Slop Deflection

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University of Ottawa Fall 2013

CVG 4148 – Theory of Structures II

Theory of Structures II

CVG 4148

Instructor: A. Abdulridha

Slop Deflection Method

CVG 4148 – Theory of Structures II Slop Deflection-1






CVG 4148 – Theory of Structures II Slop Deflection-2

Degrees of Freedom:When a structure is loaded, specified points on it,

called nodes, will undergo unknown displacements.

These displacements are referred to as the degrees

of freedom for the structure, and in the displacement

method of analysis it is important to specify these

degrees of freedom since they become the unknowns

when the method is applied.

Any load P applied to the beam will cause node A only to

rotate (neglecting axial deformation), while node B is

completely restricted from moving. Hence the beam has

only one unknown degree of freedom, θ A






CVG 4148 – Theory of Structures II Slop Deflection - 3

Again, if we neglect axialdeformation of the

members, an arbitrary

loading P applied to the

frame can cause nodes B

and C to rotate, and these

nodes can be displaced

horizontally by an equal

amount. The frame

therefore has three

degrees of freedom θB. θC

and∆C

Slope-Deflection Equations:General Case:







In order to develop the general form of the slope-

deflection equations, we will consider the typical span AB

of a continuous beam which is subjected to the arbitraryloading and has a constant EI. We wish to relate the

beam’s internal end moments and in terms of its three

degrees of freedom, namely, its angular displacements θA

and θB and linear displacement ∆ which could be caused

by a relative settlement between the supports.

Moments and angular displacements will be considered

positive when they act clockwise on the span

Furthermore, the linear displacement ∆ is considered

positive as shown, since this displacement causes the

cord of the span and the span’s cord angle to rotateclockwise.

The slope-deflection equations can be obtained byusing the principle of superposition

Angular Displacement at A, θA

negative

why M exist?

shear by rot?







The deflection of the “real beam” is to be zero at A and B, and

therefore the corresponding sum of the moments at each end and

of the conjugate beam must also be zero.

Angular Displacement at B, θB

In a similar manner forget fixed? actuse vertical?







Relative Linear Displacement, ∆If the far node B of the member is displaced relative to A, so that

the cord of the member rotates clockwise (positive displacement)

Due to the displacement of the real beam at B, the

moment at the end of the conjugate beam must have a

magnitude of ∆

why =?







Fixed-End Moments:In order to develop the slope-deflection equations, we must transform

any span loadings into equivalent moments acting at the nodes.

Since we require the slope at each end to be zero

For general span loading

-rot cause fixed?




CVG 4148 – Theory of Structures II Slop Deflection- 8

Fixed End Moments (FEM):







Slope-Deflection Equation:If the end moments due to each displacement and the loading

are added together, the resultant moments at the ends







Pin-Supported End Span:

we can modify the general slope-deflection equation so that it

has to be applied only once to the span rather than twice







Example:Draw the shear and moment diagrams for the beam shown. EI

is constant.

Example:

Determine the moment at A and B for the beam shown. Thesupport at B is displaced (settles) 80 mm. Take I = 511062

mm4.







Analysis of Frames: No Sidesway A frame will not sidesway, or be displaced to the left or

right, provided it is properly restrained.

Also, no sidesway will occur in an unrestrained frame provided

it is symmetric with respect to both loading and geometry

For both cases the term ψ in the slope-deflection equations is

equal to zero, since bending does not cause the joints to have

a linear displacement.







Example:Determine the moments at each joint of the frame shown. EI is

constant.

Analysis of Frames: Sidesway

A frame will sidesway,

or be displaced to the

side, when it or the

loading acting on it is

nonsymmetric.







Example:Determine the moments at each joint of the frame shown. The

supports at A and D are fixed and joint C is assumed pin

connected. EI is constant for each member.

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Slop Deflection