SLOPEDEFLECTION METHOD

 Peregrine Dean
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1 SLOPEDEFLECTION ETHOD The slopedeflection method uses displacements as unknowns and is referred to as a displacement method. In the slopedeflection method, the moments at the ends of the members are expressed in terms of displacements and end rotations of these ends. An important characteristic of the slopedeflection method is that it does not become increasingly complicated to apply as the number of unknowns in the problem increases. In the slopedeflection method the individual equations are relatively easy to construct regardless of the number of unknowns. DERIVATION OF THE SLOPEDEFLECTION EQUATION When the loads are applied to a frame or to a continuous beam, the member will develop end moments and become deformed as indicated. The notation used in the figure will be followed. ij i φ L j ij ji  The moments at the ends of the member are designated as ij and ji indicating that they act at ends I and j of member ij.  Rotations of ends I and j of the member are denoted by Θ i and Θ j. Since the rotations of all members of a rigid frame meeting at a common joint are equal, it is customary to refer to each of them as the joint rotation. The term ij represents the translation of one end of the member relative to the other end in a direction normal to the axis of the member. Sometimes the rotation of the axis of the member Φ ij ij /L is used in place of ij. The moments, the rotations at the ends of the member and the rotation of the axis of the member are positive when clockwise.
2 ij i i φ j ji jil L ij L L ql L L j jil L ij L L ql L L i j i φ L i j φ L ij j ji ij ( i j φ ) ql i j ij L ji L ql φ 6 ij L ji L ql φ 6 ql L ql ji ( i j φ ) L Now we wrote ij and ji in terms of the deformations Θ i, Θ j Φ ij and the external load q acting on the member. These equations are referred to as SLOPEDEFLECTION EQATIONS. Slopedeflection equations consider only bending deformations. Deformations due to shear forces and axial forces in bending members are ignored. nf ± L L FE n f nf
3 Example: It is required to determine the support moments for the continuous beam kn 0 kn/m I I x.m kn/m F F 00* 6.kNm F F 0*7. 9.7kNm Slope Deflection ( ) ( ) Equilibrium. equations. of. joınts E I Substitude. these. results.in. slope. deflection. equations 6.7 knm, 9.7kNm 9.7 knm, 0kNm
4 kn/m m Shear Force Diagram Bending oment Diagram
5 Example: A continuous beam is supported and loaded as shown in the figure. During loading support sinks by 0 mm. Analyze the beam for support moments and reactions. 0 kn 0 kn/m m 6 m 0.0 φ φ * 0.kNm F F 0* 0kNm F F 0*6 0kNm Slope Deflection. E 6 kn 00*0... m I 00*0... m knm E I...0 rad rad Substitude. these. results in. slope. deflection. equations 7.9 knm,.77knm.77 knm, 0kNm
6 kn/m m Shear Force Diagram Bending oment Diagram 7.9 0
7 ANALYSIS OF FRAES WITH NO SIDESWAY A frame will not side sway, or be displaced to the left or right, provided it is properly restrained. Also no side sway will occur in an unrestrained frame provided it is symmetric with respect to both loading and geometry. Example: It is required to analyze the frame for moments at the ends of members. is constant for all members. 0. kn / m 0kN FixedEnd oments 0kN m F F 0* 6.67kNm F F 0* 0kNm F F 0* 0 knm
8 Slope Deflection ( ) 0 ( ) 0 ( ) 0. ( ) Equilibrium. equations. of. joınts 0 0 ( 6 ) ( ) 0 0 E I Substitude. these. results.in. slope. deflection. equations
9 Example: Find ember end moments and draw shear and moment diagrams 60. kn / m is constant F F 60* 0. knm Slope Deflection. ( ) 0 ( ) 0 ( ) ( ) Equilibrium equations of joints E I Substitude. these. results.in. slope. deflection. equations knm, 6kNm 6 knm, 6kNm 6 knm, knm Normal Force Shear (kn) oment (kn.m)
10 Example: Find member end moments and draw the diagrams of the frame kn m 0 7.m F q0 * L * 0 0 F q0 * L * 0kNm 0 0 F Pab 0**7 7.kNm L 0 F Pa b 0* *7.kNm L 0 Slope Deflection. 0 ( ) 7. 0 ( ) Substitude. these. results.in. slope. deflection. equations 7.6 knm,.9knm.9 knm, 0kNm Shear oment
11 ANALYSIS OF FRAES WITH SIDESWAY A frame will side sway or be displaced to the side when the frame or loading acting on it is nonsymmetric. In the analysis of frames with side sway it is necessary to consider the shear forces at the base of the columns and the horizontal external load must be in equilibrium (force equilibrium equation) in addition to the equilibrium of joints. Example: Using the slopedeflection method determine the end moments of the members and draw the shear force and bending moment diagrams of the frame. is constant throughout the frame. 60.kN kn m.m Q Q F F 60* 0kNm F F 0*.kNm Equilibrium Q 0 0 Q 60 0 Axial deformation is neglected (no change in length of the members) so the lateral displacement of joint and are equal. Slope Deflection. 0 0 ( ). ( ).
12 Shear forces at the base of columns Q Q 60 Q Q 0 Q * Q E I Substitude. these. results.in slope. deflection. equations.6 knm, 6.7kNm 6.79 knm, 0.kNm 0.6 knm, 0.6kNm 76.7 Shear (kn) oment (kn.m)
13 Example: Determine the member end moments of the frame and draw the shear and moment diagrams. 0kN α sinα tanα sinα 0. tanα / sin β sin β 0. tan β / β tan β m. Slope Deflection. ( 0.7 ) sin α ( 0.7 ) sin α ( 0.9 ) tan α tan β ( 0.9 ) tan α tan β ( ).7.7sin β.7 ( ).7.7sin β.7
14 0 Q * ( ) 0 0 Q *. ( ) 0 F 0 Q Q 0 0 x Equilibrium Q Q Q 0 0 Q Substitude. these. results.in slope. deflection. equations.6 knm,.knm. 0 knm, 0kNm 0 knm, knm
15 Example: Determine the member end moments and draw the shear and moment diagrams of given continuous beam. kn m kn m hinge hinge 0 0 Rotations at the left and right side of the internal hinges are different from each other Slope Deflection R ( ) R ( ) L L R L R L Equilibrium R kn m 0 0 * Shear forces at each side of the hinge must be equal to each other 0 6 Force Eq. Equation 0 0
16 L R L L 00 R 00 L kn m kn m hinge hinge Shear Force Diagram Bending oment Diagram
17 Example: Find the member end moments and draw the shear force and bending moment diagrams of the given frame. 0. kn m 0. kn m Frame is symmetrical both loading and geometry. Half of the frame can be analyzed 0 0 kn m Slope Deflection Shear Force Diagram 00 0 Equilibrium Q Q Bending oment Diagram Q
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