Test: Structural Analysis- 1

Moment Distribution Method was developed by Hardy Cross
Slope Deflection is also known as Displacement Method.
Knai’s method involves use of rotation factor.
Force Method [Force] [flexibility matrix] = Deflection  Therefore, it involves use of flexibility matrix.

In the virtual work method of plastic analysis of steel structure, the virtual quantity is displacement.

Virtual work arises in the application of the principle of least action to the study of forces and movement of a mechanical system. The work of a force acting on a particle as it moves along a displacement will be different for different displacements. Among all the possible displacements that a particle may follow, called virtual displacements, one will minimize the action. This displacement is therefore the displacement followed by the particle according to the principle of least action. The work of a force on a particle along a virtual displacement is known as the virtual work.

Principle of virtual work: (unit-load Method)

Developed by Bernoulli:

To find Δ at point A du to loads P₁, P₂, P₃. Remove all loads, apply virtual load P’ on point A.

For simplicity P’ = 1

It creates internal load u on representative element.

Now remove this load, apply P₁, P₂, P₃ due to which pt. A will be displaced by Δ

∴ External virtual work = 1.Δ

Internal virtual work = u.dL


P^' = 1 = external virtual unit load in direction of Δ.

u = internal virtual load acting on element in direction of a dL.

Δ = external displacement caused by real loads.

dL = internal deformation caused by real loads.

The Maxwell reciprocal theorem, states that the deflection at a point A in the direction of  1 due to load at point B in the direction of 2 is equal in the magnitude to the deflection of point B in the direction of 2 produced by a load applied at A in direction 1. 

Hence,

δ₁₂ = δ₂₁

The fixed end moment at end A

The beam has internal hinges at C & D, so the beam can be break down as shown

In AC section moment applied = M

So the reaction will be = M/L/2 = 2/ML

The carry – over moment at B = 2M/L × L/4 = M/2

Deflection, δ = ∂U/∂F

In which beam deflection is higher, strain energy stored will be maximum.

In cantilever, the deflection will be highest.

The deflection of a simply supported beam is higher than propped cantilever as one of the support is fixed.

In a fixed beam, the deflection will be minimum.

So,

δ_cantilever >  δ_{simple-support} > δ_{propped cantilever} > δ_fixed

∴ U_fixed < U_{proped-cantilever} < U_{simple-support} < U_cantliver 

BM at section X-X, M_x = W.x

Now, for


For,


∴ For 

U will be less than but greater than 

The FBD of the beam will be

Taking moment about A,

R × L = 10 × 2L

R = 20 KN

So the extension of the spring will be

R = Kx

20 = 1. x

x = 20 mm

From similar triangle

δ_B = 20 × 2 = 40 mm

Concept :-

Stiffness value, K when far end is fixed = 4EI/L

Stiffness value, K when far end is roller = 3EI/L

Stiffness value, K when far end is guided roller = EI/L

Calculation:-

For stiffness coefficient when far end is guided roller support refer :-

Advanced Structural Analysis, Prof. Devdas MenonDepartment of Civil Engineering, Indian Institute of Technology, Madras

Module - 5.3 ,Lecture - 29Matrix Analysis of Beams and GridsPage No. - 9