BEAMS: DEFORMATION BY SUPERPOSITION

Transcription

1 ETURE EMS: EFORMTION Y SUPERPOSITION Third Edition. J. lark School of Engineering epartment of ivil and Environmental Engineering 19 hapter b r. Ibrahim. ssakkaf SPRING 00 ENES 0 Mechanics of Materials epartment of ivil and Environmental Engineering Universit of Marland, ollege Park ETURE 19. EMS: EFORMTION Y SUPERPOSITION ( ) Slide No. 1 Method of Superposition When a beam is subjected to several loads (see Fig. 18) at various positions along the beam, the problem of determining the slope and the deflection usuall becomes quite involved and tedious. This is true regardless of the method used. Hoever, man comple loading conditions are merel combinations of

2 ETURE 19. EMS: EFORMTION Y SUPERPOSITION ( ) Slide No. Method of Superposition relativel simple loading conditions P 1 P a b Figure 18 (a) (b) ETURE 19. EMS: EFORMTION Y SUPERPOSITION ( ) Slide No. Method of Superposition ssumptions: The beam behaves elasticall for the combined loading. The beam also behaves elasticall for the each of the individual loads. Small deflection theor.

3 ETURE 19. EMS: EFORMTION Y SUPERPOSITION ( ) Slide No. 4 Method of Superposition If it is assumed that the beam behaves elasticall for the combined loading, as ell as for the individual loads, the resulting final deflection of the loaded beam is simpl the sum of the deflections caused b each of the individual loads. ETURE 19. EMS: EFORMTION Y SUPERPOSITION ( ) Slide No. 5 Method of Superposition This sum ma be an algebraic one (Figure 19) or it might be a vector sum as shon in Figure 0, the tpe depending on hether or not the individual deflection lie in the same plane. The superposition method can illustrated b various practical eamples.

4 ETURE 19. EMS: EFORMTION Y SUPERPOSITION ( ) Slide No. 6 Method of Superposition P P 1 Figure 19 a + P P P u b + + t ETURE 19. EMS: EFORMTION Y SUPERPOSITION ( ) Slide No. 7 Method of Superposition Principle of Superposition: eformations of beams subjected to combinations of loadings ma be obtained as the linear combination of the deformations from the individual loadings Procedure is facilitated b tables of solutions for common tpes of loadings and supports.

5 ETURE 19. EMS: EFORMTION Y SUPERPOSITION ( ) Slide No. 8 Method of Superposition Figure 0 P δ δ + δ δ z δ z z ETURE 19. EMS: EFORMTION Y SUPERPOSITION ( ) Slide No. 9 Illustrative Eample for the Use of Superposition Figure 1 onsider the beam shon in Fig. 1, ith a fleural rigidit of EI 100 MN m. m 150 kn 8 m 0 kn/m

6 ETURE 19. EMS: EFORMTION Y SUPERPOSITION ( ) Slide No. 10 Illustrative Eample for the Use of Superposition If e are interested on finding the slope and the deflection, sa of point, then e can use the superposition method to do that as illustrated in the folloing slides. First e find the slope and deflection due the effect of each load, i.e.,, P, etc. ETURE 19. EMS: EFORMTION Y SUPERPOSITION ( ) Slide No. 11 Illustrative Eample for the Use of Superposition The resulting final slope and deflection of point of the loaded beam is simpl the sum of the slopes and deflections caused b each of the individual loads as shon in Figure. We need to find both the slope and deflection caused b the concentrated load (10 kn) and distributed load (0 kn/m)

7 ETURE 19. EMS: EFORMTION Y SUPERPOSITION ( ) Slide No. 1 Illustrative Eample for the Use of Superposition 150 kn m 0 kn/m 150 kn m + 8 m 8 m 8 m 0 kn/m Figure. Original oading is roken into To Individual oads ETURE 19. EMS: EFORMTION Y SUPERPOSITION ( ) Slide No kn m Illustrative Eample for the Use of Superposition150 kn 8 m δ 0 kn/m m 8 m P + δ m 8 m 0 kn/m δ δ δ δ due to P + due to Figure. Original eflection is roken into To Individual eflections

8 ETURE 19. EMS: EFORMTION Y SUPERPOSITION ( ) Slide No. 14 Illustrative Eample for the Use of Superposition Slope and eflection caused b P either the direct integration or the singularit functions method, it can be seen that the slope and deflection (due to P) of point of this particular loaded beam are given, respectivel, as P P ( θ ) and ( ) P P EI 56EI ETURE 19. EMS: EFORMTION Y SUPERPOSITION ( ) Slide No. 15 Illustrative Eample for the Use of Superposition Slope and eflection caused b P Therefore, P P EI ( θ ) P ( ) ( )() 8 P 56EI (8) 0.00 rad 6 ( ) m 6 56( ) (5a) (5b)

9 ETURE 19. EMS: EFORMTION Y SUPERPOSITION ( ) Slide No. 16 Illustrative Eample for the Use of Superposition Slope and eflection caused b either the direct integration or the singularit functions method, it can be seen that the slope and deflection (due to ) of point of this particular loaded beam are given, respectivel, as ( θ ) ( ) (6a) P 4EI 4 ( ) ( + ) (6b) P 4EI ETURE 19. EMS: EFORMTION Y SUPERPOSITION ( ) Slide No. 17 Illustrative Eample for the Use of Superposition Slope and eflection caused b With 0 kn/m, m, and 8 m, thus ( ) ( ) θ P 4EI 4 ( ) ( + ) P 4EI ( ) ( 56 ) rad ( ) ( 91) m 6

10 ETURE 19. EMS: EFORMTION Y SUPERPOSITION ( ) Slide No. 18 Illustrative Eample for the Use of Superposition θ ombining the slopes and deflections produced b the concentrated (P) and distributed () loads, the results are ( θ ) + ( θ ) rad ( ) + ( ) m 16.6 mm P P ETURE 19. EMS: EFORMTION Y SUPERPOSITION ( ) Slide No kn m Illustrative Eample for the Use of Superposition 150 kn 0 kn/m m m 8 m 8 m δ ( ) P + θ ( ) P 0 kn/m 8 m δ δ ( ) θ θ ( θ ) p + ( θ ) and ( ) p + ( ) Figure 4. Total Slope and eflection of Point ( θ )

11 ETURE 19. EMS: EFORMTION Y SUPERPOSITION ( ) Slide No. 0 General Procedure of Superposition It is evident from the last results that the slope or deflection of a beam is the sum of the slopes or deflections produced b the individual loads. Once the slopes or deflections produced b a fe tpical individual loads have been determined b one of the methods alread ETURE 19. EMS: EFORMTION Y SUPERPOSITION ( ) Slide No. 1 General Procedure of Superposition Presented, the superposition method provides a means of quickl solving a ide range of more complicated problems b various combinations of knon results. s more data become available, et a ider range of problems can be solved b the method of superposition.

12 ETURE 19. EMS: EFORMTION Y SUPERPOSITION ( ) Slide No. Slope and eflection Tables To facilitate the task of practicing engineers, most structural and mechanical handbooks include tables giving the deflections and slopes of beams for various loadings and tpes of support. Such a table can be found in the tetbook (Table 19) and provided herein in the net fe viegraphs (Table 1 and ). ETURE 19. EMS: EFORMTION Y SUPERPOSITION ( ) Slide No. Slopes and eflection Tables Table 1a

13 ETURE 19. EMS: EFORMTION Y SUPERPOSITION ( ) Slide No. 4 Slopes and eflection Tables Table 1b ETURE 19. EMS: EFORMTION Y SUPERPOSITION ( ) Slide No. 5 Slopes and eflection Tables Table 1c

14 ETURE 19. EMS: EFORMTION Y SUPERPOSITION ( ) Slide No. 6 Slopes and eflection Tables Table 1d ETURE 19. EMS: EFORMTION Y SUPERPOSITION ( ) Slide No. 7 Slopes and eflection Tables Table a (eer and Johnston 199)

15 ETURE 19. EMS: EFORMTION Y SUPERPOSITION ( ) Slide No. 8 Slopes and eflection Tables Table b (eer and Johnston 199) ETURE 19. EMS: EFORMTION Y SUPERPOSITION ( ) Slide No. 9 Slopes and eflection Tables (eer and Johnston 199) Table c

16 ETURE 19. EMS: EFORMTION Y SUPERPOSITION ( ) Slide No. 0 Use of Slopes and eflection Tables Notice that the slope and deflection of the beam of Figures 1 and 4 (repeated here) of the illustrative eample could have been determined from the table (Table 1) Figure 1 m 150 kn 0 kn/m 8 m ETURE 19. EMS: EFORMTION Y SUPERPOSITION ( ) Slide No. 1 Use of Slopes and eflection Tables 150 kn m 150 kn 0 kn/m m m 8 m 8 m δ ( ) P + θ ( ) P 0 kn/m 8 m δ δ ( ) θ θ ( θ ) p + ( θ ) and ( ) p + ( ) Figure 4. Total Slope and eflection of Point ( θ )

17 ETURE 19. EMS: EFORMTION Y SUPERPOSITION ( ) Slide No. Use of Slopes and eflection Tables Indeed, given the information given under cases 5 and 6 of Tables c, the slope and deflection for an value /4 could have been epressed analticall. Taking the derivative of the epression obtained in this a, ould have ielded the slope of the beam over the same interval. ETURE 19. EMS: EFORMTION Y SUPERPOSITION ( ) Slide No. Use of Slopes and eflection Tables The slope at both ends of the beam ma be obtained b simpl adding the corresponding values given in the table. Hoever, the maimum deflection of the beam of Fig. 1 cannot be obtained b adding the maimum deflections of cases 5 and 6 (Table c), since these deflections occur at different points of the beam.

18 ETURE 19. EMS: EFORMTION Y SUPERPOSITION ( ) Slide No. 4 Use of Slopes and eflection Tables ppling case 5 on the illustrative eample to find both the slope and deflection of point of the beam (Fig. 1), ields Pb ( ) [ ( b ) ] 6EI d Pb ( θ ) [ ( b )] P P d 6EI (6) ( [ ( 8 6 )( ) ] 6 10 )() (6) ( 10 )() m [ () ( 8 6 )] 0.00 rad These values confirm the results obtained using Eq. 5 of the integration method. ETURE 19. EMS: EFORMTION Y SUPERPOSITION ( ) Slide No. 5 Eample 6 Use the method of superposition to find the slope and deflection at point of the beam.

19 ETURE 19. EMS: EFORMTION Y SUPERPOSITION ( ) Slide No. 6 Eample 6 (cont d) The given loading can be obtained b superposing the loadings shon in the folloing picture equation (Fig. 5). The beam is, of course, the same in each part of the figure. For each the loadings 1 and, the slope and deflection at can be determined b using the Tables 1 or. (Tetbook Table -19) ETURE 19. EMS: EFORMTION Y SUPERPOSITION ( ) Slide No. 7 Eample 6 (cont d) Figure 5 + oading 1 oading + θ ( ) θ ( ) 1 ( θ ) 1 ( )

20 ETURE 19. EMS: EFORMTION Y SUPERPOSITION ( ) Slide No. 8 Problem 6 (cont d) For the beam and loading shon, determine the slope and deflection at point. SOUTION: Superpose the deformations due to oading I and oading II as shon. ETURE 19. EMS: EFORMTION Y SUPERPOSITION ( ) Slide No. 9 ( θ ) Eample 6 (cont d) ( / ) + 6EI oading 1: From Table 1a or Table a (also Table -19 of the tetbook), 4 ( θ ) and ( ) (7a) 1 1 6EI 8EI oading : From the same tables: + 48EI ( / ) 4 and ( ) + 8EI 4 18EI (7b)

21 ETURE 19. EMS: EFORMTION Y SUPERPOSITION ( ) Slide No. 40 Eample 6 (cont d) Figure 6 ( θ ) oading ( θ ) In portion, the bending moment for loading is zero, thus the elastic curve is a straight line: ( θ ) ( θ ) + (8) 48EI ( ) ( ) + ( θ ) (9) Slope ( θ ) ( ) ( ) ( ) ( ) ETURE 19. EMS: EFORMTION Y SUPERPOSITION ( ) Slide No. 41 Eample 6 (cont d) Total slope and deflection: Slope of Point : θ ( θ ) + ( ) θ + 6EI 48EI EI eflection of Point : θ EI 48EI 84EI ( ) ( ) ( ) ( ) + ( ) EI 84EI 84EI

22 ETURE 19. EMS: EFORMTION Y SUPERPOSITION ( ) Slide No. 4 Eample 7 Use the method of superposition, determine the deflection at the free end of the cantilever beam shon in Fig. 7 in terms of,, E, and I. Figure 7 ETURE 19. EMS: EFORMTION Y SUPERPOSITION ( ) Slide No. 4 Eample 7 (cont d) Figure 8 oading 1 oading + + Straight ine δ θ ( δ ) 1 ( θ ) 1 ( δ ) ( θ )

23 ETURE 19. EMS: EFORMTION Y SUPERPOSITION ( ) Slide No. 44 Eample 7 (cont d) Using the solutions listed in Table 1a. ases 1 and (Tetbook Table -19) ith P δ ( δ ) + ( δ ) ( δ ) + ( δ ) + ( ) θ 1 P() EI ( ) EI 4 + 8EI 6EI EI 6EI 4EI 4 ETURE 19. EMS: EFORMTION Y SUPERPOSITION ( ) Slide No. 45 Eample 8 For the simpl supported beam of Fig. 9, use the method of superposition to determine the total deflection at point in terms of P,, E, and I. P P Figure 9 /4 /4

24 ETURE 19. EMS: EFORMTION Y SUPERPOSITION ( ) Slide No. 46 Eample 8 (cont d) /4 P P /4 Figure 0 oading 1 oading P P /4 /4 + /4 /4 From Table 1c (Tet -19) ase 6 / P center 48EI From Table 1b (Tet -19) ase 5 a / 4, b / 4 Pb( 4b ) center 48EI ETURE 19. EMS: EFORMTION Y SUPERPOSITION ( ) Slide No. 47 Eample 8 (cont d) Table 1b

25 ETURE 19. EMS: EFORMTION Y SUPERPOSITION ( ) Slide No. 48 Eample 8 (cont d) Table 1c ETURE 19. EMS: EFORMTION Y SUPERPOSITION ( ) Slide No. 49 Eample 8 (cont d) eflection due to oading 1: P 1 48EI ( ) eflection due to oading : [ ] Pb ( ) ( 4b ) P( / 4) 4( / 4) 11P 48EI 48EI 768 EI Therefore, total deflection of point ( ) + ( ) P 11P 48EI 768EI 1 9P 56EI

26 ETURE 19. EMS: EFORMTION Y SUPERPOSITION ( ) Slide No. 50 Eample 9 Using the method of superposition, find the deflection at a point mida beteen the supports of the beam shon in the figure in terms of,, E, and I. /4 Figure 1 / /4 ETURE 19. EMS: EFORMTION Y SUPERPOSITION ( ) Slide No. 51 Eample 9 (cont d) The deflection at a point mida beteen the supports can be determined b considering the beam shon in Fig.. Note that since the shear forces V and V do not contribute to the deflection at an point in span, the mid-span deflection can be epressed as δ δ M + δ mid (8) M

27 ETURE 19. EMS: EFORMTION Y SUPERPOSITION ( ) Slide No. 5 Eample 9 (cont d) M 8 / 4 V Figure /4 V 4 /4 M ETURE 19. EMS: EFORMTION Y SUPERPOSITION ( ) Slide No. 5 Eample 9 (cont d) Using the solutions listed in Table 1, Table, or Table -19 of the tetbook ith M /8 and M /16 δ mid δ + δ M M EI ( / 8)( ) ( /16)( ) 16EI + 16EI

28 ETURE 19. EMS: EFORMTION Y SUPERPOSITION ( ) Slide No. 54 Eample 9 (cont d) Table 1d ETURE 19. EMS: EFORMTION Y SUPERPOSITION ( ) Slide No. 55 Eample 10 For the beam in Fig., determine the fleural stress at point and the deflection of the left-hand end. 5 psi 6 in Figure 4 in P 600 lb 80 in E psi

29 ETURE 19. EMS: EFORMTION Y SUPERPOSITION ( ) Slide No. 56 Eample 10 (cont d) The stress at point is a combination of compressive fleural stress due to the concentrated load and a tensile fleural stress due to the distributed load, hence, M z σ I ( ) M ( ) [ 5( 4)( 80) / ]( ) I 4( 6) /1 z ( 80)( ) ( 4) /1, psi (compression) ETURE 19. EMS: EFORMTION Y SUPERPOSITION ( ) Slide No in Eample 10 (cont d) E psi 5 psi 4 in P 600 lb 80 in Figure 4 δ δ z z δ + δ δ z

30 ETURE 19. EMS: EFORMTION Y SUPERPOSITION ( ) Slide No. 58 Eample 10 (cont d) The deflection at the end of a cantilever beam ith uniforml distributed load is given b (see Table 1a, case ) 4 4 5( 4)( 80) 8EI z 8 [ /1] 6 (.4 10 ) 4( 6) in and ith concentrated load at the end is given b (see Table 1a, case 1) P 600( 80) z EI.4 [ /1] 6 ( 10 ) 6( 4) 0 1. in ETURE 19. EMS: EFORMTION Y SUPERPOSITION ( ) Slide No. 59 Eample 10 (cont d) Table 1a

31 ETURE 19. EMS: EFORMTION Y SUPERPOSITION ( ) Slide No. 60 Eample 10 (cont d) Superimposing the results for the deflections due to the concentrated and distributed loads, the deflection at the free end is the vector sum: δ + z 0 0 ( 0.566) + ( 1.) in ETURE 19. EMS: EFORMTION Y SUPERPOSITION ( ) Slide No. 61 Staticall Indeterminate Transversel oaded eams The Superposition Method The concept of the superposition, hich states that a slope or deflection due to several loads is the algebraic sum of the slopes or deflections to each individual loads acting alone can be applied to staticall indeterminate beams. The superposition can provide the additional equations needed in the analsis.

32 ETURE 19. EMS: EFORMTION Y SUPERPOSITION ( ) Slide No. 6 Staticall Indeterminate Transversel oaded eams The Superposition Method Procedure Selected restraints are removed and replaced b unknon loads, e.g., forces and couples. Sketching of the deformation (deflection) diagrams corresponding to individual loads (both knon and unknon). dding up algebraicall the individual of components of slopes or deflections to produce the knon configuration. ETURE 19. EMS: EFORMTION Y SUPERPOSITION ( ) Slide No. 6 Staticall Indeterminate Transversel oaded eams Illustrative Eample using Superposition etermine the reactions at the supports for the simpl supported cantilever beam (Fig.5) presented earlier for the integration method.

33 ETURE 19. EMS: EFORMTION Y SUPERPOSITION ( ) Slide No. 64 Staticall Indeterminate Transversel oaded eams Illustrative Eample using Superposition Method (cont d) First consider the reaction at as redundant and release the beam from the support (remove restraint). The reaction R is no considered as an unknon load (see Fig. 9) and ill be determined from the condition that the deflection at must be zero. ETURE 19. EMS: EFORMTION Y SUPERPOSITION ( ) Slide No. 65 Staticall Indeterminate Transversel oaded eams Illustrative Eample using Superposition Method (cont d) R + ( ) (a) (b) (c) R ( ) R Figure 9. Original oading is roken into To oads

34 ETURE 19. EMS: EFORMTION Y SUPERPOSITION ( ) Slide No. 66 pplication of Superposition to Staticall Indeterminate eams Method of superposition ma be applied to determine the reactions at the supports of staticall indeterminate beams. esignate one of the reactions as redundant and eliminate or modif the support. etermine the beam deformation ithout the redundant support. Treat the redundant reaction as an unknon load hich, together ith the other loads, must produce deformations compatible ith the original supports. ETURE 19. EMS: EFORMTION Y SUPERPOSITION ( ) Slide No. 67 Staticall Indeterminate Transversel oaded eams Illustrative Eample using Superposition Method (cont d) In reference to Table 1a cases 1 and (Table 19 of Tetbook): R EI ( ) + and ( ) 4 EI R 8 The deflection at in the original structural configuration must equal to zero, that is ( ) + ( ) 0 R (7) (8)

35 ETURE 19. EMS: EFORMTION Y SUPERPOSITION ( ) Slide No. 68 Staticall Indeterminate Transversel oaded eams Slopes and eflection Tables Table 1a ETURE 19. EMS: EFORMTION Y SUPERPOSITION ( ) Slide No. 69 Staticall Indeterminate Transversel oaded eams Illustrative Eample using Superposition Method (cont d) Substituting Eq. 7 into Eq. 8, gives 4 R + 0 (9) EI 8EI Solving for R, the result is R + 8 (40)

36 ETURE 19. EMS: EFORMTION Y SUPERPOSITION ( ) Slide No. 70 Staticall Indeterminate Transversel oaded eams Illustrative Eample using Superposition Method (cont d) From the free-bod diagram for entire beam (Figure 40), the equations of equilibrium are used to find the rest of the reactions. + F 0; R + R 0 R R (41) ETURE 19. EMS: EFORMTION Y SUPERPOSITION ( ) Slide No. 71 Staticall Indeterminate Transversel oaded eams Illustrative Eample using Superposition Method (cont d) R M / R R Figure 40. Free-bod iagram for the Entire eam

37 ETURE 19. EMS: EFORMTION Y SUPERPOSITION ( ) Slide No. 7 Staticall Indeterminate Transversel oaded eams Illustrative Eample using Superposition Method (cont d) ut R from Eq. 40, therefore 8 5 R (4) M 0; - M R + ( ) M R (4) 8 ETURE 19. EMS: EFORMTION Y SUPERPOSITION ( ) Slide No. 7 Staticall Indeterminate Transversel oaded eams Illustrative Eample using Superposition Method (cont d) From Eqs.40, 4, and 4, R M Which confirms the results found b using the integration method. R R

38 ETURE 19. EMS: EFORMTION Y SUPERPOSITION ( ) Slide No. 74 Staticall Indeterminate Transversel oaded eams Eample 1 beam is loaded and supported as shon in the figure. etermine (a) the reaction at supports and in terms of and, and (b) the deflection at the left end of the distributed load in terms of,, E, and I. ETURE 19. EMS: EFORMTION Y SUPERPOSITION ( ) Slide No. 75 Staticall Indeterminate Transversel oaded eams Eample 1 (cont d) δ θ δ θ δ P δ R θ R δ R + + θ P δ P θp R Figure 41 The portion in Figs. 41a and is a straight line. θ R (a) (b) (c)

39 ETURE 19. EMS: EFORMTION Y SUPERPOSITION ( ) Slide No. 76 Staticall Indeterminate Transversel oaded eams Eample 1 (cont d) δ Note that the portion of the beam in Figs. 41a and 41b is a straight line, therefore (a) Using the solution listed in Table 1a ith P δ + θ( ) + δ P + θp ( ) + δ R 0 4 ( ) ( ) ( ) ( ) ( ) ( ) R ( ) 8EI 6EI EI EI + EI 0 (44) ETURE 19. EMS: EFORMTION Y SUPERPOSITION ( ) Slide No. 77 Staticall Indeterminate Transversel oaded eams Slopes and eflection Tables Table 1a

40 R ETURE 19. EMS: EFORMTION Y SUPERPOSITION ( ) Slide No. 78 Staticall Indeterminate Transversel oaded eams F Eample 1 (cont d) From hich (Eq. 44), 8 R + 7 R M Equilibrium equations give + F 0; R P ( ) + R R 0 EI 70 R 7 R ETURE 19. EMS: EFORMTION Y SUPERPOSITION ( ) Slide No. 79 Staticall Indeterminate Transversel oaded eams F Eample 1 (cont d) + M 0; R ( ) P( ) ( )( ) 8 7 ( ) 4 M M R 9 (b) eflection at left end of distributed load (at ): 8 R R 7 8 M R δ δ R ( ) ( ) + M + M ( 8 / 7)( ) ( 8 / 7)( ) ( ) ( ) EI + δ M 7 + δ + P δ EI EI 8EI EI 4

41 ETURE 19. EMS: EFORMTION Y SUPERPOSITION ( ) Slide No. 80 Staticall Indeterminate Transversel oaded eams Slopes and eflection Tables Table 1b