The spring constant is a unit of measurement. Spring rate

19.11.2019

RIGIDITY

A measure of the compliance of a body of deformation for a given type of load: the more Zh., the less . In the resistance of materials and the theory of elasticity, rigidity is characterized by a coefficient (or total internal force) and a characteristic deformation of an elastic body. body. In the case of tension-compression of the rod Zh. coefficient ES in the ratio e=P/(ES) between the tensile (compressive) force P and refers. elongation k of the rod (5 - cross-sectional area, E - Young's modulus, (see ELASTIC MODULUS). In case of torsion deformation of a round rod, Zh. - polar section, M - torque, q - relative twisting angle of the bar When the bar is bent, EI enters into the ratio c=M/E1 between the bending moment M (the moment of normal stresses in the cross section) and the curvature c of the curved axis of the bar (/ is the axial moment of inertia of the cross section). In the theory of plates and shells, the concept of cylindrical. W. is used: D \u003d Eh3 12 (1-v2), where h is the thickness (of the shell), v is the Poisson coefficient. W. is also determined for some complex structures.

Physical encyclopedic Dictionary. - M.: Soviet Encyclopedia. . 1983 .

RIGIDITY

The ability of a body or structure to resist formation deformations. If the material is subject Hook to the law then the characteristics of Zh. are modulus of elasticity E - in tension, compression, bending and G- at shift. ES in relation e= F/ES between tensile (compressive) force F and relates. elongation e of the rod with a cross-sectional area S. During torsion of a rod with a circular cross section, a rod is characterized by the value GI p(where Ip- polar moment of inertia of the section) in the ratio q=M/GI p , between the torque M and relates. the angle of twist of the rod q. When the beam is bent, Zh., equal to the value EI, is included in the ratio ( =M/EI between bending moment M(the moment of normal stresses in the cross section) and the curvature of the curved axis of the beam (, (where I- axial moment of inertia of the cross section), and when bending plates and shells, under Zh. understand a value equal to Eh 3 / 12 (l - n 2), where h is the thickness of the plate (shell), n is the coefficient. Poisson. J. has creatures. importance in the calculation of structures for stability.

Physical encyclopedia. In 5 volumes. - M.: Soviet Encyclopedia. Editor-in-Chief A. M. Prokhorov. 1988 .

Synonyms:

Antonyms:

See what "HARDNESS" is in other dictionaries:

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radiation hardness- hardness of water - [A.S. Goldberg. English Russian Energy Dictionary. 2006] Topics energy in general Synonyms water hardness EN radiation hardnesshardnessHh …

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A set of properties due to the content of Ca2+ and Mg2+ ions in water. The total concentration of Ca2+ ions (calcium L. century) and Mg2+ (magnesium L. century) is called the total L. century. Distinguish Zh. carbonate and non-carbonate. Carbonate Zh. in. ... ... Great Soviet Encyclopedia

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

First, let's define the basic terms that will be used in this article. It is known that if you act on a body from the outside, it will either gain acceleration or deform. Deformation is a change in the size or shape of a body under the influence of external forces. If the object is fully restored after the termination of the load, then such a 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 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 absolute value to the applied force, but directed in the opposite direction. The elastic force for a spring will be proportional to its elongation. Mathematically, this can be written like this:

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 that we need. 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 has arisen in an elastic body will be proportional to the force that is applied to this body." It is valid only when we are talking about small deformations (stretching or compression is much less than the length of the original body).

Determination of the stiffness factor

Stiffness factor(it also has the names of the coefficient of elasticity or proportionality) is most often written with the letter k, but sometimes you can see 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, by 1 meter). The formula for finding the elasticity coefficient is derived from a special case of Hooke's law:

The greater the value of rigidity, the greater will be the resistance of the body to its deformation. The Hooke coefficient also shows how stable the body is to the action of an external load. This parameter depends on the geometric parameters (wire diameter, number of turns and winding diameter from the wire axis) and on the material from which it is made.

The unit of stiffness in SI is N/m.

System Stiffness Calculation

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

Serial connection of the spring system

When connected in series, the overall rigidity of the system is reduced. The formula for calculating the coefficient of elasticity will be as follows:

1/k = 1/k1 + 1/k2 + … + 1/ki,

where k is the total 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 the spring system

When the springs are connected in parallel, the value of the total coefficient of elasticity of the system will increase. The calculation formula will look like this:

k = k1 + k2 + … + ki.

Measuring the stiffness of the spring empirically - in this video.

Calculation of the stiffness coefficient by experimental method

With the help of a simple experiment, you can independently calculate, what will be the Hooke coefficient. For the experiment you will need:

ruler;
spring;
cargo with a known mass.

The sequence of actions for experience is as follows:

It is necessary to fix the spring vertically, hanging it from any convenient support. The bottom edge must remain free.
Using a ruler, its length is measured and written as x1.
At the free end, you need to hang a load with a known mass m.
The length of the spring is measured in the loaded state. Denoted by x2.
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.
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 (translated into kg), and g is the free acceleration value, which is approximately 9.8.
After the calculations, it remains to find only the stiffness coefficient itself, the formula of which was indicated above: k = F / x.

Examples of tasks for finding stiffness

Task 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.

We calculate the length of the absolute elongation: x = 14-10 = 4 cm = 0.04 m.
According to 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.

Task 2

A load of mass 10 kg, when suspended on a spring, stretched it by 4 cm. Calculate how long another load of mass 25 kg will stretch it.

Let's find the force of gravity that deforms the spring: F = mg = 10 9.8 = 98 N.
Let's determine the coefficient of elasticity: k = F/x = 98 / 0.04 = 2450 N/m.
Calculate the force with which the second load acts: F = mg = 25 9.8 = 245 N.
According to Hooke's law, we write the formula for absolute elongation: x = F/k.
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

This video will show you how to determine the stiffness of a spring.

The greater the deformation of the body, the greater the elastic force arises in it. This means that the deformation and the elastic force are interrelated, and a change in one value can be used to judge a change in the other. So, knowing the deformation of the body, it is possible to calculate the elastic force arising in it. Or, knowing the force of elasticity, determine the degree of deformation of the body.

If suspended from a spring different amount weights of the same mass, then the more of them are suspended, the more the spring will stretch, that is, it will deform. The more the spring is stretched, the greater the elastic force arises in it. Moreover, experience shows that each subsequent suspended weight increases the length of the spring by the same amount.

So, for example, if the original length of the spring was 5 cm, and hanging one weight on it increased it by 1 cm (i.e., the spring became 6 cm long), then hanging two weights would increase it by 2 cm (total length will be 7 cm ), and three - by 3 cm (the length of the spring will be 8 cm).

Even before the experiment, it is known that the weight and the elastic force arising under its action are directly proportional to each other. A multiple increase in weight will increase the strength of elasticity by the same amount. Experience shows that the deformation also depends on the weight: a multiple increase in weight increases the change in length by the same factor. This means that by eliminating the weight, it is possible to establish a directly proportional relationship between the elastic force and deformation.

If we denote the elongation of the spring as a result of its stretching as x or as ∆l (l 1 - l 0, where l 0 is the initial length, l 1 is the length of the stretched spring), then the dependence of the elastic force on tension can be expressed by the following formula:

F control \u003d kx or F control \u003d k∆l, (∆l \u003d l 1 - l 0 \u003d x)

The formula uses the coefficient k . It shows the exact relationship between the elastic force and elongation. Indeed, elongation for each centimeter can increase the elastic force of one spring by 0.5 N, the second by 1 N, and the third by 2 N. For the first spring, the formula will look like F control \u003d 0.5x, for the second - F control \u003d x, for the third - F control = 2x.

The coefficient k is called rigidity springs. The stiffer the spring, the harder it is to stretch, and the greater the value of k. And the more k, the greater will be the elastic force (F control) with equal elongations (x) of different springs.

The stiffness depends on the material from which the spring is made, its shape and size.

The unit of stiffness is N/m (newton per meter). Rigidity shows how many newtons (how many forces) must be applied to a spring in order to stretch it 1 m. Or how many meters a spring will stretch if a force of 1 N is applied to stretch it. For example, a force of 1 N was applied to a spring, and it stretched by 1 cm (0.01 m). This means that its rigidity is 1 N / 0.01 m = 100 N / m.

Also, if you pay attention to the units of measurement, it becomes clear why the stiffness is measured in N / m. The elastic force, like any force, is measured in newtons, and the distance is measured in meters. In order to level the left and right sides of the equation F control = kx in units of measurement, it is necessary to reduce the meters on the right side (that is, divide by them) and add newtons (that is, multiply by them).

Relation between elastic force and deformation elastic body, described by the formula F control \u003d kx, was discovered by the English scientist Robert Hooke in 1660, therefore this ratio bears his name and is called Hooke's law.

Elastic deformation is such when, after the termination of the action of forces, the body returns to its original state. There are bodies that almost cannot be subjected to elastic deformation, while for others it can be quite large. For example, placing a heavy object on a piece of soft clay will change its shape, and this piece will not return to its original state by itself. However, if you stretch the rubber band, then after you release it, it will return to its original size. It should be remembered that Hooke's law is applicable only for elastic deformations.

The formula F control \u003d kx makes it possible to calculate the third from the known two quantities. So, knowing the applied force and elongation, you can find out the rigidity of the body. Knowing the stiffness and elongation, find the elastic force. And knowing the elastic force and stiffness, calculate the change in length.

Definition

The force that occurs as a result of the deformation of the body and trying to return it to its original state is called elastic force.

Most often it is denoted by $(\overline(F))_(upr)$. The elastic force appears only when the body is deformed and disappears if the deformation disappears. If, after removing the external load, the body completely restores its size and shape, then such a deformation is called elastic.

R. Hooke, a contemporary of I. Newton, established the dependence of the elastic force on the magnitude of the deformation. Hooke doubted the validity of his conclusions for a long time. In one of his books, he gave an encrypted formulation of his law. Which meant: "Ut tensio, sic vis" in Latin: what is the stretch, such is the strength.

Consider a spring subject to a tensile force ($\overline(F)$) that is directed vertically downwards (Fig. 1).

The force $\overline(F\ )$ is called the deforming force. Under the influence of a deforming force, the length of the spring increases. As a result, an elastic force ($(\overline(F))_u$) appears in the spring, balancing the force $\overline(F\ )$. If the deformation is small and elastic, then the elongation of the spring ($\Delta l$) is directly proportional to the deforming force:

\[\overline(F)=k\Delta l\left(1\right),\]

where in the coefficient of proportionality is called the stiffness of the spring (coefficient of elasticity) $k$.

Rigidity (as a property) is a characteristic of the elastic properties of a body that is being deformed. Rigidity is considered the ability of a body to resist an external force, the ability to maintain its geometric parameters. The greater the stiffness of the spring, the less it changes its length under the influence of given force. The stiffness coefficient is the main characteristic of stiffness (as a property of a body).

The spring constant depends on the material of which the spring is made and its geometric characteristics. For example, the stiffness coefficient of a coiled coil spring, which is wound from round wire and subjected to elastic deformation along its axis, can be calculated as:

where $G$ is the shear modulus (value depending on the material); $d$ - wire diameter; $d_p$ - spring coil diameter; $n$ is the number of coils of the spring.

The unit of measure for the stiffness coefficient in the International System of Units (SI) is the newton divided by the meter:

\[\left=\left[\frac(F_(upr\ ))(x)\right]=\frac(\left)(\left)=\frac(H)(m).\]

The stiffness coefficient is equal to the amount of force that must be applied to the spring to change its length per unit distance.

Spring stiffness formula

Let $N$ springs be connected in series. Then the stiffness of the entire joint is equal to:

\[\frac(1)(k)=\frac(1)(k_1)+\frac(1)(k_2)+\dots =\sum\limits^N_(\ i=1)(\frac(1) (k_i)\left(3\right),)\]

where $k_i$ is the stiffness of the $i-th$ spring.

When the springs are connected in series, the stiffness of the system is determined as:

Examples of problems with a solution

Example 1

Exercise. The spring in the absence of load has a length $l=0.01$ m and a stiffness equal to 10 $\frac(N)(m).\ $What will be the stiffness of the spring and its length if the force acting on the spring is $F$= 2 N ? Assume that the deformation of the spring is small and elastic.

Solution. The stiffness of the spring under elastic deformations is a constant value, which means that in our problem:

Under elastic deformations, Hooke's law is fulfilled:

From (1.2) we find the elongation of the spring:

\[\Delta l=\frac(F)(k)\left(1.3\right).\]

The length of the stretched spring is:

Calculate the new length of the spring:

Answer. 1) $k"=10\ \frac(Н)(m)$; 2) $l"=0.21$ m

Example 2

Exercise. Two springs with stiffnesses $k_1$ and $k_2$ are connected in series. What will be the elongation of the first spring (Fig. 3) if the length of the second spring is increased by $\Delta l_2$?

Solution. If the springs are connected in series, then the deforming force ($\overline(F)$) acting on each of the springs is the same, that is, it can be written for the first spring:

For the second spring we write:

If the left parts of expressions (2.1) and (2.2) are equal, then the right parts can also be equated:

From equality (2.3) we obtain the elongation of the first spring:

\[\Delta l_1=\frac(k_2\Delta l_2)(k_1).\]

Answer.$\Delta l_1=\frac(k_2\Delta l_2)(k_1)$

Powerelasticity is that power which occurs when the body is deformed and which seeks to restore the former shape and dimensions of the body.

The elastic force arises as a result of the electromagnetic interaction between the molecules and atoms of a substance.

The simplest version of deformation can be considered using the example of compression and extension of a spring.

In this picture (x > 0) — tensile strain; (x< 0) — compression deformation. (FX) is an external force.

In the case when the deformation is the most insignificant, i.e. small, the elastic force is directed to the side, which is opposite in the direction of the moving particles of the body and is proportional to the deformation of the body:

Fx = Fcontrol = - kx

Using this relation, Hooke's law is expressed, which was established experimental method. Coefficient k commonly referred to as the rigidity of the body. The stiffness of a body is measured in newtons per meter (N/m) and depends on the size and shape of the body, as well as on what materials the body is made of.

Hooke's law in physics for determining the compressive or tensile deformation of a body is written in a completely different form. In this case, the relative deformation is called

Robert Hooke

(18.07.1635 - 03.03.1703)

English naturalist, encyclopedist

attitude ε = x / l . At the same time, stress is the cross-sectional area of the body after relative deformation:

σ = F / S = -Fcontrol / S

In this case, Hooke's law is formulated as follows: the stress σ is proportional to the relative strain ε . In this formula, the coefficient E called Young's modulus. This module does not depend on the shape of the body and its dimensions, but at the same time, it directly depends on the properties of the materials that make up the given body. For different materials, Young's modulus fluctuates over a fairly wide range. For example, for rubber E ≈ 2 106 N/m2, and for steel E ≈ 2 1011 N/m2 (i.e. five orders of magnitude more).

It is quite possible to generalize Hooke's law in cases where more complex deformations are performed. For example, consider bending deformation. Consider a rod that rests on two supports and has a significant deflection.

From the side of the support (or suspension), an elastic force acts on this body, this is the reaction force of the support. The reaction force of the support at the contact of the bodies will be directed to the contact surface strictly perpendicular. This force is called the force of normal pressure.

Let's consider the second option. The way the body lies on still horizontal table. Then the reaction of the support balances the force of gravity and it is directed vertically upwards. Moreover, the weight of the body is considered the force with which the body acts on the table.