MODULE-3.1[full].pdf

STURUCTURAL ANALYSIS
18AT45/SA

 Slope deflection method:
 Introduction
 Derivation of slope deflection equations
 Analysis of continuous beam by slope deflection method subjected to point loads and UDL only.
Numerical problems
MODULE-3.1

 Slope deflection method:
• The credit of giving final touch to slope deflection method and publishing it (1915) goes to prof
G A Maney of the university of Minnesota.
• This method is ideally suited for the analysis of continuous beams and rigid jointed frames.
• Using this method basic unknown like slopes and deflection of joints can be calculated.
 Assumptions:
• All joints are rigid, i,e the angle between any two members in a joint does not change even
after deformation due to loading. Thus at joint A in the fig the angle between members AB
and BC remains ‘θ’ only, even after deformation.

• Distortions due to axial deformations are neglected. Thus, in the frame shown in fig
BB’=CC’=Δ
• Shear deformations are neglected.

 Sign Conventions:
• Moments: Clockwise end moments are positive and anticlockwise end moments are negative.
In fig MBA=20Knm and MBC= -20Knm.
• Rotations: Clockwise rotation of a tangent to elastic curve is positive rotation and
anticlockwise rotation is a negative rotation. In fig θ1 and θ4 are positive rotations and θ2 and θ3
are negative rotation.
• Settlement: Settlement ‘Δ’ is positive if right side support is below left side support. ‘Δ’ is
negative if left side support is below right side support. Thus in fig for beam AB, Δ is positive,
for beam BC Δ is negative.

 Derivation of Slope deflection Equations:
Let AB, shown in fig 1.6(a) be a member of a rigid structure. After loading it undergoes deformation.
Fig 1.6(b) shows deformed shape with all displacements (θA, θB and Δ) taken in their positive senses.
Final moments at end A and B are MAB and MBA.
Now our aim is to derive the relationship between these final end moments and their
displacements θA, θB and Δ.

The development of final moments and deformations are visualized as undergoing in the
following stages:
1. Due to given loadings MFAB and MFBA develop without any rotation at ends. These moments
are similar in a fixed beam and hence called fixed end moments. In fig 1.6(c) these are shown
in their positive senses.
2. Settlement Δ takes place without any rotations. This is similar to the settlement of supports
in fixed beams. From analysis of fixed beams we know, the end moments developed are
6EIΔ / L2 as shown in fig 1.6(d).

3. Moment M’AB comes into play in simply supported beam as shown in fig 1.6(e) to cause end
rotations θA1 and θB1 at A and B respectively.
4. Moments M’BA comes into play in simply supported beam AB, as shown in fig 1.6(f). The end
rotations developed are θA2 and θB2

Moment M’AB and M’BA give final rotations θA and θB to the beam AB. To find the rotations
due to applied moment M in a beam without end rotation (fig 1.7(a)) conjugate beam method
may be used fig 1.7(b)
It may be seen that the rotation at loaded end is ML / 3EI and at unloaded end is ML / 6EI.

 Applications of slope deflection equations:
Using slope deflection equations, rigid jointed structures can be analysed. The method is
illustrated by applying it to the following types of structures:
• Continuous beams
• Frames without side sway
• Frames with side sway

 The following steps are involved in the analysis of continuous beam by slope deflection method:
1. Each span of the continuous beam is taken as fixed beam and fixed end moments are noted.
The standard expressions for finding end moments in fixed beams are used. [ref table 1.1].
Clockwise end moments are to be noted as positive moments and anticlockwise as negative
moments.
2. Using slope deflection equations write all the end moments. In these equations, some of the
rotations and deflections will be unknown.
3. Write the joint equilibrium equations.
4. Solve the joint equilibrium equations to get the unknown rotations and deflections.
5. Substituting the values of unknowns in slope deflection equations and determine the end
moments.
6. Treating each member of the continuous beam as simply supported beam subjected to a given
loading and end moments determine the end reactions and draw shear force and bending
moment diagrams.

 Expressions for finding fixed end moments:

Numerical 1: Analyse the two span continuous beam shown in fig by slope deflection
method and draw bending moment, shear force diagrams. ( Youngs modulus is the same
throughout)

MODULE-3.1[full].pdf

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