Lecture 2 SLOPE DEFLECTION

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STRUCTURAL ANALYSIS VBB 2053

LECTURE 2

SLOPE-DEFLECTION METHOD


Learning Objectives Learn to analyse statically indeterminate

structures using slope-deflection method. Knowing the slope-deflection equations. Discuss the application of slope-deflection

method to beams and frames.


Outline Introduction to slope-deflection method. General procedure of slope-deflection method

of analysis. Derivation of slope-deflection equations. Work examples on slope-deflection method of

analysis: beams and frames.


Statically Indeterminate Structures: Method of Analysis Analyzing any indeterminate structure, it is

necessary to satisfy: Equilibrium – Equilibrium is satisfied when the

reactive forces hold the structure at rest. Compatibility – Compatibility is satisfied when the

various segments of the structure fit together without intentional breaks or overlaps.

Force-displacement – This requirements depend upon the way the material responds – linear elastic response.


Statically Indeterminate Structures: Method of Analysis There are two ways to analyze a statically

indeterminate structure Force Method Displacement Method

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Force Method of Analysis Identifying unknown redundant forces. Satisfying structure’s compatibility equations. This is done by expressing the displacements

in terms of the loads by using the load-displacement relations.

The solution of the resultant equations yields the redundant reactions, and then the equilibrium equations are used to determine the remaining reactions on the structure.

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Displacement Method of Analysis Satisfying equilibrium equations for the

structure. Unknown displacements are written in terms

of the loads by using the load-displacement relations, then solved for the displacements.

Once the displacement obtained, the unknown loads are determined from the compatibility equations using the load-displacement relations.

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Degree of Freedom When a structure is loaded, the node will

undergo unknown displacements. These displacements are referred to as the

“degree of freedom”. Specify degree of freedom is a necessary 1st step when apply displacement method.

In 3D, each node on a beam or frame can have at most 3 linear displacements & 3 rotational displacements.

In 2D, each node can have at most 2 linear displacements & 1 rotational displacement.

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Degree of Freedom The number of these unknowns are referred

to as the degree in which the structure is kinematically indeterminate.

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Any load applied to the beam will cause node A to rotate.

Node B is completely restricted from moving.

Hence, the beam has only one unknown degree of freedom.

Therefore kinematically indeterminate to 1st degree.


Degree of Freedom

10

The beam has nodes at A, B & C

There are 4 degrees of freedom A, B, C and C.

It is kinematically indeterminate to 4th degree.

The frame has 3 degrees of freedom, B, C and B.

It is kinematically indeterminate to 3rd degree.


Slope-Deflection Method The slope-deflection method uses displacements as

unknowns and is referred to as a displacement method.

In the slope-deflection 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 slope-deflection method is that it does not become increasingly complicated to apply as the number of unknowns in the problem increases.

In the slope-deflection method the individual equations are relatively easy to construct regardless of the number of unknowns.


Derivation of Slope-Deflection Equations To derive the general form of the slope-deflection

equation, let us consider the typical span AB of the continuous beam shown below when subjected to arbitrary loading.

The slope-deflection equation can be obtained using the principle of superposition

By considering separately the moments developed at each support due to θA, θB, Δ, P.

Assume clockwise is positive.


Derivation of Slope-Deflection Equations Moment due to angular displacement at A, θA

To determine the moment MAB needed to cause this displacement, we will use the conjugate beam method.

From which we obtain the following relationships.

AAB L

EI4M ABA L

EI2M

03

L2L

EI

M

2

1

3

LL

EI

M

2

1

0M

BAAB

'A

0L3

L2L

EI

M

2

1

3

LL

EI

M

2

1

0M

AABBA

'B


Derivation of Slope-Deflection Equations Moment due to angular displacement at B, θB

The applied moment MBA to the angular displacement B & the reaction moment MAB at the wall can be related as:

BBA L

EI4M BAB L

EI2M


Derivation of Slope-Deflection Equations Moment due to relative linear displacement, Δ If the far node B of the member is displaced relative to A, so that

the cord of the member rotates clockwise & yet both ends do not rotate, then equal but opposite moment and shear reactions are developed in the member.

Moment M can be related to the displacement using conjugate beam method.

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2BAAB

'B

L

EI6MMM

03

LL

EI

M

2

1

3

L2L

EI

M

2

1

0M


Derivation of Slope-Deflection Equations Moment due to loading (P or w) These moments are called Fixed End Moment (FEM). In general, linear and angular displacement of the nodes are

caused by loadings acting on the span of the member. To develop the slope-deflection equation, we must transform

these span loadings into equivalent moment acting at the nodes and then use the load-displacement relationships just derived.

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Derivation of Slope-Deflection Equations Slope-Deflection Equation The end moments due to each displacement and loadings are

added together, the resultant moments at the ends can be written as:

MAB = MθA + MθB + MΔ + Mload

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0FEMLΔ

3θ2θL

2EM

0FEMLΔ

3θ2θL

2EM

BAABBA

ABBAAB

I

I


Derivation of Slope-Deflection Equations Slope-Deflection Equation The slope-deflection equations can be generalized as follows:

MN = internal moment in the near end of the span, E = modulus of elasticity k = span stiffness, k = I/L N, F = near- and far-end slopes or angular displacements of the span

at the supports = span rotation of its cord due to a linear displacement,

= /L FEMN = fixed end moment at the near end support.

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NFNN FEMψ3θθ2Ek2M


Slope-Deflection Equation Pin supported end span Sometimes an end span of a

beam or frame is supported by a pin or roller at its far end.

The moment at the roller or pin is zero provided the angular displacement at this support does not have to be determined.

Use the generalised slope-deflection equation, we have:

Since far end is pinned, FEMF = 0

FEMN can be obtained

0ψ3θθ2Ek20

FEMψ3θθ2Ek2M

FN

NFNN


Slope-Deflection Equation Simplifying, we get:

Only applicable for end span with far end pinned or roller supported

NNN FEMψθEk3M


Work Example 1 Draw the shear & moment diagrams for the

beam shown below.

22


Work Example 2 Draw the shear & moment diagrams for the

beam shown below.

23

A

B

C

40 kN/m 60 kN

6 m1 m 1 m


Analysis of Frames: No Sidesway Slope-deflection method can also be used to analyse

frames. A frame will not sidesway to the left or right provided it

is properly restrained.


Analysis of Frames: No Sidesway No sidesway will occur in an unrestrained frame

provided it is symmetric with respect to both loading and geometry.


Work Example 6: Frame With No Sidesway Determine the moments at each joint of the

frame shown below. EI is constant. 3 spans must be considered in this

case: AB, BC & CD

Step 1: FEM

m.kN8096

wL5FEM

m.kN8096

wL5FEM

2

CB

2

BC

0ψψψ and 0θ θ that Note CDBCABDA


Work Example 6: Frame With No Sidesway Step 2: Slope-Deflection Equations

CDC

CCD

BCCB

CBBC

BBA

BAB

θEI1667.0M

θEI333.0M

80θEI25.0θEI5.0M

80θEI25.0θEI5.0M

θEI333.0M

θEI1667.0M

NFNN FEMψ3θθ2L

IE2M


Work Example 6: Frame With No Sidesway Step 3: Equilibrium Equations

Moment equilibrium at joints B & C, we have:

Substitute MBA, MBC and MCB, MCD into above equations, we get:

Solving simultaneously yields

80θEI25.0θEI833.0

80θEI25.0θEI833.0

BC

CB

0MM ;0M BCBAB 0MM ;0M CDCBC

EI

1.137θθ CB


Work Example 6: Frame With No Sidesway Step 4: Moment Calculation

Using these results, the reactions at the ends of each member can be determined

using equations of equilibrium. The moment diagram for the frame can

be drawn.

kNm9.22M

kNm7.45M

kNm7.45M

kNm7.45M

kNm7.45M

kNm9.22M

DC

CD

CB

BC

BA

AB


Work Example 7: Frame With No Sidesway

Determine the internal moments at each joint of the frame shown below. The moment of inertia for each member is given in the figure. Take E = 200 Gpa.

Take as self study at home….Refer to Hibbeler 7th edition, Example 11-6

30

4.5 m

30 kN 50 kN/m

A

B C E

D

2.4 m 2.4 m

3.6 m320 (106) mm4

260 (106) mm4

160

(106 )

mm

4

80 (106) mm4


Analysis of Frames: Sidesway A frame will sidesway when it or the loading acting on

it is nonsymmetric. Consider the frame shown here.

The loading P causes an unequal moments at joint B and C.

MBC tends to displace joint B to the right.

MCB tends to displace joint C to the left.

Since MBC > MCB, the net result is a sidesway of both joint B & C to the right.


Analysis of Frames: Sidesway When applying the slope-deflection equation to each

column, we must consider the column rotation ( = /L) as an unknown in the equation.

As a result, an extra equilibrium equation must be included in the solution.

The techniques for solving problems for frames with sidesway is best illustrated by work examples.


Work Example 8: Frame With Sidesway Determine the moments at each joint of the

frames shown below. Ends A & D are fixed. Sidesway occurs here. AB = /4 (+ ve)

DC = /6 (+ ve)

AB = (6/4)DC


Work Example 8: Frame With Sidesway Slope-Deflection Equations

(6) )ψ0.1θ333.0(EIM

(5) )ψ0.1θ667.0(EIM

(4) )θ4.0θ8.0(EIM

(3) )θ4.0θ8.0(EIM

(2) )ψ25.2θ0.1(EIM

(1) )ψ25.2θ5.0(EIM

DCCDC

DCCCD

BCCB

CBBC

DCCBA

DCBAB


Work Example 8: Frame With Sidesway Equilibrium Equation

9 unknowns and 6 equations. Thus, we need 3 equilibrium equations.

2 moment equilibrium equations for joints B & C.

Another equilibrium equation obtained from summation of forces (entire frame) in x-direction, since there is horizontal displacement .

(7) 0MM ;0M BCBAB (8) 0MM ;0M CDCBC

(9) 0VV200 ;0F DAX



Considering the free-body diagram of each column separately.

(9) 06

MM

4

MM200

6

MMV0M

4

MMV0M

CDDCBAAB

CDDCDC

BAABAB



Substitute equation (2) and (3) into equation (7). Substitute equation (4) and (5) into equation (8). Equation (1), (2), (5) and (6) into equation (9).

Solve simultaneously, we have

48.208ψEI

66.75θEI

78.243θEI

DC

C

B


Work Example 8: Frame With Sidesway Moment calculation

Using these results & solving for equation (1) to (6)

kNm183M

kNm158M

kNm158M

kNm225M

kNm225M

kNm347M

DC

CD

CB

BC

BA

AB

Lecture 2 SLOPE DEFLECTION

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