- Discontinuity Functions Table.
- The first case in the discontinuity function table is a beam under a clockwise moment.
- The second case in the discontinuity function table is a beam under a vertical load Po.
- The third case in the discontinuity function table is a beam under a uniform load wo.
- The fourth case in the discontinuity function table is a beam under a triangular load with maximum intensity wo.
- The fifth case in the discontinuity function table is a beam with an upward limited width load of wo.
- The sixth case in the discontinuity function table is a beam with an upward limited width load of maximum intensity wo.
- The 7th case in the discontinuity function table is a beam with an upward limited width load of maximum intensity wo.

## Discontinuity Functions Table.

This post concerns the Discontinuity Functions tables representing Load, Shear, and Moment. These tables are included in Prof. Timothy Philpot’s book, An Integrated Learning System, Mechanics of Materials.

It is important to establish a positive direction guide for the positive directions of loads and moments. For horizontal loads, a right arrow is made from left to right. For vertical loads, the upward direction is considered a positive direction.

For moments, the positive direction is the anticlockwise direction. This post introduces seven cases.

### The first case in the discontinuity function table is a beam under a clockwise moment.

Three equilibrium equations exist for the sum of x-forces, the sum of y-forces, and the moments acting on a beam.

The first case is a moment Mo acting on a beam in the clockwise direction, but this moment is shifted by a distance away from the left part of the beam and acting at a distance of a.

The first expression of w(x) or the load function is w(x)=M0<x-a>^-2, which is a discontinuity function. If we check the units, we have a moment, as expressed by Kips. Ft, for instance, divided by Ft^2 will produce kips/ft, which match the units of the load function.

We integrate one to get the shear function value of V(x). The value of V(x)=M0<x-a>^-1, as explained earlier, the integration is done by adding 1 to -2, but we do not divide by (-2+1) or -1 for the discontinuity function.

We integrate again to get the Moment function value of M(x). The value of M(x)=M0<x-a>^0; for values bigger than a, the term<x-a> is a positive raised to the power of zero equals 1.

If the term ( x-a ) is negative, it will be equated to zero. This is a step function where the step value is M0 or Maculauy’s function where n=0.

There is a match with the result. If we take a section cut in the beam at the Left of the moment applied, we get a zero value of M(x).

But if we take the section at the right of distance a from the left end of the beam, consider the y-axis represents M(x) and sum the moments equal zero, we can equate M(x)-M0=0, then M(x)=M0, if the section is taken near the end of the beam at the right the same result is obtained which will lead to a step function at x=a.

Please note the moment M0 is acting in the clockwise direction.

### The second case in the discontinuity function table is a beam under a vertical load Po.

The second case is an upward load Po acting on a beam shifted from the left part of the beam but acting at a distance of a.

The first expression of w(x) or the load function is w(x)=P0<x-a>^-1, which is a discontinuity function. If we check the units, we have a load, as expressed by Kips.

For instance, dividing by Ft will produce kips/ft, which match the load function units.

We integrate once to get the shear function value of V(x). The value of V(x)=P0<x-a>^0, as explained earlier, the integration is done by adding to -1, but we do not divide by(-1+1) or 0 for the discontinuity function. This is a step function where V(x)=0 at x<a and a value of P0 at x>=a.

We integrate again to get the Moment function value of M(x). The value of M(x)=P0<x-a>^1.

### The third case in the discontinuity function table is a beam under a uniform load wo.

The third case is an upward uniform load of intensity wo, in the upward direction, acting on a beam shifted from the left part of the beam but acting at a distance of a. The uniform load continues to act till the beam ends, with a length of (L-a), where L is the beam length.

The first expression of w(x) or the load function is w(x)=w0<x-a>^0, which is a step function or Macaulay’s function. If we check the units, we will have a load expressed by Kips/inch. Taking a section at the right distance a, will give a value of V(x)=w(x-a) and moment function M(x)=w0(x-a)^2/2.

### The fourth case in the discontinuity function table is a beam under a triangular load with maximum intensity wo.

In the next slide, if we check case number 4. This is the case of an upward triangular load acting on a beam, with maximum intensity wo but shifted from the left part by a distance of a. The load continues till the beam ends with a distance of (L-a).

The load intensity is w and corresponds to a base distance b, the slope of the load is wo/b, and the load function w(x)=wo/b*<x-a>^1, there is a match in units since w/b is for instance N/m divided by M to be multiplied by m this will yield N/m.

The load intensity is wo and corresponds to a base distance b, the slope of the load is wo/b, and the load function w(x)=wo/b*<x-a>^1; there is a match in units since w/b is for instance N/m divided by M to be multiplied by m this will yield N/m.

The load function w(x)=w0/b*<x-a>^1 is a Macaulay’s function that is to be integrated once to get the shear function V(x) as w/b*<x-a>^1*(1/2) if we check the value of shear at a section of distance x. We will get a triangular load with an intensity of *w0/(x-a)/b.

The shear value equals the area of the triangular load as equal to ½*(x-a)*(x-a)*w0/b, which will lead to ½*w0*(x-a)^2/b.

The moment value will equal to ½*w0*(x-a)^2/b multiplied by (1/3)*(x-a) which will give (1/6b)*(w0)*(x-a)^3. The value represents a third-degree function.

### The fifth case in the discontinuity function table is a beam with an upward limited width load of wo.

Case number 5 is the case of an upward uniform load acting on a beam shifted from the left part of the beam and acting at a distance of a1. The uniform load continues to act only a limited distance a2 from the w(x) vertical axis, with a length of (a2-a1).

To get the proper expression for w(x), consider the load continues till the beam ends, but deduct a uniform load acting downwards with a distance of (L-a2). Consider x is the distance from the vertical axis W(x).

The first expression of w(x) or the load function is w(x)=W0<x-a1>^0 -w0<x-a2>^0, which is a step function or Macaulay’s function.

Integrate one time to get the shear function V(x) equals W0<x-a1>^1 -w0<x-a2>^1. Integrate one time more to get the moment function M(x) equals W0*(1/2)*<x-a1>^2 -w0 *(1/2)*<x-a2>^2.

### The sixth case in the discontinuity function table is a beam with an upward limited width load of maximum intensity wo.

This is the case of an upward triangular load acting on a beam but shifted from the left part by a distance of a1. The load continues to act only a limited distance a2 from the w(x) vertical axis, with a length of (a2-a1).

The load intensity is w and corresponds to a base distance b; the slope of the load is w/b. Consider x is the distance from the vertical axis W(x). To get the proper expression for w(x), consider the load continues until the beam ends. The added area is a trapezoidal area of width (x-a2) with a left side height of w0 and a right side height of ( w0+(w0*(x-a2)/b).

Deduct a uniform load acting downwards with a distance of (X-a2). We will also deduct a downward triangular load with a base of (x-a2) and w0*(x-a2)/b intensity.

The first expression of w(x) or the load function is w(x)=W0/b*<x-a1>^1 – W0/b*<x-a2>^1- w0<x-a2>^0. Integrate one time to get the shear function V(x) equals W0/b*(1/2)*<x-a1>^2 – W0/b*(1/2)*<x-a2>^2- w0<x-a2>^1.

Integrate one time more to get the moment function M(x) equals W0/b*(1/6)*<x-a1>^3 – W0/b*(1/6)*<x-a2>^3- w0*(1/2)*<x-a2>^2.

### The 7th case in the discontinuity function table is a beam with an upward limited width load of maximum intensity wo.

This is the case of an upward triangular load acting on a beam but declining to the right side; the left side intensity equals w0. The load is shifted by a distance a1 from the vertical axis of w(x).

Consider the load composed of a rectangular load acting upwards with an intensity of w0 and a base of (a2-a1) minus a triangular downward load with a base of (a2-a1). Consider extending both two loads to the beam end.

We have added a rectangular load of intensity w0 on the upper side. Due to the extension, a trapezoidal load is added to the lower part, covering a larger area, with left ordinate w0 and right ordinate(w0+W0/b*(x-a2).

Add an upward load of w0*(1/b)*<x-a2> to compensate for the extra triangular load at the bottom side.

The table contains the three expressions for load expression w(x), shear expression (x), and moment expression. M(x).

The previous post is an introductory link to Macaulay’s function.

There is a video for this post; this is the link.

This is a link to my “

This is a list of published posts for practice problems: 3-Part 1 of Practice problem-7-32 Discontinuity Functions.

The second post 3a: Part 2 of Practice problem-7-32 Discontinuity Functions.

The third post is post 4- Part 1 of Practice Problem-7-35 Discontinuity Functions.

The fourth post is post 4a- Part 2 of Practice Problem-7-35 Discontinuity Functions.

The fifth post is post 5- Part 1 of Practice Problem-7-38 Discontinuity Functions.

The sixth post is post 5a- Part 2 of Practice Problem-7-38 Discontinuity Functions.