Without knowing what the tensile force of the spring is, it is impossible to calculate its stiffness coefficient, so find the tensile force. That is, Fcontrol = kx, where k is the stiffness coefficient. In this case, the weight of the load will be equal to the elastic force acting on the body whose stiffness coefficient needs to be found, for example, a spring.

With a parallel connection, the stiffness increases, with a series connection it decreases. Physics 7th grade, topic 03. Forces around us (13+2 hours) Force and dynamometer. Types of forces. Balanced forces and resultant. Physics 7th grade, topic 06. Introduction to thermodynamics (15+2 hours) Temperature and thermometers.

This relationship expresses the essence of Hooke's law. This means that in order to find the spring stiffness coefficient, the tensile force of the body should be divided by the elongation of a given spring

When a body is deformed, a force arises that tends to restore the previous size and shape of the body. This force arises due to the electromagnetic interaction between atoms and molecules of a substance.

Hooke's law can be generalized to the case of more complex deformations. Spiral springs are often used in technology (Fig. 1.12.3). It should be borne in mind that when a spring is stretched or compressed, complex torsional and bending deformations occur in its coils.

Unlike springs and some elastic materials (rubber), the tensile or compressive deformation of elastic rods (or wires) obey Hooke's linear law within very narrow limits. Secure one end of the spring vertically and leave the other end free. Rigidity is the ability of a part or structure to resist an external force applied to it, while maintaining its geometric parameters if possible.

Various springs are designed to work in compression, tension, torsion or bending. At school, during physics lessons, children are taught to determine the stiffness coefficient of a tension spring. To do this, a spring is suspended vertically on a tripod in a free state.

Calculation of Archimedes' force. Amount of heat and calorimeter. Heat of fusion/crystallization and vaporization/condensation. Heat of fuel combustion and efficiency of heat engines. For example, during bending deformation, the elastic force is proportional to the deflection of the rod, the ends of which lie on two supports (Fig. 1.12.2).

Therefore, it is often called the normal pressure force. Spring extension deformation. For metals, the relative deformation ε = x / l should not exceed 1%. With large deformations, irreversible phenomena (fluidity) and destruction of the material occur. From the point of view of classical physics, a spring can be called a device that accumulates potential energy by changing the distance between the atoms of the material from which the spring is made.

The main characteristic of rigidity is the stiffness coefficient

For steel, for example, E ≈ 2·1011 N/m2, and for rubber E ≈ 2·106 N/m2, i.e. five orders of magnitude less. The elastic force acting on the body from the side of the support (or suspension) is called the support reaction force. When bodies come into contact, the support reaction force is directed perpendicular to the contact surface.

In order to experimentally determine the coefficient of elasticity of the spring you have prepared for the trolley, it will need to be compressed. First find the extension of the spring in meters. The simplest type is tensile and compressive deformation. Calculate the stiffness coefficient by dividing the product of the mass m and the acceleration of gravity g≈9.81 m/s² by the elongation of the body x, k=m g/x. When connecting several elastically deformable bodies (hereinafter referred to as springs for brevity), the overall rigidity of the system will change.

Has the dimension / or kg/s 2 (in SI), din/cm or g/s 2 (in GHS).

The elasticity coefficient is numerically equal to the force that must be applied to the spring in order for its length to change per unit distance.

Definition and properties

The elasticity coefficient, by definition, is equal to the elastic force divided by the change in spring length: $k\; =\; F\_\backslash mathrm(e)\; /\; \backslash Delta\; l.$ The elasticity coefficient depends both on the properties of the material and on the dimensions of the elastic body. Thus, for an elastic rod one can distinguish the dependence on the dimensions of the rod (cross-sectional area $S$ and length $L$), writing the elasticity coefficient as $k\; =\; E\backslash cdot\; S\; /\; L.$ Magnitude $E$ is called Young's modulus and, unlike the coefficient of elasticity, depends only on the properties of the material of the rod.

Stiffness of deformable bodies when they are connected

When connecting several elastically deformable bodies (hereinafter referred to as springs for brevity), the overall rigidity of the system will change. With a parallel connection, the stiffness increases, with a series connection it decreases.

Parallel connection

In parallel connection $n$ $k\_1,\; k\_2,\; k\_3,...,k\_n,$ the rigidity of the system is equal to the sum of the rigidities, that is $k=\; k\_1\; +\; k\_2\; +\; k\_3\; +\; ...\; +\; k\_n.$

Proof

In parallel connection there is $n$ springs with stiffnesses $k\_1,\; k\_2,\; ...\; ,\; k\_n.$ From Newton's third law, $F\; =\; F\_1\; +\; F\_2\; +\; ...\; +\; F\_n.$(Force is applied to them $F$. In this case, a force is applied to spring 1 $F\_1,$ to spring 2 force $F\_2,$..., to the spring $n$ force $F\_n.$)

Now from Hooke's law ( $F\; =\; -k\; x$, where x is the elongation) we derive: $F\; =\; k\; x;\; F\_1\; =\; k\_1\; x;\; F\_2\; =\; k\_2\; x;\; ...;\; F\_n\; =\; k\_n\; x.$ Let's substitute these expressions into equality (1): $k\; x\; =\; k\_1\; x\; +\; k\_2\; x\; +\; ...\; +\; k\_n\; x;$ reducing by $x,$ we get: $k\; =\; k\_1\; +\; k\_2\; +\; ...\; +\; k\_n,$ Q.E.D.

Serial connection

For serial connection $n$ springs with stiffnesses equal to $k\_1,\; k\_2,\; k\_3,...,k\_n,$ The overall stiffness is determined from the equation: $1/k=(1\; /\; k\_1\; +\; 1\; /\; k\_2\; +\; 1\; /\; k\_3\; +\; ...\; +\; 1\; /\; k\_n).$

Proof

In a serial connection there is $n$ springs with stiffnesses $k\_1,\; k\_2,\; ...\; ,\; k\_n.$ From Hooke's law ( $F\; =\; -kl$, where l is the elongation) it follows that $F\; =\; k\backslash cdot\; l.$ The sum of the elongations of each spring is equal to the total elongation of the entire connection $l\_1\; +\; l\_2+\; ...\; +\; l\_n\; =\; l.$

Each spring experiences the same force $F.$ According to Hooke's law, $F\; =\; l\_1\; \backslash cdot\; k\_1\; =\; l\_2\; \backslash cdot\; k\_2\; =\; ...\; =\; l\_n\; \backslash cdot\; k\_n\; .$ From the previous expressions we derive: $l\; =\; F/k,\; \backslash quad\; l\_1\; =\; F\; /\; k\_1,\; \backslash quad\; l\_2\; =\; F\; /\; k\_2,\; \backslash quad\; ...,\; \backslash quad\; l\_n\; =\; F\; /\; k\_n.$ Substituting these expressions into (2) and dividing by $F,$ we get $1\; /\; k\; =\; 1\; /\; k\_1\; +\; 1\; /\; k\_2\; +\; ...\; +\; 1\; /\; k\_n,$ Q.E.D.

Stiffness of some deformable bodies

Constant cross-section rod

A homogeneous rod of constant cross-section, elastically deformed along the axis, has a stiffness coefficient

$k=\backslash frac(E\backslash ,\; S)(L\_0),$ E- Young's modulus, which depends only on the material from which the rod is made; S- cross-sectional area; L 0 - length of the rod.

Cylindrical coil spring

A twisted cylindrical compression or tension spring, wound from a cylindrical wire and elastically deformed along the axis, has a stiffness coefficient

$k\; =\; \backslash frac(G\; \backslash cdot\; d\_\backslash mathrm(D)^4)(8\; \backslash cdot\; d\_\backslash mathrm(F)^3\; \backslash cdot\; n),$ d D - wire diameter; d F - winding diameter (measured from the wire axis); n- number of turns; G- shear modulus (for ordinary steel G≈ 80 GPa, for spring steel G≈ 78500 MPa, for copper ~ 45 GPa).

see also

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An excerpt characterizing the Elasticity Coefficient

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6. Sound research methods in medicine: percussion, auscultation. Phonocardiography.

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Properties of a sample mean

2. Types of deformation. Hooke's law. Hardness coefficient. Elastic modulus. Properties of bone tissue.

Deformation- change in the size, shape and configuration of the body as a result of the action of external or internal forces. types of deformation:

tension-compression is a type of deformation of a body that occurs when a load is applied to it along its longitudinal axis

shear – deformation of a body caused by shear stresses

bending is a deformation characterized by curvature of the axis or gray surface of a deformable object under the influence of external forces.

torsion occurs when a load is applied to a body in the form of a pair of forces in its transverse plane.

Hooke's law- an equation of the theory of elasticity that relates the stress and strain of an elastic medium. In verbal form the law reads as follows:

The elastic force that arises in a body during its deformation is directly proportional to the magnitude of this deformation

For a thin tensile rod, Hooke's law has the form:

Here F is the tension force of the rod, Δl is the absolute elongation (compression) of the rod, and k is called the elasticity (or rigidity) coefficient.

Elasticity coefficient depends both on the properties of the material and on the dimensions of the rod. We can distinguish the dependence on the dimensions of the rod (cross-sectional area S and length L), writing the elasticity coefficient as

The stiffness coefficient is the force that causes a single displacement at a characteristic point (most often at the point of application of force).

Elastic modulus- a general name for several physical quantities that characterize the ability of a solid body (material, substance) to deform elastically when a force is applied to it.

There are no absolutely solid bodies in nature; real solid bodies can “spring” a little - this is elastic deformation. Real solids have a limit of elastic deformation, i.e. such a limit after which the mark from the pressure will already remain and will not disappear on its own.

Properties of bone tissue. Bone is a solid body whose main properties are strength and elasticity.

Bone strength is the ability to withstand external destructive forces. Strength is quantitatively determined by the tensile strength and depends on the design and composition of the bone tissue. Each bone has a specific shape and complex internal structure that allows it to withstand the load in a certain part of the skeleton. Changes in the tubular structure of the bone reduce its mechanical strength. The composition of the bone also significantly affects strength. When minerals are removed, the bone becomes rubbery, and when organic matter is removed, it becomes brittle.

Bone elasticity is the property of regaining its original shape after the cessation of exposure to environmental factors. It, just like strength, depends on the design and chemical composition of the bone.

3. Muscle tissue. The structure and functions of muscle fiber. Energy conversion during muscle contraction. Efficiency of muscle contraction.

Muscle tissue call tissues that are different in structure and origin, but similar in their ability to undergo pronounced contractions. They provide movement in space of the body as a whole, its parts and the movement of organs within the body and consist of muscle fibers.

A muscle fiber is an elongated cell. The composition of the fiber includes its shell - sarcolemma, liquid contents - sarcoplasm, nucleus, mitochondria, ribosomes, contractile elements - myofibrils, and also containing Ca 2+ ions - sarcoplasmic reticulum. The surface membrane of the cell forms transverse tubes at regular intervals through which the action potential penetrates into the cell when it is excited.

The functional unit of muscle fiber is the myofibril. The repeating structure within the myofibril is called a sarcomere. Myofibrils contain 2 types of contractile proteins: thin filaments of actin and twice as thick filaments of myosin. Muscle fiber contraction occurs due to the sliding of myosin filaments along actin filaments. In this case, the overlap of filaments increases and the sarcomere shortens.

home muscle fiber function- ensuring muscle contraction.

Energy conversion during muscle contraction. To contract a muscle, energy is used that is released during the hydrolysis of ATP by actomyosin, and the hydrolysis process is closely associated with the contractile process. By the amount of heat generated by the muscle, one can evaluate the efficiency of energy conversion during contraction. When a muscle shortens, the rate of hydrolysis increases in accordance with the increase in work performed. The energy released during hydrolysis is sufficient to provide only the work performed, but not the full energy production of the muscle.

Efficiency(efficiency) of muscle work ( r) is the ratio of the magnitude of external mechanical work ( W) to the total amount released in the form of heat ( E) energy:

The highest efficiency value of an isolated muscle is observed with an external load of about 50% of the maximum external load. Work productivity ( R) in humans is determined by the amount of oxygen consumption during work and recovery using the formula:

where 0.49 is the proportionality coefficient between the volume of oxygen consumed and the mechanical work performed, i.e. at 100% efficiency for performing work equal to 1 kgf․m(9,81J), required 0.49 ml oxygen.

Motor action / efficiency

Walking/23-33%; Running at average speed/22-30%; Cycling/22-28%; Rowing/15-30%;

Shot put/27%; Throwing/24%; Barbell lift/8-14%; Swimming/ 3%.

"

Definition 1

A spring is an elastic object that is purposefully subjected to compression or stretching, as a result of which it can store energy and then, when the external deforming force weakens, return it. Springs under normal conditions should not be subject to residual (plastic) deformations, i.e. such influences after which the shape of the product is no longer restored due to disruption of the structure of their material.

Types of springs

Springs can be classified according to the direction of the applied load:

• extension springs; designed to work in stretching mode; when deformed, their length increases; As a rule, such devices have a zero step, i.e. wound "turn to turn"; an example would be springs in balance scales, springs for automatically closing doors, etc.;
• Compression springs, on the contrary, shorten under load; in the initial state there is some distance between their turns, as, for example, in car suspension shock absorbers.

This article discusses springs, which are cylindrical spirals. Many other types of elastic devices are used in technology: springs in the form of flat spirals (used in mechanical watches), in the form of strips (springs), torsion springs (in precision scales), disk springs (compressible conical surfaces), etc. A kind of springs are shock-absorbing products made of polymer elastic materials, primarily rubber. All these devices use the same principle - to store the energy of elastic deformation and return it.

Physical characteristics of springs

Coil springs are characterized by a number of parameters, the combination of which determines their rigidity - the ability to resist deformation:

1. material; springs are most often made of steel wire, and the steel used in them is special; it is characterized by medium or high carbon content, low content of other impurities (low-alloy alloy) and special heat treatment (hardening), which gives the material additional elasticity;
2. The diameter of the wire; the smaller it is, the more elastic the spring, but the less its ability to store energy; Compression springs are usually made of thicker wire than extension springs;
3. wire section shape; the wire from which the spring is wound does not always have a round cross-section; compression springs have a flattened section, so that when the length is reduced to a maximum (the coil “sits” on the adjacent coil), the structure is more stable;
4. spring length and diameter; the length of the spring should be distinguished from the length of the wire from which it is wound; these two parameters are consistent through the number of turns and the diameter of the spring, which, in turn, should not be confused with the diameter of the wire.

There are other physical characteristics that affect the performance of springs. For example, when the temperature rises, the metal becomes less elastic, and when it decreases significantly, it can become brittle. During intensive use, the spring loses some of its elasticity over time due to the gradual destruction of the bonds between the atoms of the crystal lattice.

Concept of rigidity

Definition 2

Stiffness as a physical quantity characterizes the force that must be applied to a spring to achieve a certain degree of extension or compression.

The stiffness coefficient is calculated using Hooke's formula:

$F = -k \cdot x$,

where $F$ is the force developed by the spring, $k$ is the stiffness coefficient depending on its characteristics (see above) and measured in newtons per meter, $x$ is the absolute increment in the distance by which the length of the spring has changed after the application of an external force . The minus sign on the right side of the formula indicates that the force generated by the spring acts in the opposite direction to the load.

The stiffness coefficient can be calculated experimentally by hanging weights with a known mass on a spring located vertically and attached to the upper end. In this case there is a dependence

$m \cdot g - k \cdot x = 0$,

where $m$ is mass, $g$ is gravitational acceleration. From here

$k = \frac(m \cdot g)(x)$

Calculation of the stiffness of a cylindrical spring

It's quite easy to understand how a flat spring works. If you place a ruler on the edge of the desk and press one end of it with your hand to the surface, the other can be elastically bent, storing and releasing energy. It is obvious that at the moment of bending, the distances between the molecules of the material in some fragments of the ruler increase, in others they decrease. Electromagnetic bonds operating between molecules tend to return the substance to its previous geometric state.

The situation is somewhat more complicated with a cylindrical spring. Energy is stored in it not due to bending deformation, but due to twisting of the wire from which the spring is wound relative to the longitudinal axis of this wire.

Let us imagine a greatly enlarged cross-section of the wire from which a cylindrical spring is wound, made by a plane perpendicular to its axis. With this consideration, one can abstract from the spiral shape and mentally divide the entire volume of the wire into a set of “cylinders” touching their end surfaces, the diameter of which is equal to the diameter of the wire, and the height tends to zero. Molecular forces act between the contacting ends, preventing deformation.

When the spring is stretched or compressed, the angle of inclination between the coils changes. Neighboring “cylinders” rotate relative to each other in opposite directions around a common axis. Energy is stored in each such section. It follows that the longer the piece of wire the spring is wound from (the diameter and height of the cylinder, as well as the pitch of the coil play a role here), the greater the amount of energy it can store. Increasing the diameter of the wire also increases its energy intensity. In general, the formula that takes into account the main factors of spring stiffness looks like this:

$k = \frac(r^4)(4R^3) \cdot \frac(G)(n)$,

• $R$ is the radius of the spring cylinder,
• $n$ - number of turns of wire with radius $r$,
• $G$ is a coefficient depending on the material.

Let’s substitute numerical values ​​into the formula, simultaneously converting them to SI units:

$k = \frac((10^(-3))^4)(4 \cdot (2 \cdot 10^(-2))^3) \cdot \frac(8 \cdot 10^(10))( 25) = \frac(8 \cdot 10^(-2))(10^2 \cdot 2^3 \cdot 10^(-6)) = 100$

Answer: $100 \frac(N)(m)$

Sooner or later, when studying a physics course, pupils and students are faced with problems on the force of elasticity and Hooke's law, in which the spring stiffness coefficient appears. What is this quantity, and how is it related to the deformation of bodies and Hooke’s law?

First, let's define some basic terms., which will be used in this article. It is known that if you influence a body from the outside, it will either acquire acceleration or become deformed. Deformation is a change in the size or shape of a body under the influence of external forces. If the object is completely restored after the load is removed, then such deformation is considered elastic; if the body remains in an altered state (for example, bent, stretched, compressed, etc.), then the deformation is plastic.

Examples of plastic deformations are:

• clay crafting;
• bent aluminum spoon.

In its turn, Elastic deformations will be considered:

• elastic band (you can stretch it, after which it will return to its original state);
• spring (after compression it straightens again).

As a result of elastic deformation of a body (in particular, a spring), an elastic force arises in it, equal in magnitude to the applied force, but directed in the opposite direction. The elastic force for a spring will be proportional to its elongation. Mathematically it can be written this way:

where F is the elastic force, x is the distance by which the length of the body has changed as a result of stretching, k is the stiffness coefficient necessary for us. The above formula is also a special case of Hooke's law for a thin tensile rod. In general form, this law is formulated as follows: “The deformation that occurs in an elastic body will be proportional to the force that is applied to this body.” It is valid only in cases when we are talking about small deformations (tension or compression is much less than the length of the original body).

Determination of the stiffness coefficient

Hardness coefficient(it is also called the coefficient of elasticity or proportionality) is most often written with the letter k, but sometimes you can find the designation D or c. Numerically, the stiffness will be equal to the magnitude of the force that stretches the spring per unit length (in the case of SI - 1 meter). The formula for finding the elasticity coefficient is derived from a special case of Hooke’s law:

The greater the stiffness value, the greater will be the resistance of the body to its deformation. Hooke's coefficient also shows how resistant a body is to external loads. This parameter depends on geometric parameters (wire diameter, number of turns and winding diameter on the wire axis) and on the material from which it is made.

The SI unit of measurement for hardness is N/m.

System stiffness calculation

There are more complex problems in which calculation of total stiffness is required. In such applications, the springs are connected in series or in parallel.

Series connection of spring system

With a series connection, the overall rigidity of the system decreases. The formula for calculating the elasticity coefficient will be as follows:

1/k = 1/k1 + 1/k2 + … + 1/ki,

where k is the overall stiffness of the system, k1, k2, …, ki are the individual stiffnesses of each element, i is the total number of all springs involved in the system.

Parallel connection of spring system

In the case when the springs are connected in parallel, the value of the overall elasticity coefficient of the system will increase. The formula for calculation will look like this:

k = k1 + k2 + … + ki.

Measurement of spring stiffness experimentally - in this video.

Calculation of the stiffness coefficient using the experimental method

With the help of simple experiment, you can independently calculate what is Hooke's coefficient?. To carry out the experiment you will need:

• ruler;
• spring;
• load with known mass.

The sequence of actions for the experiment is as follows:

1. It is necessary to secure the spring vertically, hanging it from any convenient support. The bottom edge should remain free.
2. Using a ruler, its length is measured and recorded as x1.
3. A load with a known mass m must be suspended from the free end.
4. The length of the spring is measured when loaded. Denoted by x2.
5. The absolute elongation is calculated: x = x2-x1. In order to get the result in the international system of units, it is better to immediately convert it from centimeters or millimeters to meters.
6. The force that caused the deformation is the force of gravity of the body. The formula for calculating it is F = mg, where m is the mass of the load used in the experiment (converted to kg), and g is the value of free acceleration, equal to approximately 9.8.
7. After the calculations, all that remains is to find the stiffness coefficient itself, the formula of which was indicated above: k = F/x.

Examples of problems for finding rigidity

Problem 1

A force F = 100 N acts on a spring 10 cm long. The length of the stretched spring is 14 cm. Find the stiffness coefficient.

1. We calculate the absolute elongation length: x = 14-10 = 4 cm = 0.04 m.
2. Using the formula, we find the stiffness coefficient: k = F/x = 100 / 0.04 = 2500 N/m.

Answer: The spring stiffness will be 2500 N/m.

Problem 2

A load weighing 10 kg, when suspended on a spring, stretched it by 4 cm. Calculate the length to which another load weighing 25 kg will stretch it.

1. Let's find the force of gravity deforming the spring: F = mg = 10 · 9.8 = 98 N.
2. Let's determine the elasticity coefficient: k = F/x = 98 / 0.04 = 2450 N/m.
3. Let's calculate the force with which the second load acts: F = mg = 25 · 9.8 = 245 N.
4. Using Hooke's law, we write the formula for absolute elongation: x = F/k.
5. For the second case, we calculate the stretching length: x = 245 / 2450 = 0.1 m.

Answer: in the second case, the spring will stretch by 10 cm.

Video

In this video you will learn how to determine spring stiffness.