Slope Deflection Method
-
Upload
nurul-abdul-jalil -
Category
Documents
-
view
478 -
download
7
Transcript of 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
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
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
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
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
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
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
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
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
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
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
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
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
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
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�
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
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
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
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 +
-
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
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
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����
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
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 - -
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
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�
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
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
+
+ +
+ +
- -
- -
-
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
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
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
+
+
+ +
-
-
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
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
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
-
- +
+