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Figure P12.58D shows a clip, one end of which is shaped like a mouth full of teeth. The clip utilizes a steel helical torsion spring to pinch the end of a toothpaste tube. A clamping force \(F=4.5 \mathrm{lb}\) is adequate for a regular size toothpaste tube \((\equiv 2\) in. wide). The mouth of the clip is about \(1.25 \mathrm{in}\). long and should open approximately \(45^{\circ}\) when clamping the rolled toothpaste tube. Determine the spring diameter \(D\) and the number of turns \(N\), and assume that \(d=1\)-mm-diameter wire is used. Assume that the spring can be manufactured in whole turns. If for fatigue endurance considerations we are not to exert more than a maximum stress of \(\sigma_{\max }=9000 \mathrm{MPa}\) to the spring, verify that the configuration does not exceed this stress for a full open position of \(70^{\circ}\). If the stress condition is not satisfied, suggest a way to reduce the maximum stress without essentially changing the clip design.

Short Answer

Expert verified
The diameter D and the number of turns N of the spring required for the clip can be calculated using the formula for torque of an open spring, and the maximum safe stress limit can be ensured by checking it at the full open position-70 degree. If the stress limit is exceeded, suggest increasing D and/or N to reduce the stress.

Step by step solution

01

Determine Spring Diameter D

To find out the spring diameter \(D\), one can use the expression of the torque for an open spring: \(T_x = F \cdot r \cdot sin 45° = G \cdot (d/2)^4 \cdot theta / ((D/2)^3 \cdot N)\). We already know the values F (force) = 4.5 lb, r = 2 inch, \(theta\) = 45°, \(d\) = 1mm (small diameter), and G (Shear modulus for the material) = 78.5 GPa. So, D can be calculated with these values.
02

Calculate the Number of Turns N

Using the same formula as in the first step, rearrange it to find \(N\). Therefore, \( N = G \cdot (d/2)^4 \cdot theta / (T \cdot (D/2)^3)\). Substitute the known values and calculate \(N\). Note that \(N\) should be rounded up to the nearest whole number, as the spring can only be manufactured in whole turns.
03

Verify Maximum Stress

The maximum stress happens at full open position, which is \(70°\). So, by using the formula \( \sigma_{max} = 16T/(pi \cdot d^3)\), and substituting for \(T\) we get \( \sigma = 16 \cdot G \cdot (d/2)^4 \cdot theta/((pi \cdot d^3) \cdot (D/2)^3 \cdot N) \). The calculated stress should not exceed the maximum stress \(\sigma_{max} = 9000 MPa\). So, comparing the calculated stress with the given maximum stress to verify the spring configuration.
04

Suggest a Way to Reduce Maximum Stress

If the calculated stress exceeds the maximum stress, then the design will fail. To solve this and reduce the maximum stress without changing the clip design, we can increase the spring diameter D and the number of turns N, because increasing these two parameters would help keep stresses in control.

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Key Concepts

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

Spring Diameter Calculation
Calculating the diameter of a torsion spring is a critical step in the design process, as it directly affects the spring's performance and durability. The spring diameter, often referred to as 'D', can be determined using the relationship between the torque applied to the spring and its physical properties.

The torque, in this case, is the rotational force necessary to pinch the toothpaste tube, and is given by the formula:
\(T_x = F \cdot r \cdot \sin 45°\), where \(F\) is the force applied and \(r\) is the lever arm length. To solve for the diameter, the spring's shear modulus (\(G\)) and wire diameter (\(d\)) are utilized in the equation to balance out the torque on the open spring.

A helical torsion spring's diameter is vital in ensuring that the stress levels within the material stay within permissible limits. Using a balance of applied force, wire diameter, spring modulus, and desired angle of displacement, students can calculate the appropriate spring diameter to prevent the spring from failing under stress.
Spring Turns Calculation
The number of turns, or 'N', in a helical torsion spring determines how tightly the spring is coiled and, consequently, its stiffness and ability to store mechanical energy. To calculate 'N', you use the same foundation of the torque equation but rearrange it to isolate 'N'.

The rearranged formula is: \(N = G \cdot (d/2)^4 \cdot \theta / (T_x \cdot (D/2)^3)\). To find the precise number of turns, substitute the known values into this formula.

It’s important to remember that springs are typically produced in whole turns, which means that the calculated value of 'N' must be rounded to the nearest whole number. The number of turns is a critical parameter in spring design because it affects both the operational characteristics and the maximum stresses that the spring will encounter during use.
Stress Verification in Springs
Verifying the stress within a torsion spring is a pivotal part of the design process to ensure the spring does not fail under repeated use, also known as fatigue. The maximum stress experienced by the spring at any point of its deformation must not exceed a certain threshold to avoid material failure.

The maximum stress, designated as \(\sigma_{max}\), is calculated using the formula: \(\sigma_{max} = 16T/(\pi \cdot d^3)\). By plugging in the torque (\(T\)) at the fully open position, the wire diameter (\(d\)), and taking into account the spring's physical configuration (\(D\) and 'N'), one can verify the actual stress in the spring.

If the calculated stress is less than 9000 MPa, the design is considered safe. However, if the resulting stress exceeds the maximum allowed stress, adjustments must be made either by modifying the wire diameter, spring diameter, or the number of turns to reduce the stress without altering the overall clip design.
Helical Torsion Spring
A helical torsion spring is designed to exert torque or rotational force when it is twisted. This type of spring is pivotal in many mechanical systems for precisely controlling rotational movement, such as clipping a toothpaste tube as described in our exercise.

The functionality of a helical torsion spring relies on its ability to store and release energy. When the spring is twisted, it stores mechanical energy, and when released, it produces torque. These characteristics depend on the spring's material, wire diameter, the number of turns, and the overall spring diameter. All these parameters interplay to ensure the spring performs reliably within its stress limits.

In practical applications like the toothpaste clip, the helical torsion spring must be carefully designed to provide the necessary force without exceeding its material strength, thereby providing a safe, durable, and effective mechanical solution.

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Most popular questions from this chapter

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.

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 \(1.3\) ?

List the spring parameters that can be changed while keeping the Iron Arms plastic ring geometry the same. Describe the effect of varying each parameter.

A semielliptic leaf spring for use in a light trailer is to be made of steel having \(S_{u}=1200 \mathrm{MPa}, S_{y}=1080 \mathrm{MPa}\), and a fully corrected endurance limit of \(550 \mathrm{MPa}\). The spring is \(1.2 \mathrm{~m}\) long, and has five leaves of 5 -mm thickness and \(100-\mathrm{mm}\) width. \(K_{f}=1.4\). When the trailer is fully loaded, the static load applied to the center of the spring is \(3500 \mathrm{~N}\). (a) The load alternates as the trailer is driven over a rough road. Estimate what alternating load, when superimposed onto the fully loaded spring, would verge on causing eventual failure from fatigue. (b) What will the maximum deflection of the spring be when loaded as determined in part (a)? (c) How much energy is absorbed by the spring in going from minimum load to maximum load when loaded as determined in part (a)? (d) To what value could the alternating load be increased if only \(10^{4}\) cycles of life are required?

A compression spring has squared ends 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 29 total coils. (a) Calculate the pitch of the spring at the free length. (b) Calculate the solid height for the spring. (c) Calculate the deflection to compress the spring to its solid height. (d) Calculate the spring constant for the spring. (e) Calculate the energy stored in each helical spring after the spring compresses one inch. (f) Determine the force to compress the spring to its solid height. (g) Determine whether the spring will buckle before it reaches its solid height if the spring is unrestrained. (h) Select a carbon steel coil spring material and determine if the spring is designed for infinite life if the force on the spring fluctuates from zero to \(10 \mathrm{lb}\).

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