Question

In the following exercises, substitute the given quantitics into the indicated formula from technology and finance. A bar (Fig. \(1-13\) ) whose length \(L\) is \(15.2 \mathrm{m}\) has a cross- sectional area \(a\) of 12.7 \(\mathrm{cm}^{2} .\) It has an elongation \(e\) of \(2.75 \mathrm{mm}\) when it is subjected to a tensile load of \(22,500 \mathrm{N}\). Use the equation \(E=\frac{P L}{a e}\) to find the modulus of elasticity \(E,\) in newtons per square centimeter.

Short answer

The modulus of elasticity (E) is approximately 9.79 x 10^6 N/cm^2.

Step-by-step solution

Identify Known Quantities

Recognize the known variables provided for substitution: the length of the bar (L) is 15.2 meters (15.2 m), the cross-sectional area (a) is 12.7 square centimeters (12.7 cm^2), the elongation (e) is 2.75 millimeters (2.75 mm), and the tensile load ( P) is 22,500 newtons (22,500 N).

Convert Units

Before substituting into the formula, ensure that all units are consistent. Convert the elongation from millimeters to centimeters by dividing by 10 (1 cm = 10 mm) resulting in e = 2.75 mm = 0.275 cm. Similarly, convert length from meters to centimeters by multiplying by 100 (1 m = 100 cm), getting L = 15.2 m = 1520 cm.

Substitute into the Formula

Substitute the known values into the modulus of elasticity formula E = P * L / (a * e), which gives E = 22,500 N * 1520 cm / (12.7 cm^2 * 0.275 cm).

Solve the Equation

Now, perform the calculation: E = 22,500 * 1520 / (12.7 * 0.275).First, multiply the numerator (22,500 * 1520) to get 34,200,000. Then, multiply the terms in the denominator (12.7 * 0.275) to get 3.4925. Finally, divide the numerator by the denominator: 34,200,000 / 3.4925 to find the modulus of elasticity E.

Calculate and Express in Correct Units

Upon solving 34,200,000 / 3.4925, the modulus of elasticity E is approximately 9,790,186 N/cm^2. Thus, the modulus of elasticity is 9.79 x 10^6 N/cm^2 when expressed in scientific notation.

Key concepts

Tensile Load

Tensile load refers to the force exerted along the length of a material that results in tension, potentially leading to elongation or stretching. In practical terms, when a material is subjected to a tensile load, it experiences a stress that tries to pull it apart.

Understanding tensile load is crucial in the field of material science and engineering, as it directly impacts the design and safety considerations of structures. It is this tensile force that engineers use to calculate stress (\f\(\frac{Force}{Area}\f\)), which then contributes to determining the modulus of elasticity—a measure of a material's ability to resist deformation under load.

In the given exercise, a bar is placed under a tensile load of 22,500 N. To assess the mechanical response of the bar, this tensile force is crucial. The modulus of elasticity calculated from this tensile load gives us insight into how much the bar will stretch or elongate—which, in this case, is 2.75 mm, when the force is applied.

Unit Conversion

Unit conversion is essential in scientific calculations to maintain consistency and ensure accuracy. Different aspects of physics and engineering often use specific units depending on the context or convention, but equations require that all units match to cancel out appropriately and yield the correct result.

In our example, the elongation was originally given in millimeters, but for the formula to work correctly, it was necessary to convert this measurement to centimeters because the other length measurement and the cross-sectional area were in centimeters and square centimeters, respectively. Similarly, we converted the bar's length from meters to centimeters. This step is crucial; without consistent units, the modulus of elasticity could be calculated incorrectly, leading to potentially dangerous errors in material selection or structural design.

It is a good practice to double-check the units used in calculations, ensuring that they are compatible across all factors in a given formula. Unit conversion mistakes are common but avoidable errors in scientific and engineering exercises.

Mechanical Properties of Materials

The mechanical properties of materials relate to the behavior of materials under various loading conditions including tensile, compressive, and shear forces. One such property is the modulus of elasticity, or Young's modulus, which is a measure of the stiffness of a material. It is defined as the ratio of stress (force per unit area) to strain (deformation or elongation per unit length).

The modulus of elasticity is critical for engineers and architects as it determines how much a material will deform under a given load. A high modulus indicates a stiffer material that does not deform much, whereas a low modulus indicates a more flexible material. These properties dictate the applications for which various materials are suitable.

In the demonstrated exercise, the modulus of elasticity is calculated based on the tensile load and resulting elongation, taking into account the cross-sectional area and the original length of the material. This property provides valuable insights into the material's ability to sustain loads while undergoing minimal change in its shape or size. Understanding such mechanical properties is paramount in material selection for safety-critical applications like building construction, automotive engineering, and aerospace design.