160 Budynas-Nisbett:Shigley's I.Basics 4.Deflection and Stiffness T©The McGraw-Hill Mechanical Engineering Companies,2008 Design,Eighth Edition 156 Mechanical Engineering Design In Sec.4-9,we will demonstrate the usefulness of singularity functions in solving statically indeterminate problems. 4-7 Strain Energy The external work done on an elastic member in deforming it is transformed into strain, or potential,energy.If the member is deformed a distance y,and if the force-deflection relationship is linear,this energy is equal to the product of the average force and the deflection,or F F2 0=2=9 2k (a) This equation is general in the sense that the force F can also mean torque,or moment, provided,of course,that consistent units are used for k.By substituting appropriate expressions for k,strain-energy formulas for various simple loadings may be obtained. For tension and compression and for torsion,for example,we employ Eqs.(4-4)and (4-7)and obtain F21 U二2AE tension and compression (4-15) U= T2 torsion (4-16) 2GJ To obtain an expression for the strain energy due to direct shear,consider the element with one side fixed in Fig.4-8a.The force F places the element in pure shear, and the work done is U=F8/2.Since the shear strain is y=8/1=/G=F/AG. we have U= F21 direct shear (4-17刀 2AG The strain energy stored in a beam or lever by bending may be obtained by refer- ring to Fig.4-8b.Here AB is a section of the elastic curve of length ds having a radius of curvature p.The strain energy stored in this element of the beam is dU=(M/2)d6. Since pde =ds,we have Mds dU= (6 2p Figure 4-8 dx+ (a)Pure shear element (b)Beam bending element
Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition I. Basics 4. Deflection and Stiffness 160 © The McGraw−Hill Companies, 2008 156 Mechanical Engineering Design In Sec. 4–9, we will demonstrate the usefulness of singularity functions in solving statically indeterminate problems. 4–7 Strain Energy The external work done on an elastic member in deforming it is transformed into strain, or potential, energy. If the member is deformed a distance y, and if the force-deflection relationship is linear, this energy is equal to the product of the average force and the deflection, or U = F 2 y = F2 2k (a) This equation is general in the sense that the force F can also mean torque, or moment, provided, of course, that consistent units are used for k. By substituting appropriate expressions for k, strain-energy formulas for various simple loadings may be obtained. For tension and compression and for torsion, for example, we employ Eqs. (4–4) and (4–7) and obtain U = F2l 2AE tension and compression (4–15) U = T 2l 2G J torsion (4–16) To obtain an expression for the strain energy due to direct shear, consider the element with one side fixed in Fig. 4–8a. The force F places the element in pure shear, and the work done is U = Fδ/2. Since the shear strain is γ = δ/l = τ/G = F/AG, we have U = F2l 2AG direct shear (4–17) The strain energy stored in a beam or lever by bending may be obtained by referring to Fig. 4–8b. Here AB is a section of the elastic curve of length ds having a radius of curvature ρ. The strain energy stored in this element of the beam is dU = (M/2)dθ. Since ρ dθ = ds, we have dU = M ds 2ρ (b) dx A O B ds d (a) Pure shear element (b) Beam bending element F F F l Figure 4–8
Budynas-Nisbett:Shigley's I.Basics 4.Deflection and Stiffness ©The McGraw-Hil 161 Mechanical Engineering Companies,2008 Design,Eighth Edition Deflection and Stiffness 157 We can eliminate p by using Eq.(4-8).Thus M2ds dU 2EI d For small deflections,ds=dx.Then,for the entire beam M2dx U= 2EI bending (4-18) Equation(4-18)is exact only when a beam is subject to pure bending.Even when shear is present,Eq.(4-18)continues to give quite good results,except for very short beams.The strain energy due to shear loading of a beam is a complicated problem.An approximate solution can be obtained by using Eq.(4-17)with a correction factor whose value depends upon the shape of the cross section.If we use C for the correction factor and V for the shear force,then the strain energy due to shear in bending is the integral of Eq.(4-17).or CV2dx U= bending shear (4-191 2AG Values of the factor C are listed in Table 4-1. Table 4-1 Beam Cross-Sectional Shape Factor C Strain-Energy Correction Factors for Shear Rectangular 1.2 Source:Richard G.Budynas, Circular 1.11 Advanced Strength and Thin-walled tubular,round 2.00 Applied Stress Anolysis Box sectionsT 1.00 2nd ed.,McGrawHill, New York,1999. Structural sectionst 1.00 Copyright©l999The McGraw-Hill Companies. tUserf web ony. EXAMPLE 4-8 Find the strain energy due to shear in a rectangular cross-section beam,simply sup- ported,and having a uniformly distributed load. Solution Using Appendix Table A-9-7.we find the shear force to be V=号-x wl Substituting into Eq.(4-19),with C=1.2,gives Answer
Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition I. Basics 4. Deflection and Stiffness © The McGraw−Hill 161 Companies, 2008 Deflection and Stiffness 157 We can eliminate ρ by using Eq. (4–8). Thus dU = M2 ds 2E I (c) For small deflections, ds . = dx. Then, for the entire beam U = M2 dx 2E I bending (4–18) Equation (4–18) is exact only when a beam is subject to pure bending. Even when shear is present, Eq. (4–18) continues to give quite good results, except for very short beams. The strain energy due to shear loading of a beam is a complicated problem. An approximate solution can be obtained by using Eq. (4–17) with a correction factor whose value depends upon the shape of the cross section. If we use C for the correction factor and V for the shear force, then the strain energy due to shear in bending is the integral of Eq. (4–17), or U = CV2 dx 2AG bending shear (4–19) Values of the factor C are listed in Table 4–1. Table 4–1 Strain-Energy Correction Factors for Shear Source: Richard G. Budynas, Advanced Strength and Applied Stress Analysis, 2nd ed., McGraw-Hill, New York, 1999. Copyright © 1999 The McGraw-Hill Companies. Beam Cross-Sectional Shape Factor C Rectangular 1.2 Circular 1.11 Thin-walled tubular, round 2.00 Box sections† 1.00 Structural sections† 1.00 † Use area of web only. EXAMPLE 4–8 Find the strain energy due to shear in a rectangular cross-section beam, simply supported, and having a uniformly distributed load. Solution Using Appendix Table A–9–7, we find the shear force to be V = wl 2 − wx Substituting into Eq. (4–19), with C = 1.2, gives Answer U = 1.2 2AG l 0 wl 2 − wx 2 dx = w2l 3 20AG
162 Budynas-Nisbett:Shigley's I.Basics 4.Deflection and Stiffness T©The McGraw-Hil Mechanical Engineering Companies,2008 Design,Eighth Edition 158 Mechanical Engineering Design EXAMPLE 4-9 A cantilever has a concentrated load F at the end.as shown in Fig.4-9.Find the strain energy in the beam by neglecting shear. Figure 4-9 Solution At any point x along the beam,the moment is M=-Fx.Substituting this value of M into Eg.(4-18),we find Answer F2x2 dx F213 U= 2EI 6EI 4-8 Castigliano's Theorem A most unusual,powerful,and often surprisingly simple approach to deflection analy- sis is afforded by an energy method called Castigliano's theorem.It is a unique way of analyzing deflections and is even useful for finding the reactions of indeterminate struc- tures.Castigliano's theorem states that when forces act on elastic systems subject to small displacements,the displacement corresponding to any force,in the direction of the force,is equal to the partial derivative of the total strain energy with respect to that force.The terms force and displacement in this statement are broadly interpreted to apply equally to moments and angular displacements.Mathematically,the theorem of Castigliano is 4= (4-201 aF where is the displacement of the point of application of the force Fi in the direction of F;.For rotational displacement Eg.(4-20)can be written as aU G二8Mi (4-21) where 6;is the rotational displacement,in radians,of the beam where the moment Mi exists and in the direction of Mi. As an example,apply Castigliano's theorem using Eqs.(4-15)and (4-16)to get the axial and torsional deflections.The results are 8F21 FI (a) a (T21 TI 2GJ GJ (6
Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition I. Basics 4. Deflection and Stiffness 162 © The McGraw−Hill Companies, 2008 158 Mechanical Engineering Design EXAMPLE 4–9 A cantilever has a concentrated load F at the end, as shown in Fig. 4–9. Find the strain energy in the beam by neglecting shear. Solution At any point x along the beam, the moment is M = −Fx . Substituting this value of M into Eq. (4–18), we find Answer U = l 0 F2x2 dx 2E I = F2l 3 6E I 4–8 Castigliano’s Theorem A most unusual, powerful, and often surprisingly simple approach to deflection analysis is afforded by an energy method called Castigliano’s theorem. It is a unique way of analyzing deflections and is even useful for finding the reactions of indeterminate structures. Castigliano’s theorem states that when forces act on elastic systems subject to small displacements, the displacement corresponding to any force, in the direction of the force, is equal to the partial derivative of the total strain energy with respect to that force. The terms force and displacement in this statement are broadly interpreted to apply equally to moments and angular displacements. Mathematically, the theorem of Castigliano is δi = ∂U ∂Fi (4–20) where δi is the displacement of the point of application of the force Fi in the direction of Fi . For rotational displacement Eq. (4–20) can be written as θi = ∂U ∂ Mi (4–21) where θi is the rotational displacement, in radians, of the beam where the moment Mi exists and in the direction of Mi . As an example, apply Castigliano’s theorem using Eqs. (4–15) and (4–16) to get the axial and torsional deflections. The results are δ = ∂ ∂F F2l 2AE = Fl AE (a) θ = ∂ ∂T T 2l 2G J = T l G J (b) ymax F l x Figure 4–9
Budynas-Nisbett:Shigley's I.Basics 4.Deflection and Stiffness T©The McGraw-Hil 163 Mechanical Engineering Companies,2008 Design,Eighth Edition Deflection and Stiffness 159 Compare Eqs.(a)and (b)with Eqs.(4-3)and(4-5).In Example 4-8,the bending strain energy for a cantilever having a concentrated end load was found.According to Castigliano's theorem,the deflection at the end of the beam due to bending is au a F23) FB y=aF=aF 6EI 3EI (d) which checks with Table A-9-1. Castigliano's theorem can be used to find the deflection at a point even though no force or moment acts there.The procedure is: 1 Set up the equation for the total strain energy U by including the energy due to a fictitious force or moment Oi acting at the point whose deflection is to be found. 2 Find an expression for the desired deflection 8,in the direction of i,by taking the derivative of the total strain energy with respect to i. 3 Since O;is a fictitious force,solve the expression obtained in step 2 by setting Oi equal to zero.Thus, aU 6= (4-22) a0ilo=0 EXAMPLE 4-10 The cantilever of Ex.4-9 is a carbon steel bar 10 in long with a 1-in diameter and is loaded by a force F=100 Ibf. (a)Find the maximum deflection using Castigliano's theorem,including that due to shear. (b)What error is introduced if shear is neglected? Solution (a)From Eq.(4-19)and Example 4-9 data,the total strain energy is Cv2 dx 6E7+ Jo 2AG 1 For the cantilever,the shear force is constant with repect to x V=F.Also,C=1.11, from Table 4-1.Performing the integration and substituting these values in Eq.(1) gives,for the total strain energy. F231.11F21 U= 6EI+2AG (2 Then,according to Castigliano's theorem,the deflection of the end is aUF131.11F1 y= aF=3EI AG (3) We also find that 1=IdT(ly 64 =0.0491in4 64 πd2π(1)2 A= 4三 =0.7854in2 4
Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition I. Basics 4. Deflection and Stiffness © The McGraw−Hill 163 Companies, 2008 Deflection and Stiffness 159 Compare Eqs. (a) and (b) with Eqs. (4–3) and (4–5). In Example 4–8, the bending strain energy for a cantilever having a concentrated end load was found. According to Castigliano’s theorem, the deflection at the end of the beam due to bending is y = ∂U ∂F = ∂ ∂F F2l 3 6E I = Fl3 3E I (c) which checks with Table A–9–1. Castigliano’s theorem can be used to find the deflection at a point even though no force or moment acts there. The procedure is: 1 Set up the equation for the total strain energy U by including the energy due to a fictitious force or moment Qi acting at the point whose deflection is to be found. 2 Find an expression for the desired deflection δi , in the direction of Qi , by taking the derivative of the total strain energy with respect to Qi . 3 Since Qi is a fictitious force, solve the expression obtained in step 2 by setting Qi equal to zero. Thus, δi = ∂U ∂ Qi Qi=0 (4–22) EXAMPLE 4–10 The cantilever of Ex. 4–9 is a carbon steel bar 10 in long with a 1-in diameter and is loaded by a force F = 100 lbf. (a) Find the maximum deflection using Castigliano’s theorem, including that due to shear. (b) What error is introduced if shear is neglected? Solution (a) From Eq. (4–19) and Example 4–9 data, the total strain energy is U = F2l 3 6E I + l 0 CV2 dx 2AG (1) For the cantilever, the shear force is constant with repect to x, V = F. Also, C = 1.11, from Table 4–1. Performing the integration and substituting these values in Eq. (1) gives, for the total strain energy, U = F2l 3 6E I + 1.11F2l 2AG (2) Then, according to Castigliano’s theorem, the deflection of the end is y = ∂U ∂F = Fl3 3E I + 1.11Fl AG (3) We also find that I = πd4 64 = π(1)4 64 = 0.0491 in4 A = πd2 4 = π(1)2 4 = 0.7854 in2
164 Budynas-Nisbett:Shigley's I.Basics 4.Deflection and Stiffness ©The McGraw-Hil Mechanical Engineering Companies,2008 Design,Eighth Edition 160 Mechanical Engineering Design Substituting these values,together with F=100 lbf,/=10 in,E =30 Mpsi,and G=11.5 Mpsi,in Eq.(3)gives Answer y=0.02263+0.00012=0.02275in Note that the result is positive because it is in the same direction as the force F. Answer (b)The error in neglecting shear for this problem is found to be about 0.53 percent. In performing any integrations,it is generally better to take the partial derivative with respect to the load Fi first.This is true especially if the force is a fictitious force Oi,since it can be set to zero as soon as the derivative is taken.This is demonstrated in the next example.The forms for deflection can then be rewritten.Here we will assume, for axial and torsional loading,that material and cross section properties and loading can vary along the length of the members.From Eqs.(4-15),(4-16),and (4-18), =() tension and compression (4-23) aU = torsion (4-24 aU bending (4-25) EXAMPLE 4-11 Using Castigliano's method,determine the deflections of points A and B due to the force F applied at the end of the step shaft shown in Fig.4-10.The second area moments for sections AB and BC are I and 21.respectively. Solution With cantilever beams we normally set up the coordinate system such that x starts at the wall and is directed towards the free end.Here,for simplicity,we have reversed that.With the coordinate system of Fig.4-10 the bending moment expression is simpler than with the usual coordinate system,and does not require the support reactions.For 0≤x≤l,the bending moment is M=-Fx (1) Since F is at A and in the direction of the desired deflection,the deflection at A from Eq.(4-25)is Figure 4-10 -x h 211
Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition I. Basics 4. Deflection and Stiffness 164 © The McGraw−Hill Companies, 2008 160 Mechanical Engineering Design Substituting these values, together with F = 100 lbf, l = 10 in, E = 30 Mpsi, and G = 11.5 Mpsi, in Eq. (3) gives Answer y = 0.022 63 + 0.000 12 = 0.022 75 in Note that the result is positive because it is in the same direction as the force F. Answer (b) The error in neglecting shear for this problem is found to be about 0.53 percent. F A B C l/2 l/2 I1 2I1 Qi x y Figure 4–10 In performing any integrations, it is generally better to take the partial derivative with respect to the load Fi first. This is true especially if the force is a fictitious force Qi , since it can be set to zero as soon as the derivative is taken. This is demonstrated in the next example. The forms for deflection can then be rewritten. Here we will assume, for axial and torsional loading, that material and cross section properties and loading can vary along the length of the members. From Eqs. (4–15), (4–16), and (4–18), δi = ∂U ∂Fi = 1 AE F ∂F ∂Fi dx tension and compression (4–23) θi = ∂U ∂ Mi = 1 G J T ∂T ∂ Mi dx torsion (4–24) δi = ∂U ∂Fi = 1 E I M ∂ M ∂Fi dx bending (4–25) EXAMPLE 4–11 Using Castigliano’s method, determine the deflections of points A and B due to the force F applied at the end of the step shaft shown in Fig. 4–10. The second area moments for sections AB and BC are I1 and 2I1, respectively. Solution With cantilever beams we normally set up the coordinate system such that x starts at the wall and is directed towards the free end. Here, for simplicity, we have reversed that. With the coordinate system of Fig. 4–10 the bending moment expression is simpler than with the usual coordinate system, and does not require the support reactions. For 0 ≤ x ≤ l, the bending moment is M = −Fx (1) Since F is at A and in the direction of the desired deflection, the deflection at A from Eq. (4–25) is