Longitudinal elastic deformation

Storyboard

When a force is applied to the surface of a body, it creates a zone where the material either compresses or expands, resulting in a motion parallel to the normal vector of the surface. This is what is referred to as longitudinal deformation.

>Model

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Mechanisms

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Code
Concept

Mechanisms

ID:(15370, 0)



Model

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Parameters

Symbol
Text
Variable
Value
Units
Calculate
MKS Value
MKS Units
$L$
L
Body length
m
$S$
S
Body Section
m^2
$\epsilon$
epsilon
Deformación
-
$E$
E
Modulo de elasticidad
Pa
$E$
E
Modulus of Elasticity
Pa
$\nu$
nu
Poisson coefficient
-
$\sigma$
sigma
Tensión
Pa

Variables

Symbol
Text
Variable
Value
Units
Calculate
MKS Value
MKS Units
$\epsilon$
e
Deformation
-
$W$
W
Deformation energy
J
$w$
w
Deformation energy density
$\epsilon_i$
e_i
Deformation in the coordinate $i$
-
$\epsilon_{\perp}$
e_e
Deformation in the direction perpendicular to the force
-
$\epsilon_j$
e_j
Deformation in the perpendicular coordinate $j$
-
$u_i$
u_i
Displacement in i
m
$F_k$
F_k
Elastic Force
N
$u$
u
Elongation
m
$x_i$
x_i
Position in i
m
$S$
S
Section
m^2
$\sigma$
sigma
Strain
Pa
$\sigma_i$
sigma_i
Stress on axis $i$
Pa
$V$
V
Volume
m^3

Calculations


First, select the equation: to , then, select the variable: to

Calculations

Symbol
Equation
Solved
Translated

Calculations

Symbol
Equation
Solved
Translated

Variable Given Calculate Target : Equation To be used




Equations

#
Equation

$ \epsilon =\displaystyle\frac{ u }{ L }$

e = u / L


$ \epsilon_{\perp} =- \nu \epsilon $

e_e =- nu * e


$ \epsilon_i =\displaystyle\frac{\partial u_i }{\partial x_i }$

e_i = du_i / dx_i


$ F_k =\displaystyle\frac{ E S }{ L } u $

F_k = E * S * u / L


$ \sigma = E \epsilon $

sigma = E * epsilon


$ \sigma =\displaystyle\frac{ F_k }{ S }$

sigma = F / S


$ \sigma_i = E \epsilon_i $

s_i = E * e_i


$ U =\displaystyle\frac{1}{2} E \epsilon ^2$

U = E * e ^2/2


$ V = S L $

V = S * L


$ w =\displaystyle\frac{ E }{2(1+ \nu )}\left( \epsilon ^2+2 \epsilon_{\perp} ^2+\displaystyle\frac{ \nu }{1-2 \nu }( \epsilon +2 \epsilon_{\perp} )^2\right)$

w = E *( e ^2+2* e_e ^2 + nu*( e + 2* e_e )^2/(1-2* nu ))/(2*(1+ nu ))


$ W =\displaystyle\frac{1}{2} V E \epsilon ^2$

W = V * E * e ^2/2


$ W =\displaystyle\frac{1}{2 E } V \sigma ^2$

W = V * E * sigma ^2/2


$ w =\displaystyle\frac{ W }{ V }$

w = W / V

ID:(15371, 0)



Hooke force of an object

Equation

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As Hooke's Law relates the elastic Force ($F_k$) through the hooke Constant ($k$) and the elongation ($u$) in the following manner:

$ F_k = k u $



you can replace the hooke Constant ($k$) with the microscopic expression and using the definition of the modulus of Elasticity ($E$), you obtain with the body length ($L$) and the body Section ($S$) that:

$ F_k =\displaystyle\frac{ E S }{ L } u $

$L$
Body length
$m$
5355
$F_k$
Elastic Force
$N$
4978
$u$
Elongation
$m$
5343
$E$
Modulus of Elasticity
$Pa$
5357
$S$
Section
$m^2$
10335

With Hooke's Law for the elastic Force ($F_k$), the hooke Constant ($k$), and the elongation ($u$) as follows:

$ F_k = k u $



and the expression for the hooke Constant ($k$) in terms of the body length ($L$), the body Section ($S$), the microscopic length of spring ($l$), the microscopic section of spring ($s$), and the microscopic Hook constant ($k_m$):

$ k =\displaystyle\frac{ S }{ L }\displaystyle\frac{ l }{ s } k_m $



combined with the expression for the modulus of Elasticity ($E$):

$ E =\displaystyle\frac{ l }{ s } k_m $



the result is:

$ F_k =\displaystyle\frac{ E S }{ L } u $

ID:(3209, 0)



Deformation

Equation

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The elastic Force ($F_k$) is a function of the modulus of Elasticity ($E$), the body Section ($S$), the elongation ($u$), and the body length ($L$).

$ F_k =\displaystyle\frac{ E S }{ L } u $



In this case, the ratio between the elongation ($u$) and the body length ($L$) is represented by the deformation ($\epsilon$), which can be defined as follows:

$ \epsilon =\displaystyle\frac{ u }{ L }$

$L$
Body length
$m$
5355
$\epsilon$
Deformation
$-$
5358
$u$
Elongation
$m$
5343

ID:(3762, 0)



Deformation as continuum

Equation

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In general, the deformation ($\epsilon$) is defined as the variation of the elongation ($u$) in proportion to the body length ($L$):

$ \epsilon =\displaystyle\frac{ u }{ L }$



This concept can be generalized in the microscopic limit, where the deformation in the coordinate $i$ ($\epsilon_i$) is introduced as the variation of displacement in i ($\partial u_i$) over the length of an element in i ($\partial x_i$) in the direction $i$, and it would be expressed as:

$ \epsilon_i =\displaystyle\frac{\partial u_i }{\partial x_i }$

$\epsilon_i$
Deformation in the coordinate $i$
$-$
5359
$u_i$
Displacement in i
$m$
10232
$x_i$
Position in i
$m$
10233



The reason for using a different symbol to denote the differential

$d \rightarrow \partial$

is that there are several differentials that affect different variables in the model. The use of the symbol $\partial$ indicates that one should perform one variation at a time, meaning that when considering one variable, the remaining variables are assumed to have their initial values.

ID:(3763, 0)



Stress

Equation

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The elastic Force ($F_k$) is a function that depends on the modulus of Elasticity ($E$), the body Section ($S$), the elongation ($u$), and the body length ($L$).

$ F_k =\displaystyle\frac{ E S }{ L } u $



Similarly, just as the deformation ($\epsilon$) is introduced to avoid using the dimension the body length ($L$), we can construct a factor that expresses the elastic Force ($F_k$) in terms of the body Section ($S$) as the strain ($\sigma$).

$ \sigma =\displaystyle\frac{ F_k }{ S }$

$ \sigma =\displaystyle\frac{ F }{ S }$

$S$
Body Section
$m^2$
5352
$F$
$F_k$
Elastic Force
$N$
4978
$\sigma$
Strain
$Pa$
5387

ID:(3210, 0)



Hooke's law in the continuous limit

Equation

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The elastic Force ($F_k$) is a function that depends on the modulus of Elasticity ($E$), the body Section ($S$), the elongation ($u$), and the body length ($L$).

$ F_k =\displaystyle\frac{ E S }{ L } u $



This function can be rewritten using the definitions of the strain ($\sigma$) and the deformation ($\epsilon$), resulting in the continuous version of Hooke's Law:

$ \sigma = E \epsilon $

$\epsilon$
Deformación
$-$
8838
$E$
Modulo de elasticidad
$Pa$
8843
$\sigma$
Tensión
$Pa$
8845

The elastic Force ($F_k$) is a function that depends on the modulus of Elasticity ($E$), the body Section ($S$), the elongation ($u$), and the body length ($L$).

$ F_k =\displaystyle\frac{ E S }{ L } u $



This function can be expressed using the definition of the strain ($\sigma$)

$ \sigma =\displaystyle\frac{ F_k }{ S }$



and the definition of the deformation ($\epsilon$)

$ \epsilon =\displaystyle\frac{ u }{ L }$



resulting in

$ \sigma = E \epsilon $

ID:(8100, 0)



Hooke's law continues by direction

Equation

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The Hooke's Law for tensión ($\sigma$), modulo de elasticidad ($E$), and deformación ($\epsilon$) is expressed as:

$ \sigma = E \epsilon $



This law can be generalized for the stress on axis $i$ ($\sigma_i$) and the deformation in the coordinate $i$ ($\epsilon_i$) as:

$ \sigma_i = E \epsilon_i $

$\epsilon_i$
Deformation in the coordinate $i$
$-$
5359
$E$
Modulus of Elasticity
$Pa$
5357
$\sigma_i$
Stress on axis $i$
$Pa$
5361

ID:(3764, 0)



Body volume

Equation

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The total mass the volume ($V$) of the body is calculated using the body Section ($S$) and the body length ($L$):

$ V = S L $

$L$
Body length
$m$
5355
$S$
Section
$m^2$
10335
$V$
Volume
$m^3$
10334

ID:(15374, 0)



Strain Energy

Equation

>Top, >Model


Similar to a spring, deforming a material requires energy. The energy the work ($W$) required to compress or expand the material is calculated as the integral of the elastic Force ($F_k$) along the path $ds$ during deformation:

$W=\displaystyle\int_0^u \vec{F}\cdot d\vec{s}$



In the case of the continuous Hooke's Law, this reduces to:

$ W =\displaystyle\frac{1}{2} V E \epsilon ^2$

$\epsilon$
Deformation
$-$
5358
$W$
Deformation energy
$J$
5368
$E$
Modulus of Elasticity
$Pa$
5357
$V$
Volume
$m^3$
10334

If we use the equation to calculate the work ($W$) as the integral of the elastic Force ($F_k$) along the path during deformation:

$W=\displaystyle\int_0^u \vec{F}\cdot d\vec{s}$



And we employ the equation for the elastic Force ($F_k$) with the modulus of Elasticity ($E$), the body Section ($S$), the elongation ($u$), the body length ($L$), and the elongation ($u$)

$ F_k =\displaystyle\frac{ E S }{ L } u $



where we sum along the path traveled. In the case of elastic deformation, the relationship is linear and becomes:

$W=\displaystyle\frac{ES}{L}\displaystyle\int_0^u d\vec{s}\cdot\vec{s}$



This leads to:

$W=\displaystyle\frac{ES}{2L}u^2$



By using the equation for the deformation ($\epsilon$)

$ \epsilon =\displaystyle\frac{ u }{ L }$



and the equation for the volume ($V$)

$ V = S L $



we obtain:

$ W =\displaystyle\frac{1}{2} V E \epsilon ^2$

ID:(3206, 0)



Energy deformation and stress

Equation

>Top, >Model


As the deformation energy ($W$) is related to the volume ($V$), the modulus of Elasticity ($E$), and the deformation ($\epsilon$), it can be expressed as:

$ W =\displaystyle\frac{1}{2} V E \epsilon ^2$



Using Hooke's Law, we can replace the deformation ($\epsilon$) in terms of the strain ($\sigma$), resulting in:

$ W =\displaystyle\frac{1}{2 E } V \sigma ^2$

$W$
Deformation energy
$J$
5368
$E$
Modulus of Elasticity
$Pa$
5357
$\sigma$
Strain
$Pa$
5387
$V$
Volume
$m^3$
10334

Since the deformation energy ($W$) is related to the volume ($V$), the modulus of Elasticity ($E$), and the deformation ($\epsilon$) as follows:

$ W =\displaystyle\frac{1}{2} V E \epsilon ^2$



If we replace the deformation ($\epsilon$) with the strain ($\sigma$) in the equation:

$ \sigma = E \epsilon $



We obtain:

$ W =\displaystyle\frac{1}{2 E } V \sigma ^2$

ID:(3790, 0)



Energy density

Equation

>Top, >Model


For the deformation energy ($W$) contained within a volume ($V$), we can define the deformation energy density ($w$) as:

$ w =\displaystyle\frac{ W }{ V }$

$W$
Deformation energy
$J$
5368
$w$
Deformation energy density
$J/m^3$
5375
$V$
Volume
$m^3$
10334

ID:(3770, 0)



Potential energy density

Equation

>Top, >Model


The deformation energy ($W$) as a function of the volume ($V$), the modulus of Elasticity ($E$), and the deformation ($\epsilon$) is equal to

$ W =\displaystyle\frac{1}{2} V E \epsilon ^2$



So, if we divide by the volume ($V$), we obtain the deformation energy density ($w$), which is defined as

$ U =\displaystyle\frac{1}{2} E \epsilon ^2$

$\epsilon$
Deformación
$-$
8838
$w$
Deformation energy density
$J/m^3$
5375
$E$
Modulo de elasticidad
$Pa$
8843

The deformation energy ($W$) is expressed as a function of the volume ($V$), the modulus of Elasticity ($E$), and the deformation ($\epsilon$) as follows:

$ W =\displaystyle\frac{1}{2} V E \epsilon ^2$



And with the deformation energy density ($w$) defined as:

$ w =\displaystyle\frac{ W }{ V }$



We obtain:

$ U =\displaystyle\frac{1}{2} E \epsilon ^2$

ID:(8104, 0)



Poisson's ratio

Equation

>Top, >Model


Lateral deformation is directly proportional to the deformation it causes. The proportionality coefficient is denoted as the poisson coefficient ($\nu$) [1] and typically falls within the range of 0.15 to 0.4.

If the original deformation is the deformation ($\epsilon$) and the generated one is the deformation in the direction perpendicular to the force ($\epsilon_{\perp}$), the following relationship is established:

In the linear approximation, the Poisson's coefficient represents the relationship between lateral and longitudinal deformations.

$ \epsilon_{\perp} =- \nu \epsilon $

$\epsilon_i$
Deformation in the coordinate $i$
$-$
5359
$\epsilon_j$
Deformation in the perpendicular coordinate $j$
$-$
5360
$\nu$
Poisson coefficient
$-$
5365

where the sign indicates that the deformation is in the opposite direction to the cause.

[1] This concept was introduced by Siméon Denis Poisson in a statistical analysis work, in which he mentioned, among other unrelated topics to mechanics, what was later referred to as the Poisson's coefficient in an elasticity example. The work is titled "Recherches sur la Probabilité des Jugements en Matière Criminelle et en Matière Civile" (Research on the Probability of Judgments in Criminal and Civil Matters), authored by Siméon Denis Poisson (1837).

ID:(3765, 0)



General potential energy density

Equation

>Top, >Model


densidad de energía elástica ($U$) as a function of modulo de elasticidad ($E$) and deformación ($\epsilon$) is equal to

$ U =\displaystyle\frac{1}{2} E \epsilon ^2$



This equation expresses densidad de energía elástica ($U$) without considering the deformation in the direction perpendicular to the force ($\epsilon_{\perp}$), which is associated with deformación ($\epsilon$) through the Poisson's coefficient. densidad de energía elástica ($U$) can be expressed as a function of deformación ($\epsilon$) and the deformation in the direction perpendicular to the force ($\epsilon_{\perp}$) using the following equation:

$ w =\displaystyle\frac{ E }{2(1+ \nu )}\left( \epsilon ^2+2 \epsilon_{\perp} ^2+\displaystyle\frac{ \nu }{1-2 \nu }( \epsilon +2 \epsilon_{\perp} )^2\right)$

$E$
Deformation
$-$
5358
$w$
Deformation energy density
$J/m^3$
5375
$\epsilon_{\perp}$
Deformation in the direction perpendicular to the force
$-$
10236
$\epsilon$
Modulus of Elasticity
$Pa$
5357
$\nu$
Poisson coefficient
$-$
5365

densidad de energía elástica ($U$) with modulo de elasticidad ($E$), deformación ($\epsilon$), the deformation in the direction perpendicular to the force ($\epsilon_{\perp}$), and the poisson coefficient ($\nu$) is expressed as:

$ w =\displaystyle\frac{ E }{2(1+ \nu )}\left( \epsilon ^2+2 \epsilon_{\perp} ^2+\displaystyle\frac{ \nu }{1-2 \nu }( \epsilon +2 \epsilon_{\perp} )^2\right)$



If we replace the deformation in the direction perpendicular to the force ($\epsilon_{\perp}$) using the equation

$ \epsilon_{\perp} =- \nu \epsilon $



We obtain the initial expression:

$ U =\displaystyle\frac{1}{2} E \epsilon ^2$

ID:(15375, 0)