Slope Deflection Method

34
CHAPTER 2: SLOPE DEFLECTION METHOD 1.0 INTRODUCTION 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 2.0 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. Fundamental Slope-Deflection Equations: ܯ ଶாூ ቂʹ ߠ ߠ ͵ ο ܯܧܨ ܯ ଶாூ ቂʹ ߠ ߠ ͵ ο ܯܧܨ Where : Case A : Fixed End Moment Ǣ M AB θB A B w θA M BA Δ FEM AB FEM BA A B w

Transcript of Slope Deflection Method

Page 1: Slope Deflection Method

CHAPTER 2: SLOPE DEFLECTION METHOD 1.0 INTRODUCTION

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

2.0 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.

Fundamental Slope-Deflection Equations:

��� �� ��� ��� ��� � � �� ������

��� �� ��� ��� ��� � � �� ������

Where :

Case A : Fixed End Moment

��� ��������� ���� �������

MAB

θB

A B

w

θA

MBA

Δ

FEMAB FEMBA

A B

w

Page 2: Slope Deflection Method

CHAPTER 2: SLOPE DEFLECTION METHOD

����� ������

�� �� ������ �� �����

����� ����� �

�� � � ������ �� � ����

Case B: rotation at A

��� � !�"� �#��� ���� ����"� #��� $%&�$'()*�)+*, # � -�

B A

w

L

�����

�����

�.�

�.�

P

a b

L

A B

� �/� ��0�1

�� ���

� ����

���/� � 0�1

A

MAB

MBA B θA

Page 3: Slope Deflection Method

CHAPTER 2: SLOPE DEFLECTION METHOD Case D: displacement of end B related to end A

��� � �2�"�� ������� � ��2�"�� �

3.0 ANALYSIS OF BEAMS – SLOPE DEFLECTION METHOD

General Procedure:

Step 1: Scan the beam and identify the number of (a) segments and (b) kinematic unknowns. A segment is the portion of the beam between two nodes. Kinematic unknowns are those rotations and displacements that are not zero and must be computed. The support or end conditions of the beam will help answer the question.

Step 2: For each segment, generate the two governing equations. Check the end conditions to see whether one of the end rotations is zero or not (it is not possible for both the end rotations and other deflection components to be zero). If there are no element loads, the FEM term is zero. If there are one or more element loads, use the appropriate formula to compute the FEM for each element load and then sum all the FEMs. If one end of the segment displace relative to the other, compute the chord rotation; otherwise it is zero.

Step 3: For each kinematic unknown, generate an equilibrium condition using the free-body diagram.

Step 4: Solve for all unknowns by combining all the equations from steps 2 and 3. Now the equations are entirely in terms of the kinematic unknowns.

Step 5: Compute the support reactions with appropriate FBDs.

MBA

MAB

A

B

Δ

Page 4: Slope Deflection Method

CHAPTER 2: SLOPE DEFLECTION METHOD Example 1:

A continuous beam is supported and loaded as shown in the figure. Analyze the beam for support moments and reactions.

Solution :

����� � ��345�

6�

����� � ��78 9 � 9 7�:� � �; 9 7 9 ��

:� � �:7<7=�>?@

����� ��354�

6�

����� ��78 9 7 9 ��:� �; 9 � 9 7�

:� � =�<���>?@

Slope-deflection equation: hence θA = 0, θB = ?, ∆ = 0

��� ��� ��A6 B��� ��� � ��6C ������ ����A: B��� ��� � ��6C � �:7<7= � -< ���"#� � �D2< 2EFFF/�0

��� ��� ��A6 B��� ��� � ��6C ������ ����A: B��� ��� � ��6C =�<�� � -< !!�"#� �E1< 11FFF < /�0

A B

60 kN 25 kN

3 m 3 m 3 m

60 kN 25 kN FEMAB FEMBA

A B

0 0

0 0

a

a

b

b

Page 5: Slope Deflection Method

CHAPTER 2: SLOPE DEFLECTION METHOD Hence, ∑MB = 0;

MBA = 0; 8<GG�A�� �=�<�� � 8

�A�� ��� HI<IIJ<KK ���L77<7; ……………..(3)

Substituting eq. (3) into eq. (1) & (2);

��� � ��8<���A�� � �:7<7= � 8<��/�L77<7;0 � :7<7= � �L��<���>?@

��� � ��8<GG�A�� �=�<�� � 8<GG/� L77<7;0 =�<�� � 8�>?@

Support reaction;

∑MB = 0;

�L��<�� /78� 9 �0 /�; 9 70 � /M� �9 :0 � 8

N� � ��< OP�QR

∑V = 0;

M� �L<S; � 78 � �; � 8

N� � 21< �P�QR

A B

60 kN 25 kN

3 m 3 m 3 m

-133.33 kNm

Page 6: Slope Deflection Method

CHAPTER 2: SLOPE DEFLECTION METHOD Shear force and bending moment diagram;

A B

60 kN 25 kN

3 m 3 m 3 m

-133.33 kNm

63.15 kN 21.85 kN

56.12 kNm

65.57 kNm

133.33 kNm

63.15 kN

3.15 kN

21.85 kN

+ve

-ve

+ve

-ve

Page 7: Slope Deflection Method

CHAPTER 2: SLOPE DEFLECTION METHOD Example 2:

Solution:

����� � �T6�L� � �L8 9 G�

L� � �L�<���>?@

����� � T6�L� � L8 9 G�

L� � L�<���>?@

����U � �36 � �; 9 � � �L8�>?

���U� � 8�>?

2I I

A B

10 kN/m

5 kN

C

4 m 2 m

10 kN/m

A B

FEMAB FEMBA

L = 4 m

B

5 kN

C

FEMBC

L =2 m

Page 8: Slope Deflection Method

CHAPTER 2: SLOPE DEFLECTION METHOD Slope-deflection equation: hence θA = 0, θB =? ∆ = 0

��� ��� ��A6 B��� ��� � ��6C ������ ��/�0��A

G B��� ��� � ��6C � �L�<�� � �"#� � ��1< 11FFF/�0

��� ��� ��A6 B��� ��� � ��6C ������ ��/�0��A

G B��� ��� � ��6C L�<�� � ��"#� ��1< 11FFF < /�0

��V ����-�QRW

�U� � �-�QRW

Hence, ∑MB = 0;

MBA + MBc = 0; ��"#� ��1< 11 � �- � 8

�A�� ��� I<II� ���L<7= …………….. (3)

Substituting eq. (3) into eq. (1) & (2);

��� � �A�� � �L�<�� � �L<7= � L�<�� � ��P�QRW

��� � ��A�� �L�<�� � /� 9 �L<7=0 L�<�� � �-QRW

��V ����-�QRW

�U� � �-�QRW

∑MB = 0;

�L; � /L8� 9 G 9 �0 /; 9 �0 /M� �9 G0 � 8

N� � ��< �P�QR

0 0

0 0

2I I

A B

10 kN/m

5 kN

C

4 m 2 m

-15 kNm 10 kNm -10 kNm

Page 9: Slope Deflection Method

CHAPTER 2: SLOPE DEFLECTION METHOD ∑V = 0;

M� �L<�; � /L8 9 G0 � ; � 8

N� � �1< EP�QR

Shear force and bending moment diagram;

2I I

A B

10 kN/m

5 kN

C

4 m 2 m

-15 kNm 10 kNm -10 kNm

23.75 kN 21.25 kN

21.25 kN

18.75 kN

5 kN

+ve +ve

-ve

2.13 m

1.87 m

7.63 kNm

9.9 kNm

15 kNm

+ve

-ve -ve

Page 10: Slope Deflection Method

CHAPTER 2: SLOPE DEFLECTION METHOD Exercise 1:

By using Slope Deflection Method, determine: i. Fixed end moment value ii. Final bending moment diagram iii. Shear force and bending moment diagrams

Solution:

����� � ��8/�0�L� � �L;�>?@

����� � �8/�0�L� � L;�>?@

����U � �G8/70�L� � G8/�0/G0�

7� � �L;;<;7�>?@

���U� � G8/70�L� G8/G0/�0�

7� � L�=<=S�>?@

Hence θA = θc = 0 and ∆ = 0;

��� ��� ��A6 B��� ��� � ��6C ������ ����A� B��� ��� � ��6C � �L; � -< 2E�"#� � ��PFFF/�0

��� ��� ��A6 B��� ��� � ��6C ������ ����A� B��� ��� � ��6C � �L; � �< 11�"#� ��PFFF/�0

��U ��� ��A6 B��� ��U � ��6C �����U ����A7 B��� ��U � ��6C � �L;;<;7 � -< 2E�"#� � ��PP< P2FFF/10

�U� ��� ��A6 B��� ��U � ��6C ����U� ����A7 B��U ��� � ��6C L�=<=S � -< 11�"#� �1E< EOFFF /!0

40 kN

40 kN/m 20 kN/m

3 m 2 m 4 m

A B C

0 0

0 0

0 0

0 0

Page 11: Slope Deflection Method

CHAPTER 2: SLOPE DEFLECTION METHOD Hence, ∑MB = 0;

MBA + MBc = 0;

L<���A�� �L; 8<7=�A�� � �L;;<;7 � 8

�"#� � E-< �O…………..(5)

Substituting eq. (5) into eq. (1) - (4);

��� � ��8<7=�A�� � �L; � 1�< -D�QRW

��� � L<���A�� �L; � �-O< !E�QRW

��X � 8<7=�A�� � �L;;<;7 � ��-O< !E�QRW

�X� � ��8<���A�� L�=<=S � �2-< DE�QRW

Support reaction;

20 kN/m

A B

108.47 kNm 32.09 kNm

3 m

VA VB1

Y�� � 8� ��<8: L8S<G= � �8/�0 Z��[ M�/�0 � 8

������������������������������������������������������������������N� � ��2< OP�QR

YM � 8� �L7<S; � �8/�0 M�\ � 8 �������������������������������������N�� � E2< OP�QR

40 kN/m

B C

160.97 kNm 108.47 kNm

2 m

VB2 VC

40 kN

4 m

Y�U � 8� �L8S<G= L78<:= � G8/70 Z7�[ � G8/G0 M��/70 � 8

�����������������������������������������������������������������������������������������N�� � �1E< D��QR

YM � 8� L�=<:� � G8/70 � G8 MU � 8 ����������������������������������������������NX � �!�< -O�QR

Page 12: Slope Deflection Method

CHAPTER 2: SLOPE DEFLECTION METHOD Shear force & bending moment diagram;

40 kN

40 kN/m 20 kN/m

3 m 2 m 4 m

A B C

32.09 kNm 160.97 kNm

16.85 kN 214.77 kN 142.08 kN

3.55 m

0.45 m

16.85

76.85

137.92

57.92

17.92

142.08

161.02

31.85

108.7

91.17

87.14

-ve -ve

+ve

-ve -ve

+ve

+ve

Page 13: Slope Deflection Method

CHAPTER 2: SLOPE DEFLECTION METHOD Example 3

Solution

1. Fixed end moment

������ � 8

������ � 3] T]�� � /L7 9 �<;0 /L8 9 �<;�0

� � =L<�;�>?@

�����U � �T]�L� � 345�

]� � �L8/;0�L� � S8/�<;0/�<;0�

;� � �=8<S��>?@

����U� � T]�L� 345�

]� � L8/;0�L� S8/�<;0/�<;0�

;� � =8<S��>?@

����U^ � �3] � �G8�>?@

����^U � 8

2. Supports B and C do not settle, nor are they displaced up or down,_� � _U � 8. The slopes at B, (ѲB)

and C, (ѲC) are unknown. The moment equation:

�`a � ���`a ��"/�#` #a0� � 2�"b

��� �� 8�>?@

��� �� =L<�;�>?@

��U � �=8<S� ���/�cdecf0 � g��h

�� �=8<S� ���/�cdecf0i � g��/J0

i (1)

�U� � =8<S� ���/�cfecd0 � g��h

�� =8<S� ���/�cdecf0i � g��/J0

i (2)

�U^ �� �G8�>?@

A B C D

16 kN 80 kN 16 kN

2.5 m 2.5 m 2.5 m 2.5 m

10 kN/m

Page 14: Slope Deflection Method

CHAPTER 2: SLOPE DEFLECTION METHOD

�^U � �8

3. From the equilibrium equation, we know that:

Y�� � Y�U � �8� Y�� � �8������������ ��U � 8�

�����������=L<�;� =8<S� ���/�cdecf0i � 8

������������������������������������A/��� �U0; � �8<G�

���������������������������������������������������������������������������������U � � �<\��� � ���������������������������������������������/�0

Y�U � �8�����������U �U^ � 8�

�������������=8<S� ��A/��j �k0; � G8 � 8

������������������������������������A/��U �^0; � ��8<S�

�������������������������������������������������������������U �� � � \iK<\i��� ����������������������������������������/G0

4. Substituting (3) into (4);

���������������������������������� � �K<ll�� ������������������������������������������/;0

5. Substituting (5) into (3);

���������������������������������V � �;L<8��A ������������������������������������

6. Substituting ��� and ��V into (1) and (2)

��U � �=L<�;�>?@ �U� � G8�>?@

A B C D

16 kN 80 kN 16 kN

10 kN/m 71.25 kNm 40 kNm

RB RC

Page 15: Slope Deflection Method

CHAPTER 2: SLOPE DEFLECTION METHOD 7. Support reaction;

16 kN

10 kN/m 71.25 kNm

RB1

A B

m�n � �8�

�L7 � L8/�<;0 o�\ � 8

p�� � !��QR�

80 kN

10 kN/m

40 kNm

RC1

B C

RB2

71.25 kNm

2.5 m 2.5 m

2.5 m

m�� � �8�

G8 � =L<�; /S8 9 �<;0 /L8 9 ; 9 �<;0 � oU\/;0 � 8

pX� � PO< EP�QR�

m�U � �8�

G8 � =L<�; � /S8 9 �<;0�/L8 9 ; 9 �<;0 o��/;0 � 8

p�� � E�< �P�QR�

40 kNm

RC2

C D

16 kN

2.5 m

m�n � �8�

L7 oU� � 8

pX� � �2�QR�

Page 16: Slope Deflection Method

CHAPTER 2: SLOPE DEFLECTION METHOD 8. Shear force and bending moment diagram

+

- -

+

+

-

75.63 kN

71.25 kN

40 kN

+

16 kN

71.25 kN

46.25 kN 16 kN

41 kN

33.75 kN

58.75 kN

Page 17: Slope Deflection Method

CHAPTER 2: SLOPE DEFLECTION METHOD Example 4

Draw the shear and bending moment diagrams for the beam shown. Support B settles 10 mm, and EI is constant. Take E = 200 GPa, I = 400x106 mm4.

Solution:

Fixed end moment

����� � �7<�;/G0�L� � �S<���>?@

����� � 7<�;/G0�L� � S<���>?@

����U � �;8/�<;0/�<;0�;� � ��L<�;�>?@

���U� � ;8/�<;0/�<;0�;� � �L<�;�>?@

Calculate the settlement at support B

Span AB; I� �� I/J<J\J0K � =<;� 9 L8qI

Span BA; I� �� I/J<J\J0i � �7� 9 L8qI

Supports A and C do not settle, nor are they displaced up or down,��� �U� 8. The slopes at B, (ѲB) and C, (ѲC) are unknown. The moment equation:

��� ���/�88 9 L8g0r� 9 /G88 9 L8qg0s Gt ��� ��� � =<;� 9 L8���� – �S<�� � ��-�---#� � �D-O< 11FFF/�0

��� ���/�88 9 L8g0r� 9 /G88 9 L8qg0s

G ��� ��� � =<;� 9 L8���� S<�� � �!-�---#� � �OD�< 2EFFF /�0

��U ���/�88 9 L8g0r� 9 /G88 9 L8qg0s

; ��� ��U 7� 9 L8���� – ��L<�;

���������� ��O�---#� 2!�---#X 1P�< EPFFF /10�������������������������

�U� ���/�88 9 L8g0r� 9 /G88 9 L8qg0s

; ��U ��� 7� 9 L8���� ��L<�;

���������� ��O�---#X 2!�---#� !�P< �PFFF /!0�������������������������

6.25 kN/m 50 kN

4 m 2.5 m 2.5 m

3I 2I A

B C

0

0

Page 18: Slope Deflection Method

CHAPTER 2: SLOPE DEFLECTION METHOD From the equilibrium equation, we know that:

m�� � �8������������ ��U � 8�

�G8�888�� � �S:L<7= L�S�888�� 7G�888�U �;�<=; � 8

�7S�888�� 7G�888�U � ;�S<:�FFFF < < /;0

m�U � �8���������

L�S�888�U 7G�888�� GL;<�; � 8

�U ���7G�888�� � �GL;<�;L�S�888

�U � �8<;�� � ��<�G 9 L8qIFFFF < /70 Substituting (6) into (5)

�7S�888�� 7G�888/�8<;�� � ��<�G 9 L8qI0 � ;�S<:�

��7�888�� � =G7<�S

#� � �< �� 9 �-q1…….. (7)

Substituting (7) into (6)

�U � 8<;�� � ��<�G 9 L8qI �U � 8<;/�<�� 9 L8qI0 � ��<�G 9 L8qI #X ���!< 1P 9 �-q1…… (8)

Substituting (7) & (8) into (1)-(4)

��� � �7GL<:��>?@

��� � ��;S<S=�>?@

��U � �;S<;L>?@

�U� � 8<;��>?@

p� p� pX

6.25 kN/m 50 kN

4 m 2.5 m 2.5 m

3I 2I

7GL<:�� �;S<S=� �;S<;L

Page 19: Slope Deflection Method

CHAPTER 2: SLOPE DEFLECTION METHOD

m�� � �8����

�;S<S= �;8/�<;0 � oU/;0 � 8����� pX � D2< E2�QR���

m�n � �8�

o�� :7<=7 � ;8 � 8

p�� � �!2< E2�QR

Span BC

m�� � �8����

�7GL<:� � �;S<S= �7<�;/G�0

� � o�/G0 � 8�����

p�� � ��1E< 2O�QR���

m�n � �8�

o� � ��=<7S � 7<�;/G0 � 8

p� � �2�< E�QR

Span AB

262.68

237.68

46.76

96.76

+

-

-

641.93

358.79

241.89 +

-

Page 20: Slope Deflection Method

CHAPTER 2: SLOPE DEFLECTION METHOD 4.0 ANALYSIS OF FRAMES WITH NO SIDESWAY

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

Page 21: Slope Deflection Method

CHAPTER 2: SLOPE DEFLECTION METHOD Example 5

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

1. Fixed End Moment (FEM)

����� � 8�vwx

����� � 8�vwx

yz{|} � q/\�J0/~0�\� � �7G8�vwx

yz{}| � /\�J0/~0�\� � 7G8�vwx

���U^ � 8�vwx

���^U � 8�vwx

2. Hence ѲA = ѲD = 0, and displacement, ∆ = 0 (no sidesway)

�`a ���� �������� �R�� ���� � � � ���"� /�/�/�/�ѲѲѲѲX�X�X�X� ����ѲѲѲѲY�Y�Y�Y� �� �� �� ��ΔΔΔΔ0000����

��� � �� 8� ����i /�/80� �ѲB� ��/800�� �� 8<GzIѲB� � � � /L0�

���� �� 8� ����i /�ѲB� �8� ��/800� � �� 8<SzIѲB� � � � /�0�

��U � �� -7G8� ����~ /�ѲB� �ѲC� ��/800� �� -7G8� �8<;zIѲB� �8<�;zIѲC� /�0�

�U� � �� 7G8� ����~ /�ѲC� �ѲB� ��/800� �� 7G8� �8<;zIѲC� �8<�;IѲB� � /G0�

�U^ � �� 8� ����i /�ѲC� �8� ��/800� � �� 8<SzIѲC� � � � /;0�

120 kN/m

A

B C

8 m

5 m

D

Page 22: Slope Deflection Method

CHAPTER 2: SLOPE DEFLECTION METHOD �^U � �� 8� ����i /�/80� �ѲC� ��/800� �� 8<GzIѲC� � � � /70�

3. From the equilibrium equation, we know that:

�Σ{B���Σ{C���8:��Σ{B���8:� 8<SzIѲB� �8<;zIѲB� �8<�;zIѲC���7G8 (2) + (3)

�������������������������������������ѲC���/�;78/zI0�–�;<�ѲB� (7)

Σ{C���8:� 7G8� �8<;zIѲC� �8<�;zIѲB� �8<SzIѲC���8 (4) + (5)

L<�zIѲC� �8<�;zIѲB���-7G8 (8)

4. Substituting (7) into (8)

L<�zIѲC� �8<�;zIѲB���-7G8�����������L<�zI�[/�;78/zI0�–�;<�ѲB]� �8<�;zIѲB���-7G8� ����������������ѲB������78:<;�/zI (9)

5. Substituting (9) into (7)

ѲC���/�;78/zI0�–�;<�ѲB���� ������������/�;78/zI0�–�;<�[78:<;�/zI]�ѲC���-78:<;/zI� (10)

6. Substituting (9), & (10) into (1) – (6) ���� �� 8<GzIѲB�� � � ���G�<S�vwx� � � � ����� �� 8<SzIѲB�� � � ��GS=<7�vwx� � � � ���U � �� -7G8� �8<;zIѲB� �8<�;zIѲC� ��-GS=<7�vwx� � ��U�� �� 7G8� �8<;zIѲC� �8<�;IѲB� ��GS=<7�vwx� � ��U^� �� 8<SzIѲC�� � � ��-GS=<7�vwx� � ��^U � �� 8<GzIѲC�� � � ��-�G�<S�vwx��

7. Support reaction

Span BC

120 kN/m

8 m RBC RCB

487.6 kNm 487.6 kNm

Σ{C���8:�-GS=<7� �GS=<7�–�/L�8�x�S�x�G0� �RRRRBCBCBCBC�/S0���8�� � � ��������RRRRBCBCBCBC���� ������GS8�vw��GS8�vw��GS8�vw��GS8�vw���� ����ΣRV���8:�GS8�–�/L�8�x�S0� �RRRRCBCBCBCB� �� 8�� � RRRRCBCBCBCB���� �������� GS8�vwGS8�vwGS8�vwGS8�vw����

Page 23: Slope Deflection Method

CHAPTER 2: SLOPE DEFLECTION METHOD

Span AB

Span CD

487.6 kNm

243.8 kNm

B

A RAB

ΣRV = 0:

RAB = 0 kN

487.6 kNm

243.8 kNm

C

D RDC

ΣRV= 0:

RDC = 0 kN

Page 24: Slope Deflection Method

CHAPTER 2: SLOPE DEFLECTION METHOD

480 kN

480 kN

472.4 kNm

243.8 kNm 243.8 kNm

487.6 kNm 487.6 kNm

8. Shear force and bending moment diagram

B

A

+

-

C

D

+

+ -

487.6 kNm 487.6 kNm - -

Page 25: Slope Deflection Method

CHAPTER 2: SLOPE DEFLECTION METHOD Example 6

1. Fixed End Moment (FEM)

����� � � /I�0/\0/I0�K� � �LS�>?@

����� � /I�0/I0/\0�K� � 7�>?@

����U � ���\� � q/�K0/K0�

\� � ����vwx

���U� � ���\� � q/�K0/K0�

\� � ����vwx

����^ = 0 kNm

���^� = 0 kNm

2. Hence ѲC = ѲD = 0, and displacement, ∆ = 0 (no sides way frame). And Slope Deflection equation:

�`a = �R�� + ��"� (2ѲX + ѲY + 3δ)

��� � �� -LS� ����K /�ѲA� �ѲB� ��/800� �� -LS� �zIѲA� �8<;zIѲB (1)

���� �� 7� ����K /�ѲB� �ѲA� ��/800�� �� 7� �zIѲB� �8<;zIѲA (2)

��U � �� -��� ����K /�ѲB� �8� ��/800� �� -��� �zI�ѲB� (3)

��^ � �� 8� ����K /�ѲB� �8� ��/800� � �� zIѲB (4)

1 m 4 m 3 m

4 m

32 kN 24 kN/m

C A

B

D

Page 26: Slope Deflection Method

CHAPTER 2: SLOPE DEFLECTION METHOD

�U� � �� ��� ����K /�/80� �ѲB� ��/800� �� ��� �8<;zIѲB (5)

�^�� �� 8� ����K /�/80� �ѲB� ��/800� �� 8<;zIѲB (6)

3. Since ΣMA = ΣMB = 0:

Σ{A���8: -LS� �zIѲA� �8<;zIѲB� �� 8 ѲA� ��� /LS/zI0�–�8<;ѲB (7)

Σ{B���8:� 7� �zIѲB� �8<;zIѲA�-��� �zI�ѲB� �zIѲB� �� 8

������������ ��8<;zIѲA� ��zIѲB� ��� �7 (8)

4. Substituting (7) into (8)

�������������8<;zI�[/LS/zI0�–�8<;ѲB]� ��zIѲB� � ��� �7�� � � -8<�;zIѲB� ��zIѲB� ��� L=�

ѲѲѲѲBBBB���� �������� 7<LS�/zI7<LS�/zI7<LS�/zI7<LS�/zI (9)

5. Substituting (9) into (7)

ѲA� ��� /LS/zI0�–�8<;[7<LS�/zI]�ѲѲѲѲAAAA���� �������� LG<:8G/zILG<:8G/zILG<:8G/zILG<:8G/zI���� (10)

6. Substituting (9) and (10) into (1) – (6)

��� � �� 8�vwx� � � � �

���� �� L:<7G�vwx� � � � �� ��U � �� -�;<S��vwx� � �

��^ � �� 7<LS�vwx� � ��U� � �� �;<8:�vwx� � ��^�� �� �<8:�vwx�

Page 27: Slope Deflection Method

CHAPTER 2: SLOPE DEFLECTION METHOD 7. Support reaction

Span AB

Span BC

Span BD

Σ{B��8:�L:<7G�–�/���x��0� �RRRRABABABAB�/G0� �� 8�� � ������������RRRRABABABAB���� ������������ L:<8:�vwL:<8:�vwL:<8:�vwL:<8:�vw����ΣRV���8:�L:<8:�–���� �RRRRBABABABA�� �� 8�� � RRRRBABABABA���� �������� L�<:L�vwL�<:L�vwL�<:L�vwL�<:L�vw�������

Σ{c��8:�-�;<S�� ��;<8:�–�/�G�x�G�x��0� �RRRRBCBCBCBC�/G0� ��������8�� � �����������������������������RRRRBCBCBCBC���� �������G;<7S�vw�������G;<7S�vw�������G;<7S�vw�������G;<7S�vw����ΣRV���8:�G;<7S�–�/�G�x�G0� �RRRRCBCBCBCB� �� 8�� � RRRRCBCBCBCB���� �������� ;8<���vw;8<���vw;8<���vw;8<���vw����

24 kN/m

4 m RBC RCB

25.82 kNm 35.09 kNm

6.18 kNm

3.09 kNm

B

D RDB

ΣRV = 0:

RDB = 0 kN

3 m RAB RBA

0 kNm 19.64 kNm 32 kN

1 m

Page 28: Slope Deflection Method

CHAPTER 2: SLOPE DEFLECTION METHOD 8. Shear force and bending moment diagram

19.09

12.91

45.68

50.32

1.9 m

2.1 m

19.09

19.64

25.82

35.26

17.57

3.09

6.18

+

+ +

+ +

- -

- -

-

Page 29: Slope Deflection Method

CHAPTER 2: SLOPE DEFLECTION METHOD Example 7

1. Fixed End Moment (FEM)

������ �� q��\� � � �� � /I0/I0�

\� � � �� -�<�;�vwx� � �

������ �� ��\� � � �� /I0/I0�

\� � � � �� �<�;�vwx� �

� ��U��� �� q��\� ��-�PL� �� q/I0/I0�

\� �-�/G�x��0� �� -���vwx� � ����̂�� �� 8�vwx� ��^���� �� 8�vwx� �

2. Hence ѲA = ѲD = 0, and displacement, ∆ = 0 (no sides way frame). And Slope Deflection equation:

�`a = �R�� + ��"� (2ѲX + ѲY + 3δ)

��� � �� -�<�;� ����I /�/80� �ѲB� ��/800� �� -�<�;� �8<77=zIѲB (1)

���� �� �<�;� ����I /�ѲB� �8� ��/800� �� �<�;� �L<���zIѲB� (2)

��U � �� -LG (3)

��^ � �� 8� ����K /�ѲB� �8� ��/800� � �� zIѲB� (4)

�^�� �� 8� ����K /�/80� �ѲB� ��/800� �� 8<;zIѲB (5)

B

3 kN/m

C A

D

2 m 3 m

4 m

4 kN

Page 30: Slope Deflection Method

CHAPTER 2: SLOPE DEFLECTION METHOD 3. Since ΣMB = 0:

Σ{B���8:� � �<�;� �L<���zIѲA� �zIѲB�-�LG�� �� 8� � � � �� � ������������������������������ � � ѲB� ��� /;<8G/zI0� (6)

4. Substituting (6) into (1) – (5) ��� � �� L<LL�vwx� � � � �� ���� �� S<:=�vwx� � � � �� ��U � �� -LG�vwx� � �� ��^ � �� ;<8��vwx� � �� � �^�� �� �<;��vwx�

5. Support reaction

Span AB

Span BC

Span BD

3 kN/m

3 m RAB RBA

1.11 kNm 8.97 kNm

Σ{B��8:�L<LL� �S<:=�–�/��x���x�L<;0� �RRRRABABABAB�/�0���8�� � ������������������������������RRRRAB��AB��AB��AB����L<LG�vw��L<LG�vw��L<LG�vw��L<LG�vw����ΣRV���8:�L<LG�–�/��x��0� �RRRRBABABABA� �� 8�� � RRRRBABABABA���� �������� =<S7�vw=<S7�vw=<S7�vw=<S7�vw����

3 kN/m

2 m RBc

4 kN

C

ΣRV���8:�G�–�/��x��0� �RRRRBcBcBcBc���� �� 8�� � RRRRBcBcBcBc���� �������� L8�vwL8�vwL8�vwL8�vw�����

5.03 kNm

2.52 kNm

B

D

ΣRV = 0:

RDB = 0 kN

RDB

14 kNm

Page 31: Slope Deflection Method

CHAPTER 2: SLOPE DEFLECTION METHOD 6. Shear force and bending moment diagram

1.14

7.86

10

4

0.38 m

2.62 m

1.11

8.97

1.33

14

2.52

5.03

+

+

+ +

-

-

Page 32: Slope Deflection Method

CHAPTER 2: SLOPE DEFLECTION METHOD Example 8

1. Fixed End Moment (FEM)

����� ���/:0/G�0L� � ��L��>?@

����� ��/:0/G�0L� � �L��>?@

����U ���/;0/L0/��0

�� ����<���>?@

���U� ���/;0/�0/L�0

�� � �L<LL�>?@

2. Hence ѲA = ѲC = 0, and displacement, ∆ = 0 (no sides way frame). And Slope Deflection equation:

��� �� ���K /��� ��0 � L� � -< P�"#� � �� (1)

��� ����AG /��� ��0 L� � �"#� ���������������������������������������������������/�0

��U ����A� /��� �U0 � �<�� � �< 11�"#� � ��< ��������������������������������/�0

�U� ����A� /��U ��0 �L<LL � -< 2E�"#� ��< �������������������������������/G0

3. Since ΣMB = 0:

Y��� �Y��U � 8�

4 m

A B

C

9 kN/m

1m

2 m

5 kN

0

0

0

0

Page 33: Slope Deflection Method

CHAPTER 2: SLOPE DEFLECTION METHOD

�A�� L� L<���A�� � ��<�� � 8 ��������������������������������������������������A�� � �G<L:�������������������/;0

4. Substituting (5) into (1) – (4) ��� � �LG<L8�>?@ ��� � =<SL�>?@ ��U � �=<SL>?@ �U� � �L<7:�>?@

5. Support reaction Span AB Span BC

14.10 7.81 9 kN/m

4 m

RA RB1

Y�� � 8� �LG<L8 =<SL � /: 9 G 9 �0 o�/G0 � 8

o� � L:<;=�>?

Y�n � 8� L:<;= o�\ � /:0/G0 � 8

o�\ � L7<G��>?

Y�U � 8� �=<SL � L<7: � /; 9 �0 o��/�0 � 8

o�� � 7<;�>?

Y�� � 8� 7<; oU � ; � 8

oU � �L<;�>?

RB2

RC

1m

1.69

5 kN

7.81

2 m

Page 34: Slope Deflection Method

CHAPTER 2: SLOPE DEFLECTION METHOD 6. Shear force and bending moment diagram

+

+ -

19.57

16.43

6.5

1.5

14.10

7.4

7.4

7.81

1.31

1.69

-

- +

+