Material Properties

About Invar

Invar is an alloy of nickel and iron. This alloy is also known generically as FeNi36 (64FeNi in the US). Invar is notable for its uniquely low coefficient of thermal expansion (CTE or α). The name Invar comes from the word invariable, referring to its relative lack of expansion or contraction with temperature changes. Invar has a near-zero coefficient of thermal expansion, making it useful in constructing precision instruments whose dimensions need to remain constant in spite of varying temperature. The discovery of the alloy was made in 1896 by Swiss physicist Charles Édouard Guillaume for which he received the Nobel Prize in Physics in 1920.

Summary

Name	Invar
Phase at STP	solid
Density	8100 kg/m3
Ultimate Tensile Strength	445 MPa
Yield Strength	280 MPa
Young’s Modulus of Elasticity	135 GPa
Brinell Hardness	200 BHN
Melting Point	1687 °C
Thermal Conductivity	12 W/mK
Heat Capacity	505 J/g K
Price	29 $/kg

Density of Invar

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 Invar is 8100 kg/m³.

 

Example: Density

Calculate the height of a cube made of Invar, 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 = 0.498 m.

Mechanical Properties of Invar

Strength of Invar

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. In designing structures and machines, it is important to consider these factors, in order that the material selected will have adequate strength to resist applied loads or forces and retain its original shape.

Strength of a material is its ability to withstand this applied load without failure or plastic deformation. For tensile stress, the capacity of a material or structure to withstand loads tending to elongate is known as ultimate tensile strength (UTS). 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. In case of tensional stress of a uniform bar (stress-strain curve), the Hooke’s law describes behaviour of a bar in the elastic region. 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.

See also: Strength of Materials

Ultimate Tensile Strength of Invar

Ultimate tensile strength of Invar is 445 MPa.

Yield Strength of Invar

Yield strength of Invar is 280 MPa.

Modulus of Elasticity of Invar

The Young’s modulus of elasticity of Invar is 135 GPa.

Hardness of Invar

In materials science, hardness is the ability to withstand surface indentation (localized plastic deformation) and scratching. 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 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:

Brinell hardness of Invar is approximately 200 BHN (converted).

See also: Hardness of Materials

 

Example: Strength

Assume a plastic rod, which is made of Invar. 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 = 445 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 = 445 x 10⁶ x 0.0001 = 44 500 N

Thermal Properties of Invar

Invar – Melting Point

Melting point of Invar is 1687 °C.

Note that, these points are associated with the standard atmospheric pressure. 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. For various chemical compounds and alloys, it is difficult to define the melting point, since they are usually a mixture of various chemical elements.

Invar – Thermal Conductivity

Thermal conductivity of Invar is 12 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.

Invar – Specific Heat

Specific heat of Invar is 505 J/g K.

Specific heat, or specific heat capacity, is a property related to internal energy that is very important in thermodynamics. The intensive properties c_v and c_p are defined for pure, simple compressible substances as partial derivatives of the internal energy u(T, v) and enthalpy h(T, p), respectively:

where the subscripts v and p denote the variables held fixed during differentiation. The properties c_v and c_p are referred to as specific heats (or heat capacities) because under certain special conditions they relate the temperature change of a system to the amount of energy added by heat transfer. Their SI units are J/kg K or J/mol 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 Invar with the thermal conductivity of k₁ = 12 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 + 0.15/12 + 1/30) = 6.86 W/m²K

The heat flux can be then calculated simply as: q = 6.86 [W/m²K] x 30 [K] = 205.71 W/m²

The total heat loss through this wall will be: q_loss = q . A = 205.71 [W/m²] x 30 [m²] = 6171.43 W

About Malleable Cast Iron

Malleable cast iron is white cast iron that has been annealed. Through an annealing heat treatment, the brittle structure as first cast is transformed into the malleable form. Therefore, its composition is very similar to that of white cast iron, with slightly higher amounts of carbon and silicon. Malleable iron contains graphite nodules that are not truly spherical as they are in ductile iron, because they are formed as a result of heat treatment rather than forming during cooling from the melt. Malleable iron is made by first casting a white iron so that flakes of graphite are avoided and all the undissolved carbon is in the form of iron carbide. Malleable iron starts as a white iron casting that is then heat treated for a day or two at about 950 °C (1,740 °F) and then cooled over a day or two. As a result, the carbon in iron carbide transforms into graphite nodules surrounded by a ferrite or pearlite matrix, depending on cooling rate. The slow process allows the surface tension to form the graphite nodules rather than flakes. . Malleable iron, like ductile iron, possesses considerable ductility and toughness because of its combination of nodular graphite and low carbon metallic matrix. Like ductile iron, malleable iron also exhibits high resistance to corrosion, excellent machinability. The good damping capacity and fatigue strength of malleable iron are also useful for long service in highly stressed parts. There are two types of ferritic malleable iron: blackheart and whiteheart.

Summary

Name	Malleable iron
Phase at STP	solid
Density	7150 kg/m3
Ultimate Tensile Strength	580 MPa
Yield Strength	480 MPa
Young’s Modulus of Elasticity	172 GPa
Brinell Hardness	250 BHN
Melting Point	1260 °C
Thermal Conductivity	40 W/mK
Heat Capacity	465 J/g K
Price	3 $/kg

Density of Malleable Iron

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 Malleable Iron is 7150 kg/m³.

 

Example: Density

Calculate the height of a cube made of Malleable Iron, 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 = 0.519 m.

Mechanical Properties of Malleable Iron

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 Malleable Cast Iron

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 malleable cast iron – ASTM A220 is 580 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 malleable cast iron – ASTM A220 is 480 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 malleable cast iron – ASTM A220 is 172 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 Malleable Cast Iron

Brinell hardness of malleable cast iron – ASTM A220 is approximately 250 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 Malleable Iron. 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 = 580 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 = 580 x 10⁶ x 0.0001 = 58 000 N

Thermal Properties of Malleable Iron

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 Malleable Cast Iron – ASTM A220

Melting point of malleable cast iron – ASTM A220 is around 1260°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 Malleable Cast Iron – ASTM A220

The thermal conductivity of malleable cast iron is approximately 40 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 Malleable Iron with the thermal conductivity of k₁ = 40 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 + 0.15/40 + 1/30) = 7.29 W/m²K

The heat flux can be then calculated simply as: q = 7.29 [W/m²K] x 30 [K] = 218.85 W/m²

The total heat loss through this wall will be: q_loss = q . A = 218.85 [W/m²] x 30 [m²] = 6565.35 W

About High-speed Steel – AISI M2

For example, molybdenum high-speed steel – AISI M2 is the “standard” and most widely used industrial HSS. Molybdenum high speed steels are designated as Group M steels according to the AISI classification system. M2 HSS has small and evenly distributed carbides giving high wear resistance, though its decarburization sensitivity is a little bit high. It is usually used to manufacture a variety of tools, such as drill bits, taps and reamers.

Summary

Name	High-speed steel
Phase at STP	solid
Density	78160 kg/m3
Ultimate Tensile Strength	1200 MPa
Yield Strength	1000 MPa
Young’s Modulus of Elasticity	200 GPa
Brinell Hardness	720 BHN
Melting Point	1430 °C
Thermal Conductivity	41 W/mK
Heat Capacity	470 J/g K
Price	8 $/kg

Density of High-speed 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 High-speed Steel is 78160 kg/m³.

 

Example: Density

Calculate the height of a cube made of High-speed 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 = 0.234 m.

Mechanical Properties of High-speed 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 High-speed 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 high-speed steel – AISI M2 depends on heat treatment process, but it is about 1200 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 high-speed steel – AISI M2 depends on heat treatment process, but it is about 1000 MPa. Compressive yield strength is about 3250 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 high-speed steel – AISI M2 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 High-speed Steel

Rockwell hardness of high-speed steel – AISI M2 depends on heat treatment process, but it is approximately 65 HRC.

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 High-speed 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 = 1200 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 = 1200 x 10⁶ x 0.0001 = 120 000 N

Thermal Properties of High-speed Steel

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 High-speed Steel – AISI M2

Melting point of high-speed steel – AISI M2 steel is around 1430°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 High-speed Steel – AISI M2

The thermal conductivity of high-speed steel – AISI M2 is 41 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 High-speed Steel with the thermal conductivity of k₁ = 41 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 + 0.15/41 + 1/30) = 7.29 W/m²K

The heat flux can be then calculated simply as: q = 7.29 [W/m²K] x 30 [K] = 218.99 W/m²

The total heat loss through this wall will be: q_loss = q . A = 218.99 [W/m²] x 30 [m²] = 6569.73 W

About PH Stainless Steel

PH stainless steels (precipitation-hardening) contain around 17% chromium and 4% nickel. These steels can develop very high strength through additions of aluminum, titanium, niobium, vanadium, and/or nitrogen, which form coherent intermetallic precipitates during a heat treatment process referred to as heat aging. As the coherent precipitates form throughout the microstructure, they strain the crystalline lattice and impede the movement of dislocations, or defects in a crystal’s lattice. Since dislocations are often the dominant carriers of plasticity, this serves to harden the material. Precipitation-hardening stainless steels have high toughness, strength, and corrosion resistance. Precipitation-hardening stainless steels have been increasingly used for a variety of applications in marine construction, aircraft and gas turbines, chemical industries and nuclear power plants.

17-4PH Stainless Steel

For example, precipitation-hardened stainless steel 17-4 PH (AISI 630) have an initial microstructure of austenite or martensite. Austenitic grades are converted to martensitic grades through heat treatment (e.g. throung heat treatment at about 1040 °C followed by quenching) before precipitation hardening can be done. Subsequent ageing treatment at about 475 °C precipitates Nb and Cu-rich phases that increase the strength up to above 1000 MPa yield strength. In all heat treatments performed the predominant microstructure is lath martensite. Unlike austenitic alloys, however, heat treatment strengthens PH steels to levels higher than martensitic alloys. Precipitation-hardening stainless steels are designated by the AISI 600-series. Of all of the available stainless grades, they generally offer the greatest combination of high strength coupled with excellent toughness and corrosion resistance.  They are as corrosion resistant as austenitic grades. Common uses are in the aerospace and some other high-technology industries.

Summary

Name	PH Stainless Steel
Phase at STP	solid
Density	7750 kg/m3
Ultimate Tensile Strength	1000 MPa
Yield Strength	850 MPa
Young’s Modulus of Elasticity	200 GPa
Brinell Hardness	400 BHN
Melting Point	1450 °C
Thermal Conductivity	18 W/mK
Heat Capacity	460 J/g K
Price	9 $/kg

Density of PH 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 PH Stainless Steel is 7750 kg/m³.

 

Example: Density

Calculate the height of a cube made of PH 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 = 0.505 m.

Mechanical Properties of PH 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 PH 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 precipitation hardening steels – 17-4PH stainless steel depends on heat treatment process, but it is about 1000 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 precipitation hardening steels – 17-4PH stainless steel depends on heat treatment process, but it is about 850 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 precipitation hardening steels – 17-4PH stainless steel 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 PH Stainless Steel

Brinell hardness of precipitation hardening Steels – 17-4PH stainless steel is approximately 353 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 PH 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 = 1000 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 = 1000 x 10⁶ x 0.0001 = 100 000 N

Thermal Properties of PH Stainless Steel

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 PH Stainless Steel

Melting point of precipitation hardening steels – 17-4PH stainless steel 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 PH Stainless Steel

The thermal conductivity of precipitation hardening steels – 17-4PH stainless steel is 18 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 PH Stainless Steel with the thermal conductivity of k₁ = 18 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 + 0.15/18 + 1/30) = 7.06 W/m²K

The heat flux can be then calculated simply as: q = 7.06 [W/m²K] x 30 [K] = 211.77 W/m²

The total heat loss through this wall will be: q_loss = q . A = 211.77 [W/m²] x 30 [m²] = 6352.94 W

About Amalgam

An amalgam is an alloy of mercury with another metal. It may be a liquid, a soft paste or a solid, depending upon the proportion of mercury.

Summary

Name	Amalgam
Phase at STP	solid
Density	14000 kg/m3
Ultimate Tensile Strength	70 MPa
Yield Strength	N/A
Young’s Modulus of Elasticity	40 GPa
Brinell Hardness	120 BHN
Melting Point	227 °C
Thermal Conductivity	23 W/mK
Heat Capacity	210 J/g K
Price	150 $/kg

Density of Amalgam

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 Amalgam is 14000 kg/m³.

 

Example: Density

Calculate the height of a cube made of Amalgam, 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 = 0.415 m.

Mechanical Properties of Amalgam

Strength of Amalgam

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. In designing structures and machines, it is important to consider these factors, in order that the material selected will have adequate strength to resist applied loads or forces and retain its original shape.

Strength of a material is its ability to withstand this applied load without failure or plastic deformation. For tensile stress, the capacity of a material or structure to withstand loads tending to elongate is known as ultimate tensile strength (UTS). 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. In case of tensional stress of a uniform bar (stress-strain curve), the Hooke’s law describes behaviour of a bar in the elastic region. 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.

See also: Strength of Materials

Ultimate Tensile Strength of Amalgam

Ultimate tensile strength of Amalgam is 70 MPa.

Yield Strength of Amalgam

Yield strength of Amalgam is N/A.

Modulus of Elasticity of Amalgam

The Young’s modulus of elasticity of Amalgam is 40 GPa.

Hardness of Amalgam

In materials science, hardness is the ability to withstand surface indentation (localized plastic deformation) and scratching. 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 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:

Brinell hardness of Amalgam is approximately 120 BHN (converted).

See also: Hardness of Materials

 

Example: Strength

Assume a plastic rod, which is made of Amalgam. 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 = 70 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 = 70 x 10⁶ x 0.0001 = 7 000 N

Thermal Properties of Amalgam

Amalgam – Melting Point

Melting point of Amalgam is 227 °C.

Note that, these points are associated with the standard atmospheric pressure. 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. For various chemical compounds and alloys, it is difficult to define the melting point, since they are usually a mixture of various chemical elements.

Amalgam – Thermal Conductivity

Thermal conductivity of Amalgam is 23 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.

Amalgam – Specific Heat

Specific heat of Amalgam is 210 J/g K.

Specific heat, or specific heat capacity, is a property related to internal energy that is very important in thermodynamics. The intensive properties c_v and c_p are defined for pure, simple compressible substances as partial derivatives of the internal energy u(T, v) and enthalpy h(T, p), respectively:

where the subscripts v and p denote the variables held fixed during differentiation. The properties c_v and c_p are referred to as specific heats (or heat capacities) because under certain special conditions they relate the temperature change of a system to the amount of energy added by heat transfer. Their SI units are J/kg K or J/mol 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 Amalgam with the thermal conductivity of k₁ = 23 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 + 0.15/23 + 1/30) = 7.15 W/m²K

The heat flux can be then calculated simply as: q = 7.15 [W/m²K] x 30 [K] = 214.51 W/m²

The total heat loss through this wall will be: q_loss = q . A = 214.51 [W/m²] x 30 [m²] = 6435.23 W

About Beryllium Bronze

Copper beryllium, also known as beryllium bronze, is a copper alloy with 0.5—3% beryllium. Copper beryllium is the hardest and strongest of any copper alloy (UTS up to 1,400 MPa), in the fully heat treated and cold worked condition. It combines high strength with non-magnetic and non-sparking qualities and it is similar in mechanical properties to many high strength alloy steels but, compared to steels, it has better corrosion resistance (similar to pure copper). It has good thermal conductivity (210 W/m°C) 3-5 times more than tool steel. These high performance alloys have long been used for non-sparking tools in the mining (coal mines), gas and petrochemical industries (oil rigs). Beryllium copper screwdrivers, pliers, wrenches, cold chisels, knives, and hammers are available for these environments. Because of the excellent fatigue resistance, copper beryllium is widely used for springs, spring wire, load cells, and other parts that must retain their shape under cyclic loads.

Summary

Name	Beryllium copper
Phase at STP	solid
Density	8250 kg/m3
Ultimate Tensile Strength	1110 MPa
Yield Strength	1110 MPa
Young’s Modulus of Elasticity	131 GPa
Brinell Hardness	150 BHN
Melting Point	866 °C
Thermal Conductivity	115 W/mK
Heat Capacity	420 J/g K
Price	13 $/kg

Density of Beryllium Copper

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 Beryllium Copper is 8250 kg/m³.

 

Example: Density

Calculate the height of a cube made of Beryllium Copper, 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 = 0.495 m.

Mechanical Properties of Beryllium Bronze

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 Beryllium Bronze

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 copper beryllium – UNS C17200 is about 1110 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 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 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 Beryllium Bronze

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 Beryllium Copper. 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 = 1110 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 = 1110 x 10⁶ x 0.0001 = 111 000 N

Thermal Properties of Beryllium Bronze

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 Beryllium Bronze

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 Beryllium Bronze

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 Beryllium Copper with the thermal conductivity of k₁ = 115 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 + 0.15/115 + 1/30) = 7.43 W/m²K

The heat flux can be then calculated simply as: q = 7.43 [W/m²K] x 30 [K] = 222.82 W/m²

The total heat loss through this wall will be: q_loss = q . A = 222.82 [W/m²] x 30 [m²] = 6684.61 W

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.

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

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: q_loss = q . A = 221.67 [W/m²] x 30 [m²] = 6650.25 W

About Stellite

Stellite alloys are a group of cobalt-chromium ‘super-alloys’ consisting of complex carbides in an alloy matrix predominantly designed for high wear resistance and superior chemical and corrosion performance in hostile environments. Unlike other superalloys, cobalt-base alloys are characterized by a solid-solution-strengthened austenitic (fcc) matrix in which a small quantity of carbide is distributed. While not used commercially to the extent of Ni-based superalloys, alloying elements found in research Co-based alloys are C, Cr, W, Ni, Ti, Al, Ir, and Ta. They possess better weldability and thermal fatigue resistance as compared to nickel based alloy. Moreover, they have excellent corrosion resistance at high temperatures (980-1100 °C) because of their higher chromium contents.

Summary

Name	Stellite
Phase at STP	solid
Density	8690 kg/m3
Ultimate Tensile Strength	1200 MPa
Yield Strength	1050 MPa
Young’s Modulus of Elasticity	230 GPa
Brinell Hardness	550 BHN
Melting Point	1297 °C
Thermal Conductivity	14.8 W/mK
Heat Capacity	423 J/g K
Price	50 $/kg

Density of Stellite

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 Stellite is 8690 kg/m³.

 

Example: Density

Calculate the height of a cube made of Stellite, 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 = 0.486 m.

Mechanical Properties of Stellite

Strength of Stellite

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. In designing structures and machines, it is important to consider these factors, in order that the material selected will have adequate strength to resist applied loads or forces and retain its original shape.

Strength of a material is its ability to withstand this applied load without failure or plastic deformation. For tensile stress, the capacity of a material or structure to withstand loads tending to elongate is known as ultimate tensile strength (UTS). 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. In case of tensional stress of a uniform bar (stress-strain curve), the Hooke’s law describes behaviour of a bar in the elastic region. 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.

See also: Strength of Materials

Ultimate Tensile Strength of Stellite

Ultimate tensile strength of Stellite is 1200 MPa.

Yield Strength of Stellite

Yield strength of Stellite is 1050 MPa.

Modulus of Elasticity of Stellite

The Young’s modulus of elasticity of Stellite is 230 GPa.

Hardness of Stellite

In materials science, hardness is the ability to withstand surface indentation (localized plastic deformation) and scratching. 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 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:

Brinell hardness of Stellite is approximately 550 BHN (converted).

See also: Hardness of Materials

 

Example: Strength

Assume a plastic rod, which is made of Stellite. 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 = 1200 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 = 1200 x 10⁶ x 0.0001 = 120 000 N

Thermal Properties of Stellite

Stellite – Melting Point

Melting point of Stellite is 1297 °C.

Note that, these points are associated with the standard atmospheric pressure. 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. For various chemical compounds and alloys, it is difficult to define the melting point, since they are usually a mixture of various chemical elements.

Stellite – Thermal Conductivity

Thermal conductivity of Stellite is 14.8 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.

Stellite – Specific Heat

Specific heat of Stellite is 423 J/g K.

Specific heat, or specific heat capacity, is a property related to internal energy that is very important in thermodynamics. The intensive properties c_v and c_p are defined for pure, simple compressible substances as partial derivatives of the internal energy u(T, v) and enthalpy h(T, p), respectively:

where the subscripts v and p denote the variables held fixed during differentiation. The properties c_v and c_p are referred to as specific heats (or heat capacities) because under certain special conditions they relate the temperature change of a system to the amount of energy added by heat transfer. Their SI units are J/kg K or J/mol 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 Stellite with the thermal conductivity of k₁ = 14.8 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 + 0.15/14.8 + 1/30) = 6.97 W/m²K

The heat flux can be then calculated simply as: q = 6.97 [W/m²K] x 30 [K] = 209.11 W/m²

The total heat loss through this wall will be: q_loss = q . A = 209.11 [W/m²] x 30 [m²] = 6273.16 W

 

About Ductile Cast Iron

Ductile iron, also known as nodular iron or spheroidal graphite iron, is very similar to gray iron in composition, but during solidification the graphite nucleates as spherical particles (nodules) in ductile iron, rather than as flakes. Ductile iron is not a single material but part of a group of materials which can be produced with a wide range of properties through control of their microstructure. The matrix phase surrounding these particles is either pearlite or ferrite, depending on heat treatment. Ductile iron is stronger and more shock resistant than gray iron, so although it is more expensive due to alloyants, it may be the preferred economical choice because a lighter casting can perform the same function.

Summary

Name	Ductile iron
Phase at STP	solid
Density	7300 kg/m3
Ultimate Tensile Strength	414 MPa
Yield Strength	276 MPa
Young’s Modulus of Elasticity	170 GPa
Brinell Hardness	180 BHN
Melting Point	1150 °C
Thermal Conductivity	36 W/mK
Heat Capacity	460 J/g K
Price	3 $/kg

Density of Ductile Cast Iron

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 Ductile Cast Iron is 7300 kg/m³.

 

Example: Density

Calculate the height of a cube made of Ductile Cast Iron, 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 = 0.515 m.

Mechanical Properties of Ductile Cast Iron

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 Ductile Cast Iron – ASTM A536 – 60-40-18

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 ductile cast Iron – ASTM A536 – 60-40-18 is 414 MPa (>60 ksi).

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 ductile cast Iron – ASTM A536 – 60-40-18 is  276 MPa (>40 ksi).

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 ductile cast Iron – ASTM A536 – 60-40-18 is 170 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 Ductile Cast Iron – ASTM A536 – 60-40-18

Brinell hardness of ductile cast Iron – ASTM A536 – 60-40-18 is approximately 150 – 180 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 Ductile Cast Iron. 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 = 414 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 = 414 x 10⁶ x 0.0001 = 41 400 N

Thermal Properties of Ductile Cast Iron

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 Ductile Cast Iron – ASTM A536 – 60-40-18

Melting point of ductile cast Iron – ASTM A536 – 60-40-18 steel is around 1150°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 Ductile Cast Iron – ASTM A536 – 60-40-18

The thermal conductivity of ductile cast iron is 36 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 Ductile Cast Iron with the thermal conductivity of k₁ = 36 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 + 0.15/36 + 1/30) = 7.27 W/m²K

The heat flux can be then calculated simply as: q = 7.27 [W/m²K] x 30 [K] = 218.18 W/m²

The total heat loss through this wall will be: q_loss = q . A = 218.18 [W/m²] x 30 [m²] = 6545.46 W

About Tool Steel

Tool steel refers to a variety of carbon and alloy steels that are particularly well-suited to be made into tools (punches, dies, molds, tools for cutting, blanking, forming, drawing, steering and slitting tools). Their suitability comes from their distinctive hardness, resistance to abrasion and deformation, and their ability to hold a cutting edge at elevated temperatures. With a carbon content between 0.5% and 1.5%, tool steels are manufactured under carefully controlled conditions to produce the required quality. The presence of carbides in their matrix plays the dominant role in the qualities of tool steel.

They are generally grouped into two classes:

• Plain carbon steels containing a high percentage of carbon, about 0.80-1.50%
• Alloy tool steels, in which other elements (chromium, molybdenum, vanadium, tungsten and cobalt) are added to provide greater strength, toughness, corrosion and heat resistance of steel.

One of subgroups of tool steels is high-speed steels (HSS), which were named primarily for their ability to machine and cut materials at high speeds (high hot hardness). It is often used in power-saw blades and drill bits. This group of tool steels is described in a separate article.

Alloying Agents in Alloy Tool Steels

Pure iron is too soft to be used for the purpose of structure, but the addition of small quantities of other elements (carbon, manganese or silicon for instance) greatly increases its mechanical strength. The synergistic effect of alloying elements and heat treatment produces a tremendous variety of microstructures and properties. The four major alloying elements that form carbides in tool and die steel are: tungsten, chromium, vanadium and molybdenum. These alloying elements combine with carbon to form very hard and wear-resistant carbide compounds.

• Tungsten. Produces stable carbides and refines grain size so as to increase hardness, particularly at high temperatures. Tungsten is used extensively in high-speed tool steels and has been proposed as a substitute for molybdenum in reduced-activation ferritic steels for nuclear applications.
• Chromium. Chromium increases hardness, strength, and corrosion resistance. The strengthening effect of forming stable metal carbides at the grain boundaries and the strong increase in corrosion resistance made chromium an important alloying material for steel. Generally speaking, the concentration specified for most grades is approximately 4%. This level appears to result in the best balance between hardness and toughness. Chromium plays an important role in the hardening mechanism and is considered irreplaceable. At higher temperatures, chromium contributes increased strength. It is ordinarily used for applications of this nature in conjunction with molybdenum.
• Molybdenum. Molybdenum (about 0.50-8.00%) when added to a tool steel makes it more resistant to high temperature. Molybdenum increases hardenability and strength, particularly at high temperatures due to the high melting point of molybdenum. Molybdenum is unique in the extent to which it increases the high-temperature tensile and creep strengths of steel. It retards the transformation of austenite to pearlite far more than it does the transformation of austenite to bainite; thus, bainite may be produced by continuous cooling of molybdenum-containing steels.
• Vanadium. Vanadium is generally added to steel to inhibit grain growth during heat treatment. In controlling grain growth, it improves both the strength and toughness of hardened and tempered steels. The size of the grain determines the properties of the metal. For example, smaller grain size increases tensile strength and tends to increase ductility. A larger grain size is preferred for improved high-temperature creep properties.

Example of Tool Steel – A2 Steel

A2 tool steel is an air hardening, cold work steel of group A steels containing molybdenum and chromium. A2 steel contains 5% of chromium steel which provides high hardness after heat treatment with good dimensional stability. The carbon content in A2 tool steels is high. A2 offers good toughness with medium wear resistance and is relatively easy to machine. A2 tool steel can be used in many applications which require good wear resistance as well as good toughness.

Summary

Name	Tool Steel
Phase at STP	solid
Density	7810 kg/m3
Ultimate Tensile Strength	1860 MPa
Yield Strength	1400 MPa
Young’s Modulus of Elasticity	200 GPa
Brinell Hardness	630 BHN
Melting Point	1420 °C
Thermal Conductivity	26 W/mK
Heat Capacity	465 J/g K
Price	2 $/kg

Density of Tool 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 Tool Steel is 7810 kg/m³.

 

Example: Density

Calculate the height of a cube made of Tool 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.

Mechanical Properties of Tool Steel – A2 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 Tool Steel – A2 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 tool steel – A2 steel depends on heat treatment process, but it is about 1860 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 tool steel – A2 steel depends on heat treatment process, but it is about 1400 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 tool steel – A2 steel 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 Tool Steel – A2 Steel

Rockwell hardness of tool steel – A2 steel depends on heat treatment process, but it is approximately 60 HRC.

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 Tool 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 = 1860 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 = 1860 x 10⁶ x 0.0001 = 186 000 N

Thermal Properties of Tool Steel – A2 Steel

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 Tool Steel – A2 Steel

Melting point of tool steel – A2 steel is around 1420°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 Tool Steel – A2 Steel

The thermal conductivity of tool steel – A2 steel is 26 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 Tool Steel with the thermal conductivity of k₁ = 26 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 + 0.15/26 + 1/30) = 7.19 W/m²K

The heat flux can be then calculated simply as: q = 7.19 [W/m²K] x 30 [K] = 215.67 W/m²

The total heat loss through this wall will be: q_loss = q . A = 215.67 [W/m²] x 30 [m²] = 6470.05 W