Introduction: The Force Method of Analysis - 2 | Structural Analysis - Civil Engineering (CE) PDF Download

Solution with R_B as the redundant 

2) After selecting  R_B as redundant, express all other reactions in terms of the redundant R_B . This can be accomplished with the help of equilibrium equations.

Thus,

R_A = wL − RB                        (7.1a)   

and  

             (7.1b)

3) Now release the restraint corresponding to redundant reaction R_B . Releasing restraint in the present case amounts to removing the support at B. Now on the resulting cantilever beam (please note that the released structure is statically determinate structure), apply uniformly distributed load w and the redundant reaction R_B as as shown in Fig. 7.2b. The released structure with the external loads is also sometimes referred as the primary structure.  

4) The deflection at B of the released structure (cantilever beam, in the present case) due to uniformly distributed load and due to redundant reaction R_B could be easily computed from any one of the known methods (moment area method or unit load method). However it is easier to compute deflection at B due to uniformly distributed load and R_B in two steps. First, consider only uniformly distributed load and evaluate deflection at B, which is denoted by (Δ_B)₁ as shown in Fig. 7.2c. Since R_B is redundant, calculate the deflection at B due to unit load at B acting in the direction of R_B and and is denoted by (Δ_B)₂ as shown in

In the present case the positive direction of redundant and deflections are assumed to act upwards. For the present case, (Δ_B)₁ and (Δ_B)₂ are given by,

                                      (7.2a)

and 

                                                  (7.2b)

From the principle of superposition, the deflection at B, (Δ_B), is the sum of deflection due to uniformly distributed load (Δ_B)₁ and deflection R_B(Δ_B)₂ due to redundant R_B . Hence,

Δ_B =  (Δ_B)₁ + R_B(Δ_B)₂                                (7.2c)

5) It is observed that, in the original structure, the deflection at B is zero. Hence the compatibility equation can be written as,

Δ_B =  (Δ_B)₁ + R_B(Δ_B)₂   = 0                                  (7.3a)

Solving the above equation, the redundant R_B can be evaluated as,

                               (7.3b)

Substituting values of (Δ_B)₁ and (Δ_B)₂ , the value of R_B is obtained as,

\(R_B = {3wL\over8}\)                             (7.3d)

The displacement at B due to unit load acting at B in the direction of R_B is known as the flexibility coefficient and is denoted in this course by a_BB.

6) Once R_B is evaluated, other reaction components can be easily determined from equations of statics. Thus,

                                        (7.4a)

                         (7.4b)

7) Once the reaction components are determined, the bending moment and shear force at any cross section of the beam can be easily evaluated from equations of static equilibrium. For the present case, the bending moment and shear force diagram are shown in Fig. 7.2e.

Solution with M_A as redundant

1) As stated earlier, in the force method the choice of redundant is arbitrary. Hence, in the above problem instead of R_B one could choose M_A as the redundant reaction. In this section the above problem is solved by taking M_A as redundant reaction.

2) Now release (remove) the restraint corresponding to redundant reaction M_A. This can be done by replacing the fixed support at A by a pin. While releasing the structure, care must be taken to see that the released structure is stable and statically determinate.

3) Calculate the slope at A due to external loading and redundant moment M_A. This is done in two steps as shown in Fig. 7.3b and Fig.7.3c. First consider only uniformly distributed load (see Fig. 7.3b) and compute slope at A, i.e.(θ_A)₁ from force displacement relations. Since M_A is redundant, calculate the slope at A due to unit moment acting at A in the direction of M_A which is denoted by (θ_A)₂ as in Fig. 7.3c. Taking anticlockwise moment and anticlockwise rotations as positive, the slope at A, due to two different cases may be written as,

                    (7.5a)

(θ_A)₂ = L/3EI                           (7.5b)

From the principle of superposition, the slope at A, θ_A is the sum of slopes (θ_A)₁ due to external load and M_A (θ_A)₂ due to redundant moment M_A. Hence

 M_A = (θ_A)₁ + M_A (θ_A)₂               (7.5c)

4) From the geometry of the original structure, it is seen that the slope at A is zero. Hence the required compatibility equation or geometric condition may be written as,

(θ_A) = (θ_A)₁ + M_A (θ_A)₂ = 0                      (7.5d)

Solving fo M_A,

               (7.5e)

Substituting the values of (θ_A)₁ , and (θ_A)₂ in in equation (7.5e), the value of M_A is calculated as

                      (7.5f)

5) Now other reaction components can be evaluated using equilibrium equations. Thus,

R_A = 5wL/8                         (7.6a)

R_B = 3wL/8                          (7.6b)