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Thermal expansion is a fundamental concept in the mechanics of materials. It refers to the way materials change in size due to changes in temperature. Brass, like many materials, expands when heated and contracts when cooled. This property is described by the coefficient of thermal expansion, which for brass is given as \(19.5 \times 10^{-6}\, \text{K}^{-1}\). This means that for every degree Celsius change in temperature, a brass object will expand or contract by a factor of \(19.5 \times 10^{-6}\) of its original length.

In the context of the exercise, the brass wire is subjected to temperature changes that can alter its tensile stress. The relationship is given by \(\Delta \sigma = E \alpha \Delta T\), where \(\Delta \sigma\) is the change in stress, \(E\) is the modulus of elasticity, and \(\Delta T\) is the temperature change. This formula underscores how significant thermal expansion can be, affecting the overall mechanical performance of the material under different thermal conditions.

The modulus of elasticity, also known as Young’s modulus, is a measure of a material's elasticity. It gives us an idea of how much a material will deform under a certain amount of stress. For brass, \(E = 110 \, \text{GPa}\). This is equivalent to \(110 \times 10^3 \, \text{MPa}\), indicating that brass is a relatively stiff material.

In the exercise, the modulus of elasticity plays a crucial role. It connects the change in tensile stress with the change in temperature through the formula \(\Delta \sigma = E \alpha \Delta T\). This relationship means that for a given change in temperature, the amount of stress change can be predicted if \(E\), the modulus of elasticity, is known. The stiffer the material (higher \(E\)), the more stress change is expected for the same temperature change.

Tensile stress refers to the amount of stretching force (tensile force) experienced by a material divided by its cross-sectional area. It is quantified in units such as pascals (Pa) or megapascals (MPa).

In this exercise, we calculated the tensile stress in the wire using the formula \(\sigma = \frac{T}{A}\), where \(T = 98 \, \text{N}\) is the tensile force, and \(A\) is the cross-sectional area of the wire (approximately \(4.6 \times 10^{-6} \, \text{m}^2\)). This calculation results in a tensile stress of approximately \(21.3 \, \text{MPa}\).

This stress level is well below the maximum allowable shear stress of \(60 \, \text{MPa}\) specified in the exercise. This means the wire can safely take this tensile load without yielding. However, changes in thermal conditions must also be considered, as they can lead to additional stress changes.

The cross-sectional area of a wire is vital in determining the resistance to mechanical stresses. For cylindrical objects like wires, the cross-sectional area \(A\) is calculated using the formula: \(A = \frac{\pi d^2}{4}\), where \(d\) is the diameter of the wire.

In the case of the brass wire with a diameter of \(2.42\, \text{mm}\), it is first converted to meters: \(2.42 \times 10^{-3} \, \text{m}\). Substituting this diameter into the formula yields an area of approximately \(4.6 \times 10^{-6}\, \text{m}^2\).

This calculated cross-sectional area is essential for further calculations of tensile stress. It allows for the determination of how much force the wire can withstand per unit area. A larger cross-sectional area means that the wire can shoulder more force without reaching its tensile strength limit.