slope deflection method

SLOPE-DEFLECTION METHOD The slope-deflection method uses displacements displacements as unknowns and 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 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 THE SLOPE-DEFLECTION EQUATION When the loads are applied to a frame or to a continuous beam, the member will develop end moments and become deformed as indicated. The notation used in the figure will be followed. 1- The moments moments at the the ends of the member member are are designated as Mij and M ji indicating that they act at ends I and j of member ij.

ij

θ i

φ  ∆ij θ  j  L

 ji

2- Rotations Rotations of ends ends I and j of the the member are denoted by Θi and Θ j. Since the rotations of all members of a rigid frame meeting at a common  joint are equal, it is customary to refer to each of them as the joint rotation. 3-The term  ∆ij represents the translation of one end of the member relative to the other end in a direction normal to the axis of the member. Sometimes the rotation of the axis of the member Φij= ∆ij/L is used in place of  ∆ij . The moments, the rotations at the ends of the member and the rotation of the axis of the member

∆i ij

M

∆  j = − θ i

∆i = −

φ 

∆

M

− φ  =

θ  j

− φ  =

 ji

ij

θi

+

θ

−

 ji

 j

M 

q L2 8

+

M 

− − ij

φ 

=

+

+

E I3

2

θi

M L i2j L

2 E I3 E I3 L 2j i L M L iLj

2

θ  j

∆  j

L j iL

E3 I

2

q L2 2 L L

+

E I3   2

8

q L2 2 L L

+

E I3   2

8

∆  j  L

∆i  L M

ij

L

−

3EI

M

φ 

= −

=

2 EI

=j i

2 EI

 L  L

M

ij

ji

L

+

6 EI

L

+

6 EI

M

ji

L

3 EI

q L3 24 EI  

−

q L3 24 EI  

( 2 θ + θ − 3 φ  ) − i

j

(θ +

2θ

i

−j

3 φ 

)+

q L2 12

q L2 12

Now we wrote Mij and M ji in terms of the deformations Θi , Θ j Φij and the external load q acting on the member. These equations are referred to as SLOPE-DEFLECTION EQATIONS. Slope-deflection equations consider only bending deformations. Deformations due to shear forces and axial forces in bending members are ignored.

M

n f 

=

2  E I 

 L

  

2θ

n

+

θ  f

−

3

∆  ± M  nFfE M   L 

 

 

Example: It is required to determine the support moments for the continuous beam. 100 kN 1

20 kN/m 2

− M 1 F2 = M 2F 1 =

3

 I 

3 I 

2x2.5m

7.50

−

 F 23

=M = F  32

100*5 8

= 62.5kNm

20*7.52 12

= 93.75kNm

S lo l o p e − D e fl f l e ct c t i o n . E q u a ti tio n s

93.75

100

46.875

40.625

59.375

20 kN/m

87.5

M  1 2 = M  2 1 =

62.5

M  2 3 = M  3 2 =

12

1

Equilibrium Equilibrium.equations equations.of . joı n ts

+ M 23 = 0 M 32 = 0 M 21

21

23

2 E I  5 2 E I  5 6 E I  7 .5 6 E I  7 .5

θ 2 − 6 2 . 5 = 2 θ 2 + 6 2 . 5 =

( 2θ 2 + θ 3 ) − 9 3 . 7 5 = (θ 2 +

2 θ 3 ) + 9 3 . 7 5 =

32

2

3

2 .4

θE2I + 0 . 8

θ  E3I =  3 1 . 2 5

0 .8

θE2I + 1 . 6

θE3I = 

− 9 3 .7 5 →

θ2

=

3 9 .0 6 2 5

→

θ3

=

− 7 8 .1 2 5

I S u b s t it it u d e.t h e se s e. r e su s u l ts t s. i n . s lo l o p e. d e fl f l ec e c ti t i o n. e q u a t io io n s M 12

= − 4 6 .8 7 5 k N m , → M 21 =

9 3 .7 5 k N m

 

EI

 

100

46.875

20 kN/m 1

2

40.625

3

87.5

59.375

62.5

59.375

62.5

+

+ 3.125 m

-

40.62 5

Shear Force Diagram

87.5 97.66 54.69 + +

46.875

Bending Moment Diagram 93.75

Example: A continuous continuous beam is supported supported and loaded loaded as shown in the figure. During loading support 2 sinks by 10 mm. Analyze the beam for support moments and reactions.  E  = 200*10 6... kN  m 2

40 kN 1

20

10 kN/m 2

2

−

 F 12

=M =

−

 F 23

=M =

F  21

3

2m

F  32

40*4 8 10*6 2 12

= 20kNm

 I = 100 10 0 *10 *10 − 6. ..m 4 I =E20000...

= 30kNm

2

6m

S l o p e − D e f l e c ti ti o n. E q u a t i o n s 1

3

φ 12 = 21

0.01 4

2

23

32

2

+ M 23 = 0 M 32 − 4 0 = 0 M 21

φ 23 =

+ 0 . 3 3 3 3θ 3 =

0 . 3 3 3 3θ 2

6

20*2 = 40.kNm

θ 2 θ 3

= =

 E I 

=

− 56.111

M  2 1 =

2  E I   0 .0 1  2 θ  2 − 3   + 20 = 4  4 

M  2 3 =

2  E I   0 .0 1  2 θ 2 + θ  3 + 3   − 30 = 6  6 

M  3 2 =

2  E I   0 .0 1  θ 2 + 2 θ  3 + 3   + 30 = 6  6 

51.667

 E I 

+ 0 . 6 6 6 7θ 3 = −

42.222

2  E I   0 .0 1  θ  2 − 3   − 20 = 4  4 

−0.01

3

1 . 6 6 7θ 2

M  1 2 =

2.111.10

23.333

−3

 E I  r a d 

Sub stitude stitude .these .results i n .slope .d eflection eflect ion .equ ations 12

= − 2 . 8 0 5 5 . 1 0 − 3 r a d 

M

23

= − 7 3 .8 8 9 k N m , → M 21 = − 1 2 .7 7 8 k N m = 1 2 .7 7 8 k N m , → M 32 = 4 0 k N m

k2N m

40

73.889

10 kN/m

12.778

20

40

1

3

20

20

30

8.8

21.67

38.8 + −

41.67

1.67

-

−

3.125 m

Shear Force Diagram

21.2 35.25

9.45

12.778 +

-

73.889

+ −

40

Bending Moment Diagram

ANALYSIS OF FRAMES WITH NO SIDESWAY A frame will not side sway, or be displaced to the the left or right, provided provided it is properly restrained. restrained. Also no side sway will occur in an unrestrained frame provided it is symmetric with respect to both loading and geometry.

Example: It is required to analyze the frame for moments at the ends of members. EI is constant for all members. members.

20.kN / m

40kN  3

1 2

Fixed-End Moments

2 −

20kN 

2

 F 12

=M = F  21

− M  F23 = M 3F 2 =

4

− M  F42 = M 2F 4 =

4m

2

2

20*4 2 12 40*4 8 20*4 8

= 26.67 kNm = 20kNm = 10kNm

S lo l o p e − D e fl f l e ct ctio n. E q u a t io n s M  1 2 = M  2 1 = M  2 3 = M  3 2 = M  4 2 = M  2 4 =

2 E I  4 2 E I  4 2 E I  4 2 E I  4 2 E I  4 2 E I  4

θ 2 − 2 6 . 6 7 =

−27.88

2 θ 2 + 2 6 . 6 7 =

24.245

( 2 θ 2 + θ 3 ) − 2 0 = (θ 2 +

( 2 θ 2 ) + 1 0 =

24

Equilibrium Equilibrium.equations.of . joı n ts M 2 1 + M 2 3 + M  2 4 = 0

−11.21

M 32 = 0 2 E I 

7.575

4

( 6θ 2 + θ 3 ) + 1 6 . 6 7 = (θ 2 + 2θ 3 ) + 2 0 =

0

0

→

40

θ2

=

− 2 .4 2 5

→

θ3

=

− 1 8 .7 8 7

  I EI S u b s t it it u d e.t h e se s e. r e su s u l ts t s. i n . s lo l o p e. d e fl f l ec e c ti t i o n. e q u a t io io n s 4

20

3

−31.82

2 E I 

31.82

32

2

2 θ 3 ) + 2 0 = 0

(θ 2 ) − 1 0 =

24.25

23

21

24.08 27.88

7.57

+

39.09

20

12.04

+ 11.21

− 40.91

+ −

+

−

7.57

− −−

24.25

31.82 27.88

10.61

+

27.96

9.09

−

−

11.21

10.91

+

 

Example: Find Member end moments moments and draw shear and moment moment diagrams

M 2 1 + M  2 3 = 0

60.kN / m Equilibrium equations of joints

2

3

1 .5

4

EI is constant

0 .2 5

M 12

8 −

 F 23

=M = F  32

S lo lo p e M  1 2

=

2 E I  4

M  2 1

=

M  2 3

=

2 E I 

=

2 E I 

=

2 E I 

M  3 4 M 

=

12

θE2I + 1 . 5

θE3I = 

4 8 8 4 2 E I 

θ  2

−320 →

= 320.kNm

M

23

M

34

256

→

θ3

=

−256

240

+

256 −

=

2 θ 2

=

θ2

= 1 2 8 k N m , → M 21 = 2 5 6 k N m = − 2 5 6 k N m , → M 32 = 2 5 6 k N m = − 2 5 6 k N m , → M 43 = − 1 2 8 k N m

− D e fl f l e c t i o n . E q u a t i o n s 256

2 E I 

M  3 2

60*82

θ E3 I=   3 2 0

  EI EI   S u b s t it it u d e.t h es e s e. r e su s u l ts t s. i n . s lo l o p e. d e fl f l ec e c ti t i o n. e q u a t io io n s

4

1

θE2 I+ 0 . 2 5

M 3 2 + M 3 4 = 0

96

=

+

240

Shear

128

128

2 θ 3 ) + 3 2 0

224

(θ  ) =

+

256

=

−

− −

96

( 2θ 3 ) =

−

(kN)

( 2θ 2 + θ 3 ) − 3 2 0 = (θ 2 +

96

−

Normal 240

Force

− 240

+ 128

256 −

Moment

−

(kN.m)

128

+

Example: Find member end moments and draw the diagrams of the frame

− M  =

50 2

M

3

2 I 

 EI  12

kN 

M  2 1 = M  2 3 = M  3 2 =

5 2 E I  5 4 E I  10 4 E I  10

=

7.m

S lo l o p e − D e fl fl e c t i o n . E q u a t i o n s 2 E I 

 F  23

 F  32

3

M  1 2 =

=

 F  21

−

5

1

m

q0 * L2 30

=

22.59

 Pab 2  L2

 Pa2 b  L2

M 32 = 0

→

θ 3

M

=

102

50*32 *7 102

= 73.5kNm

= 31.5kNm

−

−

7.41

=

− 6 7 .6 8

7 .6 4 k N m , → M 55 29 kN m

21

=

7.64

→ M

5 5 .2 9 k N m 0 kN m

+

55.29

EI EI   S u b s t it it u d e.t h es e s e. r e su s u l ts t s. i n . s lo l o p e. d e fl f l ec e c ti t i o n. e q u a t io io n s M 12

50*3*72

66.3 Shear

2 θ 3 ) + 3 1 . 5 = 5 6 .6 1

= 15

= 10kNm

30

−

+ M 2 1 + M 2 3 = 0

=

=

12*52

40.53

( 2θ 2 + θ 3 ) − 7 3 . 5 =

=

=

20

+

+

2 θ 2 + 1 0 =

θ2

=

20

12*52

9.47

θ 2 − 1 5 =

(θ 2 +

q0 * L2

 F  12

−

Moment

ANALYSIS OF FRAMES WITH SIDESWAY A frame will side sway or be displaced to the side when the frame frame or loading acting acting on it is non-symmetric. non-symmetric. In the analysis of frames with side sway it is necessary to consider the shear forces at the base of the columns and the horizontal external load must be in equilibrium (force equilibrium equation) in addition to the equilibrium of joints. Example: Using the slope-deflection method determine the end moments of the members and draw the shear force and bending moment diagrams of the frame. EI is constant throughout the frame. ∆

40. kN  m

2

∆

θ 2

3

θ 3

2

S lo l o p e − D e fl f l e ct ctio n. E q u a t io n s

60.kN  4 1

1

1

4.m

Q12

M  1 2 =

− M 1 F2 = M 2F 1 = −

60 Q43

Axial deformation is neglected (no change in length of the members) so the lateral displacement of joint2 and 3 are equal.

 F 23

= M 3F 2 =

60*4 8 40*4

= 30kNm 2

12

+ 30 =

M  2 3

5 3 .3 3 =

Equilibrium Equations

M  3 2

M 2 1 + M  2 3 = 0

M  3 4

Q1 2 + Q 4 3 + 6 0 = 0

∆  2 E I    2 θ 2 − 3  4  4  2 E I  = ( 2 θ 2 + θ 3 ) − 4 2 E I  = (θ 2 + 2θ 3 ) + 4 ∆  2 E I   = − 2 θ  3 3   3  3 

M  2 1 =

= 53.33kNm

M 3 2 + M 3 4 = 0

∆  2 E I    θ 2 − 3  − 3 0 = 4  4 

M  4 3 =

∆  2 E I   − θ  3  3 = 3  3 

5 3 .3 3 =

=

Shear forces at the base of columns

21

2

3 4  1 

34

3

43

Q12

1

Q43

4

= 0 → M 2 1 + M 1 2 − 4Q1 2 − 2 * 6 0 = 0

3

= 0 → M 3 4 + M 4 3 − 3Q 4 3 = 0

Q 43 =

M 2 1 + M 1 2 4

2 3 .9 6

→ θ 3 =

− 1 4 .8 5 7

4 5 .9 8

EI

 

M  2 3 34

= − 3 5 .2 6 k N m , → M 21 = 3 6 .7 2 k N m = − 3 6 .7 9 k N m , → M 32 = 5 0 .4 5 k N m = − 5 0 .4 6 k N m , → M 43 = − 4 0 .5 6 k N m 83.43

− 30

+

M 3 4 + M 4 3 3

30.37

40

−

+

76.57

50.45

Shear

30.34 −

(kN)

−

36.75

36.53

29.63

+

36.75

60 36.26

→ ∆ =

s l o p e .d e f l e c ti t i o n .e q u a t i o n s 12

2

Q12 =

4 .6 6 7

  −240  1    4 6 . 6 7 =   E I     − 1 0 6 .6 7     

EI EI S u b s t i t u d e . th t h e s e . r e s u l ts ts. i n

12

∑M ∑M

1

θ2 =

60

− 5 . 0 6   θ 2  − 0 . 7 5   θ 3 − 1 . 3 3   ∆

5 .3 3 3

40.56

24

−

−

Moment

+

50.45 −

(kN.m) −

35.26

40.56

+

Example: Determine the member end moments of the frame and draw the shear and moment diagrams.

∆

∆

3′

∆

50kN 

sin β  sin β 

2 ∆ sinα  sin α 

∆

∆

3

tanα  tan α 

3

2′

α 

1

3m

tan β  tan β 

 β 

sin α  = 0.8

sin β  = 0.8

tan α  = 4 / 3

tan β  = 4 / 3

5

4

1 2.25

S lo p e − D e fl e c tio n . E q u a tio n s M  1 2 = M 

21

M 

23

= =

M  3 2 = M  3 4 = M 

43

=

∆ 2 EI    = θ2 − 3 θ  2 − 0 . 7 5 ∆ ) = (   5  5 s i n α   5 ∆ 2 E I 2 EI    = 2θ 2 − 3 2 θ  2 − 0 . 7 5 ∆ ) = (   5  5 s i n α   5  2 E I 1 2 EI ∆  1   2 θ 2 + θ 3 + 0 . 9  ∆ + (  2θ 2 + θ 3 + 3  =   5  5  ta n α t a n β    5    ∆  1 2  E I    1 2  E I  + (θ 2 + 2 θ  3 + 0 . 9 ∆  θ 2 + 2 θ  3 + 3  =   5   5  ta n α t a n β     5  2 E I 2 EI ∆   = 2θ 3 − 3 2 θ  3 − ∆ ) = (   3 .7 5  3 . 7 5 s i n  β   3 .7 5  ∆ 2 E I 2 EI   = θ3 − 3 θ  3 − ∆ ) = (   3 75  3 7 5 i  β   3 75

2 E I

)= )=

23

32

∑M ∑ ∑

M 23 + M 32 5 21

34

2

2

= 0 → 4Q12 +

M 3 = 0 → 3Q43 +

M 23 + M 32 5

M 23 + M 32 5

* 3 − ( M 12 + M 21 ) = 0  * 2.25 − ( M 43 + M 34 ) = 0 

F x = 0 → Q12 + Q43 + 50 = 0

3

Equilibrium Equations

Q43 12

Q12

43

1

M 3 2 + M 3 4 = 0 Q1 2 + Q 4 3 + 5 0 = 0

M 23 + M 32 5

11.02

M 23 + M 32 5

4

M 2 1 + M  2 3 = 0

+

17.06 −

−

9.67

25.1 +

EI EI S u b s t i t u d e . th th e s e .r e s u l t s . i n

+

M +

34

12

M  2 3 M

23.26

EI

s l o p e .d e f le l e c t i o n .e q u a t i o n s

− −

30 −

1 0 . 1 5   θ 2   4 0  1     0 .2   0 .9 3 3 0 . 0 8 7  θ 3  = 0 −      − 0 . 0 3 0 . 0 8 6 7 − 0 . 3 6 1   ∆   E I   − 2 5       7 .6 4 6 7 1 .4 1 − 4 .5 9 θ2 = → θ 3 = → ∆ =

34

= − 2 3 .2 6 k N m , → M 21 = − 2 5 .1k N m = 2 5 . 1 0 k N m , → M 32 = 3 0 k N m = − 3 0 k N m , → M 43 = − 3 4 k N m

 

Example: Determine the member end moments moments and draw the shear and moment diagrams of given continuous beam.

3 kN  m 1

2 kN  m

12 2 hinge

10

5

3

M  2 1

5

10

Equilibrium Equations

∆  2 E I    L − θ  3  2  − 25 = 10  10 

M 34 = 0 → θ 3 R = 41.667

M  3 2 = M  3 4 = M  4 3 =

∆  2 E I   2 θ 2 R + θ 3L + 3   − 15 = 10  10 

M 23 = 0 M 32 = 0 12

3 kN  m

21

∆  2 E I    R L  θ 2 + 2θ 3 + 3  + 15 = 10  10  2 E I  10 2 E I  10

( 2θ  ) − 1 6 . 6 6 7 =  R 3

(θ  ) + 1 6 . 6 6 7 =  R 3

 I 

M 21 = 0

∆  2 E I    L = − 2 θ  3 2   + 25 = 10  10 

M  2 3 =

4

hinge

S lo l o p e − D e fl f l e c t i o n . E q u a t io io n s M  1 2 =

Rotations at the left and right side of the internal hinges are different from each other

23

12

M 12 + M 21 + 150

M 23 + M 32 − 12*5

10

10

32

Shear forces at each side of the hinge must be equal to each e ach other

M 12 + M 21 10

+ 15 =

M 23 + M 32 10

−6

Force Eq. Equation

− 3 − 3 − 1 .2   θ 2 L   − 1 05 0    R    − 125  0 0 − 0 .3  θ 2  1    = 2 1 0 .3   θ 3 L   EI   7 5      1 2 0 .3   ∆   − 75  → θ 2 =R − 5 0 0 → θ 3 =L − 6 5 0 θ 2 =L 8 0 0 I EI 3 2  0  0

∆ = 5750

EI

 

 EI   E I 

3 kN  m 1

2 kN  m

12 2 hinge

10

3

5

4

hinge

5

10

6 + −

36

−

6

7.5

+

210

Shear Force Diagram

39.06

30 −

12.5

+

+ −

25

Bending Moment Diagram

Example: Find the member end moments and draw the shear force and bending moment diagrams of the given give n frame.

20.

kN 

20. kN  m m

2

3 1

4

10

S lo p e − D ef le c ti o n .E q u a ti o n s M  1 2 = M  M  M 

21

23

32

= =

2  E I  5 2  E I  5

Frame is symmetrical both loading and geometry. Half of the frame can be analyzed

1

4

4

3

2

3

Equilibrium Equations

θ  2 =

Q32 2

= 0 → Q32 = 0

M 2 3 + M 3 2 + 2 5 0 = 0 M 12 = 6 6 . 6 6 → M  2 1 = 1 3 3 . 3 3

θ 2 = 166.667  EI  ∆ = 937.5 EI 

2  E I   ∆  θ  2 − 3 =   + 4 1 .6 6 7 = 5  5 

3

∑M

Q32 = 0

2  E I   ∆  2 θ  2 − 3   − 4 1 .6 6 7 = 5  5 

32

2

M 2 1 + M 2 3 = 0

2 θ  2 =

20 kN  m

23

M 2 3 = − 1 3 3 .3 3 → M 3 2 = − 1 1 6 .6 7 116.66

100

40

2

+

2

3

−

−

+

100 1

Shear Force Diagram

+

40

4

+

− −

133.33

66.67

−

−

1

3

Bending Moment Diagram

133.33 4 +

66.67