Chapter 11: Chapter 11: Displacement Method of Analysis: Slope-Deflection Equations Displacement Method of Analysis: Slope-Deflection Equations

Chapter 11:

Chapter 11:

Displacement Method of Analysis: Slope-Deflection Equations

Displacement Method of Analysis: Slope-Deflection Equations

Displacement Method of Analysis: General Displacement Method of Analysis: General

Procedures Procedures

• Disp method requires satisfying eqm eqn for the Disp method requires satisfying eqm eqn for the structures

structures

• The unknowns disp are written in terms of the The unknowns disp are written in terms of the loads by using the load-disp relations

loads by using the load-disp relations

• These eqn are solved for the disp These eqn are solved for the disp

• Once the disp are obtained, the unknown loads are Once the disp are obtained, the unknown loads are determined from the compatibility eqn using the

determined from the compatibility eqn using the load disp relations

load disp relations

Displacement Method of Analysis: General Displacement Method of Analysis: General

Procedures Procedures

• When a structure is loaded, specified points on it When a structure is loaded, specified points on it called nodes, will undergo unknown disp

called nodes, will undergo unknown disp

• These disp are referred to as the degree of These disp are referred to as the degree of freedom

freedom

• The no. of these unknowns is referred to as the The no. of these unknowns is referred to as the degree in which the structure is kinematically degree in which the structure is kinematically

indeterminate indeterminate

• We will consider some e.g.s We will consider some e.g.s

Displacement Method of Analysis: General Displacement Method of Analysis: General

Procedures Procedures

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

rotate

• Node B is completely restricted from moving Node B is completely restricted from moving

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

freedom

• The beam has nodes at A, B & C The beam has nodes at A, B & C

• There are 4 degrees of freedom There are 4 degrees of freedom   _{A A} , ,   _{B B} , ,   _{C C} and and   _{C C}

Slope-deflection equation Slope-deflection equation

• Slope deflection method requires less work both to Slope deflection method requires less work both to write the necessary eqn for the solution of a

write the necessary eqn for the solution of a

problem& to solve these eqn for the unknown disp problem& to solve these eqn for the unknown disp

& associated internal loads

& associated internal loads

• General Case General Case

• To develop the general form of the slope-deflection To develop the general form of the slope-deflection eqn, we will consider the

eqn, we will consider the typical span AB of the

typical span AB of the

continuous beam when

continuous beam when

Slope-deflection equation Slope-deflection equation

• Angular Disp Angular Disp

• Consider node A of the member to rotate Consider node A of the member to rotate   _{A A} while while its while end node B is held fixed

its while end node B is held fixed

• To determine the moment M To determine the moment M _{AB AB} needed to cause this needed to cause this disp, we will use the conjugate beam method

disp, we will use the conjugate beam method

Slope-deflection equation Slope-deflection equation

• Angular Disp Angular Disp

3 0 2 2

1 3

2 1

0

3 0 2 2

1 3

2 1

0

' '



 



 



 



 



 

 

 



 



 









 



 



 



 



 

 

 



 



 









L L EI L

L M EI L

M M

L L EI

L M EI L

M M

AB A BA

B

BA AB

A



Slope-deflection equation Slope-deflection equation

• Angular Disp Angular Disp

• From which we obtain the following: From which we obtain the following:

• Similarly, end B of the beam rotates to its final Similarly, end B of the beam rotates to its final position while end A is held fixed

position while end A is held fixed

• We can relate the applied moment M We can relate the applied moment M _{BA BA} to the to the angular disp

angular disp   _{B B} & the reaction moment M & the reaction moment M _{AB AB} at the at the wall wall

2

4

A BA

A

AB L

M EI L

M  EI   

B AB

B BA

M EI

M 4 EI  ² 





Slope-deflection equation Slope-deflection equation

• Relative linear disp Relative linear disp

• If the far node B if the member is displaced relative If the far node B if the member is displaced relative to A, so that the cord of the member rotates

to A, so that the cord of the member rotates

clockwise & yet both ends do not rotate then equal clockwise & yet both ends do not rotate then equal

but opposite moment and shear reactions are but opposite moment and shear reactions are

developed in the member developed in the member

• Moment M can be related to the disp using Moment M can be related to the disp using conjugate beam method

conjugate beam method

Slope-deflection equation Slope-deflection equation

• Relative linear disp Relative linear disp

• The conjugate beam is free at both ends since the The conjugate beam is free at both ends since the real member is fixed support

real member is fixed support

• The disp of the real beam at B, the moment at end The disp of the real beam at B, the moment at end B’ of the conjugate beam must have a magnitude of B’ of the conjugate beam must have a magnitude of

  as indicated as indicated

   

 











 

 

 

 

 





6

3 0 2

1 3

2 2

1

0

M EI M

M

L L EI M L L

EI

M

M

Slope-deflection equation Slope-deflection equation

• Fixed end moment Fixed end moment

• In general, linear & angular disp of the nodes are In general, linear & angular disp of the nodes are caused by loadings acting on the span of the

caused by loadings acting on the span of the member

member

• To develop the slope-deflection eqn, we must To develop the slope-deflection eqn, we must transform these span loadings into equivalent transform these span loadings into equivalent

moment acting at the nodes & then use the load- moment acting at the nodes & then use the load-

disp relationships just derived

disp relationships just derived

Slope-deflection equation Slope-deflection equation

• Slope-deflection eqn Slope-deflection eqn

• If the end moments due to each disp & loadings are If the end moments due to each disp & loadings are added together, the resultant moments at the ends added together, the resultant moments at the ends

can be written as:

can be written as:

0 3

2 2

0 3

2 2



 



 



 



 



  

 



 



 



 



 



 



 



  

 



 



 

BA A

B BA

AB B

A AB

L FEM L

E I M

L FEM L

E I M









Slope-deflection equation Slope-deflection equation

• Slope-deflection eqn Slope-deflection eqn

• The results can be expressed as a single eqn The results can be expressed as a single eqn

 

support end

near at the

moment end

fixed

disp linear a

to due cord its

of rotation span

supports at the

span the

of disp angular or

slopes end

far and near ,

stiffness span

&

elasticity of

modulus ,

span the

of end near at the

moment internal

0 3

2 2





















FEM N F N

k E M N

N F

N

N Ek FEM

M













Slope-deflection equation Slope-deflection equation

• Pin supported end span Pin supported end span

• Sometimes an end span of a beam or frame is Sometimes an end span of a beam or frame is supported by a pin or roller at its far end

supported by a pin or roller at its far end

• The moment at the roller or pin is zero provided the The moment at the roller or pin is zero provided the angular disp at this support does not have to be

angular disp at this support does not have to be determined

determined

 

 ^{2 3}  ⁰

2 0

3 2

2





























F N

N F

N N

Ek

FEM Ek

M

Slope-deflection equation Slope-deflection equation

• Pin supported end span Pin supported end span

• Simplifying, we get: Simplifying, we get:

• This is only applicable for end span with far end This is only applicable for end span with far end pinned or roller supported

pinned or roller supported

 _N  _N

N Ek FEM

M  3    

Draw the shear and moment diagrams for the beam where EI is constant.

Example 11.1

Example 11.1

Slope-deflection equation

2 spans must be considered in this problem Using the formulas for FEM, we have:

Solution Solution

m wL kN

FEM

m wL kN

FEM

CB BC

. 8

. 20 10

) 6 ( 6 ) 20

(

. 2

. 30 7

) 6 ( 6 ) 30

(

2 2

2 2



















Slope-deflection equation

•Note that (FEM)

is –ve and

•(FEM)

= (FEM)

since there is no load on span AB

•Since A & C are fixed support, 

= 

=0

•Since the supports do not settle nor are they displaced up or down, 

= 

= 0

Solution Solution

 

 

(1)

0 ) 0 ( 3 )

0 ( 8 2 2

3 2

2

B AB

N F

N N

EI E I M

L FEM E I

M

















 



 



 





 



 



 

Slope-deflection equation

Solution Solution

(4) 3 10.8

(3) 7.2

3 - 2

(2)

2 Similarly,









B CB

B BC

B BA

M EI M EI M EI







Equilibrium eqn

The necessary fifth eqn comes from the condition of moment equilibrium at support B

Here M

& M

are assumed to act in the +ve direction to be consistent with the slope-deflection eqn

Solution Solution

(5)

 0

 _BC

BA M

M

Equilibrium equation

Solution Solution

kNm M

kNm .

M

EI θ .

BA AB

B

; 09

. 3

; 54

1

: gives (4)

to (1) eqn

into value

this sub

- Re

17 6

: gives (5)

eqn into

(3) and

(2) eqn

Sub





 





Equilibrium equation

•Using these results, the shears at the end spans are determined.

•The free-body diagram of the entire beam & the shear &

moment diagrams are shown.

Solution

Solution

Analysis of Frames: No sidesway Analysis of Frames: No sidesway

• A frame will not sidesway to the left or right A frame will not sidesway to the left or right provided it is properly restrained

provided it is properly restrained

• No sidesway will occur in an unrestrained frame No sidesway will occur in an unrestrained frame provided it is symmetric wrt both loading and provided it is symmetric wrt both loading and

geometry

geometry

Determine the moments at each joint of the frame. EI is constant.

Example 11.6

Example 11.6

Slope-deflection eqn

3 spans must be considered in this case: AB, BC & CD

Solution Solution

0 and

0 that

Note

. 96 80

) 5 (

. 96 80

) 5 (

2 2























CD BC

AB D

A CB BC

m wL kN

FEM

m wL kN

FEM











Slope-deflection eqn We have

Solution Solution

 

C DC

C CD

B C

CB

C B

BC

B BA

B AB

N F

N N

EI

M EI

M EI EI

M EI EI

M EI

M EI

M

L FEM E I

M

  

    







1667 .

0 . 333

0 . 5 0 . 25 80 0 0 . . 333 5 0 . 25 80 0 0 . 1667

3 2

2

   

   

 





 



 



 

Slope-deflection eqn

•The remaining 2 eqn come from moment equlibrium at joints B

& C, we have:

•Solving for these 8 eqn, we get:

Solution Solution

0 0









CD CB

BC BA

M M

M M

80 25

. 0 833

. 0

80 25

. 0 833

. 0











B C

C B

EI EI

EI EI









Slope-deflection eqn

Solving simultaneously yields:

Solution Solution

kNm M

kNm M

kNm M

kNm M

kNm M

kNm M

EI

DC CD

CB BC

BA AB

C B

9 . 22

; 7

. 45

7 . 45

; 7

. 45

7 . 45

; 9

. 22

1 . 137























 



Analysis of Frames: Sidesway Analysis of Frames: Sidesway

• A frame will sidesway when it or the loading acting on it is A frame will sidesway when it or the loading acting on it is nonsymmetric

nonsymmetric

• The loading P causes an unequal moments at joint B & C The loading P causes an unequal moments at joint B & C

• M M _{BC BC} tends to displace joint B tends to displace joint B to the right

to the right

• M M _{CB CB} tends to displace joint C tends to displace joint C to the left

to the left

• Since M Since M _{BC BC} > M > M _{CB CB} , the net result , the net result is a sidesway of both joint B & C is a sidesway of both joint B & C

to the right

to the right

Analysis of Frames: Sidesway Analysis of Frames: Sidesway

• When applying the slope-deflection eqn to each When applying the slope-deflection eqn to each column, we must consider the column rotation, column, we must consider the column rotation,  

as an unknown in the eqn as an unknown in the eqn

• As a result, an extra eqm eqn must be included in As a result, an extra eqm eqn must be included in the solution

the solution

• The techniques for solving problems for frames The techniques for solving problems for frames with sidesway is best illustrated by e.g.

with sidesway is best illustrated by e.g.

Analysis of Frames: Sidesway Analysis of Frames: Sidesway

• When applying the slope-deflection eqn to each When applying the slope-deflection eqn to each column, we must consider the column rotation, column, we must consider the column rotation,  

as an unknown in the eqn as an unknown in the eqn

• As a result, an extra eqm eqn must be included in As a result, an extra eqm eqn must be included in the solution

the solution

• The techniques for solving problems for frames The techniques for solving problems for frames with sidesway is best illustrated by e.g.

with sidesway is best illustrated by e.g.

Explain how the moments in each joint of the two-story frame. EI is constant.

Example 11.9

Example 11.9

Slope-deflection eqn

We have 12 equations that contain 18 unknowns

Solution

Solution

Slope-deflection eqn

No FEMs have to be calculated since the applied loading acts at the joints

Members AB & FE undergo rotations of 

= 

/5 Members AB & FE undergo rotations of 

= 

/5 Moment eqm of joints B, C, D and E, requires

Solution Solution

(16)

0

(15)

0

(14)

0

(13)

0





















DE DC

CD CB

BC BE

BA

M M

M

M M

M M

M M

M

Slope-deflection eqn

Similarly, shear at the base of columns must balance the applied horizontal loads

Solution Solution

(18)

0 120

0 V

V 80

40 0

(17)

5 0 40 5

0 V

V 40

0

FE AB

x

ED BC

x

 

 

















 

 















FE EF

BA AB

DE ED

CB BC

M M

M M

F

M M

M M

F

Slope-deflection eqn

•Sub eqn (1) to (12) into eqn (13) to (18)

•These eqn can be solved simultaneously

•The results are resub into eqn (1) to (12) to obtain the moments

at the joints

Solution

Solution