Skip to main content

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:

                           


Labels:

Comments

Post a Comment

Popular posts from this blog

CENTRE OF GRAVITY

6.1 Centre of gravity: The point at which whole weight of the body is assumed to be concentrated, is known as the centre of gravity of that body. 6.2 Centroid: The point at which whole area of a plane(circle, rectangle, triangle, quadrilateral etc) is assumed to be concentrated is known as the centroid of that area. It is represented by C.G or simply G. •  The centroid and centre of gravity lie at the same point. 6.3  C.G of simple plane figures: C.G of a uniform rod lies at it's middle point C.G of a rectangle or a parallelogram lies at a point where it's both the diagonals meet.  C.G of a triangle lies at a point at which the three medians of the triangle meet. C.G of a circle lies at the centre of the circle. 6.4 Centroid of plane areas by Method of Moments: Consider a plane area with a total area "A" whose C.G is to be determined.  Let's divide this area into a nu...
Post a Comment
Read more

CONCEPT OF SHEAR FORCE AND BENDING MOMENT

1.1 Shear force and Bending moment: When a beam is subjected to a set of loads and reactions, as a result of which the internal forces and moments tend to setup within the beam This internal force is the shear force  and the internal moments are the bending moments.  For illustration,  consider a beam with a given set of loading supported at two points A &  B giving rise to reactions at the two supports.                        (fig.  1.1) Let's cut the beam into two parts &  let's assume that the resultant of loads (F) and reactions to the left of section xx is vertically upwards.                         (fig.  1.2) For equilibrium of the section, the resultant of loads and reactions to the right of section xx has to be equal to ''F" but vertically downwards. This force is known as shear force...
2 comments
Read more