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Chapter 12: Problem 55

A full-elliptic leaf spring operates normally with a load that fluctuates between 100 and \(200 \mathrm{lb}\), but is to be designed for an overload that fluctuates between 100 and 300 lb. Total spring length is 24 in., \(h=0.1\) in., \(K_{f}=1.3\), and the steel to be used has \(S_{u}=180 \mathrm{ksi}, S_{y}=160 \mathrm{ksi}\), and \(S_{n}=80 \mathrm{ksi}\) (this figure pertains to the actual size and surface). (a) Determine the total width \(b\) required. (b) Show, on a \(\sigma_{m}-\sigma_{a}\) diagram the "operating points" for (1) machine turned off and spring supporting a static load of 100 lb only, (2) normal loads applied, and (3) design overloads applied. (c) Determine the spring rate.

Short Answer

Expert verified
The total width of the spring required is 225 inches. For different operating conditions, corresponding \(\sigma_{m}-\sigma_{a}\) points can be plotted on the diagram. In the case of design overload, the spring rate comes out to be 200 lb/in, while for normal operating loads, it is 100 lb/in.

Step by step solution

01

Total Width of Spring

Let's start by finding the total width \(b\) of the spring. From the provided fatigue failure theory, the total width can be calculated using the equation \(b = \frac{6W}{S_nh}\), where \(W\) is the maximum load, \(S_n\) is the endurance limit, and \(h\) is the thickness. Substituting the given values, we can find \[b = \frac{6*300}{80*0.1} = 225 \text{inches}\]
02

Plotting the Operating Points

Next, we'll plot the operating points on a \(\sigma_{m}-\sigma_{a}\) diagram. The mean stress \(\sigma_{m}\) and amplitude stress \(\sigma_{a}\) can be calculated using the equations \(\sigma_{m} = \frac{\sigma_{max}+\sigma_{min}}{2}\) and \(\sigma_{a} = \frac{\sigma_{max}-\sigma_{min}}{2}\) respectively. \n1. For machine turned off and spring supporting 100 lb only, \(\sigma_{m}\) and \(\sigma_{a}\) equals to the stress caused by 100 lb both. \n2. For normal loads applied, \(\sigma_{m} = \frac{(100+200)}{2}\) and \(\sigma_{a} = \frac{(200-100)}{2}\). \n3. For design overloads applied, \(\sigma_{m} = \frac{(100+300)}{2}\) and \(\sigma_{a} = \frac{(300-100)}{2}\). \nThese points can now plotted on the \(\sigma_{m}-\sigma_{a}\) diagram.
03

Calculation of Spring Rate

Finally, let's determine the spring rate. The spring rate \(k\) can be calculated based on the ratio of the change in force (\(F\)) to the change in length (\(x\)) according to Hooke's Law. Since the lower and upper loads are given, \[k = \frac{F_{2}-F_{1}}{x_{2}-x_{1}}\]. Given that the spring remains within its elastic range, we can assume \(x_{2}-x_{1}\) is constant, making the rate directly proportional to the load change. As we are only given forces, in this case the spring rate would translate to \[k = F_{2} - F_{1},\] so in context of the design overload, \[k = 300 - 100 = 200 lb/in.\].In relation to the normal operating loads, \[k = 200 - 100 = 100 lb/in.\].

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Key Concepts

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

Fatigue Failure Theory
Fatigue failure theory is crucial in designing components like leaf springs that are subject to fluctuating loads. When a material undergoes repeated stress cycles, it can eventually fail even if the stresses are below its tensile strength. This phenomenon is known as fatigue failure.
Understanding the stresses involved is key. The mean stress \( \sigma_{m} \) represents the average of the maximum and minimum stresses, while the amplitude stress \( \sigma_{a} \) reflects half the difference. Designing for fatigue involves ensuring these stresses stay within a safe limit, usually below the material's endurance limit \( S_n \).
For this leaf spring, the steel used has an endurance limit of 80 ksi, which guides decisions to avoid fatigue-induced failure.
Leaf Spring Design
Leaf spring design is a fascinating aspect of mechanical engineering that combines the principles of elasticity and strength. A full-elliptic leaf spring, like the one in this exercise, is a traditional choice for various applications due to its load-bearing capabilities.
The design process involves determining dimensions like the total width and thickness to ensure it can support the expected loads without deforming or breaking. Factors such as the intended load range, material properties, and the shape of the spring all play roles.
In this case, the total width \( b \) is calculated using the formula \( b = \frac{6W}{S_nh} \), where \( W \) is the maximum load, ensuring the spring can handle the design overload without failure.
Stress Analysis
Stress analysis is an essential part of evaluating how a leaf spring will perform under varying loads. It involves examining both the static and dynamic stresses within the spring.
A \( \sigma_{m}-\sigma_{a} \) diagram helps visualize the stress conditions. Operating points on this diagram show the mean and amplitude stresses during different loading scenarios, enabling engineers to assess the risk of fatigue.
  • Point 1: Static load of 100 lb when the machine is off.
  • Point 2: Normal operation with loads fluctuating between 100 and 200 lb.
  • Point 3: Design overload condition between 100 and 300 lb.

These analyses ensure the spring remains safe and reliable across its operational range.
Spring Rate Calculation
Spring rate calculation helps determine how stiff or soft a leaf spring is. It reflects how much force is required to compress the spring by a certain distance.
Using Hooke's Law, the spring rate \( k \) is calculated as the change in load divided by the change in length. Here, given the forces, \( k = F_{2} - F_{1} \), where \( F_{2} \) and \( F_{1} \) are the higher and lower load limits, respectively.
  • For design overloads, the spring rate is 200 lb/in, reflecting a more robust construction.
  • In normal operation, the spring rate is 100 lb/in, offering a smoother ride.
Understanding the spring rate helps in designing systems that respond optimally to varying loads, maintaining both performance and safety.

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Most popular questions from this chapter

A semielliptic leaf spring similar to the one in Figure \(12.25\) has four leaves, each made from \(0.1 \times 2\)-in. steel having properties of \(S_{u}=180 \mathrm{ksi}, S_{y}=160 \mathrm{ksi}\), and \(S_{n}=80 \mathrm{ksi}\). The latter figure includes appropriate corrections for size and surface. The fatigue stress concentration factor \(K_{f}\) (which is due to stress concentration at the clips and the center hole) is 1.3. Use the simplified "triangular plate" model. (a) What total spring length is needed to give a spring rate of \(75 \mathrm{lb} /\) in. (i.e., \(75 \mathrm{lb}\) applied at the center causes a l-in. deflection at the center)? (b) In service, the spring will carry a static load (applied to the center) of \(P\), plus a superimposed dynamic load that varies from \(+P / 2\) to \(-P / 2\). What is the highest value of \(P\) that will give infinite life, with a safety factor of \(1.3\) ?

Search the Internet for calculators useful for designing compression springs. Select a spring calculator based on the following characteristics: (a) potentially useful, (b) easy to use, \((c)\) results are accurate and correct. Write a short description of the spring calculator.

Review the web site http://www.acxesspring.com. (a) List the common spring materials presented. (b) Which materials are listed as highly susceptible to hydrogen embrittlement? (c) What processes are stated as causing hydrogen embrittlement?

The Iron Arms TM rotating forearm grips exercises the forearms by resisting rotation of the handle grips inward and outward-see Figure P12.32 and Figures P2.2 and P2.3 (Chapter 2 ). The rotation of the gripped handle, about its center, is opposed by the force on the free end of the spring. The handle assembly is shaped like a "D." The curved portion of the D slides smoothly inside the hollow ring; the straight portion of the \(D\) is for hand gripping. The handle grip length is approximately \(4.0\) in. long and \(1.25\) in. in diameter. The rings have an outer diameter of \(7.75\) inches and an inner diameter of \(5.375\) in. The entire Iron Arms assembly has an overall length of \(15.40\) in., a width of \(7.75\) in. (outer ring diameter), and a thickness of \(1.25\) in. Each spring has squared ends with 29 total coils and is made of carbon steel wire. The mean coil diameter is \(.812\) in., and the wire diameter, \(d=0.088\) in. The coil outer diameter is \(0.90\) in. The spring has a free length of \(10.312\) in., and a compressed length of \(9.75\) in. when assembled and in the "starting position." (a) Calculate the energy stored in each helical spring after the spring is rotated \(90^{\circ}\) from the starting position. (b) Calculate the initial force on the fore-end of the spring to start rotation of one handle and the force on the free end of the spring when the handle has rotated \(90^{\circ}\). (c) Calculate the spring constant, \(k\), for the spring. (d) Determine (i) the height of the spring when compressed solid, (ii) the corresponding force on the free end of the spring, and (iii) the rotation of the handle in degrees from the handle start position. (e) Select a carbon steel coil spring material and determine if the spring is designed for infinite life.

Review the web site http://www.mwspring.com. List the information needed for the manufacturer to provide a torsion spring.
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