Slope-Deflection Method: Framesfast10.vsb.cz/koubova/SDM_frame.pdf · Slope-Deflection Method:...

Slope-Deflection

Method: Frames 1) Without Side-sway

2) With Side-sway

Slope-Deflection Method:

Frames Without Frames Without Frames Without Frames Without SideSideSideSide----sssswaywaywayway� In frames axial deformations are much smaller than the

bending deformations and are neglected in the analysis.

� With this assumption some frames will not side-sway

(the frames will not be displaced to the right or left).

� Frames do not side-sway if:

� They are restrained against side-sway.

� The frame geometry and loading is symmetrical.

� The analysis of such rigid frames by slope deflection

equation essentially follows the same steps as that of

continuous beams.

1. Degrees of freedom

� The frame is kinematically indeterminate to first

degree. Only one joint rotation ϕϕϕϕb is unknown.

Slope-Deflection Method: Frames Without Side-sway,

Example 1

a

2

EI = const.

2 2

b

c

F1 = 5 kN

d

F2 = 10 kN

4


Example 1

2. Fixed end moments are calculated referring to the table.

2

2

1

8

1

8

0

bd bd

db bd

cb bc

M F l

M F l

M M

= − ⋅ ⋅

= + ⋅ ⋅

= =

a

2

EI = const.

2 2

b

c

F1 = 5 kN

d

F2 = 10 kN

4


Example 1

a

2

EI = const.

2 2

b

c

F1 = 5 kN

d

F2 = 10 kN

4

3. Express internal end moments

by slope-deflection equations.

( )

( )

( )

( )

22

22

22

22

cbcb c bcb

bcbc b ccb

bdbd b dbd

dbdb d bbd

EIM M

l

EIM M

l

EIM M

l

EIM M

l

ϕ ϕ

ϕ ϕ

ϕ ϕ

ϕ ϕ

= + ⋅ +

= + ⋅ +

= + ⋅ +

= + ⋅ +


Example 1

a

2

EI = const.

2 2

b

c

F1 = 5 kN

d

F2 = 10 kN

4

4. Equilibrium equations

� End moments are expressed in

terms of unknown rotation ϕb.

Now, the required equation to

solve for the rotation ϕb is the

moment equilibrium equation

at rigid joint b.

b

211 ⋅= FM F

b bdM

bcM

10 : 0Fb bc bd bM M M M= + + =∑

5. End moments

� After evaluating ϕb, substitute it

to evaluate beam end moments.


Example 16. Shear forces

� Now, reactions at supports are evaluated using equilibrium

equations. Shear forces are equal to plus/minus this reactions.

( )

( )

1

1

cb cb cb bcbc

bc bc cb bcbc

V R M Ml

V R M Ml

= = − −

= − = − + +

2

2

1

2

1

2

bdbd bd bd db

bd

bddb db bd db

bd

lV R F M M

l

lV R F M M

l

= = ⋅ − −

= − = − ⋅ + +

4c b

Mcb Mbc

Rcb=Vcb Rbc=-Vbc

2b d

Mbd Mbd

Rbd=Vbd Rdb=-Vdb

F2 = 10 kN

2


Example 1

7. Normal forces

� The required equations to

solve for normal forces are

the force equilibrium

equations at rigid joint b.

0 : 0xb bd bcF N V= − =∑

10 : 0zb bc bdF N V F= + + =∑b

1F Vbd

Nbd

Vbc

Nbc

a

2

EI = const.

2 2

b

c

F1 = 5 kN

d

F2 = 10 kN

4


Example 1

8. Draw normal and shear force diagrams.

Ncb

= N

bc

Nbd = Ndb0

-F1

Vcb

= V

bc

Vbd

Vdb


Example 1

8. Draw bending moment diagram

Mcb

-Mbc

-F1·2 Mbd -Mbd

MF


Frames With Frames With Frames With Frames With SideSideSideSide----sssswaywaywayway� In frames axial deformations are much smaller than the

bending deformations and are neglected in the analysis.

� The frames will side-sway (the frames will be displaced

to the right or left or will be displaced up or down).

� Frames side-sway if:

� They are not restrained against side-sway.

� The frame geometry and loading is not symmetrical.


� Frame is kinematically

indeterminate to third

degrees (three unknown

displacements - two

rotations ϕϕϕϕb, ϕϕϕϕc and

one translation ub = uc).

Slope-Deflection Method: Frames With Side-sway,

Example 2

a

2I

4

b c

F=

12

kN

d

3

I

I

3


Example 2


1

8

1

8

0

0

ab ab

ba ab

bc cb

cd dc

M F l

M F l

M M

M M

= − ⋅ ⋅

= + ⋅ ⋅

= =

= =

a

2I

4

b c

F=

12

kN

d

3

I

I

3


Example 2

3. Express internal end moments


( )abbaab

ababab l

EIMM ψϕϕ ⋅−+⋅+= 32

2

bar rotation

tan

tan

b c

ab abab

cd cdcd

u u

l

l

ψ ψ

ψ ψ

= = ∆

∆≈ =

∆≈ = a

b c

d

∆∆∆∆ ∆∆∆∆

ψψψψab

ψψψψcd

chords of

the elastic curve


Frames With Side-sway,

Example 23. Express internal end moments


( )

( )

( )

( )

( )

( )

22 3

22 3

22 3

22 3

22 3

22 3

ababab a b ab

ab

abbaba b a ab

ab

bcbcbc b c bc

bc

bccbcb c b bc

bc

cdcdcd c d cd

cd

cddcdc d c cd

cd

EIM M

l

EIM M

l

EIM M

l

EIM M

l

EIM M

l

EIM M

l

ϕ ϕ ψ

ϕ ϕ ψ

ϕ ϕ ψ

ϕ ϕ ψ

ϕ ϕ ψ

φ φ ψ

= + ⋅ + − ⋅

= + ⋅ + − ⋅

= + ⋅ + − ⋅

= + ⋅ + − ⋅

= + ⋅ + − ⋅

= + ⋅ + − ⋅

a

2I

4

b c

F=

12

kN

d

3

I

I

3


Example 2


� End moments are expressed in terms of unknown rotations ϕb, ϕc

and translation ∆. Now, the required equation to solve are the moment equilibrium equations at rigid joints b and c and then so-called sideway equation.

0 : 0b ba bcM M M= + =∑

a

2I

4

b c

F=

12

kN

d

3

I

I

3

0 : 0c cb cdM M M= + =∑


Example 2


� Sideway equation Sideway equation Sideway equation Sideway equation (the horizontal equilibrium)

0 : 0x ba cdF V V= + =∑

b c

baV cdV

( )

1

2

1

abba ab ba

ab

cd cd dccd

lV F M M

l

V M Ml

= − ⋅ + +

= − −

a

2I

4

b c

F=

12

kN

d

3

I

I

3

Part of the frame containing

the joints with the same

translation ∆ “cut off”.


Example 2

4. Solving equilibrium equations

EI

EI

EI

EIEIEI

EIEIEI

EIEIEI

d

c

cb

cb

cb

99,14

88,4

76,2

6556,067,033,0

067,033,25,0

933,05,033,2

=∆

=

−=

=∆⋅⋅+⋅⋅−⋅⋅−=∆⋅⋅−⋅⋅+⋅⋅

−=∆⋅⋅−⋅⋅+⋅⋅

ϕ

ϕ

ϕϕϕϕϕϕ


Example 2

a

2I

4

b c

F=

12

kN

d

3

I

I

3

kNm75,63

2

3

2

kNm50,33

2

3

4

kNm50,32

kNm32,02

kNm32,033

49

kNm84,1533

29

−=∆⋅⋅−⋅⋅=

−=∆⋅⋅−⋅⋅=

=⋅+⋅=

−=⋅+⋅=

=∆⋅−⋅+=

−=∆⋅−⋅+−=

EIEIM

EIEIM

EIEIM

EIEIM

EIEIM

EIEIM

cdc

ccd

bccb

cbbc

bba

bab

ϕ

ϕ

ϕϕ

ϕϕ

ϕ

ϕ

5. End moments

� After evaluating ϕb, ϕc, ∆, substitute them to evaluate end moments.


Example 2

6. Shear forces

( )

( )

( )

( ) kN79,050,332,04

1

1

kN79,050,332,04

1

1

−=+−−=

++−=−=

−=−+=

−−==

cb

cbbcbc

cbcb

bc

cbbcbc

bcbc

V

MMl

RV

V

MMl

RV

( )

( ) kN41,332,084,153126

1

2

1

kN59,832,084,153126

1

2

1

−=+−⋅−=

++⋅−=−=

=−+⋅=

−−⋅==

ba

baabab

abbaba

ab

baabab

ababab

V

MMl

Fl

RV

V

MMl

Fl

RV

4b c

Mbc Mcb

Rbc=Vbc Rcb=-Vcb

3a b

MabMba

Rab=Vab Rba=-Vba

F = 12 kN

3 3c d

Mcd Mdc

Rcd=Vcd Rdc=-Vdc

( )

( )

( )

( ) kN41,375,650,33

1

1

kN41,375,650,33

1

1

=−−−=

++−=−=

=++=

−−==

dc

dccdcd

dcdc

cd

dccdcd

cdcd

V

MMl

RV

V

MMl

RV


Example 2

7. Normal forces

� The required equations to solve for normal forces are the

force equilibrium equations at rigid joint b and c.

kN41,3

0

0

−==

=−=∑

bc

babc

babc

xb

N

VN

VN

F

kN79,0

0

0

=−=

=+=∑

ba

bcba

bcba

zb

N

VN

VN

F

kN41,3

0

0

−=−=

=−−=∑

cb

cdcb

cdcb

xc

N

VN

VN

F

kN79,0

0

0

−==

=−=∑

cd

cbcd

cbcd

zc

N

VN

VN

F

b cVbc

Vba

Vcb

Vcd

Nbc

Nba

Ncb

Ncd


Example 2

8. Draw normal and shear force diagrams

Nab

= N

ba=

0,7

9

Nbc= Ncb = -3,41 Ncd =

Ndc =

-0,79

Vcd =

Vdc =

3,41

V ab=

8,5

9

Vba

= -

3,41

Vbc= Vcb = -0,79


Example 2

8. Draw bending moment diagram

Mcd =

-3,50

Mab

= -

15,8

4Mbc= -0,32

kNm92,9359,884,15

3

=⋅+−=⋅+=

F

ababF

M

VMM

-Mcb = -3,50

-Mba

= -

0,32

-Mdc =

6,75MF

= 9

,92


Example 3


� Two unknown displacements

- one rotation ϕϕϕϕb- one translation ua = ub = ud.

� Support rotation ϕϕϕϕa is not necessary to solution because

the moment Ma is known (Ma = 0) ⇒ beam portion ab is taken as hinged-fixed.

a

EI = const.

24

b

c

d

F2 = 5 kN

4

q = 6 kN/m

F1 = 10 kN


Example 3


27

120

0

0

ba ab

bc

cb

M q l

M

M

= + ⋅ ⋅

=

=

a

EI = const.

24

b

c

d

F2 = 5 kN

4

q = 6 kN/m

F1 = 10 kN


Example 3

3. Express internal end moments by slope-deflection equations.

tan

a b d

bc bcbc

u u u

lψ ψ

= = = ∆

∆≈ =

∆∆∆∆∆∆∆∆

ψψψψbc

chords of

the elastic curve

ab

c

d

∆∆∆∆

( )

( )abaab

ababab

abbaab

ababab

l

EIMM

l

EIMM

ψϕ

ψϕϕ

−+=

⋅−+⋅+=

3

322

fixed end rotation


Example 3

3. Express internal end

moments by slope-

deflection equations.

( )

( )

( )

3

22 3

22 3

abbaba b ab

ab

bcbcbc b c bc

bc

bccbcb c b bc

bc

EIM M

l

EIM M

l

EIM M

l

ϕ ψ

ϕ ϕ ψ

ϕ ϕ ψ

= + −

= + ⋅ + − ⋅

= + ⋅ + − ⋅

a

EI = const.

24

b

c

d

F2 = 5 kN

4

q= 6 kN/m

F1 = 10 kN


Example 3

a

EI = const.

24

b

c

d

F2 = 5 kN

4

q = 6 kN/m

F1 = 10 kN


� End moments are expressed in terms of unknown rotations ϕband translation ∆. Required equation to solve are the moment

equilibrium equations at rigid joints b and then sideway equation.

20 : 0Fb ba bc bM M M M= + − =∑

b

222 ⋅= FM F

b

baMbcM


Example 34. Equilibrium equations

� sideway equation

(the horizontal equilibrium)

10 : 0x bcF F V= − =∑

( )1bc bc cb

bc

V M Ml

= − −

a b dF1 = 10 kN

bcV

a

EI = const.

24

b

c

d

F2 = 5 kN

4

q= 6 kN/m

F1 = 10 kN

Part of the frame containing the joints

with the same translation ∆ “cut off”.


Example 3

4. Solving equilibrium equations?

1,75 0,375 4,4

?0,375 0,1875 10

bb

b

EI EI EI

EI EIEI

ϕϕ

ϕ

=⋅ ⋅ − ⋅ ⋅∆ =

⋅ ⋅ − ⋅ ⋅∆ = − ∆ =

5. End moments

� After evaluating ϕb, ∆, substitute them to evaluate end moments.

35,6

4

3

8

3

2 8

ba b

bc b

cb b

EIM

EIM EI

EI EIM

ϕ

ϕ

ϕ

= + ⋅

= ⋅ − ⋅∆

= ⋅ − ⋅ ∆


Example 3

6. Shear forces

( )

( )

1

1

bc bc bc cbbc

cb cb bc cbbc

V R M Ml

V R M Ml

= = − −

= − = − + +

21

2 3

1

2 3

ab abab ab ab ba

ab

ab abba ba ab ba

ab

q l lV R M M

l

q l lV R M M

l

⋅ ⋅ = = ⋅ − −

⋅ = − = − ⋅ + +

4b c

Mbc Mcb

Rbc=Vbc Rcb=-Vcb

a b

Mab Mba

Rab=Vab Rba=-Vba

q = 6 kN/m

4


Example 3

7. Normal forces

� The required equations to solve for normal forces are the

force equilibrium equations at rigid joint b.

0 : 0xb ba bcF N V= − − =∑

20 : 0zb bc baF N V F= − + =∑b

Vba

Vbc

Nba

Nbc

F2


Example 3

8. Draw normal and shear force diagrams.

Nbc =

Ncb =

-14,975

Nab= Nba = -10

0

Vbc =

Vcb =

10

F2= 5Vab= 2,025

Vba= -9,975

2°

xn= 3,647

m647,36

42975,92=⋅⋅=

⋅⋅=

q

lVx abba

n


Example 3

kNm35,046

647,36647,3975,99,23

63

3

=⋅

⋅−⋅+−=

⋅⋅−⋅−−=

max

max

ab

nnbabamax

M

M

l

xqxVMM

8. Draw bending moment diagram.M

bc =-13,9

-F2·2 = -10

-Mba= -23,9

3°

-Mcb =

26,1

Mmax= 0,35

Slope-Deflection Method: Framesfast10.vsb.cz/koubova/SDM_frame.pdf · Slope-Deflection Method:...

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