Module 1. Energy Methods in Structural Analysis

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1 Module 1 Energy Methods in Structurl Anlysis

2 Lesson 4 Theorem of Lest Work

3 Instructionl Objectives After reding this lesson, the reder will be ble to: 1. Stte nd prove theorem of Lest Work.. Anlyse stticlly indeterminte structure. 3. Stte nd prove Mxwell-Betti s Reciprocl theorem. 4.1 Introduction In the lst chpter the Cstiglino s theorems were discussed. In this chpter theorem of lest work nd reciprocl theorems re presented long with few selected problems. We know tht for the stticlly determinte structure, the prtil derivtive of strin energy with respect to externl force is equl to the displcement in the direction of tht lod t the point of ppliction of lod. This theorem when pplied to the stticlly indeterminte structure results in the theorem of lest work. 4. Theorem of Lest Work According to this theorem, the prtil derivtive of strin energy of stticlly indeterminte structure with respect to stticlly indeterminte ction should vnish s it is the function of such redundnt forces to prevent ny displcement t its point of ppliction. The forces developed in redundnt frmework re such tht the totl internl strin energy is minimum. This cn be proved s follows. Consider bem tht is fixed t left end nd roller supported t right end s shown in Fig Let P P,..., be the forces cting t distnces 1, x 1, x,..., x n from the left end of the bem of spn L. Let u, 1 u,..., u n be the displcements t the loding points P P,..., respectively s shown in Fig , This is stticlly indeterminte structure nd choosing R s the redundnt rection, we obtin simple cntilever bem s shown in Fig. 4.1b. Invoking the principle of superposition, this my be treted s the superposition of two cses, viz, cntilever bem with lods P P,..., nd cntilever bem with redundnt force R (see Fig. 4. nd Fig. 4.b) 1, P n P n P n

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5 In the first cse (4.), obtin deflection below A due to pplied lods P, 1 P,..., P n. This cn be esily ccomplished through Cstiglino s first theorem s discussed in Lesson 3. Since there is no lod pplied t A, pply fictitious lod Q t A s in Fig. 4.. Let u be the deflection below A. Now the strin energy U s stored in the determinte structure (i.e. the support A removed) is given by, U S = Pu P u Pn un + Qu (4.1) u P1 1 It is known tht the displcement 1 below point is due to ction of P, P,..., Pn cting t x 1, x,..., x n respectively nd due to Q t A. Hence, u 1 my be expressed s,

6 u1 = 11P1+ 1P npn + 1 Q (4.) where, ij is the flexibility coefficient t i due to unit force pplied t j. Similr equtions my be written for u, u3,..., un nd u. Substituting for u, u 3,..., un nd u in eqution (4.1) from eqution (4.), we get, 1 1 US = P1 [ 11P1 + 1P npn + 1Q] + P [ 1P1 + P +... npn + Q] Pn [ n 1P1+ np +... nnpn + nq] + Q[ 1P1+ P npn + Q] (4.3) Tking prtil derivtive of strin energy t A. U s with respect to Q, we get deflection s Q = 1P1+ P P + Q n n (4.4) Substitute Q = s it is fictitious in the bove eqution, s Q = u = 1P1+ P np (4.5) n Now the strin energy stored in the bem due to redundnt rection R A is, U r 3 R L = (4.6) 6EI Now deflection t A due to R is r = u = R L 3EI 3 (4.7) The deflection due to R should be in the opposite direction to one cused by superposed lods P1, P,..., Pn, so tht the net deflection t A is zero. From eqution (4.5) nd (4.7) one could write, s Q = u = r (4.8) Since Q is fictitious, one could s well replce it by R. Hence,

7 ( Us + Ur) = (4.9) or, = (4.1) This is the sttement of theorem of lest work. Where U is the totl strin energy of the bem due to superimposed lods P, P,..., Pn nd redundnt rection R. Exmple 4.1 Find the rections of propped cntilever bem uniformly loded s shown in Fig Assume the flexurl rigidity of the bem EI to be constnt throughout its length. 1 Version CE IIT, Khrgpur

8 There three rections R, Rb nd M b s shown in the figure. We hve only two eqution of equilibrium viz., F = nd M = y. This is stticlly indeterminte structure nd choosing R b s the redundnt rection, we obtin simple cntilever bem s shown in Fig. 4.3b. Now, the internl strin energy of the bem due to pplied lods nd redundnt rection, considering only bending deformtions is, U = L M dx EI (1) According to theorem of lest work we hve,

9 b L Bending moment t distnce x from B, M M = = () EI b wx M = Rb x (3) M b = x (4) Hence, U L b b = ( R x wx EI / ) x dx (5) b RBL = 3 3 wl EI = (6) Solving for, we get, R b 3 R B = 8 wl R 5 = wl Rb = wl nd 8 M wl = (7) 8 Exmple 4. A ring of rdius R is loded s shown in figure. Determine increse in the dimeter AB of the ring. Young s modulus of the mteril is E nd second moment of the re is I bout n xis perpendiculr to the pge through the centroid of the cross section.

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11 The free body digrm of the ring is s shown in Fig Due to symmetry, the slopes t C nd D is zero. The vlue of redundnt moment M is such s to mke slopes t C nd D zero. The bending moment t ny section θ of the bem is, PR M = M (1 cosθ ) (1) Now strin energy stored in the ring due to bending deformtions is, π M R U = dθ () EI Due to symmetry, one could consider one qurter of the ring. According to theorem of lest work, M π M M = = Rdθ (3) EI M M M = 1 = 4R EI π [ M M = π M Rdθ EI PR (1 cosθ )] dθ (4) (5) Integrting nd solving for M, M 1 1 = PR π (6) M =. 18PR Now, increse in dimeter Δ, my be obtined by tking the first prtil derivtive of strin energy with respect to P. Thus, Δ= P

12 Now strin energy stored in the ring is given by eqution (). Substituting the vlue of nd eqution (1) in (), we get, M / = R π PR PR U { ( 1) (1 cosθ )} d EI π θ (7) Now the increse in length of the dimeter is, P / = R π PR PR R R { ( 1) (1 cosθ )}{ ( 1) (1 cosθ )} d EI π π θ (8) After integrting, 3 3 PR π PR Δ= { ) =.149 (9) EI 4 π EI 4.3 Mxwell Betti Reciprocl theorem Consider simply supported bem of spn L s shown in Fig Let this bem be loded by two systems of forces P1 nd P seprtely s shown in the figure. Let u1 be the deflection below the lod point P when only lod P1 is cting. Similrly let u1 be the deflection below lod P1, when only lod P is cting on the bem.

13 The reciprocl theorem sttes tht the work done by forces cting through displcement of the second system is the sme s the work done by the second system of forces cting through the displcements of the first system. Hence, ccording to reciprocl theorem, Now, u1 nd u the vlues of u 1 1 P (4.11) 1 u1 = P u1 cn be clculted using Cstiglino s first theorem. Substituting nd u in eqution (4.7) we get, P L 5P1 L P1 = P (4.1) 48EI 48EI Hence it is proved. This is lso vlid even when the first system of forces is P P,..., nd the second system of forces is given by Q Q,..., Q. Let, P, 1 n 1 n 1, u u n be the displcements cused by the forces P, 1 P,..., Pn, δ δ n Q, 1 Q,..., Q n u,..., only nd δ 1,..., be the displcements due to system of forces only cting on the bem s shown in Fig. 4.6.

14 Now the reciprocl theorem my be stted s, Pi δ i = Qi ui i = 1,,..., n (4.13) Summry In lesson 3, the Cstiglino s first theorem hs been stted nd proved. For stticlly determinte structure, the prtil derivtive of strin energy with respect to externl force is equl to the displcement in the direction of tht lod t the point of ppliction of the lod. This theorem when pplied to the stticlly indeterminte structure results in the theorem of Lest work. In this chpter the theorem of Lest Work hs been stted nd proved. Couple of problems is solved to illustrte the procedure of nlysing stticlly indeterminte structures. In the

15 end, the celebrted theorem of Mxwell-Betti s reciprocl theorem hs been sted nd proved.

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