Budynas-Nisbett:Shigley's I.Basics 4.Deflection and Stiffness I©The McGraw-Hil 155 Mechanical Engineering Companies,2008 Design,Eighth Edition Deflection and Stiffness 151 EXAMPLE 4-5 Consider the beam of Table A-9-6.which is a simply supported beam having a con- centrated load F not in the center.Develop the deflection equations using singularity functions. Solution First,write the load intensity equation from the free-body diagram, q=R1x)-1-Fx-a)-1+R2K-1)-1 ) Integrating Eg.(1)twice results in V=R1x)°-F(x-a)°+R2(x-I)° (2 M=Ri(x)-F(x-a)+R2(x-1) 3 Recall that as long as the q equation is complete,integration constants are unnecessary for Vand M:therefore,they are not included up to this point.From statics,setting V=M=0 for x slightly greater than I yields R1 =Fb/l and R2=Fa/l.Thus Eq.(3) becomes M=中-F-a+c- Integrating Eqs.(4-12)and(4-13)as indefinite integrals gives 鉴--5-aP+-+G E=-专-a+-+G+G: Note that the first singularity term in both equations always exists,so (x)2=x2 and(x)3=x3.Also,the last singularity term in both equations does not exist until x=1,where it is zero,and since there is no beam forx>/we can drop the last term. Thus 2--专-r+G (4 =)x3-二(r-a)3+C1x+C, Ely=- 5) The constants of integration Ci and C2 are evaluated by using the two boundary con- ditions y=0 atx=0 andy=0 atx=I.The first condition,substituted into Eq.(5), gives C2=0(recall that(0-a)3=0).The second condition,substituted into Eq.(5). yields -6(-a)+Cl Fb2 Fb3 0=F_ 6-6+Cl Solving for C1, G=--的
Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition I. Basics 4. Deflection and Stiffness © The McGraw−Hill 155 Companies, 2008 Deflection and Stiffness 151 EXAMPLE 4–5 Consider the beam of Table A–9–6, which is a simply supported beam having a concentrated load F not in the center. Develop the deflection equations using singularity functions. Solution First, write the load intensity equation from the free-body diagram, q = R1x −1 − Fx − a −1 + R2x − l −1 (1) Integrating Eq. (1) twice results in V = R1x 0 − Fx − a 0 + R2x − l 0 (2) M = R1x 1 − Fx − a 1 + R2x − l 1 (3) Recall that as long as the q equation is complete, integration constants are unnecessary for V and M; therefore, they are not included up to this point. From statics, setting V = M = 0 for x slightly greater than l yields R1 = Fb/l and R2 = Fa/l. Thus Eq. (3) becomes M = Fb l x 1 − Fx − a 1 + Fa l x − l 1 Integrating Eqs. (4–12) and (4–13) as indefinite integrals gives E I dy dx = Fb 2l x 2 − F 2 x − a 2 + Fa 2l x − l 2 + C1 EIy = Fb 6l x 3 − F 6 x − a 3 + Fa 6l x − l 3 + C1x + C2 Note that the first singularity term in both equations always exists, so x2 = x2 and x3 = x3. Also, the last singularity term in both equations does not exist until x = l, where it is zero, and since there is no beam for x > l we can drop the last term. Thus E I dy dx = Fb 2l x2 − F 2 x − a 2 + C1 (4) EIy = Fb 6l x3 − F 6 x − a 3 + C1x + C2 (5) The constants of integration C1 and C2 are evaluated by using the two boundary conditions y = 0 at x = 0 and y = 0 at x = l. The first condition, substituted into Eq. (5), gives C2 = 0 (recall that 0 − a3 = 0). The second condition, substituted into Eq. (5), yields 0 = Fb 6l l 3 − F 6 (l − a) 3 + C1l = Fbl2 6 − Fb3 6 + C1l Solving for C1, C1 = − Fb 6l (l 2 − b2 )
15 Budynas-Nisbett:Shigley's I.Basics 4.Deflection and Stiffness T©The McGraw-Hill Mechanical Engineering Companies,2008 Design,Eighth Edition 152 Mechanical Engineering Design Finally,substituting Ci and C2 in Eq.(5)and simplifying produces y=6E7bx2+b2-月--a F (61 Comparing Eq.(6)with the two deflection equations in Table A-9-6,we note that the use of singularity functions enables us to express the deflection equation with a single equation. EXAMPLE 4-6 Determine the deflection equation for the simply supported beam with the load distrib- ution shown in Fig.4-6. Solution This is a good beam to add to our table for later use with superposition.The load inten- sity equation for the beam is q=R1x)-1-w(x)0+wx-a)°+R2K-)-1 where the w(x-a)o is necessary to"turn off'the uniform load atx=a. From statics,the reactions are R=-o)l =0a2 21 (2 For simplicity,we will retain the form of Eq.(1)for integration and substitute the values of the reactions in later. Two integrations of Eq.(1)reveal V=R1x)°-wx)'+wx-a)+R2x-° (3) M=R-2+-o2+R-y (4 As in the previous example,singularity functions of order zero or greater starting at x=0 can be replaced by normal polynomial functions.Also,once the reactions are determined,singularity functions starting at the extreme right end of the beam can be omitted.Thus,Eq.(4)can be rewritten as M=风-分+-aP (5 I Figure 4-6
Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition I. Basics 4. Deflection and Stiffness 156 © The McGraw−Hill Companies, 2008 152 Mechanical Engineering Design Figure 4–6 x y R1 w l a B A C R2 Finally, substituting C1 and C2 in Eq. (5) and simplifying produces y = F 6EIl [bx(x2 + b2 − l 2 ) − lx − a 3 ] (6) Comparing Eq. (6) with the two deflection equations in Table A–9–6, we note that the use of singularity functions enables us to express the deflection equation with a single equation. EXAMPLE 4–6 Determine the deflection equation for the simply supported beam with the load distribution shown in Fig. 4–6. Solution This is a good beam to add to our table for later use with superposition. The load intensity equation for the beam is q = R1x −1 − wx 0 + wx − a 0 + R2x − l −1 (1) where the wx − a0 is necessary to “turn off” the uniform load at x = a. From statics, the reactions are R1 = wa 2l (2l − a) R2 = wa2 2l (2) For simplicity, we will retain the form of Eq. (1) for integration and substitute the values of the reactions in later. Two integrations of Eq. (1) reveal V = R1x 0 − wx 1 + wx − a 1 + R2x − l 0 (3) M = R1x 1 − w 2 x 2 + w 2 x − a 2 + R2x − l 1 (4) As in the previous example, singularity functions of order zero or greater starting at x = 0 can be replaced by normal polynomial functions. Also, once the reactions are determined, singularity functions starting at the extreme right end of the beam can be omitted. Thus, Eq. (4) can be rewritten as M = R1x − w 2 x2 + w 2 x − a 2 (5)
Budynas-Nisbett:Shigley's I.Basics 4.Deflection and Stiffness ©The McGraw-Hil 157 Mechanical Engineering Companies,2008 Design,Eighth Edition Deflection and Stiffness 153 Integrating two more times for slope and deflection gives E1空=2-x+"x-a2+G (6) dx 6 6 =g-贺+员-a+G+6 Ely= 6 7 The boundary conditions are y=0 at x=0 and y=0 at x=1.Substituting the first condition in Eq.(7)shows C2=0.For the second condition +Cul 6 Solving for Ci and substituting into Eq.(7)yields E1y= g2-月-元t2-月-知0-a+0-a Finally,substitution of R from Eq.(2)and simplifying results gives D Answer y= 2ETI2ax(I-a)G2-P)-xl-P)-xl-a+ As stated earlier,singularity functions are relatively simple to program,as they are omitted when their arguments are negative,and the ()brackets are replaced with() parentheses when the arguments are positive. EXAMPLE 4-7 The steel step shaft shown in Fig.4-7a is mounted in bearings at A and F A pulley is centered at C where a total radial force of 600 Ibf is applied.Using singularity functions evaluate the shaft displacements at-in increments.Assume the shaft is simply supported. Solution The reactions are found to be R=360 lbf and R2=240 lbf.Ignoring R2.using singularity functions,the moment equation is M=360x-600(x-8)1 (1) This is plotted in Fig.4-7b. For simplification,we will consider only the step at D.That is,we will assume sec- tion AB has the same diameter as BC and section EF has the same diameter as DE. Since these sections are short and at the supports,the size reduction will not add much to the deformation.We will examine this simplification later.The second area moments for BC and DE are 高15=02485m1oc=175=0404n 64
Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition I. Basics 4. Deflection and Stiffness © The McGraw−Hill 157 Companies, 2008 Deflection and Stiffness 153 Integrating two more times for slope and deflection gives E I dy dx = R1 2 x2 − w 6 x3 + w 6 x − a 3 + C1 (6) EIy = R1 6 x3 − w 24 x4 + w 24 x − a 4 + C1x + C2 (7) The boundary conditions are y = 0 at x = 0 and y = 0 at x = l. Substituting the first condition in Eq. (7) shows C2 = 0. For the second condition 0 = R1 6 l 3 − w 24 l 4 + w 24(l − a) 4 + C1l Solving for C1 and substituting into Eq. (7) yields EIy = R1 6 x(x2 − l 2 ) − w 24 x(x3 − l 3 ) − w 24l x(l − a) 4 + w 24x − a 4 Finally, substitution of R1 from Eq. (2) and simplifying results gives Answer y = w 24EIl [2ax(2l − a)(x2 − l 2 ) − xl(x3 − l 3 ) − x(l − a) 4 + lx − a 4 ] As stated earlier, singularity functions are relatively simple to program, as they are omitted when their arguments are negative, and the brackets are replaced with ( ) parentheses when the arguments are positive. EXAMPLE 4–7 The steel step shaft shown in Fig. 4–7a is mounted in bearings at A and F. A pulley is centered at C where a total radial force of 600 lbf is applied. Using singularity functions evaluate the shaft displacements at 1 2 - in increments. Assume the shaft is simply supported. Solution The reactions are found to be R1 = 360 lbf and R2 = 240 lbf. Ignoring R2, using singularity functions, the moment equation is M = 360x − 600x − 8 1 (1) This is plotted in Fig. 4–7b. For simplification, we will consider only the step at D. That is, we will assume section AB has the same diameter as BC and section EF has the same diameter as DE. Since these sections are short and at the supports, the size reduction will not add much to the deformation. We will examine this simplification later. The second area moments for BC and DE are IBC = π 64 1.54 = 0.2485 in4 IDE = π 64 1.754 = 0.4604 in4
158 Budynas-Nisbett:Shigley's I.Basics 4.Deflection and Stiffness T©The McGraw-Hil Mechanical Engineering Companies,2008 Design,Eighth Edition 154 Mechanical Engineering Design Figure 4-7 600 Ibf 1.500 1.750 Dimensions in inches. 1000 D 1000 A 4 +0.5 8 5 19.5- (a) .20 2880 Ibf-in 2760 Ibf-in M c (c) A plot of M/I is shown in Fig.4-7c.The values at points band c.and the step change are 2760 =11106.61bf/im3 M 2760 0.2485 0.4604 =5994.8lbf/im3 △() =5994.8-11106.6=-5111.81bf/im3 The slopes for ab and cd,and the change are 360-600 mab 0.2485 =-965.81bf/im4 md=二594.8 11.5 =-521.31bf/im4 △m=-521.3-(-965.8)=444.51bf/im4 Dividing Eq.(1)by Igc and,at x=8.5 in,adding a step of-5 111.8 Ibf/in3 and a ramp of slope 444.5 Ibf/in,gives 7=1448.7x-2414.5-8)1-511.8-85)0+4445x-85) M (21 Integrating twice gives =724.35x2-1207.3x-8)2-5111.8-8.5) dx +222.3x-8.5)2+C1 (3) and Ey=241.5x3-402.4x-8)3-2555.9x-8.5)2+74.08(x-8.5)3+C1x+C2 (4 Atx=0,y=0.This gives C2=0(remember,singularity functions do not exist until the argument is positive).Atx=20 in,y =0,and 0=241.5(20)3-402.4(20-8)3-2555.9(20-8.5)2+74.08(20-8.5)3+C1(20)
Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition I. Basics 4. Deflection and Stiffness 158 © The McGraw−Hill Companies, 2008 154 Mechanical Engineering Design A plot of M/I is shown in Fig. 4–7c. The values at points b and c, and the step change are M I b = 2760 0.2485 = 11 106.6 lbf/in3 M I c = 2760 0.4604 = 5 994.8 lbf/in3 M I = 5 994.8 − 11 106.6 = −5 111.8 lbf/in3 The slopes for ab and cd, and the change are mab = 360 − 600 0.2485 = −965.8 lbf/in4 mcd = −5 994.8 11.5 = −521.3 lbf/in4 m = −521.3 − (−965.8) = 444.5 lbf/in4 Dividing Eq. (1) by IBC and, at x 8.5 in, adding a step of −5 111.8 lbf/in3 and a ramp of slope 444.5 lbf/in4, gives M I = 1 448.7x − 2 414.5x − 8 1 − 5 111.8x − 8.5 0 + 444.5x − 8.5 1 (2) Integrating twice gives E dy dx = 724.35x2 − 1207.3x − 8 2 − 5 111.8x − 8.5 1 +222.3x − 8.5 2 + C1 (3) and Ey = 241.5x3 − 402.4x − 8 3 − 2 555.9x − 8.5 2 + 74.08x − 8.5 3 + C1x + C2 (4) At x = 0, y = 0. This gives C2 = 0 (remember, singularity functions do not exist until the argument is positive). At x = 20 in, y = 0, and 0 = 241.5(20)3 − 402.4(20 − 8)3 − 2 555.9(20 − 8.5)2 + 74.08(20 − 8.5)3 + C1(20) M/I M a b D E B C A R2 F x c d 2880 lbf-in 2760 lbf-in 1.000 1.750 1.500 0.5 8 600 lbf 1.000 8.5 19.5 20 R1 y (a) (b) (c) Figure 4–7 Dimensions in inches.
Budynas-Nisbett:Shigley's I.Basics 4.Deflection and Stiffness ©The McGraw-Hil 159 Mechanical Engineering Companies,2008 Design,Eighth Edition Deflection and Stiffness 155 Solving,gives C1=-50 565 Ibf/in2.Thus,Eq.(4)becomes,with E=30(10)6 psi, y=3010241.5r3-4024-83-2555.90-852 +74.08x-8.5)3-50565x) (5) When using a spreadsheet,program the following equations: y 30109241.5r3-50565m 0≤x≤8in 1 y 3010241.5r3-402.4ex-83-50565] 8≤x≤8.5in y=3010241.5r3-402.4cx-83-255.9-8.52 +74.08(x-8.5)3-50565x] 8.5≤x≤20in The following table results. + X y 0 0.000000 4.5 -0.006851 9 -0.009335 13.5 -0.007001 18 -0.002377 0.5 -0.000842 5 -0.007421 9.5 -0.009238 14 -0.006571 18.5 -0.001790 1 -0.001677 5.5 -0.007931 10 -0.009096 14.5 -0.006116 19 -0.001197 1.5 -0.002501 6 -0.008374 10.5 -0.008909 15 -0.005636 19.5 -0.000600 2 -0.003307 6.5 -0.008745 11 -0.008682 15.5 -0.005134 20 0.000000 2.5 -0.004088 7 -0.009037 11.5 -0.008415 16 -0.004613 3 -0.004839 7.5 -0.009245 12 -0.008112 16.5 -0.004075 3.5 -0.005554 8 -0.009362 12.5 -0.007773 > -0.003521 4 -0.006227 8.5 -0.009385 13 -0.007403 17.5 -0.002954 where x and y are in inches.We see that the greatest deflection is atx=8.5 in,where y=-0.009385in. Substituting C into Eq.(3)the slopes at the supports are found to be=1.686(10-3) rad =0.09657 deg.and 0F =1.198(10-3)rad =0.06864 deg.You might think these to be insignificant deflections,but as you will see in Chap.7,on shafts,they are not. A finite-element analysis was performed for the same model and resulted in ylr=85n=-0.009380in0A=-0.09653°0r=0.06868° Virtually the same answer save some round-off error in the equations. If the steps of the bearings were incorporated into the model,more equations result, but the process is the same.The solution to this model is yx=85in=-0.009387in0A=-0.09763° 6r=0.06973° The largest difference between the models is of the order of 1.5 percent.Thus the sim- plification was justified
Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition I. Basics 4. Deflection and Stiffness © The McGraw−Hill 159 Companies, 2008 Deflection and Stiffness 155 Solving, gives C1 = −50 565 lbf/in2 . Thus, Eq. (4) becomes, with E = 30(10)6 psi, y = 1 30(106) (241.5x3 − 402.4x − 8 3 − 2 555.9x − 8.5 2 +74.08x − 8.5 3 − 50 565x) (5) When using a spreadsheet, program the following equations: y = 1 30(106) (241.5x3 − 50 565x) 0 ≤ x ≤ 8 in y = 1 30(106) [241.5x3 − 402.4(x − 8)3 − 50 565x] 8 ≤ x ≤ 8.5 in y = 1 30(106) [241.5x3 − 402.4 (x − 8) 3 − 2 555.9 (x − 8.5) 2 +74.08 (x − 8.5) 3 − 50 565x] 8.5 ≤ x ≤ 20 in The following table results. xy x y x y x y x y 0 0.000000 4.5 0.006851 9 0.009335 13.5 0.007001 18 0.002377 0.5 0.000842 5 0.007421 9.5 0.009238 14 0.006571 18.5 0.001790 1 0.001677 5.5 0.007931 10 0.009096 14.5 0.006116 19 0.001197 1.5 0.002501 6 0.008374 10.5 0.008909 15 0.005636 19.5 0.000600 2 0.003307 6.5 0.008745 11 0.008682 15.5 0.005134 20 0.000000 2.5 0.004088 7 0.009037 11.5 0.008415 16 0.004613 3 0.004839 7.5 0.009245 12 0.008112 16.5 0.004075 3.5 0.005554 8 0.009362 12.5 0.007773 17 0.003521 4 0.006227 8.5 0.009385 13 0.007403 17.5 0.002954 where x and y are in inches. We see that the greatest deflection is at x = 8.5 in, where y = −0.009385 in. SubstitutingC1 into Eq. (3) the slopes at the supports are found to be θA = 1.686(10−3) rad = 0.09657 deg, and θF = 1.198(10−3) rad = 0.06864 deg. You might think these to be insignificant deflections, but as you will see in Chap. 7, on shafts, they are not. A finite-element analysis was performed for the same model and resulted in y|x = 8.5 in = −0.009380 in θA = −0.09653◦ θF = 0.06868◦ Virtually the same answer save some round-off error in the equations. If the steps of the bearings were incorporated into the model, more equations result, but the process is the same. The solution to this model is y|x = 8.5 in = −0.009387 in θA = −0.09763◦ θF = 0.06973◦ The largest difference between the models is of the order of 1.5 percent. Thus the simplification was justified.