{"id":5,"date":"2017-11-01T14:00:22","date_gmt":"2017-11-01T18:00:22","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/somtest0\/2017\/11\/01\/chapter-1\/"},"modified":"2017-11-17T15:04:42","modified_gmt":"2017-11-17T20:04:42","slug":"stress-and-strain","status":"web-only","type":"chapter","link":"https:\/\/pressbooks.bccampus.ca\/somtest0\/chapter\/stress-and-strain\/","title":{"raw":"Stress and Strain","rendered":"Stress and Strain"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3 itemprop=\"educationalUse\">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<\/li>\r\n \t<li>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=\"bcc-box bcc-success\">\r\n<h3 itemprop=\"educationalUse\">Mechanical Stress<\/h3>\r\nThis section discusses the effects of mechanical loads (forces) acting on members.\u00a0 Chapter 3 will cover the effects of thermal loads (thermal expansion).\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">Normal, tensile and compressive stresses<\/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\/somtest0\/wp-content\/uploads\/sites\/273\/2017\/11\/Ch2Fig1-300x54.jpg\" alt=\"\" class=\"alignnone size-medium wp-image-107\" width=\"300\" height=\"54\" \/>\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\/somtest0\/wp-content\/uploads\/sites\/273\/2017\/11\/Ch2Fig2-300x51.jpg\" alt=\"\" class=\"alignnone size-medium wp-image-108\" width=\"300\" height=\"51\" \/>\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<img src=\"https:\/\/pressbooks.bccampus.ca\/somtest0\/wp-content\/uploads\/sites\/273\/2017\/11\/Ch2Mott1-300x237.jpg\" alt=\"\" class=\"alignnone wp-image-116\" width=\"571\" height=\"452\" \/>\r\n\r\n<em><strong>Hint<\/strong><\/em>:\u00a0 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.\r\n<div class=\"bcc-box bcc-success\">\r\n<h3 itemprop=\"educationalUse\">Strain and Modulus of Elasticity<\/h3>\r\n<\/div>\r\n&nbsp;\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\/somtest0\/wp-content\/uploads\/sites\/273\/2017\/11\/Ch2Fig3-300x54.jpg\" alt=\"\" class=\"alignnone wp-image-113\" width=\"317\" height=\"57\" \/>\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\/somtest0\/wp-content\/uploads\/sites\/273\/2017\/11\/Ch2Fig4-300x58.jpg\" alt=\"\" class=\"alignnone size-medium wp-image-120\" width=\"300\" height=\"58\" \/>\r\n\r\n&nbsp;\r\n<div class=\"bcc-box bcc-success\">\r\n<h3 itemprop=\"educationalUse\">Reasonable Answers<\/h3>\r\nWhen solving normal stress - strain problems, especially in the SI system, you may be able to judge if your answers are reasonable or not.\u00a0 Consider the following example:\r\n\r\n<\/div>\r\nA 1 m long, 20 mm diameter, A 36 Carbon Steel bar (<em>Yield strength 248 MPa, Ultimate strength 400 MPa, Elastic Modulus =200 GPa<\/em>) suspends a 6 metric tons load.\u00a0 Evaluate the stress and the strain in the bar.\r\n\r\n&nbsp;","rendered":"<div class=\"textbox learning-objectives\">\n<h3 itemprop=\"educationalUse\">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<\/li>\n<li>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=\"bcc-box bcc-success\">\n<h3 itemprop=\"educationalUse\">Mechanical Stress<\/h3>\n<p>This section discusses the effects of mechanical loads (forces) acting on members.\u00a0 Chapter 3 will cover the effects of thermal loads (thermal expansion).<\/p>\n<\/div>\n<div class=\"textbox shaded\">Normal, tensile and compressive stresses<\/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\/somtest0\/wp-content\/uploads\/sites\/273\/2017\/11\/Ch2Fig1-300x54.jpg\" alt=\"\" class=\"alignnone size-medium wp-image-107\" width=\"300\" height=\"54\" srcset=\"https:\/\/pressbooks.bccampus.ca\/somtest0\/wp-content\/uploads\/sites\/273\/2017\/11\/Ch2Fig1-300x54.jpg 300w, https:\/\/pressbooks.bccampus.ca\/somtest0\/wp-content\/uploads\/sites\/273\/2017\/11\/Ch2Fig1-768x138.jpg 768w, https:\/\/pressbooks.bccampus.ca\/somtest0\/wp-content\/uploads\/sites\/273\/2017\/11\/Ch2Fig1-1024x184.jpg 1024w, https:\/\/pressbooks.bccampus.ca\/somtest0\/wp-content\/uploads\/sites\/273\/2017\/11\/Ch2Fig1-65x12.jpg 65w, https:\/\/pressbooks.bccampus.ca\/somtest0\/wp-content\/uploads\/sites\/273\/2017\/11\/Ch2Fig1-225x40.jpg 225w, https:\/\/pressbooks.bccampus.ca\/somtest0\/wp-content\/uploads\/sites\/273\/2017\/11\/Ch2Fig1-350x63.jpg 350w, https:\/\/pressbooks.bccampus.ca\/somtest0\/wp-content\/uploads\/sites\/273\/2017\/11\/Ch2Fig1.jpg 1026w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/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\/somtest0\/wp-content\/uploads\/sites\/273\/2017\/11\/Ch2Fig2-300x51.jpg\" alt=\"\" class=\"alignnone size-medium wp-image-108\" width=\"300\" height=\"51\" srcset=\"https:\/\/pressbooks.bccampus.ca\/somtest0\/wp-content\/uploads\/sites\/273\/2017\/11\/Ch2Fig2-300x51.jpg 300w, https:\/\/pressbooks.bccampus.ca\/somtest0\/wp-content\/uploads\/sites\/273\/2017\/11\/Ch2Fig2-768x131.jpg 768w, https:\/\/pressbooks.bccampus.ca\/somtest0\/wp-content\/uploads\/sites\/273\/2017\/11\/Ch2Fig2-1024x174.jpg 1024w, https:\/\/pressbooks.bccampus.ca\/somtest0\/wp-content\/uploads\/sites\/273\/2017\/11\/Ch2Fig2-65x11.jpg 65w, https:\/\/pressbooks.bccampus.ca\/somtest0\/wp-content\/uploads\/sites\/273\/2017\/11\/Ch2Fig2-225x38.jpg 225w, https:\/\/pressbooks.bccampus.ca\/somtest0\/wp-content\/uploads\/sites\/273\/2017\/11\/Ch2Fig2-350x60.jpg 350w, https:\/\/pressbooks.bccampus.ca\/somtest0\/wp-content\/uploads\/sites\/273\/2017\/11\/Ch2Fig2.jpg 1135w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/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><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/somtest0\/wp-content\/uploads\/sites\/273\/2017\/11\/Ch2Mott1-300x237.jpg\" alt=\"\" class=\"alignnone wp-image-116\" width=\"571\" height=\"452\" srcset=\"https:\/\/pressbooks.bccampus.ca\/somtest0\/wp-content\/uploads\/sites\/273\/2017\/11\/Ch2Mott1-300x237.jpg 300w, https:\/\/pressbooks.bccampus.ca\/somtest0\/wp-content\/uploads\/sites\/273\/2017\/11\/Ch2Mott1-768x607.jpg 768w, https:\/\/pressbooks.bccampus.ca\/somtest0\/wp-content\/uploads\/sites\/273\/2017\/11\/Ch2Mott1-1024x810.jpg 1024w, https:\/\/pressbooks.bccampus.ca\/somtest0\/wp-content\/uploads\/sites\/273\/2017\/11\/Ch2Mott1-65x51.jpg 65w, https:\/\/pressbooks.bccampus.ca\/somtest0\/wp-content\/uploads\/sites\/273\/2017\/11\/Ch2Mott1-225x178.jpg 225w, https:\/\/pressbooks.bccampus.ca\/somtest0\/wp-content\/uploads\/sites\/273\/2017\/11\/Ch2Mott1-350x277.jpg 350w, https:\/\/pressbooks.bccampus.ca\/somtest0\/wp-content\/uploads\/sites\/273\/2017\/11\/Ch2Mott1.jpg 1334w\" sizes=\"auto, (max-width: 571px) 100vw, 571px\" \/><\/p>\n<p><em><strong>Hint<\/strong><\/em>:\u00a0 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 class=\"bcc-box bcc-success\">\n<h3 itemprop=\"educationalUse\">Strain and Modulus of Elasticity<\/h3>\n<\/div>\n<p>&nbsp;<\/p>\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\/somtest0\/wp-content\/uploads\/sites\/273\/2017\/11\/Ch2Fig3-300x54.jpg\" alt=\"\" class=\"alignnone wp-image-113\" width=\"317\" height=\"57\" srcset=\"https:\/\/pressbooks.bccampus.ca\/somtest0\/wp-content\/uploads\/sites\/273\/2017\/11\/Ch2Fig3-300x54.jpg 300w, https:\/\/pressbooks.bccampus.ca\/somtest0\/wp-content\/uploads\/sites\/273\/2017\/11\/Ch2Fig3-768x139.jpg 768w, https:\/\/pressbooks.bccampus.ca\/somtest0\/wp-content\/uploads\/sites\/273\/2017\/11\/Ch2Fig3-65x12.jpg 65w, https:\/\/pressbooks.bccampus.ca\/somtest0\/wp-content\/uploads\/sites\/273\/2017\/11\/Ch2Fig3-225x41.jpg 225w, https:\/\/pressbooks.bccampus.ca\/somtest0\/wp-content\/uploads\/sites\/273\/2017\/11\/Ch2Fig3-350x63.jpg 350w, https:\/\/pressbooks.bccampus.ca\/somtest0\/wp-content\/uploads\/sites\/273\/2017\/11\/Ch2Fig3.jpg 950w\" sizes=\"auto, (max-width: 317px) 100vw, 317px\" \/><\/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\/somtest0\/wp-content\/uploads\/sites\/273\/2017\/11\/Ch2Fig4-300x58.jpg\" alt=\"\" class=\"alignnone size-medium wp-image-120\" width=\"300\" height=\"58\" srcset=\"https:\/\/pressbooks.bccampus.ca\/somtest0\/wp-content\/uploads\/sites\/273\/2017\/11\/Ch2Fig4-300x58.jpg 300w, https:\/\/pressbooks.bccampus.ca\/somtest0\/wp-content\/uploads\/sites\/273\/2017\/11\/Ch2Fig4-768x149.jpg 768w, https:\/\/pressbooks.bccampus.ca\/somtest0\/wp-content\/uploads\/sites\/273\/2017\/11\/Ch2Fig4-65x13.jpg 65w, https:\/\/pressbooks.bccampus.ca\/somtest0\/wp-content\/uploads\/sites\/273\/2017\/11\/Ch2Fig4-225x44.jpg 225w, https:\/\/pressbooks.bccampus.ca\/somtest0\/wp-content\/uploads\/sites\/273\/2017\/11\/Ch2Fig4-350x68.jpg 350w, https:\/\/pressbooks.bccampus.ca\/somtest0\/wp-content\/uploads\/sites\/273\/2017\/11\/Ch2Fig4.jpg 855w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/p>\n<p>&nbsp;<\/p>\n<div class=\"bcc-box bcc-success\">\n<h3 itemprop=\"educationalUse\">Reasonable Answers<\/h3>\n<p>When solving normal stress &#8211; strain problems, especially in the SI system, you may be able to judge if your answers are reasonable or not.\u00a0 Consider the following example:<\/p>\n<\/div>\n<p>A 1 m long, 20 mm diameter, A 36 Carbon Steel bar (<em>Yield strength 248 MPa, Ultimate strength 400 MPa, Elastic Modulus =200 GPa<\/em>) suspends a 6 metric tons load.\u00a0 Evaluate the stress and the strain in the bar.<\/p>\n<p>&nbsp;<\/p>\n","protected":false},"author":239,"menu_order":2,"template":"","meta":{"pb_show_title":"","pb_short_title":"Stress and 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