CE601-Structura Anaysis I UNIT-IV SOPE-DEFECTION METHOD 1. What are the assumptions made in sope-defection method? (i) Between each pair of the supports the beam section is constant. (ii) The joint in structure may rotate or defect as a whoe, but the anges between the members meeting at that joint remain the same.. How many sope defection equations are avaiabe for a two span continuous beam? There wi be nos. of sope-defection equations, two for each span.. What is the moment at a hinged end of a simpe beam? Moment at the hinged ends of a simpe beam is zero.. What are the quantities in terms of which the unknown moments are expressed in sope-defection method? In sope-defection method, unknown moments are expressed in terms of (i) sopes (θ) and (ii) defections ( ). The beam shown in Fig. is to be anaysed by sope-defection method. What are the unknowns and, to determine them, what are the conditions used? A B C Unknowns: θa, θb, θc Equiibrium equations used: (i) MAB = 0 (ii) MBA + MBC = 0 (iii) MCB = 0 6. How do you account for sway in sope defection method for porta frames? Because of sway, there wi br rotations in the vertica members of a frame. This causes moments in the vertica members. To account for this, besides the equiibrium, one more equation namey shear equation connecting the joint-moments is used. 7. Write down the equation for sway correction for the porta frame shown in Fig. The shear equation (sway correction) is MAB + MBA + MCD + MDC = 0 0 01-016

CE601-Structura Anaysis I D A. Write down the sope defection equation for a fixed end support. B A C The sope defection equation for end A is MAB D = M AB + EI θa + θb + Here θa= 0. Since there is no support settement, = 0. MAB = M AB + EI θb + 9. Write down the equiibrium equations for the frame shown in Fig. B C Unknowns : θ B, θc Equiibrium equations : At B, MBA + MBC = 0 h At C, MCB + MCD = 0 P MDC + P Shear equation : MAB + MBA Ph + =0 A MCD + D 10. Who introduced sope-defection method of anaysis? Sope-defection method was introduced by Prof. George A.Maney in 191. 11. Write down the genera sope-defection equations and state what each term represents? A B 1 01-016

CE601-Structura Anaysis I Genera sope-defection equations: MAB = M AB + EI θa + θb + MBA = M BA + EI θb + θa + where, M AB, M BA = Fixed end moment at A and B respectivey due to the given oading. θa, θb = Sopes at A and B respectivey = Sinking of support A with respect to B 1. Mention any three reasons due to which sway may occur in porta frames. Sway in porta frames may occur due to (i) unsymmetry in geometry of the frame (ii) unsymmetry in oading or (iii) Settement of one end of a frame. 1. How many sope-defection equations are avaiabe for each span? Two numbers of sope-defection equations are avaiabe for each span, describing the moment at each end of the span. 1. Write the fixed end moments for a beam carrying a centra cockwise moment. M A B / / Fixed end moments : M AB = M BA = M 1. State the imitations of sope defection method. (i) It is not easy to account for varying member sections (ii) It becomes very cumbersome when the unknown dispacements are arge in number. 16.Why is sope-defection method caed a dispacement method? 01-016

CE601-Structura Anaysis I In sope-defection method, dispacements (ike sopes and dispacements) are treated as unknowns and hence the method is a dispacement method. 17. Define degrees of freedom. In a structure, the number of independent joint dispacements that the structure can undrgo are known as degrees of freedom. 1. In a continuous beam, one of the support sinks. What wi happen to the span and support moments associated with the sinking of support. C D E 1 et support D sinks by. This wi not affect span moments. Fixed end moments (support moments) wi get deveoped as under M CD = M DC = -6EI 1 M DE = M ED = -6EI 1 19. A rigid frame is having totay 10 joints incuding support joints. Out of sope-defection and moment distribution methods, which method woud you prefer for anaysis? Why? Moment distribution method is preferabe. If we use sope-defection method, there woud be 10 (or more) unknown dispacements and an equa number of equiibrium equations. In addition, there woud be unknown support momentsper span and the same number of sope-defection equations. Soving them is difficut. 0. What is the basis on which the sway equation is formed for a structure? Sway is deat with in sope-defection method by considering the horizonta equiibrium of the whoe frame taking into account the shears at the base eve of coumns and externa horizonta forces. The shear condition is MAB + MBA Ph + MCD + MDC + P =0 01-016

CE601-Structura Anaysis I Sope defection equations The sope defection equations express the member end moments in terms of rotations anges. The sope defection equations of member ab of fexura rigidity EIab and ength ab are: where θa, θb are the sope anges of ends a and b respectivey, Δ is the reative atera dispacement of ends a and b. The absence of cross-sectiona area of the member in these equations impies that the sope defection method negects the effect of shear and axia deformations. The sope defection equations can aso be written using the stiffness factor chord rotation and the : Derivation of sope defection equations When a simpe beam of ength ab and fexura rigidity E Iab is oaded at each end with cockwise moments Mab and Mba, member end rotations occur in the same direction. These rotation anges can be cacuated using the unit dummy force method or the moment-area theorem. Rearranging these equations, the sope defection equations are derived. Equiibrium conditions 01-016

CE601-Structura Anaysis I Joint equiibrium Joint equiibrium conditions impy that each joint with a degree of freedom shoud have no unbaanced moments i.e. be in equiibrium. Therefore, Here, Mmember are the member end moments, Mf are the fixed end moments, and Mjoint are the externa moments directy appied at the joint. Shear equiibrium When there are chord rotations in a frame, additiona equiibrium conditions, namey the shear equiibrium conditions need to be taken into account. Degrees of freedom Rotation anges θa, θb, θc, θd of joints A, B, C, D respectivey are taken as the unknowns. There are no chord rotations due to other causes incuding support settement. Fixed end moments Fixed end moments are: 01-016

CE601-Structura Anaysis I Sope defection equations The sope defection equations are constructed as foows: Joint equiibrium equations Joints A, B, C shoud suffice the equiibrium condition. Therefore 6 01-016

CE601-Structura Anaysis I Rotation anges The rotation anges are cacuated from simutaneous equations above. Member end moments Substitution of these vaues back into the sope defection equations yieds the member end moments (in knm): Exampe: Anayze the propped cantiever shown by using sope defection method. Then draw Bending moment and shear force diagram. 7 01-016

CE601-Structura Anaysis I Soution: End A is fixed hence A =0 End B is Hinged hence B 0 Assume both ends are fixed and therefore fixed end moments are FAB w w, FBA 1 1 The Sope defection equations for fina moment at each end are EI A B w EI B 1 EI B A MBA FBA w EI B 1 MAB FAB (1) ( ) In the above equations there is ony one unknown B. To sove we have boundary condition at B; Since B is simpy supported, the BM at B is zero ie. MBA=0. From equation () MBA EI B w EI B 0 1 w - ve sign indicates the rotation is anticockw ise Substituting the vaue of EI B in equation (1) and () we have end moments MAB MBA w w w - ve sign indicates moment is anticockw ise 1 w w 0 1 01-016

CE601-Structura Anaysis I MBA has to be zero, because it is hinged. Now consider the free body diagram of the beam and find reactions using equations of equiibrium. MB 0 R A M AB w w w w R A w V 0 R A R B w R B w R A w w w Probem can be treated as 9 01-016

CE601-Structura Anaysis I The bending moment diagram for the given probem is as beow 60 01-016

CE601-Structura Anaysis I The max BM occurs where SF=0. Consider SF equation at a distance of x from right support w wx 0 X SX Hence the max BM occurs at Mmax from support B w MX w 9 w 1 And point of contra fexure occurs where BM=0, Consider BM equation at a distance of x from right support. X MX wx w 0 X For shear force diagram, consider SF equation from B w wx S X 0 SB w S X S A w SX Exampe: Anayze continuous beam ABCD by sope defection method and then draw bending moment diagram. Take EI constant. 61 01-016

CE601-Structura Anaysis I Soution: A 0, B 0, C 0 FEMS FAB Wab 100 -. KN M 6 FBA Wab 100. KNM 6 w 0 F BC - 1.67 KNM 1 1 w 0 F CB 1.67 KNM 1 1 FCD 0 1. - 0 KN M Sope defection equations: MAB F AB EI A B. 1 EI B - - - - - - - -- 1 MBA FBA EI B A.9 EI B - - - - - - - -- EI B C 1.67 EI B EI C EI C B 1.67 EI C EI B MCB FCB MBC FBC - - - - - - - - - - - - - - - - MCD 0 KNM 6 01-016

CE601-Structura Anaysis I In the above equations we have two unknown rotations B and C, accordingy the boundary conditions are: MBA MBC 0 MCB MCD 0 EI B 1.67 EI B EI C 7. EI B EI C 0 1 Now, MBA MBC.9 EI C EI B 0 11.67 EI B EI C - - - - - - - - And, MCB MCD 1.67 6 Soving () and (6) we get EI B.67 Rotation @ B anticockw ise EI C 1.7 Rotation @ B cockwise Substituting vaue of EI B and EI C in sope defection equations we have 1.67 61.00 KNM MBA.9.67 67.11 KNM MBC 1.67.67 1.7 67.11 KNM MCB 1.67 1.7.67 0.00 KNM MCD 0 KNM MAB. 6 01-016

CE601-Structura Anaysis I Reactions: Consider free body diagram of beam AB, BC and CD as shown Span AB 6 01-016

CE601-Structura Anaysis I RB 6 100 67.11 61 RB 67.69 KN R A 100 RB.1 KN Span BC RC 0 0 67.11 RC. KN RB 0 RB 7. KN Maximum Bending Moments: Span AB: Occurs under point oad 67.11 61 Max 1. 61 6.6 KNM 6 Span BC: where SF=0, consider SF equation with C as reference S X. 0 x 0 x..1 m 0 Mmax..1 0.1 0 1.6 KN M Exampe: Anayse the continuous beam ABCD shown in figure by sope defection method. The support B sinks by 1mm. 6 Take E 00 10 KN / m and I 10 10 m 6 01-016

CE601-Structura Anaysis I Soution: In this probem A =0, B 0, C 0, =1mm FEMs: FAB Wab. KNM FBA Wa b.9 KNM FBC w 1.67 KNM FCB w 1.67 KNM FEM due to yied of support B For span AB: 6EI 6 00 1 10 10 10 6 6 KNM 6 1000 mab mba For span BC: 6EI 6 00 1 10 10 10 6.6KNM 1000 mbc mcb Sope defection equation 66 01-016

CE601-Structura Anaysis I EI ( A B ) EI 6EI F AB A B 1 -. EI B 6 1 0. EI B EI 6EI MBA FBA ( B A ).9 EI B 6.9 EI B EI 6EI MBC FBC ( B C ) - 1.67 EI B C.6.0 EI B EI C EI 6EI MCB FCB ( C B ) 1.67 EI C B.6 0.1 EI C EI B MCD 0 KNM MAB F AB - - - - - - - -- 1 - - - - - - - -- - - - - - - - -- - - - - - - - -- - - - - - - - -- There are ony two unknown rotations B and C. Accordingy the boundary conditions are MBA MBC 0 MCB MCD 0 Now, M M 9.6 EI EI 0 BA BC B C 1 MCB MCD 0.1 EI B EI C 0 Soving these equations we get 67 01-016

CE601-Structura Anaysis I EI B 1. Anticockwise EI C 9.71 Anticockwise Substituting these vaues in sope defections we get the fina moments: 1 1. 60.9 KNM MBA.9 1. 61.99 KNM MBC.0 1. 9.71 61.99 KNM MCB 0.1 9.71 1. 0.00 KNM MCD 0 KNM MAB 0. Consider the free body diagram of continuous beam for finding reactions Reactions: Span AB: RB 6 = 100 x + 61.99 60.9 RB = 66. RA = 100 RB =.1 KN Span BC: 6 01-016

CE601-Structura Anaysis I RB = 0 x x + 61.99 0 RB = 6.0 KN RC = 0 x - RB =.60 KN Exampe: Three span continuous beam ABCD is fixed at A and continuous over B, C and D. The beam subjected to oads as shown. Anayse the beam by sope defection method and draw bending moment and shear force diagram. 69 01-016

CE601-Structura Anaysis I Soution: Since end A is fixed A 0, B 0, c 0, D 0 FEMs: F AB W 60-0 KNM FBA W 60 0 KNM FBC M 1. KNM FCB M 1. KNM FCD w 10-1. KNM 1 1 w 10 FDC 1. KNM 1 1 Sope defection equations: M AB F AB EI A B - 0 EI 0 B - - - - - - - - 1-0 0.EI B MBA F BA EI B A 70 01-016

CE601-Structura Anaysis I 0 EI B 0 - - - - - - - -- 0 EI B MBC F BC EI B C 1. EI B C - - - - - - - -- 1. EI B 0.EI C M CB F CB EI C B 1. EI C B - - - - - - - -- 1. EI C 0. EI B M CD F CD EI C D - 1. EI C D 1. EI C 0.EI D MDC F DC - - - - - - - - - - EI D C 1. EI D C - - - - - - - - - - 6 1. 0.EI C EI D In the above Equations there are three unknowns, EI B,EI C & EI D, accordingy the boundary conditions are: i MBA MBC 0 ii MCB MCD 0 iii MDC 0 ( hinged) 71 01-016

CE601-Structura Anaysis I Now MBA MBC 0 0 EI B 1. EI B 0.EI C 0 EI B 0.EI C. 0 7 MCB MBC 0 1. EI C 0.EI B 1. EI C 0.EI D 0 0.EI B EI C 0.EI D 0. 0 MDC 0 9 1. 0.EI C EI D 0 By soving (7), () & (9), we get EI B.0 EI C 11.1 EI D 1.90 By substituting the vaues of B, c and D in respective equations we get MAB 0 0..0.0 KNM MBA 0.0.96 KNM MBC 1. -.0 0. 11.1 -.96 KNM MCB 1. 11.1 0..0 11.6 KNM MCD 1. 11.1 0. 1.90 11.6 KNM MDC 1. 0. 11.1 1.90 0 KNM Reactions: Consider the free body diagram of beam. Beam AB: 7 01-016

CE601-Structura Anaysis I 60.96.0 0.9 KN R A 60 RB 0.01 KN RB Beam BC: 11.6 0.96 1.9 KN RB RC 1.9 KN RB is downward RC Beam CD: 10 11.6 17.09 KN RC 10 RD.91 KN RD 7 01-016

CE601-Structura Anaysis I Exampe: Anayse the continuous beam shown using sope defection method. Then draw bending moment and shear force diagram. Soution: In this probem A 0, end A is fixed FEMs: FAB w 10 -. KNM 1 1 w FBA. KNM 1 FBC FCD W 0 6 -.0 KNM W.0 KNM Sope defection equations: M AB F AB EI A B -. E I 0 B 7 01-016

CE601-Structura Anaysis I MBA F BA EI B A.. MBC F BC E I B 0 - - - - - - - - EI B EI B C -. EI B C 6 -. EI B EI C M CB F CB - - - - - - - - 1 EI B -. - - - - - - - - EI C B. EI C B 6. EI C EI B - - - - - - - - In the above equation there are two unknown B and C, accordingy the boundary conditions are: i MBA MBC 0 ii MCB 0 EI B. EI B EI C 17. EI B EI C 0 6 Now, MBA MBC. 7 01-016

CE601-Structura Anaysis I and MCB. EI C EI B 0 1 EI C 11. EI B - - - - - - - - - -- (6) Substituting in eqn. () 17 1 EI B 11. EI B 0 6 1. EI B 0 6. 6 EI B 17. rotation anticockwise 1. from equation (6) 1 11. 17..19 rotation anticockwise EI C Substituting EI B 17. and EI C.19 in the sope defection equation we get Fina Moments: - 17. -66.0 KNM MBA. 17. 7.1 KNM MAB. 17..19 1.1 KNM MCB..19 ( 17.) 0.00 MBC. Reactions: Consider free body diagram of beams as shown 76 01-016

CE601-Structura Anaysis I Span AB: 7.1 66.0 10.1 KN R A 10 RB.7 KN RB Span BC: 1.1 0. KN 6 RC 0 RB 6.7 KN RB 77 01-016

CE601-Structura Anaysis I Max BM Span AB: Max BM occurs where SF=0, consider SF equation with A as origin S x.7-10x 0 x.7 m.7 M max.7.7 10 6 6.67 KNM Span BC: Max BM occurs under point oad BC Mmax 1.1 19.1 KN M 7 01-016