10CHAPTER1Tension, Compression,andShearfromthe stressconcentrationas thelargestlateraldimension of thebar.Inother words,the stress distribution in the steel eyebar of Fig.1-3 isuniform atdistancesborgreaterfromtheenlarged ends,wherebisthewidthofthebar,and thestressdistributionintheprismaticbar of Fig.1-2is uniform atdistances d orgreater from the ends,where d is thediameterof the bar (Fig.1-2d).More detailed discussions of stress concentrationsproduced byaxial loads aregiven inSection 2.10.Of course, even whenthe stress is notuniformlydistributed,theequation =P/A may still be useful because it gives the averagenormal stress on the cross section.NormalStrainAs already observed, a straight bar will change in length when loadedaxially,becominglongerwhenintensionandshorterwhenincompressionFor instance,consider again theprismaticbarofFig.1-2.The elongationof thisbar(Fig.1-2c)isthecumulativeresultof thestretchingof allelements ofthe material throughoutthevolume ofthebar.Letus assumethatthematerialisthesameeverywhereinthebar.Then,ifweconsiderhalfof the bar (length L/2), it will have an elongation equal to /2, and if weconsiderone-fourthofthebar, it willhavean elongation equal to/4In general,the elongation of a segment is equal to its length dividedbythetotal lengthLand multiplied by thetotal elongation .Therefore,aunit length of thebar will have an elongation equal to 1/L times .Thisquantity is called the elongation per unitlength,or strain,and is denotedby the Greek letter e (epsilon). We see that strain is given by the equation8(1-2)LIf the bar is in tension,the strain is called a tensile strain,representing anelongation or stretching ofthematerial.If the bar is in compression,thestrain is a compressive strain and the bar shortens.Tensile strain isusuallytaken as positiveand compressive strain as negative.The strain eis called anormal strainbecause it is associated with normal stresses.Because normal strain is the ratio of two lengths, it is a dimension-lessquantity,thatis,ithas nounits.Therefore,strain is expressedsimply as a number, independent of any system of units.Numericalvalues of strainareusuallyvery small,becausebars made of structuralmaterials undergo only small changes in length when loaded.Asanexample,consider a steel bar havinglengthL equal to2.0mWhen heavily loaded in tension, this bar might elongate by 1.4 mm,whichmeans that the strain is81.4 mm=0.0007=700×10-6L2.0m
from the stress concentration as the largest lateral dimension of the bar. In other words, the stress distribution in the steel eyebar of Fig. 1-3 is uniform at distances b or greater from the enlarged ends, where b is the width of the bar, and the stress distribution in the prismatic bar of Fig. 1-2 is uniform at distances d or greater from the ends, where d is the diameter of the bar (Fig. 1-2d). More detailed discussions of stress concentrations produced by axial loads are given in Section 2.10. Of course, even when the stress is not uniformly distributed, the equation s P/A may still be useful because it gives the average normal stress on the cross section. Normal Strain As already observed, a straight bar will change in length when loaded axially, becoming longer when in tension and shorter when in compression. For instance, consider again the prismatic bar of Fig. 1-2. The elongation d of this bar (Fig. 1-2c) is the cumulative result of the stretching of all elements of the material throughout the volume of the bar. Let us assume that the material is the same everywhere in the bar. Then, if we consider half of the bar (length L/2), it will have an elongation equal to d/2, and if we consider one-fourth of the bar, it will have an elongation equal to d/4. In general, the elongation of a segment is equal to its length divided by the total length L and multiplied by the total elongation d. Therefore, a unit length of the bar will have an elongation equal to 1/L times d. This quantity is called the elongation per unit length, or strain, and is denoted by the Greek letter e (epsilon). We see that strain is given by the equation (1-2) If the bar is in tension, the strain is called a tensile strain, representing an elongation or stretching of the material. If the bar is in compression, the strain is a compressive strain and the bar shortens. Tensile strain is usually taken as positive and compressive strain as negative. The strain e is called a normal strain because it is associated with normal stresses. Because normal strain is the ratio of two lengths, it is a dimensionless quantity, that is, it has no units. Therefore, strain is expressed simply as a number, independent of any system of units. Numerical values of strain are usually very small, because bars made of structural materials undergo only small changes in length when loaded. As an example, consider a steel bar having length L equal to 2.0 m. When heavily loaded in tension, this bar might elongate by 1.4 mm, which means that the strain is e 5 L d 1 2 .4 .0 m m m 0.0007 700 106 e 5 L d 10 CHAPTER 1 Tension, Compression, and Shear
11SECTION1.2 Normal StressandStrainIn practice,the original units of and L are sometimes attached to thestrain itself, and then the strain is recorded in forms such as mm/m,μm/m, and in./in.For instance,the strain ein the preceding illustrationcould be given as 700 μm/m or 700x10-in./in.Also, strain is some-times expressed as a percent, especially when the strains are large. (Intheprecedingexample,thestrainis0.07%.)Uniaxial Stress and StrainThe definitions of normal stress and normal strain are based upon purelystatic and geometric considerations, which means that Eqs. (1-i) and(1-2)canbeused for loadsof anymagnitude and for anymaterial.Theprincipal requirement is that the deformation of the bar be uniformthroughout its volume, which in turn requires that the bar be prismatic,the loads act through the centroids ofthe cross sections,and the materialbe homogeneous (that is, the same throughout all parts of the bar). Theresulting stateof stress and strain is called uniaxial stressand strain.Further discussions of uniaxial stress,including stresses in direc-tions other than the longitudinal direction of the bar, are given later inSection 2.6.We will also analyze more complicated stress states, suchasbiaxial stressandplanestress, inChapter7.Line of Action of the Axial Forcesfor a Uniform Stress DistributionThroughouttheprecedingdiscussion of stress and strain inaprismaticbar,weassumedthatthe normal stress owasdistributeduniformlyoverthe cross section.Now we will demonstrate that this condition is met ifthe line of action of the axial forces is throughthe centroid of the cross-sectional area.Consider a prismatic bar of arbitrary cross-sectional shape subjectedto axial forces P that produce uniformly distributed stresses o (Fig.1-4a).Also, let Prrepresent the point in the cross section wherethe line ofaction of theforces intersects the cross section (Fig.1-4b).We constructa set of xy axes in the plane of the cross section and denote the coordi-natesofpointP,byxandy.Todeterminethesecoordinates,weobservethatthemomentsMandM,of theforcePaboutthexandyaxes,respectively,must be equal to the corresponding moments of theuniformly distributed stresses.Themomentsof theforcePareM,=-Px(a,b)M,= Pyin which a moment is considered positive when its vector (using theright-handrule)acts inthepositivedirectionof thecorrespondingaxis."To visualize the right-hand rule, imagine that you grasp an axis of coordinates with yourright hand so that your fingers fold around the axis and your thumb points in the positivedirection 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.
12CHAPTER1Tension, Compression,andShearP(a)FIG.1-4 Uniform stress distribution inxa prismatic bar:(a)axial forcesPand(b)crosssectionofthebar(b)The moments of the distributed stresses are obtained by integratingover the cross-sectional area A.The differential force acting on anelement of area dA (Fig.1-4b)is equal to cdA.Themoments of thiselemental force about thexand y axes are oydA and-oxdA,respectivelyinwhichxandydenotethecoordinatesof theelementdA.Thetotalmoments areobtained by integrating over the cross-sectional area:M,=oydAM,=-oxdA(c,d)Theseexpressions givethemoments producedbythe stresses .Next,we equate themoments M,and Myas obtained from theforceP(Eqs.a andb)tothemoments obtainedfromthedistributedstresses (Eqs.c and d):Py =oydAPx =oxdABecausethestresses areuniformlydistributed,weknowthattheyareconstantoverthecross-sectional areaA andcanbeplacedoutsidetheintegral signs.Also,weknowthat is equal to P/A.Therefore,weobtain thefollowingformulas forthecoordinates of pointp:lydAxdAy=(1-3a,b)xAAThese equations arethe same as theequations definingthe coordinatesofthecentroidofanarea(seeEgs.12-3aandbinChapter12).Therefore,wehavenowarrived atanimportantconclusion:Inordertohaveuniformtension or compression in a prismatic bar,the axial force mustact through the centroid of the cross-sectional area.As explained previ-ously, we always assume that these conditions are met unless it isspecifically stated otherwise.The following examples illustratethe calculation of stresses andstrains in prismatic bars. In the first example we disregard the weight ofthe bar and in the second we include it. (It is customary when solvingtextbookproblemstoomittheweightofthestructureunless specificallyinstructedtoincludeit.)
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
13SECTION1.2NormalStressandStrainExample1-1A short post constructed from a hollow circular tube of aluminum supports acompressive load of 26kips(Fig-1-5).Theinner and outerdiameters ofthetube are d, = 4.0 in.and d,= 4.5 in., respectively,and its length is 16 in.Theshortening of the post due to the load is measured as 0.012 in.Determine the compressive stress and strain in the post. (Disregard theweight of the post itself, and assume that the post does not buckle under theload.)26k16in.FIG.1-5Example1-1.HollowaluminumpostincompressionSolutionAssuming that the compressive load acts at the center of the hollow tubewe can usetheequation o=P/A (Eq.1-I)to calculatethenormal stress.TheforcePequals26k(or26.000lb).andthecross-sectionalareaAisA = (adz - d) = [(4.5 in.)2 - (4.0 in)] = 3.338 in.24Therefore, the compressive stress in the post isP26,0001b=7790psiA3.338in.3Thecompressivestrain(fromEg.1-2)isS0.012 in.=750X10-6L16in.Thus, the stress and strain in the post have been calculated.Note: As explained earlier,strain is a dimensionless quantity and no unitsare needed. For clarity, however, units are often given. In this example, e couldbe written as 750 × 10- in./in. or 750 μuin./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.
14CHAPTER1Tension, Compression,andShearExample1-2A circular steel rod of length L and diameter d hangs in a mine shaft and holdsan orebucketof weightWat its lowerend (Fig.l-6).(a) Obtain a formula for the maximum stress Omax in the rod, taking intoaccount the weight of the rod itself.(b) Calculate the maximum stress if L = 40 m, d = 8 mm, and W = 1.5 kN.FIG.1-6Example1-2.Steel rodsupportingaweightwSolution(a) The maximum axial force Fmax in the rod occurs at the upper end and isequal to the weight W of the ore bucket plus the weight Wo of the rod itself.Thelatter is equal to the weight density yof the steel times the volume Vof the rod,or(1-4)Wo=V=yALin which A is the cross-sectional area of the rod. Therefore, the formula for themaximumstress(fromEg.1-1)becomesFmaxW+ALW+yL(1-5)OmaxAAA(b) To calculate themaximum stress, we substitute numerical values into thepreceding equation.The cross-sectional area A equals md-/4,where d =8mm,and the weight density y of steel is 77.0 kN/m (from Table H-1 in Appendix H)Thus,1.5kN+ (77.0 kN/m2)(40 m)Omax(8mm)/4=29.8MPa+3.1MPa=32.9MPaIn this example, the weight of the rod contributes noticeably to the maximumstress 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.