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Energy methods
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- 2. • Concept ofstrain energy • Strain energy stored due to axial forces • Strain energy stored due to bending stresses. • Maxwell’s reciprocal deflection theorem or law of reciprocal deflections • Betti’s law • Castigliano’s theorem-I • Castigliano’s theorem-II • Unit load method. • Analysis of trusses having redundancy 1 and more than 1 • Lack of fit in indeterminate frames • Temperature stresses in redundant frames 2v shiva
- 3. STRAIN ENERGY • Whenan elastic member is deformed under the action of an external loading the member is said to have possessed or stored energy which is called the strain energy of the member or the resilience of the member. • Strain energy stored by a member = work done by the external forces to produce deformation. • If the elastic limit is not exceeded the external work is transformed into elastic strain energy, which is recoverable as member returns to its original condition as the load is removed. 3v shiva
- 4. STRAIN ENERGY STOREDDUE TO AXIAL forces An elastic member of Length = l; C/S area = A; Subjected to external axial load W. Extension due to load = δ; External work done We = ½.W.δ strain energy stored = Wi; External work done = strain energy stored. (We = Wi) Let tension in member = S For equilibrium,W = S; Tensile stress, σ = S/A; ε = σ/E; δ = S.l/A.E We = 1/2 W.δ = 1/2(S.S.l/AE)= S2l/2A.E We = S2l/2A.E. (strain energy stored due to axial loading on the member). Strain energy stored per unit volume of the member = (S2l/2A.E)/A.l = S2/2A2.E = σ2/2E 4v shiva
- 5. • Strain energystored due to bending: • Consider a beam subjected to uniform moment M. • Consider elemental length ds of the beam between two sections 1-1 and 2-2. • In this elemental length consider infinite number of elemental cylinders of length ds and C/S area da. • Intensity of stress in elemental cylinder, σ = M.y/I • Energy stored by the elemental cylinder = energy stored per unit volume X volume of the cylinder. • = (σ2/2E) X ds.da • = (M2 y2/2 I2E) X ds.da • = Σ(M2 y2/2 I2E) X ds.da • = (M2 ds / 2 I2E) X Σda .y2 • energy stored by ds length of beam = (M2 ds / 2 IE) • Total energy stored by whole beam = ∫(M2 / 2 IE)ds 5v shiva
- 6. Find the deflectionof the free end of a cantilever carrying a concentrated load P at the free end EI pl EI lp p p EI lp dx EI xp l 3 6 . 2 1 . 2 1 loadexternalby thedonework 62 3 32 32 0 22 = = = = = EI dxM 2 Wi,cantileverbystoredenergystrain 2 6v shiva
- 7. Find the deflectionof a simply supported beam carrying a concentrated load P at mid-span EI pl EI lp p p EI lp dx EI xp l 48 96 . 2 1 . 2 1 loadexternalby thedonework 962 2 . 2 3 32 322/ 0 2 = = = = = = EI dxM 2 Wibeam,by wholestoredenergystrain 2 7v shiva
- 8. A portal frameABCD has its end A hinged while the end D is placed on rollers. A horizontal force P is applied on the end D as shown fig. determine movement of D. Assume all members have the same flexural rigidity. )32( 3 )32( 6 . 2 1 . 2 1 loadexternalby thedonework .beDofmovementhorizontal )32( 626 .2 2 )( 2 )( 2 beambystoredenergycolumnsbystoredenergystrain Wiframe,bystoredenergystrain 2 22 222232 0 2 0 2 bh EI ph bh EI hp p p bh EI hp EI bhp EI hp dx EI ph dy EI py bh += += = +=+= += += = 8v shiva
- 9. Work done bya force on a member • If a load acts on a member and produces a deflection δ in its line of action by virtue of its own direct action, the work done by the load W = ½.W.δ. • If a member subjected to a load K is given a deformation y in the line of action of K by virtue of some other external agency, the work done by the load K due to the above deformation = Ky. 9v shiva
- 10. MAXWELL’S RECIPROCAL DEFLECTIONTHEOREM OR LAW OF RECIPROCAL DEFLECTIONS • James Clerk Maxwell, a Cavendish professor at Cambridge University offered this law in 1864. • In any beam or truss the deflection at any point D due to a load W at any other point C is the same as the deflection at C due to the same load W applied at D. • Fig(i) shows a structure AB carrying load W at C. Due to this load, deflection at point C = ∆c and deflection at any other point D = ∆d • Work done on the structure = 1/2.W. ∆c 10v shiva
- 11. • Now fig(ii)shows the same structure AB loaded with same load(W) at points C and D. • Now there will be further deflection of δc and δd at C and D respectively. • Total work done = 1/2.W.∆c + 1/2.W.δd + W.δc 11v shiva
- 12. • Fig(iii) showsa structure AB carrying load W at D. Due to this load, deflection at point D = δd and deflection at any other point C = δc • Work done on the structure = 1/2.W. δd 12v shiva
- 13. • Now fig(iv)shows the same structure AB loaded with same load(W) at points C and D. • Now there will be further deflection of ∆c and ∆d at C and D respectively. • Total work done = 1/2.W.δd + 1/2.W.∆c + W.∆d 13v shiva
- 14. • Equating thetwo expressions obtained for the total work done when both the loads are present on the structure, we have = 1/2.W.∆c + 1/2.W.δd + W.δc = 1/2.W.δd + 1/2.W.∆c + W.∆d W.δc = W.∆d ∴ δc = ∆d i.e., The deflection at C due to the load W at D = The deflection at D due to the same load W at C. 14v shiva
- 15. Maxwell’s law inthe most general form is stated as follows; δc = ∆d In any structure whose material is elastic and obeys Hooke’s law and in which the supports remain unyielding and the temperature remains unchanged, the deflection of any point 1 in a direction ab due to a load W acting at a point 2 acting in a direction cd is numerically equal to the deflection of the point 2 in the direction cd due to load W acting at the point 1 in the direction ab. 15v shiva
- 16. BETTI’S LAW This isgeneralized form of Maxwell’s law offered by Italian professor, E.Betti in 1872. The law is stated as follows: In any structure whose material is elastic and obeys Hooke’s law and in which the supports remain unyielding and the temperature remains constant, the virtual work done by a system of forces P1, P2, P3 …. during the distortion caused by a system of forces Q1, Q2, Q3 … is equal to the virtual work done by the system of forces Q1, Q2, Q3 …during the distortion caused by the system of forces, P1, P2, P3 … 16v shiva
- 17. In any structurewhose materialis elastic and obeys Hooke’s law and in which the supports remain unyielding and the temperatureremains constant, the virtual work done by a system of forces P1, P2, P3 …. during the distortion caused by a system of forces Q1, Q2, Q3 … is equal to the virtual work done by the system of forces Q1, Q2, Q3 …during the distortion caused by the system of forces, P1, P2, P3 … P1y1 + P2y2 + P3y3 = Q1Y1 + Q2Y2 + Q3Y3 17v shiva
- 18. THE FIRST THEOREMOF CASTIGLIANO • The first theorem of Castigliano states that the partial derivative of total strain energy in any structure with respect to the applied force or moment gives the displacement or rotation respectively at the point of application of force or moment in the direction of applied force or moment. 18v shiva
- 19. • Let thedeflection at r be yr. • We = external work done by the given load system. • Wi = correspondingstrain energy stored. ∴We = Wi Let ∆Pr aloneis applied getting deformation of ∆yr. 19v shiva
- 20. Pr quantity.smallinfinitelyanisPrwhenjustifiedisrelationaboveThe Pr .Pr thatknowwe .Pr bestructureby thestoredenergystrainingcorrespondLet the .PrdoneworkExternalTotal structure.onthe....PnP3,P2,P1,systemloadgiventheaddnowr,atPrcarryingisstructuretheAs ignored.becanquantity,smallofisproducttheAs .Pr. 2 1 donework = = = = +=+ + += = Wi yr Wi yr Wiyr WiWe WiWiyrWe WiWi yrWe yr Pr Wi Rotationat a point where couple Mo acts = Mo Wi 20v shiva
- 21. • In casethe deflection is to be determined at a point where no load acts introduce a fictitious force Q at the point in the direction of the desired deflection. The partial derivative of strain energy stored with respect to Q is determined, and Q is put equal to zero. 21v shiva
- 22. Example: A simplysupported beam carries a point load P eccentrically on the span. Find the deflection under the load. Assume uniform flexural rigidity • Reaction at A = Va = Pb/l • Reaction at B = Vb = Pa/l • The strain energy stored by the beam, • = = = = + = + = . 3 6 2bygivenisPloadunder thedeflection 6 )( 6 22 . 2 22 22 222 2 222 0 2 0 22 EIl bPa EIl ba P P Wi EIl baP ba EIl baP EI dx l Pax EI dx l xPb EI dxM ba 22v shiva
- 23. The bend ABCshown in fig. carries a concentrated vertical load P at A. find the vertical and horizontal deflections of A. Assume uniform flexural rigidity ( ) )3( 3 P,respect towithstoredenergystraintotaltheatedifferenti A,ofdeflectionverticalthefindTo )3( 6 2 )( 22 2 22 0 2 0 2 2 ha EI Pa P Wi ha EI aP EI dy Pa EI dx Px EI dxM v ha + == = + = += 23v shiva
- 24. ( ) () ( ) EIl Pah QhPah EI EI dy yQyPa Q Wi EI dy QyPa EI dx Px EI dxM Wi h h h ha 2 0;Qputting 32 1 2 20 haveweQ,respect tohenergy witstrainatingdifferenti 222 bygivennowisstructureby thestoredenergystrain A.atlyhorizontalQloadfictitiousagintroducin byfoundisA,ofdeflectionHorizontal 2 32 0 0 2 0 2 2 = = += ++= = ++== 24v shiva
- 25. THE SECOND THEOREMOF CASTIGLIANO The second theorem of castigliano states that the work done by external forces in a structure will be minimum. The theorem is very useful in analysis of indeterminate structures. Let W be the work done by the external forces on the structure, U be the strain energy stored in the structure and W1 be the work done by the reactive forces. Strain energy U = W + W1 W = U – W1 By castigliano’s second theorem W should be minimum. Thus the partial derivative of the work done with respect to external forces will be zero. In case the supports are unyielding the work done by reactive forces will be zero. Strain energy stored will be equal to the work done by external forces and will be the minimum. Thus the partial derivative of strain energy with respect to redundant reaction will be zero. 25v shiva
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- 27. To find thereaction at the free end of propped cantilever carrying UDL using Castigliano’s second theorem 8 3 83 0 83 1 0 2 ).2/.(2 zero,toequatingandQw.r.tatingdifferentiso, 0 Q Wi minimum,betodeflectionfor thetheoremsecondso'castigliantoAccording 2 )2/.( 2 Q,loadfictituousaintroduce end,freeatdeflectionthefindTo 43 43 0 2 0 222 wl Q wlQl wlQl EI EI xwxxQ EI dxwxxQ EI dxM l l = == = −= = − = = − = 27v shiva
- 28. Find the verticalreaction at C for the frame shown in figure ( ) ( ) ( ) 13 4 0 12448 0 2 . 22 . .. )2( 0 22 . 22 . .2. )2(2 2 R Wi ;0 R Wi i.e.,minimum,is storedenergystrainthatconditionfromobtainedbecanR 222 . )2(22 columninstoredenergystrainbeaminstoredenergystrain frameby thestoredenergyStrain 433 0 22/ 0 2 0 22/ 0 0 222/ 0 2 2 WL R WLRL EI RL EI dyLwyLR IE dx Rx EI dyLwyLR x IE dx Rx EI dywyLR IE dx Rx EI dxM wi ll ll ll = =−+= = −+= = −+= = −+== += 28v shiva
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- 32. steps to analyzethe redundant frame Step-I: consider a member(say BC) as redundant and remove BC and find member forces P1,P2,P3,.. For the given loading system. Step-II: Remove the given loading systemand apply a pair of unit forces at B and C. find the member forces K1,K2,K3,.. Step-III: Tabulate the above results as follows: Step-IV: find a correcting factor X given by member P K L A PKL/A K2 L/A TOTAL Σ(PKL/A) Σ(K2 L/A) = A LK A LKP X . .. 2 32v shiva
- 33. Step-V: Now findthe force in any member by the relation S = P + X.K 33v shiva
- 34. Example: Determine theforces in the members of the frame shown in fig. sectional areas in millimeters • Solution: Let BD is redundant, so remove BD and find forces. If we resolve and solve we get force in members as; Pca = 15√41KN(compressive) Pcd = 60KN(Tensile) Pda = 75KN(Tensile) Pcb = 0 Pba = 0 34v shiva
- 35. • Now removethe given load and apply a pair of unit loads at B and D in place of member BD. Kdc = 4/√41(compressive) Kda = 5/√41(compressive) Kcb = 5/√41(compressive) Kba = 4/√41(compressive) Kca = 1(Tensile) The actual force in the member is given by S = P + X.K, = A LK A LKP X . .. 2 35v shiva
- 36. member P KL A PKL/A K2 L/A AB 0 4/√41 4000 900 0 1.7344 BC 0 5/√41 5000 1200 0 2.5407 CD -60 4/√41 4000 900 -166.585 1.7344 DA -75 5/√41 5000 1200 -244.022 2.5407 AC 15√41 -1 1000√41 1500 -410 4.2687 BD 0 -1 1000√41 1500 0 4.2687 TOTAL -820.607 17.0876 36v shiva
- 37. • X =820.607/17.0876 ∴The actual force in the members are calculated as below; • Sab = Pab + X.Kab = 0 + 48.0235X4/√41 = 30KN(compressive) • Sbc = Pbc + X.Kbc = 0 + 48.0235X5/√41 = 37.5KN(compressive) • Scd = Pcd + X.Kcd = -60 + 48.0235X4/√41 = -30KN (tensile) • Sda = Pda + X.Kda = -75 + 48.0235X5/√41 = -37.5KN(tensile) • Sac = Pac + X.Kac = 48.0235(-1)+15√41 = 48.023KN(compressive) • Sbd = Pbd + X.Kbd = 0 + 48.0235(-1)= -48.023KN(tensile) 37v shiva
- 38. Frame with tworedundant members • Fig.1 shows a truss with two redundant members. • Let S1,S2,S3,… be the actual forces. • Let BF and DF be the redundant members. • Let P1,P2,P3,..be the member forces after removing BF and DF shown in fig.2. • Let the member forces be K1,K2,K3,..due to unit load at B and F as shown in fig.3 • Let the member forces be K’1,K’2,K’3,..due to unit load at D and F as shown in fig.4 • The actual forces in the members of the truss are given by, • S1=P1+XK1+YK’1; • S2=P2+XK2+YK’3; and so on… 38v shiva
- 39. (ii)and(i)equationssolvingbydeterminedbecanYandXunknownsThe ).....(0 1 11'1 1 11' 1 11'1 0 12 11')1'11(2 Y Wi , )......(0 1 11'1 1 11 1 111 0 12 11)1'11(2 X Wi 0 Wi ;0 X Wi conditionsby thei.e.,minimum,isstoredenergystrainthat the conditionbythegivenareYandXquantitiesredundantThe 12 1)1'11( 12 11 trussby thestoredenergystrainThe 2 2 2 2 ii EA LKK Y EA LK X EA LKP EA LKYKXKP similarly i EA LKK Y EA LK X EA LKP EA LKYKXKP Y EA LYKXKP Wi EA LS Wi =++ = ++ = =++ = ++ = = = ++ = == 39v shiva
- 40. Find the forcesin the members FD and DH of the frame shown in figure. The ratio of length to the cross-sectional area for all the members is the same • Given l/A is constant for all members. • Consider FD and DH to be redundant members and assume them to be removed. • Let, • X = Force in member FD • Y = force in member DH 40v shiva
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- 43. P K K’PK PK' K 2 K ’2 KK’ AC -42.42 0 0 0 0 0 0 0 CD -60 -0.707 0 42.42 0 0.5 0 0 DE -60 0 -0.707 0 42.42 0 0.5 0 EB -70.71 0 0 0 0 0 0 0 AF 30 0 0 0 0 0 0 0 FG 30 -0.707 0 -21.21 0 0.5 0 0 GH 50 0 -0.707 0 -35.35 0 0.5 0 HB 50 0 0 0 0 0 0 0 CF 0 -0.707 0 0 0 0.5 0 0 CG 42.42 1 0 42.42 0 1 0 0 DG 0 -0.707 -0.707 0 0 0.5 0.5 0.5 EG 14.14 0 1 0 14.14 0 1 0 EH 40 0 -0.707 0 -28.28 0 0.5 0 DF 0 1 0 0 0 1 0 0 DH 0 0 1 0 0 0 1 0 TOTAL 63.63 -7.07 3.999 3.999 0.5 43v shiva
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- 47. member P KK’ L PKL PK’L K2L K’2L KK’L TOTAL -25.25 -36 8.44 8.64 0.636 47v shiva
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- 50. LACK OF FIT: •Lack of fit is an occurance in trussesin which there is a differencebetween the lengthof a member and the distance between the nodes it is supposedto fit. • This happens because the connecting member is either too long or too short as compared to the distance between corresponding nodes. • It must also be noted here that lack of fit occurs only in indeterminate trusses because in determinate trusses the variation in member length automatically adjusts the spacing between the nodes, hence, lack of fit cannot occur. • A lack of fit may be unintentionally introduced in a truss due to fabrication errors or it may be deliberately introduced as a part of the design by the engineer. • In cases of lack of fit, the member is either forcibly lengthened or forcibly shortened to fit the truss depending on whether it is too short or too long respectively. This causes the various members meeting at these joints to develop stresses of opposite nature. This pre-stress occurring due to lack of fit may be used advantageously in design of truss. • The members that develop tension due to loading may be introduced with a compressive pre-stress and the members that develop compression may be introduced with a tensile pre-stress. This effectively increases the stress range over which a member may be loaded and the maximum load carrying capacity of the truss. 50v shiva
- 51. Consider a staticallyindeterminateframe in which one of the members BC is shorterthan its length by amount δ. To fit this member in position, the joints will have to be brought closer and member itself will get elongated and thus will be in tension. Let ‘x’ be the force in the member having lack of fit To analyze the frame, it is made determinateby removing the member having lack of fit Unit forces are applied at the joints of the members having lack of fit as shown in fig • Let ‘k’ be the force in the member due to unit forces. • The force in the member due to force ‘x’will be ‘kx’. • The strain energy in all the members of determinateframe will be 51v shiva
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- 53. + −= + = AoE l AE lK X AoE l AE lK X O O 2 2 tensile.belmember wilin theforceshort,tooismembertheIf e,compressivbelmemberwilin theforcelong,tooismembertheIf 53v shiva
- 54. • If anymember in a redundant frame is subjected to change in temperature, there will be change in the length of the member and this will produce stress in all the other members of the frame. • Consider figure(a), • Let member BC be subjected to temperature rise of T°C. • Change in length of member due to temperature change will be δl = loᾱT. 54v shiva
- 55. • The membersubjected to temperature change is removed and the frame is made determinate. • Unit loads are applied at the joints of the members as shown in figure(b) and forces in various members calculated. • Let the force in a member due to unit loads be k. • The force in the member due to force ‘X’ in redundant member BC will be kX. • Total strain energy in all the members of determinate frame will be 55v shiva
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- 57. A triangularframeshown infigurehas sectionalarea equal to 35X10-4 m2 for all the members. The member AB was found to be 1X10-3 m shorterthan its correctlengthat the timeof assembling.Find the forces in the members, if the member AB is forced in position.Take Young’smodulusas 2.1X108 KN/m2 Given: Indeterminatestructureby degree of static indeterminacy = 1 area of members = 35X10-4 m2 Error in length(δ)= 1X10-3 m Young’s modulus, E = 2.1X108 KN/m2 Let T = force in member AB due to lack of fit. Remove member AB and apply unit tensile force at jointsA and B as shown in figure(b) 57v shiva
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- 59. member Length in m AREAE(KN/m2) k K 2 l/AE Actual force kT in KN AB 10 0.0035 210000000 1 1.36054E-05 -8.97 BC 10 0.0035 210000000 1 1.36054E-05 -8.97 AC 10 0.0035 210000000 1 1.36054E-05 -8.97 CO 5.77 0.0035 210000000 -1.732 2.35496E-05 15.53604 BO 5.77 0.0035 210000000 -1.732 2.35496E-05 15.53604 AO 5.77 0.0035 210000000 -1.732 2.35496E-05 15.53604 TOTAL 0.000111465 59v shiva
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- 61. Problem:In the planebracedframeshown in figure(a)all the membershave same cross sectionalarea of 8 cm2 and are made of same material. The member AC in the framewas initiallyshort by 0.25 cm. determinetheforce in each member. In the aboveframedeterminetheforce in each member when member AC is subjected to a risein temperatureof 20°C. ᾱ = 1.1 x 10-5 /°C and E = 2 x 105 N/mm2 61v shiva
- 62. • Given: • Area,A = 8 cm2 = 0.0008m2 • E = constant= 2 x 105 N/mm2 = 2 x 1011 N/m2 • ᾱ = 1.1 x 10-5 /°C • Error in length of member AC = 0.25 cm = 0.0025m • Member AC is removed and unit loadsare applied as shown in figure(b). • Forces in the members are calculatedas, 62v shiva
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- 64. memb er L AE K K2L/AE Force in member due to lack of fit(KX) Force in member due to temperat ure(KX1) Final force AB 4 0.0008 200000000 -0.707 1.24962E- 05 14.64197 -7.2821 7.35987 BC 4 0.0008 200000000 -0.707 1.24962E- 05 14.64197 -7.2821 7.35987 CD 4 0.0008 200000000 -0.707 1.24962E- 05 14.64197 -7.2821 7.35987 DA 4 0.0008 200000000 -0.707 1.24962E- 05 14.64197 -7.2821 7.35987 BD 5.656 0.0008 200000000 1 0.00003535 -20.71 10.3 -10.41 FB 2.828 0.0008 200000000 0 0 0 0 0 FC 2.828 0.0008 200000000 0 0 0 0 0 EA 2.828 0.0008 200000000 0 0 0 0 0 ED 2.828 0.0008 200000000 0 0 0 0 0 AC 5.6535 0.0008 200000000 1 3.53344E- 05 -20.71 10.3 -10.41 TOTAL 0.00012066 9 64v shiva
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- 66. The diagonal ACof the frame work, shown in figure was found to be 0.625x10-3 m longer than its correct length, at the time of assembling. The member AC was forced in position. Find the forces induced in the members of the frame. Take E = 2X108 KN/m2 and diameter of each member = 2X10-2 m 66v shiva
- 67.