Betti's Law

Betti's theorem states: For a structure acted upon by two system of loads and corresponding displacements, the work done by the first system of loads moving through the displacements of the second system is equal to the work done by this second system of loads undergoing the displacements produced by the first load system.

https://en.wikipedia.org/wiki/Betti%27s_theorem

F1 * D2 = F2 * D1

                 D1= (F1/ F2) *D2

Example:

Solution:

    The differential equation also known as Eular-Bernaulli equation  which is used to determine the small deflections of beam is;

EI y'' = - M

    This equation relates elastic curvature of beam with bending moment. The sign convention for this equation is as follows;

    positive x axis is towards right and;

    positive y axis is towards downward

    

    So, using this equation one can write a equation for deflection by integrating twice which will have two integrating constants and can be evaluated by substituting known boundary conditions of slope and deflection. Further, when the equation of bending moment is written in multiple expressions for entire domain (beam span), the continuity equations of slope and deflections are used to evaluate the coefficients of integration.

    In this particular example, if we directly apply this method it will take larger time, since the equation of bending moment will have four equations which will get eight integration constants to get evaluate in order to get solved to write equation for deflection.

    So, lets do some engineer's hack to get rid of this riddle. 

    And the key trick is, Betti's law. 

Wow!!!

But,

How???

   Let's do it;

Step 1: Referencing points. 

Step 2: Defining first system of loading and displacement.

Step 3: Defining second system of loads and displacements. The choice for positioning second system of loads is not fixed and user can put them on those places where it's easier to solve the equations and get result; however the  positions of second system of displacements has to match the reference points.

Step 4: Now you can write equation for deflection quit easily.

                  M = - P ( L - x )

And using Euler-Bernaulli's equation for elastic curve 

EI y'' = - M

                   = P L - P x

                                    EI y' = P L x - ( P x^2 ) / 2 + C1

                          

                                                                EI y = ( P L x^2 ) / 2 - ( P x^3 ) / 6 + x C1 + C2

Using boundary conditions at x = 0; y' = 0 and at x = 0; y = 0, we'll obtain, C1 = 0 and C2 = 0,

So,

                                                                          EI y = ( P L x^2 ) / 2 - ( P x^3 ) / 6  

    DA = (11 /384 EI )  P L^3

  DB = ( 5 / 48 EI ) P L^3

       DC =  ( 27 / 128 EI ) P L^3

DD = ( 1 / 3 EI ) P L^3

Step 5: Using Betti's Law, and substituting F1, F2 and D2 and obtaining D1 as follows:

                            F1 = P + 2 P + 3 P + 4 P

          = 10 P

F2 = P

                               D2 = DA + DB + DC + DD

                                                 = ( 260 /384 EI )  P L^3

So;

                                                     D1 = ( 10 P / P ) ( 260 /384 EI )  P L^3

                                          = ( 2600 / 384 EI ) P L^3