So now we can write

 

d²v/dx² = M / (E I)

 

If we allow the use of  ‘ to represent d/dx, then

 

E I v’’ = M

 

We also know from our work with shear and bending moment diagrams that

 

dM/dx = V        and        dV/dx = w

 

So we can further say that

 

d/dx (E I v’’) = V         and        d²/dx² (E I v’’) = w

 

If we assume that E and I are constant over the length of the beam, then

 

E I v’’’ = V                    and        E I v’’’’ = w

 

So we have a choice between using a second, third or fourth order differential equation to determine the deflection in a beam.  Generally, we choose to use either the second or fourth order equation.  These equations require that we be able to express either the distributed load, w, or the bending moment, M, in the beam as a function of x.  We know from our work with shear and bending moment diagrams that for an arbitrary problem, neither w nor M can be expressed as a single function over the entire length.  Rather, the beam has sections of its length where M or w is written as a function of x.  Let’s ignore this fact for now and assume that we have a statically determinant beam composed of one section and w and M can be expressed as a function of x.

 

If we use the second order differential equation, we must first draw a free body diagram of the beam and determine the reactions using equilibrium.  Then, we use the method of sections to cut the beam at an arbitrary distance x from the left hand end, draw a free body diagram of one part of the beam and use equilibrium to solve for M.  At this point, we substitute the function for M into the differential equation and integrate twice to find a function for deflection.

 

E I v’’ = M

 

E I v’ =  ∫  M dx + A

 

 

E I v =   ∫ ∫  M dx  dx + A x + B

 

 

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