## Lesson 32. SOLVING NUMERICAL

 Module 7. Springs Lesson 32SOLVING NUMERICAL 32.1 Numerical A helical spring of wire diameter 6mm and spring index 6 is acted by an initial load of 800N. After compressing it further by 10mm the stress in the wire is 500MPa. Find the number of active coils. G = 84000MPa. Solution: D = Spring index (C) x d = 36mm &amp;amp;amp;amp;lt;!-- /* Font Definitions */ @font-face {font-family:&amp;amp;amp;amp;quot;Cambria Math&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;quot;&amp;amp;amp;amp;quot;; margin:0in; margin-bottom:.0001pt; mso-pagination:widow-orphan; font-size:12.0pt; font-family:&amp;amp;amp;amp;quot;Times New Roman&amp;amp;amp;amp;quot;,&amp;amp;amp;amp;quot;serif&amp;amp;amp;amp;quot;; mso-fareast-font-family:&amp;amp;amp;amp;quot;Times New Roman&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; mso-ascii-font-family:Calibri; mso-fareast-font-family:Calibri; mso-hansi-font-family:Calibri;} @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;gt; &amp;amp;amp;amp;lt;!-- /* Font Definitions */ @font-face {font-family:&amp;amp;amp;amp;quot;Cambria Math&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;quot;&amp;amp;amp;amp;quot;; margin:0in; margin-bottom:.0001pt; mso-pagination:widow-orphan; font-size:12.0pt; font-family:&amp;amp;amp;amp;quot;Times New Roman&amp;amp;amp;amp;quot;,&amp;amp;amp;amp;quot;serif&amp;amp;amp;amp;quot;; mso-fareast-font-family:&amp;amp;amp;amp;quot;Times New Roman&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; mso-ascii-font-family:Calibri; mso-fareast-font-family:Calibri; mso-hansi-font-family:Calibri;} @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;gt; τmax = (Kw) x ( 8FD / &amp;amp;amp;amp;lt;!-- /* Font Definitions */ @font-face {font-family:&amp;amp;amp;amp;quot;Cambria Math&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;quot;&amp;amp;amp;amp;quot;; margin:0in; margin-bottom:.0001pt; mso-pagination:widow-orphan; font-size:12.0pt; font-family:&amp;amp;amp;amp;quot;Times New Roman&amp;amp;amp;amp;quot;,&amp;amp;amp;amp;quot;serif&amp;amp;amp;amp;quot;; mso-fareast-font-family:&amp;amp;amp;amp;quot;Times New Roman&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; mso-ascii-font-family:Calibri; mso-fareast-font-family:Calibri; mso-hansi-font-family:Calibri;} @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;gt; Πd3 ) Kw = (4C-1 / 4C-4) + ( 0.615/ C) = 1.2525 Or 500 = 1.2525 x (8F x 36/ π x F) Therefore F = 940.86 N K = F / δ= (940.6 - 800) / 10= 14 N/mmK = (Gd4) / (8D4N)OrN = (Gd4) / (K8D3N)= (84000 x 64) / 14 x 8 x 3 x 63= ~ 21tuns &amp;amp;amp;amp;lt;!-- /* Font Definitions */ @font-face {font-family:&amp;amp;amp;amp;quot;Cambria Math&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;quot;&amp;amp;amp;amp;quot;; margin:0in; margin-bottom:.0001pt; mso-pagination:widow-orphan; font-size:12.0pt; font-family:&amp;amp;amp;amp;quot;Times New Roman&amp;amp;amp;amp;quot;,&amp;amp;amp;amp;quot;serif&amp;amp;amp;amp;quot;; mso-fareast-font-family:&amp;amp;amp;amp;quot;Times New Roman&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; mso-ascii-font-family:Calibri; mso-fareast-font-family:Calibri; mso-hansi-font-family:Calibri;} @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;} 32.2 Problem Design a leaf spring for the following specifications: Total load = 140 kN ; Number of springs supporting the load = 4 ; Maximum number of leaves= 10; Span of the spring = 1000 mm ; Permissible deflection = 80 mm.Take Young’s modulus, E = 200 kN/mm2 and allowable stress in spring material as 600 MPa. Solution: Given: Total load = 140 kN ; No. of springs = 4; n = 10 ; 2L = 1000 mm or L = 500 mm; δ = 80 mm; E = 200 kN/mm2 = 200 × 103 N/mm2; σ = 600 MPa = 600 N/mm2 We know that load on each spring, 2W =Total load/No. of springs =140/4 =35 kN There fore, W = 35 / 2 = 17.5 kN = 17500 N Let t = Thickness of the leaves, and b = Width of the leaves.