A steel alloy to be used for a spring application must have a modulus of resilience of at least \(2.07 \mathrm{MPa}(300 \mathrm{psi}) .\) What must be its minimum yield strength?

Short Answer

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Answer: The minimum yield strength of the steel alloy for a spring application is 910 MPa.

Step by step solution

01

Understanding the Modulus of Resilience formula

The modulus of resilience, represented as \(U_{R}\), is defined as the amount of strain energy per unit volume that a material can absorb without permanent deformation. For a linear elastic material, modulus of resilience can be calculated using the following formula: $$ U_{R} = \frac{1}{2} * \frac{\sigma_{y}^2}{E} $$ Where \(U_{R}\) is the modulus of resilience, \(\sigma_{y}\) is the yield strength, and \(E\) is the modulus of elasticity (Young's modulus). In this exercise, we are given the modulus of resilience and need to find the minimum yield strength.
02

Rearranging the formula to find yield strength

Since we need to calculate the yield strength from the modulus of resilience and Young's modulus, we can rearrange the modulus of resilience formula: $$ \sigma_{y} = \sqrt{2 * U_{R} * E} $$
03

Convert given modulus of resilience to SI unit

Since we are given the modulus of resilience in psi (pound per square inch), let's convert it to the SI unit MPa (Mega Pascal) to keep our calculation consistent: 1 psi = 0.006895 MPa So, 300 psi = 300 * 0.006895 = 2.07 MPa Now we have modulus of resilience, \(U_{R}\) = 2.07 MPa.
04

Finding the modulus of elasticity for steel alloy

The modulus of elasticity (Young's modulus) for steel alloys typically ranges between 190 to 210 GPa. For this example, let us assume the modulus of elasticity for the steel alloy to be 200 GPa. It is important to note that the exact value of modulus of elasticity may vary depending on the specific alloy used. Now we have, \(E\) = 200 GPa (Giga Pascal) = 200,000 MPa.
05

Calculating the minimum yield strength

Now that we have the needed values, we can plug them into the formula from Step 2 and solve for the yield strength: $$ \sigma_{y} = \sqrt{2 * (2.07 \,\text{MPa}) * (200,000 \,\text{MPa})} $$ $$ \sigma_{y} = \sqrt{828,000 \,\text{MPa}} $$ $$ \sigma_{y} = 910 \,\text{MPa} $$ The minimum yield strength for this steel alloy to be used in a spring application must be 910 MPa.

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Key Concepts

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

Yield Strength
Yield strength is a fundamental property of materials that marks the onset of plastic deformation. It is the stress at which a material begins to deform permanently. Before a material reaches its yield point, it behaves elastically, meaning it will return to its original shape when the applied stress is removed.

Understanding yield strength is crucial for engineers and designers as it determines how a material will behave under load before it starts to deform irreversibly. It is especially important in the design of structures and mechanical components that must withstand specific load conditions without failing.

In the context of our exercise where we determine the minimum yield strength for a steel alloy used in a spring, the yield strength represents the maximum stress that the spring material can handle without being permanently bent or stretched.
Modulus of Elasticity
The modulus of elasticity, also known as Young's modulus, is a measure of the stiffness of a material. It defines the relationship between stress (force per unit area) and strain (deformation) in the elastic region of the material’s stress-strain curve.

A higher modulus of elasticity indicates a stiffer material that does not deform much under load. Materials with a high modulus of elasticity, such as steel or diamond, are ideal for applications where rigidity and maintaining shape under stress are important.

In our exercise, knowing the modulus of elasticity, assumed to be 200 GPa for the steel alloy, allows us to calculate the modulus of resilience, which further enables us to derive the minimum yield strength necessary for the spring application. It's a key factor in determining how much energy the material can store elastically.
Strain Energy
Strain energy refers to the energy stored in a material due to elastic deformation. When a material is deformed within its elastic limit, the work done on it is stored in the form of strain energy. Once the stress is removed, this energy is released as the material returns to its original shape.

The concept of strain energy is particularly relevant in spring design. Springs are designed to store and release energy. The modulus of resilience, which is the area under the stress-strain curve up to the yield point, represents the maximum strain energy per unit volume that can be absorbed without permanent deformation.

In the steel alloy spring from our problem, the modulus of resilience specified at 2.07 MPa was used to calculate the minimum yield strength required to make sure the spring can perform as needed without undergoing plastic deformation.

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Most popular questions from this chapter

In Section 2.6 it was noted that the net bonding energy \(E_{N}\) between two isolated positive and negative ions is a function of interionic distance \(r\) as follows: $$E_{N}=-\frac{A}{r}+\frac{B}{r^{n}}\quad\quad\quad (6.25)$$ where \(A, B,\) and \(n\) are constants for the particular ion pair. Equation 6.25 is also valid for the bonding energy between adjacent ions in solid materials. The modulus of elasticity \(E\) is proportional to the slope of the interionic force-separation curve at the equilibrium interionic separation; that is, $$E \propto\left(\frac{d F}{d r}\right)_{r_{0}}$$ Derive an expression for the dependence of the modulus of elasticity on these \(A, B,\) and \(n\) parameters (for the two-ion system) using the following procedure: 1\. Establish a relationship for the force \(F\) as a function of \(r,\) realizing that $$F=\frac{d E_{N}}{d r}$$ 2\. Now take the derivative \(d F / d r\). 3\. Develop an expression for \(r_{0},\) the equilibrium separation. since \(r_{0}\) corresponds to the value of \(r\) at the minimum of the \(E_{N^{-} \text {versus- } r}\) curve (Figure \(2.8 b\) ), take the derivative \(d E_{N} / d r,\) set it equal to zero, and solve for \(r\) which corresponds to \(r_{0}\). 4\. Finally, substitute this expression for \(r_{0}\) into the relationship obtained by taking \(d F / d r\).

Consider a cylindrical specimen of a steel alloy (Figure \(6.21) 8.5 \mathrm{mm}(0.33 \text { in. })\) in diameter and \(80 \mathrm{mm}(3.15 \text { in. })\) long that is pulled in tension. Determine its elongation when a load of \(65,250 \mathrm{N}\left(14,500 \mathrm{lb}_{\mathrm{f}}\right)\) is applied.

Cite five factors that lead to scatter in measured material properties.

A cylindrical specimen of a nickel alloy having an elastic modulus of \(207 \mathrm{GPa}\left(30 \times 10^{6}\right.\) psi) and an original diameter of \(10.2 \mathrm{mm}\) \((0.40 \text { in. })\) will experience only elastic deformation when a tensile load of \(8900 \mathrm{N}\left(2000 \mathrm{lb}_{\mathrm{f}}\right)\) is applied. Compute the maximum length of the specimen before deformation if the maximum allowable elongation is \(0.25 \mathrm{mm}\) \((0.010 \text { in. })\).

(a) A 10 -mm-diameter Brinell hardness indenter produced an indentation \(2.50 \mathrm{mm}\) in diameter in a steel alloy when a load of \(1000 \mathrm{kg}\) was used. Compute the HB of this material. (b) What will be the diameter of an indentation to yield a hardness of \(300 \mathrm{HB}\) when a \(500-\mathrm{kg}\) load is used?

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