**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, A_{S} = 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

l_{P} = 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, 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**

*δ*_{21} = *δ*_{12}

_{}

where, *δ*_{21 }= deflection in the direction (2) due to applied load in the direction (1).

*δ*_{12} = 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, *δ _{V} =* Vertical deflection (deflection in the direction of load P) at end A

*δ _{H} =* Horizontal deflection of end A.

(iv) A portal frame as shown in figure carries a load P at A

where, *δ _{H }*&

*δ*is horizontal & vertical deflection at end A respectively.

_{V}(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, *δ _{V} =* 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 M_{1}, M_{2}, M_{3} and M_{4} 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. M_{AB} & M_{BA} are final moments at A & B respectively. θ_{A} and θ_{B} 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.

## Comments

write a commentVikash KumarJun 16, 2017

Nikhil JindeSep 5, 2017

Vivek SavaliyaJan 1, 2018

Satrudhan Mukhiya Satrudhan MukhiyaApr 26, 2019