# Displacement Methods: Slope Deflection and Moment Distribution Methods Notes

By Vishwajeet Sinha|Updated : October 5th, 2020

### Strain Energy Method

Strain energy stored due to axial load where, P = Axial load

dx = Elemental length

AE = Axial rigidity

Strain energy stored due to bending where, MX = Bending moment at section x-x

ds = Elemental length

El = Flexural rigidity

or E = Modulus of elasticity

l = Moment of inertia

Strain energy stored due to shear where, q = Shear stress

G = Modulus of rigidity

dv = Elemental volume

Strain energy stored due to shear force where, AS = Area of shear

S = Shear force

G = Modulus of rigidity

ds = Elemental length

Strain energy stored due to torsion where, T = Torque acting on circular bar

dx = Elemental length

G = Modulus of rigidity

lP = Polar moment of inertia

Strain energy stored in terms of maximum shear stress where, Maximum shear stress at the surface of rod under twisting.

G = Modulus of rigidity

V = Volume

Strain energy stored in hollow circular shaft is, Where, D = External dia of hollow circular shafts

d = Internal dia of hollow circular shaft Maximum shear stress

Castigliano's first theorem where, U = Total strain energy

Δ = Displacement in the direction of force P.

θ = Rotation in the direction of moment M.

Castiglianos Second Theorem Betti's Law  where, Pm = Load applied in the direction m.

Pn = Load applied in the direction n.

δmn = Deflection in the direction 'm' due to load applied in the direction 'n'.

δnm = 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

(i) 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  where, δ = Deflection of D in the direction of load P.

(ii) Semicircular arch whose one end is hinged and other supported on roller carried a load P as shown in the figure.  where, δ = Deflection at B in the direction of load P.

(iii) A quadrantal ring AB of radius r support a concentrated load P as shown  where, δV = Vertical deflection (deflection in the direction of load P) at end A

δH = Horizontal deflection of end A.

(iv) A portal frame as shown in figure carries a load P at A  where, δH & δV is horizontal & vertical deflection at end A respectively.

(v) 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  where, δV = Vertical deflection at 'O'.

Moment distribution method (Hardy Cross method)

Stiffness: It is the force/moment required to be applied at a joint so as to produce unit deflection/rotation at that joint. Where, K = Stiffness

F = Force required to produce deflection Δ

M = Moment required to produce rotation θ.

Stiffness of beam

(i) Stiffness of member BA when farther end A is fixed.  Where, 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  where, K = Stiffness of BA at joint B. When farther end is hinged

Carry over factor:

Carry over factor = COF may greater than, equal to or less than 1.

Standard Cases:  (ii) COF = 0   Distribution Factor (D.F.):  where M1, M2, M3 and M4 are moment induced in member OA, OB, OC and OD respectively.

Relative Stiffness:

(i) When farther end is fixed

Relative stiffness for member (ii) When farther end is hinged

Relative stiffness for member Where, l = MOl and L = Length of Beam Stiffness of OA Stiffness of OB Stiffness of OC Stiffness of OD = 0

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. MAB & MBA 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. GradeStack Learning Pvt. Ltd.Windsor IT Park, Tower - A, 2nd Floor, Sector 125, Noida, Uttar Pradesh 201303 help@byjusexamprep.com