Slope deflection equation with numerical problems on frames with sway and without sway

1. DEPARTMENT OF CIVIL ENGINEERING
SHIVALIK COLLEGE OF ENGINEERING
Dehradun, Uttrakhand, India, Email: info@sce.org.in, Website: www.sce.org.in
SLOPE DEFLECTION METHOD
ON PORTAL FRAMES
NAME – ASHU KUMAR KUSHWAHA
ROLL NO – 180410107004
BRANCH – B.TECH 3RD YEAR (C.E)
FACULTY NAME – DR.SUJEET KUMAR
DESIGNATION - RISP

2. INTRODUCTION
This method was developed by G.A Maney in
Germany in 1914. This method is applicable to
all types of statically indeterminate beams &
frames and in this method, we solve for
unknown joint rotations, which are expressed in
terms of the applied loads and the bending
moments. Deflections due to shear and axial
stresses are not considered as the effect are
small.

3. In slope deflection method, we shall find out the relation
between moments and slope at the joints. Then total
rotation due to applied loads and settlement of supports
may be found out and then by using the relations between
moments and slopes the net moments induced at the joint
may be found out.
a) The beam is having a constant section between each
pair of supports.
b) The joint in a structure as whole, may rotate or deflect,
but the angles between members meeting at the joint
remain the same.

4. ASSUMPTIONS
This method is based on the following simplified
assumptions:
 All the joints of the frame are rigid.
 Distortion, due to axial and shear stresses, being
very small, are neglected.

5. APPLICATIONS
1. Continuous Beams
2. Frames without side sway
3. Frames with side sway

6. SIGN CONVENTION:–
(1) ROTATIONS:– Clockwise joint rotations are
considered as (+ve).
(2) END MOMENTS:– Clockwise end moments
are considered as (+ve).
(3) SETTLEMENT:- When leads to clockwise
rotation considered as (+ve).

7. DERIVATION
Considering a continuous beam where A,B are the intermediate supports,
the final end moments developed at support A,B (MAB,MBA) will be due to
effect of
i) Effect of loading
ii) Rotation of joint A
iii) Rotation of joint B
iv) Effect of settlement of support
A B

8. By using superposition theorem the final end moment will be the summation of all the effects.
i) Effect of loading
A B
ii) Rotation at A
A B
ӨA
MFAB
= - PL/8
MFBA
= PL/8
MFAB1
= 4EIӨA/L
MFBA1
= 2EI ӨA/L
P
L/2 L/2

9. iii) Rotation at B
A B
ӨB
iv) Support B sinks by ∆
A B
ӨA
∆
MFAB2
= 2EIӨB/L
MFBA2
= 4EIӨB/L
MFAB3
= -6EI∆/L2
MFBA3
= -6EI∆/L2

10. Final end moments
MAB
= MFAB
+ MFAB1
+ MFAB2
+ MFAB3
MAB
= MFAB
+ 4EIӨA + 2EIӨB - 6EI∆
L L L2
MAB
= MFAB
+ 2EI (2ӨA +ӨB - 3∆ )
L L
Similarly,
MBA
= MFBA
+ MFBA1
+ MFBA2
+ MFBA3
MBA
= MFBA
+ 2EI (ӨA +2ӨB - 3∆ )
L L

11. PROCEDURE
i) Find fixed end moments, considering all span as a fixed beam.
* Do not consider effect of sinking in this step.
ii) Write slope deflection equation, considering the displacement as unknowns (∆,Ө).
iii) Using equilibrium conditions find the unknown displacement
iv) Substitute the values of (Ө,∆) in equation and determine the final end moments.
v) Draw the final shear force diagram, bending moment diagram, elastic curve.

12. FIXED END MOMENTS

14. EXAMPLE 1 :–
Analyse the structure by slope deflection method. Sketch the bending moment and shear force
diagram.
SOLUTION:-
a) Fixed end moments:
Span AB:
MFAB = MFBA = 0
Span BC:
MFBC = - WL2 = -10x36 = -30 KNm
12 12
MFCB = WL2 = 10x36 = 30 KNm
12 12
Span CD:
MFCD = MFDC = 0
FRAMES WITHOUT SIDE SWAY

15. b) Slope deflection equation:
MAB
= MFAB
+ 2EI (2ӨA +ӨB - 3∆ ) = 0 + 2EI (0 +ӨB + 0) (ӨA = 0)
L L 3
M
AB = 2EI ӨB / 3
MBA
= MFBA
+ 2EI (ӨA +2ӨB - 3∆ ) = 0 + 2EI (0 +2ӨB +0)
L L 3
MBA
= 4EI ӨB / 3
MBC
= MFBC
+ 2EI (2ӨB +ӨC - 3∆ ) = -30 + 2EI (2ӨB - ӨB + 0) (ӨC = -ӨB )
L L 6
M
BC
= = -30 + EI ӨB / 3

16. c) Joint equilibrium equations:
At joint B,
∑M=0 MBA + MBC = 0
4EI ӨB – 30 + EI ӨB = 0
3 3
ӨB = -30 x 3 = 18 ӨC = - ӨB = -18
5EI EI EI
d) Final moments:
M
AB = 2EI (18) = 12 KNm
3 EI
M
BA 4EI (18) = 24 KNm
3 EI
M
BC = -30 + EI (18) = -24 KNm
3 EI
M
CB = 24 KNm
M
CD = -24 KNm
M
DC = -12 KNm

17. e) To draw shear force diagram:
Span AB:
Taking moment about A,
-FBA (3) + MAB + MBA = 0
24 + 12 = FBA (3)
FBA = 12 KN
FAB = 12 KN
Span BC:
RC = 10 x 6 = 30 KN
2
RB = 30 KN
Span CD:
FCD = 12 KN
FDC = 12 KN

18. f) Draw bending moment diagram:
Simply supported bending moment in BC = WL2 = 10 x 62 = 45 KNm
8 8

19. EXAMPLE 2:–
Analyse the structure by slope deflection method. Sketch the bending
moment and shear force diagram.
SOLUTION:-
a) Fixed end moments:
MFAB = - WL2 = -10x16 = -13.33 KNm
12 12
MFBA = WL2 = 10x16 = 13.33 KNm
12 12
MFBC = -10x2 = -20 KNm
MFBE = -WL = 20x4 = -10 KNm
8 8
MFEB = WL = 20x4 = 10 KNm
8 8

20. b) Slope deflection equation:
MAB
= MFAB
+ 2EI (2ӨA +ӨB - 3∆ ) = -13.33 + 2Ex2I (0 +ӨB + 0)
L L 4
M
AB = -13.33 + EI ӨB
MBA
= MFBA
+ 2EI (ӨA +2ӨB - 3∆ ) = 13.33 + 2Ex2I (0 +2ӨB +0)
L L 4
MBA
= 13.33 + 2EI ӨB
MBE
= MFBE
+ 2EI (2ӨB +ӨE - 3∆ ) = -10 + 2EI (2ӨB +0 + 0)
L L 4
M
BE = -10 + EI ӨB
MEB
= MFEB
+ 2EI (2ӨE +ӨB - 3∆ ) = 10 + 2EI (0 + ӨB + 0)
L L 4
M
BE
= = 10 + 0.5EI ӨB

21. c) Joint equilibrium equations:
At joint B,
∑M=0 MBA + MBC + MBE = 0
13.33 + 2EI ӨB – 10 + EIӨB -20 = 0
3EI ӨB – 16.67 = 0
ӨB = 16.67 ӨC = 5.557
3EIӨC EI
d) Final moments:
M
AB = -13.33 + EI (5.557) = -7.773 KNm
EI
M
BA = 13.33 + 2EI (5.557) = 24.447 KNm
EI
M
BC = -20 KNm
M
BE = -10 + EI (5.557) = -4.443 KNm
EI
M
EB = 10 + EI (5.557) = 12.778 KNm
2 EI

22. e) To draw shear force diagram:
Span AB:
taking moment about B,
RA x 4 – 10 x 4(4/2) - 7.773 + 24.447 = 0
RA = 15.832 KN
RB1 = 10 x 4 – RA = 24.168 KN
Span BE:
Taking moment about B,
-RE (4) – 4.443 + 12.778 + 20(2) = 0
RE = 12.084 KN
RB2 = total load – RE = 7.916 KN

23. f) Draw bending moment diagram:
Simply supported bending moment in AB = WL2 = 10 x 16 = 20 KNm
8 8
Simply supported bending moment in BE = WL = 20 x 4 = 20 KNm
4 4

24. PORTAL FRAMES WITH SIDE SWAY
 Portal frames may sway due to following reasons:
1. Eccentric or unsymmetrical loading on the portal frames.
2. Unsymmetrical shape of the frames.
3. Different end conditions of the column of the portal
frame.
4. Non uniform section of the members of the frame.
5. Horizontal loading on the column of the frame.
6. Settlement of the supports of the frame.

25. EXAMPLE 3 :–
Analyse the frame by slope deflection method and draw the bending moment diagram.
SOLUTION:-
a) Fixed end moments:
MFAB = -WL = -8 x 3 = -3 KNm
8 8
MFBA = WL = 8 x 3 = 3 KNm
8 8
MFBC = - WL2 = -9x22 = -3 KNm
12 12
MFCB = WL2 = 9x22 = 3 KNm
12 12
MFCD = MFDC = 0

26. b) Slope deflection equation:
MAB
= MFAB
+ 2EI (2ӨA +ӨB - 3∆ ) = -3 + 2Ex2I (ӨB - 3∆) = -3 + 4EI (ӨB - ∆)
L L 3 3 3
MBA
= MFBA
+ 2EI (ӨA +2ӨB - 3∆ ) = 3 + 2Ex2I (0 +2ӨB - 3∆) = 3 + 4EI (2ӨB - ∆)
L L 3 3 3
MBC
= MFBC
+ 2EI (2ӨB +ӨC - 3∆ ) = -3 + 2EI (2ӨB + ӨC) = -3 + EI (2ӨB + ӨC)
L L 2
MCB
= MFCB
+ 2EI (2ӨC +ӨB - 3∆ ) = 3 + 2EI (2ӨC + ӨB) = 3 + EI (2ӨC + ӨB)
L L 2
MCD
= MFCD
+ 2EI (2ӨC +ӨD - 3∆ ) = 0 + 2E x 1.5I (2ӨC - 3∆) = 1.5EI (2ӨC - 3∆)
L L 2 2 2
MDC
= MFDC
+ 2EI (2ӨD +ӨC - 3∆ ) = 0 + 2E x 1.5I (ӨC -3∆) = 1.5EI (ӨC -3∆)
L L 2 2 2

27. c) Equilibrium and shear equations:
At joint B,
∑M=0 MBA + MBC = 0
3 + 4EI (2ӨB - ∆) – 3 + EI(2ӨB + ӨC ) = 0
3
14 ӨB - 4 ∆ + 3 ӨC = 0
MCB + MCD = 0
3 + EI (2 ӨC + ӨB ) + 1.5EI(2 ӨC - 3∆) = 0
2
ӨB + 5 ӨC - 2.25EI∆ = - 3
EI
MAB + MBA – Ph + MCD + MDC + P = 0
L L
EI ( 8 ӨB + 13.5 ӨC – 18.82∆ ) + 24 = 0
On solving,
ӨB = 0.4148
EI
ӨC = - 0.044
EI
∆ = 1.42
EI

28. d) Final moments:
M
AB = -3 + 4EI (0.4148 – 1.42) = -4.34 KNm
3 EI EI
M
BA = 3 + 4EI (2 x 0.4148 – 1.42) = 2.21 KNm
3 EI EI
M
BC = -3 + EI( 2 x 0.4148 - 0.044) = -2.21 KNm
EI EI
M
CB = 3 + EI (2 x -0.044 – 0.4148) = 3.33 KNm
EI EI
M
CD = 1.5EI (2 x -0.044 – 3 x 1.42) = -3.33 KNm
EI 2 x EI
M
DC = 1.5EI(-0.044 – 3 x 1.42) = -3.26 KNm
EI 2 x EI

29. e) Draw bending moment diagram:
Simply supported bending moment in AB = WL = 8 x 3 = 6 KNm
4 4
Simply supported bending moment in BC = WL2 = 9 x 22 = 4.5 KNm
8 8

30. THANK
YOU