Exp.5 Helical Spring PDF

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Tafila Technical University Faculty of Engineering Department of Mechanical Engineering Strength of Materials Laboratory “Helical Spring”

Student Name: Ahmad Nader Ghawanmeh. Student ID: 320180101053. Experiment NO.: 5 Date: 5/5/2020 Instructor Name: Audai AL-Akailah.

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Contents: Background…………………………………………………………………………………………………………………………3 Objective…………………………………………………………………………………………………………………………….3 Calculation and Discussion………………………………………………………………………………………… (4-5-6) Conclusion………………………………………………………………………………………………………………………….6 References………………………………………………………………………………………………………………………….7

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- Background: “Mechanical springs are used in machines to exert force, to provide flexibility, and to store or absorb energy. In general, springs may be classified as either wire springs, flat springs, or specially shaped springs, and there are variations within these divisions. Wire springs include helical springs of round or square wire and are made to resist tensile, compressive or torsional loads. Under flat springs are include the cantilever and elliptical types, the wound motor- or clock-type power springs, and the flat springs washers, usually called Bellville springs.

Fig.1: Different types of springs_ compression, Tension and Torsion In this laboratory, we are interested in modeling and testing only helical springs under tension. A typical helical spring is shown if figure 1. The spring shown has ground, plain ends on both ends. The figure 1(b) makes an imaginary cut of the helical spring and shows that the internal reaction includes a shear force, F, and a torque, T, which acts across the section. The distance, D, is the mean spring diameter, the small distance, d, is the wire diameter, and is the applied force acting at both ends of the spring. The maximum stress for the cut across section can be derived by the superposition of the shear stresses caused by the sheer force and the torque, and may be given as the equation:

 max =

8 FD 4 F + d 3 d 2

Eq. 1

- Objectives: 1- To test the performance of helical springs under tension and compare their comparing their theoretical deformation and spring constants to the actual deformation and spring constant. 2- To find the spring constant of the given Helical Spring using Load - Extension method also to determine the mass of the given body. 3

- Calculation and Discussion:

Fig.2: Compression Spring FBD The stress equation is for the maximum shear stress which occurs at the inner fiber of the spring. At the outside fiber, the shear stress is reduced because the stresses due to torsion and the shear force act in opposite directions. This equation neglects the curvature of the spring, which usually acts as a stress concentrator. If the curvature were included the shear stress distribution deviates from the approximate linear model used above. The curvature stress distribution is shown in figure 2(d).

Spring deflection is usually the most common measurement associated with spring. Hooke’s Law as given in most elementary physics books is presents the spring deflection equation as:

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F =k x

Eq 2.

where is the spring constant, F is the applied force, and x is the spring deflection. For the helical springs a relation similar to k can be derived which takes the form of

ktheor

d 4G = 8D 3N

Eq. 3

Where d, is the wire diameter, D is the means spring diameter, N is the number of active coils in the spring, and G, is the shear modulus of the material. It should be noted that the active coils in the spring is usually less than actual coils since the coils at the ends do not usually support a full load in a compression spring.

Experiment Data: Table 1: Experiment Data Load Extension 0 0 2 0.8 5 2 8 3.2 9 3.6 13 5.2 17 6.8 19 7.6

Spring Constant(K)

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y = 2.5x - 2E-09

Spring Constant (K)= 2.5(N/mm)

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Load(N)

•

10 5 0 0

1

2

3

4

5

6

7

-5

Extension(mm)

Fig.3: Graph represents the load vs extension 5

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• •

A wire of diameter 4 mm is used to form a helical spring of 80 turns and 30 mm outer diameter. Outer diameter D out = 30 mm, then R = 13 mm.

• The calculation of the modulus of rigidity of the spring’s material: G= 64KnR^3/d^4 G= (64*2.5*80*13^3)/ (4^4) G= 109850 Pa • The calculation of the total shear stress developed in the spring when P = 186.39N: Total Sheer stress= 16PR (1+d/4R) Πd^3 Total Sheer stress= (16*186.39*13/3.14*4^3) *(1+4/4*13) Total Sheer stress= 2699.5 Pa

- Conclusion: In this experiment, the stretch of a spring changes as the force applied on the spring changes. As the stretch increases, the force increases, and it is a constant increase. During the conduction of this lab, several new phrases that were defined came up: extension, Hooke’s law, the spring constant, the percent difference, and the Eel equation. Extension is the stretch of the spring—how far it stretches. Hooke’s law is a principle of physics. It states that the force needed to extend (or compress) a spring is proportional to the distance stretched—there is a linear correlation. This ties in with the spring constant, which is the force needed per meter of stretch of a spring. This is also known as “K,” and shows the linear and constant relationship between force and distance stretched.

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- References: 1- https://www.academia.edu/39000133/Experiment_6_-_Helical_Spring 2- https://www.scribd.com/doc/100413138/EGR-236-Lab-10-Helical-Spring-Test

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