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M1l4 Document Transcript

  • 1. Module 1Energy Methods inStructural Analysis Version 2 CE IIT, Kharagpur
  • 2. Lesson 4Theorem of Least Work Version 2 CE IIT, Kharagpur
  • 3. Instructional ObjectivesAfter reading this lesson, the reader will be able to:1. State and prove theorem of Least Work.2. Analyse statically indeterminate structure.3. State and prove Maxwell-Betti’s Reciprocal theorem.4.1 IntroductionIn the last chapter the Castigliano’s theorems were discussed. In this chaptertheorem of least work and reciprocal theorems are presented along with fewselected problems. We know that for the statically determinate structure, thepartial derivative of strain energy with respect to external force is equal to thedisplacement in the direction of that load at the point of application of load. Thistheorem when applied to the statically indeterminate structure results in thetheorem of least work.4.2 Theorem of Least WorkAccording to this theorem, the partial derivative of strain energy of a staticallyindeterminate structure with respect to statically indeterminate action shouldvanish as it is the function of such redundant forces to prevent any displacementat its point of application. The forces developed in a redundant framework aresuch that the total internal strain energy is a minimum. This can be proved asfollows. Consider a beam that is fixed at left end and roller supported at right endas shown in Fig. 4.1a. Let P1 , P2 ,...., Pn be the forces acting at distances x1 , x 2 ,......, x n from the left end of the beam of span L . Let u1 , u 2 ,..., u n be thedisplacements at the loading points P1 , P2 ,...., Pn respectively as shown in Fig. 4.1a.This is a statically indeterminate structure and choosing Ra as the redundantreaction, we obtain a simple cantilever beam as shown in Fig. 4.1b. Invoking theprinciple of superposition, this may be treated as the superposition of two cases,viz, a cantilever beam with loads P1 , P2 ,...., Pn and a cantilever beam with redundantforce Ra (see Fig. 4.2a and Fig. 4.2b) Version 2 CE IIT, Kharagpur
  • 4. Version 2 CE IIT, Kharagpur
  • 5. In the first case (4.2a), obtain deflection below A due to applied loads P1 , P2 ,...., Pn .This can be easily accomplished through Castigliano’s first theorem as discussedin Lesson 3. Since there is no load applied at A , apply a fictitious load Q at A as inFig. 4.2. Let u a be the deflection below A .Now the strain energy U s stored in the determinate structure (i.e. the support Aremoved) is given by, 1 1 1 1 US = P1u1 + P2 u 2 + .......... + Pn u n + Qu a (4.1) 2 2 2 2It is known that the displacement u1 below point P1 is due to action of P , P2 ,...., Pn 1acting at x1 , x 2 ,......, x n respectively and due to Q at A . Hence, u1 may beexpressed as, Version 2 CE IIT, Kharagpur
  • 6. u1 = a11 P + a12 P2 + .......... + a1n Pn + a1a Q 1 (4.2)where, aij is the flexibility coefficient at i due to unit force applied at j . Similarequations may be written for u2 , u3 ,...., un and ua . Substituting for u2 , u3 ,...., un and uain equation (4.1) from equation (4.2), we get, 1 1US = P [a11 P + a12 P2 + ... + a1n Pn + a1a Q ] + P2 [a21 P + a22 P2 + ...a2 n Pn + a2 a Q ] + ....... 1 1 1 2 2 (4.3) 1 1 + Pn [an1 P + an 2 P2 + ...ann Pn + ana Q] + Q[aa1 P + aa 2 P2 + .... + aan Pn + aaa Q] 1 1 2 2Taking partial derivative of strain energy U s with respect to Q , we get deflectionat A . ∂U s = aa1 P + aa 2 P2 + ........ + aan Pn + aaa Q (4.4) ∂Q 1Substitute Q = 0 as it is fictitious in the above equation, ∂U s = ua = aa1 P + aa 2 P2 + ........ + aan Pn (4.5) ∂Q 1Now the strain energy stored in the beam due to redundant reaction RA is, Ra L3 2 Ur = (4.6) 6 EINow deflection at A due to Ra is ∂U r R L3 = −ua = a (4.7) ∂Ra 3EIThe deflection due to Ra should be in the opposite direction to one caused bysuperposed loads P , P2 ,...., Pn , so that the net deflection at A is zero. From 1equation (4.5) and (4.7) one could write, ∂Us ∂U = ua = − r (4.8) ∂Q ∂RaSince Q is fictitious, one could as well replace it by Ra . Hence, Version 2 CE IIT, Kharagpur
  • 7. ∂ (U s + U r ) = 0 (4.9) ∂Raor, ∂U =0 (4.10) ∂RaThis is the statement of theorem of least work. Where U is the total strain energyof the beam due to superimposed loads P , P2 ,...., Pn and redundant reaction Ra . 1Example 4.1Find the reactions of a propped cantilever beam uniformly loaded as shown in Fig.4.3a. Assume the flexural rigidity of the beam EI to be constant throughout itslength. Version 2 CE IIT, Kharagpur
  • 8. There three reactions Ra , Rb and M b as shown in the figure. We have only twoequation of equilibrium viz., ∑F y = 0 and ∑ M = 0 . This is a staticallyindeterminate structure and choosing Rb as the redundant reaction, we obtain asimple cantilever beam as shown in Fig. 4.3b.Now, the internal strain energy of the beam due to applied loads and redundantreaction, considering only bending deformations is, L M2 U =∫ dx (1) 0 2 EIAccording to theorem of least work we have, Version 2 CE IIT, Kharagpur
  • 9. ∂U M ∂M L =0=∫ (2) ∂Rb 0 EI ∂Rb wx 2Bending moment at a distance x from B , M = Rb x − (3) 2 ∂M =x (4) ∂RbHence, ∂U ( R x − wx 2 / 2) x L =∫ b dx (5) ∂Rb 0 EI ∂U ⎡ RB L3 wL4 ⎤ 1 =⎢ − ⎥ =0 (6) ∂Rb ⎣ 3 8 ⎦ EISolving for Rb , we get, 3 RB = wL 8 5 wL2 Ra = wL − Rb = wL and M a = − (7) 8 8Example 4.2A ring of radius R is loaded as shown in figure. Determine increase in thediameter AB of the ring. Young’s modulus of the material is E and secondmoment of the area is I about an axis perpendicular to the page through thecentroid of the cross section. Version 2 CE IIT, Kharagpur
  • 10. Version 2 CE IIT, Kharagpur
  • 11. The free body diagram of the ring is as shown in Fig. 4.4. Due to symmetry, theslopes at C and D is zero. The value of redundant moment M 0 is such as to makeslopes at C and D zero. The bending moment at any section θ of the beam is, PR M = M0 − (1 − cos θ ) (1) 2Now strain energy stored in the ring due to bending deformations is, 2π M 2R U = ∫ 2 EI dθ 0 (2)Due to symmetry, one could consider one quarter of the ring. According totheorem of least work, ∂U 2π M ∂M =0=∫ Rdθ (3) ∂M 0 0 EI ∂M 0 ∂M =1 ∂M 0 2π ∂U M ∂M 0 = ∫ EI Rdθ 0 (4) π 4R 2 PR 0= ∫ [M 0 − 2 (1 − cosθ )] dθ EI 0 (5)Integrating and solving for M 0 , ⎛1 1⎞ M 0 = PR ⎜ − ⎟ (6) ⎝2 π ⎠ M 0 = 0.182 PRNow, increase in diameter Δ , may be obtained by taking the first partial derivativeof strain energy with respect to P . Thus, ∂U Δ= ∂P Version 2 CE IIT, Kharagpur
  • 12. Now strain energy stored in the ring is given by equation (2). Substituting the valueof M 0 and equation (1) in (2), we get, π /2 2R PR 2 PR U= EI ∫{ 2 0 ( − 1) − π 2 (1 − cosθ )}2 dθ (7)Now the increase in length of the diameter is, π /2 ∂U 2 R PR 2 PR R 2 R = ∂P EI ∫ 2{ 2 0 ( − 1) − π 2 (1 − cosθ )}{ ( − 1) − (1 − cosθ )}dθ 2 π 2 (8)After integrating, PR 3 π 2 PR 3 Δ= { − ) = 0.149 (9) EI 4 π EI4.3 Maxwell–Betti Reciprocal theoremConsider a simply supported beam of span L as shown in Fig. 4.5. Let this beambe loaded by two systems of forces P1 and P2 separately as shown in the figure.Let u 21 be the deflection below the load point P2 when only load P1 is acting.Similarly let u12 be the deflection below load P1 , when only load P2 is acting on thebeam. Version 2 CE IIT, Kharagpur
  • 13. The reciprocal theorem states that the work done by forces acting throughdisplacement of the second system is the same as the work done by the secondsystem of forces acting through the displacements of the first system. Hence,according to reciprocal theorem, P1 × u12 = P2 × u 21 (4.11)Now, u12 and u 21 can be calculated using Castiglinao’s first theorem. Substitutingthe values of u12 and u 21 in equation (4.27) we get, 5 P2 L3 5 P L3 P1 × = P2 × 1 (4.12) 48 EI 48 EIHence it is proved. This is also valid even when the first system of forces isP1 , P2 ,...., Pn and the second system of forces is given by Q1 , Q2 ,...., Qn . Letu1 , u 2 ,...., u n be the displacements caused by the forces P1 , P2 ,...., Pn only andδ 1 , δ 2 ,...., δ n be the displacements due to system of forces Q1 , Q2 ,...., Qn only actingon the beam as shown in Fig. 4.6. Version 2 CE IIT, Kharagpur
  • 14. Now the reciprocal theorem may be stated as, Pi δ i = Qi u i i = 1,2,...., n (4.13)SummaryIn lesson 3, the Castigliano’s first theorem has been stated and proved. Forstatically determinate structure, the partial derivative of strain energy with respectto external force is equal to the displacement in the direction of that load at thepoint of application of the load. This theorem when applied to the staticallyindeterminate structure results in the theorem of Least work. In this chapter thetheorem of Least Work has been stated and proved. Couple of problems is solvedto illustrate the procedure of analysing statically indeterminate structures. In the Version 2 CE IIT, Kharagpur
  • 15. end, the celebrated theorem of Maxwell-Betti’s reciprocal theorem has been satedand proved. Version 2 CE IIT, Kharagpur