Page 1 ENGINEERING MECHANICS FOR STRUCTURES 4.3 CHAPTER 4 Stress, Strain, Stress-Strain 4.1 The Concept of Stress — An Introduction We have talked about internal forces, distributed them uniformly over an area and they became a normal stress acting perpendicular to some internal surface, or a shear stress acting tangentially, in plane. Up to now, we have said little about how these normal and shear stresses might vary with position throughout a solid. 1 Up to now, the choice of planes, their orientation within a solid, was dictated by the geometry of the solid and the nature of the loading. We have said nothing about how these stress components might change if we looked at a set of planes of another orientation. Now we consider a more general situation, an arbitrarily shaped solid. We are going to lift our gaze up from the world of crude structural elements such as truss bars in tension, shafts in torsion, or beams in bending to view these “solids” from a more abstract perspective. They all become special cases of more gen- eral stuff we call a solid continuum. We will address two questions: •How might stresses vary from one point to another throughout a continuum; •How do the normal and shear components of stress acting on a plane at a given point change as we change the orientation of the plane at the point. The ?rst bullet introduces the notion of stress ?eld; the second concerns the transformation of com- ponents of stress at a point. To begin with the ?rst bullet we re-examine the case of a bar suspended vertically and loaded by its own weight, a case considered in section 3.2, page 62. (Note, I have changed the orientation of the reference 1. The beam is the one exception. There we explored how different normal stress distributions over a rectangular cross-section could be equivalent to a bending moment and zero resultant force. Page 2 ENGINEERING MECHANICS FOR STRUCTURES 4.3 CHAPTER 4 Stress, Strain, Stress-Strain 4.1 The Concept of Stress — An Introduction We have talked about internal forces, distributed them uniformly over an area and they became a normal stress acting perpendicular to some internal surface, or a shear stress acting tangentially, in plane. Up to now, we have said little about how these normal and shear stresses might vary with position throughout a solid. 1 Up to now, the choice of planes, their orientation within a solid, was dictated by the geometry of the solid and the nature of the loading. We have said nothing about how these stress components might change if we looked at a set of planes of another orientation. Now we consider a more general situation, an arbitrarily shaped solid. We are going to lift our gaze up from the world of crude structural elements such as truss bars in tension, shafts in torsion, or beams in bending to view these “solids” from a more abstract perspective. They all become special cases of more gen- eral stuff we call a solid continuum. We will address two questions: •How might stresses vary from one point to another throughout a continuum; •How do the normal and shear components of stress acting on a plane at a given point change as we change the orientation of the plane at the point. The ?rst bullet introduces the notion of stress ?eld; the second concerns the transformation of com- ponents of stress at a point. To begin with the ?rst bullet we re-examine the case of a bar suspended vertically and loaded by its own weight, a case considered in section 3.2, page 62. (Note, I have changed the orientation of the reference 1. The beam is the one exception. There we explored how different normal stress distributions over a rectangular cross-section could be equivalent to a bending moment and zero resultant force. The Concept of Stress — An Introduction 4.4 ENGINEERING MECHANICS FOR STRUCTURES axes). We will construct a differential equation which governs how the axial stress varies as we move up and down the bar. We will solve this differential equation, not forgetting to apply an appropriate boundary con- dition and determine the axial stress ?eld. We see that for equilibrium of the differential element of the bar, of planar cross-sectional area A and of weight density ?, we have If we assume the tensile force is uniformly distributed over the cross-sectional area, and dividing by the area (which does not change with the independent spatial coordinate y) we can write Chanting “...going to the limit, letting ? y go to zero”, we obtain a differential equation ?xing how s(y), a function of y, varies throughout our continuum, namely We solve this ordinary differential equation easily, integrating once and obtain The Constant is ?xed by a prescribed condition at some y surface; If the end of the bar is stress free, we indi- cate this writing If, on another occasion, a weight of magnitude P 0 is suspended from the free end, we would have Here then are two stress ?elds for two different loading conditions 1 . Each stress ?eld describes how the normal stress s(x,y,z) varies throughout the continuum at every point in the continuum. I show the stress as a function of x and z as well as y to emphasize that we can evaluate its value at every point in the continuum, although it only varies with y. That the stress does not vary with x and z was implied when we 1. A third loading condition is obtained by setting the weight density ? to zero; our bar then is assumed weightless relative to the end-load P 0. F(y) x y z y ? y F(y) + ? F ? w(y) = ? A? y ? y + ?s s(y) s(y) F ? F ? A ? y ·· F – –+0 = s?s ? ? y ·s – –+0 = where s FA / = y d ds ?–0 = s y () ? y · Cons t tan + = at y = 0 s 0 = so s y () ? y · = at y = 0 s P 0 A / = and s y () ? y · P 0 A / + = Page 3 ENGINEERING MECHANICS FOR STRUCTURES 4.3 CHAPTER 4 Stress, Strain, Stress-Strain 4.1 The Concept of Stress — An Introduction We have talked about internal forces, distributed them uniformly over an area and they became a normal stress acting perpendicular to some internal surface, or a shear stress acting tangentially, in plane. Up to now, we have said little about how these normal and shear stresses might vary with position throughout a solid. 1 Up to now, the choice of planes, their orientation within a solid, was dictated by the geometry of the solid and the nature of the loading. We have said nothing about how these stress components might change if we looked at a set of planes of another orientation. Now we consider a more general situation, an arbitrarily shaped solid. We are going to lift our gaze up from the world of crude structural elements such as truss bars in tension, shafts in torsion, or beams in bending to view these “solids” from a more abstract perspective. They all become special cases of more gen- eral stuff we call a solid continuum. We will address two questions: •How might stresses vary from one point to another throughout a continuum; •How do the normal and shear components of stress acting on a plane at a given point change as we change the orientation of the plane at the point. The ?rst bullet introduces the notion of stress ?eld; the second concerns the transformation of com- ponents of stress at a point. To begin with the ?rst bullet we re-examine the case of a bar suspended vertically and loaded by its own weight, a case considered in section 3.2, page 62. (Note, I have changed the orientation of the reference 1. The beam is the one exception. There we explored how different normal stress distributions over a rectangular cross-section could be equivalent to a bending moment and zero resultant force. The Concept of Stress — An Introduction 4.4 ENGINEERING MECHANICS FOR STRUCTURES axes). We will construct a differential equation which governs how the axial stress varies as we move up and down the bar. We will solve this differential equation, not forgetting to apply an appropriate boundary con- dition and determine the axial stress ?eld. We see that for equilibrium of the differential element of the bar, of planar cross-sectional area A and of weight density ?, we have If we assume the tensile force is uniformly distributed over the cross-sectional area, and dividing by the area (which does not change with the independent spatial coordinate y) we can write Chanting “...going to the limit, letting ? y go to zero”, we obtain a differential equation ?xing how s(y), a function of y, varies throughout our continuum, namely We solve this ordinary differential equation easily, integrating once and obtain The Constant is ?xed by a prescribed condition at some y surface; If the end of the bar is stress free, we indi- cate this writing If, on another occasion, a weight of magnitude P 0 is suspended from the free end, we would have Here then are two stress ?elds for two different loading conditions 1 . Each stress ?eld describes how the normal stress s(x,y,z) varies throughout the continuum at every point in the continuum. I show the stress as a function of x and z as well as y to emphasize that we can evaluate its value at every point in the continuum, although it only varies with y. That the stress does not vary with x and z was implied when we 1. A third loading condition is obtained by setting the weight density ? to zero; our bar then is assumed weightless relative to the end-load P 0. F(y) x y z y ? y F(y) + ? F ? w(y) = ? A? y ? y + ?s s(y) s(y) F ? F ? A ? y ·· F – –+0 = s?s ? ? y ·s – –+0 = where s FA / = y d ds ?–0 = s y () ? y · Cons t tan + = at y = 0 s 0 = so s y () ? y · = at y = 0 s P 0 A / = and s y () ? y · P 0 A / + = The Concept of Stress — An Introduction ENGINEERING MECHANICS FOR STRUCTURES 4.5 stipulated or assumed that the internal force, F, acting upon any y plane was uniformly distributed over that plane. This example is a special case in another way; not only is it one-dimensional in its dependence upon spatial position, but it is the simplest example of stress at a point in that it is described fully by a single com- ponent of stress, the normal stress acting on a plane perpendicular to the y axis. The ?gure below is meant to illustrate a more general, indeed, the most general state of stress at a point. It requires some explanation: The odd looking structural element, ?xed to the ground at bottom and to the left, and carrying what appears to be a uniformly distributed load over a portion of its bottom and a concentrated load on its top, is meant to symbolize an arbitrarily loaded, arbitrarily constrained, arbitrarily shaped solid continuum. It could be a beam, a truss, a thin-walled cylinder though it looks more like a potato — which too is a solid contin- uum. At any arbitrarily chosen point inside this object we can ask about the internal stress state. But what stress component? For there are more than one; in fact they come in sets of three. One set acts upon what we call an x plane, another upon a y plane, a third set upon a z plane. Which plane is which is de?ned by its normal:An x plane has its normal in the x direction, etc. Each set includes three scalar components, one normal stress component acting perpendicular to its reference plane, with its direction along one coordinate axis, and two shear stress components acting in plane in the direction of the other two coordinate axes. That’s a grand total of nine stress components to de?ne the stress at a point. To fully de?ne the stress ?eld throughout a continuum you need to specify how these nine scalar components vary from one point in the continuum to another. That’s a tall order. Fortunately, equilibrium requirements applied to a differential element of the continuum, what we will call a “micro-equilibrium” consideration, will reduce the number of independent stress components at a point from nine to six. We will ?nd that the shear stress component s xy acting on the x face must equal its neighbor around the corner s yx acting on the y face and that s zy = s yz and s xz = s zx accordingly. Fortunately too, in most of the engineering structures you will encounter, diagnose or design, only two or three of these now six components will matter, will be signi?cant. And, as in the example just treated of a bar suspended vertically and loaded by its own weight and/or by an end load, often variations of the stress components in one, or more, of the three coordinate directions may be uniform. But perhaps the most important simpli?cation is a simpli?cation in modeling, made at the outset of our encounter. One particu- larly useful model, applicable to many structural elements is called Plane Stress and, as you might infer from the label alone, it restricts our attention to variations of stress in two dimensions. x y z s x s z s xz s zy s zx s yx yx P s y s yz x y z x y z x y z s xy Page 4 ENGINEERING MECHANICS FOR STRUCTURES 4.3 CHAPTER 4 Stress, Strain, Stress-Strain 4.1 The Concept of Stress — An Introduction We have talked about internal forces, distributed them uniformly over an area and they became a normal stress acting perpendicular to some internal surface, or a shear stress acting tangentially, in plane. Up to now, we have said little about how these normal and shear stresses might vary with position throughout a solid. 1 Up to now, the choice of planes, their orientation within a solid, was dictated by the geometry of the solid and the nature of the loading. We have said nothing about how these stress components might change if we looked at a set of planes of another orientation. Now we consider a more general situation, an arbitrarily shaped solid. We are going to lift our gaze up from the world of crude structural elements such as truss bars in tension, shafts in torsion, or beams in bending to view these “solids” from a more abstract perspective. They all become special cases of more gen- eral stuff we call a solid continuum. We will address two questions: •How might stresses vary from one point to another throughout a continuum; •How do the normal and shear components of stress acting on a plane at a given point change as we change the orientation of the plane at the point. The ?rst bullet introduces the notion of stress ?eld; the second concerns the transformation of com- ponents of stress at a point. To begin with the ?rst bullet we re-examine the case of a bar suspended vertically and loaded by its own weight, a case considered in section 3.2, page 62. (Note, I have changed the orientation of the reference 1. The beam is the one exception. There we explored how different normal stress distributions over a rectangular cross-section could be equivalent to a bending moment and zero resultant force. The Concept of Stress — An Introduction 4.4 ENGINEERING MECHANICS FOR STRUCTURES axes). We will construct a differential equation which governs how the axial stress varies as we move up and down the bar. We will solve this differential equation, not forgetting to apply an appropriate boundary con- dition and determine the axial stress ?eld. We see that for equilibrium of the differential element of the bar, of planar cross-sectional area A and of weight density ?, we have If we assume the tensile force is uniformly distributed over the cross-sectional area, and dividing by the area (which does not change with the independent spatial coordinate y) we can write Chanting “...going to the limit, letting ? y go to zero”, we obtain a differential equation ?xing how s(y), a function of y, varies throughout our continuum, namely We solve this ordinary differential equation easily, integrating once and obtain The Constant is ?xed by a prescribed condition at some y surface; If the end of the bar is stress free, we indi- cate this writing If, on another occasion, a weight of magnitude P 0 is suspended from the free end, we would have Here then are two stress ?elds for two different loading conditions 1 . Each stress ?eld describes how the normal stress s(x,y,z) varies throughout the continuum at every point in the continuum. I show the stress as a function of x and z as well as y to emphasize that we can evaluate its value at every point in the continuum, although it only varies with y. That the stress does not vary with x and z was implied when we 1. A third loading condition is obtained by setting the weight density ? to zero; our bar then is assumed weightless relative to the end-load P 0. F(y) x y z y ? y F(y) + ? F ? w(y) = ? A? y ? y + ?s s(y) s(y) F ? F ? A ? y ·· F – –+0 = s?s ? ? y ·s – –+0 = where s FA / = y d ds ?–0 = s y () ? y · Cons t tan + = at y = 0 s 0 = so s y () ? y · = at y = 0 s P 0 A / = and s y () ? y · P 0 A / + = The Concept of Stress — An Introduction ENGINEERING MECHANICS FOR STRUCTURES 4.5 stipulated or assumed that the internal force, F, acting upon any y plane was uniformly distributed over that plane. This example is a special case in another way; not only is it one-dimensional in its dependence upon spatial position, but it is the simplest example of stress at a point in that it is described fully by a single com- ponent of stress, the normal stress acting on a plane perpendicular to the y axis. The ?gure below is meant to illustrate a more general, indeed, the most general state of stress at a point. It requires some explanation: The odd looking structural element, ?xed to the ground at bottom and to the left, and carrying what appears to be a uniformly distributed load over a portion of its bottom and a concentrated load on its top, is meant to symbolize an arbitrarily loaded, arbitrarily constrained, arbitrarily shaped solid continuum. It could be a beam, a truss, a thin-walled cylinder though it looks more like a potato — which too is a solid contin- uum. At any arbitrarily chosen point inside this object we can ask about the internal stress state. But what stress component? For there are more than one; in fact they come in sets of three. One set acts upon what we call an x plane, another upon a y plane, a third set upon a z plane. Which plane is which is de?ned by its normal:An x plane has its normal in the x direction, etc. Each set includes three scalar components, one normal stress component acting perpendicular to its reference plane, with its direction along one coordinate axis, and two shear stress components acting in plane in the direction of the other two coordinate axes. That’s a grand total of nine stress components to de?ne the stress at a point. To fully de?ne the stress ?eld throughout a continuum you need to specify how these nine scalar components vary from one point in the continuum to another. That’s a tall order. Fortunately, equilibrium requirements applied to a differential element of the continuum, what we will call a “micro-equilibrium” consideration, will reduce the number of independent stress components at a point from nine to six. We will ?nd that the shear stress component s xy acting on the x face must equal its neighbor around the corner s yx acting on the y face and that s zy = s yz and s xz = s zx accordingly. Fortunately too, in most of the engineering structures you will encounter, diagnose or design, only two or three of these now six components will matter, will be signi?cant. And, as in the example just treated of a bar suspended vertically and loaded by its own weight and/or by an end load, often variations of the stress components in one, or more, of the three coordinate directions may be uniform. But perhaps the most important simpli?cation is a simpli?cation in modeling, made at the outset of our encounter. One particu- larly useful model, applicable to many structural elements is called Plane Stress and, as you might infer from the label alone, it restricts our attention to variations of stress in two dimensions. x y z s x s z s xz s zy s zx s yx yx P s y s yz x y z x y z x y z s xy Plane Stress 4.6 ENGINEERING MECHANICS FOR STRUCTURES 4.2 Plane Stress If we assume our continuum has the form of a thin plate of uniform thickness but of arbi- trary closed contour in the x-y plane, our previous arbitrarily loaded, arbitrarily con- strained continuum (we don’t show these again) takes the planar form below. Because the plate is thin in the z direction, (h/L <<1)we will assume that variations of the stress components with z is uniform or, in other words, our stress components will be at most functions of x and y, s(x,y). We also take it that the z boundary planes are unloaded, stressfree. These two together imply that the set of three “z” stress components that act upon any arbitrarily located z plane within the interior must also vanish. We will also take advantage of the micro-equilibrium consequences, yet to be explored but noted previously, and set s yz and s xz to zero. Our state of stress at a point is then as it is shown on the exploded view of the point - the block in the middle of the ?gure - and again from the point of view of look- ing normal to a z plane at the far right. This special model is called Plane Stress. A Word about Sign Convention: The ?gure at the far right seems to include more stress compo- nents than necessary; after all, if, in modeling, we eliminate the stress components acting on a z face and s yz and s xz as well, that should leave, at most, four components acting on the x and y faces. Yet there appear to be eight in the ?gure. No, there are only at most four components; we must learn to read the ?gure. To do so, we make use of another sketch of stress at a point, the point A. The ?gure at the top is meant to indicated that we are looking at four faces or planes simultaneously. When we look at the x face from the right we are looking at the stress components on a positive x face — it has its outward normal in the positive x direction — and a positive normal stress, by con- vention, is directed in the positive x direction.A positive shear stress component, acting in plane, also acts, by convention, in a positive coordinate direction - in this case the positive y direc- tion. On the positive y face, we follow the same convention; a positive s y acts on a positive y face in the positive y coordinate direction; a positive s yx acts on a positive y face in the positive x coordi- nate direction. h A point x y s x s xy s y s yx z no stresses here on this z face A point s x s x s y s y s xy s yx s xy s yx L A point s x s x s y s y s xy s yx s xy s yx s x s xy A A A A s x s xy s y s yx s y s yx A x y Page 5 ENGINEERING MECHANICS FOR STRUCTURES 4.3 CHAPTER 4 Stress, Strain, Stress-Strain 4.1 The Concept of Stress — An Introduction We have talked about internal forces, distributed them uniformly over an area and they became a normal stress acting perpendicular to some internal surface, or a shear stress acting tangentially, in plane. Up to now, we have said little about how these normal and shear stresses might vary with position throughout a solid. 1 Up to now, the choice of planes, their orientation within a solid, was dictated by the geometry of the solid and the nature of the loading. We have said nothing about how these stress components might change if we looked at a set of planes of another orientation. Now we consider a more general situation, an arbitrarily shaped solid. We are going to lift our gaze up from the world of crude structural elements such as truss bars in tension, shafts in torsion, or beams in bending to view these “solids” from a more abstract perspective. They all become special cases of more gen- eral stuff we call a solid continuum. We will address two questions: •How might stresses vary from one point to another throughout a continuum; •How do the normal and shear components of stress acting on a plane at a given point change as we change the orientation of the plane at the point. The ?rst bullet introduces the notion of stress ?eld; the second concerns the transformation of com- ponents of stress at a point. To begin with the ?rst bullet we re-examine the case of a bar suspended vertically and loaded by its own weight, a case considered in section 3.2, page 62. (Note, I have changed the orientation of the reference 1. The beam is the one exception. There we explored how different normal stress distributions over a rectangular cross-section could be equivalent to a bending moment and zero resultant force. The Concept of Stress — An Introduction 4.4 ENGINEERING MECHANICS FOR STRUCTURES axes). We will construct a differential equation which governs how the axial stress varies as we move up and down the bar. We will solve this differential equation, not forgetting to apply an appropriate boundary con- dition and determine the axial stress ?eld. We see that for equilibrium of the differential element of the bar, of planar cross-sectional area A and of weight density ?, we have If we assume the tensile force is uniformly distributed over the cross-sectional area, and dividing by the area (which does not change with the independent spatial coordinate y) we can write Chanting “...going to the limit, letting ? y go to zero”, we obtain a differential equation ?xing how s(y), a function of y, varies throughout our continuum, namely We solve this ordinary differential equation easily, integrating once and obtain The Constant is ?xed by a prescribed condition at some y surface; If the end of the bar is stress free, we indi- cate this writing If, on another occasion, a weight of magnitude P 0 is suspended from the free end, we would have Here then are two stress ?elds for two different loading conditions 1 . Each stress ?eld describes how the normal stress s(x,y,z) varies throughout the continuum at every point in the continuum. I show the stress as a function of x and z as well as y to emphasize that we can evaluate its value at every point in the continuum, although it only varies with y. That the stress does not vary with x and z was implied when we 1. A third loading condition is obtained by setting the weight density ? to zero; our bar then is assumed weightless relative to the end-load P 0. F(y) x y z y ? y F(y) + ? F ? w(y) = ? A? y ? y + ?s s(y) s(y) F ? F ? A ? y ·· F – –+0 = s?s ? ? y ·s – –+0 = where s FA / = y d ds ?–0 = s y () ? y · Cons t tan + = at y = 0 s 0 = so s y () ? y · = at y = 0 s P 0 A / = and s y () ? y · P 0 A / + = The Concept of Stress — An Introduction ENGINEERING MECHANICS FOR STRUCTURES 4.5 stipulated or assumed that the internal force, F, acting upon any y plane was uniformly distributed over that plane. This example is a special case in another way; not only is it one-dimensional in its dependence upon spatial position, but it is the simplest example of stress at a point in that it is described fully by a single com- ponent of stress, the normal stress acting on a plane perpendicular to the y axis. The ?gure below is meant to illustrate a more general, indeed, the most general state of stress at a point. It requires some explanation: The odd looking structural element, ?xed to the ground at bottom and to the left, and carrying what appears to be a uniformly distributed load over a portion of its bottom and a concentrated load on its top, is meant to symbolize an arbitrarily loaded, arbitrarily constrained, arbitrarily shaped solid continuum. It could be a beam, a truss, a thin-walled cylinder though it looks more like a potato — which too is a solid contin- uum. At any arbitrarily chosen point inside this object we can ask about the internal stress state. But what stress component? For there are more than one; in fact they come in sets of three. One set acts upon what we call an x plane, another upon a y plane, a third set upon a z plane. Which plane is which is de?ned by its normal:An x plane has its normal in the x direction, etc. Each set includes three scalar components, one normal stress component acting perpendicular to its reference plane, with its direction along one coordinate axis, and two shear stress components acting in plane in the direction of the other two coordinate axes. That’s a grand total of nine stress components to de?ne the stress at a point. To fully de?ne the stress ?eld throughout a continuum you need to specify how these nine scalar components vary from one point in the continuum to another. That’s a tall order. Fortunately, equilibrium requirements applied to a differential element of the continuum, what we will call a “micro-equilibrium” consideration, will reduce the number of independent stress components at a point from nine to six. We will ?nd that the shear stress component s xy acting on the x face must equal its neighbor around the corner s yx acting on the y face and that s zy = s yz and s xz = s zx accordingly. Fortunately too, in most of the engineering structures you will encounter, diagnose or design, only two or three of these now six components will matter, will be signi?cant. And, as in the example just treated of a bar suspended vertically and loaded by its own weight and/or by an end load, often variations of the stress components in one, or more, of the three coordinate directions may be uniform. But perhaps the most important simpli?cation is a simpli?cation in modeling, made at the outset of our encounter. One particu- larly useful model, applicable to many structural elements is called Plane Stress and, as you might infer from the label alone, it restricts our attention to variations of stress in two dimensions. x y z s x s z s xz s zy s zx s yx yx P s y s yz x y z x y z x y z s xy Plane Stress 4.6 ENGINEERING MECHANICS FOR STRUCTURES 4.2 Plane Stress If we assume our continuum has the form of a thin plate of uniform thickness but of arbi- trary closed contour in the x-y plane, our previous arbitrarily loaded, arbitrarily con- strained continuum (we don’t show these again) takes the planar form below. Because the plate is thin in the z direction, (h/L <<1)we will assume that variations of the stress components with z is uniform or, in other words, our stress components will be at most functions of x and y, s(x,y). We also take it that the z boundary planes are unloaded, stressfree. These two together imply that the set of three “z” stress components that act upon any arbitrarily located z plane within the interior must also vanish. We will also take advantage of the micro-equilibrium consequences, yet to be explored but noted previously, and set s yz and s xz to zero. Our state of stress at a point is then as it is shown on the exploded view of the point - the block in the middle of the ?gure - and again from the point of view of look- ing normal to a z plane at the far right. This special model is called Plane Stress. A Word about Sign Convention: The ?gure at the far right seems to include more stress compo- nents than necessary; after all, if, in modeling, we eliminate the stress components acting on a z face and s yz and s xz as well, that should leave, at most, four components acting on the x and y faces. Yet there appear to be eight in the ?gure. No, there are only at most four components; we must learn to read the ?gure. To do so, we make use of another sketch of stress at a point, the point A. The ?gure at the top is meant to indicated that we are looking at four faces or planes simultaneously. When we look at the x face from the right we are looking at the stress components on a positive x face — it has its outward normal in the positive x direction — and a positive normal stress, by con- vention, is directed in the positive x direction.A positive shear stress component, acting in plane, also acts, by convention, in a positive coordinate direction - in this case the positive y direc- tion. On the positive y face, we follow the same convention; a positive s y acts on a positive y face in the positive y coordinate direction; a positive s yx acts on a positive y face in the positive x coordi- nate direction. h A point x y s x s xy s y s yx z no stresses here on this z face A point s x s x s y s y s xy s yx s xy s yx L A point s x s x s y s y s xy s yx s xy s yx s x s xy A A A A s x s xy s y s yx s y s yx A x y Plane Stress ENGINEERING MECHANICS FOR STRUCTURES 4.7 We emphasize that we are looking at a point, point A, in these ?gures. More precisely we are look- ing at two mutually perpendicular planes intersecting at the point and from two vantage points in each case. We draw these two views of the two planes as four planes in order to more clearly illustrate our sign conven- tion. But you ought to imagine the square having zero height and width: the s x acting to the left, in the nega- tive x direction, upon the negative x face at the left, with its outward normal pointing in the negative x direction is a positive component at the point, the equal and opposite reaction to the s x acting to the right, in the positive x direction, upon the positive x face at the right, with its outward normal pointing in the posi- tive x direction. Both are positive as shown; both are the same quantity. So too the shear stress component s xy shown acting down, in the negative y direction, on the negative x face is the equal and opposite internal reaction to s xy shown acting up, in the positive y direction, on the positive x face A general statement of our sign convention, which holds for all nine components of stress, even in 3D, is as follows: A positive component of stress acts on a positive face in a positive coordinate direc- tion or on a negative face in a negative coordinate direction. *** An Example: We might model the end-loaded cantilever with rel- atively thin rectangular cross-section as a plane stress prob- lem. In this, b is the “thin” dimension, i.e., b/L <1. If we assume a normal stress distribution over an x face is proportional to some odd power of y, as we did in sec- tion 3.2, exercise 3.5, our state of stress at a point might look like that shown in ?gure (c). In this, s x would have the form where C(n,b,h) is a constant which depends upon the cross- sectional dimensions of the beam and the odd exponent n. (See page 57). The factor W(L-x) is the magnitude of the internal bending moment at the location x measured from the root. See ?gure (b). But this is only one component of our stress ?eld. What are the other eight components of stress at point A? Our plane stress model allows us to claim that the three z face components are zero and if we take s yz and s xz to be zero, that still leaves s xy , s yx , and s y in addition to s x . Now, in fact, we are doomed from the start; we know that the problem is statically indeterminate so we are not going to be able to construct a unique solution to the equilibrium requirements and specify all nine components of our stress ?eld. Still we can experiment taking advantage of the indeterminacy at our disposal. For example, we know that a shear force of magni- tude W acts at any x section. In this case it does not vary with x. We might assume, then, that the shear force is uniformly distributed over the cross-section and set Our stress at a point at point A would then look like ?gure (d). h W L x y z (a) point A s x (x,y) point A b point A s x s x s xy s xy (c) s x (x,y) W W(L-x) x W point A (b) x y (d) s x xy , () C nbh ,, () WL x – ()y n · = s xy W –bh () / =Read More

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