Analysis of non sway frame portal frames 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. Analysis of Frames:- • When all the Joint should change their position due to deformation is called as Sway Frames. • Frame is Non –Sway if it is Symmetrical frame but symmetrical, in all respect i.e. about Vertical axis, supports, sectional properties and loads etc if not it is a Sway frame. • In Sway Frame Horizontal translation i.e Δ is additional unknown with rotation at the Joint. • Sway Moment 6EI Δ /L2 we have to consider in Slope deflection Equation. • Note :- Only Vertical member are subjected to sway moment not horizontal.
- 9. Numerical On Non-Sway Frames Analyse the frame shown in Fig 3 by Slope and Deflection Method and Draw Bending Moment diagram. Degree of Freedom=01 ( i.e. ΘB) Solution:-1) Fixed End Moment:- 2 2 F F AB BA F F BC CB BD WL 2.4X4 M =-M = = =3.2KNm. 12 12 WL 12X3 M =-M = = =4.5KNm. 8 8 M =10X1.5=15KNm. Sign Convention: Clockwise moments are Negative, Anticlockwise moments are Positive.
- 10. 2) Apply slope Deflection Equation. 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 2EI M =3.2+ Θ ( As their is fixed support at A,so Θ =0) 4 M =3.2+0.5EIΘ ------ (I). 2EI 4EI 6EI M =M + Θ + Θ ± Δ L L L 4EI M =-3.2+ Θ . 4 M =-3.2+EIΘ ------ (II) F BC BC B C 2 BC B BC B F CB CB B C 2 CB B CB B 4EI 2EI 6EI M =M + Θ + Θ ± Δ. L L L 4EI M = 4.5+ Θ . 3 ( As their is fixed support at C,so Θ =0) M = 4.5+1.33EIΘ .------- (III) 2EI 4EI 6EI M =M + Θ + Θ ± Δ. L L L 2EI M =-4.5+ Θ .. 3 M =-4.5+0.67EIΘ ---------(IV). C
- 11. 3) Equilibrium Equation B BA BC BD B B B B M =0. M +M +M =0. -3.2+EIΘ 4.5+1.33EIΘ 15 0 2.33EIΘ =-16.33 EIΘ = -6.99
- 12. 4) Substituting the values of in Slope Deflection Equation, We get, AB B AB AB BA B BA BA M =3.2+0.5EIΘ M =3.2+0.5(-6.99) M =-0.295KNm. M =-3.2+EIΘ M =-3.2+(-6.99) M =-10.19KNm. BC B BC BC CB B CB CB M = 4.5+1.33EIΘ M = 4.5+1.33(-6.99) M = -4.79KNm. M =-4.5+0.67EIΘ . M =-4.5+0.67(-6.99) M = -9.18 KNm.
- 14. Analyse the portal Frame shown in Fig 3A. Draw Bending Moment diagram. Solution:- 1) Fixed End Moment:- F F AB BA F F BC CB F F CD DC WL 32X4 M =-M = = =16KNm. 8 8 WL 48X3 M =-M = = =18KNm. 8 8 M =M =0(NoLoad). Sign Convention: Clockwise moments are Negative, Anticlockwise moments are Positive.
- 15. 2) Apply slope Deflection Equation. F AB AB A B 2 AB B A AB B F BA BA A B 2 BA B BA B F BC BC B C 4EI 2EI 6EI M =M + Θ + Θ ± Δ L L L 2E(2I) M =16+ Θ (As their is fixed support at A,so Θ =0) 4 M =16+EIΘ ------ (I). 2EI 4EI 6EI M =M + Θ + Θ ± Δ L L L 4E(2I) M =-16+ Θ . 4 M =-16+2EIΘ ------ (II) 4EI 2EI M =M + Θ + Θ L L 2 BC B C BC B C 6EI ± Δ. L 4EI 2EI M = 18+ Θ + Θ 3 3 M 18+1.33EIΘ +0.67EIΘ .------- (III)
- 16. Apply slope Deflection Equation. F CB CB B C 2 CB B C CB B C F CD CD C D 2 CD C D CD C 2EI 4EI 6EI M =M + Θ + Θ ± Δ. L L L 2EI 4EI M =-18+ Θ + Θ . 3 3 M =-18+0.67EIΘ +1.334EIΘ .---------(IV). 4EI 2EI 6EI M =M + Θ + Θ ± Δ. L L L 4E(2I) M =0+ Θ .(As their is fixed support at D,so Θ =0) 4 M = 2EIΘ ---------(V F DC DC C D 2 DC C D DC C ). 2EI 4EI 6EI M =M + Θ + Θ ± Δ. L L L 2E(2I) M =0+ Θ .(As their is fixed support at D,so Θ =0) 4 M = EIΘ ---------(VI).
- 17. 3) Equilibrium Equation B BA BC B B C B C C CB CD B C C B C M =0. M +M =0. -16+2EIΘ +18+1.33EIΘ +0.67EIΘ =0 3.33EIΘ +0.67EIΘ = -2 ----- (1) M =0. M +M =0. 0.67EIΘ +1.334EIΘ +2EIΘ = 18------ (2) solving Equation 1 and 2 we get, EIΘ =-1.75 EIΘ =5.76
- 18. 4) Substituting the values of in Slope Deflection Equation, We get, AB B AB AB BA B BA BA BC B C BC BC M =16+EIΘ ------ (I). M =16-1.75 M = 14.25 KNm. M =-16+2EIΘ ------ (II) M =-16+2(-1.75) M = -19.53KNm. M = 18+1.33EIΘ +0.67EIΘ .------- (III) M = 18+1.33(-1.75)+0.67(5.76). M = 19.53KNm.
- 19. 4) Substituting the values of in Slope Deflection Equation, We get, CB B C CB CB CD C CD CD DC C DC M =-18+0.67EIΘ +1.334EIΘ .---------(IV). M = -18+0.67(-1.75)+1.334(5.76) M =-11.51KNm. M = 2EIΘ ---------(V). M = 2 X5.76 M = 11.52 KNm. M = EIΘ ---------(VI). M = 5.76 KNm.