to steel enabled its application in aircraft construction. Although the addition of copper improves strength, it also makes these alloys susceptible to corrosion. The electrical and heat conductivity of duralumin is less than that of pure aluminium and more than that of steel.
Density of Duralumin
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 Duralumin is 2780 kg/m3.
Example: Density
Calculate the height of a cube made of Duralumin, 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 = 0.711 m.
Density of Materials
Mechanical Properties of Duralumin
Strength of Duralumin
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 – Duralumin
Ultimate tensile strength of 2024 aluminium alloy depends greatly on the temper of the material, but it is about 450 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.
Example: Strength
Assume a plastic rod, which is made of Duralumin. 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 = 450 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 = 450 x 106 x 0.0001 = 45 000 N
Yield Strength
Yield strength of 2024 aluminium alloy depends greatly on the temper of the material, but it is about 300 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 2024 aluminium alloy is about 76 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 Aluminium Alloys – Duralumin
Brinell hardness of 2024 aluminium alloy depends greatly on the temper of the material, but it is approximately 110 MPa.
Rockwell hardness test is one of the most common indentation hardness tests, that has been developed for hardness testing. In contrast to Brinell test, the Rockwell tester measures the depth of penetration of an indenter under a large load (major load) compared to the penetration made by a preload (minor load). The minor load establishes the zero position. The major load is applied, then removed while still maintaining the minor load. The difference between depth of penetration before and after application of the major load is used to calculate the Rockwell hardness number. That is, the penetration depth and hardness are inversely proportional. The chief advantage of Rockwell hardness is its ability to display hardness values directly. The result is a dimensionless number noted as HRA, HRB, HRC, etc., where the last letter is the respective Rockwell scale.
The Rockwell C test is performed with a Brale penetrator (120°diamond cone) and a major load of 150kg.
Thermal Properties of Aluminium Alloys – Duralumin
Thermal properties of materials refer to the response of materials to changes in their temperature and to the application of heat. As a solid absorbs 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 Aluminium Alloys
Melting point of 2024 aluminium alloy is around 570°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 Aluminium Alloys
The thermal conductivity of 2024 aluminium alloy is 140 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 Duralumin with the thermal conductivity of k1 = 140 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 + 0.15/140 + 1/30) = 7.44 W/m2K
The heat flux can be then calculated simply as: q = 7.44 [W/m2K] x 30 [K] = 223.21 W/m2
The total heat loss through this wall will be: qloss = q . A = 223.21 [W/m2] x 30 [m2] = 6696.19 W
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