Beam Deflections: 4th Order Method and Additional Topics

Transcription

1 11 eam Deflections: 4th Order Method and dditional Topics 11 1

2 ecture 11: EM DEFECTIONS: 4TH ORDER METHOD ND DDITION TOICS TE OF CONTENTS age Fourth Order Method Description Example 1: Cantilever under Triangular Distributed oad Superposition Example : Cantilever Under To oad Cases Example 3: Staticall Indeterminate eam Continuit Conditions Example 4: Simpl Supported eam Under Midspan oint oad

3 11.1. Fourth Order Method Description 11.1 FOURTH ORDER METHOD DESCRITION The fourth-order method to find beam deflections gets its name from the order of the ODE to be integrated: EI zz v IV (x) = p(x) is a fourth order ODE. The procedure can be broken don into the folloing steps. 1. Express the applied load p(x) as function of x, using positive-upard convention. This step ma involve changing load signs as necessar, as in the example belo. 3. Integrate p(x) tice to get V (x) and M z (x) 4. ause. Determine integration constants from static Cs, and replace in M z (x). (If the constants are too complicated hen expressed in terms of the data, the might be kept in smbolic form until later.) 5 8. From here on, same as the second order method. n example of this technique follos Example 1: Cantilever under Triangular Distributed oad This example has been orked out in the previous lecture using the second order method. It is defined in Figure 11.1, hich reproduces the figure of the previous Chapter for convenience. Constant EI zz x (x) = x / Figure eam problem for Example 1. The applied load (x) = x/ is considered positive if it goes donard, that is, if > 0. This is converted to a negative load p (x) = x/ to insert in the ODEs. From inspection the applied load is p(x) = x. (11.1) Notice the minus sign to pass from the user s convention: (x) >0 if directed donard, to the generic load convention: p(x) >0 if directed upard. Integrating p(x) tice ields V (x) = p(x) dx = x + C 1, M z (x) = V (x) dx = (11.) x 3 6 C 1x + C. ppl no the static Cs at the free end : V = V (0) = C 1 = 0 and M z = M z (0) = C = 0. Hence M z (x) = 1 (x) x ( 1 3 x) = x 3 (11.3)

4 ecture 11: EM DEFECTIONS: 4TH ORDER METHOD ND DDITION TOICS Constant EI zz (x) = x / x (a) Original problem = x + x (b) Decomposition into to load cases and superposition Figure 11.. eam problem for Example. (x) = x / From here on the steps are the same as in the second order method orked out in ecture 10. The deflection curve is v(x) = 10EI zz (x x ) (11.4) The maximum deflection occurs at the cantilever tip, and is given b v = v(0) = 4 (11.5) 30EI zz The negative sign indicates that the beam deflects donard if > Superposition ll equations of the beam theor e are using are linear. This makes possible to treat complicated load cases b superposition of the solutions of simpler ones. Simple beam configurations and load cases ma be compiled in textbooks and handbooks; for example ppendix D of eer-johnston- DeWolf. The folloing example illustrates the procedure Example : Cantilever Under To oad Cases Consider the problem shon in Figure 11.(a). The cantilever beam is subject to a tip point force as ell as a triangular distributed load. This combination can be decomposed into the to load cases shon in Figure 11.(b). oth of these have been separatel solved previousl as Examples 1 and of ecture 10 (the latter also as the example in the previous section). The deflection curves for these cases ill be distinguished as v (x) and v (x), respectivel. We had obtained v (x) = ( x) ( + x), v (x) = 6EI zz 10EI zz (x x ) (11.6) 11 4

5 11. SUEROSITION Constant EI zz x (x) = x / M (a) Original (staticall indeterminate) problem R (b) Support reactions R = x + x (x) = x / R (c) Decomposition into to load cases and superposition Figure eam problem for Example 3. The deflection under the combined loading is obtained b adding the foregoing solutions: v(x) = v (x) + v (x) = ( x) ( + x) 6EI zz The tip deflection is 10EI zz (x x ). (11.7) v = v(0) = 3 4 = 3 (10 + ) (11.8) 3EI zz 30EI zz 30EI zz Superposition can be also used for an other quantit of interest, for example transverse shear forces, bending moments and deflection curve slopes. n application to staticall indeterminate beam analsis is given next Example 3: Staticall Indeterminate eam The problem is defined in Figure 11.3(a). The beam is simpl supported at and clamped at. If the supports are removed 3 reactions are activated: R, R and M, as pictured in Figure 11.3(b). ut there are onl to nontrivial static equilibrium equations: F = 0 and M an point = 0 because F x = 0 is triviall satisfied. Consequentl the beam is staticall indeterminate because the reactions cannot be determined b statics alone. One additional kinematic equation is required to complete the analsis. We select reaction R as redundant force to be carried along as a fictitious applied load. Removing the support at and including R makes the beam staticall determinate. See Figure 11.3(b). This beam ma be vieed as being loaded b a combination of to load cases: (1) the actual triangular load (x), and () a point load R at. ut this is exactl the problem solved in Example, if e replace b R. The deflection curve of this beam is v(x) = R 6EI zz ( x) ( + x) EI zz (x x ). (11.9)

6 ecture 11: EM DEFECTIONS: 4TH ORDER METHOD ND DDITION TOICS No the tip deflection must be zero because there is a simple support at. Setting v = v(0) = 0 provides the value of R : v = v(0) = R 3 4 = 0 R = 3EI zz 30EI zz 10 (11.10) This reaction value can be substituted to complete the solution. For example, the bending moment is M z (x) = R x x 3 6 = x x = x 30 (3 5x ) (11.11) The moment is zero at (x = 0), becomes positive for 0 < x < 3/5.7746, crosses zero at x = and reaches M z = /15 at the fixed end. The deflection is v(x) = hich ma be simplified to 60EI zz ( x) ( + x) 10EI zz (x x ), (11.1) v(x) = 10 EI zz x ( x ) (11.13) Continuit Conditions If the applied load is discontinuous, i.e., not a smooth function of x, it is necessar to divide the beam into segments separated b the discontinuit points. The ODEs are integrated over each segment. These solutions are patched b continuit conditions expressing that the slope v (x) and the deflection v(x) are continuous beteen segments. This matching results in extra relations beteen integration constants, hich permits elimination of all integration constants except those that can be determined b the standard Cs. The procedure is illustrated ith the next example Example 4: Simpl Supported eam Under Midspan oint oad The problem is defined in Figure 11.4(a). The calculation of the deflection curve ill be done b the second order method. Divide the beam into to segments: C, hich extends over 0 x 1, and C, hich extends over 1 x. For brevit, these are identified as segments 1 and, respectivel, in the equations belo. The expression of the bending moment over each segment is easil obtained from statics. From smmetr, the support reactions are obviousl R = R = 1 as shon in Figure 11.4(b). inspection one obtains that the bending moment M z (x), diagrammed in Figure 11.4(c), is M z1 (x) = x M z (x) = M z (x) = ( x) 11 6 over segment 1 (C), over segment (C). (11.14)

7 11.3 CONTINUITY CONDITIONS Constant EI zz x C segment 1 C segment / / (a) roblem definition R = R = (b) Support reactions and division into to segments R = M x z1 (x)= segment 1 C + + segment (c) ending moment diagram Figure eam problem for Example 4. R = M (-x) z(x)= Integrate M z /EI zz over each segment: EI EI zz v zz v x 1 (x) = (x) = 4 + C 1 over segment 1 (C), EI zz v (x) = x( x) 4 + Ĉ 1 over segment (C). (11.15) It is convenient to stop here and get rid of Ĉ 1 to avoid proliferation of integration constants. To do that, note that the midspan slope v C must be the same from both expressions: v C = v 1 ( 1 ) = v ( 1 ). Else the beam ould have a kink at C. This is called a continuit condition. Equating /16 + C 1 = (3/16) + Ĉ 1 ields Ĉ 1 = C 1 /8, hich is replaced in the second expression above: EI EI zz v zz v x 1 (x) = (x) = 4 + C 1 over segment 1 (C), (11.16) EI zz v (x) = x( x) C 1 over segment (C). No Ĉ 1 is gone. Integrate again both segments: EI zz v(x) = EI zz v 1 (x) = x3 1 + C 1 x + C over segment 1 (C), EI zz v (x) = x (3 x) 1 x 8 + C 1 x + Ĉ over segment (C). (11.17) To get rid of Ĉ e sa that the midspan deflection v C must be the same from both expressions: v C = v 1 ( 1 ) = v ( 1 ). This continuit condition gives Ĉ = C + 3 /48, hich replaced 11 7

8 ecture 11: EM DEFECTIONS: 4TH ORDER METHOD ND DDITION TOICS ields EI zz v(x) = EI zz v 1 (x) = x3 1 + C 1 x + C over segment 1 (C), EI zz v (x) = x (3 x) 1 x C 1 x + C over segment (C). (11.18) We have no onl to integration constants. To determine C 1 and C use the kinematic Cs at and. v = v 1 (0) = C = 0 and v = v () = 0 C 1 = /16. Substitution gives, after some simplifications, EI zz v 1 (x) = x EI zz v(x) = 48 (4x 3 ) over segment 1 (C), EI zz v (x) = 48 (4x 3 1x + 9 x 3 ) over segment (C). The midspan deflection, obtainable from either segment, is (11.19) v C = v 1 ( 1 ) = v ( 1 3 ) = (11.0) 48EI zz s can be seen the procedure is elaborate and error prone, even for this ver simple problem. It can be streamlined b using Discontinuit Functions (DFs), hich are covered in ecture