About Bronze
The bronzes are a family of copper-based alloys traditionally alloyed with tin, but can refer to alloys of copper and other elements (e.g. aluminum, silicon, and nickel). Bronzes are somewhat stronger than the brasses, yet they still have a high degree of corrosion resistance. Generally they are used when, in addition to corrosion resistance, good tensile properties are required. For example, beryllium copper attains the greatest strength (to 1,400 MPa) of any copper-based alloy.
Historically, alloying copper with another metal, for example tin to make bronze, was first practiced about 4000 years after the discovery of copper smelting, and about 2000 years after “natural bronze” had come into general use. An ancient civilization is defined to be in the Bronze Age either by producing bronze by smelting its own copper and alloying with tin, arsenic, or other metals. Bronze, or bronze-like alloys and mixtures, were used for coins over a longer period. Bronzes are still widely used today for springs, bearings, bushings, automobile transmission pilot bearings, and similar fittings, and is particularly common in the bearings of small electric motors. Brass and bronze are common engineering materials in modern architecture and primarily used for roofing and facade cladding due to their visual appearance.
Summary
| Name | Bronze |
| Phase at STP | solid |
| Density | 8770 kg/m3 |
| Ultimate Tensile Strength | 310 MPa |
| Yield Strength | 150 MPa |
| Young’s Modulus of Elasticity | 103 GPa |
| Brinell Hardness | 75 BHN |
| Melting Point | 1000 °C |
| Thermal Conductivity | 75 W/mK |
| Heat Capacity | 435 J/g K |
| Price | 4 $/kg |
Density of Bronze
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 Bronze is 8770 kg/m3.
Example: Density
Calculate the height of a cube made of Bronze, 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 = 0.485 m.
Density of Materials
Mechanical Properties of Bronzes
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 Bronzes
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 aluminium bronze – UNS C95400 is about 550 MPa.
Ultimate tensile strength of tin bronze – UNS C90500 – gun metal is about 310 MPa.
Ultimate tensile strength of copper beryllium – UNS C17200 is about 1380 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 aluminium bronze – UNS C95400 is about 250 MPa.
Yield strength of tin bronze – UNS C90500 – gun metal is about 150 MPa.
Yield strength of copper beryllium – UNS C17200 is about 1100 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 aluminium bronze – UNS C95400 is about 110 GPa.
Young’s modulus of elasticity of tin bronze – UNS C90500 – gun metal is about 103 GPa.
Young’s modulus of elasticity of copper beryllium – UNS C17200 is about 131 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 Bronzes
Brinell hardness of aluminium bronze – UNS C95400 is approximately 170 MPa. The hardness of aluminum bronzes increases with aluminum (and other alloy) content as well as with stresses caused through cold working.
Brinell hardness of tin bronze – UNS C90500 – gun metal is approximately 75 BHN.
Rockwell hardness of copper beryllium – UNS C17200 is approximately 82 HRB.
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.
Example: Strength
Assume a plastic rod, which is made of Bronze. 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 = 310 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 = 310 x 106 x 0.0001 = 31 000 N
Strength of Materials
Elasticity of Materials
Hardness of Materials
Thermal Properties of Bronzes
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 Bronzes
Melting point of aluminium bronze – UNS C95400 is around 1030°C.
Melting point of tin bronze – UNS C90500 – gun metal is around 1000°C.
Melting point of copper beryllium – UNS C17200 is around 866°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 Bronzes
The thermal conductivity of aluminium bronze – UNS C95400 is 59 W/(m.K).
The thermal conductivity of tin bronze – UNS C90500 – gun metal is 75 W/(m.K).
The thermal conductivity of copper beryllium – UNS C17200 is 115 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 Bronze with the thermal conductivity of k1 = 75 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 + 0.15/75 + 1/30) = 7.39 W/m2K
The heat flux can be then calculated simply as: q = 7.39 [W/m2K] x 30 [K] = 221.67 W/m2
The total heat loss through this wall will be: qloss = q . A = 221.67 [W/m2] x 30 [m2] = 6650.25 W
Melting Point of Materials
Thermal Conductivity of Materials
Heat Capacity of Materials
