2-Discontinuity Functions Table.

Last Updated on January 27, 2026 by Maged kamel

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, upward is considered the positive direction.

For some moments, the positive direction is anticlockwise. 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 a distance from the left part of the beam and acts at a distance 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, expressed in kips. Ft, for instance, divided by Ft^2 will produce kips/ft, which match the units of the load function.

We integrate once to obtain the shear function value 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, which equals 1.

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

There is a match between the result and the expected value. 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 expressed in kips.

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

We integrate once to obtain the shear function V(x). The value of V(x) = P0<x-a>^0, as explained earlier, is computed by integrating from -1 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 for x<a and P0 for 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 acting on a beam shifted from the left part of the beam, at a distance a. The uniform load continues to act over the length (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 see a load expressed in 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.

1-Discontinuity functions table for cases from one to three.

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, shifted to the left by a distance 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, multiplied by m, which yields 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, which 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, which is ½*(x-a)*(x-a)*w0/b, leading 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), assume the load continues to the beam ends, but deduct a uniform downward load over a distance of (L-a2). Consider x is the distance from the vertical axis W(x).

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, shifted to the left by a distance 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 the base distance is 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).

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.

2-Discontinuity table for cases from 4-6

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, declining to the right; 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 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 the load expression w(x), the shear expression (x), and the moment expression. M(x).

3-Discontinuity table for case 7

You can download the content of this post in PDF format from the document below.

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 “ YouTube playlist” for Singularity functions.

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.