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**Solution with R _{B} as the redundant **

2) After selecting **R _{B }**as redundant, express all other reactions in terms of the redundant

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

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)

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