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SLOPE DEFLECTION METHOD

Slope deflection method:
Slope deflection method is a displacement method used for the analysis of beams and frames. This method was introduced by George A Maney and was significantly used for more than a decade until the development of the Moment distribution method.

Steps involved in solving a problem by slope deflection method:
1.Determine the fixed end moments by considering each span as a fixed beam
2. Write the slope deflection equations for each span
3. Write the equilibrium equations for each joint
4. Solve the equilibrium equations and slope deflection equations for finding rotations(θ) and settlements (Δ)
5. Substitute the values of Δ and θ and find the final end moments (M)
6. Draw BMD and SFD

 Slope deflection equations for any member AB of span L.

MAB = MFAB+2EI/L[2θΑ+θΒ-3δ/L]

 MBA= MFBA+2EI/L[θA+2θB-3δ/L]

where,
MAB & MBA= Final end moments at ends A & B respectively in any member AB.

MFAB  & MFBA = Fixed end moments at ends A and B respectively in any member AB.

θA & θB=Rotations at ends A & B respectively in any member AB.

L = Span of member AB

δ = settlement

Solved example for illustration:

Q: Analyse the continuous beam ABCD shown in figure by slope deflection method.
    



Solution:
Step 1: Fixed end moments    

MFAB = -wL²/12
 = -(30×4²)/12 = -40KNm

MFBA = wL²/12 
= (30×4²)/12 = 40KNm

MFBC = -wab²/L² 
= -(60×2×2²)/4² = -30KNm

MFCB = wba²/L²
 = (60×2×2²)/4² = 30KNm

MFCD = -wab²/L² 
= -(60×2×4²)/6² = -53.33KNm

MFDC = wba²/L² 
= (60×4×2²)/6² = 26.667KNm

Step 2: Slope deflection equations   

MAB = MFAB+2EI/L[2θΑ+θΒ-3δ/L]
(since θA=0 & δ=0)

MAB = -40+2EI/4[2(0)+θB-0]
= -40+0.5EIθB

MBA= MFBA+2EI/L[θA+2θB-3δ/L]
(since θA=0)

MBA=40+2EI/4[0+2θB-0]
=40+EIθB

MBC= -30+2EI/4[2θB+θC-0]
=-30+EIθB+0.5EIθC

MCB= 30+2EI/4[θΒ+2θC-0]
=30+0.5EIθΒ+EIθC

MCD= -55.33+2EI/6[2θC+θD-0]
= -55.33+0.667EIθC+0.33EIθD

MDC= 26.667+2EI/6[θC+2θD-0]
=26.667+0.33EIθC+0.667EIθD
      

Step 3: Equilibrium equations   
ΣMB=0
MBA+MBC=0
or 40+EIθB-30+EIθB+0.5EIθC =0

or 2EIθB+0.5EIθC= -10 _____(1)

ΣMC=0
MCB+MCD=0
or 30+0.5EIθΒ+EIθC-55.33+0.667EIθC+0.33EIθD

or 1.5EIθB+5EIθC+EIθD+70 ___(2)

ΣMD=0
MDC=26.667+0.33EIθC+0.667EIθD=0

or EIθC+2EIθD= -80 _______(3)

on solving equations 1,2 & 3 we get,
EIθC= 28.485
EIθB= -12.121
EIθD -54.243
          

Step 4: Final end moments    
 Final end moments are calculated by substituting the values of EIθC, EIθB & EIθD in the slope deflection equations.

Hence,
MAB=-40+0.5(-12.121)
=-46.061KNm
MBA = 40-12.121=27.879KNm
MBC = -27.879KNm
MCB=30+0.5(-12.121)+28.485
=52.425KNm
MCD= -52.425KNm
MDC=0

   

Bending moment diagram:

                           


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