Module 1 Energy Methods in Structural Analysis Version 2 CE IIT, Kharagpur Lesson 4 Theorem of Least Work Version 2 CE IIT, Kharagpur Instructional Objectives After 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 Introduction In the last chapter the Castigliano’s theorems were discussed. In this chapter theorem of least work and reciprocal theorems are presented along with few selected problems. We know that for the statically determinate structure, the partial derivative of strain energy with respect to external force is equal to the displacement in the direction of that load at the point of application of load. This theorem when applied to the statically indeterminate structure results in the theorem of least work. 4.2 Theorem of Least Work According to this theorem, the partial derivative of strain energy of a statically indeterminate structure with respect to statically indeterminate action should vanish as it is the function of such redundant forces to prevent any displacement at its point of application. The forces developed in a redundant framework are such that the total internal strain energy is a minimum. This can be proved as follows. Consider a beam that is fixed at left end and roller supported at right end as 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 the displacements 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 redundant reaction, we obtain a simple cantilever beam as shown in Fig. 4.1b. Invoking the principle 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 redundant force Ra (see Fig. 4.2a and Fig. 4.2b) Version 2 CE IIT, Kharagpur Version 2 CE IIT, Kharagpur 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 discussed in Lesson 3. Since there is no load applied at A , apply a fictitious load Q at A as in Fig. 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 A removed) is given by, US = 1 1 1 1 P1u1 + P2 u 2 + .......... + Pn u n + Qu a 2 2 2 2 (4.1) It is known that the displacement u1 below point P1 is due to action of P1 , P2 ,...., Pn acting at x1 , x 2 ,......, x n respectively and due to Q at A . Hence, u1 may be expressed as, Version 2 CE IIT, Kharagpur u1 = a11 P1 + a12 P2 + .......... + a1n Pn + a1a Q (4.2) where, aij is the flexibility coefficient at i due to unit force applied at j . Similar equations may be written for u2 , u3 ,...., un and ua . Substituting for u2 , u3 ,...., un and ua in equation (4.1) from equation (4.2), we get, 1 1 P1[a11 P1 + a12 P2 + ... + a1n Pn + a1a Q ] + P2 [a21 P1 + a22 P2 + ...a2 n Pn + a2 a Q ] + ....... 2 2 (4.3) 1 1 + Pn [an1 P1 + an 2 P2 + ...ann Pn + ana Q] + Q[aa1 P1 + aa 2 P2 + .... + aan Pn + aaa Q] 2 2 US = Taking partial derivative of strain energy U s with respect to Q , we get deflection at A . ∂U s = aa1 P1 + aa 2 P2 + ........ + aan Pn + aaa Q ∂Q (4.4) Substitute Q = 0 as it is fictitious in the above equation, ∂U s = ua = aa1 P1 + aa 2 P2 + ........ + aan Pn ∂Q (4.5) Now the strain energy stored in the beam due to redundant reaction RA is, Ur = Ra2 L3 6 EI (4.6) Now deflection at A due to Ra is R L3 ∂U r = −ua = a ∂Ra 3EI (4.7) The deflection due to Ra should be in the opposite direction to one caused by superposed loads P1 , P2 ,...., Pn , so that the net deflection at A is zero. From equation (4.5) and (4.7) one could write, ∂U ∂Us = ua = − r ∂Q ∂Ra (4.8) Since Q is fictitious, one could as well replace it by Ra . Hence, Version 2 CE IIT, Kharagpur ∂ (U s + U r ) = 0 ∂Ra (4.9) ∂U =0 ∂Ra (4.10) or, This is the statement of theorem of least work. Where U is the total strain energy of the beam due to superimposed loads P1 , P2 ,...., Pn and redundant reaction Ra . Example 4.1 Find 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 its length. Version 2 CE IIT, Kharagpur There three reactions Ra , Rb and M b as shown in the figure. We have only two equation of equilibrium viz., ∑F y = 0 and ∑ M = 0 . This is a statically indeterminate structure and choosing Rb as the redundant reaction, we obtain a simple cantilever beam as shown in Fig. 4.3b. Now, the internal strain energy of the beam due to applied loads and redundant reaction, considering only bending deformations is, L U =∫ 0 M2 dx 2 EI (1) According to theorem of least work we have, Version 2 CE IIT, Kharagpur ∂U M ∂M =0=∫ ∂Rb EI ∂Rb 0 L Bending moment at a distance x from B , M = Rb x − (2) wx 2 2 ∂M =x ∂Rb (3) (4) Hence, ( R x − wx 2 / 2) x ∂U =∫ b dx ∂Rb 0 EI (5) ∂U ⎡ RB L3 wL4 ⎤ 1 =⎢ − =0 ⎥ ∂Rb ⎣ 3 8 ⎦ EI (6) L Solving for Rb , we get, 3 RB = wL 8 Ra = wL − Rb = wL2 5 wL and M a = − 8 8 (7) Example 4.2 A ring of radius R is loaded as shown in figure. Determine increase in the diameter AB of the ring. Young’s modulus of the material is E and second moment of the area is I about an axis perpendicular to the page through the centroid of the cross section. Version 2 CE IIT, Kharagpur Version 2 CE IIT, Kharagpur The free body diagram of the ring is as shown in Fig. 4.4. Due to symmetry, the slopes at C and D is zero. The value of redundant moment M 0 is such as to make slopes at C and D zero. The bending moment at any section θ of the beam is, M = M0 − PR (1 − cos θ ) 2 (1) Now strain energy stored in the ring due to bending deformations is, 2π U = M 2R ∫0 2 EI dθ (2) Due to symmetry, one could consider one quarter of the ring. According to theorem of least work, 2π M ∂M ∂U =0=∫ Rdθ 0 EI ∂M ∂M 0 0 (3) ∂M =1 ∂M 0 ∂U = ∂M 0 2π M ∫ EI Rdθ (4) 0 π 0= 4R 2 PR [M 0 − (1 − cosθ )] dθ ∫ EI 0 2 (5) Integrating and solving for M 0 , ⎛1 1⎞ M 0 = PR ⎜ − ⎟ ⎝2 π ⎠ (6) M 0 = 0.182 PR Now, increase in diameter Δ , may be obtained by taking the first partial derivative of strain energy with respect to P . Thus, Δ= ∂U ∂P Version 2 CE IIT, Kharagpur Now strain energy stored in the ring is given by equation (2). Substituting the value of M 0 and equation (1) in (2), we get, U= 2R EI π /2 PR 2 PR ( − 1) − (1 − cosθ )}2 dθ π 2 ∫{ 2 0 (7) Now the increase in length of the diameter is, ∂U 2 R = ∂P EI π /2 PR 2 PR R 2 R ( − 1) − (1 − cosθ )}{ ( − 1) − (1 − cosθ )}dθ π 2 2 π 2 ∫ 2{ 2 0 (8) After integrating, Δ= 4.3 PR 3 π 2 PR 3 { − ) = 0.149 EI 4 π EI (9) Maxwell–Betti Reciprocal theorem Consider a simply supported beam of span L as shown in Fig. 4.5. Let this beam be 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 the beam. Version 2 CE IIT, Kharagpur The reciprocal theorem states that the work done by forces acting through displacement of the second system is the same as the work done by the second system 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. Substituting the values of u12 and u 21 in equation (4.27) we get, P1 × 5 P2 L3 5 P L3 = P2 × 1 48 EI 48 EI (4.12) Hence it is proved. This is also valid even when the first system of forces is P1 , P2 ,...., Pn and the second system of forces is given by Q1 , Q2 ,...., Qn . Let u1 , u 2 ,...., u n be the displacements caused by the forces P1 , P2 ,...., Pn only and δ 1 , δ 2 ,...., δ n be the displacements due to system of forces Q1 , Q2 ,...., Qn only acting on the beam as shown in Fig. 4.6. Version 2 CE IIT, Kharagpur Now the reciprocal theorem may be stated as, Pi δ i = Qi u i i = 1,2,...., n (4.13) Summary In lesson 3, the Castigliano’s first theorem has been stated and proved. For statically determinate structure, the partial derivative of strain energy with respect to external force is equal to the displacement in the direction of that load at the point of application of the load. This theorem when applied to the statically indeterminate structure results in the theorem of Least work. In this chapter the theorem of Least Work has been stated and proved. Couple of problems is solved to illustrate the procedure of analysing statically indeterminate structures. In the Version 2 CE IIT, Kharagpur end, the celebrated theorem of Maxwell-Betti’s reciprocal theorem has been sated and proved. Version 2 CE IIT, Kharagpur
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