{"id":2599,"date":"2024-08-06T19:03:39","date_gmt":"2024-08-06T13:33:39","guid":{"rendered":"https:\/\/study.madeeasy.in\/?p=2599"},"modified":"2025-07-16T15:25:16","modified_gmt":"2025-07-16T09:55:16","slug":"hooke-law","status":"publish","type":"post","link":"https:\/\/www.madeeasy.in\/study\/ce\/strength-of-material\/hooke-law","title":{"rendered":"Hooke Law for Plane Stress"},"content":{"rendered":"

Consider a plane stress element is shown in figure.
\nIn this stress element
\n\u03c3z<\/sub> = 0
\n\u03c3yz<\/sub> = 0
\n\u03c4xz<\/sub> = 0<\/p>\n

Let the normal strains in x, y and z directions are represented by \u03b5x<\/sub>, \u03b5y<\/sub>, \u03b5z<\/sub> respectively.\"Plane<\/p>\n

Shear strain in x-y plane, x-z plane, y-z plane are represented by \u03c6xy<\/sub>, \u03c6xz<\/sub> and \u03c6yz<\/sub> respectively.
\nAs discussed in earlier chapters,<\/p>\n

\u03b5x <\/sub>= \u03c3x<\/sub>\/E – \u00b5\u03c3y<\/sub>\/E …..(i)<\/p>\n

\u03b5y <\/sub>= \u03c3y<\/sub>\/E – \u00b5\u03c3x<\/sub>\/E …..(ii)<\/p>\n

\u03b5z<\/sub> = -\u00b5\u03c3y<\/sub>\/E – -\u00b5\u03c3z<\/sub>\/E …..(iii)<\/p>\n

Normal strains in x, y and z directions can be found
\nout by using these equations if stresses are known.<\/p>\n

Also, \u03c6xy<\/sub> = \u03c4xy<\/sub>\/G<\/p>\n

Note that axial stress \u03c3x and \u03c3y does not produce shear strain.
\nSolving (i) and (ii), we get<\/p>\n

\"HOOKE\u2019S<\/p>\n

All the above 7 equations are known as Hooke\u2019s law for plane stress.
\nIn these equation,
\nE = Young\u2019s modulus of elasticity
\nG = Shear modulus
\n\u00b5 = Poisson\u2019s ratio<\/p>\n

Volume Change<\/strong><\/p>\n

As discussed in earlier chapter<\/p>\n

Volume strain for a body, \u03b5v<\/sub> = \u03c3x<\/sub> + \u03c3y<\/sub> + \u03c3z<\/sub>\/E (1-2\u00b5 )<\/span><\/p>\n

But for plane stress, \u03c3z<\/sub> = 0<\/p>\n

So, dilatation or volumetric strain, \u03b5v<\/sub> = \u03c3x<\/sub> + \u03c3y<\/sub> \/E (1-2\u00b5 )<\/span><\/p>\n

For uniaxial stress, \u03c3y<\/sub> = \u03c3z<\/sub> = 0<\/p>\n

So, volumetric strain, \u03b5v<\/sub> = \u03c3x <\/sub>\/E (1-2\u00b5 )<\/span><\/p>\n

Triaxial stress<\/strong><\/p>\n

Consider a stress element as shown in figure subjected to normal stresses \u03c3x<\/sub>, \u03c3y<\/sub> and \u03c3z<\/sub> only. As the stress element is subjected to axial stresses in three mutually perpendicular direction, it is said to be a case of triaxial stress.<\/p>\n

\"Triaxial<\/p>\n

It is to be noted that as there are only normal stress on stress element and no shear stress, hence these normal stresses are principal stresses for the element.<\/p>\n

Cut an inclined plane parallel to z-axis through element and normal to plane parallel to xy plane as shown in figure a(ii).<\/p>\n

On face ACGH, only stresses are normal stress \u03c3x\u2032 and shear stress \u03c4x\u2032y\u2032<\/sub> which can be determined by using transformation equations. Note that these are independent of stress \u03c3z<\/sub>.<\/p>\n

Thus, in triaxial state also, transformation equations or Mohr\u2019s circle can be used to find normal and shear stress on an inclined plane.
\nSimilar analysis can be done if planes parallel to x and y axis are cut.<\/p>\n

Maximum Shear Stress<\/h3>\n

As studied earlier, plane of maximum shear stress occur at 45\u00b0 to principal planes.
\nHence, if we rotate the stress element about z-axis by 45\u00b0, the maximum value of shear stress,<\/p>\n

(\u03c4max<\/sub>)z<\/sub> = \u00b1 \u03c3x<\/sub> – \u03c3y<\/sub>\/2 ——(i)<\/p>\n

(\u03c4max<\/sub>)y<\/sub> = \u00b1 \u03c3x<\/sub> – \u03c3z<\/sub>\/2 ——(ii)<\/p>\n

(\u03c4max<\/sub>)x<\/sub> = \u00b1 \u03c3y<\/sub> – \u03c3z<\/sub>\/2 ——(iii)<\/p>\n

Absolute maximum shear stress is largest of shear stress obtained from above 3 equations.<\/p>\n

Hooke\u2019s law for triaxial stress<\/h3>\n

Let the strains produced by normal stresses on stress element in figure are \u03b5x<\/sub>,\u03b5y<\/sub> ,\u03b5z<\/sub> in x, y and z-axis respectively.<\/p>\n

\"Triaxial<\/p>\n

Volume change<\/h4>\n

As discussed in earlier chapters<\/p>\n

Volumetric strains, \u03b5v = \u03c3x<\/sub> + \u03c3y<\/sub>+ \u03c3z<\/sub>\u00a0<\/sub>\/E (1 – 2\u00b5)<\/p>\n

Special case I: Spherical stress<\/strong><\/p>\n

A spherical stress element is a special type of triaxial stress in which<\/p>\n

\u03c3x<\/sub> = \u03c3y <\/sub>= \u03c3z<\/sub>\u00a0<\/sub>= \u03c3o<\/sub><\/p>\n

In this type of stress element, any plane inclined at an angle \u03b8 cut through the element will only have normal stress<\/p>\n

\u03c3o\u00a0<\/sub>and no shear stress.<\/p>\n

The normal strain in all the direction is same. So, normal strain, \u03b5 = \u03c30 <\/sub>\/E (1 – 2\u00b5)<\/p>\n

Volumetric strains, \u03b5v = \u03c3x<\/sub> + \u03c3y<\/sub>+ \u03c3z<\/sub>\u00a0<\/sub>\/E (1 – 2\u00b5)<\/p>\n

\u03c3x<\/sub> = \u03c3y <\/sub>= \u03c3z<\/sub>\u00a0<\/sub>= \u03c3o<\/sub><\/p>\n

\u03b5 = 3\u03c30 <\/sub>\/E (1 – 2\u00b5) = 3\u03b50<\/sub><\/p>\n

k = E \/ 3(1 – 2\u00b5)<\/p>\n

k = Bulk modulus of elasticity = \u03c3o<\/sub>\/\u03b5v<\/sub><\/p>\n\n\n\n\n\n\n\n
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\n Must Read: <\/strong> What is Strength of Material?<\/strong><\/a><\/p>\n

 <\/p>\n","protected":false},"excerpt":{"rendered":"

Consider a plane stress element is shown in figure. In this stress element \u03c3z = 0 \u03c3yz = 0 \u03c4xz<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[685,2],"tags":[715,717,716],"class_list":["post-2599","post","type-post","status-publish","format-standard","hentry","category-strength-of-material","category-ce","tag-maximum-shear-stress","tag-spherical-stress","tag-triaxial-stress"],"_links":{"self":[{"href":"https:\/\/www.madeeasy.in\/study\/wp-json\/wp\/v2\/posts\/2599","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.madeeasy.in\/study\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.madeeasy.in\/study\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.madeeasy.in\/study\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.madeeasy.in\/study\/wp-json\/wp\/v2\/comments?post=2599"}],"version-history":[{"count":0,"href":"https:\/\/www.madeeasy.in\/study\/wp-json\/wp\/v2\/posts\/2599\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.madeeasy.in\/study\/wp-json\/wp\/v2\/media?parent=2599"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.madeeasy.in\/study\/wp-json\/wp\/v2\/categories?post=2599"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.madeeasy.in\/study\/wp-json\/wp\/v2\/tags?post=2599"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}