上海交通大学：《Design and Manufacturing》课程教学资源（参考文献）Shigley's Mechanical Engineering Design ch4 deflection and stiffness

Budynas-Nisbett:Shigley's I.Basics 4.Deflection and Stiffness ©The McGraw-Hfll Mechanical Engineering Companies,2008 Design,Eighth Edition Deflection and Stiffness Chapter Outline 41 Spring Rates 142 4-2 Tension,Compression,and Torsion 143 4-3 Deflection Due to Bending 144 4-4 Beam Deflection Methods 146 4-5 Beam Deflections by Superposition 147 4-6 Beam Deflections by Singularity Functions 150 4-7 Strain Energy 156 4-8 Castigliano's Theorem 158 4-9 Deflection of Curved Members 163 4-10 Statically Indeterminate Problems 168 4-11 Compression Members-General 173 4-12 Long Columns with Central Loading 173 4-13 Intermediate-Length Columns with Central Loading 176 4-14 Columns with Eccentric Loading 176 4-15 Struts or Short Compression Members 180 4-16 Elastic Stability 182 4-17 Shock and Impact 183 4-18 Suddenly Applied Loading 184 141

Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition I. Basics 4. Deflection and Stiffness © The McGraw−Hill 145 Companies, 2008 4 Deflection and Stiffness Chapter Outline 4–1 Spring Rates 142 4–2 Tension, Compression, and Torsion 143 4–3 Deflection Due to Bending 144 4–4 Beam Deflection Methods 146 4–5 Beam Deflections by Superposition 147 4–6 Beam Deflections by Singularity Functions 150 4–7 Strain Energy 156 4–8 Castigliano’s Theorem 158 4–9 Deflection of Curved Members 163 4–10 Statically Indeterminate Problems 168 4–11 Compression Members—General 173 4–12 Long Columns with Central Loading 173 4–13 Intermediate-Length Columns with Central Loading 176 4–14 Columns with Eccentric Loading 176 4–15 Struts or Short Compression Members 180 4–16 Elastic Stability 182 4–17 Shock and Impact 183 4–18 Suddenly Applied Loading 184 141

146 Budynas-Nisbett:Shigley's I.Basics 4.Deflection and Stiffness T©The McGraw-Hill Mechanical Engineering Companies,2008 Design,Eighth Edition 142 Mechanical Engineering Design All real bodies deform under load,either elastically or plastically.A body can be suffi- ciently insensitive to deformation that a presumption of rigidity does not affect an analy- sis enough to warrant a nonrigid treatment.If the body deformation later proves to be not negligible,then declaring rigidity was a poor decision,not a poor assumption.A wire rope is flexible,but in tension it can be robustly rigid and it distorts enormously under attempts at compressive loading.The same body can be both rigid and nonrigid. Deflection analysis enters into design situations in many ways.A snap ring,or retain- ing ring,must be flexible enough to be bent without permanent deformation and assembled with other parts,and then it must be rigid enough to hold the assembled parts together.In a transmission,the gears must be supported by a rigid shaft.If the shaft bends too much,that is,if it is too flexible,the teeth will not mesh properly,and the result will be excessive impact,noise,wear,and early failure.In rolling sheet or strip steel to pre- scribed thicknesses,the rolls must be crowned,that is,curved,so that the finished product will be of uniform thickness.Thus,to design the rolls it is necessary to know exactly how much they will bend when a sheet of steel is rolled between them.Sometimes mechanical elements must be designed to have a particular force-deflection characteristic.The suspension system of an automobile,for example,must be designed within a very narrow range to achieve an optimum vibration frequency for all conditions of vehicle loading, because the human body is comfortable only within a limited range of frequencies. The size of a load-bearing component is often determined on deflections,rather than limits on stress. This chapter considers distortion of single bodies due to geometry(shape)and loading,then,briefly,the behavior of groups of bodies. 4-1 Spring Rates Elasriciry is that property of a material that enables it to regain its original configuration after having been deformed.A spring is a mechanical element that exerts a force when deformed.Figure 4-la shows a straight beam of length simply supported at the ends and loaded by the transverse force F.The deflection y is linearly related to the force,as long as the elastic limit of the material is not exceeded,as indicated by the graph.This beam can be described as a linear spring. In Fig.4-1b a straight beam is supported on two cylinders such that the length between supports decreases as the beam is deflected by the force F.A larger force is required to deflect a short beam than a long one,and hence the more this beam is deflected,the stiffer it becomes.Also,the force is not linearly related to the deflection, and hence this beam can be described as a nonlinear stiffening spring. Figure 4-1 (a)A linear spring;(b)a stiffening spring;(da softening spring

Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition I. Basics 4. Deflection and Stiffness 146 © The McGraw−Hill Companies, 2008 142 Mechanical Engineering Design All real bodies deform under load, either elastically or plastically. A body can be suffi- ciently insensitive to deformation that a presumption of rigidity does not affect an analy￾sis enough to warrant a nonrigid treatment. If the body deformation later proves to be not negligible, then declaring rigidity was a poor decision, not a poor assumption. A wire rope is flexible, but in tension it can be robustly rigid and it distorts enormously under attempts at compressive loading. The same body can be both rigid and nonrigid. Deflection analysis enters into design situations in many ways. A snap ring, or retain￾ing ring, must be flexible enough to be bent without permanent deformation and assembled with other parts, and then it must be rigid enough to hold the assembled parts together. In a transmission, the gears must be supported by a rigid shaft. If the shaft bends too much, that is, if it is too flexible, the teeth will not mesh properly, and the result will be excessive impact, noise, wear, and early failure. In rolling sheet or strip steel to pre￾scribed thicknesses, the rolls must be crowned, that is, curved, so that the finished product will be of uniform thickness. Thus, to design the rolls it is necessary to know exactly how much they will bend when a sheet of steel is rolled between them. Sometimes mechanical elements must be designed to have a particular force-deflection characteristic. The suspension system of an automobile, for example, must be designed within a very narrow range to achieve an optimum vibration frequency for all conditions of vehicle loading, because the human body is comfortable only within a limited range of frequencies. The size of a load-bearing component is often determined on deflections, rather than limits on stress. This chapter considers distortion of single bodies due to geometry (shape) and loading, then, briefly, the behavior of groups of bodies. 4–1 Spring Rates Elasticity is that property of a material that enables it to regain its original configuration after having been deformed. A spring is a mechanical element that exerts a force when deformed. Figure 4–1a shows a straight beam of length l simply supported at the ends and loaded by the transverse force F. The deflection y is linearly related to the force, as long as the elastic limit of the material is not exceeded, as indicated by the graph. This beam can be described as a linear spring. In Fig. 4–1b a straight beam is supported on two cylinders such that the length between supports decreases as the beam is deflected by the force F. A larger force is required to deflect a short beam than a long one, and hence the more this beam is deflected, the stiffer it becomes. Also, the force is not linearly related to the deflection, and hence this beam can be described as a nonlinear stiffening spring. Figure 4–1 (a) A linear spring; (b) a stiffening spring; (c) a softening spring. l F F y l F y d F y (a) y F (b) y F (c) y

Budynas-Nisbett:Shigley's I.Basics 4.Deflection and Stiffness I©The McGraw-Hil Mechanical Engineering Companies,2008 Design,Eighth Edition Deflection and Stiffness 143 Figure 4-1c is an edge-view of a dish-shaped round disk.The force necessary to flatten the disk increases at first and then decreases as the disk approaches a flat con- figuration,as shown by the graph.Any mechanical element having such a characteristic is called a nonlinear softening spring. If we designate the general relationship between force and deflection by the equation F=F(y) (a) then spring rate is defined as △FdF aymo Ay=dy k(y)=lim (4-1) where y must be measured in the direction of F and at the point of application of F.Most of the force-deflection problems encountered in this book are linear,as in Fig.4-1a.For these,k is a constant,also called the spring constant;consequently Eq.(4-1)is written F k= (4-2) We might note that Eqs.(4-1)and(4-2)are quite general and apply equally well for torques and moments,provided angular measurements are used for y.For linear dis- placements,the units of k are often pounds per inch or newtons per meter,and for angular displacements,pound-inches per radian or newton-meters per radian. 4-2 Tension,Compression,and Torsion The total extension or contraction of a uniform bar in pure tension or compression, respectively,is given by 6= (4-3) Ae This equation does not apply to a long bar loaded in compression if there is a possibil- ity of buckling (see Secs.4-11 to 4-15).Using Egs.(4-2)and (4-3),we see that the spring constant of an axially loaded bar is k=AE (4-4 The angular deflection of a uniform round bar subjected to a twisting moment T was given in Eq.(3-35),and is TI 0= GJ (4-5 where 0 is in radians.If we multiply Eq.(4-5)by 180/and substitute J=d4/32 for a solid round bar,we obtain 583.6Tl 0= (4-61 Gd4 where 0 is in degrees. Equation (4-5)can be rearranged to give the torsional spring rate as T GJ k= =7 (4-7刀

Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition I. Basics 4. Deflection and Stiffness © The McGraw−Hill 147 Companies, 2008 Deflection and Stiffness 143 Figure 4–1c is an edge-view of a dish-shaped round disk. The force necessary to flatten the disk increases at first and then decreases as the disk approaches a flat con- figuration, as shown by the graph. Any mechanical element having such a characteristic is called a nonlinear softening spring. If we designate the general relationship between force and deflection by the equation F = F(y) (a) then spring rate is defined as k(y) = lim y→0 F y = d F dy (4–1) where y must be measured in the direction of F and at the point of application of F. Most of the force-deflection problems encountered in this book are linear, as in Fig. 4–1a. For these, k is a constant, also called the spring constant; consequently Eq. (4–1) is written k = F y (4–2) We might note that Eqs. (4–1) and (4–2) are quite general and apply equally well for torques and moments, provided angular measurements are used for y. For linear dis￾placements, the units of k are often pounds per inch or newtons per meter, and for angular displacements, pound-inches per radian or newton-meters per radian. 4–2 Tension, Compression, and Torsion The total extension or contraction of a uniform bar in pure tension or compression, respectively, is given by δ = Fl AE (4–3) This equation does not apply to a long bar loaded in compression if there is a possibil￾ity of buckling (see Secs. 4–11 to 4–15). Using Eqs. (4–2) and (4–3), we see that the spring constant of an axially loaded bar is k = AE l (4–4) The angular deflection of a uniform round bar subjected to a twisting moment T was given in Eq. (3–35), and is θ = T l G J (4–5) where θ is in radians. If we multiply Eq. (4–5) by 180/π and substitute J = πd4/32 for a solid round bar, we obtain θ = 583.6T l Gd4 (4–6) where θ is in degrees. Equation (4–5) can be rearranged to give the torsional spring rate as k = T θ = G J l (4–7)

148 Budynas-Nisbett:Shigley's I.Basics 4.Deflection and Stiffness T©The McGraw-Hil Mechanical Engineering Companies,2008 Design,Eighth Edition 144 Mechanical Engineering Design 4-3 Deflection Due to Bending The problem of bending of beams probably occurs more often than any other loading problem in mechanical design.Shafts,axles,cranks,levers,springs,brackets,and wheels, as well as many other elements,must often be treated as beams in the design and analy- sis of mechanical structures and systems.The subject of bending,however,is one that you should have studied as preparation for reading this book.It is for this reason that we include here only a brief review to establish the nomenclature and conventions to be used throughout this book. The curvature of a beam subjected to a bending moment M is given by 1 M D-EI (4-8) where p is the radius of curvature.From studies in mathematics we also learn that the curvature of a plane curve is given by the equation dy/dx2 1+(dy/dx) (4-91 where the interpretation here is that y is the lateral deflection of the beam at any point x along its length.The slope of the beam at any point x is o= dx (a) For many problems in bending,the slope is very small,and for these the denominator of Eq.(4-9)can be taken as unity.Equation(4-8)can then be written El=dx 6 Noting Eqs.(3-3)and(3-4)and successively differentiating Eq.(b)yields v d3y El=dx3 (c g dy EI=dxi (d) It is convenient to display these relations in a group as follows: 品 dy (4-10) v dy EI=dx3 (4-11) M dy EIdx (4-12) dy (4-13) y=f(x) (4-14

Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition I. Basics 4. Deflection and Stiffness 148 © The McGraw−Hill Companies, 2008 144 Mechanical Engineering Design 4–3 Deflection Due to Bending The problem of bending of beams probably occurs more often than any other loading problem in mechanical design. Shafts, axles, cranks, levers, springs, brackets, and wheels, as well as many other elements, must often be treated as beams in the design and analy￾sis of mechanical structures and systems. The subject of bending, however, is one that you should have studied as preparation for reading this book. It is for this reason that we include here only a brief review to establish the nomenclature and conventions to be used throughout this book. The curvature of a beam subjected to a bending moment M is given by 1 ρ = M E I (4–8) where ρ is the radius of curvature. From studies in mathematics we also learn that the curvature of a plane curve is given by the equation 1 ρ = d2 y/dx2 [1 + (dy/dx)2]3/2 (4–9) where the interpretation here is that y is the lateral deflection of the beam at any point x along its length. The slope of the beam at any point x is θ = dy dx (a) For many problems in bending, the slope is very small, and for these the denominator of Eq. (4–9) can be taken as unity. Equation (4–8) can then be written M E I = d2 y dx2 (b) Noting Eqs. (3–3) and (3–4) and successively differentiating Eq. (b) yields V E I = d3 y dx3 (c) q E I = d4 y dx4 (d) It is convenient to display these relations in a group as follows: q E I = d4 y dx4 (4–10) V E I = d3 y dx3 (4–11) M E I = d2 y dx2 (4–12) θ = dy dx (4–13) y = f (x) (4–14)

Budynas-Nisbett:Shigley's I.Basics 4.Deflection and Stiffness T©The McGraw-Hil 149 Mechanical Engineering Companies,2008 Design,Eighth Edition Deflection and Stiffness 145 I Figure 4-2 1=20in -x Loading.w w=80 Ibf/in (a) R1= R,=号 Shear,V V=+800 lbf V=-8001bf (b) M—x Moment M (c) M=M=0 Ele Slope,El0 82=0 Ely Deflection,Ely y0=%=0 (e) The nomenclature and conventions are illustrated by the beam of Fig.4-2.Here,a beam of length /=20 in is loaded by the uniform load w=80 Ibf per inch of beam length. The x axis is positive to the right,and the y axis positive upward.All quantities- loading,shear,moment,slope,and deflection-have the same sense as y:they are pos- itive if upward,negative if downward. The reactions R1=R2=+800 lbf and the shear forces Vo=+800 Ibf and Vi=-800 Ibf are easily computed by using the methods of Chap.3.The bending moment is zero at each end because the beam is simply supported.For a simply- supported beam,the deflections are also zero at each end. EXAMPLE 4-1 For the beam in Fig.4-2,the bending moment equation,for 0sxsl.is M=- Using Eq.(4-12).determine the equations for the slope and deflection of the beam,the slopes at the ends,and the maximum deflection. Solution Integrating Eq.(4-12)as an indefinite integral we have E1 =∫Md=x2-”x+C1 4 1) dx 6 where Ci is a constant of integration that is evaluated from geometric boundary conditions. We could impose that the slope is zero at the midspan of the beam,since the beam and

Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition I. Basics 4. Deflection and Stiffness © The McGraw−Hill 149 Companies, 2008 Deflection and Stiffness 145 The nomenclature and conventions are illustrated by the beam of Fig. 4–2. Here, a beam of length l = 20 in is loaded by the uniform load w = 80 lbf per inch of beam length. The x axis is positive to the right, and the y axis positive upward. All quantities— loading, shear, moment, slope, and deflection—have the same sense as y; they are pos￾itive if upward, negative if downward. The reactions R1 = R2 = +800 lbf and the shear forces V0 = +800 lbf and Vl = −800 lbf are easily computed by using the methods of Chap. 3. The bending moment is zero at each end because the beam is simply supported. For a simply￾supported beam, the deflections are also zero at each end. Figure 4–2 (a) (b) (c) (d) (e) l = 20 in R1 = wl 2 R2 = wl 2 y Loading, w w = 80 lbf/in x + + – V0 M0 Ml Vl V M x – – + EI EI0 EIl x x Shear, V V0 = +800 lbf Vl = –800 lbf EIy x Deflection, EIy y0 = yl = 0 Moment, M M0 = Ml = 0 Slope, EI l/2 = 0 w EXAMPLE 4–1 For the beam in Fig. 4–2, the bending moment equation, for 0 ≤ x ≤ l, is M = wl 2 x − w 2 x2 Using Eq. (4–12), determine the equations for the slope and deflection of the beam, the slopes at the ends, and the maximum deflection. Solution Integrating Eq. (4–12) as an indefinite integral we have E I dy dx = M dx = wl 4 x2 − w 6 x3 + C1 (1) where C1 is a constant of integration that is evaluated from geometric boundary conditions. We could impose that the slope is zero at the midspan of the beam, since the beam and

150 Budynas-Nisbett:Shigley's I.Basics 4.Deflection and Stiffness T©The McGraw-Hill Mechanical Engineering Companies,2008 Design,Eighth Edition 146 Mechanical Engineering Design loading are symmetric relative to the midspan.However,we will use the given bound- ary conditions of the problem and verify that the slope is zero at the midspan.Integrating Eq.(1)gives w=-+G+G (2 The boundary conditions for the simply supported beam are y=0 atx=0 and I. Applying the first condition,y=0 atx=0,to Eq.(2)results in C2=0.Applying the second condition to Eq.(2)with C2=0, 0=-+G1=0 Solving for Ci yields C=-wl3/24.Substituting the constants back into Egs.(1)and (2)and solving for the deflection and slope results in y=24E72r2-x3-月 (3) 0= 在=24E76r2-43-月 dy w (4 Comparing Eq.(3)with that given in Table A-9,beam 7,we see complete agreement. For the slope at the left end,substitutingx=0 into Eq.(4)yields wl3 0lx=0=-24E1 and atx=, wl3 0lk=1=24E1 At the midspan,substitutingx=1/2 gives dy/dx=0,as earlier suspected. The maximum deflection occurs where dy/dx=0.Substituting x=1/2 into Eq.(3)yields 5wl4 ymax=一 384E1 which again agrees with Table A-9-7. The approach used in the example is fine for simple beams with continuous loading.However,for beams with discontinuous loading and/or geometry such as a step shaft with multiple gears,flywheels,pulleys,etc.,the approach becomes unwieldy.The following section discusses bending deflections in general and the techniques that are provided in this chapter. 4-4 Beam Deflection Methods Equations(4-10)through(4-14)are the basis for relating the intensity of loading q. vertical shear V,bending moment M,slope of the neutral surface 6,and the trans- verse deflection y.Beams have intensities of loading that range from q=constant

Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition I. Basics 4. Deflection and Stiffness 150 © The McGraw−Hill Companies, 2008 146 Mechanical Engineering Design loading are symmetric relative to the midspan. However, we will use the given bound￾ary conditions of the problem and verify that the slope is zero at the midspan. Integrating Eq. (1) gives EIy = M dx = wl 12 x3 − w 24 x4 + C1x + C2 (2) The boundary conditions for the simply supported beam are y = 0 at x = 0 and l. Applying the first condition, y = 0 at x = 0, to Eq. (2) results in C2 = 0. Applying the second condition to Eq. (2) with C2 = 0, EIy(l) = wl 12 l 3 − w 24 l 4 + C1l = 0 Solving for C1 yields C1 = −wl 3/24. Substituting the constants back into Eqs. (1) and (2) and solving for the deflection and slope results in y = wx 24E I (2lx2 − x3 − l 3 ) (3) θ = dy dx = w 24E I (6lx2 − 4x3 − l 3 ) (4) Comparing Eq. (3) with that given in TableA–9, beam 7, we see complete agreement. For the slope at the left end, substituting x = 0 into Eq. (4) yields θ|x=0 = − wl 3 24E I and at x = l, θ|x=l = wl 3 24E I At the midspan, substituting x = l/2 gives dy/dx = 0, as earlier suspected. The maximum deflection occurs where dy/dx = 0. Substituting x = l/2 into Eq. (3) yields ymax = − 5wl 4 384E I which again agrees with Table A–9–7. The approach used in the example is fine for simple beams with continuous loading. However, for beams with discontinuous loading and/or geometry such as a step shaft with multiple gears, flywheels, pulleys, etc., the approach becomes unwieldy. The following section discusses bending deflections in general and the techniques that are provided in this chapter. 4–4 Beam Deflection Methods Equations (4–10) through (4–14) are the basis for relating the intensity of loading q, vertical shear V, bending moment M, slope of the neutral surface θ, and the trans￾verse deflection y. Beams have intensities of loading that range from q = constant

Budynas-Nisbett:Shigley's I.Basics 4.Deflection and Stiffness T©The McGraw-Hil 151 Mechanical Engineering Companies,2008 Design,Eighth Edition Deflection and Stiffness 147 (uniform loading),variable intensity g(x),to Dirac delta functions (concentrated loads). The intensity of loading usually consists of piecewise contiguous zones,the expressions for which are integrated through Egs.(4-10)to (4-14)with varying degrees of difficulty.Another approach is to represent the deflection y(x)as a Fourier series,which is capable of representing single-valued functions with a finite number of finite discontinuities,then differentiating through Eqs.(4-14)to(4-10),and stopping at some level where the Fourier coefficients can be evaluated.A complication is the piecewise continuous nature of some beams(shafts)that are stepped-diameter bodies. All of the above constitute,in one form or another,formal integration methods, which,with properly selected problems,result in solutions for q.V,M.0,and y.These solutions may be 1 Closed-form,or 2 Represented by infinite series,which amount to closed form if the series are rapidly convergent,or 3 Approximations obtained by evaluating the first or the first and second terms. The series solutions can be made equivalent to the closed-form solution by the use of a computer.Roark's'formulas are committed to commercial software and can be used on a personal computer. There are many techniques employed to solve the integration problem for beam deflection.Some of the popular methods include: Superposition(see Sec.4-5) The moment-area method2 Singularity functions (see Sec.4-6) Numerical integration3 The two methods described in this chapter are easy to implement and can handle a large array of problems. There are methods that do not deal with Egs.(4-10)to(4-14)directly.An energy method,based on Castigliano's theorem,is quite powerful for problems not suitable for the methods mentioned earlier and is discussed in Secs.4-7 to 4-10.Finite element programs are also quite useful for determining beam deflections. 4-5 Beam Deflections by Superposition The results of many simple load cases and boundary conditions have been solved and are available.Table A-9 provides a limited number of cases.Roark's*provides a much more comprehensive listing.Superposition resolves the effect of combined loading on a structure by determining the effects of each load separately and adding Warren C.Young and Richard G.Budynas.Roark's Formulas for Stress and Strain.7th ed..McGraw-Hill. New York,2002. 2See Chap.9.F.P.Beer,E.R.Johnston Jr.,and J.T.DeWolf,Mechanics of Materials,4th ed..McGraw-Hill, New York,2006. 3See Sec.4-4,J.E.Shigley and C.R.Mischke,Mechanical Engineering Design,6th ed.,McGraw-Hill, New York,2001. Warren C.Young and Richard G.Budynas,Roark's Formulas for Stress and Strain.7th ed.,McGraw-Hill, New York,2002

Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition I. Basics 4. Deflection and Stiffness © The McGraw−Hill 151 Companies, 2008 Deflection and Stiffness 147 (uniform loading), variable intensity q(x), to Dirac delta functions (concentrated loads). The intensity of loading usually consists of piecewise contiguous zones, the expressions for which are integrated through Eqs. (4–10) to (4–14) with varying degrees of difficulty. Another approach is to represent the deflection y(x) as a Fourier series, which is capable of representing single-valued functions with a finite number of finite discontinuities, then differentiating through Eqs. (4–14) to (4–10), and stopping at some level where the Fourier coefficients can be evaluated. A complication is the piecewise continuous nature of some beams (shafts) that are stepped-diameter bodies. All of the above constitute, in one form or another, formal integration methods, which, with properly selected problems, result in solutions for q, V, M, θ, and y. These solutions may be 1 Closed-form, or 2 Represented by infinite series, which amount to closed form if the series are rapidly convergent, or 3 Approximations obtained by evaluating the first or the first and second terms. The series solutions can be made equivalent to the closed-form solution by the use of a computer. Roark’s1 formulas are committed to commercial software and can be used on a personal computer. There are many techniques employed to solve the integration problem for beam deflection. Some of the popular methods include: • Superposition (see Sec. 4–5) • The moment-area method2 • Singularity functions (see Sec. 4–6) • Numerical integration3 The two methods described in this chapter are easy to implement and can handle a large array of problems. There are methods that do not deal with Eqs. (4–10) to (4–14) directly. An energy method, based on Castigliano’s theorem, is quite powerful for problems not suitable for the methods mentioned earlier and is discussed in Secs. 4–7 to 4–10. Finite element programs are also quite useful for determining beam deflections. 4–5 Beam Deflections by Superposition The results of many simple load cases and boundary conditions have been solved and are available. Table A–9 provides a limited number of cases. Roark’s4 provides a much more comprehensive listing. Superposition resolves the effect of combined loading on a structure by determining the effects of each load separately and adding 1 Warren C. Young and Richard G. Budynas, Roark’s Formulas for Stress and Strain, 7th ed., McGraw-Hill, New York, 2002. 2 See Chap. 9, F. P. Beer, E. R. Johnston Jr., and J. T. DeWolf, Mechanics of Materials, 4th ed., McGraw-Hill, New York, 2006. 3 See Sec. 4–4, J. E. Shigley and C. R. Mischke, Mechanical Engineering Design, 6th ed., McGraw-Hill, New York, 2001. 4 Warren C. Young and Richard G. Budynas, Roark’s Formulas for Stress and Strain, 7th ed., McGraw-Hill, New York, 2002

152 Budynas-Nisbett:Shigley's I.Basics 4.Deflection and Stiffness T©The McGraw-Hil Mechanical Engineering Companies,2008 Design,Eighth Edition 148 Mechanical Engineering Design the results algebraically.Superposition may be applied provided:(1)each effect is linearly related to the load that produces it,(2)a load does not create a condition that affects the result of another load,and(3)the deformations resulting from any spe- cific load are not large enough to appreciably alter the geometric relations of the parts of the structural system. The following examples are illustrations of the use of superposition. EXAMPLE 4-2 Consider the uniformly loaded beam with a concentrated force as shown in Fig.4-3. Using superposition,determine the reactions and the deflection as a function of x. Solution Considering each load state separately,we can superpose beams 6 and 7 of Table A-9. For the reactions we find Fb wl Answer R1= 7+2 Answer Fa,wl R=+2 The loading of beam 6 is discontinuous and separate deflection equations are given for regions AB and BC.Beam 7 loading is not discontinuous so there is only one equa- tion.Superposition yields Fbx Answer yAB= 72+-乃+0Q2--内 Answer f2+2-2)+4e7r-- WX 6EIl I Figure 4-3 If we wanted to determine the maximum deflection in the previous example,we would set dy/dx =0 and solve for the value of x where the deflection is a maximum. If a=1/2,the maximum deflection would obviously occur at x=1/2 because of symmetry.However,if a<1/2,where would the maximum deflection be?It can be shown that as the force Fmoves toward the left support,the maximum deflection moves toward the left support also,but not as much as F(see Prob.4-34).Thus,we would set dygc/dx =0 and solve for x. Sometimes it may not be obvious that we can use superposition with the tables at hand,as demonstrated in the next example

Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition I. Basics 4. Deflection and Stiffness 152 © The McGraw−Hill Companies, 2008 148 Mechanical Engineering Design the results algebraically. Superposition may be applied provided: (1) each effect is linearly related to the load that produces it, (2) a load does not create a condition that affects the result of another load, and (3) the deformations resulting from any spe￾cific load are not large enough to appreciably alter the geometric relations of the parts of the structural system. The following examples are illustrations of the use of superposition. Figure 4–3 x y R1 F w a b l B A C R2 If we wanted to determine the maximum deflection in the previous example, we would set dy/dx = 0 and solve for the value of x where the deflection is a maximum. If a = l/2, the maximum deflection would obviously occur at x = l/2 because of symmetry. However, if a < l/2, where would the maximum deflection be? It can be shown that as the force F moves toward the left support, the maximum deflection moves toward the left support also, but not as much as F (see Prob. 4–34). Thus, we would set dyBC/dx = 0 and solve for x. Sometimes it may not be obvious that we can use superposition with the tables at hand, as demonstrated in the next example. EXAMPLE 4–2 Consider the uniformly loaded beam with a concentrated force as shown in Fig. 4–3. Using superposition, determine the reactions and the deflection as a function of x. Solution Considering each load state separately, we can superpose beams 6 and 7 of Table A–9. For the reactions we find Answer R1 = Fb l + wl 2 Answer R2 = Fa l + wl 2 The loading of beam 6 is discontinuous and separate deflection equations are given for regions AB and BC. Beam 7 loading is not discontinuous so there is only one equa￾tion. Superposition yields Answer yAB = Fbx 6EIl (x2 + b2 − l 2 ) + wx 24E I (2lx2 − x3 − l 3 ) Answer yBC = Fa(l − x) 6EIl (x2 + a2 − 2lx) + wx 24E I (2lx2 − x3 − l 3 )

Budynas-Nisbett:Shigley's I.Basics 4.Deflection and Stiffness T©The McGraw-Hill 153 Mechanical Engineering Companies,2008 Design,Eighth Edition Deflection and Stiffness 149 EXAMPLE 4-3 Consider the beam in Fig.44a and determine the deflection equations using superposition. Solution For region AB we can superpose beams 7 and 10 of Table A-9 to obtain Answer a=72--内+602-的 WX For region BC,how do we represent the uniform load?Considering the uniform load only,the beam deflects as shown in Fig.4-4b.Region BC is straight since there is no bending moment due to w.The slope of the beam at B is 68 and is obtained by taking the derivative of y given in the table with respect to x and setting x=1.Thus, 24E76r2-4r3-月 D Substituting x=I gives w/3 a=2476-4-A=4E7 The deflection in region BC due to w is0g(x-1).and adding this to the deflection due to F,in BC,yields Answer c=ga-0+f《ex-P-aB-】 Figure 4-4 lo)Beam with uniformly distributed load and overhang force;(b)deflections due to uniform load only. (a) (b) EXAMPLE 4-4 Figure 4-5a shows a cantilever beam with an end load.Normally we model this prob- lem by considering the left support as rigid.After testing the rigidity of the wall it was found that the translational stiffness of the wall was k,force per unit vertical deflection, and the rotational stiffness was k,moment per unit angular(radian)deflection (see Fig.4-5b).Determine the deflection equation for the beam under the load F

Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition I. Basics 4. Deflection and Stiffness © The McGraw−Hill 153 Companies, 2008 Deflection and Stiffness 149 EXAMPLE 4–3 Consider the beam in Fig. 4–4a and determine the deflection equations using superposition. Solution For region AB we can superpose beams 7 and 10 of Table A–9 to obtain Answer yAB = wx 24E I (2lx2 − x3 − l 3 ) + Fax 6EIl (l 2 − x2 ) For region BC, how do we represent the uniform load? Considering the uniform load only, the beam deflects as shown in Fig. 4–4b. Region BC is straight since there is no bending moment due to w. The slope of the beam at B is θB and is obtained by taking the derivative of y given in the table with respect to x and setting x = l. Thus, dy dx = d dx wx 24E I (2lx2 − x3 − l 3 )  = w 24E I (6lx2 − 4x3 − l 3 ) Substituting x = l gives θB = w 24E I (6ll2 − 4l 3 − l 3 ) = wl 3 24E I The deflection in region BC due to w is θB(x − l), and adding this to the deflection due to F, in BC, yields Answer yBC = wl 3 24E I (x − l) + F(x − l) 6E I [(x − l) 2 − a(3x − l)] x y R1 F w l a B A C R2 (a) x y w x l B B yBC = B(x – l) A C (b) Figure 4–4 (a) Beam with uniformly distributed load and overhang force; (b) deflections due to uniform load only. EXAMPLE 4–4 Figure 4–5a shows a cantilever beam with an end load. Normally we model this prob￾lem by considering the left support as rigid. After testing the rigidity of the wall it was found that the translational stiffness of the wall was kt force per unit vertical deflection, and the rotational stiffness was kr moment per unit angular (radian) deflection (see Fig. 4–5b). Determine the deflection equation for the beam under the load F.

154 Budynas-Nisbett:Shigley's I.Basics 4.Deflection and Stiffness T©The McGraw-Hil Mechanical Engineering Companies,2008 Design,Eighth Edition 150 I Mechanical Engineering Design Solution Here we will superpose the modes of deflection.They are:(1)translation due to the compression of spring k.(2)rotation of the spring k,.and(3)the elastic deformation of the beam given by Table A-9-1.The force in spring k,is R=F,giving a deflec- tion from Eq.(4-2)of F y=-k The moment in spring k,is M=Fl.This gives a clockwise rotation of0=Fl/k. Considering this mode of deflection only,the beam rotates rigidly clockwise,leading to a deflection equation of (2 Finally,the elastic deformation of the beam from Table A-9-1 is Fx2 为=6E7-30 (3) Adding the deflections from each mode yields Answer y= 6E7-30-F、F1 I Figure 4-5 (a) 4-6 Beam Deflections by Singularity Functions Introduced in Sec.3-3,singularity functions are excellent for managing discontinuities,and their application to beam deflection is a simple extension of what was presented in the ear- lier section.They are easy to program,and as will be seen later,they can greatly simplify the solution of statically indeterminate problems.The following examples illustrate the use of singularity functions to evaluate deflections of statically determinate beam problems

Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition I. Basics 4. Deflection and Stiffness 154 © The McGraw−Hill Companies, 2008 150 Mechanical Engineering Design Solution Here we will superpose the modes of deflection. They are: (1) translation due to the compression of spring kt , (2) rotation of the spring kr, and (3) the elastic deformation of the beam given by Table A–9–1. The force in spring kt is R1 = F, giving a deflec￾tion from Eq. (4–2) of y1 = − F kt (1) The moment in spring kr is M1 = Fl. This gives a clockwise rotation of θ = Fl/kr . Considering this mode of deflection only, the beam rotates rigidly clockwise, leading to a deflection equation of y2 = − Fl kr x (2) Finally, the elastic deformation of the beam from Table A–9–1 is y3 = Fx2 6E I (x − 3l) (3) Adding the deflections from each mode yields Answer y = Fx2 6E I (x − 3l) − F kt − Fl kr x 4–6 Beam Deflections by Singularity Functions Introduced in Sec. 3–3, singularity functions are excellent for managing discontinuities, and their application to beam deflection is a simple extension of what was presented in the ear￾lier section. They are easy to program, and as will be seen later, they can greatly simplify the solution of statically indeterminate problems. The following examples illustrate the use of singularity functions to evaluate deflections of statically determinate beam problems. x y R1 l F (a) M1 x kr kt F (b) R1 Figure 4–5