Roymech engineering encyclopedia

Singularity Functions for Beam Analysis


Singularity functions provide a compact method for representing concentrated forces, applied moments and distributed loads acting at different positions along a beam. They are particularly useful where several loading discontinuities would otherwise require separate equations for different regions of the beam.

Need help choosing a beam calculation? See the Beam Design and Analysis Guide for guidance on beam types, loading, bending, shear, deflection and selecting the appropriate RoyMech calculation method.

Once the loading is represented by a singularity-function expression, successive integration can be used to obtain shear force, bending moment, slope and deflection throughout the beam.

Contents

Introduction

For simple beam-loading arrangements, shear force and bending moment can usually be determined directly from equilibrium. More complicated loading may require different equations in different regions because concentrated loads, applied moments and changes in distributed loading create discontinuities.

Singularity functions provide a convenient method of incorporating these discontinuities into a single mathematical expression. The notation is closely related to the Macaulay method commonly used in elementary beam analysis.

The sign convention used for loads, shear force and bending moment must be applied consistently. See Shear Force and Bending Moment Diagrams for the beam sign convention used in the RoyMech beam reference.

Singularity Function Notation

A singularity function may be expressed in the general form:

fn (x) = ⟨ x − a ⟩ n

where:

Symbol Description
x Position along the beam measured from the selected origin
a Position of the load or other discontinuity measured from the selected origin
n Order of the singularity function

For positive integer values of n, the function is zero before x = a and behaves as the normal algebraic expression (x − a)n after x reaches a.

⟨ x − a ⟩ n = { 0 for  x < a ( x − a ) n for  x ≥ a

Rules for Singularity Functions

Positive Orders

For n > 0:

⟨ x − a ⟩ n = 0 for x < a
⟨ x − a ⟩ n = ( x − a ) n for x ≥ a

Zero-Order Function

For n = 0 the singularity function represents a unit step:

⟨ x − a ⟩ 0 = { 0 for  x < a 1 for  x ≥ a

Negative Orders

Negative-order singularity functions are used to represent concentrated actions at x = a. In beam analysis the two particularly useful forms are:

Function Beam Action
⟨ x − a ⟩ − 2 Concentrated moment at x = a
⟨ x − a ⟩ − 1 Concentrated force at x = a

These negative-order terms are generalized functions concentrated at x = a and should not be interpreted simply as ordinary negative powers of (x − a).

Integration

For n ≥ 0:

∫ ⟨ x − a ⟩ n dx = ⟨ x − a ⟩ n + 1 n + 1

This is the same integration rule as for an ordinary algebraic power.

For the negative-order beam singularities:

∫ ⟨ x − a ⟩ − 2 dx = ⟨ x − a ⟩ − 1
∫ ⟨ x − a ⟩ − 1 dx = ⟨ x − a ⟩ 0

Differentiation

For n ≥ 1:

d dx ⟨ x − a ⟩ n = n ⟨ x − a ⟩ n − 1

Singularity Functions Used for Beam Loading

The following forms are commonly encountered when representing beam loads. The diagrams illustrate how the singularity function becomes active at x = a.

Concentrated Moment

Unit concentrated moment represented by a negative second-order singularity function

A concentrated moment M at x = a may be represented by:

M ⟨ x − a ⟩ − 2

Integrating once:

∫ M ⟨ x − a ⟩ − 2 dx = M ⟨ x − a ⟩ − 1

Concentrated Force

Unit concentrated force represented by a negative first-order singularity function

A concentrated force F at x = a may be represented by:

F ⟨ x − a ⟩ − 1

Integrating:

∫ F ⟨ x − a ⟩ − 1 dx = F ⟨ x − a ⟩ 0

Uniform Distributed Load – Unit Step

Unit step singularity function representing the start of a uniform distributed load

A uniform distributed load of intensity w0 beginning at x = a is represented by:

w0 ⟨ x − a ⟩ 0

Integrating:

∫ w0 ⟨ x − a ⟩ 0 dx = w0 ⟨ x − a ⟩ 1

Linearly Increasing Distributed Load – Unit Ramp

Unit ramp singularity function representing a linearly increasing distributed load

A linearly increasing distributed load beginning at x = a can be represented by a first-order singularity function:

w0 ⟨ x − a ⟩ 1

where the coefficient must be selected to give the required load gradient.

Linearly Decreasing Distributed Load

Combination of step and negative ramp singularity functions representing a linearly decreasing load

A decreasing linear load can be constructed by combining a step function with a negative ramp. In unit form:

⟨ x − a ⟩ 0 − ⟨ x − a ⟩ 1

The coefficients and length scale must be selected to reproduce the required physical load distribution.

Quadratic Singularity Function

Second-order singularity function showing quadratic variation from x equals a

For n = 2:

⟨ x − a ⟩ 2 = { 0 for  x < a ( x − a ) 2 for  x ≥ a

Its integral is:

∫ ⟨ x − a ⟩ 2 dx = ⟨ x − a ⟩ 3 3

Example – Simply Supported Beam with Central Load

Consider a simply supported beam of span L carrying a concentrated load F at mid-span. By symmetry, each support reaction is F/2.

Simply supported beam of span L carrying a central concentrated load F

Taking the left-hand support as x = 0, the loading can be represented by:

fn (x) = F2 ⟨ x ⟩ −1 − F ⟨ x − L2 ⟩ −1 + F2 ⟨ x − L ⟩ −1

Shear Force

Integrating once gives the shear-force expression:

S (x) = F2 ⟨ x ⟩ 0 − F ⟨ x − L2 ⟩ 0 + F2 ⟨ x − L ⟩ 0

Thus the shear force is +F/2 between the left support and the central load, and −F/2 between the central load and the right support.

Bending Moment

Integrating again gives:

M (x) = F2 ⟨ x ⟩ 1 − F ⟨ x − L2 ⟩ 1 + F2 ⟨ x − L ⟩ 1

Within the beam span the final support term becomes active only at x = L. The expression therefore automatically describes the two bending-moment regions without requiring separate initial equations.

The maximum bending moment occurs at the centre:

Mmax = F L 4

Example – Beam with Partial Uniformly Distributed Load

Consider the simply supported beam shown below. A uniformly distributed load of intensity w acts from the left-hand support A to point B over a length a. The total beam span is L.

The objective is to obtain expressions for shear force, bending moment, slope and deflection at any position along the beam.

Simply supported beam of span L carrying a uniform distributed load w over length a

Support Reactions

The resultant of the distributed load is wa and acts at a/2 from support A.

Taking moments about A:

R2 L = w a a2

Therefore:

R2 = w a2 2 L

and from vertical equilibrium:

R1 = wa − R2 = w a ( L − a2 ) L

Load Function

The beam loading may be represented by:

fn (x) = R1 ⟨ x ⟩ −1 − w ⟨ x ⟩ 0 + w ⟨ x − a ⟩ 0 + R2 ⟨ x − L ⟩ −1

The negative UDL term begins at x = 0. The corresponding positive term beginning at x = a cancels the distributed load beyond the end of the loaded region.

Shear Force

Integrating the load function gives:

S (x) = R1 ⟨ x ⟩ 0 − w ⟨ x ⟩ 1 + w ⟨ x − a ⟩ 1 + R2 ⟨ x − L ⟩ 0

For positions within the span, 0 < x < L:

S (x) = R1 − wx + w ⟨ x − a ⟩

Thus, over the loaded region 0 ≤ x ≤ a:

S (x) = R1 − wx

and beyond the distributed load, a ≤ x < L:

S (x) = R1 − wa = − R2
The original RoyMech page showed a factor w/2 in the simplified shear expression at this point. The integrated load function gives S(x) = R1 − wx + w⟨x − a⟩. The factor 1/2 first appears on integrating shear force to obtain bending moment.

Bending Moment

Integrating the shear-force expression gives:

M (x) = R1 ⟨ x ⟩ 1 − w2 ⟨ x ⟩ 2 + w2 ⟨ x − a ⟩ 2 + R2 ⟨ x − L ⟩ 1

For 0 ≤ x ≤ L:

M (x) = R1 x − w2 x2 + w2 ⟨ x − a ⟩ 2

Beam Slope

For constant flexural rigidity EI, using EI d2y/dx2 = M(x), integration gives:

EI dy dx = R1 2 x2 − w6 x3 + w6 ⟨ x − a ⟩ 3 + C1

Beam Deflection

Integrating once more:

EIy = R1 6 x3 − w24 x4 + w24 ⟨ x − a ⟩ 4 + C1 x + C2

Boundary Conditions

For the simply supported beam:

y = 0 at x = 0
y = 0 at x = L

The first condition gives:

C2 = 0

Applying y = 0 at x = L:

0 = R1 6 L3 − w24 L4 + w24 ( L − a ) 4 + C1 L

Therefore:

C1 = − R1 6 L2 + w24 L3 − w 24 L ( L − a ) 4

General Deflection Equation

Substitution of the integration constants gives:

EIy = R1 6 x ( x2 − L2 ) − w24 x ( x3 − L3 ) − w 24 L x ( L − a ) 4 + w24 ⟨ x − a ⟩ 4

Finally, substituting R1 = wa(L − a/2)/L and rearranging:

y = w 24 E I L [ 2 a x ( 2L − a ) ( x2 − L2 ) − xL ( x3 − L3 ) − x ( L − a ) 4 + L ⟨ x − a ⟩ 4 ]

This expression gives the beam deflection throughout the span 0 ≤ x ≤ L. The singularity term automatically becomes active when x reaches the end of the distributed load at x = a.

Related Beam References



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