Nick Connor

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

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

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

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

Melting Point of 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 thermodynamics/thermodynamic-properties/what-is-temperature-physics/”>temperature

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

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

Melting Point of 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

About Duplex Stainless Steel

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

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

Duplex Stainless Steels – SAF 2205 – 1.4462

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

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

Summary

Name	Duplex Stainless Steel
Phase at STP	solid
Density	7800 kg/m3
Ultimate Tensile Strength	620 MPa
Yield Strength	440 MPa
Young’s Modulus of Elasticity	200 GPa
Brinell Hardness	217 BHN
Melting Point	1450 °C
Thermal Conductivity	19 W/mK
Heat Capacity	460 J/g K
Price	7 $/kg

Density of Duplex Stainless Steel

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

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

Density of Duplex Stainless Steel is 7800 kg/m³.

 

Example: Density

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

Solution:

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

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

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

Properties of Duplex Stainless Steel

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

Mechanical Properties of Duplex Stainless Steel

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

Strength of Duplex Stainless Steel

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

Ultimate Tensile Strength

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

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

Yield Strength

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

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

Young’s Modulus of Elasticity

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

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

Hardness of Duplex Stainless Steel

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

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

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

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

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

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

 

Example: Strength

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

Solution:

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

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

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

Thermal Properties of Duplex Stainless Steel

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

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

Melting Point of Duplex Stainless Steel

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

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

Thermal Conductivity of Duplex Stainless Steel

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

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

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

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

 

Example: Heat transfer calculation

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

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

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

Solution:

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

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

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

 

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

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

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

About Chromoly Steel

Chromoly steel is medium-carbon ultrahigh-strength low alloy steel that gets its name from a combination of the words “chromium” and “molybdenum” – two of the major alloying elements. Chromoly steel is often used when more strength is required than that of mild carbon steel, though it often comes at an increase in cost. Chromoly falls under the AISI 41xx steel designations (ASTM A519). Examples of applications for 4130, 4140 and 4145 include structural tubing, bicycle frames, crank shafts, chain links, drill collars, gas bottles for transportation of pressurized gases, firearm parts, clutch and flywheel components, and roll cages.

Summary

Name	Chromoly Steel
Phase at STP	solid
Density	7850 kg/m3
Ultimate Tensile Strength	700 MPa
Yield Strength	500 MPa
Young’s Modulus of Elasticity	205 GPa
Brinell Hardness	200 BHN
Melting Point	1427 °C
Thermal Conductivity	41 W/mK
Heat Capacity	477 J/g K
Price	4 $/kg

Density of Chromoly 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 Chromoly Steel is 7850 kg/m³.

 

Example: Density

Calculate the height of a cube made of Chromoly 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 = 0.503 m.

Mechanical Properties of Chromoly 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 Chromoly 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 41xx steel – chromoly steel depends on certain grade, but it is about 700 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 41xx steel – chromoly steel depends on certain grade, but it is about 500 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 41xx steel – chromoly steel is 205 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 Chromoly Steel

Brinell hardness of 41xx steel – chromoly steel is approximately 200 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 Chromoly 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 = 700 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 = 700 x 10⁶ x 0.0001 = 70 000 N

Thermal Properties of Chromoly Steel

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

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

Melting Point of Chromoly Steel

Melting point of 41xx steel – chromoly steel is around 1427°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 Chromoly Steel

The thermal conductivity of 41xx steel – chromoly steel is around 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 Chromoly 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 7068 Alloy

7068 aluminium alloy is a heat treatable wrought alloy with good fatigue strength, good anodizing response, and high thermal conductivity. It is alloyed with zinc, magnesium and copper. 7068 aluminium alloy is one of the strongest commercially available aluminium alloys, with a tensile strength comparable to that of some steels.

Summary

Name	7068 Alloy
Phase at STP	solid
Density	2850 kg/m3
Ultimate Tensile Strength	640 MPa
Yield Strength	590 MPa
Young’s Modulus of Elasticity	73 GPa
Brinell Hardness	190 BHN
Melting Point	597 °C
Thermal Conductivity	190 W/mK
Heat Capacity	1050 J/g K
Price	7 $/kg

Density of 7068 Alloy

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 7068 Alloy is 2850 kg/m³.

 

Example: Density

Calculate the height of a cube made of 7068 Alloy, 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 = 0.705 m.

Mechanical Properties of 7068 Alloy

Strength of 7068 Alloy

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 7068 Alloy

Ultimate tensile strength of 7068 Alloy is 640 MPa.

Yield Strength of 7068 Alloy

Yield strength of 7068 Alloy is 590 MPa.

Modulus of Elasticity of 7068 Alloy

The Young’s modulus of elasticity of 7068 Alloy is 73 GPa.

Hardness of 7068 Alloy

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 7068 Alloy is approximately 190 BHN.

See also: Hardness of Materials

 

Example: Strength

Assume a plastic rod, which is made of 7068 Alloy. 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 = 640 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 = 640 x 10⁶ x 0.0001 = 64 000 N

Thermal Properties of 7068 Alloy

7068 Alloy – Melting Point

Melting point of 7068 Alloy is 597 °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.

7068 Alloy – Thermal Conductivity

Thermal conductivity of 7068 Alloy is 190 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.

7068 Alloy – Specific Heat

Specific heat of 7068 Alloy is 1050 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 7068 Alloy with the thermal conductivity of k₁ = 190 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 + 0.15/190 + 1/30) = 7.46 W/m²K

The heat flux can be then calculated simply as: q = 7.46 [W/m²K] x 30 [K] = 223.68 W/m²

The total heat loss through this wall will be: q_loss = q . A = 223.68 [W/m²] x 30 [m²] = 6710.27 W

About Grade 5 Titanium Alloy – Ti-6Al-4V

Grade 5 is the most commonly used alloy and it is an alpha + beta alloy. Grade 5 alloy accounts for 50% of total titanium usage the world over. It has a chemical composition of 6% aluminum, 4% vanadium, 0.25% (maximum) iron, 0.2% (maximum) oxygen, and the remainder titanium. Generally, Ti-6Al-4V is used in applications up to 400 degrees Celsius. It has a density of roughly 4420 kg/m³. It is significantly stronger than commercially pure titanium (grades 1-4) due to its possibility to be heat treated. This grade is an excellent combination of strength, corrosion resistance, weld and fabricability.

Summary

Name	Ti-6Al-4V
Phase at STP	solid
Density	4500 kg/m3
Ultimate Tensile Strength	1170 MPa
Yield Strength	1100 MPa
Young’s Modulus of Elasticity	114 GPa
Brinell Hardness	375 BHN
Melting Point	1660 °C
Thermal Conductivity	6.7 W/mK
Heat Capacity	560 J/g K
Price	50 $/kg

Density of Ti-6Al-4V

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 Ti-6Al-4V is 4500 kg/m³.

 

Example: Density

Calculate the height of a cube made of Ti-6Al-4V, 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 = 0.606 m.

Mechanical Properties of Grade 5 Titanium Alloy – Ti-6Al-4V

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 Grade 5 Titanium Alloy – Ti-6Al-4V

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 Ti-6Al-4V – Grade 5 titanium alloy is about 1170 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 Ti-6Al-4V – Grade 5 titanium alloy 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 Ti-6Al-4V – Grade 5 titanium alloy is about 114 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 Grade 5 Titanium Alloy – Ti-6Al-4V

Rockwell hardness of Ti-6Al-4V – Grade 5 titanium alloy is approximately 41 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 Ti-6Al-4V. 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 = 1170 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 = 1170 x 10⁶ x 0.0001 = 117 000 N

Thermal Properties of Grade 5 Titanium Alloy – Ti-6Al-4V

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

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

Melting Point of Grade 5 Titanium Alloy – Ti-6Al-4V

Melting point of Ti-6Al-4V – Grade 5 titanium alloy is around 1660°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 Grade 5 Titanium Alloy – Ti-6Al-4V

The thermal conductivity of Ti-6Al-4V – Grade 5 titanium alloy is 6.7 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 Ti-6Al-4V with the thermal conductivity of k₁ = 6.7 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 + 0.15/6.7 + 1/30) = 6.42 W/m²K

The heat flux can be then calculated simply as: q = 6.42 [W/m²K] x 30 [K] = 192.65 W/m²

The total heat loss through this wall will be: q_loss = q . A = 192.65 [W/m²] x 30 [m²] = 5779.55 W

About Soft Tin Solder

Tin is a post-transition metal in group 14 of the periodic table. It is obtained chiefly from the mineral cassiterite, which contains tin dioxide. The first alloy used on a large scale was bronze, made of tin and copper, from as early as 3000 BC. Tin is one of the first metals known to humans, it is nontoxic, soft and pliable, and suitable for cold rolling. Tin resists corrosion, making it an ideal coating for other metals. Tin has a low coefficient of friction, and the addition of alloying elements such as copper, antimony, bismuth, cadmium, and silver increase its hardness. Tin has long been used in alloys with lead as solder. Tin itself has a very low melting point, tin alloyed with lead forms a eutectic mixture at the weight proportion of 61.9% tin and 38.1% lead with melting temperature of 183 °C (361.4 °F). Such solders are primarily used for joining pipes or electric circuits.

Soldering is a technique for joining metals using a filler metal alloy that has a melting temperature less than about 425°C (800°F). Because of this lower temperature and different alloys used as fillers, the metallurgical reaction between filler and work piece is minimal, resulting in a weaker joint. In electronics assembly, the eutectic alloy with 63% tin and 37% lead (or 60/40, which is almost identical in melting point) has been the alloy of choice. This eutectic alloy has melting point lower than those of either tin or lead.

Tin is an important constituent in solders because it wets and adheres to many common base metals at temperatures considerably below their melting points. Small amounts of various metals, notably antimony and silver, are added to tin-lead solders to increase their strength. 60-40 solder provides strong and reliable joints under a variety of environmental conditions. There are also high-tin solders, which are used for joining parts of electrical apparatuses because their electrical conductivity is higher than that of high-lead solders. These solders are also used where lead may be a hazard, for example, in contact with drinking water or food.

Summary

Name	Soft Tin Solder
Phase at STP	solid
Density	8600 kg/m3
Ultimate Tensile Strength	56 MPa
Yield Strength	N/A
Young’s Modulus of Elasticity	30 GPa
Brinell Hardness	16 BHN
Melting Point	183 °C
Thermal Conductivity	50 W/mK
Heat Capacity	167 J/g K
Price	70 $/kg

Density of Soft Tin Solder

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 Soft Tin Solder is 8600 kg/m³.

 

Example: Density

Calculate the height of a cube made of Soft Tin Solder, 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 = 0.488 m.

Mechanical Properties of Soft Tin Solder

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 Soft Tin Solder

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 soft solder – 60-40 solder depends greatly on the temperature, but for 19°C is about 56 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.

Young’s Modulus of Elasticity

Young’s modulus of elasticity of soft solder – 60-40 solder is about 30 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 Soft Tin Solder

Brinell hardness of soft solder – 60-40 solder approximately 16 HB.

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 Soft Tin Solder. 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 = 56 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 = 56 x 10⁶ x 0.0001 = 5 600 N

Thermal Properties of Soft Tin Solder

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

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

Melting Point of Soft Solder – 60-40 Solder

Melting point of soft solder – 60-40 solder is around 183°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 Soft Solder – 60-40 Solder

The thermal conductivity of soft solder – 60-40 solder is 50 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 Soft Tin Solder with the thermal conductivity of k₁ = 50 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 + 0.15/50 + 1/30) = 7.33 W/m²K

The heat flux can be then calculated simply as: q = 7.33 [W/m²K] x 30 [K] = 220.05 W/m²

The total heat loss through this wall will be: q_loss = q . A = 220.05 [W/m²] x 30 [m²] = 6601.47 W

About Brass

Brass is is the generic term for a range of copper-zinc alloys. Brass can be alloyed with zinc in different proportions, which results in a material of varying mechanical, corrosion and thermal properties. Increased amounts of zinc provide the material with improved strength and ductility. Brasses with a copper content greater than 63% are the most ductile of any copper alloy and are shaped by complex cold forming operations. Brass has higher malleability than bronze or zinc. The relatively low melting point of brass and its fluidity make it a relatively easy material to cast. Brass can range in surface color from red to yellow to gold to silver depending on the zinc content. Some of the common uses for brass alloys include costume jewelry, locks, hinges, gears, bearings, hose couplings, ammunition casings, automotive radiators, musical instruments, electronic packaging, and coins. 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	Brass
Phase at STP	solid
Density	8530 kg/m3
Ultimate Tensile Strength	315 MPa
Yield Strength	95 MPa
Young’s Modulus of Elasticity	110 GPa
Brinell Hardness	100 BHN
Melting Point	677 °C
Thermal Conductivity	120 W/mK
Heat Capacity	380 J/g K
Price	5 $/kg

Density of Brass

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 Brass is 8530 kg/m³.

 

Example: Density

Calculate the height of a cube made of Brass, 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 = 0.489 m.

Mechanical Properties of Brass

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 Brass

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 cartridge brass – UNS C26000 is about 315 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 cartridge brass – UNS C26000 is about 95 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 cartridge brass – UNS C26000 is about 110 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 Brass

Brinell hardness of cartridge brass – UNS C26000 is approximately 100 MPa.

Rockwell hardness test is one of the most common indentation hardness tests, that has been developed for hardness testing. In contrast to Brinell test, the Rockwell tester measures the depth of penetration of an indenter under a large load (major load) compared to the penetration made by a preload (minor load). The minor load establishes the zero position. The major load is applied, then removed while still maintaining the minor load. The difference between depth of penetration before and after application of the major load is used to calculate the Rockwell hardness number. That is, the penetration depth and hardness are inversely proportional. The chief advantage of Rockwell hardness is its ability to display hardness values directly. The result is a dimensionless number noted as HRA, HRB, HRC, etc., where the last letter is the respective Rockwell scale.

The Rockwell C test is performed with a Brale penetrator (120°diamond cone) and a major load of 150kg.

 

Example: Strength

Assume a plastic rod, which is made of Brass. 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 = 315 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 = 315 x 10⁶ x 0.0001 = 31 500 N

Thermal Properties of Brass

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

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

Melting Point of Brass

Melting point of cartridge brass – UNS C26000 is around 950°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 Brass

The thermal conductivity of cartridge brass – UNS C26000 is 120 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 Brass with the thermal conductivity of k₁ = 120 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 + 0.15/120 + 1/30) = 7.43 W/m²K

The heat flux can be then calculated simply as: q = 7.43 [W/m²K] x 30 [K] = 222.91 W/m²

The total heat loss through this wall will be: q_loss = q . A = 222.91 [W/m²] x 30 [m²] = 6687.31 W