# Analysis of indeterminate beam by slopeand deflection method

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### Analysis of indeterminate beam by slopeand deflection method

• 2. • Slope Deflection Method was presented by George A Money in 1815. • Use to Analyse Statically Indeterminate Beams and Frames. • Slope Deflection method is a Displacement method i.e. Equation Method. • This Method mainly involves slope and Deflection of the member at it’s joint and hence the named as Slope deflection method. • The Basic unknown in slope Deflection method are the non zero joint displacement of a structure. i.e.., Degree of Kinematic Indeterminacy (Degree of freedom).
• 3. Slope and Deflection Equation for member AB F AB AB A B 2 F BA BA A B 2 4EI 2EI 6EI M = M + Θ + Θ ± Δ L L L 2EI 4EI 6EI M = M + Θ + Θ ± Δ L L L FixedEndmomentduetoTransverseLoad. NearEndRotatio F FM =M = AB BA 4EIΘ= ncontri L bution. 2EIΘ= L 6EI Far EndRotationcontribution. Relative TranslationContributΔ= 2L ion.
• 4. • ΘA= End slope at A or the rotation of the joint A. • ΘB= End slope at B or the rotation of the joint B. • ΔA=End deflection at A or the translation of Joint A. • ΔB=End deflection at B or the translation of Joint B. • ΔB/A=Relative Deflection or Translation of Joint B with respect to A. • MF AB=Fixed end Moment at A for member AB. • MF BA=Fixed End moment at B for member BA.
• 5. Clockwise moments are Negative, Anticlockwise moments are Positive.
• 6. Clockwise moments are Negative, Anticlockwise moments are Positive.
• 7. 1)Analyse the Two span continuous beam shown in Fig.No.1 by slope and Deflection method ,draw bending moment and shear force Diagram. Solution:- Fixed End Moment 2 2 F AB 2 2 F BA F F BC CB CD WL 20X6 M = = = 60KNm. 12 12 WL 20X6 M = - = - = -60KNm. 12 12 WL 80X4 M = -M = = = 40KNm. 8 8 M = 40X2 = 80KNm. TotalRestrainedmoment at C = -40 + 80 = 40KNm. Sign Convention: Clockwise moments are Negative, Anticlockwise moments are Positive.
• 8. Application of slope and Deflection equation to member AB F AB AB A B 2 AB B A AB B F BA BA A B 2 BA B BA B 4EI 2EI 6EI M = M + Θ + Θ ± Δ L L L 2E(2I) M = 60+ Θ 6 (Astheiris fixedsupportat A,soΘ = 0&noSinking of support,so Δ = 0) 2EI M = 60+ Θ ------(I). 3 2EI 4EI 6EI M = M + Θ + Θ ± Δ L L L 4E(2I) M = -60+ Θ . 6 4EI M = -60+ Θ - 3 -----(II)
• 9. Application of slope and Deflection equation to member BC F BC BC B C 2 BC B C BC B C F CB CB B C 2 CB B C CB B C 4EI 2EI 6EI M = M + Θ + Θ ± Δ. L L L 4EI 2EI M = 40 + Θ + Θ . 4 4 M = 40 + EIΘ + 0.5EIΘ .- - - - - - - (III) 2EI 4EI 6EI M = M + Θ + Θ ± Δ. L L L 2EI 4EI M = -40 + Θ + Θ . 4 4 M = -40 + 0.5EIΘ + EIΘ .- - - - - - - - - (IV).
• 10.  C CB CD B C B C M = 0. M + M = 0. -40 + 0.5EIΘ + EIΘ + 80 = 0 0.5EIΘ + EIΘ = -40 - - - - - - (2) M = 0.B M +M = 0.BA BC 4EI -60+ Θ + 40+EIΘ + 0.5EIΘ = 0B B C3 2.34EIΘ + 0.5EIΘ = 20 - - - - - (1)B C solving Equation1 and2 we get, 19.14Θ =B EI -49.56Θ = C EI
• 11. Substituting the values of in slope deflection equation 2EIM = 60 + ΘAB B3 2M = 60 + 19.14.AB 3 M = 72.76 KNm.AB  4EIM = -60 + ΘBA B3 4M = -60 + X19.14BA 3 M = -34.4 KNm.BA
• 12. M = 40+EIΘ +0.5EIΘBBC C M = 40+19.14+0.5X(-49.56) BC M =34.36KNm. BC Substituting the values of in slope deflection equation M =-40+0.5EIΘ +EIΘ . BCB C M =-40+0.5(19.14)+(-49.56) CB M =-80KNm. CB M =80KNm. CD
• 16. Analyse the continuous beam shown in Fig.No.2 by slope and Deflection method if Joint B Sinks by 10mm ,Given EI=4000KM-m2.Draw bending moment and shear force Diagram. Solution:-1) Fixed End Moment:- 2 2WL 20X8F FM =-M = = =106.67KNm.AB BA 12 12 WL 60X4F FM =-M = = =30KNm. BC CB 8 8 Sign Convention: Clockwise moments are Negative, Anticlockwise moments are Positive.
• 17. 4EI 2EI 6EIFM =M + Θ + Θ ± ΔAB AB A B 2L L L 2E(2I) 6X4000X2 10M =106.67+ Θ + XAB B 28 10008 (As their is fixedsupport at A,so Θ = 0)A M =106.67+0.5EIΘ +7.5AB B M =114.17+0.5EIΘ ------ (I).AB B 2EI 4EI 6EIFM =M + Θ + Θ ± ΔBA BA A B 2L L L M = -106.67+BA 4E(2I) 6X4000X2 10Θ + X .B 28 10008 M = -99.17+EIΘ ------ (II)BA B Application of slope and Deflection equation to member AB
• 18. Application of slope and Deflection equation to member BC 4EI 2EI 6EIFM =M + Θ + Θ ± Δ. BBC BC CL L 2L 4EI 2EI 6X4000 10M =30+ Θ + Θ - X . BBC C4 4 2 10008 M =15+EIΘ +0.5EIΘ .-------(III) BBC C 2EI 4EI 6EIFM =M + Θ + Θ ± Δ. BCB CB CL L 2L 2EI 4EI 6X4000 10M =-30+ Θ + Θ - X . BCB C4 4 2 10008 M =-45+0.5EIΘ +EIΘ .--------- BCB C (IV).
• 19. M = 0. B M +M = 0. BA BC -99.17+EIΘ +15+EIΘ +0.5EIΘ = 0 B B C 2EIΘ +0.5EIΘ = 84.17 ----- (1) B C M =0. C M =0. CB 0.5EIΘ +EIΘ = 45 ------ (2) B C Solving Equation1 and2 we get, EIΘ =35.24 B EIΘ =27.38 C
• 20. M =114.17+0.5EIΘ AB B M =114.17+0.5 X 35.24 AB M =131.79KNm. AB M = -99.17+EIΘ BA B M = -99.17+35.24 BA M = -63.93KNm. BA M = 15+EIΘ +0.5EIΘ . BBC C M = 15+35.24+0.5X27.38. BC M = 63.93KNm. BC M = -45+0.5EIΘ +EIΘ . BCB C M = -45+0.5X35.24+27.38 CB M = 0 CB
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