Module 3. Analysis of Statically Indeterminate Structures by the Displacement Method. Version 2 CE IIT, Kharagpur

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1 odule 3 nalysis of Statically Indeterminate Structures by the Displacement ethod

2 Lesson 15 The Slope-Deflection ethod: eams (ontinued)

3 Instructional Objectives fter reading this chapter the student will be able to 1. Derive slope-deflection equations for the case beam with yielding supports. 2. Estimate the reactions induced in the beam due to support settlements. 3. nalyse the beam undergoing support settlements and subjected to external loads. 4. Write joint equilibrium equations in terms of moments. 5. Relate moments to joint rotations and support settlements Introduction In the last lesson, slope-deflection equations were derived without considering the rotation of the beam axis. In this lesson, slope-deflection equations are derived considering the rotation of beam axis. In statically indeterminate structures, the beam axis rotates due to support yielding and this would in turn induce reactions and stresses in the structure. Hence, in this case the beam end moments are related to rotations, applied loads and beam axes rotation. fter deriving the slope-deflection equation in section 15.2, few problems are solved to illustrate the procedure. onsider a beam as shown in Fig.15.1.The support is at a higher elevation compared to by an amount Δ. Hence, the member axis has rotated by an amount ψ from the original direction as shown in the figure. Let L be the span of the beam and flexural rigidity of the beam EI, is assumed to be constant for the beam. The chord has rotated in the counterclockwise direction with respect to its original direction. The counterclockwise moment and rotations are assumed to be positive. s stated earlier, the slopes and rotations are derived by superposing the end moments developed due to (1) Externally applied moments on beams. (2) Displacements θ and Δ (settlement)

4 The given beam with initial support settlement may be thought of as superposition of two simple cases as shown in Fig.15.1 (b) and in Fig. 15.1(c). In Fig.15.1b, the kinematically determinate beam is shown with the applied load. For this case, the fixed end moments are calculated by force method. Let φ and φ be the end rotations of the elastic curve with respect to rotated beam axis ' ' (see Fig.15.1c) that are caused by end moments and. ssuming that rotations and displacements shown in Fig.15.1c are so small that Δ tan ψ = ψ = (15.1) l lso, using the moment area theorem, φ and φ are written as ' L ' L φ = θ ψ = (15.2a) 3EI 6EI

5 ' L ' L φ = θ ψ = (15.2b) 3EI 6EI ' ' Now solving for and in terms ofθ andψ, 2EI ' = (2θ + θ 3ψ ) (15.3a) L 2EI ' = (2θ + θ 3ψ ) (15.3b) L Now superposing the fixed end moments due to external load and end moments due to displacements, the end moments in the actual structure is obtained.thus (see Fig.15.1) F = + (15.4a) ' F = + ' (15.4b) ' ' Substituting for and in equation (15.4a) and (15.4b), the slopedeflection equations for the general case are obtained. Thus, F 2EI = + (2θ + θ 3ψ ) (15.5a) L F 2EI = + (2θ + θ 3ψ ) (15.5b) L In the above equations, it is important to adopt consistent sign convention. In the above derivation Δ is taken to be negative for downward displacements. Example 15.1 alculate the support moments in the continuous beam (see Fig.15.2a) having constant flexural rigidity EI throughout,due to vertical settlement of the 4 4 support by 5mm. ssume E =200 GPa and I = 4 m.lso plot quantitative elastic curve.

6 In the continuous beam, two rotations θ and θ need to be evaluated. Hence, beam is kinematically indeterminate to second degree. s there is no external load on the beam, the fixed end moments in the restrained beam are zero (see Fig.15.2b). For each span, two slope-deflection equations need to be written. In span, is below. Hence, the chord rotates in clockwise direction. Thus, ψ is taken as negative.

7 5 ψ = = 1 (1) 5 Writing slope-deflection equation for span, 2EI L For span = 0, Hence, EI = θ 5 ( = 2θ ) + θ 3ψ 3 ( + 3 ) 2 = O. 4EIθ EI (2) Similarly, for beam-end moment at end, in span ( ) = 0.4EI θ = 0.8EIθ EI (3) In span, the support is above support, Hence the chord joining rotates in anticlockwise direction. ψ = ψ = 1 (4) Writing slope-deflection equations for span, = 0.8EIθ + 0.4EIθ 1.2 EI = 0.8EIθ + 0.4EIθ 1.2 EI (5) Now, consider the joint equilibrium of support (see Fig.15.2c)

8 = 0 (6) + Substituting the values of and in equation (6), 0.8EIθ Simplifying, EI + 0.8EIθ + 0.4EIθ 1.2 EI = 0 1.6θ + 0.4θ = 1.2 (7) lso, the support is simply supported and hence, = 0 = 0 = 0.8θ + 0.4θ 1.2 EI 0.8θ + 0.4θ = 1.2 (8) We have two unknowns θ and θ and there are two equations in θ andθ. Solving equations (7) and (8) θ = radians θ = radians (9) Substituting the values of θ and EI in slope-deflection equations, = kn.m = kn.m = kn.m = 0 kn.m () Reactions are obtained from equations of static equilibrium (vide Fig.15.2d)

9 In beam, = 0, = kn( ) R R L R R R = kn( ) = kn( ) = kn( ) The shear force and bending moment diagram is shown in Fig.15.2e and elastic curve is shown in Fig.15.2f.

10 Example 15.2 continuous beam D is carrying a uniformly distributed load of 5 kn/m as shown in Fig.15.3a. ompute reactions and draw shear force and bending moment diagram due to following support settlements.

11 Support 0.005m vertically downwards Support 0.01 m vertically downwards 4 ssume E =200 GPa, I = 1.35 m In the above continuous beam, four rotationsθ and θ D are to be evaluated. One equilibrium equation can be written at each support.hence, solving the four equilibrium equations, the rotations are evaluated and hence the moments from slope-deflection equations. Now consider the kinematically restrained beam as shown in Fig.15.3b. Referring to standard tables the fixed end moments may be evaluated.otherwise one could obtain fixed end moments from force method of analysis. The fixed end moments in the present case are (vide fig.15.3b) F F F = kn.m = kn.m (clockwise) = kn.m (counterclockwise) F = kn.m (clockwise)

12 F D = kn.m (counterclockwise) F D = kn.m (clockwise) (1) In the next step, write slope-deflection equations for each span. In the span, is below and hence the chord joining rotates in the clockwise direction (see Fig.15.3c) ψ = = radians (negative as the chord rotates in the clockwise direction from the original direction) ψ = radians (negative as the chord rotates in the clockwise direction) 0.01 ψ D = = radians (positive as the chord D rotates in the counter clockwise direction from the original direction) (2) Now, writing the expressions for the span end moments, for each of the spans, = = EI(2θ + θ ) EI(2θ + θ ) = = EI(2θ + θ EI(2θ + θ ) )

13 D D EI(2θ + θ 0.001) = D = EI(2 D + θ 41 θ 0.001) (3) For the present problem, four joint equilibrium equations can be written, one each for each of the supports. They are = 0 = 0 = 0 + = 0 = 0 + = 0 D D = 0 = 0 D (4) Substituting the values of beam end moments from equations (3) in equation (4), four equations are obtained in four unknown rotations θ and θ D.They are, ( EI = = 270,000 kn.m ) 2θ θ + = θ + 4θ + θ = θ + D 4 θ + θ = θ + 2θ D = (5) Solving the above sets of simultaneous equations, values ofθ and θ D are evaluated. θ θ θ =.9629 = = radians radians radians θ D = radians (6)

14 Substituting the values in slope-deflection equations the beam end moments are evaluated. = ,000{2( ) + ( ) )} = 0 = ,000{2( ) } = 5.40 kn.m D = ,000{2( = ,000{2( = ,000{2 ( ) + ( ) ) ) } = kn.m } = kn.m 0.001} = kn.m 4 = ,000{ } = 0 kn.m (7) D Reactions are obtained from equilibrium equations. Now consider the free body diagram of the beam with end moments and external loads as shown in Fig.15.3d. R = kn( )

15 R L R R R L R R R D = kn( ) = 27.7 kn( ) = 22.3 kn( ) = kn( ) = kn( ) The shear force and bending moment diagrams are shown in Fig.15.5e. Summary In this lesson, slope-deflection equations are derived for the case of beam with yielding supports. oments developed at the ends are related to rotations and support settlements. The equilibrium equations are written at each support. The continuous beam is solved using slope-deflection equations. The deflected shape of the beam is sketched. The bending moment and shear force diagrams are drawn for the examples solved in this lesson.