slope deflection method

(1)

SLOPE-DEFLECTION METHOD

The slope-deflection

The slope-deflection method uses displamethod uses displacements as cements as unknowns anunknowns and is referredd is referred to as to as a displacement a displacement method. method. In theIn the slope-deflection method, the moments at the ends of

slope-deflection method, the moments at the ends of the members are expressed the members are expressed in terms of displacements and endin terms of displacements and end rotations of these ends.

rotations of these ends.

An important characteristic of the slope-deflection method is

An important characteristic of the slope-deflection method is that it does not become that it does not become increasingly complicated toincreasingly complicated to apply as the num

apply as the number of unknowns in the ber of unknowns in the problem increases. problem increases. In the slope-deflection method the In the slope-deflection method the individual equationsindividual equations are relatively easy to construct regardless of the number of unknowns.

are relatively easy to construct regardless of the number of unknowns. DERIVATION OF THE SLOPE-DEFLECTION EQUATION

DERIVATION OF THE SLOPE-DEFLECTION EQUATION When the loads are applied to a frame or to

When the loads are applied to a frame or to a continuous beam, the member will develop end moments and becomea continuous beam, the member will develop end moments and become deformed as indicated.

deformed as indicated. The notation The notation used in the figure used in the figure will be followed.will be followed.

ii

θ

j j

θ

ji ji ij ij ij ij

∆

φ

L

1-1- The momThe moments at thents at the ends of the me ends of the member arember aree designated as M

designated as M_ij_ijand Mand M_ji_ji indicating that they act atindicating that they act at ends I and j of member ij.

ends I and j of member ij.

2-2- RotaRotations of etions of ends I and j of tnds I and j of the member he member areare denoted by

denoted by ΘΘ

iiandandΘΘ j j. . Since the Since the rotations rotations of allof all

members of a rigid frame meeting at

members of a rigid frame meeting at a commona common joint are equal, it is customary to refer to each of joint are equal, it is customary to refer to each of

them as the joint rotation. them as the joint rotation. 3-The term

3-The term ∆ ∆

ij

ij represents the translation of onerepresents the translation of one

end of the member relative to the other end in a end of the member relative to the other end in a direction normal to the axis of the member. direction normal to the axis of the member. Sometimes the rotation of the axis of the

Sometimes the rotation of the axis of the membermember Φ

Φ_ij_ij₌₌ ∆ ∆_ij_ij_{/L is used in place of}_{/L is used in place of} ∆ ∆_ij_ij_..

The moments, the rotations at the ends of The moments, the rotations at the ends of thethe member and the rotation of the axis of the

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2 2 2 2 2 2 22 2 2 3 3 2 2 3 3 8 8 3 3 22 2 2 22 2 2 3 3 2 2 3 3 8 8 3 3 22 jjii iijj j j jjii iijj ii j j ii ii j j

M

M L

L

_L

M

M L

L

_L

_{L q}

_{q L}

_{L L}

_L

E

E I

I

E

E I

I

E

E I

I

M

M L

L

_L

M

M L

L

_L

_{L q}

_{q L}

_{L L}

_L

E

E I

I

E

E I

I

E

E I

I

L

θ

φ

θ

φ

∆

=

−

+

++

∆

=

−

+

++

∆

−

− ==

∆

−

− ==

ii

θ

j j

θ

ji ji ij ij

∆

φ

j j

∆

ii

∆

++

−−

++

ij ij ji ji 2 2 8 8 q q LL

(

))

(

))

3 3 3 3 2 2 2 2 3 3 66 22 44 6 6 33 2244 2 2 2 2 33 1 2 1 2 2 2 2 2 33 1 1 22 ii jj jj ii ii ii jj jj ii j j ii jj ii jj jj ii ii jj

M

M L

L

M

M L

L

_{q L}

E

I

E

I

E

I

M

M L

L

M

M L

L

_{q L}

E

I

E

I

E

I

E

I

q

L

M

L

E

I

q

L

M

L

θ

φ

θ

φ

θ

φ

θ

φ

−

=

−

++

−

− =

=

−

+

−−

=

+

−

−−

=

+

−

++

Now we wrote M

Now we wrote M_ij_ijand Mand M_ji_ji in terms of the deformationsin terms of the deformations ΘΘ

ii,,ΘΘ j j ΦΦijijand the external load q acting on the member.and the external load q acting on the member.

These equations

These equations are referred are referred to as to as SLOPE-DEFLECTION SLOPE-DEFLECTION EQATIONS. EQATIONS. Slope-deflection Slope-deflection equations equations consider onlyconsider only bending deformations.

bending deformations. Deformations due to shear Deformations due to shear forces and axial forces forces and axial forces in bending members are ignored.in bending members are ignored.

n f n f 2 2 2 2 _n_n _f_f 33 _n_nF E M F E _f_f M

E

E I

I

M

L

θ

L

∆





=

_

+

−

_

±±





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Example: It is required

Example: It is required to determine the support to determine the support moments for themoments for the continuous beam.continuous beam. 20 kN/m 20 kN/m 11 ₂₂ ₃₃ 100 kN 100 kN

3

3 I

I

1 12 2 2211 2 2 2 23 3 3322 100*5 100*5 62.5 62.5 8 8 20*7.5 20*7.5 93.75 93.75 12 12 F F F F F F F F M M MM kkNNmm M M kkNNmm

−

=

−

=

2x2.5m 2x2.5m 7.507.50

(

))

(

))

1 1 2 2 22 2 2 1 1 22 2 2 33 22 33 3 3 22 22 33 .. 2 2 6 2 . 5 6 2 . 5 5 5 2 2 2 2 66 22 ..55 5 5 6 6 2 2 99 33 ..77 55 7 7 ..55 6 6 2 2 99 33 ..77 55 7 7 ..55 S l

S lo p e o p e DD e fe flle ce cttiio n o n E q u a tE q u a tiio n so n s E

E I I M

M

E E I I M

M

E E I I M

M

E E I I M M θ θ θ θ θ θ θ θ θ θ θ θ

−

=

−

=

+

=

+

−

=

+

=

20 kN/m 20 kN/m 93.75 93.75 100 100

46.875

8

87 7..5

5

6

62 2..5

5

59.375

40.625

12 12

1

2

3232

₃

23 23 21 21 2 2 33 2 2 33 22 33 1 1 22 22 11 2 2 ..4 4 00 ..8 8 33 11 ..22 55 3 3 99 ..00 66 22 5 5 77 88 ..11 22 55 0 0 ..8 8 11 ..6 6 99 33 ..77 55 . . .. .. iinn.. .. .. 4 4 66 ..88 77 5 5 , , 99 33 ..77 55 E E I I EE I I E E I I EE I I I I EE II S u b s t i S u b s t it u d e t u d e t h e st h e se e r e sr e su l tu l ts s s ls lo p e o p e d e fd e fl el ec tc ti o n i o n e q u a t ie q u a t io n so n s M M kkNN mm MM kkNN mm θ θ θ θ θ θ θθ θθ θθ

+

==

−−

+

=

−

→

=

→

==

=

−

→

==

2 21 1 2233 32 32 . . . . .. 0 0 0 0 Equilibri

Equilibrium um equatiequations ons of of jojoıınntsts

M M M M M M

+

==

(4)

20 kN/m 20 kN/m 11 ₂₂ ₃₃ 46.875 46.875 100100 40.625 40.625 _59.375_59.375 8877..55 6622..55 62.5 62.5 --++ 3.125 m

3.125 m Shear Force DiagramShear Force Diagram 59.375 59.375 ++ 40.62 40.62 5 5 87.587.5 --++ ++ 93.75 93.75 97.66 97.66 54.69 54.69

Bending Moment Diagram Bending Moment Diagram 46.875

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Example:

Example: A continuoA continuous beam is suppus beam is supported and lorted and loaded as shown in oaded as shown in the figure. the figure. During loading During loading support 2 support 2 sinks by 10 sinks by 10 mm.mm. Analyze the beam for

Analyze the beam for support moments and reactions.support moments and reactions.

2 2 6 6 6 6 44 2 2 200*10 ... 200*10 ... 10 1000 *1*10 0 ... 20000... 20000... kN kN m m E E I I mm E EI I kkNNmm − − = = = = = =

20

40 kN 40 kN 1 12 2 2211 2 2 2 23 3 3322 40*4 40*4 20 20 8 8 10*6 10*6 30 30 12 12 F F F F F F F F M M kkNNmm M M kkNNmm

−

− =

= =

=

−

=

10 kN/m 10 kN/m 11 ₂₂ ₃₃ 6 m 6 m 2m 2m

2

1 1 2 2 22 2 2 1 1 22 2 2 33 22 33 3 3 22 22 33 .. 2 2 00 ..00 11 3 3 22 00 4 4 44 2 2 00 ..00 11 2 2 33 22 00 4 4 44 2 2 00 ..00 11 2 2 33 33 00 6 6 66 2 2 00 ..00 11 2 2 33 33 00 6 6 66 S l o p e S l o p e D D e fe f ll e c te c ti o n i o n E E q u aq u a tt i o n si o n s E

E I I M M E I E I M M E I E I M M E I E I M M θ θ θ θ θ θ θ θ θ θ θ θ

−





=

_

−

_

−

=









=

_

−

_

+

=









=

_

+

_

−

=









=

_

+

_

+

=





1 1 0.01 0.01 12 12 44

φ

=

φ

2323 0.010.0166 − −

=

2 2 3 3 32 32

3

2 20*2

0*2 4

=

40 0..

kkN

Nm

m

2

23 23 21 21 2 2 33 2 2 33 3 3 2 2 3 3 3 3 5 1 . 6 6 7 5 1 . 6 6 7 1 1 ..66 66 7 7 00 ..33 33 33 33 2 3 . 3 3 3 2 3 . 3 3 3 0 0 ..33 33 33 3 3 00 ..66 66 66 77 4 2 . 2 2 2 4 2 . 2 2 2 2 . 1 1 1 . 1 0 2 . 1 1 1 . 1 0 5 6 . 1 1 1 5 6 . 1 1 1 2 . 8 0 5 5 . 1 0 2 . 8 0 5 5 . 1 0 E E I I

E E I I rraa d d E

E I I

rraa d d θ θ θ θ θ θ θ θ θ θ θ θ −− −−

+

==

+

=

= −−

=

==

−−

=

= −−

2 21 1 2233 32 32 0 0 4 40 0 00 M M M M M M

+

==

−

==

1 1 22 22 11 2 2 33 33 22 . . .. iinn.. .. .. 7 7 33 ..88 88 9 9 , , 11 22 ..77 77 88 1 1 22 ..77 77 8 8 , , 44 00 Sub

Sub stistitude these resultstude these results slope d

slope d eflecteflection equion equ ationsations

kkNN m m M M kkNN mm M M kkNN m m M M kkNN mm

=

−

→

=

−−

=

→

==

(6)

40 40 73.889 73.889 3 3

12.778

10 kN/m10 kN/m

40

20

11 20 20 --++ 3.125 m 3.125 m

21.2

1.67 1.67 38.8 38.8 --++ ++ 9.45 9.45

_12.778

35.25

40

− − Bending Moment Bending Moment Diagram Diagram Shear Force Shear Force Diagram Diagram − − 41.67 41.67

20

30 30

8.8

21.67

− −

73.889

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ANALYSIS OF FRAMES WITH NO SIDESWAY

A frame will not

A frame will not side sway, or be displaced to tside sway, or be displaced to the left or right, prhe left or right, provided it is properly rovided it is properly restrained. estrained. Also no side sway willAlso no side sway will occur in an unrestrained frame provided it is symmetric with respect to both loading and geometry.

occur in an unrestrained frame provided it is symmetric with respect to both loading and geometry.

Example:

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

40 40kN

kN

3

2

20

0. .

kN

kN m

//

m

Fixed-End Moments Fixed-End Moments

1

2

2 2 1 12 2 2211 2 23 3 3322 4 42 2 2244 20*4 20*4 26.67 26.67 12 12 40*4 40*4 20 20 8 8 20*4 20*4 10 10 8 8 F F F F F F F F F F F F M M kkNNmm M M MM kkNNmm M M MM kkNNmm

−

=

−

− =

= =

=

−

− =

= =

=

2

20 20kN

kN

2

4

4 m

m

2

(8)

(

))

(

))

(

( ))

(

))

1 1 2 2 22 2 2 1 1 22 2 2 33 22 33 3 3 22 22 33 4 4 2 2 22 2 2 4 4 22 .. 2 2 2 6 . 6 7 2 6 . 6 7 4 4 2 2 2 2 22 66 ..66 77 4 4 2 2 2 2 22 00 4 4 2 2 2 2 22 00 4 4 2 2 1 1 00 4 4 2 2 2 2 11 00 4 4 S l

S lo p e o p e DD e fe flle ce cttiio n o n E q u a tE q u a t iio n so n s E

E I I M

M

E E I I M

M

E E I I M

M

E E I I M

M

E E I I M

M

E E I I M M θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ

−

=

−

=

+

=

+

−

=

+

=

−

=

+

=

27.88 27.88 − − 24.245 24.245 31.82 31.82 − − 0 0 7.575 7.575 11.21 11.21 − −

2

3232 21 21

3

23 23 24 24 2 21 1 223 3 2244 32 32 . . . . .. 0 0 0 0 Equilibr

Equilibrium ium equations equations of of jojoıınntsts

M M M M M M M M

+

=

(

))

(

))

2 2 33 2 2 33 22 33 2 2 6 6 11 66 ..66 7 7 00 4 4 2 2 22 ..44 22 55 11 88 ..77 88 77 2 2 22 0 0 00 4 4 . . .. .. iinn.. .. .. E E I I E E I I

I I EE II S u b s t i S u b s t it u d e t u d e t h e st h e se e r e sr e su l tu l ts s s ls lo p e o p e d e fd e fl el ec tc ti o n i o n e q u a t ie q u a t io n so n s θ θ θ θ θ θ θθ θθ θθ

+

==

−

−−

+

+ =

= →

→ =

=

→

→ ==

40 40 27.88 27.88 24.25 24.25 20 20 31.8231.82 7.57 7.57 24.08 24.08 − − −− −− −−

−−

++ ++ ++ 27.88 27.88 24.25 24.25 31.82 31.82 7.57 7.57 10.61 10.61 11.21 11.21 39.09 39.09 2 200 1122..0044 40.91 40.91 9.099.09 −− ++ ++ ++ 11.21 11.21 −− 27.96 27.96 −− 10.91 10.91

(9)

Example:

Example: Find Member end momentFind Member end moments and draw s and draw shear and momshear and moment diagramsent diagrams

2 21 1 2233 3 32 2 3344 0 0 0 0 M M M M M M M M

+

=

+

=

6 600. .

kN

kN m

//

m

Equilibrium equations of joints Equilibrium equations of joints

2

2 ₃

₃

8

4

1

4

EI is constant EI is constant 2 2 2 23 3 3322 60*8 60*8 320. 320. 12 12 F F F F M M kkNNmm

−

=

2 2 33 2 2 33 22 33 1 1 22 22 11 2 2 33 33 22 3 3 44 44 33 1 1 ..5 5 00 ..22 5 5 33 22 00 2 2 55 6 6 22 55 66 0 0 ..22 5 5 11 ..5 5 33 22 00 . . .. .. iinn.. .. .. 1 1 22 8 8 , , 22 55 66 2 2 55 6 6 , , 22 55 66 2 2 55 6 6 , , 11 22 88 E E I I EE I I E E I I EE I I E E I I EE I I S u b s t i S u b s t it u d e t u d e t h et h es e s e r e sr e su l tu l ts s s ls lo p e o p e d e fd e fl el ec tc ti o n i o n e q u a t ie q u a t io n so n s M M kkNN m m M M kkNN mm M M kkNN m m M M kkNN mm M M kkNN m m M M kkNN mm θ θ θ θ θ θ θθ θθ θθ

+

==

−−

+

=

− →

− → =

=

→

→ ==

=

→

==

=

−

→

==

=

−

→

=

−−

240 240 −− −− 240 240 96 96

++

(

))

(

))

(

))

(

( ))

1 1 2 2 22 2 2 1 1 22 2 2 33 22 33 3 3 22 22 33 3 3 4 4 33 .. 2 2 4 4 2 2 2 2 4 4 2 2 2 2 33 22 00 8 8 2 2 2 2 33 22 00 8 8 2 2 2 2 4 4 2 2 S l

E I I M

M

E E I I M

M

E E I I M

M

E E I I M

M

E E I I M

M

−−

=

==

=

==

=

+

−

==

=

+

==

=

==

=

==

2

25

56

6

2

25

56

6

Shear Shear (kN) (kN) 96 96

128

128128

224

−− −− −−

++

1 1228 8 112288

2

25

56

6

2

25

56

6

−− −− 240 240 −− 240 240 −− 96 96 Moment Moment (kN.m) (kN.m) Normal Normal Force Force

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Example: Find member end moments and draw the

Example: Find member end moments and draw the diagrams of the framediagrams of the frame ₂₂ ₂₂

0 0 12 12 2 2 22 0 0 21 21 2 2 22 23 23 2 2 22 2 2 22 32 32 2 2 22 * * 12*512*5 15 15 2 20 0 2200 * * 12*512*5 10 10 3 30 0 3300 50*3*7 50*3*7 73.5 73.5 10 10 50*3 *7 50*3 *7 31.5 31.5 10 10 F F F F F F F F q q LL M M q q LL M M kkNNmm Pab Pab kNm kNm L L Pa b Pa b kNm kNm L L

−

=

−

=

50 50 12 12kN kN m m 3 3

2

2 I

I

2 2

EI

5

1 1

3

7. 7.m

m

(

))

(

))

1 1 2 2 22 2 2 1 1 22 2 2 33 22 33 3 3 22 22 33 .. 2 2 1 1 55 5 5 2 2 2 2 11 00 5 5 4 4 2 2 77 33 ..55 1 1 00 4 4 2 2 33 11 ..55 1 1 00 S l

E I I M

M

E E I I M

M

E E I I M

M

E E I I M M θ θ θ θ θ θ θ θ θ θ θ θ

−

=

−

=

+

=

+

−

=

+

=

2 21 1 2233 32 32 0 0 0 0 M M M M M M

+

=

2 2 33 1 1 22 22 11 5 5 66 ..66 1 1 66 77 ..66 88 . . .. .. iinn.. .. .. 7 7 ..66 4 4 , , 55 55 ..22 99 5 5 55 22 9 9 00 E E I I EE I I S u b s t i S u b s t it u d e t u d e t h et h es e s e r e sr e su l tu l ts s s ls lo p e o p e d e fd e fl el ec tc ti o n i o n e q u a t ie q u a t io n so n s M M kkNN m m M M kkNN mm M M kkNN m m M M kkNN mm θ θ

=

→

θ θ

==

−−

=

→

==

→

−− −− ++ −− −− 40.53 40.53

9.47

++ ++ ++ 22.59 22.59 7.41 7.41

7.64

55.29

66.3

Shear Shear Moment Moment

(11)

ANALYSIS

ANALYSIS OF

OF FRAMES

FRAMES WITH

WITH SIDESWAY

SIDESWAY

A frame will side sway or

A frame will side sway or be displaced to the side when the frabe displaced to the side when the frame or loading actime or loading acting on it is non-ng on it is non-symmetric. symmetric. In theIn the analysis of frames with side

analysis of frames with side sway it is necessary to consider the sway it is necessary to consider the shear forces at the base of shear forces at the base of the columns and thethe columns and the horizontal external load must be in equilibrium (force equilibrium equation) in addition to the equilibrium of joints. 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

Example: Using the slope-deflection method determine the end moments of the members and draw the moments of the members and draw the shear force andshear force and bending moment diagrams of the frame. EI is

bending moment diagrams of the frame. EI is constant throughout the frame.constant throughout the frame.

40. 40.kN kN m m 1 1 2 2 33 4 4 60. 60.kN kN 2 2

θ

₃₃ ∆ ∆ ∆∆

Axial deformation is neglected (no change Axial deformation is neglected (no change in length of the members) so

in length of the members) so the lateralthe lateral displacement of joint2 and 3 are equal. displacement of joint2 and 3 are equal.

2

4. 4.m

m

1 1

1

₁₁₂₂ ₂₂₁₁ 2 2 2 23 3 3322 60*4 60*4 30 30 8 8 40*4 40*4 53.33 53.33 12 12 F F F F F F F F M M M M kkNNmm M M kkNNmm

−

− =

= =

=

−

=

(

))

(

))

1 1 2 2 22 2 2 1 1 22 2 2 33 22 33 3 3 22 22 33 3 3 4 4 33 4 4 3 3 33 .. 2 2 3 3 33 00 4 4 44 2 2 2 3 2 3 33 00 4 4 44 2 2 2 2 55 33 ..33 33 4 4 2 2 2 2 55 33 ..33 33 4 4 2 2 2 2 33 3 3 33 2 2 3 3 3 3 33 S l

E I I M

M

E E I I M

M

E E I I M

M

E E I I M

M

E E I I M

M

−

∆





=

_

−

_

−

=





∆





=

_

−

_

+

=





=

+

−

=

+

=

∆





=

_

−

_

=





∆





=

_

−

_

=





Equilibrium Equations Equilibrium Equations 2 21 1 2233 3 32 2 3344 1 12 2 4433 0 0 0 0 6 60 0 00 M M M M M M M M Q Q QQ

+

=

+

=

+

=

60

43 43

Q

12 12

Q

(12)

Shear forces at the base of

Shear forces at the base of columnscolumns

34 34 1 1 2 2 12 12

Q

43 43

Q

43 43 21 21 12 12

60

3 3 4 4 2 2 2211 1122 1122 3 3 3344 4433 4433 2 21 1 1122 12 12 3 34 4 4433 43 43 0 0 4 4 22 ** 660 0 00 0 0 3 3 00 3 300 4 4 3 3 M M MM MM QQ M M MM MM QQ M M M M Q Q M M M M Q Q

=

= →

→

+ −

−

=

= →

→

+

−

=

+

=

−

+

=

∑

2 2 3 3 2 2 33 1 1 22 22 11 2 2 33 3 3 55 ..33 33 3 3 55 ..00 6 6 22 44 00 1 1 4 4 1 1 00 ..77 5 5 44 66 ..66 77 1 1 44 ..66 66 7 7 11 ..33 3 3 11 00 66 ..66 77 2 2 33 ..99 6 6 11 44 ..88 55 7 7 44 55 ..99 88 . . . . .. iinn . . .. 3 3 55 ..22 6 6 , , 33 66 ..77 22 E

E I I

E E II EE II EE II S u S u b s t i t u d e tb s t i t u d e th e s e r e s uh e s e r e s u l tl tss s l o p e d s l o p e d e f l e c te f l e c ti o n e q ui o n e q u a t i o na t i o n ss kk N N m m M M kk N N mm M M θ θ θ θ θ θ θ θ

−





−















₋



₌



















₋





_∆





₋















−

=

→

=

→

∆

=

−

→

=

3 3 22 3 3 44 44 33 3 3 66 ..77 9 9 , , 55 00 ..44 55 5 5 00 ..44 6 6 , , 44 00 ..55 66 kk N N m m M M kk N N mm kk N N m m M M kk N N mm

=

−

→

=

−

→

=

−

36.75

50.45

36.26

40.5640.56 − − − − − − − − − − 76.57 76.57 30.37 30.37 30.34 30.34

+

35.26 35.26 24 24

60

50.45

36.53

− − Moment Moment (kN.m) (kN.m) 40 40 83.43 83.43 29.63 29.63 − − 36.75 36.75 40.56 40.56 Shear Shear (kN) (kN)

(13)

Example: Determine the member end moments of

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

1 1 2 2 3 3 4 4

α

β

3

1

1 ∆

∆

2

2 ′′

3 3

′′

tan tanα α ∆ ∆ tantan β β ∆ ∆ sin sinα α ∆ ∆ sin sin β β ∆ ∆ 50 50kN kN ssiin n 00..88 ttaan n 44 // 33 α α α α = = = = ssiin n 00..88 ttaan n 44 // 33 β β β β = = = =

5

3 3mm 2.252.25

( (

))

( (

))

( (

))

1 122 22 22 2 211 22 22 2 233 22 33 22 33 3 322 22 33 .. 2 2 22 3 3 00.. 7755 5 5 55 ss ii nn 55 2 2 22 2 2 33 22 00..7755 5 5 55 ss ii nn 55 2 2 11 11 22 2 2 33 22 00.. 99 5 5 55 tt aa nn tt aa nn 55 2 2 1 1 11 2 2 33 5 5 55 tt aa nn tt aa nn S

S lo lo p p e e D D ee flfl ee cc tio tio n n E E q q u u a a tio tio n n ss E EII EEII M M E EII EEII M M E EII EEII M M E E I I M M θ θ θ θ α α θ θ θ θ α α θ θ θθ θθ θθ α α β β θ θ θ θ α α β β − − ∆ ∆    = = __ −− _ == −− ∆∆ ==    ∆ ∆    = = __ −− _ == −− ∆∆ ==      ∆∆     = = __ ++ ++ __ ++ _ _ == ++ + + ∆∆ ==         ∆∆    = = _ ++ ++ __ ++ _     

( (

))

( (

))

( (

))

2 2 33 3 344 33 33 4 433 33 33 2 2 2 2 0 0 ..99 5 5 2 2 22 2 2 3 3 22 3 3 ..7 7 5 5 3 3 ..7 7 5 5 ss ii n n 3 3 ..7 7 55 2 2 22 3 3 3 3 7 7 5 5 3 3 7 7 5 5 ii 3 3 7 7 55 E E I I

E EII EEII M M E EII EEII M M θ θ θ θ θ θ θ θ β β θ θ θ θ β β   = = ++ ++ ∆∆ = =      ∆∆  = = __ −− _ == −− ∆∆ ==      ∆∆  = = __ −− _ == −− ∆∆ ==   

(14)

2 233 3322 2 23 3 3322 5 5 M M ++M M 21 21 12 12 34 34 43 43 1 1 2 2 ₃₃ 4 4 43 43

Q

12 12

Q

2 23 3 3322 5 5 M M ++M M 2 23 3 3322 5 5 M M ++M M 2 23 3 3322 2 2 1122 1122 2211 2 23 3 3322 3 3 4433 4433 3344 1 12 2 4433 0 0 44 ** 33 (( )) 0 0 5 5 0 0 3 3 ** 22..225 5 ( ( ) ) 00 5 5 0 0 5500 0 0 x x M M M M M M QQ MM MM M M M M M M QQ MM MM F F Q Q QQ

+

=

→

+

−

+

=

+

=

→

+

−

+

=

= →

→ +

+ + =

+ =

∑

2 21 1 2233 3 32 2 3344 1 12 2 4433 0 0 0 0 5 50 0 00 M M M M M M M M Q Q QQ

+

=

+

=

+

=

2 2 3 3 2 2 33 1 1 22 22 11 2 2 33 4 1 4 1 00 ..11 55 00 1 1 0 0 ..2 2 00 ..99 33 3 3 00 ..00 88 7 7 00 0 0 ..00 3 3 00 ..00 88 66 7 7 00 ..33 66 1 1 22 55 4 4 ..55 9 9 77 ..66 44 6 6 77 11 ..44 11 . . . . .. ii nn . . .. 2 2 33 ..22 6 6 , , 22 55 ..11 2 5 . 1 2 5 . 1 E E I I

E E II E IEI EE II S u S u b s t i t u d e tb s t i t u d e th e s e r e s u l t sh e s e r e s u l t s s l o p e d s l o p e d e f le f le c t i o n e q ue c t i o n e q u a t i o na t i o n ss M M kk NNmm MM kk NNmm M M θ θ θ θ θ θ θ θ



















₋



₌



















₋





_∆





₋















−

=

→

→ =

=

→

∆

=

−

→

=

−

=

3 3 22 3 3 44 44 33 0 0 ,, 33 00 3 3 00 ,, 33 44 kk N N m m M M kk N N mm M M kk N N m m M M kk N N mm

→

=

−

→

=

−

− − + + 9.67 9.67 11.02 11.02 17.06 17.06 + + + + + +

34

30

25.1 25.1 23.26 23.26 − − − − − − − − Equilibrium Equations Equilibrium Equations

(15)

Example: Determine the member end m

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

3

kN kN m m

₁₂

2

kN kN mm 10 10 ₅₅ ₅₅ ₁₀₁₀ 1 1 22 33 44

Rotations at the left Rotations at the left and

and right right side side of of thethe internal hinges are internal hinges are different from each different from each other other h hiinnggee hhiinnggee Equilibrium Equations Equilibrium Equations

( (

))

( ( ))

1 1 2 2 22 2 2 1 1 22 2 2 33 22 33 3 3 22 22 33 3 3 4 4 33 4 4 3 3 33 .. 2 2 3 3 22 55 1 1 0 0 11 00 2 2 2 3 2 3 22 55 1 1 00 11 00 2 2 2 2 33 11 55 1 1 00 11 00 2 2 2 2 33 11 55 1 1 00 11 00 2 2 2 2 11 66 ..66 66 77 1 1 00 2 2 1 6 . 6 6 7 1 6 . 6 6 7 1 1 00 L L L L R R LL R R LL R R R R S l S lo p e o p e DD e fe flle ce cttiio n o n E q u a t iE q u a t io n so n s E

E I I M

M

E E I I M

M

E E I I M

M

E E I I M

M

E E I I M

M

−

∆





=

_

−

_

−

=





∆





=

_

−

_

+

=





∆





=

_

+

_

−

=





∆





=

_

+

_

+

=





=

−

=

+

=

34 34 33 21 21 23 23 32 32 41.667 41.667 0 0 0 0 0 0 0 0 R R M M I I M M M M M M θ θ

=

= →

→

=

1 122

3

kN kN _m_m 2121 2323 3232

12

2 23 3 3322 12*512*5 10 10 M M

+

M M

−

1 12 2 2211 150150 10 10 M M

+

M M

+

Shear forces at each side

Shear forces at each side of the hinge must be equal to eof the hinge must be equal to each otherach other

2 23 3 3322 1 12 2 2211 ₁₅₁₅ ₆₆ 1 100 1100 M M M M M M

+

M M

₊

₌

+

₋

Force Eq. Equation Force Eq. Equation

(16)

2 2 2 2 3 3 2 2 2 2 33 3 3 3 3 3 3 11..2 2 11005500 2 2 0 0 0 0 00..3 3 11 112255 0 0 2 2 1 1 00..3 3 7755 0 0 1 1 2 2 00..3 3 7755 8 8000 0 55000 0 665500 5750 5750 L L R R L L L L RR LL EI EI I I EEII EEII E EI I θ θ θ θ θ θ θ θ θ θ θ θ

−





−















₋





₋











₌

























_

_





−

∆













−

=

→

=

→

=

∆

∆ =

=

3

kN kN m m

₁₂

2

kN kN mm 1 1 22 _h_hiinnggee 33 _h_hiinnggee 44 10 10 ₅₅ ₅₅ ₁₁₀₀ − − −− + + + +

6

12.512.5 Shear Shear Force Force Diagram Diagram − − + + 7.5 7.5 + + − − 6 6

36

39.06

30

Bending Bending Moment Moment Diagram Diagram

25

210 210

(17)

Example: Find the member end moments and

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

20. 20.kN kN m m 20. 20.kN kN m m 4 4 1 1 2 2 33

10

4

3

1 1 2 2 33 1 1 2 2 33 4 4 1 1 2 2 3 3 4 4 1 2 1 2 22 2 1 2 1 22 2 3 2 3 22 3 2 3 2 22 .. 2 2 5 5 2 2 2 2 5 5 2 2 2 2 3 3 4 4 1 1 ..6 6 6 6 77 5 5 55 2 2 3 3 4 4 1 1 .. 6 6 6 6 77 5 5 55 S S l o l o p p e e D e fD e f l el e c c t it i o o n n E E q q u u a a t it i o o n n ss E

E I I M

M

E E I I M

M

E E I I M

M

E E I I M M θ θ θ θ θ θ θ θ − − = = == = = == ∆ ∆    = = __ −− −_ − ==    ∆ ∆    = = __ −− +_ + ==    32 32 Q Q 2 21 1 2233 32 32 0 0 0 0 M M M M Q Q

+

=

2 2 3232 2 23 3 3322 0 0 00 2 2550 0 00 M M QQ M M M M

=

= →

→

=

+

=

∑

2 2 33 1 122 2211 2 233 3322 6 666..666 6 113333..3333 1 13333..333 3 111166..6677 M M M M M M M M

=

→

=

−

→

=

−

2 2 166.667166.667 937.5 937.5 EI EI EI EI θ θ == ∆ = ∆ = + + + + − − − − 40 40 40 40 100 100 100 100 + +

+

++ − − −− −− ₋₋ 116.66 116.66

133.33

133.33 _133.33

_133.33

66.67 66.67 _66.67_66.67

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

geometry. Half of the frame can be analyzedanalyzed

2 233

20

kN kN _m_m 3322 Equilibrium Equations Equilibrium Equations Shear Shear Force Force Diagram Diagram Bending Bending Moment Moment Diagram Diagram