211.2EQUILIBRIUMOFADEFORMABLEBODY1-26.Determinetheresultant internal loadingsacting on*1-28Thebrace and drill bit is used to drill a hole at O.Ifthe cross section of theframe at points F and G.The contactthe drill bit jams when the brace is subjected to the forcesatEissmooth.shown,determine the resultant internal loadings acting onthe cross section of the drill bitat A.F=30lb10Ib5f9in.30F=501bin.gin.6in.6in.6in.2ftAProb.1-283ft4ft1-29.The curved rod AD of radius r has a weight per80lblengthofw.Ifitliesinthehorizontalplane.determinetheresultant internal loadings acting on the cross section atpointB.Hint:Thedistancefrom thecentroid C of segmentABtopointOisCO=0.9745r.Prob.1-26DB1-27.The pipe has a mass of 12kg/m.If it is fixed to thewall atA,determinetheresultantinternal loadingsactingonT90°the cross section at B.Prob.1291-30.A differential element taken from a curved bar isshowninthefigure.ShowthatdN/de=V,dV/de=-N.dM/do=-T,anddT/do=M.T+dTM + dM300N400N12V+dv"N+dNC12mT500NProb.1-27Prob.1-30
1.2 Equilibrium of a Deformable Body 21 1 1–26. Determine the resultant internal loadings acting on the cross section of the frame at points F and G. The contact at E is smooth. 4 ft 1.5 ft 1.5 ft 3 ft A B C E D G 80 lb 5 ft 2 ft 2 ft 30 F Prob. 1–26 1–27. The pipe has a mass of 12 kg>m. If it is fixed to the wall at A, determine the resultant internal loadings acting on the cross section at B. 1 m 2 m 2 m B A y z x C 3 4 5 500 N 400 N 300 N Prob. 1–27 *1–28 The brace and drill bit is used to drill a hole at O. If the drill bit jams when the brace is subjected to the forces shown, determine the resultant internal loadings acting on the cross section of the drill bit at A. z x y O A 9 in. 6 in. 6 in. 6 in. 9 in. 3 in. Fx 30 lb Fy 50 lb Fz 10 lb Prob. 1–28 1–29. The curved rod AD of radius r has a weight per length of w. If it lies in the horizontal plane, determine the resultant internal loadings acting on the cross section at point B. Hint: The distance from the centroid C of segment AB to point O is CO = 0.9745r. A B C 45 90 D O r 22.5 Prob. 1–29 1–30. A differential element taken from a curved bar is shown in the figure. Show that dN>du = V, dV>du = -N, dM>du = -T, and dT>du = M. M V N d M dM T dT N dN V dV T Prob. 1–30
22CHAPTER1STRESS1.3STRESSMRoItwasstatedinSection1.2thattheforceandmomentactingataspecifiedFR4point O on the sectioned area of the body,Fig.1-8, represents theresultanteffectsofthe distributionof loading that acts overthe sectionedarea,Fig.1-9a.Obtainingthisdistributionisof primary importanceinmechanics of materials.Tosolvethisproblemit isfirst necessarytoestablishtheconceptofstress.We begin by considering the sectioned area to be subdivided intosmall areas,suchasA shown inFig.1-9a.As wereduceAtoasmaller and smaller size,we will make two assumptions regarding theproperties of thematerial.We will consider the material to becontinuous,thatis,toconsistofacontinuumoruniformdistributionofmatterhavingnovoids.Also,thematerial mustbe cohesive,meaningFthat all portions of it are connected together,without having breaks.cracks,or separations.A typical finite yetvery smallforce AF,acting onFig.1-8AA, is shown in Fig.1-9a.This force, likeall the others, will have aunique direction, but to compare it with all the other forces, we willreplace itbyits threecomponents,namely,Fx,Fy,andF.AsAAapproacheszero, so doF and its components; however,thequotientof theforce and area will approach a finite limit.Thisquotient is calledstress,and it describes the intensity of the internal force acting on aspecificplane(area)passingthroughapoint.AFAFAAH(c)(a)(b)Fig.1-9
22 Chapter 1 Stress 1 1.3 STRESS It was stated in Section 1.2 that the force and moment acting at a specified point O on the sectioned area of the body, Fig. 1–8, represents the resultant effects of the distribution of loading that acts over the sectioned area, Fig. 1–9a. Obtaining this distribution is of primary importance in mechanics of materials. To solve this problem it is first necessary to establish the concept of stress. We begin by considering the sectioned area to be subdivided into small areas, such as ∆A shown in Fig. 1–9a. As we reduce ∆A to a smaller and smaller size, we will make two assumptions regarding the properties of the material. We will consider the material to be continuous, that is, to consist of a continuum or uniform distribution of matter having no voids. Also, the material must be cohesive, meaning that all portions of it are connected together, without having breaks, cracks, or separations. A typical finite yet very small force ∆F, acting on ∆A, is shown in Fig. 1–9a. This force, like all the others, will have a unique direction, but to compare it with all the other forces, we will replace it by its three components, namely, ∆Fx, ∆Fy, and ∆Fz. As ∆A approaches zero, so do ∆F and its components; however, the quotient of the force and area will approach a finite limit. This quotient is called stress, and it describes the intensity of the internal force acting on a specific plane (area) passing through a point. F1 F2 O MRO FR Fig. 1–8 F1 F2 F1 F A F Fz z x y Fx Fy z (c) x y (b) z z x y (a) x y tyz sy tyx txz sx txy Fig. 1–9
231.3STRESSNormal Stress.The intensity of the force acting normal to A isreferred toas thenormal stress, (sigma).Since AF,is normal to theareathenAF,o,=lim=0△A(1-4)Ifthenormalforceorstress"pulls"onAasshowninFig.1-9a,itistensile stress,whereas if it"pushes"onAAitis compressivestress.Shear Stress.The intensity of force acting tangent to A is calledthe shear stress, (tau).Herewehavetwo shear stress components,AFt0Tzx=JimAA(15)AF,Ty = LimoAA-TeFig.1-10The subscript notation z specifies the orientation of the area A,Fig.1-10,and x and y indicate the axes along which each shear stress actsGeneral State of Stress.If the bodyis further sectioned byplanes parallel to the x-z plane,Fig.1-9b,and the y-z plane,Fig.1-9c, wecan then"cut out"a cubic volume element of material that representsthe state of stress acting around a chosen point in the body.This state ofstress is then characterized by three components acting on each face ofthe element,Fig.1-11.Units.Sincestressrepresents a forceperunitarea,intheInternationalStandard orSI system,themagnitudesofbothnormal and shearstressare specified in the basic units of newtons per square meter (N/m).Thiscombinationofunitsiscalledapascal(1Pa=1N/m),andbecauseitisrather small,prefixes such askilo-(103),symbolized byk,mega-(10°),Fig.1-11symbolizedbyM,orgiga-(1o,symbolized byG,areused inengineeringto represent larger,more realistic values of stress.*In the Foot-Pound-Second systemof units,engineersusuallyexpress stress inpoundspersquareinch(psi)orkilopoundsper square inch(ksi),where1kilopound(kip) = 1000 lb.*Sometimes stress is expressed inunits ofN/mm2,where1mm=10-m.However,inthe SI system, prefixes are not allowed in the denominator of a fraction, and therefore itis better to use the equivalent 1N/mm2=1 MN/m?=1 MPa
1.3 Stress 23 1 Normal Stress. The intensity of the force acting normal to ∆A is referred to as the normal stress, s (sigma). Since ∆Fz is normal to the area then sz = lim ∆AS0 ∆Fz ∆A (1–4) If the normal force or stress “pulls” on ∆A as shown in Fig. 1–9a, it is tensile stress, whereas if it “pushes” on ∆A it is compressive stress. Shear Stress. The intensity of force acting tangent to ∆A is called the shear stress, t (tau). Here we have two shear stress components, tzx = lim ∆AS0 ∆Fx ∆A tzy = lim ∆AS0 ∆Fy ∆A (1–5) The subscript notation z specifies the orientation of the area ∆A, Fig. 1–10, and x and y indicate the axes along which each shear stress acts. General State of Stress. If the body is further sectioned by planes parallel to the x–z plane, Fig. 1–9b, and the y–z plane, Fig. 1–9c, we can then “cut out” a cubic volume element of material that represents the state of stress acting around a chosen point in the body. This state of stress is then characterized by three components acting on each face of the element, Fig. 1–11. Units. Since stress represents a force per unit area, in the International Standard or SI system, the magnitudes of both normal and shear stress are specified in the basic units of newtons per square meter (N>m2 ). This combination of units is called a pascal (1 Pa = 1 N>m2 ), and because it is rather small, prefixes such as kilo- (103 ), symbolized by k, mega- (106 ), symbolized by M, or giga- (109 ), symbolized by G, are used in engineering to represent larger, more realistic values of stress.* In the Foot-PoundSecond system of units, engineers usually express stress in pounds per square inch (psi) or kilopounds per square inch (ksi), where 1 kilopound (kip) = 1000 lb. x y z sz sx sy tyz tyx txy txz tzx tzy Fig. 1–11 x y z Tzx Tzy sz Fig. 1–10 *Sometimes stress is expressed in units of N>mm2 , where 1 mm = 10-3 m. However, in the SI system, prefixes are not allowed in the denominator of a fraction, and therefore it is better to use the equivalent 1 N>mm2 = 1 MN>m2 = 1 MPa
24CHAPTER 1STRESS1.4AVERAGENORMALSTRESSINANAXIALLYLOADEDBARWe will now determine the average stress distribution acting over thecross-sectional area of an axiallyloaded bar such as the one shown inFig.1-12a.Specifically,thecross sectionisthesectiontakenperpendicularto thelongitudinal axis of the bar,and since the bar is prismatic all crosssections are the same throughout its length.Provided the material of thebar is both homogeneous and isotropic,that is, it has the same physicalandmechanical propertiesthroughoutitsvolume,andithasthesameproperties in all directions, then when the load P is applied to the barthrough the centroid of its cross-sectional area, the bar will deformuniformly throughout thecentral region of itslength,Fig.1-12b.Realize that many engineering materials may be approximated asbeingbothhomogeneous andisotropic.Steel,for example,containsthousandsofrandomlvorientedcrvstalsineachcubicmillimeterofitsvolume,and sincemost objectsmadeofthismaterialhaveaphysical sizethatisverymuchlargerthana singlecrystal,theabove assumptionregarding thematerial's composition is quiterealistic.Notethatanisotropicmaterials,suchaswood,havedifferentpropertiesin different directions; and although this is the case, if the grains of woodare oriented along thebar's axis (asfor instance in atypical wood board),thenthebarwill alsodeformuniformlywhen subjectedtotheaxial loadPAverage Normal Stress Distribution.Ifwepassasectionthroughthebar,andseparateit intotwoparts,thenequilibriumrequirestheresultantnormal forceNatthe sectiontobeequal toP,Fig.1-12c.Andbecause thematerial undergoes a uniform deformation,it is necessary thatthe cross sectionbe subjected toa constant normal stress distribution.P4P4Region ofN=Puniform4deformationInternal forceof barCross-sectionalareaExtemalforcePVPP(a)(b)(c)Fig.1-12
24 Chapter 1 Stress 1 1.4 AVERAGE NORMAL STRESS IN AN AXIALLY LOADED BAR We will now determine the average stress distribution acting over the cross-sectional area of an axially loaded bar such as the one shown in Fig. 1–12a. Specifically, the cross section is the section taken perpendicular to the longitudinal axis of the bar, and since the bar is prismatic all cross sections are the same throughout its length. Provided the material of the bar is both homogeneous and isotropic, that is, it has the same physical and mechanical properties throughout its volume, and it has the same properties in all directions, then when the load P is applied to the bar through the centroid of its cross-sectional area, the bar will deform uniformly throughout the central region of its length, Fig. 1–12b. Realize that many engineering materials may be approximated as being both homogeneous and isotropic. Steel, for example, contains thousands of randomly oriented crystals in each cubic millimeter of its volume, and since most objects made of this material have a physical size that is very much larger than a single crystal, the above assumption regarding the material’s composition is quite realistic. Note that anisotropic materials, such as wood, have different properties in different directions; and although this is the case, if the grains of wood are oriented along the bar’s axis (as for instance in a typical wood board), then the bar will also deform uniformly when subjected to the axial load P. Average Normal Stress Distribution. If we pass a section through the bar, and separate it into two parts, then equilibrium requires the resultant normal force N at the section to be equal to P, Fig. 1–12c. And because the material undergoes a uniform deformation, it is necessary that the cross section be subjected to a constant normal stress distribution. P P (a) P (b) P Region of uniform deformation of bar N P P External force Cross-sectional area Internal force (c) Fig. 1–12
251.4AVERAGENORMALSTRESSINANAXIALLYLOADEDBARAs a result, each small area AA on the cross section is subjected to aforceN=α△A,Fig.1-12d,andthesumoftheseforcesactingovertheentirecross-sectionalareamustbeequivalenttotheinternalresultantANforcePatthesection.Ifwelet△A-→dAandtherefore△N→dN,then,recognizingisconstant,wehaveAAN=GAgdAdN+FRz =ZF;N=aAN(1-6)AHereWPα=averagenormal stress atanypoint on thecross-sectionalarea(d)N=internalresultantnormal force,whichactsthroughthecentroidofthecross-sectional area.Nis determined using the method of sectionsFig.1-12 (cont.)and the equations ofequilibrium,wherefor this caseN=PA=cross-sectional area of the bar where isdeterminedEquilibrium.The stress distribution inFig.1-12 indicates that onlyanormalstressexistsonanysmall volumeelementofmaterial locatedateachpointonthecross section.Thus, ifweconsidervertical equilibriumof an element of material and then apply the equation of forceequilibrium to its free-body diagram, Fig.1-13,EF,=0:G(△A)-G'(△A)=0g=GAAAAAFree-body diagramStressonelementFig.1-13
1.4 Average Normal Stress in an Axially Loaded Bar 25 1 As a result, each small area ∆A on the cross section is subjected to a force ∆N = s ∆A, Fig. 1–12d, and the sum of these forces acting over the entire cross-sectional area must be equivalent to the internal resultant force P at the section. If we let ∆A S dA and therefore ∆N S dN, then, recognizing s is constant, we have + c FRz = ΣFz; LdN = LA s dA N = sA s = N A (1–6) Here s = average normal stress at any point on the cross-sectional area N = internal resultant normal force, which acts through the centroid of the cross-sectional area. N is determined using the method of sections and the equations of equilibrium, where for this case N = P. A = cross-sectional area of the bar where s is determined Equilibrium. The stress distribution in Fig. 1–12 indicates that only a normal stress exists on any small volume element of material located at each point on the cross section. Thus, if we consider vertical equilibrium of an element of material and then apply the equation of force equilibrium to its free-body diagram, Fig. 1–13, ΣFz = 0; s(∆A) - s′(∆A) = 0 s = s′ A s s¿ A s s¿ A A Stress on element Free-body diagram Fig. 1–13 (d) P N s A N y x x z y A s Fig. 1–12 (cont.)