قانون هوك

قانون هوك: لزيادة القوة يزيد الامتداد.

مانومترات تعتمد في عملها على قانون هوك. القوة التي تشكلت بفعل ضغط الغاز داخل الأنبوب المعدني الملفوف تتناسب مع الضغط.

The balance wheel at the core of many mechanical clocks and watches depends on Hooke's law. Since the torque generated by the coiled spring is proportional to the angle turned by the wheel, its oscillations have a nearly constant period.

قانون هوك هو مبدأ في الفيزياء ينص على أن القوة $F$ التي يتغير بها الجسم (الإجهاد) مرتبطة خطيًا بالقوة المسببة لهذا التغير (الشد). المواد التي ينطبق عليها قانون هوك تقريبًا هي مواد خطية المرونة.

سمى قانون هوك على اسم الفيزيائي الإنجليزي روبرت هوك الذي عاش في القرن السابع عشر. لقد ذكر هذا القانون في 1676 كبديل لاتيني, نشره في 1678 كجملة تعني :

"لزيادة القوة يزيد الامتداد"

من أجل الأنظمة التي يطبق عيها قانون هوك، الامتداد الناتج يتناسب مباشرة مع الحمل:

$\mathbf {F} =k\mathbf {x} \$

حيث :

$\ \mathbf {x}$ هي الفرق في المسافة بين موضع الجسم الجديد وموقعه الأصلي سواء كان مضغوطًا أو ممدودا"الازاحة الحاصلة" (عادة تقاس بالمتر)

$\ \mathbf {F}$ هي قوة الإعادة أو كما يطلق عليها القوة المشوهه للجسم اي معناها ان هذه القوة تغير من ابعاد الجسم ولو وصلت لحد معين قد تسبب تشوه للجسم اي لا يعود لشكله الاصلي قبل ان تؤثر عليه تلك القوة التي تمارسها المادة (عادة تقاس بالنيوتن)

و

$\ \mathbf {k}$هو ثابت القوة ووحدته القوة إلى الطول (يقاس بالنيوتن لكل متر)

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 نظرة عامة

يتميز كثير من الأجسام، كالسلك الزنبركي او القضيب المعدني، بخاصية تسمى المرونة، فعندما يستطيل الجسم أو ينضغط تحت تأثير قوة مسلطة فإنه يميل إلى العودة إلى طوله الأصلي عند إزالة القوة. لنفرض مثلاً ان الزنبرك المبين بالشكل (1) طوله الأصلي L0 وانه قد استطال بمقدار LΔ تحت تأثير القوة المسلطة F. بدراسة هذا السلوك وجد روبرت هوك (1635 - 1703) أن الاستطالة تتضاعف مرتين إذا تضاعفت القوة المسلطة مرتين، بشرط ألا تكون الاستطالة كبيرة جداً، أي ان L α FΔ عموماً. وقد وضع هوك اكتشافاته هذه في صورة قاعدة تعرف الآن بقانون هوك:

عندما يتمدد جسم مرن أو يتشوه بأي صورة اخرى فإن مقدار التشوه يتناسب خطياً مع القوة المشوهة.

ولكن عند امتداد (استطالة) الزنبرك بمقدار كبير بحيث يتعدى ما يعرف بحد المرونة فإن ينحرف عن هذا التناسب الطردي بين LΔ و F وعلاوة على ذلك سنلاحظ أن الزنبرك لن يعود إلى طوله الأصلي عند إزالة القوة المسلطة.

شكل 1

وعند استبدال الزنبرك المبين بالشكل (1) بقضيب مصمت سنجد أيضاً أن القضيب يتبع قانون هوك. وبالرغم من أن الاستطالة النسبية للقضيب أصغر كثيراً من قيمتها في حالة الزنبرك فإن القضيب يستطيل بانتظام بما يتفق مع قانون هولك ، ولكن قيم الاستطالة تكون أصغر مما في حالة الزنبرك؛ ويوضح الشكل (2) السلوك المشاهد عملياً في تجربة نموذجية من هذا النوع. لاحظ ان قانون هوك ينطبق في المنطقة المرنة فقط ، وسوف يفترض في المناقشة الآتية أن القوة والاستطالة صغيران بحيث لا يتعدى تشوه المادة حد مرونتها.

شكل 2

لاستخدام قانون هوك في وصف الخواص المرنة للجوامد سوف نستخدم مصطلحين هامين هما الإجهاد والانفعال ، وسنقوم بتعريف هاتين الكميتين بمساعدة تجربة الاستطالة ( او الشد) المبينة بالشكل (3). في هذه التجربة تؤثر القوة الشادة (المطيلة) F عمودياً على المساحة الطرفية A لقضيب طوله الأصلي L0 فيستطيل القضيب نتيجة لذلك بمقدار LΔ. يعرف الإجهاد الناتج عن F كالتالي:

ويعرف انفعال القضيب في الشكل 3)) كما يلي:

شكل 3: إجهاد الشد وإجهاد الضغط في حالة قضيب منتظم الإجهاد هو F/A والانفعال هو L / L0Δ.

وقد عرف الانفعال بالنسبة L / L0Δ، بدلا ً من LΔ، لأن أي جسم مرن يستطيع بمقدار يتناسب طردياً مع طوله الأصلي. وبقسمة LΔ على L0 نكون قد تخلصنا من تأثير طول الجسم على الاستطالة، وهو تأثير لا يمثل أي أهمية فيما يتعلق بخواص مادة القضيب ذاتها. ونظراً لأن الانفعال نسبة بين طولين فإنه كمية ليست لها وحدات. وسنرى مؤخراً في هذا القسم أن هناك انواعاً اخرى من الانفعال ،وهذا يتوقف على الناحية الهندسية للموقف. اما في هذه الحالة الحالية فإننا نتحدث عن انفعال شد. ولكن إذا ضغط القضيب في اتجاه مواز لطوله فإن الانفعال، طبقاً للتعريف، سيكون أيضاً هو النسبة بين التغير في الطول والطول الاصلي.

الآن يمكننا إعادة صياغة قانون هوك. ذلك أن الإجهاد مقياس للقوة المشوهة والانفعال مقياس للتشوه. وعليه يمكن كتابة قانون هوك على الصورة:

(الانفعال) (ثابت) = الإجهاد

وبهذه الصورة يمكن تطبيق قانون هوك على مواقف كثيرة تختلف عن استطالة القضيب، وقد أثبتت تجارب هوك أن هذا القانون صالح للتطبيق في حالات استطالة وانحناء وفي العديد من الزنبركات والأجسام الأخرى. وكما أوضحنا سابقاً فإن قانون هولك ينطبق طبعاً في المنطقة المرنة من التشوهات فقط. يعتمد ثابت التناسب في المعادلة (3) على طبيعة المادة ونوع التشوه الذي تعانيه، وهو يعرف بمعامل مرونة المادة. إذن ، طبقاً للتعريف:

الاجهاد/الانفاعل= معامل المرونة

وحيث أن الانفعال كمية ليس لها وحدات، فإن وحدات معامل المرونة هي نفس وحدات الإجهاد. لاحظ ان معامل المرونة يكون كبيراً عندما يسبب الإجهاد الكبير انفعالاً صغيراً فقط. وعليه فإن معامل المرونة مقياس لجسوءة المادة. وهناك، وفي الواقع، عدد انواع من معاملات المرونة ، وهذا يتوقف على تفاصيل الطريقة التي تستطيع بها المادة أو تنحني او تتشوه بأي طريقة أخرى من الطرق.

 التعريف الرسمي

 الزنبرك الخطي

Plot of applied force F vs. elongation X for a helical spring according to Hooke's law (red line) and what the actual plot might look like (dashed line). At bottom, pictures of spring states corresponding to some points of the plot; the middle one is in the relaxed state (no force applied).

Consider a simple helical spring that has one end attached to some fixed object, while the free end is being pulled by a force whose magnitude is $F$. Suppose that the spring has reached a state of equilibrium, where its length is not changing anymore. Let $X$ be the amount by which the free end of the spring was displaced from its "relaxed" position (when it is not being stretched). Hooke's law states that

$F=kX\,$

or, equivalently,

$X={\frac {1}{k}}F\,$

where $k$ is a positive real number, characteristic of the spring. Moreover, the same formula holds when the spring is compressed, with $F$ and $X$ both negative in that case. According to this formula, the graph of the applied force $F$ as a function of the displacement $X$ will be a straight line passing through the origin, whose slope is $k$.

Hooke's law for a spring is often stated under the convention that $F$ is the restoring (reaction) force exerted by the spring on whatever is pulling its free end. في تلك الحالة تصبح المعادلة:

$F=-kX\,$

since the direction of the restoring force is opposite to that of the displacement.

 الزنبرك "العددي" العام

Hooke's spring law usually applies to any elastic object, of arbitrary complexity, as long as both the deformation and the stress can be expressed by a single number that can be both positive and negative.

 صياغة المتجه

In the case of a helical spring that is stretched or compressed along its axis, the applied (or restoring) force and the resulting elongation or compression have the same direction (which is the direction of said axis). Therefore, if $F$ and $X$ are defined as vectors, Hooke's equation still holds, and says that the force vector is the elongation vector multiplied by a fixed scalar.

 General tensor form

With respect to an arbitrary Cartesian coordinate system, the force and displacement vectors can be represented by 3×1 matrices of real numbers. Then the tensor $\kappa$ connecting them can be represented by a 3×3 matrix $\kappa$ of real coefficients, that, when multiplied by the displacement vector, gives the force vector:

$F\;=\;{\begin{bmatrix}F_{1}\\F_{2}\\F_{3}\end{bmatrix}}\;=\;{\begin{bmatrix}\kappa _{1\,1}&\kappa _{1\,2}&\kappa _{1\,3}\\\kappa _{2\,1}&\kappa _{2\,2}&\kappa _{2\,3}\\\kappa _{3\,1}&\kappa _{3\,2}&\kappa _{3\,3}\end{bmatrix}}{\begin{bmatrix}X_{1}\\X_{2}\\X_{3}\end{bmatrix}}\;=\;\kappa X$

That is,

$F_{i}\;=\;\kappa _{i\,1}X_{1}+\kappa _{i\,2}X_{2}+\kappa _{i\,3}X_{3}\;=\;\sum _{j=1}^{3}\kappa _{i\,j}X_{j}$

for $i$ equal to 1,2, and 3. Therefore, Hooke's law $F=\kappa X$ can be said to hold also when $X$ and $F$ are vectors with variable directions, except that the stiffness of the object is a tensor $\kappa$, rather than a single real number $k$.

 قانون هوك للوسائط المستمرة

المقالة الرئيسية: مرونة خطية

$\sigma =-c\epsilon ,$

where $c$ is a fourth-order tensor (that is, a linear map between second-order tensors) usually called the stiffness tensor or elasticity tensor. One may also write it as

$\epsilon =-s\sigma ,$

where the tensor $s$, called the compliance tensor, represents the inverse of said linear map.

In a Cartesian coordinate system, the stress and strain tensors can be represented by 3×3 matrices

$\epsilon ={\begin{bmatrix}\epsilon _{1\,1}&\epsilon _{1\,2}&\epsilon _{1\,3}\\\epsilon _{2\,1}&\epsilon _{2\,2}&\epsilon _{2\,3}\\\epsilon _{3\,1}&\epsilon _{3\,2}&\epsilon _{3\,3}\end{bmatrix}}\quad \quad \quad \quad \sigma ={\begin{bmatrix}\sigma _{1\,1}&\sigma _{1\,2}&\sigma _{1\,3}\\\sigma _{2\,1}&\sigma _{2\,2}&\sigma _{2\,3}\\\sigma _{3\,1}&\sigma _{3\,2}&\sigma _{3\,3}\end{bmatrix}}$

Being a linear mapping between the nine numbers $\sigma _{i\,j}$ and the nine numbers $\epsilon _{k\,\ell }$, the stiffness tensor $c$ is represented by a matrix of 3×3×3×3 = 81 real numbers $c_{i\,j\,k\,\ell }$. Hooke's law then says that

$\sigma _{i\,j}=-\sum _{k=1}^{3}\sum _{\ell =1}^{3}c_{i\,j\,k\,\ell }\epsilon _{k\,\ell }$

where $i$ and $j$ are 1, 2, or 3.

 قوانين مماثلة

Since Hooke's law is a simple proportionality between two quantities, its formulas and consequences are mathematically similar to those of many other physical laws, such as those describing the motion of fluids, or the polarization of a dielectric by an electric field.

In particular, the tensor equation $\sigma =c\epsilon$ relating elastic stresses to strains is entirely similar to the equation $\tau =\mu {\dot {\epsilon }}$ relating the viscous stress tensor $\tau$ and the strain rate tensor ${\dot {\epsilon }}$ in flows of viscous fluids; although the former pertains to static stresses (related to amount of deformation) while the latter pertains to dynamical stresses (related to the rate of deformation).

 وحدات القياس

In SI units, displacements are measured in metres (m), and forces in newtons (N or kg·m/s²). Therefore the spring constant $k$, and each element of the tensor $\kappa$, is measured in newtons per metre (N/m), or kilograms per second squared (kg/s²).

For continuous media, each element of the stress tensor $\sigma$ is a force divided by an area; it is therefore measured in units of pressure, namely pascals (Pa, or N/m², or kg/m/s². The elements of the strain tensor $\epsilon$ are dimensionless (displacements divided by distances). Therefore the entries of $c_{ijk\ell }$ are also expressed in units of pressure.

 General application to elastic materials

Stress–strain curve for low-carbon steel. Hooke's law is only valid for the portion of the curve between the origin and the yield point (2).
1. Ultimate strength
2. Yield strength – corresponds to yield point
3. Rupture
4. Strain hardening region
5. Necking region
A: Engineering stress (F/A₀)
B: True stress (F/A)

Objects that quickly regain their original shape after being deformed by a force, with the molecules or atoms of their material returning to the initial state of stable equilibrium, often obey Hooke's law.

 صيغ مشتقة

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 Tensional stiffness of a uniform bar

We may view a rod of any elastic material as a linear spring. The rod has length L and cross-sectional area A. Its extension (strain) is linearly proportional to its tensile stress σ by a constant factor $\varepsilon$, the inverse of its modulus of elasticity E, such that

$E\varepsilon =[constant]=\sigma$.

In turn,

$\varepsilon ={\frac {\Delta L}{L}}$    (i.e., [change in length] as a fraction or percentage of total length),

and because

$\sigma ={\frac {F}{A}}$ ,

such that

$\varepsilon ={\frac {\sigma }{E}}={\frac {({\frac {F}{A}})}{E}}={\frac {F}{AE}}$ ,

this relationship may also be expressed as

$\Delta L=\varepsilon L={\frac {\sigma }{E}}L={\frac {F}{AE}}L={\frac {FL}{AE}}$  .

 طاقة الزنبرك

The potential energy stored in a spring is given by

$\mathrm {PE} ={1 \over 2}kx^{2}$

which comes from adding up the energy it takes to incrementally compress the spring. That is, the integral of force over displacement. Since the external force has the same general direction as the displacement, the potential energy of a spring is always non-negative.

 Harmonic oscillator

انظر أيضاً: Harmonic oscillator

A mass suspended by a spring is the classical example of a harmonic oscillator

 Rotation in Gravity-Free Space

If the mass m was attached to a spring with force constant k and rotating in free space, the spring tension (F_t) would balance the required centripetal force (F_c) as follows -

$F_{t}=kx$

$F_{c}=m\omega ^{2}r$

Since F_t = F_c and x = r, therefore:

$k=m\omega ^{2}$

Given that $\omega =2\pi f$, this leads to the same frequency equation as above -

$f={1 \over 2\pi }{\sqrt {k \over m}}$

 Linear elasticity theory for continuous media

Note: the Einstein summation convention of summing on repeated indices is used below.

 Isotropic materials

(see viscosity for an analogous development for viscous fluids.)

Thus in index notation:

$\varepsilon _{ij}=\left({\tfrac {1}{3}}\varepsilon _{kk}\delta _{ij}\right)+\left(\varepsilon _{ij}-{\tfrac {1}{3}}\varepsilon _{kk}\delta _{ij}\right)$

where $\delta _{ij}$ is the Kronecker delta. In direct tensor notation:

${\boldsymbol {\varepsilon }}=\mathrm {vol} ({\boldsymbol {\varepsilon }})+\mathrm {dev} ({\boldsymbol {\varepsilon }})~;~~\mathrm {vol} ({\boldsymbol {\varepsilon }}):={\tfrac {1}{3}}~\mathrm {tr} ({\boldsymbol {\varepsilon }})~\mathbf {I} ~;~~\mathrm {dev} ({\boldsymbol {\varepsilon }}):={\boldsymbol {\varepsilon }}-\mathrm {vol} ({\boldsymbol {\varepsilon }})$

where $\mathbf {I}$ is the second-order identity tensor. The first term on the right is the constant tensor, also known as the volumetric strain tensor, and the second term is the traceless symmetric tensor, also known as the deviatoric strain tensor or shear tensor.

The most general form of Hooke's law for isotropic materials may now be written as a linear combination of these two tensors:

$\sigma _{ij}=3K\left({\tfrac {1}{3}}\varepsilon _{kk}\delta _{ij}\right)+2G\left(\varepsilon _{ij}-{\tfrac {1}{3}}\varepsilon _{kk}\delta _{ij}\right)\,~;~~{\boldsymbol {\sigma }}=3K~\mathrm {vol} ({\boldsymbol {\varepsilon }})+2G~\mathrm {dev} ({\boldsymbol {\varepsilon }})$

where K is the bulk modulus and G is the shear modulus.

Using the relationships between the elastic moduli, these equations may also be expressed in various other ways. A common form of Hooke's law for isotropic materials, expressed in direct tensor notation, is

${\boldsymbol {\sigma }}=\lambda ~\mathrm {tr} ({\boldsymbol {\varepsilon }})~\mathbf {I} +2\mu ~{\boldsymbol {\varepsilon }}={\mathsf {c}}:{\boldsymbol {\varepsilon }}~;~~{\mathsf {c}}=\lambda ~\mathbf {I} \otimes \mathbf {I} +2\mu ~{\mathsf {I}}$

where $\lambda :=K-2/3G$ and $\mu :=G$ are the Lamé constants, $\mathbf {I}$ is the second-rank identity tensor, and ${\mathsf {I}}$ is the symmetric part of the fourth-rank identity tensor. In index notation:

$\sigma _{ij}=\lambda ~\varepsilon _{kk}~\delta _{ij}+2\mu ~\varepsilon _{ij}=c_{ijk\ell }~\varepsilon _{k\ell }~;~~c_{ijk\ell }=\lambda ~\delta _{ij}~\delta _{k\ell }+\mu ~(\delta _{ik}~\delta _{j\ell }+\delta _{i\ell }~\delta _{jk})$

The inverse relationship is

${\boldsymbol {\varepsilon }}={\tfrac {1}{2\mu }}~{\boldsymbol {\sigma }}-{\tfrac {\lambda }{2\mu (3\lambda +2\mu )}}~\mathrm {tr} ({\boldsymbol {\sigma }})~\mathbf {I} ={\tfrac {1}{2G}}~{\boldsymbol {\sigma }}+\left({\tfrac {1}{9K}}-{\tfrac {1}{6G}}\right)~\mathrm {tr} ({\boldsymbol {\sigma }})~\mathbf {I}$

Therefore the compliance tensor in the relation ${\boldsymbol {\varepsilon }}={\mathsf {s}}:{\boldsymbol {\sigma }}$ is

${\mathsf {s}}=-{\tfrac {\lambda }{2\mu (3\lambda +2\mu )}}~\mathbf {I} \otimes \mathbf {I} +{\tfrac {1}{2\mu }}~{\mathsf {I}}=\left({\tfrac {1}{9K}}-{\tfrac {1}{6G}}\right)~\mathbf {I} \otimes \mathbf {I} +{\tfrac {1}{2G}}~{\mathsf {I}}$

In terms of Young's modulus and Poisson's ratio, Hooke's law for isotropic materials can then be expressed as

$\varepsilon _{ij}={\tfrac {1}{E}}(\sigma _{ij}-\nu [\sigma _{kk}\delta _{ij}-\sigma _{ij}])~;~~{\boldsymbol {\varepsilon }}={\tfrac {1}{E}}({\boldsymbol {\sigma }}-\nu [\mathrm {tr} ({\boldsymbol {\sigma }})~\mathbf {I} -{\boldsymbol {\sigma }}])$

This is the form in which the strain is expressed in terms of the stress tensor in engineering. The expression in expanded form is

${\begin{aligned}\varepsilon _{11}&={\tfrac {1}{E}}\left[\sigma _{11}-\nu (\sigma _{22}+\sigma _{33})\right]\\\varepsilon _{22}&={\tfrac {1}{E}}\left[\sigma _{22}-\nu (\sigma _{11}+\sigma _{33})\right]\\\varepsilon _{33}&={\tfrac {1}{E}}\left[\sigma _{33}-\nu (\sigma _{11}+\sigma _{22})\right]\\\varepsilon _{12}&={\tfrac {1}{2G}}~\sigma _{12}~;~~\varepsilon _{13}={\tfrac {1}{2G}}~\sigma _{13}~;~~\varepsilon _{23}={\tfrac {1}{2G}}~\sigma _{23}\end{aligned}}$

where E is the Young's modulus and $\nu$ is Poisson's ratio. (See 3-D elasticity).

Derivation of Hooke's law in 3D

The 3-D form of Hooke's law can be derived using Poisson's ratio and the 1-D form of Hooke's law as follows. Consider the strain and stress relation as a superposition of two effects: stretching in direction of the load (1) and shrinking (caused by the load) in perpendicular directions (2 and 3), $\varepsilon _{1}'={\frac {1}{E}}\sigma _{1}$, $\varepsilon _{2}'=-{\frac {\nu }{E}}\sigma _{1}$, $\varepsilon _{3}'=-{\frac {\nu }{E}}\sigma _{1}$, where $\nu$ is the Poisson's ratio and $E$ the Young's modulus. We get similar equations to the loads in directions 2 and 3, $\varepsilon _{1}''=-{\frac {\nu }{E}}\sigma _{2}$, $\varepsilon _{2}''={\frac {1}{E}}\sigma _{2}$, $\varepsilon _{3}''=-{\frac {\nu }{E}}\sigma _{2}$, and $\varepsilon _{1}'''=-{\frac {\nu }{E}}\sigma _{3}$, $\varepsilon _{2}'''=-{\frac {\nu }{E}}\sigma _{3}$, $\varepsilon _{3}'''={\frac {1}{E}}\sigma _{3}$. Summing the three cases together ($\varepsilon _{i}=\varepsilon _{i}'+\varepsilon _{i}''+\varepsilon _{i}'''$) we get $\varepsilon _{1}={\frac {1}{E}}(\sigma _{1}-\nu (\sigma _{2}+\sigma _{3}))$ $\varepsilon _{2}={\frac {1}{E}}(\sigma _{2}-\nu (\sigma _{1}+\sigma _{3}))$ $\varepsilon _{3}={\frac {1}{E}}(\sigma _{3}-\nu (\sigma _{1}+\sigma _{2}))$ or by adding and subtracting one $\nu \sigma$ $\varepsilon _{1}={\frac {1}{E}}((1+\nu )\sigma _{1}-\nu (\sigma _{1}+\sigma _{2}+\sigma _{3}))$ $\varepsilon _{2}={\frac {1}{E}}((1+\nu )\sigma _{2}-\nu (\sigma _{1}+\sigma _{2}+\sigma _{3}))$ $\varepsilon _{3}={\frac {1}{E}}((1+\nu )\sigma _{3}-\nu (\sigma _{1}+\sigma _{2}+\sigma _{3}))$ and further we get by solving $\sigma _{1}$ $\sigma _{1}={\frac {E}{1+\nu }}\varepsilon _{1}+{\frac {\nu }{1+\nu }}(\sigma _{1}+\sigma _{2}+\sigma _{3})$. Calculating the sum $\sum _{i=1,2,3}\varepsilon _{i}={\frac {1}{E}}((1+\nu )\sum _{i=1,2,3}\sigma _{i}-3\nu (\sum _{i=1,2,3}\sigma _{i}))={\frac {1-2\nu }{E}}\sum _{i=1,2,3}\sigma _{i}$ $\sigma _{1}+\sigma _{2}+\sigma _{3}={\frac {E}{1-2\nu }}(\varepsilon _{1}+\varepsilon _{2}+\varepsilon _{3})$ and substituting it to the equation solved for $\sigma _{1}$ gives $\sigma _{1}={\frac {E}{1+\nu }}\varepsilon _{1}+{\frac {E\nu }{(1+\nu )(1-2\nu )}}(\varepsilon _{1}+\varepsilon _{2}+\varepsilon _{3})$, $\sigma _{1}=2\mu \varepsilon _{1}+\lambda (\varepsilon _{1}+\varepsilon _{2}+\varepsilon _{3})$, where $\mu$ and $\lambda$ are the Lamé parameters. Similar treatment of directions 2 and 3 gives the Hooke's law in three dimensions.

In matrix form, Hooke's law for isotropic materials can be written as

${\begin{bmatrix}\varepsilon _{11}\\\varepsilon _{22}\\\varepsilon _{33}\\2\varepsilon _{23}\\2\varepsilon _{13}\\2\varepsilon _{12}\end{bmatrix}}={\begin{bmatrix}\varepsilon _{11}\\\varepsilon _{22}\\\varepsilon _{33}\\\gamma _{23}\\\gamma _{13}\\\gamma _{12}\end{bmatrix}}={\cfrac {1}{E}}{\begin{bmatrix}1&-\nu &-\nu &0&0&0\\-\nu &1&-\nu &0&0&0\\-\nu &-\nu &1&0&0&0\\0&0&0&2(1+\nu )&0&0\\0&0&0&0&2(1+\nu )&0\\0&0&0&0&0&2(1+\nu )\end{bmatrix}}{\begin{bmatrix}\sigma _{11}\\\sigma _{22}\\\sigma _{33}\\\sigma _{23}\\\sigma _{13}\\\sigma _{12}\end{bmatrix}}$

where $\gamma _{ij}:=2\varepsilon _{ij}$ is the engineering shear strain. The inverse relation may be written as

${\begin{bmatrix}\sigma _{11}\\\sigma _{22}\\\sigma _{33}\\\sigma _{23}\\\sigma _{13}\\\sigma _{12}\end{bmatrix}}={\cfrac {E}{(1+\nu )(1-2\nu )}}{\begin{bmatrix}1-\nu &\nu &\nu &0&0&0\\\nu &1-\nu &\nu &0&0&0\\\nu &\nu &1-\nu &0&0&0\\0&0&0&(1-2\nu )/2&0&0\\0&0&0&0&(1-2\nu )/2&0\\0&0&0&0&0&(1-2\nu )/2\end{bmatrix}}{\begin{bmatrix}\varepsilon _{11}\\\varepsilon _{22}\\\varepsilon _{33}\\2\varepsilon _{23}\\2\varepsilon _{13}\\2\varepsilon _{12}\end{bmatrix}}$

which can be simplified thanks to the Lamé constants :

${\begin{bmatrix}\sigma _{11}\\\sigma _{22}\\\sigma _{33}\\\sigma _{23}\\\sigma _{13}\\\sigma _{12}\end{bmatrix}}={\begin{bmatrix}2\mu +\lambda &\lambda &\lambda &0&0&0\\\lambda &2\mu +\lambda &\lambda &0&0&0\\\lambda &\lambda &2\mu +\lambda &0&0&0\\0&0&0&\mu &0&0\\0&0&0&0&\mu &0\\0&0&0&0&0&\mu \end{bmatrix}}{\begin{bmatrix}\varepsilon _{11}\\\varepsilon _{22}\\\varepsilon _{33}\\2\varepsilon _{23}\\2\varepsilon _{13}\\2\varepsilon _{12}\end{bmatrix}}$

 Plane stress

Under plane stress conditions $\sigma _{31}=\sigma _{13}=\sigma _{32}=\sigma _{23}=\sigma _{33}=0$. In that case Hooke's law takes the form

${\begin{bmatrix}\varepsilon _{11}\\\varepsilon _{22}\\2\varepsilon _{12}\end{bmatrix}}={\cfrac {1}{E}}{\begin{bmatrix}1&-\nu &0\\-\nu &1&0\\0&0&2(1+\nu )\end{bmatrix}}{\begin{bmatrix}\sigma _{11}\\\sigma _{22}\\\sigma _{12}\end{bmatrix}}$

The inverse relation is usually written in the reduced form

${\begin{bmatrix}\sigma _{11}\\\sigma _{22}\\\sigma _{12}\end{bmatrix}}={\cfrac {E}{1-\nu ^{2}}}{\begin{bmatrix}1&\nu &0\\\nu &1&0\\0&0&{\cfrac {1-\nu }{2}}\end{bmatrix}}{\begin{bmatrix}\varepsilon _{11}\\\varepsilon _{22}\\2\varepsilon _{12}\end{bmatrix}}$

 Anisotropic materials

$\sigma _{ij}={\cfrac {\partial U}{\partial \epsilon _{ij}}}\quad \implies \quad c_{ijk\ell }={\cfrac {\partial ^{2}U}{\partial \epsilon _{ij}\partial \epsilon _{k\ell }}}~.$

The arbitrariness of the order of differentiation implies that $c_{ijk\ell }=c_{k\ell ij}\,$. These are called the major symmetries of the stiffness tensor. This reduces the number of elastic constants to 21 from 36. The major and minor symmetries indicate that the stiffness tensor has only 21 independent components.

 Matrix representation (stiffness tensor)

It is often useful to express the anisotropic form of Hooke's law in matrix notation, also called Voigt notation. To do this we take advantage of the symmetry of the stress and strain tensors and express them as six-dimensional vectors in an orthonormal coordinate system ($\mathbf {e} _{1},\mathbf {e} _{2},\mathbf {e} _{3}$) as

$[{\boldsymbol {\sigma }}]={\begin{bmatrix}\sigma _{11}\\\sigma _{22}\\\sigma _{33}\\\sigma _{23}\\\sigma _{13}\\\sigma _{12}\end{bmatrix}}\equiv {\begin{bmatrix}\sigma _{1}\\\sigma _{2}\\\sigma _{3}\\\sigma _{4}\\\sigma _{5}\\\sigma _{6}\end{bmatrix}}~;~~[{\boldsymbol {\epsilon }}]={\begin{bmatrix}\epsilon _{11}\\\epsilon _{22}\\\epsilon _{33}\\2\epsilon _{23}\\2\epsilon _{13}\\2\epsilon _{12}\end{bmatrix}}\equiv {\begin{bmatrix}\epsilon _{1}\\\epsilon _{2}\\\epsilon _{3}\\\epsilon _{4}\\\epsilon _{5}\\\epsilon _{6}\end{bmatrix}}$

Then the stiffness tensor (${\mathsf {c}}$) can be expressed as

$[{\mathsf {C}}]={\begin{bmatrix}c_{1111}&c_{1122}&c_{1133}&c_{1123}&c_{1131}&c_{1112}\\c_{2211}&c_{2222}&c_{2233}&c_{2223}&c_{2231}&c_{2212}\\c_{3311}&c_{3322}&c_{3333}&c_{3323}&c_{3331}&c_{3312}\\c_{2311}&c_{2322}&c_{2333}&c_{2323}&c_{2331}&c_{2312}\\c_{3111}&c_{3122}&c_{3133}&c_{3123}&c_{3131}&c_{3112}\\c_{1211}&c_{1222}&c_{1233}&c_{1223}&c_{1231}&c_{1212}\end{bmatrix}}\equiv {\begin{bmatrix}C_{11}&C_{12}&C_{13}&C_{14}&C_{15}&C_{16}\\C_{12}&C_{22}&C_{23}&C_{24}&C_{25}&C_{26}\\C_{13}&C_{23}&C_{33}&C_{34}&C_{35}&C_{36}\\C_{14}&C_{24}&C_{34}&C_{44}&C_{45}&C_{46}\\C_{15}&C_{25}&C_{35}&C_{45}&C_{55}&C_{56}\\C_{16}&C_{26}&C_{36}&C_{46}&C_{56}&C_{66}\end{bmatrix}}$

and Hooke's law is written as

$[{\boldsymbol {\sigma }}]=[{\mathsf {C}}][{\boldsymbol {\epsilon }}]\qquad {\text{or}}\qquad \sigma _{i}=C_{ij}\epsilon _{j}~.$

Similarly the compliance tensor (${\mathsf {s}}$) can be written as

$[{\mathsf {S}}]={\begin{bmatrix}s_{1111}&s_{1122}&s_{1133}&2s_{1123}&2s_{1131}&2s_{1112}\\s_{2211}&s_{2222}&s_{2233}&2s_{2223}&2s_{2231}&2s_{2212}\\s_{3311}&s_{3322}&s_{3333}&2s_{3323}&2s_{3331}&2s_{3312}\\2s_{2311}&2s_{2322}&2s_{2333}&4s_{2323}&4s_{2331}&4s_{2312}\\2s_{3111}&2s_{3122}&2s_{3133}&4s_{3123}&4s_{3131}&4s_{3112}\\2s_{1211}&2s_{1222}&2s_{1233}&4s_{1223}&4s_{1231}&4s_{1212}\end{bmatrix}}\equiv {\begin{bmatrix}S_{11}&S_{12}&S_{13}&S_{14}&S_{15}&S_{16}\\S_{12}&S_{22}&S_{23}&S_{24}&S_{25}&S_{26}\\S_{13}&S_{23}&S_{33}&S_{34}&S_{35}&S_{36}\\S_{14}&S_{24}&S_{34}&S_{44}&S_{45}&S_{46}\\S_{15}&S_{25}&S_{35}&S_{45}&S_{55}&S_{56}\\S_{16}&S_{26}&S_{36}&S_{46}&S_{56}&S_{66}\end{bmatrix}}$

 Change of coordinate system

If a linear elastic material is rotated from a reference configuration to another, then the material is symmetric with respect to the rotation if the components of the stiffness tensor in the rotated configuration are related to the components in the reference configuration by the relation

$c_{pqrs}=l_{pi}~l_{qj}~l_{rk}~l_{s\ell }~c_{ijk\ell }$

where $l_{ab}$ are the components of an orthogonal rotation matrix $[L]$. The same relation also holds for inversions.

In matrix notation, if the transformed basis (rotated or inverted) is related to the reference basis by

$[\mathbf {e} _{i}']=[L][\mathbf {e} _{i}]$

then

$C_{ij}~\epsilon _{i}~\epsilon _{j}=C_{ij}'~\epsilon '_{i}~\epsilon '_{j}~.$

In addition, if the material is symmetric with respect to the transformation $[L]$ then

$C_{ij}=C'_{ij}\quad \implies \quad C_{ij}~(\epsilon _{i}~\epsilon _{j}-\epsilon '_{i}~\epsilon '_{j})=0~.$

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

 Orthotropic materials

المقالة الرئيسية: Orthotropic material

Orthotropic materials have three orthogonal planes of symmetry. If the basis vectors ($\mathbf {e} _{1},\mathbf {e} _{2},\mathbf {e} _{3}$) are normals to the planes of symmetry then the coordinate transformation relations imply that

${\begin{bmatrix}\sigma _{1}\\\sigma _{2}\\\sigma _{3}\\\sigma _{4}\\\sigma _{5}\\\sigma _{6}\end{bmatrix}}={\begin{bmatrix}C_{11}&C_{12}&C_{13}&0&0&0\\C_{12}&C_{22}&C_{23}&0&0&0\\C_{13}&C_{23}&C_{33}&0&0&0\\0&0&0&C_{44}&0&0\\0&0&0&0&C_{55}&0\\0&0&0&0&0&C_{66}\end{bmatrix}}{\begin{bmatrix}\epsilon _{1}\\\epsilon _{2}\\\epsilon _{3}\\\epsilon _{4}\\\epsilon _{5}\\\epsilon _{6}\end{bmatrix}}$

The inverse of this relation is commonly written as

${\begin{bmatrix}\epsilon _{\rm {xx}}\\\epsilon _{\rm {yy}}\\\epsilon _{\rm {zz}}\\2\epsilon _{\rm {yz}}\\2\epsilon _{\rm {zx}}\\2\epsilon _{\rm {xy}}\end{bmatrix}}={\begin{bmatrix}{\tfrac {1}{E_{\rm {x}}}}&-{\tfrac {\nu _{\rm {yx}}}{E_{\rm {y}}}}&-{\tfrac {\nu _{\rm {zx}}}{E_{\rm {z}}}}&0&0&0\\-{\tfrac {\nu _{\rm {xy}}}{E_{\rm {x}}}}&{\tfrac {1}{E_{\rm {y}}}}&-{\tfrac {\nu _{\rm {zy}}}{E_{\rm {z}}}}&0&0&0\\-{\tfrac {\nu _{\rm {xz}}}{E_{\rm {x}}}}&-{\tfrac {\nu _{\rm {yz}}}{E_{\rm {y}}}}&{\tfrac {1}{E_{\rm {z}}}}&0&0&0\\0&0&0&{\tfrac {1}{G_{\rm {yz}}}}&0&0\\0&0&0&0&{\tfrac {1}{G_{\rm {zx}}}}&0\\0&0&0&0&0&{\tfrac {1}{G_{\rm {xy}}}}\\\end{bmatrix}}{\begin{bmatrix}\sigma _{\rm {xx}}\\\sigma _{\rm {yy}}\\\sigma _{\rm {zz}}\\\sigma _{\rm {yz}}\\\sigma _{\rm {zx}}\\\sigma _{\rm {xy}}\end{bmatrix}}$

where

${E}_{\rm {i}}\,$ is the Young's modulus along axis $i$

$G_{\rm {ij}}\,$ is the shear modulus in direction $j$ on the plane whose normal is in direction $i$

$\nu _{\rm {ij}}\,$ is the Poisson's ratio that corresponds to a contraction in direction $j$ when an extension is applied in direction $i$.

Under plane stress conditions, $\sigma _{zz}=\sigma _{zx}=\sigma _{yz}=0$, Hooke's law for an orthotropic material takes the form

${\begin{bmatrix}\varepsilon _{\rm {xx}}\\\varepsilon _{\rm {yy}}\\2\varepsilon _{\rm {xy}}\end{bmatrix}}={\begin{bmatrix}{\frac {1}{E_{\rm {x}}}}&-{\frac {\nu _{\rm {yx}}}{E_{\rm {y}}}}&0\\-{\frac {\nu _{\rm {xy}}}{E_{\rm {x}}}}&{\frac {1}{E_{\rm {y}}}}&0\\0&0&{\frac {1}{G_{\rm {xy}}}}\end{bmatrix}}{\begin{bmatrix}\sigma _{\rm {xx}}\\\sigma _{\rm {yy}}\\\sigma _{\rm {xy}}\end{bmatrix}}\,.$

The inverse relation is

${\begin{bmatrix}\sigma _{\rm {xx}}\\\sigma _{\rm {yy}}\\\sigma _{\rm {xy}}\end{bmatrix}}={\cfrac {1}{1-\nu _{\rm {xy}}\nu _{\rm {yx}}}}{\begin{bmatrix}E_{\rm {x}}&\nu _{\rm {yx}}E_{\rm {x}}&0\\\nu _{\rm {xy}}E_{\rm {y}}&E_{\rm {y}}&0\\0&0&G_{\rm {xy}}(1-\nu _{\rm {xy}}\nu _{\rm {yx}})\end{bmatrix}}{\begin{bmatrix}\varepsilon _{\rm {xx}}\\\varepsilon _{\rm {yy}}\\2\varepsilon _{\rm {xy}}\end{bmatrix}}\,.$

The transposed form of the above stiffness matrix is also often used.

 Transversely isotropic materials

A transversely isotropic material is symmetric with respect to a rotation about an axis of symmetry. For such a material, if $\mathbf {e} _{3}$ is the axis of symmetry, Hooke's law can be expressed as

${\begin{bmatrix}\sigma _{1}\\\sigma _{2}\\\sigma _{3}\\\sigma _{4}\\\sigma _{5}\\\sigma _{6}\end{bmatrix}}={\begin{bmatrix}C_{11}&C_{12}&C_{13}&0&0&0\\C_{12}&C_{11}&C_{13}&0&0&0\\C_{13}&C_{13}&C_{33}&0&0&0\\0&0&0&C_{44}&0&0\\0&0&0&0&C_{44}&0\\0&0&0&0&0&{\tfrac {1}{2}}(C_{11}-C_{12})\end{bmatrix}}{\begin{bmatrix}\epsilon _{1}\\\epsilon _{2}\\\epsilon _{3}\\\epsilon _{4}\\\epsilon _{5}\\\epsilon _{6}\end{bmatrix}}$

More frequently, the $x\equiv \mathbf {e} _{1}$ axis is taken to be the axis of symmetry and the inverse Hooke's law is written as

${\begin{bmatrix}\epsilon _{\rm {xx}}\\\epsilon _{\rm {yy}}\\\epsilon _{\rm {zz}}\\2\epsilon _{\rm {yz}}\\2\epsilon _{\rm {zx}}\\2\epsilon _{\rm {xy}}\end{bmatrix}}={\begin{bmatrix}{\tfrac {1}{E_{\rm {x}}}}&-{\tfrac {\nu _{\rm {yx}}}{E_{\rm {y}}}}&-{\tfrac {\nu _{\rm {yx}}}{E_{\rm {y}}}}&0&0&0\\-{\tfrac {\nu _{\rm {xy}}}{E_{\rm {x}}}}&{\tfrac {1}{E_{\rm {y}}}}&-{\tfrac {\nu _{\rm {yz}}}{E_{\rm {y}}}}&0&0&0\\-{\tfrac {\nu _{\rm {xy}}}{E_{\rm {x}}}}&-{\tfrac {\nu _{\rm {yz}}}{E_{\rm {y}}}}&{\tfrac {1}{E_{\rm {y}}}}&0&0&0\\0&0&0&{\tfrac {2(1+\nu _{\rm {yz}})}{E_{\rm {y}}}}&0&0\\0&0&0&0&{\tfrac {1}{G_{\rm {xy}}}}&0\\0&0&0&0&0&{\tfrac {1}{G_{\rm {xy}}}}\\\end{bmatrix}}{\begin{bmatrix}\sigma _{\rm {xx}}\\\sigma _{\rm {yy}}\\\sigma _{\rm {zz}}\\\sigma _{\rm {yz}}\\\sigma _{\rm {zx}}\\\sigma _{\rm {xy}}\end{bmatrix}}$

 Thermodynamic basis

Linear deformations of elastic materials can be approximated as adiabatic. Under these conditions and for quasistatic processes the first law of thermodynamics for a deformed body can be expressed as

$\delta W=\delta U\,$

where $\delta U$ is the increase in internal energy and $\delta W$ is the work done by external forces. The work can be split into two terms

$\delta W=\delta W_{s}+\delta W_{b}\,$

where $\delta W_{s}$ is the work done by surface forces while $\delta W_{b}$ is the work done by body forces. If $\delta \mathbf {u}$ is a variation of the displacement field $\mathbf {u}$ in the body, then the two external work terms can be expressed as

$\delta W_{s}=\int _{\partial \Omega }\mathbf {t} \cdot \delta \mathbf {u} ~{\rm {dS}}~;~~\delta W_{b}=\int _{\Omega }\mathbf {b} \cdot \delta \mathbf {u} ~{\rm {dV}}$

where $\mathbf {t}$ is the surface traction vector, $\mathbf {b}$ is the body force vector, $\Omega \,$ represents the body and $\partial \Omega$ represents its surface. Using the relation between the Cauchy stress and the surface traction, $\mathbf {t} =\mathbf {n} \cdot {\boldsymbol {\sigma }}$ (where $\mathbf {n}$ is the unit outward normal to $\partial \Omega$), we have

$\delta W=\delta U=\int _{\partial \Omega }(\mathbf {n} \cdot {\boldsymbol {\sigma }})\cdot \delta \mathbf {u} ~{\rm {dS}}+\int _{\Omega }\mathbf {b} \cdot \delta \mathbf {u} ~{\rm {dV}}$

Converting the surface integral into a volume integral via the divergence theorem gives

$\delta U=\int _{\Omega }[{\boldsymbol {\nabla }}\cdot ({\boldsymbol {\sigma }}\cdot \delta \mathbf {u} )+\mathbf {b} \cdot \delta \mathbf {u} ]~{\rm {dV}}~.$

Using the symmetry of the Cauchy stress and the identity

${\boldsymbol {\nabla }}\cdot ({\boldsymbol {A}}\cdot \mathbf {b} )=({\boldsymbol {\nabla }}\cdot {\boldsymbol {A}})\cdot \mathbf {b} +{\tfrac {1}{2}}[{\boldsymbol {A}}^{T}:{\boldsymbol {\nabla }}\mathbf {b} +{\boldsymbol {A}}:({\boldsymbol {\nabla }}\mathbf {b} )^{T}]$

we have the following

$\delta U=\int _{\Omega }[{\boldsymbol {\sigma }}:{\tfrac {1}{2}}\{{\boldsymbol {\nabla }}\delta \mathbf {u} +({\boldsymbol {\nabla }}\delta \mathbf {u} )^{T}\}+\{{\boldsymbol {\nabla }}\cdot {\boldsymbol {\sigma }}+\mathbf {b} \}\cdot \delta \mathbf {u} ]~{\rm {dV}}~.$

From the definition of strain and from the equations of equilibrium we have

$\delta {\boldsymbol {\epsilon }}={\tfrac {1}{2}}[{\boldsymbol {\nabla }}\delta \mathbf {u} +({\boldsymbol {\nabla }}\delta \mathbf {u} )^{T}]~;~~{\boldsymbol {\nabla }}\cdot {\boldsymbol {\sigma }}+\mathbf {b} =\mathbf {0} ~.$

Hence we can write

$\delta U=\int _{\Omega }{\boldsymbol {\sigma }}:\delta {\boldsymbol {\epsilon }}~{\rm {dV}}$

and therefore the variation in the internal energy density is given by

$\delta U_{0}={\boldsymbol {\sigma }}:\delta {\boldsymbol {\epsilon }}~.$

An elastic material is defined as one in which the total internal energy is equal to the potential energy of the internal forces (also called the elastic strain energy). Therefore the internal energy density is a function of the strains, $U_{0}=U_{0}({\boldsymbol {\epsilon }})$ and the variation of the internal energy can be expressed as

$\delta U_{0}={\cfrac {\partial U_{0}}{\partial {\boldsymbol {\epsilon }}}}:\delta {\boldsymbol {\epsilon }}~.$

Since the variation of strain is arbitrary, the stress–strain relation of an elastic material is given by

${\boldsymbol {\sigma }}={\cfrac {\partial U_{0}}{\partial {\boldsymbol {\epsilon }}}}~.$

For a linear elastic material, the quantity $\partial U_{0}/\partial {\boldsymbol {\epsilon }}$ is a linear function of ${\boldsymbol {\epsilon }}$, and can therefore be expressed as

${\boldsymbol {\sigma }}={\mathsf {c}}:{\boldsymbol {\epsilon }}$

where ${\mathsf {c}}$ is a fourth-rank tensor of material constants, also called the stiffness tensor. We can see why ${\mathsf {c}}$ must be a fourth-rank tensor by noting that, for a linear elastic material,

${\cfrac {\partial }{\partial {\boldsymbol {\epsilon }}}}[{\boldsymbol {\sigma }}({\boldsymbol {\epsilon }})]={\text{constant}}={\mathsf {c}}\,.$

In index notation

${\cfrac {\partial \sigma _{ij}}{\partial \epsilon _{k\ell }}}={\text{constant}}=c_{ijk\ell }\,.$

Clearly, the right-hand side constant requires four indices and is a fourth-rank quantity. We can also see that this quantity must be a tensor because it is a linear transformation that takes the strain tensor to the stress tensor. We can also show that the constant obeys the tensor transformation rules for fourth-rank tensors.

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 المصادر

• A.C. Ugural, S.K. Fenster, Advanced Strength and Applied Elasticity, 4th ed
• Walter Lewin explains Hooke's law. From Walter Lewin (1 October 1999). Hooke's Law, Simple Harmonic Oscillator. MIT Course 8.01: Classical Mechanics, Lecture 10 (ogg) (videotape) (in English). Cambridge, MA USA: MIT OCW. Event occurs at 1:21–10:10. Retrieved 23 December 2010. ...arguably the most important equation in all of Physics.{{cite AV media}}: CS1 maint: unrecognized language (link)
• A test of Hooke's law. From Walter Lewin (1 October 1999). Hooke's Law, Simple Harmonic Oscillator. MIT Course 8.01: Classical Mechanics, Lecture 10 (ogg) (videotape) (in English). Cambridge, MA USA: MIT OCW. Event occurs at 10:10–16:33. Retrieved 23 December 2010.{{cite AV media}}: CS1 maint: unrecognized language (link)

 وصلات خارجية

• Java Applet demonstrating Hooke's Law in motion
• Hooke's Law for General Stress/Strain Calculations

صيغ التحويل
Homogeneous isotropic linear elastic materials have their elastic properties uniquely determined by any two moduli among these; thus, given any two, any other of the elastic moduli can be calculated according to these formulas.
	$K=\,$	$E=\,$	$\lambda =\,$	$G=\,$	$\nu =\,$	$M=\,$	Notes
$(K,\,E)$	$K$	$E$	${\tfrac {3K(3K-E)}{9K-E}}$	${\tfrac {3KE}{9K-E}}$	${\tfrac {3K-E}{6K}}$	${\tfrac {3K(3K+E)}{9K-E}}$
$(K,\,\lambda )$	$K$	${\tfrac {9K(K-\lambda )}{3K-\lambda }}$	$\lambda$	${\tfrac {3(K-\lambda )}{2}}$	${\tfrac {\lambda }{3K-\lambda }}$	$3K-2\lambda \,$
$(K,\,G)$	$K$	${\tfrac {9KG}{3K+G}}$	$K-{\tfrac {2G}{3}}$	$G$	${\tfrac {3K-2G}{2(3K+G)}}$	$K+{\tfrac {4G}{3}}$
$(K,\,\nu )$	$K$	$3K(1-2\nu )\,$	${\tfrac {3K\nu }{1+\nu }}$	${\tfrac {3K(1-2\nu )}{2(1+\nu )}}$	$\nu$	${\tfrac {3K(1-\nu )}{1+\nu }}$
$(K,\,M)$	$K$	${\tfrac {9K(M-K)}{3K+M}}$	${\tfrac {3K-M}{2}}$	${\tfrac {3(M-K)}{4}}$	${\tfrac {3K-M}{3K+M}}$	$M$
$(E,\,\lambda )$	${\tfrac {E+3\lambda +R}{6}}$	$E$	$\lambda$	${\tfrac {E-3\lambda +R}{4}}$	${\tfrac {2\lambda }{E+\lambda +R}}$	${\tfrac {E-\lambda +R}{2}}$	$R={\sqrt {E^{2}+9\lambda ^{2}+2E\lambda }}$
$(E,\,G)$	${\tfrac {EG}{3(3G-E)}}$	$E$	${\tfrac {G(E-2G)}{3G-E}}$	$G$	${\tfrac {E}{2G}}-1$	${\tfrac {G(4G-E)}{3G-E}}$
$(E,\,\nu )$	${\tfrac {E}{3(1-2\nu )}}$	$E$	${\tfrac {E\nu }{(1+\nu )(1-2\nu )}}$	${\tfrac {E}{2(1+\nu )}}$	$\nu$	${\tfrac {E(1-\nu )}{(1+\nu )(1-2\nu )}}$
$(E,\,M)$	${\tfrac {3M-E+S}{6}}$	$E$	${\tfrac {M-E+S}{4}}$	${\tfrac {3M+E-S}{8}}$	${\tfrac {E-M+S}{4M}}$	$M$	$S=\pm {\sqrt {E^{2}+9M^{2}-10EM}}$ There are two valid solutions. The plus sign leads to $\nu \geq 0$. The minus sign leads to $\nu \leq 0$.
$(\lambda ,\,G)$	$\lambda +{\tfrac {2G}{3}}$	${\tfrac {G(3\lambda +2G)}{\lambda +G}}$	$\lambda$	$G$	${\tfrac {\lambda }{2(\lambda +G)}}$	$\lambda +2G\,$
$(\lambda ,\,\nu )$	${\tfrac {\lambda (1+\nu )}{3\nu }}$	${\tfrac {\lambda (1+\nu )(1-2\nu )}{\nu }}$	$\lambda$	${\tfrac {\lambda (1-2\nu )}{2\nu }}$	$\nu$	${\tfrac {\lambda (1-\nu )}{\nu }}$	Cannot be used when $\nu =0\Leftrightarrow \lambda =0$
$(\lambda ,\,M)$	${\tfrac {M+2\lambda }{3}}$	${\tfrac {(M-\lambda )(M+2\lambda )}{M+\lambda }}$	$\lambda$	${\tfrac {M-\lambda }{2}}$	${\tfrac {\lambda }{M+\lambda }}$	$M$
$(G,\,\nu )$	${\tfrac {2G(1+\nu )}{3(1-2\nu )}}$	$2G(1+\nu )\,$	${\tfrac {2G\nu }{1-2\nu }}$	$G$	$\nu$	${\tfrac {2G(1-\nu )}{1-2\nu }}$
$(G,\,M)$	$M-{\tfrac {4G}{3}}$	${\tfrac {G(3M-4G)}{M-G}}$	$M-2G\,$	$G$	${\tfrac {M-2G}{2M-2G}}$	$M$
$(\nu ,\,M)$	${\tfrac {M(1+\nu )}{3(1-\nu )}}$	${\tfrac {M(1+\nu )(1-2\nu )}{1-\nu }}$	${\tfrac {M\nu }{1-\nu }}$	${\tfrac {M(1-2\nu )}{2(1-\nu )}}$	$\nu$	$M$