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For a brass alloy, the stress at which plastic deformation begins is \(345 \mathrm{MPa}(50,000 \mathrm{psi})\), and the modulus of elasticity is \(103 \mathrm{GPa}\left(15.0 \times 10^{6} \mathrm{psi}\right)\) (a) What is the maximum load that can be applied to a specimen with a cross- sectional area of \(130 \mathrm{~mm}^{2}\left(0.2 \mathrm{in} .{ }^{2}\right)\) without plastic deformation? (b) If the original specimen length is \(76 \mathrm{~mm}(3.0\) in.), what is the maximum length to which it can] be stretched without causing plastic deformation?

Short Answer

Expert verified
Answer: The maximum load that can be applied to the specimen without plastic deformation is approximately \(44.85 \times 10^3 \, \mathrm{N}\), and the maximum length to which it can be stretched without causing plastic deformation is approximately \(76.2546 \mathrm{mm}\).

Step by step solution

01

(Step 1: Calculate the maximum load without plastic deformation)

(To calculate the maximum load without plastic deformation, we will use the given stress at which plastic deformation occurs and the cross-sectional area of the specimen. The stress is defined as the force divided by the area. We need to determine the force exerted on the specimen that will not cause plastic deformation. The formula for stress is: \(\text{stress} = \frac{\text{force}}{\text{area}}\) We know the stress at which plastic deformation begins is \(345 \mathrm{MPa}\) and the cross-sectional area is \(130 \mathrm{mm}^2\). To find the force, we can rearrange the formula: \(\text{force} = \text{stress} \times \text{area}\))
02

(Step 2: Find the force)

(Now, simply plug the stress and area values into the formula: \(\text{force} = 345 \mathrm{MPa} \times 130 \mathrm{mm}^2\) Keep in mind that \(1 \mathrm{Pa} = 1 \mathrm{N/m}^2\), and thus \(1 \mathrm{MPa} = 10^6 \mathrm{N/m}^2\). Therefore, you should convert the stress to \(\mathrm{N/m}^2\) before calculating the force: \(\text{force} = 345 \times 10^6 \mathrm{N/m}^2 \times 130 \mathrm{mm}^2\) Since \(1 \mathrm{mm} = 10^{-3}\) meters, convert the area to \(\mathrm{m^2}\): \(\text{force} = 345 \times 10^6 \mathrm{N/m}^2 \times 130 \times 10^{-6} \mathrm{m}^2\) Now, calculate the value: \(\text{force} = 44.85 \times 10^3 \, \mathrm{N}\))
03

(Step 3: Calculate the maximum length without plastic deformation)

(To find the maximum length without plastic deformation, we will use Hooke's Law and the given modulus of elasticity. Hooke's Law states that the stress is proportional to the strain. Mathematically, the proportionality constant is the modulus of elasticity: \(\text{stress} = \text{modulus of elasticity} \times \text{strain}\) Strain is defined as the change in length divided by the original length. We have the stress at which plastic deformation begins and the modulus of elasticity. So, we can calculate the strain: \(\text{strain} = \frac{\text{stress}}{\text{modulus of elasticity}}\) Then, we will use the strain to determine the maximum change in length: \(\text{change in length} = \text{strain} \times \text{original length}\) Finally, we will add the maximum change in length to the original length to find the maximum length without plastic deformation.)
04

(Step 4: Find the strain and change in length)

(Given the stress and modulus of elasticity, we can compute the strain: \(\text{strain} = \frac{345 \mathrm{MPa}}{103 \mathrm{GPa}}\) Convert the values to the same units: \(\text{strain} = \frac{345 \times 10^6 \mathrm{N/m}^2}{103 \times 10^9 \mathrm{N/m}^2}\) Now, compute the value: \(\text{strain} \approx 0.00335\) Next, we have to find the change in length. We know the original length is \(76 \mathrm{mm}\). Therefore: \(\text{change in length} \approx 0.00335 \times 76 \mathrm{mm}\) Calculate the value: \(\text{change in length} \approx 0.2546 \mathrm{mm}\))
05

(Step 5: Find the maximum length without plastic deformation)

(Now, just add the change in length to the original length to find the maximum length without plastic deformation: \(\text{maximum length} = 76 \mathrm{mm} + 0.2546 \mathrm{mm}\) Calculate the value: \(\text{maximum length} \approx 76.2546 \mathrm{mm}\)) Therefore, the maximum load that can be applied to the specimen without plastic deformation is approximately \(44.85 \times 10^3 \, \mathrm{N}\), and the maximum length to which it can be stretched without causing plastic deformation is approximately \(76.2546 \mathrm{mm}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Stress-Strain Relationship
Understanding the stress-strain relationship is essential for materials science and mechanical engineering. This relationship describes how a material deforms under various forces. When a force is applied to a material, it experiences stress, which is the internal force reaction. Simultaneously, the material undergoes a strain, which measures deformation as a ratio of change in length to original length. Stress is calculated as the force divided by the cross-sectional area (\[\text{stress} = \frac{\text{force}}{\text{area}}\]). The stress-strain curve graphically represents this relationship, initially showing a linear region where the material behaves elastically. Here, it obeys Hooke’s Law, returning to its original shape upon unloading.Once the material undergoes plastic deformation, it deforms permanently and does not revert to its original form. The point of transition from elastic to plastic behavior is called the yield point. Accurately determining this point is crucial for ensuring materials do not fail under load.
Modulus of Elasticity
The modulus of elasticity, also known as Young's modulus, is a key mechanical property. It quantifies a material's stiffness by defining the ratio of stress to strain in the linear elastic region of the stress-strain curve:\[\text{modulus of elasticity} = \frac{\text{stress}}{\text{strain}}\]Young's modulus is essential for predicting how materials will deform under various stresses. A high modulus indicates a stiffer material, which resists deformation effectively. For example, brass alloys have a modulus of elasticity of 103 GPa. This value shows they have a good resistance to deformation up to a certain limit. Beyond this elastic limit, the material will not return to its original length, marking the beginning of plastic deformation. Choosing materials with the correct modulus ensures safety and effectiveness in structural applications and machinery.
Hooke's Law
Hooke's Law is fundamental for understanding the elastic behavior of materials. It expresses the proportionality of stress and strain in the initial, linear phase of the stress-strain curve:\[\text{stress} = \text{modulus of elasticity} \times \text{strain}\]Under Hooke’s Law, the force applied to a material is directly proportional to the displacement, provided the limit of elasticity is not exceeded. Within this elastic region, materials exhibit reversible deformation. This law plays a pivotal role in engineering and physics by allowing the prediction and design of objects subjected to various forces.Materials behave elastically up to the yield point, meaning they will return to their original form when the force is removed. It is crucial to keep applied forces within this limit to avoid permanent deformations. In practical applications, the knowledge of when a material will stop obeying Hooke’s Law helps prevent structural failures and ensures reliability.

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Most popular questions from this chapter

A cylindrical rod of steel \(\left(E=207 \mathrm{GPa}, 30 \times 10^{\circ}\right.\) psi) having a yield strength of \(310 \mathrm{MPa}(45,000\) psi) is to be subjected to a load of \(11,100 \mathrm{~N}\) (2500 \(\left.\mathrm{Ib}_{i}\right)\). If the length of the rod is \(500 \mathrm{~mm}(20.0 \mathrm{in}\).), what must be the diameter to allow an elongation of \(0.38 \mathrm{~mm}(0.015\) in.)?

(a) A single crystal of a metal that has the BCC crystal structure is oriented such that a tensile stress is applied in the \([100]\) direction. If the magnitude of this stress is \(4.0 \mathrm{MPa}\), compute the resolved shear stress in the \([1 \overline{11}]\) direction on each of the \((110),(011)\), and \((10 \overline{1})\) planes. (b) On the basis of these resolved shear stress values, which slip system(s) is (are) most favorably oriented?

A cylindrical specimen of a nickel alloy having an elastic modulus of 207 GPa (30\times10^{6 psi) and an } original diameter of \(10.2 \mathrm{~mm}(0.40\) in.) experiences only elastic deformation when a tensile load of \(8900 \mathrm{~N}\left(2000 \mathrm{lb}_{i}\right)\) is applied. Compute the maximum length of the specimen before deformation if the maximum allowable elongation is \(0.25 \mathrm{~mm}\) \((0.010\) in.).

A cylindrical specimen of stainless steel having a diameter of \(12.8 \mathrm{~mm}(0.505 \mathrm{in}\).) and a gauge length of \(50.800 \mathrm{~mm}(2.000 \mathrm{in}\) ) is pulled in tension. Use he load-elongation characteristics shown in the ollowing table to complete parts (a) through (f). (a) Plot the data as engineering stress versus engineering strain. (b) Compute the modulus of elasticity. (c) Determine the yield strength at a strain offset of \(0.002\). (d) Determine the tensile strength of this alloy. (e) What is the approximate ductility, in percent elongation? (f) Compute the modulus of resilience.

Briefly explain why HCP metals are typically more brittle than FCC and BCC metals.

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