1. Alpha College of Engineering and Technology
Civil-A
Subject- Structure Analysis- 2
Group no-18
Group members-
1) Chaudhary Babubhai ( )
2) Chaudhary Arvind ( )
3) Chaudhary Madrup ( )
4) Jain Mehul ( )
Guided By :- Prof. Vivek Patel
Prof. Umesh Dobariya 1

2. SINKING FOR SLOPE DEFLECTION METHOD

3. SINKING ( SETTLEMENT):
The settlement will be
taken as positive, if it
rotates the beam as a ᵟ=
whole in the clockwise ve
direction.
The settlement will be taken
as negative, if it rotates
the beam as a whole in the ᵟ=
anticlockwise direction ve

4. In this problem A ϴ =0, ϴB≠0, ϴc≠ 0, δ =15mm
Example (1): Analyse the continuous beam ABCD shown in figure by slope deflection method. The support B sinks by 15mm.
Solution:
In this problem A ϴ =0, ϴB≠0, ϴc≠ 0, δ =15mm

5. SINKING ( SETTLEMENT):
FEM: FAB = - wab ˄2 / l˄2 = kn.m FAB = + wa˄2b / l˄2 = kn.m FBC = - wl˄2 / 8 = kn.m FCB = + wl˄2 / 8 = kn.m FEM due to yield of support B

6. SINKING ( SETTLEMENT):
Slope deflection equation:
MAB = FAB + 2EI / l ( 2ϴA+ ϴB - 3 δ/l )
= /3 EI ϴB- 6
= /3 EI ϴB (1)
MBA = FBA + 2EI / l ( 2ϴA+ ϴB - 3 δ/l )
= /3 EI ϴB- 6
= /3 EI ϴB (2)

7. SINKING ( SETTLEMENT):
MBC = FBC + 2EI / l ( 2ϴB+ ϴC - 3 δ/l )
= /5 EIϴB + 2/5 EI ϴC - 6
= /5 EIϴB + 2/5 EI ϴC (3)
MCB = FCB + 2EI / l ( 2ϴC+ ϴB - 3 δ/l )
= /5 EI ϴC + 2/5 EI ϴB
= /5 EI ϴC + 2/5 EI ϴB (4)
MCB = - 30 KN.M (5)

8. SINKING ( SETTLEMENT):
There are only two unknown rotations ϴBand ϴC.
Accordingly the boundary conditions are:
MBA + MBC = 0
MCB + MCD = 0
Now, MBA + MBC = /15 EIϴB + 2/5 EIϴC = 0
MCB + MCD = /5 EIϴB + 4/5 EIϴC = 0
Solving the equation we get,
ϴB = / EI
ϴC = / EI

9. SINKING ( SETTLEMENT):
Final Moments:
MAB= /3 (-31.35) = KN.M
MBA= /3 (-31.35) = KN.M
MBC= /5 (-31.35) + 2/5 (-9.71) = KN.M
MCB= /5 (-9.71) + 2/5 (-31.35) = KN.M
MCD = - 30 KN.M

10. SINKING ( SETTLEMENT):
Consider the free body diagram of continuous beam for finding reactions

11. SINKING ( SETTLEMENT):

12. SINKING ( SETTLEMENT):
Reactions:
Span AB:
RB × 6 = 100 x – 60.89
RB = 66.85
RA = 100 – RB
= KN
Span BC:
RB × 5 = 20 x 5 x – 30
RB = KN
RC = 20 x 5 - RB = KN

13. SINKING ( SETTLEMENT):
B.M. diagram:
Span AB, M = Wab / l = (100×4×2) / 6 = kn.m
Span BC, M = Wl˄2 / 8 = ( 20×25) / 8 = kn.m

14. Example (2):Determine the internal moments at the supports of the beam shown in Fig. The support at B is displaced (settles) 12 mm.

15. SINKING ( SETTLEMENT):
Solution
Two spans must be considered. FEMs are determined using

16. SINKING ( SETTLEMENT):
2. SLOPE DEFLECTION METHOD:

17. SINKING ( SETTLEMENT):
3. Equilibrium condition:

18. SINKING ( SETTLEMENT):
In order to obtained the rotations equations (n) & (p) may then be solved simultaneously, it may be noted that since A is fixed support. Thus,
Substituting these values into equations (i to l) yields