Page 1 1 Chapter 1 Tension and Compression in Bars Page 2 1 Chapter 1 Tension and Compression in Bars 1 Tension and Compression in Bars 1.1 Stress.............................................................. 7 1.2 Strain.............................................................. 13 1.3 Constitutive Law................................................ 14 1.4 Single Bar under Tension or Compression................. 18 1.5 Statically Determinate Systems of Bars ................... 29 1.6 Statically Indeterminate Systems of Bars ................. 33 1.7 Supplementary Examples...................................... 40 1.8 Summary ......................................................... 46 Objectives: In this textbook about the Mechanics of Ma- terials we investigate the stressing and the deformations of elastic structures subject to applied loads. In the ?rst chapter we will re- strict ourselves to the simplest structural members, namely, bars under tension or compression. In order to treat such problems, we need kinematic relations and a constitutive law to complement the equilibrium conditions which are known from Volume 1. The kinematic relations represent the geometry of the deformation, whereas the behaviour of the elastic material is described by the constitutive law. The students will learn how to apply these equations and how to solve statically determinate as well as statically indeterminate problems. D. Gross et al., Engineering Mechanics 2, DOI 10.1007/978-3-642-12886-8_1, © Springer-Verlag Berlin Heidelberg 2011 Page 3 1 Chapter 1 Tension and Compression in Bars 1 Tension and Compression in Bars 1.1 Stress.............................................................. 7 1.2 Strain.............................................................. 13 1.3 Constitutive Law................................................ 14 1.4 Single Bar under Tension or Compression................. 18 1.5 Statically Determinate Systems of Bars ................... 29 1.6 Statically Indeterminate Systems of Bars ................. 33 1.7 Supplementary Examples...................................... 40 1.8 Summary ......................................................... 46 Objectives: In this textbook about the Mechanics of Ma- terials we investigate the stressing and the deformations of elastic structures subject to applied loads. In the ?rst chapter we will re- strict ourselves to the simplest structural members, namely, bars under tension or compression. In order to treat such problems, we need kinematic relations and a constitutive law to complement the equilibrium conditions which are known from Volume 1. The kinematic relations represent the geometry of the deformation, whereas the behaviour of the elastic material is described by the constitutive law. The students will learn how to apply these equations and how to solve statically determinate as well as statically indeterminate problems. D. Gross et al., Engineering Mechanics 2, DOI 10.1007/978-3-642-12886-8_1, © Springer-Verlag Berlin Heidelberg 2011 1.1 Stress 7 1.1 1.1 Stress Let us consider a straight bar with a constant cross-sectional area A. The line connecting the centroids of the cross sections is called the axis of the bar. The ends of the bar are subjected to the forces F whose common line of action is the axis (Fig. 1.1a). The external load causes internal forces. The internal forces can be visualized by an imaginary cut of the bar (compare Volu- me 1, Section 1.4). They are distributed over the cross section (see Fig. 1.1b) and are called stresses. Being area forces, they have the dimension force per area and are measured, for example, as mul- tiples of the unit MPa (1 MPa = 1 N/mm 2 ). The unit “Pascal” (1 Pa = 1 N/m 2 ) is named after the mathematician and physicist Blaise Pascal (1623–1662); the notion of “stress” was introduced by Augustin Louis Cauchy (1789–1857). In Volume 1 (Statics) we only dealt with the resultant of the internal forces (= normal for- ce) whereas now we have to study the internal forces (= stresses). b a d e c ? t s c ? F F F F F s F F c F c FF t s A c A * = A cos ? N Fig. 1.1 In order to determine the stresses we ?rst choose an imaginary cut c- c perpendicular to the axis of the bar. The stresses are shown in the free-body diagram (Fig. 1.1b); they are denoted by s. We assume that they act perpendicularly to the exposed surface A of the cross section and that they are uniformly distributed. Since they are normal to the cross section they are called normal stresses. Their resultant is the normal force N shown in Fig. 1.1c (compare Volume 1, Section 7.1). Therefore we have N = sA and the stresses s can be calculated from the normal force N: Page 4 1 Chapter 1 Tension and Compression in Bars 1 Tension and Compression in Bars 1.1 Stress.............................................................. 7 1.2 Strain.............................................................. 13 1.3 Constitutive Law................................................ 14 1.4 Single Bar under Tension or Compression................. 18 1.5 Statically Determinate Systems of Bars ................... 29 1.6 Statically Indeterminate Systems of Bars ................. 33 1.7 Supplementary Examples...................................... 40 1.8 Summary ......................................................... 46 Objectives: In this textbook about the Mechanics of Ma- terials we investigate the stressing and the deformations of elastic structures subject to applied loads. In the ?rst chapter we will re- strict ourselves to the simplest structural members, namely, bars under tension or compression. In order to treat such problems, we need kinematic relations and a constitutive law to complement the equilibrium conditions which are known from Volume 1. The kinematic relations represent the geometry of the deformation, whereas the behaviour of the elastic material is described by the constitutive law. The students will learn how to apply these equations and how to solve statically determinate as well as statically indeterminate problems. D. Gross et al., Engineering Mechanics 2, DOI 10.1007/978-3-642-12886-8_1, © Springer-Verlag Berlin Heidelberg 2011 1.1 Stress 7 1.1 1.1 Stress Let us consider a straight bar with a constant cross-sectional area A. The line connecting the centroids of the cross sections is called the axis of the bar. The ends of the bar are subjected to the forces F whose common line of action is the axis (Fig. 1.1a). The external load causes internal forces. The internal forces can be visualized by an imaginary cut of the bar (compare Volu- me 1, Section 1.4). They are distributed over the cross section (see Fig. 1.1b) and are called stresses. Being area forces, they have the dimension force per area and are measured, for example, as mul- tiples of the unit MPa (1 MPa = 1 N/mm 2 ). The unit “Pascal” (1 Pa = 1 N/m 2 ) is named after the mathematician and physicist Blaise Pascal (1623–1662); the notion of “stress” was introduced by Augustin Louis Cauchy (1789–1857). In Volume 1 (Statics) we only dealt with the resultant of the internal forces (= normal for- ce) whereas now we have to study the internal forces (= stresses). b a d e c ? t s c ? F F F F F s F F c F c FF t s A c A * = A cos ? N Fig. 1.1 In order to determine the stresses we ?rst choose an imaginary cut c- c perpendicular to the axis of the bar. The stresses are shown in the free-body diagram (Fig. 1.1b); they are denoted by s. We assume that they act perpendicularly to the exposed surface A of the cross section and that they are uniformly distributed. Since they are normal to the cross section they are called normal stresses. Their resultant is the normal force N shown in Fig. 1.1c (compare Volume 1, Section 7.1). Therefore we have N = sA and the stresses s can be calculated from the normal force N: 8 1 Tension and Compression in Bars s = N A . (1.1) In the present example the normal force N is equal to the applied force F. Thus, we obtain from (1.1) s = F A . (1.2) In the case of a positive normal force N (tension) the stress s is then positive (tensile stress). Reversely, if the normal force is negative (compression) the stress is also negative (compressive stress). Let us now imagine the bar being sectioned by a cut which is not orthogonal to the axis of the bar so that its direction is given by the angle ? (Fig. 1.1d). The internal forces now act on the exposed surface A * = A/ cos?. Again we assume that they are uniformly distributed. We resolve the stresses into a component s perpendicular to the surface (the normal stress) and a component t tangential to the surface (Fig. 1.1e). The component t which acts in the direction of the surface is called shear stress. Equilibrium of the forces acting on the left portion of the bar yields (see Fig. 1.1e) ? : sA * cos? + tA * sin?- F =0 , ? : sA * sin ?- tA * cos?=0 . Note that we have to write down the equilibrium conditions for the forces, not for the stresses.With A * = A/ cos? we obtain s + t tan ? = F A ,s tan ?- t =0 . Solving these two equations for s and t yields s = 1 1+tan 2 ? F A ,t = tan ? 1+tan 2 ? F A . Page 5 1 Chapter 1 Tension and Compression in Bars 1 Tension and Compression in Bars 1.1 Stress.............................................................. 7 1.2 Strain.............................................................. 13 1.3 Constitutive Law................................................ 14 1.4 Single Bar under Tension or Compression................. 18 1.5 Statically Determinate Systems of Bars ................... 29 1.6 Statically Indeterminate Systems of Bars ................. 33 1.7 Supplementary Examples...................................... 40 1.8 Summary ......................................................... 46 Objectives: In this textbook about the Mechanics of Ma- terials we investigate the stressing and the deformations of elastic structures subject to applied loads. In the ?rst chapter we will re- strict ourselves to the simplest structural members, namely, bars under tension or compression. In order to treat such problems, we need kinematic relations and a constitutive law to complement the equilibrium conditions which are known from Volume 1. The kinematic relations represent the geometry of the deformation, whereas the behaviour of the elastic material is described by the constitutive law. The students will learn how to apply these equations and how to solve statically determinate as well as statically indeterminate problems. D. Gross et al., Engineering Mechanics 2, DOI 10.1007/978-3-642-12886-8_1, © Springer-Verlag Berlin Heidelberg 2011 1.1 Stress 7 1.1 1.1 Stress Let us consider a straight bar with a constant cross-sectional area A. The line connecting the centroids of the cross sections is called the axis of the bar. The ends of the bar are subjected to the forces F whose common line of action is the axis (Fig. 1.1a). The external load causes internal forces. The internal forces can be visualized by an imaginary cut of the bar (compare Volu- me 1, Section 1.4). They are distributed over the cross section (see Fig. 1.1b) and are called stresses. Being area forces, they have the dimension force per area and are measured, for example, as mul- tiples of the unit MPa (1 MPa = 1 N/mm 2 ). The unit “Pascal” (1 Pa = 1 N/m 2 ) is named after the mathematician and physicist Blaise Pascal (1623–1662); the notion of “stress” was introduced by Augustin Louis Cauchy (1789–1857). In Volume 1 (Statics) we only dealt with the resultant of the internal forces (= normal for- ce) whereas now we have to study the internal forces (= stresses). b a d e c ? t s c ? F F F F F s F F c F c FF t s A c A * = A cos ? N Fig. 1.1 In order to determine the stresses we ?rst choose an imaginary cut c- c perpendicular to the axis of the bar. The stresses are shown in the free-body diagram (Fig. 1.1b); they are denoted by s. We assume that they act perpendicularly to the exposed surface A of the cross section and that they are uniformly distributed. Since they are normal to the cross section they are called normal stresses. Their resultant is the normal force N shown in Fig. 1.1c (compare Volume 1, Section 7.1). Therefore we have N = sA and the stresses s can be calculated from the normal force N: 8 1 Tension and Compression in Bars s = N A . (1.1) In the present example the normal force N is equal to the applied force F. Thus, we obtain from (1.1) s = F A . (1.2) In the case of a positive normal force N (tension) the stress s is then positive (tensile stress). Reversely, if the normal force is negative (compression) the stress is also negative (compressive stress). Let us now imagine the bar being sectioned by a cut which is not orthogonal to the axis of the bar so that its direction is given by the angle ? (Fig. 1.1d). The internal forces now act on the exposed surface A * = A/ cos?. Again we assume that they are uniformly distributed. We resolve the stresses into a component s perpendicular to the surface (the normal stress) and a component t tangential to the surface (Fig. 1.1e). The component t which acts in the direction of the surface is called shear stress. Equilibrium of the forces acting on the left portion of the bar yields (see Fig. 1.1e) ? : sA * cos? + tA * sin?- F =0 , ? : sA * sin ?- tA * cos?=0 . Note that we have to write down the equilibrium conditions for the forces, not for the stresses.With A * = A/ cos? we obtain s + t tan ? = F A ,s tan ?- t =0 . Solving these two equations for s and t yields s = 1 1+tan 2 ? F A ,t = tan ? 1+tan 2 ? F A . 1.1 Stress 9 It is practical to write these equations in a di?erent form. Using the standard trigonometric relations 1 1+tan 2 ? =cos 2 ?, cos 2 ? = 1 2 (1 + cos2 ?), sin ?cos? = 1 2 sin 2 ? and the abbreviation s 0 = F/A (= normal stress in a section perpendicular to the axis) we ?nally get s = s 0 2 (1 + cos2 ?),t = s 0 2 sin 2?. (1.3) Thus, the stresses depend on the direction of the cut. If s 0 is known, the stresses s and t can be calculated from (1.3) for arbi- trary values of the angle ?. The maximum value of s is obtained for ? =0, in which case s max = s 0 ; the maximum value of t is found for ? = p/4forwhich t max = s 0 /2. If we section a bar near an end which is subjected to a concen- trated force F (Fig. 1.2a, section c- c) we ?nd that the normal stress is not distributed uniformly over the cross-sectional area. The concentrated force produces high stresses near its point of application (Fig. 1.2b). This phenomenon is known as stress con- centration. It can be shown, however, that the stress concentration is restricted to sections in the proximity of the point of application of the concentrated force: the high stresses decay rapidly towards the average value s 0 as we increase the distance from the end of the bar. This fact is referred to as Saint-Venant’s principle (Adh´ emar Jean Claude Barr´ e de Saint-Venant, 1797–1886). Fig. 1.2 b a c c c c c F F F F F F s sRead More

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