direct stiffness method, Lecture notes of Civil Engineering

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Module

Analysis of Statically

Indeterminate

Structures by the

Displacement Method

Lesson

The Slope-Deflection

Method: An

Introduction

In Fig.14.01b, we have nodes at A,B,C and D. Under the action of lateral loads and , this continuous beam deform as shown in the figure. Here axial

deformations are neglected. For this beam we have five degrees of freedom

P 1 , P 2 P 3

θ (^) A , θ B , θ C ,^ and^ as indicated in the figure. In Fig.14.02a, a symmetrical plane

frame is loaded symmetrically. In this case we have only two degrees of freedom

θ (^) D Δ D

θ (^) B and^ θ (^) C. Now consider a frame as shown in Fig.14.02b. It has three

degrees of freedom viz. (^) θ (^) B , (^) θ C and (^) Δ (^) D as shown. Under the action of horizontal

and vertical load, the frame will be displaced as shown in the figure. It is observed that nodes at B and C undergo rotation and also get displaced horizontally by an equal amount.

Hence in plane structures, each node can have at the most one linear displacement and one rotation. In this module first slope-deflection equations as applied to beams and rigid frames will be discussed.

Instructional Objectives

After reading this chapter the student will be able to

1. Calculate kinematic degrees of freedom of continuous beam.
2. Derive slope-deflection equations for the case beam with unyielding supports.
3. Differentiate between force method and displacement method of analyses.
4. State advantages of displacement method of analysis as compared to force method of analysis.
5. Analyse continuous beam using slope-deflection method.

14.1 Introduction

In this lesson the slope-deflection equations are derived for the case of a beam with unyielding supports .In this method, the unknown slopes and deflections at nodes are related to the applied loading on the structure. As introduced earlier, the slope-deflection method can be used to analyze statically determinate and indeterminate beams and frames. In this method it is assumed that all deformations are due to bending only. In other words deformations due to axial forces are neglected. As discussed earlier in the force method of analysis compatibility equations are written in terms of unknown reactions. It must be noted that all the unknown reactions appear in each of the compatibility equations making it difficult to solve resulting equations. The slope-deflection equations are not that lengthy in comparison. The slope-deflection method was originally developed by Heinrich Manderla and Otto Mohr for computing secondary stresses in trusses. The method as used today was presented by G.A.Maney in 1915 for analyzing rigid jointed structures.

14.2 Slope-Deflection Equations

Consider a typical span of a continuous beam AB as shown in Fig.14.1.The beam has constant flexural rigidity EI and is subjected to uniformly distributed loading and concentrated loads as shown in the figure. The beam is kinematically indeterminate to second degree. In this lesson, the slope-deflection equations are derived for the simplest case i.e. for the case of continuous beams with unyielding supports. In the next lesson, the support settlements are included in the slope-deflection equations.

The fixed end moments are required for various load cases. For ease of calculations, fixed end forces for various load cases are given at the end of this lesson. In the actual structure end A rotates by (^) θ (^) A and end B rotates by (^) θ B. Now

it is required to derive a relation relating (^) θ (^) A and (^) θ B with the end moments and

. Towards this end, now consider a simply supported beam acted by moment

M (^) AB ′ M ′ BA M (^) AB ′^ at A as shown in Fig. 14.4. The end moment M (^) AB ′^ deflects the beam as

shown in the figure. The rotations θ A ′^ and θ B ′^ are calculated from moment-area

theorem.

AB A

M L

EI

′ = (14.1a)

AB B

M L

EI

θ

′ = − (14.1b)

Now a similar relation may be derived if only M (^) BA ′^ is acting at end B (see Fig.

14.4).

BA B

M L

EI

θ

′′ = and (14.2a)

BA A

M L

EI

θ

′′ = − (14.2b)

Now combining these two relations, we could relate end moments acting at A and B to rotations produced at A and B as (see Fig. 14.2c)

EI

M L

EI

M ABL BA

A 3 6

' ' θ = − (14.3a)

EI

M L

EI

M BAL BA

B

θ = (14.3b)

Solving for M AB ′^ and M BA ′^ in terms of θ A and θ B ,

AB A B L

EI

M ′ = θ + θ (14.4)

BA L B A

EI

M ′ = θ + θ (14.5)

Now writing the equilibrium equation for joint moment at A (see Fig. 14.2).

AB

F

M AB = MAB + M ′ (14.6a)

Similarly writing equilibrium equation for joint B

BA

F

M BA = MBA + M ′ (14.6b)

Substituting the value of M (^) A ′ (^) B from equation (14.4) in equation (14.6a) one

obtains,

A B

F AB AB L

EI

M = M + θ + θ (14.7a)

Similarly substituting M (^) BA ′^ from equation (14.6b) in equation (14.6b) one obtains,

B A

F BA BA L

EI

M = M + θ + θ (14.7b)

Sometimes one end is referred to as near end and the other end as the far end. In that case, the above equation may be stated as the internal moment at the near end of the span is equal to the fixed end moment at the near end due to

external loads plus L

2 EI

times the sum of twice the slope at the near end and the

slope at the far end. The above two equations (14.7a) and (14.7b) simply referred to as slope–deflection equations. The slope-deflection equation is nothing but a load displacement relationship.

curve of the beam is drawn in Fig.14.5b in order to identify degrees of freedom. By fixing the support or restraining the support against rotation, the fixed-fixed beams area obtained as shown in Fig.14.5c.

B

(b). Fixed end moments and are calculated referring to the

Fig. 14. and following the sign conventions that counterclockwise moments are positive.

F BC

F BA

F M (^) AB , M , M F MCB

2 2 2

21 kN.m 12 6

F M (^) AB

× × ×

M (^) BAF = − 21 kN.m 4 42 5.33 kN.m 12

F M (^) BC

×

M (^) CBF = − 5.33 kN.m (1)

(c) Slope-deflection equations Since ends A and C are fixed, the rotation at the fixed supports is zero,

θ A = θ C = 0. Only one non-zero rotation is to be evaluated for this problem. Now,

write slope-deflection equations for span AB and BC.

( 2 )

A B

F AB AB l

EI

M = M + θ + θ

AB B

EI

M θ 6

BA^21 B A

l

EI

M =− + θ + θ

BA B

EI

M θ 6

MBC = 5. 33 + EI θ B (4)

MCB = − 5. 33 + 0. 5 EI θ B (5)

(d) Equilibrium equations In the above four equations (2-5), the member end moments are expressed in

terms of unknown rotation θ B. Now, the required equation to solve for the rotation

θ B is the moment equilibrium equation at support B. The free body diagram of

support B along with the support moments acting on it is shown in Fig. 14.5d. For, moment equilibrium at support (^) B , one must have,

∑ M^ B =^0 M^ BA +^ MBC =^0 (6)

Substituting the values of M (^) BA and MBC in the above equilibrium equation,

5. 33 0 6

− 21 + B + + EI B =

EI

θ θ

⇒ 1. 667 θ BEI = 15. 667

B EI EI

θ = ≅ (7)

(e) End moments After evaluating (^) θ B , substitute it in equations (2-5) to evaluate beam end moments. Thus,

Example 14.

Draw shear force and bending moment diagram for the continuous beam loaded as shown in Fig.14.6a.The relative stiffness of each span of the beam is also shown in the figure.

ABCD

For the cantilever beam portion CD , no slope-deflection equation need to be written as there is no internal moment at end D. First, fixing the supports at B and C , calculate the fixed end moments for span AB and BC. Thus,

16 kN.m 12

F M (^) AB

×

M (^) BAF = − 16 kN.m

2 2

7.5 kN.m 6

F M (^) BC

× ×

M (^) CBF = − 7.5 kN.m (1)

In the next step write slope-deflection equation. There are two equations for each span of the continuous beam.

M (^) AB 16 0.25 EI (^) B 16 0.25 EI 18.04 kN.m EI

= + θ = + × =

M (^) BA 16 0.5 EI (^) B 16 0.5 EI 11.918 kN.m EI

= − + θ = − + × = −

8.164 9. M (^) BC 7.5 1.334 EI 0.667 EI ( ) 11.918 kN.m EI EI

= + × + =

M (^) CB 7.5 0.667 EI 1.334 EI ( ) 15 kN.m EI EI

= − + × + − = − (8)

Reactions are obtained from equilibrium equations (ref. Fig. 14.6c)

RA × 8 − 18. 041 − 3 × 8 × 4 + 11. 918 = 0

RA = 12.765 kN

RBR = 5 − 0.514 kN =4.486 kN

RBL = 11.235 kN

RC = 5 + 0.514 kN =5.514 kN

The shear force and bending moment diagrams are shown in Fig. 14.6d.

For ease of calculations, fixed end forces for various load cases are given in Fig. 14.7.

Summary

In this lesson the slope-deflection equations are derived for beams with unyielding supports. The kinematically indeterminate beams are analysed by slope-deflection equations. The advantages of displacement method of analysis over force method of analysis are clearly brought out here. A couple of examples are solved to illustrate the slope-deflection equations.