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 boundary 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 transverse 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 specific 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 equation. 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 problem 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 deflection 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 earlier 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