Displacement Methods: Slope Deflection and Moment Distribution Methods | Structural Analysis - Civil Engineering (CE) PDF Download

Strain Energy Method

1. Strain energy stored due to axial load
   
   where, P = Axial load
   dx = Elemental length
   AE = Axial rigidity
2. Strain energy stored due to bending
   
   where, M_X = Bending moment at section x-x
   ds = Elemental length
   El = Flexural rigidity
   or E = Modulus of elasticity
   l = Moment of inertia
3. Strain energy stored due to shear
   
   where, q = Shear stress
   G = Modulus of rigidity
   dv = Elemental volume
4. Strain energy stored due to shear force
   
   where, A_S = Area of shear
   S = Shear force
   G = Modulus of rigidity
   ds = Elemental length
5. Strain energy stored due to torsion
   
   where, T = Torque acting on circular bar
   dx = Elemental length
   G = Modulus of rigidity
   l_P = Polar moment of inertia
6. Strain energy stored in terms of maximum shear stress
   
   where, image007 Maximum shear stress at the surface of rod under twisting.
   G = Modulus of rigidity
   V = Volume
7. Strain energy stored in hollow circular shaft is
   
   Where, D = External dia of hollow circular shafts
   d = Internal dia of hollow circular shaft
   τ_max = Maximum shear stress

Castigliano's first theorem

∂U/∂Δ = P & ∂U/∂θ = M
where, U = Total strain energy
Δ = Displacement in the direction of force P.
θ = Rotation in the direction of moment M.

Castiglianos Second Theorem

∂U/∂P = Δ, ∂U/∂M = θ

Betti's Law

∑P_mδ_mn = ∑P_nδ_nm

where, P_m = Load applied in the direction m.
P_n = Load applied in the direction n.
δ_mn = Deflection in the direction 'm' due to load applied in the direction 'n'.
δ_nm = Deflection in the direction 'n' due to load applied in the direction 'm'.

Maxwells Reciprocal Theorem

δ₂₁ = δ₁₂

where, δ₂₁ = deflection in the direction (2) due to applied load in the direction (1).
δ₁₂ = Deflection in the direction (1) due to applied load in the direction (2).

Standard Cases of Deflection

1. For the portal frame shown in the figure below horizontal deflection at D due to load P, assuming all members have same flexural rigidity is given by
   δ = Ph²(2h + 3b)/3Elwhere, δ = Deflection of D in the direction of load P.
2. Semicircular arch whose one end is hinged and other supported on roller carried a load P as shown in the figure.
   δ = π/2.PR³/Elwhere, δ = Deflection at B in the direction of load P.
3. A quadrantal ring AB of radius r support a concentrated load P as shownwhere, δ_V = Vertical deflection (deflection in the direction of load P) at end A
   δ_H = Horizontal deflection of end A.
4. A portal frame as shown in figure carries a load P at Awhere, δ_H & δ_V is horizontal & vertical deflection at end A respectively.
5. Figure shows two identical wires OA and OB each of area A and inclined at 45° from horizontal. A load P is supported at O
   δ_v = PL/AEwhere, δ_V = Vertical deflection at 'O'.

Moment distribution method (Hardy Cross method)

1. Stiffness: It is the force/moment required to be applied at a joint so as to produce unit deflection/rotation at that joint.
   k = F/Δ or M/θ
   Where, K = Stiffness
   F = Force required to produce deflection Δ
   M = Moment required to produce rotation θ.
2. Stiffness of beam
   (i) Stiffness of member BA when farther end A is fixed.
   k = 4El/LWhere, K = Stiffness of BA at joint B. When farther end is fixed.
   El = Flexural rigidity
   L = Length of the beam
   M = Moment at B.
   (ii) Stiffness of member BA when farther end A is hinged
   k = 4El/Lwhere, K = Stiffness of BA at joint B. When farther end is hinged
3. Carry over factor
   Carry over factor = Carry over moment/Applied moment
   COF may greater than, equal to or less than 1.
4. Standard Cases
   (i) COF = 1/2
   (ii) COF = 0(iii) COF = a/b
5. Distribution Factor (D.F.)
   DF = stiffness of a member/Sum of stiffness of all members at that joint 
   or
   DF= Relative stiffness of a member/Sum of relative stiffness of all member at that jointwhere M₁, M₂, M₃ and M₄ are moment induced in member OA, OB, OC and OD respectively.
6. Relative Stiffness
   (i) When farther end is fixed
   Relative stiffness for member = l/L
   (ii) When farther end is hinged
   Relative stiffness for member = 3l/4L
   Where, l = MOl and L = Length of BeamStiffness of OA = 4El/l
   Stiffness of OB
   Stiffness of OC = 34El/l
   Stiffness of OD = 0
7. Fixed Convention
   +ve → Sagging
   –ve → Hogging
   and All clockwise moment → +ve
   and All Anti clockwise moment → –ve
   Span length is l

Slope deflection Method (G.A. Maney Method)

In this method, joints are considered rigid. It means joints rotate as a whole and the angles between the tangents to the elastic curve meeting at that joint do not change due to rotation. The basic unknown are joint displacement (θ and Δ).
To find θ and Δ, joints equilibrium conditions and shear equations are established. The forces (moments) are found using force displacement relations. Which are called slope deflection equations.

Slope Deflection Equation
(i) The slope deflection equation at the end A for member AB can be written as:

(ii) The slope deflection equation at the end B for member BA can be written as:

where, L = Length of beam, El = Flexural Rigidity  are fixed end moments at A & B respectively. M_AB & M_BA are final moments at A & B respectively. θA and θB are rotation of joint A & B respectively.
Δ = Settlement of support
Sign Convention
+M → Clockwise
-M → Anti-clockwise
+θ → Clockwise
-θ → Anti-clockwise
Δ → +ve, if it produces clockwise rotation to the member & vice-versa.
The number of joint equilibrium conditions will be equal to number of ‘θ’ components & number of shear equations will be equal to number of ‘Δ’ Components.