## Lesson 25. SOLVING NUMERICAL

 Module 5. Properties of material, failures and factor of safety Lesson 25SOLVING NUMERICAL The load on a bolt consists of an axial pull of 20 KN together with a transverse Shear force of 10 KN. Find the diameter of bolt required according to Maximum principal stress theory. Maximum distortion energy theory. Maximum shear stress theory. Maximum strain energy theory; and Maximum principal strain theory.   Take Permissible Tensile stress at elastics limit=100Mpa and Poisson’s ratio=0.3 Nomenclature Load=P бt1 = normal tensile stress in x direction бt2 = normal tensile stress in y direction б1, б2= Principal stresses Shear stress= ז бt(el)= Permissible Tensile stress at elastics limit Sol. Given load=Ptensile =20KN Pshear=10KN (stress(sigma symbol))tensile(elongation)=100Mpa=100N/MM2 µ=1/m=0.3 Let d=diameter of the bolt in mm Therefore Cross sectional area of the bolt &amp;amp;amp;amp;amp;lt;!-- /* Font Definitions */ @font-face {font-family:&amp;amp;amp;amp;amp;quot;Cambria Math&amp;amp;amp;amp;amp;quot;; panose-1:2 4 5 3 5 4 6 3 2 4; mso-font-charset:0; mso-generic-font-family:roman; mso-font-pitch:variable; mso-font-signature:-1610611985 1107304683 0 0 159 0;} /* Style Definitions */ p.MsoNormal, li.MsoNormal, div.MsoNormal {mso-style-unhide:no; mso-style-qformat:yes; mso-style-parent:&amp;amp;amp;amp;amp;quot;&amp;amp;amp;amp;amp;quot;; margin:0in; margin-bottom:.0001pt; mso-pagination:widow-orphan; font-size:12.0pt; font-family:&amp;amp;amp;amp;amp;quot;Times New Roman&amp;amp;amp;amp;amp;quot;,&amp;amp;amp;amp;amp;quot;serif&amp;amp;amp;amp;amp;quot;; mso-fareast-font-family:&amp;amp;amp;amp;amp;quot;Times New Roman&amp;amp;amp;amp;amp;quot;;} .MsoChpDefault {mso-style-type:export-only; mso-default-props:yes; font-size:10.0pt; mso-ansi-font-size:10.0pt; mso-bidi-font-size:10.0pt;} @page WordSection1 {size:8.5in 11.0in; margin:1.0in 1.0in 1.0in 1.0in; mso-header-margin:.5in; mso-footer-margin:.5in; mso-paper-source:0;} div.WordSection1 {page:WordSection1;} --&amp;amp;amp;amp;amp;gt; A = (Π / 4) x d2=0.785*d2 mm2 &amp;amp;amp;amp;amp;lt;!-- /* Font Definitions */ @font-face {font-family:&amp;amp;amp;amp;amp;quot;Cambria Math&amp;amp;amp;amp;amp;quot;; panose-1:2 4 5 3 5 4 6 3 2 4; mso-font-charset:0; mso-generic-font-family:roman; mso-font-pitch:variable; mso-font-signature:-1610611985 1107304683 0 0 159 0;} /* Style Definitions */ p.MsoNormal, li.MsoNormal, div.MsoNormal {mso-style-unhide:no; mso-style-qformat:yes; mso-style-parent:&amp;amp;amp;amp;amp;quot;&amp;amp;amp;amp;amp;quot;; margin:0in; margin-bottom:.0001pt; mso-pagination:widow-orphan; font-size:12.0pt; font-family:&amp;amp;amp;amp;amp;quot;Times New Roman&amp;amp;amp;amp;amp;quot;,&amp;amp;amp;amp;amp;quot;serif&amp;amp;amp;amp;amp;quot;; mso-fareast-font-family:&amp;amp;amp;amp;amp;quot;Times New Roman&amp;amp;amp;amp;amp;quot;;} .MsoChpDefault {mso-style-type:export-only; mso-default-props:yes; font-size:10.0pt; mso-ansi-font-size:10.0pt; mso-bidi-font-size:10.0pt;} @page WordSection1 {size:8.5in 11.0in; margin:1.0in 1.0in 1.0in 1.0in; mso-header-margin:.5in; mso-footer-margin:.5in; mso-paper-source:0;} div.WordSection1 {page:WordSection1;} --&amp;amp;amp;amp;amp;gt; We know that axial tensile stress бt1 = (Pt1 / A) = (20 / 0.785d2) = (25.47 / d2) kN / mm2 And transverse shear stress Shear stress(ז) = ( Ps/ A) = (10 / 0.785d2 ) = ( 12.73/ d2) kN/mm2 &amp;amp;amp;amp;amp;lt;!-- /* Font Definitions */ @font-face {font-family:&amp;amp;amp;amp;amp;quot;Cambria Math&amp;amp;amp;amp;amp;quot;; panose-1:2 4 5 3 5 4 6 3 2 4; mso-font-charset:0; mso-generic-font-family:roman; mso-font-pitch:variable; mso-font-signature:-1610611985 1107304683 0 0 159 0;} @font-face {font-family:Calibri; panose-1:2 15 5 2 2 2 4 3 2 4; mso-font-charset:0; mso-generic-font-family:swiss; mso-font-pitch:variable; mso-font-signature:-1610611985 1073750139 0 0 159 0;} /* Style Definitions */ p.MsoNormal, li.MsoNormal, div.MsoNormal {mso-style-unhide:no; mso-style-qformat:yes; mso-style-parent:&amp;amp;amp;amp;amp;quot;&amp;amp;amp;amp;amp;quot;; margin:0in; margin-bottom:.0001pt; mso-pagination:widow-orphan; font-size:12.0pt; font-family:&amp;amp;amp;amp;amp;quot;Times New Roman&amp;amp;amp;amp;amp;quot;,&amp;amp;amp;amp;amp;quot;serif&amp;amp;amp;amp;amp;quot;; mso-fareast-font-family:&amp;amp;amp;amp;amp;quot;Times New Roman&amp;amp;amp;amp;amp;quot;;} p.MsoListParagraph, li.MsoListParagraph, div.MsoListParagraph {mso-style-unhide:no; mso-style-qformat:yes; margin-top:0in; margin-right:0in; margin-bottom:10.0pt; margin-left:.5in; mso-add-space:auto; line-height:115%; mso-pagination:widow-orphan; font-size:11.0pt; font-family:&amp;amp;amp;amp;amp;quot;Calibri&amp;amp;amp;amp;amp;quot;,&amp;amp;amp;amp;amp;quot;sans-serif&amp;amp;amp;amp;amp;quot;; 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margin-top:0in; margin-right:0in; margin-bottom:0in; margin-left:.5in; margin-bottom:.0001pt; mso-add-space:auto; line-height:115%; mso-pagination:widow-orphan; font-size:11.0pt; font-family:&amp;amp;amp;amp;amp;quot;Calibri&amp;amp;amp;amp;amp;quot;,&amp;amp;amp;amp;amp;quot;sans-serif&amp;amp;amp;amp;amp;quot;; mso-fareast-font-family:&amp;amp;amp;amp;amp;quot;Times New Roman&amp;amp;amp;amp;amp;quot;; mso-bidi-font-family:&amp;amp;amp;amp;amp;quot;Times New Roman&amp;amp;amp;amp;amp;quot;;} p.MsoListParagraphCxSpLast, li.MsoListParagraphCxSpLast, div.MsoListParagraphCxSpLast {mso-style-unhide:no; mso-style-qformat:yes; mso-style-type:export-only; margin-top:0in; margin-right:0in; margin-bottom:10.0pt; margin-left:.5in; mso-add-space:auto; line-height:115%; mso-pagination:widow-orphan; font-size:11.0pt; font-family:&amp;amp;amp;amp;amp;quot;Calibri&amp;amp;amp;amp;amp;quot;,&amp;amp;amp;amp;amp;quot;sans-serif&amp;amp;amp;amp;amp;quot;; mso-fareast-font-family:&amp;amp;amp;amp;amp;quot;Times New Roman&amp;amp;amp;amp;amp;quot;; mso-bidi-font-family:&amp;amp;amp;amp;amp;quot;Times New Roman&amp;amp;amp;amp;amp;quot;;} .MsoChpDefault {mso-style-type:export-only; mso-default-props:yes; font-size:10.0pt; mso-ansi-font-size:10.0pt; mso-bidi-font-size:10.0pt;} @page WordSection1 {size:8.5in 11.0in; margin:1.0in 1.0in 1.0in 1.0in; mso-header-margin:.5in; mso-footer-margin:.5in; mso-paper-source:0;} div.WordSection1 {page:WordSection1;} --&amp;amp;amp;amp;amp;gt; 1). According to maximum principal stress theory. We know that maximum principal stress, According to maximum principal stress theory бt1 = бt(el) (30365 / d2) = 100 d2 = ( 30365/ 100) = 303.65 = 17.42mm 2) According to maximum shear stress theory We know that maximum shear stress According to maximum shear stress theory