Analysis of fixed beams by force method

  Analysis of Fixed beams using Force method

The following video explains the force method for analysis of fixed beams

Further problems of Fixed End Moments: 

          In all problems, the basic principles are as follows
                 1) area of M/EI diagram = 0
                 2) moment of M/EI diagram about any of the support = 0

This is a problem in which the loading consists of a concentrated moment or couple

 Restating the basic principles of analysis of Fixed beams, as below

For applying these basic principles, we need the M/EI diagram of the beam. This can be drawn by considering the Fixed beam as a simply supported beam with given loading and simply supported beam with the load of (redundant) support reactions. Hence, the 1) M/EI diagram for simply supported beam with M0 loading and 2) M/EI diagram with the loading of support moments are drawn as below. Note that the M/EI diagram due to support moments is same in all problems.

 Using the above figures and their geometric properties, the area of the figures is determined and equated to zero as shown below

 Similarly, the moment of the figures is determined and equated to zero as shown below

 Now substitute equation 1 , to eliminate M_A, as follows

 By substituting this expression for M_B in equation 1, we can determine the expression for M_A also. This completes the problem

Determining Fixed End Moments under distributed load

Recall that the Fixed End moments for the following case is given as below

Using the above result, the Fixed End Moments for any case of distributed load can be determined as below. For example, any distributed loading can be considered as a series of differential concentrated loads (shown shaded in the figure below).

The concentrated load corresponding to the shaded portion of distributed load is q_xdx (=P). This load is acting at a distance of x (=a) from A and L-x (=b) from B. On substituting these values of P, a and b in the expression above, we get the end moments caused by the differential load. IN the figure below, the support moment at A is evaluated

On integrating the above expression we get the support moment due to the whole distributed load

The following example illustrates the application of the above procedure for fixed beam with UDL. Note that UDL is a distributed load having constant intensity. So, q_x=w is a constant with respect to x.

Similarly, M_B can be evaluated. It can be concluded that M_B=M_A for this case of symmetrical loading.