{"id":36,"date":"2017-11-20T13:03:30","date_gmt":"2017-11-20T18:03:30","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/powr4406\/chapter\/stress-and-strain\/"},"modified":"2018-06-25T12:00:47","modified_gmt":"2018-06-25T16:00:47","slug":"stress-and-strain","status":"publish","type":"chapter","link":"https:\/\/pressbooks.bccampus.ca\/powr4406\/chapter\/stress-and-strain\/","title":{"raw":"Stress and Strain","rendered":"Stress and Strain"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\nAfter completing this chapter you should be able to:\r\n<ul>\r\n \t<li>Define normal and shear stress and strain and discuss the relationship between design stress, yield stress and ultimate stress<\/li>\r\n \t<li>Design members under tension, compression and shear loads<\/li>\r\n \t<li>Determine members deformation under tension and compression<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div class=\"textbox shaded\" style=\"text-align: center\"><strong>Mechanical stress<\/strong><\/div>\r\n<div>This section discusses the effects of mechanical loads (forces) acting on members.\u00a0 Next chapter will cover the effects of thermal loads (thermal expansion).<\/div>\r\n<div>\r\n<div class=\"textbox shaded\">Normal, tensile and compressive stresses<\/div>\r\n<\/div>\r\nTension or compression in a member generate normal stresses; they are called \u201cnormal\u201d because the cross-section that resists the load is perpendicular (normal) to the direction of the applied forces.\u00a0 Both tensile and compressive stresses are calculated with:\r\n\r\n<img src=\"https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2Fig1-300x54.jpg\" alt=\"\" class=\"alignnone wp-image-31\" width=\"272\" height=\"49\" \/>\r\n\r\nIf a member has a variable cross-section, the area that must be used in calculations is the minimum cross-sectional area; this will give you the maximum stress in the member, which ultimately will govern the design.\r\n<div class=\"textbox shaded\">\r\n\r\nShear stresses\r\n\r\n<\/div>\r\nIn shear the cross-section area that resists the load is parallel with the direction of applied forces.\u00a0 In addition to that, when estimating the shear area you must factor in how many cross-sections contribute to the overall strength of the assembly.\r\n\r\nFor instance, if you consider the pin of a door hinge as subjected to a shear load, you have to count how many cross-sections resist the load.\r\n\r\nThe formula for calculating the shear stress is the same:\r\n\r\n<img src=\"https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2Fig2-300x51.jpg\" alt=\"\" class=\"alignnone wp-image-32\" width=\"271\" height=\"46\" \/>\r\n\r\nIn a punching operation the area that resists the shear is in the shape of a cylinder for a round hole (think of a cookie cutter).\u00a0 Therefore the area in shear will be found from multiplying the circumference of the shape by the thickness of the plate.\r\n\r\n<a href=\"https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2FigHoleShear-1.jpg\"><img src=\"https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2FigHoleShear-1-300x209.jpg\" alt=\"\" class=\"alignnone wp-image-578 size-medium\" width=\"300\" height=\"209\" \/><\/a>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3 itemprop=\"educationalUse\"><strong>Please note:<\/strong><\/h3>\r\nWhen looking at textbook figures you will observe that two forces are indicated.\u00a0 This does not mean that the force you use in the formula is (2 \u00d7 Force P), but simply indicates that one is the Action force and the second one is the Reaction.\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\" style=\"text-align: center\"><strong>Strain and modulus of elasticity<\/strong><\/div>\r\n<div class=\"textbox shaded\">\r\n\r\nNormal strain\r\n\r\n<\/div>\r\nA member in tension or compression will elastically deform proportional with, among other parameters, the original length.\u00a0 Strain, also called unit deformation, is a non-dimensional parameter expressed as:\r\n\r\n<img src=\"https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2Fig3-300x54.jpg\" alt=\"\" class=\"alignnone wp-image-34\" width=\"245\" height=\"44\" \/>\r\n\r\nIf you choose to use a negative value for compression strain (reduction in length) then you must also express the equivalent compression stress as a negative value.\r\n<div class=\"textbox shaded\">Modulus of elasticity<\/div>\r\nThe stress \u2013 strain curve is generated from the <a href=\"https:\/\/en.wikipedia.org\/wiki\/Tensile_testing\" target=\"_blank\" rel=\"noopener\">tensile test<\/a>.\u00a0 Over the elastic region of the graph the deformation is direct proportional with the load.\u00a0 Dividing the load by the cross-section area (constant) and the deformation by the original length (constant) leads to a graphical representation of Strain vs. Stress.\u00a0 The constant ratio of stress and strain is Young's Modulus or Elastic Modulus, a property of each material.\r\n\r\n<img src=\"https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2Fig4-300x58.jpg\" alt=\"\" class=\"alignnone wp-image-35\" width=\"228\" height=\"44\" \/>\r\n<div class=\"textbox shaded\">Elastic deformation<\/div>\r\nCombining the above two relations for strain and Modulus of Elasticity leads to a unified formula for elastic deformation in tension or compression.\r\n\r\n<img src=\"https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2Fig11.jpg\" alt=\"\" class=\"alignnone wp-image-105\" width=\"199\" height=\"62\" \/>\r\n\r\nThis relation is applicable to members with uniform cross-sections, homogeneous material, subject to tensile or compressive loads that results in stresses below the proportional limit (straight line in the \u03c3-\u03b5 curve).\r\n<div class=\"textbox shaded\" style=\"text-align: center\"><strong>Design stress and safety factors<\/strong><\/div>\r\nThese topics were covered in 1<sup>st<\/sup> year Strength of Materials and are presented here as a brief review.\r\n\r\nMembers subjected to an excessive stress may fail by breaking, when actual working stress is greater than the ultimate stress, or due to excessive deformation that renders then inoperable.\u00a0 Consider a heavy condensate line that sags beyond an acceptable limit and while it doesn't break, the flange connections at the end of the lines will develop leaks due to angular movement.\r\n\r\nDesign stress,\u00a0\u03c3<sub>d<\/sub>, is the maximum level of actual\/working stress that is considered acceptable from a safety point of view.\u00a0 The design stress is determined by:\r\n<ul>\r\n \t<li>Material properties, <a href=\"https:\/\/en.wikipedia.org\/wiki\/Ultimate_tensile_strength\">Ultimate Tensile Strength<\/a> or <a href=\"https:\/\/en.wikipedia.org\/wiki\/Yield_(engineering)\">Yield Strength<\/a>, depending if breakage must be avoided or deformation must be limited<\/li>\r\n \t<li><a href=\"https:\/\/en.wikipedia.org\/wiki\/Factor_of_safety\">Safety factor<\/a> (or design factor) N, ratio of maximum strength to the intended load.<\/li>\r\n<\/ul>\r\nThe safety factor is chosen by the designer based on experience, judgment AND guidelines\/rules from relevant codes and standards, based on several criteria such as risk of injuries, design data accuracy, probability, industry standards, and last but not least, cost.\u00a0 Safety factors standards were set by <a href=\"https:\/\/en.wikipedia.org\/wiki\/Structural_engineering\">structural engineers<\/a>, based on rigorous estimates and backed by years of experience.\u00a0\u00a0 Standards are continuously evolving reflecting new and improved design <a href=\"https:\/\/en.wikipedia.org\/wiki\/Allowable_Strength_Design\">philosophies<\/a>. Example:\r\n<ul>\r\n \t<li>published by <a href=\"https:\/\/www.ansi.org\/\">ANSI<\/a> \/ <a href=\"https:\/\/en.wikipedia.org\/wiki\/American_Institute_of_Steel_Construction\">AISC<\/a> , such as <a href=\"https:\/\/www.aisc.org\/globalassets\/aisc\/publications\/standards\/a360-16-spec-and-commentary.pdf\">Specification for Structural Steel Buildings<\/a><\/li>\r\n<\/ul>\r\n<div class=\"textbox shaded\" style=\"text-align: center\"><strong>Design cases<\/strong><\/div>\r\nWhen solving problems students may encounter different scenarios.\u00a0 While the theoretical concepts are the same, the paths to final answers may be different, as required by each approach.\r\n<ol>\r\n \t<li>Estimating if a design\/construction is safe or not\r\n<ol>\r\n \t<li>Given: loads magnitude and distribution, material properties, member shape and dimensions<\/li>\r\n \t<li>Find: actual stress and compare to the design stress; alternatively find the safety factor and decide if it is acceptable based on applicable standards<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Selecting a suitable material\r\n<ol>\r\n \t<li>Given: loads magnitude and distribution, member shape and dimensions<\/li>\r\n \t<li>Find: what material type or grade will provide a strength (yield or ultimate) greater than required, while considering the selected or specified safety factor<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Determining the shape and dimensions of member's cross-section\r\n<ol>\r\n \t<li>Given: loads magnitude and distribution, material properties<\/li>\r\n \t<li>Find: the shape and dimensions of the member so that actual cross-sectional area is greater than minimum required.<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Evaluating maximum allowable load on a component\r\n<ol>\r\n \t<li>Given: load type and distribution, material properties, member shape and dimensions<\/li>\r\n \t<li>Find: maximum load magnitude that leads to an acceptable stress<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n<div class=\"textbox shaded\" style=\"text-align: center\"><strong>Members made from two different materials<\/strong><\/div>\r\nThere are cases when a member under normal stresses is made out of two (or more) materials.\u00a0 One of the objective of such problems is to find the stress in each component.\r\n\r\nFor example, you may have a short column made from a steel pipe filled with concrete, as in the figure.\u00a0 Given the total load, materials properties and geometrical dimensions, we must find the individual stress in each component.\r\n\r\n<img src=\"https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2Fig8-300x300.jpg\" alt=\"\" class=\"alignnone wp-image-59\" width=\"342\" height=\"342\" \/>\r\n\r\nBoth, the steel pipe and the concrete core work together in supporting the load therefore we must find additional relations that combine the two problems into one .\u00a0 Typically, we look for:\r\n<ul>\r\n \t<li>a relation that describes the force distribution between the two materials<\/li>\r\n \t<li>a relation that correlates the deformations of each material<\/li>\r\n<\/ul>\r\nFor this particular problem we may say that:\r\n\r\nEquation 1:\u00a0\u00a0 Total load P = load supported by steel P <sub>steel<\/sub> + load supported by concrete P <sub>concrete\u00a0<\/sub>\r\n\r\ntherefore \u00a0 \u00a0 \u00a0 P = Stress<sub> steel<\/sub> \u00d7 Area <sub>steel<\/sub> + Stress <sub>concrete<\/sub> \u00d7 Area <sub>concrete\r\n<\/sub>\r\n\r\nEquation 2:\u00a0\u00a0 The deformations of both materials are the same\r\n\r\ntherefore \u00a0 \u00a0 \u00a0 Strain <sub>steel<\/sub> = Strain <sub>concrete<\/sub>\r\n\r\nConsidering that Elastic Modulus = Stress \/ Strain, equation (2) yields a relation between the stress and elasticity of both materials\r\n\r\n<img src=\"https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2Fig9-300x78.jpg\" alt=\"\" class=\"alignnone size-medium wp-image-62\" width=\"300\" height=\"78\" \/>\r\n\r\nSubstituting this last relation into equation (1) and solving for Stress <sub>concrete<\/sub> leads to a relation as follows\r\n\r\n<img src=\"https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2Fig10-300x48.jpg\" alt=\"\" class=\"alignnone wp-image-63\" width=\"469\" height=\"75\" \/>\r\n\r\nFurther, Stress <sub>steel<\/sub> can be found.\r\n\r\nNote that depending on the problem, the original two relations may be different therefore a full step-by-step derivation may be required each time.\r\n<div class=\"bcc-box bcc-success\">\r\n<h3><strong>Reasonable answers<\/strong><\/h3>\r\nWhen solving normal stress - strain problems, especially in the SI system, you should be able to judge if your answers are reasonable or not.\r\n\r\n<\/div>\r\n<strong>Example: <\/strong>A 1 m long, 20 mm diameter, A 36 Carbon Steel bar (Materials Properties in Appendix B, Table B2) suspends a 6 tons load.\u00a0 Evaluate the stress and the strain in the bar.\r\n\r\n<img src=\"https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2Fig5-300x45.jpg\" alt=\"\" class=\"alignnone wp-image-48\" width=\"447\" height=\"67\" \/>\r\n\r\n<img src=\"https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2Fig6-300x63.jpg\" alt=\"\" class=\"alignnone wp-image-49\" width=\"261\" height=\"55\" \/>\r\n\r\nNote that typically loads are in kN, cross-section areas in 10<sup>-3<\/sup> m<sup>2<\/sup> and resulting stresses in MPa.\r\n\r\nAlso, since Elastic Moduli are in GPa, the strain (non-dimensional) will be in range of 10<sup>-3<\/sup>.\u00a0 This bar will stretch 0.9 mm under the given load.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3 itemprop=\"educationalUse\"><strong>Assigned Problems<\/strong><\/h3>\r\nWhen solving these questions you are required to use the textbook Appendices.\u00a0 They are valuable references for material properties, geometrical dimensions, etc.\r\n\r\n<\/div>\r\n<strong>Problem 1: <\/strong>A condensate line 152 mm nominal size made of schedule 40 carbon steel pipe is supported by threaded rod hangers spaced at 2.5 m center-to-center.\u00a0\u00a0 The hangers are carbon steel, 50 cm long, with a root diameter of 12 mm.\u00a0 Calculate the stress and the strain in the hangers.\u00a0 Use E=200 GPa for the hangers material.\r\n\r\n<strong>Problem 2: <\/strong>A <a href=\"https:\/\/en.wikipedia.org\/wiki\/Clevis_fastener\">clevis fastener<\/a> with a 1\/2 inch pin is used in a shop lifting machine.\u00a0 If the pin is made of A36 steel determine the maximum safe load, using a safety factor of 2.5 based on the yield strength.\r\n\r\n<strong>Problem 3: <\/strong>A boiler is supported on several short columns as indicated in the figure, made out of Class 35 gray cast iron.\u00a0 Each column supports a load of 50 tonnes.\u00a0 The required safety factor for this construction is 3. Are the columns safe?\r\n\r\n<img src=\"https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2ColumnCorrected.jpg\" alt=\"\" class=\"alignnone wp-image-224\" width=\"303\" height=\"187\" \/>\r\n\r\nUse the following dimensions:\u00a0 A = 30 mm, B = 80 mm, C = 50 mm, D = 140 mm\r\n\r\n<strong>Problem 4: <\/strong>A tension member in a roof truss is subject to a load of 25 <a href=\"https:\/\/en.wikipedia.org\/wiki\/Kip_(unit)\">kips<\/a>.\u00a0 The construction requires using <a href=\"https:\/\/skyciv.com\/steel-i-beam-sizes\/\">L2x2x1\/4<\/a> angle, with a cross-section of 0.944 in<sup>2<\/sup>.\u00a0\u00a0 For building-like structures <a href=\"https:\/\/www.aisc.org\/\">American Institute of Steel Construction <\/a>recommends using a design stress of 0.60\u00d7S<sub>y<\/sub>.\u00a0 Using Appendix B table B2 specify a suitable steel material.\r\n\r\n<strong>Problem 5: <\/strong>A tie rod <a href=\"https:\/\/en.wikipedia.org\/wiki\/Hydraulic_cylinder\">hydraulic cylinder<\/a> as in the figure is made from a 6 inch Schedule 40 stainless steel pipe, 15 inches long.\u00a0 The six tie rods are 1\/2-13 UNC threaded rods with a root diameter of 0.4822 inch and a<a href=\"https:\/\/en.wikipedia.org\/wiki\/Unified_Thread_Standard\"> thread pitch of 13 TPI<\/a>.\u00a0 When assembling the cylinder a clamping force equivalent to one full nut turn from hand-tight position is required.\r\n\r\n&nbsp;\r\n\r\n<img src=\"https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2Tie_rod_cylinder-300x225.jpg\" alt=\"\" class=\"alignnone size-medium wp-image-65\" width=\"300\" height=\"225\" \/>\r\n\r\nDetermine the stress in the cylinder and in the tie rods.\u00a0 Also calculate the strain in each component using an elastic modulus of E<sub>ss<\/sub> = 28\u00d710<sup>6<\/sup> psi and E<sub>rod<\/sub> = 30\u00d710<sup>6<\/sup> psi.\r\n\r\n<strong>Problem 6: <\/strong>Suggest one improvement to this chapter.<strong>\r\n<\/strong>\r\n\r\n&nbsp;","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<p>After completing this chapter you should be able to:<\/p>\n<ul>\n<li>Define normal and shear stress and strain and discuss the relationship between design stress, yield stress and ultimate stress<\/li>\n<li>Design members under tension, compression and shear loads<\/li>\n<li>Determine members deformation under tension and compression<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox shaded\" style=\"text-align: center\"><strong>Mechanical stress<\/strong><\/div>\n<div>This section discusses the effects of mechanical loads (forces) acting on members.\u00a0 Next chapter will cover the effects of thermal loads (thermal expansion).<\/div>\n<div>\n<div class=\"textbox shaded\">Normal, tensile and compressive stresses<\/div>\n<\/div>\n<p>Tension or compression in a member generate normal stresses; they are called \u201cnormal\u201d because the cross-section that resists the load is perpendicular (normal) to the direction of the applied forces.\u00a0 Both tensile and compressive stresses are calculated with:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2Fig1-300x54.jpg\" alt=\"\" class=\"alignnone wp-image-31\" width=\"272\" height=\"49\" srcset=\"https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2Fig1-300x54.jpg 300w, https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2Fig1-768x138.jpg 768w, https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2Fig1-1024x184.jpg 1024w, https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2Fig1-65x12.jpg 65w, https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2Fig1-225x40.jpg 225w, https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2Fig1-350x63.jpg 350w, https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2Fig1.jpg 1026w\" sizes=\"auto, (max-width: 272px) 100vw, 272px\" \/><\/p>\n<p>If a member has a variable cross-section, the area that must be used in calculations is the minimum cross-sectional area; this will give you the maximum stress in the member, which ultimately will govern the design.<\/p>\n<div class=\"textbox shaded\">\n<p>Shear stresses<\/p>\n<\/div>\n<p>In shear the cross-section area that resists the load is parallel with the direction of applied forces.\u00a0 In addition to that, when estimating the shear area you must factor in how many cross-sections contribute to the overall strength of the assembly.<\/p>\n<p>For instance, if you consider the pin of a door hinge as subjected to a shear load, you have to count how many cross-sections resist the load.<\/p>\n<p>The formula for calculating the shear stress is the same:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2Fig2-300x51.jpg\" alt=\"\" class=\"alignnone wp-image-32\" width=\"271\" height=\"46\" srcset=\"https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2Fig2-300x51.jpg 300w, https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2Fig2-768x131.jpg 768w, https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2Fig2-1024x174.jpg 1024w, https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2Fig2-65x11.jpg 65w, https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2Fig2-225x38.jpg 225w, https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2Fig2-350x60.jpg 350w, https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2Fig2.jpg 1135w\" sizes=\"auto, (max-width: 271px) 100vw, 271px\" \/><\/p>\n<p>In a punching operation the area that resists the shear is in the shape of a cylinder for a round hole (think of a cookie cutter).\u00a0 Therefore the area in shear will be found from multiplying the circumference of the shape by the thickness of the plate.<\/p>\n<p><a href=\"https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2FigHoleShear-1.jpg\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2FigHoleShear-1-300x209.jpg\" alt=\"\" class=\"alignnone wp-image-578 size-medium\" width=\"300\" height=\"209\" srcset=\"https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2FigHoleShear-1-300x209.jpg 300w, https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2FigHoleShear-1-768x534.jpg 768w, https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2FigHoleShear-1-1024x712.jpg 1024w, https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2FigHoleShear-1-65x45.jpg 65w, https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2FigHoleShear-1-225x156.jpg 225w, https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2FigHoleShear-1-350x243.jpg 350w, https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2FigHoleShear-1.jpg 1099w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/a><\/p>\n<div class=\"bcc-box bcc-success\">\n<h3 itemprop=\"educationalUse\"><strong>Please note:<\/strong><\/h3>\n<p>When looking at textbook figures you will observe that two forces are indicated.\u00a0 This does not mean that the force you use in the formula is (2 \u00d7 Force P), but simply indicates that one is the Action force and the second one is the Reaction.<\/p>\n<\/div>\n<div class=\"textbox shaded\" style=\"text-align: center\"><strong>Strain and modulus of elasticity<\/strong><\/div>\n<div class=\"textbox shaded\">\n<p>Normal strain<\/p>\n<\/div>\n<p>A member in tension or compression will elastically deform proportional with, among other parameters, the original length.\u00a0 Strain, also called unit deformation, is a non-dimensional parameter expressed as:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2Fig3-300x54.jpg\" alt=\"\" class=\"alignnone wp-image-34\" width=\"245\" height=\"44\" srcset=\"https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2Fig3-300x54.jpg 300w, https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2Fig3-768x139.jpg 768w, https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2Fig3-65x12.jpg 65w, https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2Fig3-225x41.jpg 225w, https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2Fig3-350x63.jpg 350w, https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2Fig3.jpg 950w\" sizes=\"auto, (max-width: 245px) 100vw, 245px\" \/><\/p>\n<p>If you choose to use a negative value for compression strain (reduction in length) then you must also express the equivalent compression stress as a negative value.<\/p>\n<div class=\"textbox shaded\">Modulus of elasticity<\/div>\n<p>The stress \u2013 strain curve is generated from the <a href=\"https:\/\/en.wikipedia.org\/wiki\/Tensile_testing\" target=\"_blank\" rel=\"noopener\">tensile test<\/a>.\u00a0 Over the elastic region of the graph the deformation is direct proportional with the load.\u00a0 Dividing the load by the cross-section area (constant) and the deformation by the original length (constant) leads to a graphical representation of Strain vs. Stress.\u00a0 The constant ratio of stress and strain is Young&#8217;s Modulus or Elastic Modulus, a property of each material.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2Fig4-300x58.jpg\" alt=\"\" class=\"alignnone wp-image-35\" width=\"228\" height=\"44\" srcset=\"https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2Fig4-300x58.jpg 300w, https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2Fig4-768x149.jpg 768w, https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2Fig4-65x13.jpg 65w, https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2Fig4-225x44.jpg 225w, https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2Fig4-350x68.jpg 350w, https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2Fig4.jpg 855w\" sizes=\"auto, (max-width: 228px) 100vw, 228px\" \/><\/p>\n<div class=\"textbox shaded\">Elastic deformation<\/div>\n<p>Combining the above two relations for strain and Modulus of Elasticity leads to a unified formula for elastic deformation in tension or compression.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2Fig11.jpg\" alt=\"\" class=\"alignnone wp-image-105\" width=\"199\" height=\"62\" srcset=\"https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2Fig11.jpg 289w, https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2Fig11-65x20.jpg 65w, https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2Fig11-225x70.jpg 225w\" sizes=\"auto, (max-width: 199px) 100vw, 199px\" \/><\/p>\n<p>This relation is applicable to members with uniform cross-sections, homogeneous material, subject to tensile or compressive loads that results in stresses below the proportional limit (straight line in the \u03c3-\u03b5 curve).<\/p>\n<div class=\"textbox shaded\" style=\"text-align: center\"><strong>Design stress and safety factors<\/strong><\/div>\n<p>These topics were covered in 1<sup>st<\/sup> year Strength of Materials and are presented here as a brief review.<\/p>\n<p>Members subjected to an excessive stress may fail by breaking, when actual working stress is greater than the ultimate stress, or due to excessive deformation that renders then inoperable.\u00a0 Consider a heavy condensate line that sags beyond an acceptable limit and while it doesn&#8217;t break, the flange connections at the end of the lines will develop leaks due to angular movement.<\/p>\n<p>Design stress,\u00a0\u03c3<sub>d<\/sub>, is the maximum level of actual\/working stress that is considered acceptable from a safety point of view.\u00a0 The design stress is determined by:<\/p>\n<ul>\n<li>Material properties, <a href=\"https:\/\/en.wikipedia.org\/wiki\/Ultimate_tensile_strength\">Ultimate Tensile Strength<\/a> or <a href=\"https:\/\/en.wikipedia.org\/wiki\/Yield_(engineering)\">Yield Strength<\/a>, depending if breakage must be avoided or deformation must be limited<\/li>\n<li><a href=\"https:\/\/en.wikipedia.org\/wiki\/Factor_of_safety\">Safety factor<\/a> (or design factor) N, ratio of maximum strength to the intended load.<\/li>\n<\/ul>\n<p>The safety factor is chosen by the designer based on experience, judgment AND guidelines\/rules from relevant codes and standards, based on several criteria such as risk of injuries, design data accuracy, probability, industry standards, and last but not least, cost.\u00a0 Safety factors standards were set by <a href=\"https:\/\/en.wikipedia.org\/wiki\/Structural_engineering\">structural engineers<\/a>, based on rigorous estimates and backed by years of experience.\u00a0\u00a0 Standards are continuously evolving reflecting new and improved design <a href=\"https:\/\/en.wikipedia.org\/wiki\/Allowable_Strength_Design\">philosophies<\/a>. Example:<\/p>\n<ul>\n<li>published by <a href=\"https:\/\/www.ansi.org\/\">ANSI<\/a> \/ <a href=\"https:\/\/en.wikipedia.org\/wiki\/American_Institute_of_Steel_Construction\">AISC<\/a> , such as <a href=\"https:\/\/www.aisc.org\/globalassets\/aisc\/publications\/standards\/a360-16-spec-and-commentary.pdf\">Specification for Structural Steel Buildings<\/a><\/li>\n<\/ul>\n<div class=\"textbox shaded\" style=\"text-align: center\"><strong>Design cases<\/strong><\/div>\n<p>When solving problems students may encounter different scenarios.\u00a0 While the theoretical concepts are the same, the paths to final answers may be different, as required by each approach.<\/p>\n<ol>\n<li>Estimating if a design\/construction is safe or not\n<ol>\n<li>Given: loads magnitude and distribution, material properties, member shape and dimensions<\/li>\n<li>Find: actual stress and compare to the design stress; alternatively find the safety factor and decide if it is acceptable based on applicable standards<\/li>\n<\/ol>\n<\/li>\n<li>Selecting a suitable material\n<ol>\n<li>Given: loads magnitude and distribution, member shape and dimensions<\/li>\n<li>Find: what material type or grade will provide a strength (yield or ultimate) greater than required, while considering the selected or specified safety factor<\/li>\n<\/ol>\n<\/li>\n<li>Determining the shape and dimensions of member&#8217;s cross-section\n<ol>\n<li>Given: loads magnitude and distribution, material properties<\/li>\n<li>Find: the shape and dimensions of the member so that actual cross-sectional area is greater than minimum required.<\/li>\n<\/ol>\n<\/li>\n<li>Evaluating maximum allowable load on a component\n<ol>\n<li>Given: load type and distribution, material properties, member shape and dimensions<\/li>\n<li>Find: maximum load magnitude that leads to an acceptable stress<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<div class=\"textbox shaded\" style=\"text-align: center\"><strong>Members made from two different materials<\/strong><\/div>\n<p>There are cases when a member under normal stresses is made out of two (or more) materials.\u00a0 One of the objective of such problems is to find the stress in each component.<\/p>\n<p>For example, you may have a short column made from a steel pipe filled with concrete, as in the figure.\u00a0 Given the total load, materials properties and geometrical dimensions, we must find the individual stress in each component.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2Fig8-300x300.jpg\" alt=\"\" class=\"alignnone wp-image-59\" width=\"342\" height=\"342\" srcset=\"https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2Fig8-300x300.jpg 300w, https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2Fig8-150x150.jpg 150w, https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2Fig8-65x65.jpg 65w, https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2Fig8-225x225.jpg 225w, https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2Fig8-350x351.jpg 350w, https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2Fig8.jpg 471w\" sizes=\"auto, (max-width: 342px) 100vw, 342px\" \/><\/p>\n<p>Both, the steel pipe and the concrete core work together in supporting the load therefore we must find additional relations that combine the two problems into one .\u00a0 Typically, we look for:<\/p>\n<ul>\n<li>a relation that describes the force distribution between the two materials<\/li>\n<li>a relation that correlates the deformations of each material<\/li>\n<\/ul>\n<p>For this particular problem we may say that:<\/p>\n<p>Equation 1:\u00a0\u00a0 Total load P = load supported by steel P <sub>steel<\/sub> + load supported by concrete P <sub>concrete\u00a0<\/sub><\/p>\n<p>therefore \u00a0 \u00a0 \u00a0 P = Stress<sub> steel<\/sub> \u00d7 Area <sub>steel<\/sub> + Stress <sub>concrete<\/sub> \u00d7 Area <sub>concrete<br \/>\n<\/sub><\/p>\n<p>Equation 2:\u00a0\u00a0 The deformations of both materials are the same<\/p>\n<p>therefore \u00a0 \u00a0 \u00a0 Strain <sub>steel<\/sub> = Strain <sub>concrete<\/sub><\/p>\n<p>Considering that Elastic Modulus = Stress \/ Strain, equation (2) yields a relation between the stress and elasticity of both materials<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2Fig9-300x78.jpg\" alt=\"\" class=\"alignnone size-medium wp-image-62\" width=\"300\" height=\"78\" srcset=\"https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2Fig9-300x78.jpg 300w, https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2Fig9-768x199.jpg 768w, https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2Fig9-65x17.jpg 65w, https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2Fig9-225x58.jpg 225w, https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2Fig9-350x91.jpg 350w, https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2Fig9.jpg 874w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/p>\n<p>Substituting this last relation into equation (1) and solving for Stress <sub>concrete<\/sub> leads to a relation as follows<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2Fig10-300x48.jpg\" alt=\"\" class=\"alignnone wp-image-63\" width=\"469\" height=\"75\" srcset=\"https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2Fig10-300x48.jpg 300w, https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2Fig10-768x122.jpg 768w, https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2Fig10-1024x163.jpg 1024w, https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2Fig10-65x10.jpg 65w, https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2Fig10-225x36.jpg 225w, https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2Fig10-350x56.jpg 350w, https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2Fig10.jpg 1440w\" sizes=\"auto, (max-width: 469px) 100vw, 469px\" \/><\/p>\n<p>Further, Stress <sub>steel<\/sub> can be found.<\/p>\n<p>Note that depending on the problem, the original two relations may be different therefore a full step-by-step derivation may be required each time.<\/p>\n<div class=\"bcc-box bcc-success\">\n<h3><strong>Reasonable answers<\/strong><\/h3>\n<p>When solving normal stress &#8211; strain problems, especially in the SI system, you should be able to judge if your answers are reasonable or not.<\/p>\n<\/div>\n<p><strong>Example: <\/strong>A 1 m long, 20 mm diameter, A 36 Carbon Steel bar (Materials Properties in Appendix B, Table B2) suspends a 6 tons load.\u00a0 Evaluate the stress and the strain in the bar.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2Fig5-300x45.jpg\" alt=\"\" class=\"alignnone wp-image-48\" width=\"447\" height=\"67\" srcset=\"https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2Fig5-300x45.jpg 300w, https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2Fig5-768x115.jpg 768w, https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2Fig5-1024x153.jpg 1024w, https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2Fig5-65x10.jpg 65w, https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2Fig5-225x34.jpg 225w, https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2Fig5-350x52.jpg 350w, https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2Fig5.jpg 1417w\" sizes=\"auto, (max-width: 447px) 100vw, 447px\" \/><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2Fig6-300x63.jpg\" alt=\"\" class=\"alignnone wp-image-49\" width=\"261\" height=\"55\" srcset=\"https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2Fig6-300x63.jpg 300w, https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2Fig6-65x14.jpg 65w, https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2Fig6-225x48.jpg 225w, https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2Fig6-350x74.jpg 350w, https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2Fig6.jpg 757w\" sizes=\"auto, (max-width: 261px) 100vw, 261px\" \/><\/p>\n<p>Note that typically loads are in kN, cross-section areas in 10<sup>-3<\/sup> m<sup>2<\/sup> and resulting stresses in MPa.<\/p>\n<p>Also, since Elastic Moduli are in GPa, the strain (non-dimensional) will be in range of 10<sup>-3<\/sup>.\u00a0 This bar will stretch 0.9 mm under the given load.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3 itemprop=\"educationalUse\"><strong>Assigned Problems<\/strong><\/h3>\n<p>When solving these questions you are required to use the textbook Appendices.\u00a0 They are valuable references for material properties, geometrical dimensions, etc.<\/p>\n<\/div>\n<p><strong>Problem 1: <\/strong>A condensate line 152 mm nominal size made of schedule 40 carbon steel pipe is supported by threaded rod hangers spaced at 2.5 m center-to-center.\u00a0\u00a0 The hangers are carbon steel, 50 cm long, with a root diameter of 12 mm.\u00a0 Calculate the stress and the strain in the hangers.\u00a0 Use E=200 GPa for the hangers material.<\/p>\n<p><strong>Problem 2: <\/strong>A <a href=\"https:\/\/en.wikipedia.org\/wiki\/Clevis_fastener\">clevis fastener<\/a> with a 1\/2 inch pin is used in a shop lifting machine.\u00a0 If the pin is made of A36 steel determine the maximum safe load, using a safety factor of 2.5 based on the yield strength.<\/p>\n<p><strong>Problem 3: <\/strong>A boiler is supported on several short columns as indicated in the figure, made out of Class 35 gray cast iron.\u00a0 Each column supports a load of 50 tonnes.\u00a0 The required safety factor for this construction is 3. Are the columns safe?<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2ColumnCorrected.jpg\" alt=\"\" class=\"alignnone wp-image-224\" width=\"303\" height=\"187\" srcset=\"https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2ColumnCorrected.jpg 993w, https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2ColumnCorrected-300x185.jpg 300w, https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2ColumnCorrected-768x473.jpg 768w, https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2ColumnCorrected-65x40.jpg 65w, https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2ColumnCorrected-225x139.jpg 225w, https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2ColumnCorrected-350x216.jpg 350w\" sizes=\"auto, (max-width: 303px) 100vw, 303px\" \/><\/p>\n<p>Use the following dimensions:\u00a0 A = 30 mm, B = 80 mm, C = 50 mm, D = 140 mm<\/p>\n<p><strong>Problem 4: <\/strong>A tension member in a roof truss is subject to a load of 25 <a href=\"https:\/\/en.wikipedia.org\/wiki\/Kip_(unit)\">kips<\/a>.\u00a0 The construction requires using <a href=\"https:\/\/skyciv.com\/steel-i-beam-sizes\/\">L2x2x1\/4<\/a> angle, with a cross-section of 0.944 in<sup>2<\/sup>.\u00a0\u00a0 For building-like structures <a href=\"https:\/\/www.aisc.org\/\">American Institute of Steel Construction <\/a>recommends using a design stress of 0.60\u00d7S<sub>y<\/sub>.\u00a0 Using Appendix B table B2 specify a suitable steel material.<\/p>\n<p><strong>Problem 5: <\/strong>A tie rod <a href=\"https:\/\/en.wikipedia.org\/wiki\/Hydraulic_cylinder\">hydraulic cylinder<\/a> as in the figure is made from a 6 inch Schedule 40 stainless steel pipe, 15 inches long.\u00a0 The six tie rods are 1\/2-13 UNC threaded rods with a root diameter of 0.4822 inch and a<a href=\"https:\/\/en.wikipedia.org\/wiki\/Unified_Thread_Standard\"> thread pitch of 13 TPI<\/a>.\u00a0 When assembling the cylinder a clamping force equivalent to one full nut turn from hand-tight position is required.<\/p>\n<p>&nbsp;<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2Tie_rod_cylinder-300x225.jpg\" alt=\"\" class=\"alignnone size-medium wp-image-65\" width=\"300\" height=\"225\" srcset=\"https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2Tie_rod_cylinder-300x225.jpg 300w, https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2Tie_rod_cylinder-768x576.jpg 768w, https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2Tie_rod_cylinder-1024x768.jpg 1024w, https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2Tie_rod_cylinder-65x49.jpg 65w, https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2Tie_rod_cylinder-225x169.jpg 225w, https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-content\/uploads\/sites\/290\/2017\/11\/Ch2Tie_rod_cylinder-350x263.jpg 350w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/p>\n<p>Determine the stress in the cylinder and in the tie rods.\u00a0 Also calculate the strain in each component using an elastic modulus of E<sub>ss<\/sub> = 28\u00d710<sup>6<\/sup> psi and E<sub>rod<\/sub> = 30\u00d710<sup>6<\/sup> psi.<\/p>\n<p><strong>Problem 6: <\/strong>Suggest one improvement to this chapter.<strong><br \/>\n<\/strong><\/p>\n<p>&nbsp;<\/p>\n","protected":false},"author":239,"menu_order":2,"template":"","meta":{"pb_show_title":"on","pb_short_title":"Stress-Strain","pb_subtitle":"Stress-Strain","pb_authors":[],"pb_section_license":""},"chapter-type":[47],"contributor":[57],"license":[],"class_list":["post-36","chapter","type-chapter","status-publish","hentry","chapter-type-standard","contributor-alex-podut"],"part":3,"_links":{"self":[{"href":"https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-json\/pressbooks\/v2\/chapters\/36","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-json\/wp\/v2\/users\/239"}],"version-history":[{"count":25,"href":"https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-json\/pressbooks\/v2\/chapters\/36\/revisions"}],"predecessor-version":[{"id":613,"href":"https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-json\/pressbooks\/v2\/chapters\/36\/revisions\/613"}],"part":[{"href":"https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-json\/pressbooks\/v2\/parts\/3"}],"metadata":[{"href":"https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-json\/pressbooks\/v2\/chapters\/36\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-json\/wp\/v2\/media?parent=36"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-json\/pressbooks\/v2\/chapter-type?post=36"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-json\/wp\/v2\/contributor?post=36"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/powr4406\/wp-json\/wp\/v2\/license?post=36"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}