SECTION 1.2 Normal Stress and Strain 11 In practice,the original units of 6 and L are sometimes attached to the strain itself,and then the strain is recorded in forms such as mm/m, um/m,and in./in.For instance,the strain e in the preceding illustration could be given as 700 um/m or 700X10 in./in.Also,strain is some- times expressed as a percent,especially when the strains are large.(In the preceding example,the strain is 0.07%.) Uniaxial Stress and Strain The definitions of normal stress and normal strain are based upon purely static and geometric considerations,which means that Egs.(1-1)and (1-2)can be used for loads of any magnitude and for any material.The principal requirement is that the deformation of the bar be uniform throughout its volume,which in turn requires that the bar be prismatic, the loads act through the centroids of the cross sections,and the material be homogeneous(that is,the same throughout all parts of the bar).The resulting state of stress and strain is called uniaxial stress and strain. Further discussions of uniaxial stress,including stresses in direc- tions other than the longitudinal direction of the bar,are given later in Section 2.6.We will also analyze more complicated stress states,such as biaxial stress and plane stress,in Chapter 7. Line of Action of the Axial Forces for a Uniform Stress Distribution Throughout the preceding discussion of stress and strain in a prismatic bar,we assumed that the normal stress o was distributed uniformly over the cross section.Now we will demonstrate that this condition is met if the line of action of the axial forces is through the centroid of the cross- sectional area. Consider a prismatic bar of arbitrary cross-sectional shape subjected to axial forces P that produce uniformly distributed stresses o(Fig.1-4a). Also,let pi represent the point in the cross section where the line of action of the forces intersects the cross section(Fig.1-4b).We construct a set of xy axes in the plane of the cross section and denote the coordi- nates of point pi by x and y.To determine these coordinates,we observe that the moments M and M of the force P about the x and y axes, respectively,must be equal to the corresponding moments of the uniformly distributed stresses. The moments of the force P are Mx=PM,=-P阮 (a,b) in which a moment is considered positive when its vector (using the right-hand rule)acts in the positive direction of the corresponding axis. To visualize the right-hand rule,imagine that you grasp an axis of coordinates with your right hand so that your fingers fold around the axis and your thumb points in the positive direction of the axis.Then a moment is positive if it acts about the axis in the same direc- tion as your fingers
SECTION 1.2 Normal Stress and Strain 11 In practice, the original units of d and L are sometimes attached to the strain itself, and then the strain is recorded in forms such as mm/m, 'm/m, and in./in. For instance, the strain e in the preceding illustration could be given as 700 'm/m or 700106 in./in. Also, strain is sometimes expressed as a percent, especially when the strains are large. (In the preceding example, the strain is 0.07%.) Uniaxial Stress and Strain The definitions of normal stress and normal strain are based upon purely static and geometric considerations, which means that Eqs. (1-1) and (1-2) can be used for loads of any magnitude and for any material. The principal requirement is that the deformation of the bar be uniform throughout its volume, which in turn requires that the bar be prismatic, the loads act through the centroids of the cross sections, and the material be homogeneous (that is, the same throughout all parts of the bar). The resulting state of stress and strain is called uniaxial stress and strain. Further discussions of uniaxial stress, including stresses in directions other than the longitudinal direction of the bar, are given later in Section 2.6. We will also analyze more complicated stress states, such as biaxial stress and plane stress, in Chapter 7. Line of Action of the Axial Forces for a Uniform Stress Distribution Throughout the preceding discussion of stress and strain in a prismatic bar, we assumed that the normal stress s was distributed uniformly over the cross section. Now we will demonstrate that this condition is met if the line of action of the axial forces is through the centroid of the crosssectional area. Consider a prismatic bar of arbitrary cross-sectional shape subjected to axial forces P that produce uniformly distributed stresses s (Fig. 1-4a). Also, let p1 represent the point in the cross section where the line of action of the forces intersects the cross section (Fig. 1-4b). We construct a set of xy axes in the plane of the cross section and denote the coordinates of point p1 by x and y. To determine these coordinates, we observe that the moments Mx and My of the force P about the x and y axes, respectively, must be equal to the corresponding moments of the uniformly distributed stresses. The moments of the force P are Mx Py My Px (a,b) in which a moment is considered positive when its vector (using the right-hand rule) acts in the positive direction of the corresponding axis.* * To visualize the right-hand rule, imagine that you grasp an axis of coordinates with your right hand so that your fingers fold around the axis and your thumb points in the positive direction of the axis. Then a moment is positive if it acts about the axis in the same direction as your fingers.
12 CHAPTER 1 Tension,Compression,and Shear dA (a) FIG.1-4 Uniform stress distribution in 0 a prismatic bar:(a)axial forces P. and (b)cross section of the bar (b) The moments of the distributed stresses are obtained by integrating over the cross-sectional area A.The differential force acting on an element of area dA (Fig.1-4b)is equal to odA.The moments of this elemental force about the x and y axes are oydA and -oxdA,respectively, in which x and y denote the coordinates of the element dA.The total moments are obtained by integrating over the cross-sectional area: M,=oydA My =-oxdA (c,d) These expressions give the moments produced by the stresses o. Next,we equate the moments M and My as obtained from the force P (Egs.a and b)to the moments obtained from the distributed stresses(Egs.c and d): Py=aydA P=∫axdA Because the stresses o are uniformly distributed,we know that they are constant over the cross-sectional area A and can be placed outside the integral signs.Also,we know that o is equal to P/A.Therefore,we obtain the following formulas for the coordinates of point p: y dA x dA 少= A (1-3a,b) A These equations are the same as the equations defining the coordinates of the centroid of an area (see Egs.12-3a and b in Chapter 12).There- fore,we have now arrived at an important conclusion:In order to have uniform tension or compression in a prismatic bar,the axial force must act through the centroid of the cross-sectional area.As explained previ- ously,we always assume that these conditions are met unless it is specifically stated otherwise. The following examples illustrate the calculation of stresses and strains in prismatic bars.In the first example we disregard the weight of the bar and in the second we include it.(It is customary when solving textbook problems to omit the weight of the structure unless specifically instructed to include it.)
12 CHAPTER 1 Tension, Compression, and Shear The moments of the distributed stresses are obtained by integrating over the cross-sectional area A. The differential force acting on an element of area dA (Fig. 1-4b) is equal to sdA. The moments of this elemental force about the x and y axes are sydA and sxdA, respectively, in which x and y denote the coordinates of the element dA. The total moments are obtained by integrating over the cross-sectional area: Mx sydA My sxdA (c,d) These expressions give the moments produced by the stresses s. Next, we equate the moments Mx and My as obtained from the force P (Eqs. a and b) to the moments obtained from the distributed stresses (Eqs. c and d): Py sydA Px sx dA Because the stresses s are uniformly distributed, we know that they are constant over the cross-sectional area A and can be placed outside the integral signs. Also, we know that s is equal to P/A. Therefore, we obtain the following formulas for the coordinates of point p1: y x (1-3a,b) These equations are the same as the equations defining the coordinates of the centroid of an area (see Eqs. 12-3a and b in Chapter 12). Therefore, we have now arrived at an important conclusion: In order to have uniform tension or compression in a prismatic bar, the axial force must act through the centroid of the cross-sectional area. As explained previously, we always assume that these conditions are met unless it is specifically stated otherwise. The following examples illustrate the calculation of stresses and strains in prismatic bars. In the first example we disregard the weight of the bar and in the second we include it. (It is customary when solving textbook problems to omit the weight of the structure unless specifically instructed to include it.) x dA A y dA A P P (a) P A s = (b) O y y x A p1 dA x x – y – FIG. 1-4 Uniform stress distribution in a prismatic bar: (a) axial forces P, and (b) cross section of the bar
SECTION 1.2 Normal Stress and Strain 13 Example 1-1 A short post constructed from a hollow circular tube of aluminum supports a compressive load of 26 kips(Fig.1-5).The inner and outer diameters of the tube are d=4.0 in.and d2=4.5 in.,respectively,and its length is 16 in.The shortening of the post due to the load is measured as 0.012 in. Determine the compressive stress and strain in the post.(Disregard the weight of the post itself,and assume that the post does not buckle under the load.) 26k 16in. FIG.1-5 Example 1-1.Hollow aluminum post in compression Solution Assuming that the compressive load acts at the center of the hollow tube, we can use the equation o P/A(Eq.1-1)to calculate the normal stress.The force P equals 26 k (or 26,000 Ib),and the cross-sectional area A is A=-平d3-d=年45inP-40n月=3.38im2 Therefore,the compressive stress in the post is P26,0001b A3338in2=7790psi The compressive strain (from Eq.1-2)is e=9=0,012in=750x10-6 L 16 in. Thus,the stress and strain in the post have been calculated. Note:As explained earlier,strain is a dimensionless quantity and no units are needed.For clarity,however,units are often given.In this example,e could be written as 750 X 10 in./in.or 750 uin./in
A short post constructed from a hollow circular tube of aluminum supports a compressive load of 26 kips (Fig. 1-5). The inner and outer diameters of the tube are d1 4.0 in. and d2 4.5 in., respectively, and its length is 16 in. The shortening of the post due to the load is measured as 0.012 in. Determine the compressive stress and strain in the post. (Disregard the weight of the post itself, and assume that the post does not buckle under the load.) Example 1-1 SECTION 1.2 Normal Stress and Strain 13 16 in. 26 k FIG. 1-5 Example 1-1. Hollow aluminum post in compression Solution Assuming that the compressive load acts at the center of the hollow tube, we can use the equation s 5 P/A (Eq. 1-1) to calculate the normal stress. The force P equals 26 k (or 26,000 lb), and the cross-sectional area A is A p 4 d 2 2 d 2 1 p 4 (4.5 in.)2 (4.0 in.)2 3.338 in.2 Therefore, the compressive stress in the post is s P A 3 2 . 6 3 , 3 0 8 00 in l . b 2 7790 psi The compressive strain (from Eq. 1-2) is e L d 0. 1 0 6 12 in i . n. 750 106 Thus, the stress and strain in the post have been calculated. Note: As explained earlier, strain is a dimensionless quantity and no units are needed. For clarity, however, units are often given. In this example, e could be written as 750 1026 in./in. or 750 'in./in.
14 CHAPTER 1 Tension,Compression,and Shear Example 1-2 A circular steel rod of length L and diameter d hangs in a mine shaft and holds an ore bucket of weight W at its lower end (Fig.1-6). (a)Obtain a formula for the maximum stress omax in the rod,taking into account the weight of the rod itself. (b)Calculate the maximum stress if L=40 m,d=8 mm,and W=1.5 kN. FIG.1-6 Example 1-2.Steel rod supporting a weight W Solution (a)The maximum axial force Fx in the rod occurs at the upper end and is equal to the weight W of the ore bucket plus the weight Wo of the rod itself.The latter is equal to the weight density y of the steel times the volume V of the rod, or Wo=YV YAL (1-4) in which A is the cross-sectional area of the rod.Therefore,the formula for the maximum stress(from Eq.1-1)becomes Fmax W+yAL W yL (1-5) A A A (b)To calculate the maximum stress,we substitute numerical values into the preceding equation.The cross-sectional area A equals md-/4,where d =8 mm, and the weight density yof steel is 77.0 kN/m3(from Table H-1 in Appendix H). Thus, 1.5kN Omax r(8mm)24 +(77.0kN/m3(40m) =29.8MPa+3.1MPa=32.9MPa In this example,the weight of the rod contributes noticeably to the maximum stress and should not be disregarded
14 CHAPTER 1 Tension, Compression, and Shear A circular steel rod of length L and diameter d hangs in a mine shaft and holds an ore bucket of weight W at its lower end (Fig. 1-6). (a) Obtain a formula for the maximum stress smax in the rod, taking into account the weight of the rod itself. (b) Calculate the maximum stress if L 40 m, d 8 mm, and W 1.5 kN. Example 1-2 L W d FIG. 1-6 Example 1-2. Steel rod supporting a weight W Solution (a) The maximum axial force Fmax in the rod occurs at the upper end and is equal to the weight W of the ore bucket plus the weight W0 of the rod itself. The latter is equal to the weight density g of the steel times the volume V of the rod, or W0 gV gAL (1-4) in which A is the cross-sectional area of the rod. Therefore, the formula for the maximum stress (from Eq. 1-1) becomes smax Fm A ax W A gAL W A gL (1-5) (b) To calculate the maximum stress, we substitute numerical values into the preceding equation. The cross-sectional area A equals pd2 /4, where d 8 mm, and the weight density g of steel is 77.0 kN/m3 (from Table H-1 in Appendix H). Thus, smax p(8 1. m 5k m N ) 2 /4 (77.0 kN/m3 )(40 m) 29.8 MPa 3.1 MPa 32.9 MPa In this example, the weight of the rod contributes noticeably to the maximum stress and should not be disregarded.
SECTION 1.3 Mechanical Properties of Materials 15 1.3 MECHANICAL PROPERTIES OF MATERIALS The design of machines and structures so that they will function prop- erly requires that we understand the mechanical behavior of the materials being used.Ordinarily,the only way to determine how materials behave when they are subjected to loads is to perform experiments in the laboratory.The usual procedure is to place small specimens of the material in testing machines,apply the loads,and then measure the resulting deformations(such as changes in length and changes in diameter). Most materials-testing laboratories are equipped with machines capable of loading specimens in a variety of ways,including both static and dynamic loading in tension and compression. A typical tensile-test machine is shown in Fig.1-7.The test spec- imen is installed between the two large grips of the testing machine and then loaded in tension.Measuring devices record the deformations,and the automatic control and data-processing systems (at the left in the photo)tabulate and graph the results. A more detailed view of a tensile-test specimen is shown in Fig.1-8 on the next page.The ends of the circular specimen are enlarged where they fit in the grips so that failure will not occur near the grips them- selves.A failure at the ends would not produce the desired information about the material,because the stress distribution near the grips is not uniform,as explained in Section 1.2.In a properly designed specimen, failure will occur in the prismatic portion of the specimen where the stress distribution is uniform and the bar is subjected only to pure tension.This situation is shown in Fig.1-8,where the steel specimen has just fractured under load.The device at the left,which is attached by ■睡 FIG.1-7 Tensile-test machine with automatic data-processing system. (Courtesy of MTS Systems Corporation)
SECTION 1.3 Mechanical Properties of Materials 15 1.3 MECHANICAL PROPERTIES OF MATERIALS The design of machines and structures so that they will function properly requires that we understand the mechanical behavior of the materials being used. Ordinarily, the only way to determine how materials behave when they are subjected to loads is to perform experiments in the laboratory. The usual procedure is to place small specimens of the material in testing machines, apply the loads, and then measure the resulting deformations (such as changes in length and changes in diameter). Most materials-testing laboratories are equipped with machines capable of loading specimens in a variety of ways, including both static and dynamic loading in tension and compression. A typical tensile-test machine is shown in Fig. 1-7. The test specimen is installed between the two large grips of the testing machine and then loaded in tension. Measuring devices record the deformations, and the automatic control and data-processing systems (at the left in the photo) tabulate and graph the results. A more detailed view of a tensile-test specimen is shown in Fig. 1-8 on the next page. The ends of the circular specimen are enlarged where they fit in the grips so that failure will not occur near the grips themselves. A failure at the ends would not produce the desired information about the material, because the stress distribution near the grips is not uniform, as explained in Section 1.2. In a properly designed specimen, failure will occur in the prismatic portion of the specimen where the stress distribution is uniform and the bar is subjected only to pure tension. This situation is shown in Fig. 1-8, where the steel specimen has just fractured under load. The device at the left, which is attached by FIG. 1-7 Tensile-test machine with automatic data-processing system. (Courtesy of MTS Systems Corporation)