Singularity Function

Singularity functions.

When dealing with bending moment, shear force and beam deflection a more generalized and powerful way is using singularity functions.

This functions besides their application in computer programming, they give a huge advantage for students and practitioners dealing with the above mentioned problems.

The process starts with defining the function representing the loading. Once that is done successfully and the boundary conditions are set successive integarals are done to arrive at the conservative shear, moments , slope and deflection equations. and each integral will yeild constants.

Using boundary conditions each constant is solved for and the equations are set. Then the resulting functions can be used cautiously since rules are associated with the functions.

One such rule is shown In the above picture.

Some of the singularity functions with their respective integrals are listed in this picture.

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loads with singularity functions and respective shear and moment.

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Here is a beam with a concentrated load of 'P' at 'a' distance from the left support. Let's call the left support A and the right support B. From equilibrium of the simple beam the reactions at the left and the right supports are P(1-a/L) and Pa/L.

Now the singularity function for the load would be, P^(-1).

The shear force on the beam is the integral of the load singularity function. And by definition of integral of singularity function( I have a reason for saying this, u can ask want to understand) the integral of ^(-1) is just ^(0). So the shear force equation would be

V(x)=P^(0)+c1

And the moment equation can be obtained by simply integration of the shear function obtained above.

M(X)=P^(1)+c1*x+c2

Now that we have the equations all we need to worry about is the constants of integration. For them we have boundary conditions which goes:

The shear at x=0 is the reaction at the support and the moment at x=0 is 0 since the support is pin. Two boundary conditions would give two equation

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Now substituting into (1)

V(0)=0, we can get C1=Ay=P(1-a/L)

And substituting into (2)

M(0)=0, we can get C2=0

Now the equations would simplify to:

V(x)= P^(0)+p(1-a/L)
M(x)=P^(1)+p(1-a/L)*x

Here we use <> ( macauleys brackets after the person who developed them) because it has meaning. If x

We can proceed further with it and determine the slope and deflection equations.

Since y"( I was tempted to write d^2y/dx2 gene see how messy it is ) =M/EI :

y"=(P^(1)+P(1-a/L)*x)/EI

Integrating the above equation gives

y'=(P^(2)/2+P(1-a/L)*x^2/2)/EI +C3

And integration of the above equation would be:
y=(P^(3)/6+P(1-a/L)x^(3)/6)/EI+C3x+C4

In order to calculate the constants of integration we need to identify the boundary conditions.

Since the system is made of pin and roller support we know the deflection at the two supports is zero. Meaning:

y(0)=0 and y(L)=0

Substituting the boundary conditions into the above equations:

y'(0)=0, C4=0 and
y'(L)=0, C3=-P(1-a/L)L^2 /3EI

Therefore the equation of slope and deflection with singularity functions become:

y'=(P^(2)/2+P(1-a/L)*x^2/2)/EI -P(1-a/L)L^2 /3EI

y=(P^(3)/6+P(1-a/L)x^(3)/6)/EI-(P(1-a/L)L^2 /3EI )x

The power of singularity functions isn't observed with a beam supporting a concentrated load at some distance from its support just like the above one. It is usually useful when multiple loads are assigned over the span of the beam and it would help avoid the numerous sections that we take in order to write the shear and moment equations.

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