Delta-function introduced by one of the founders of quantum electrodynamics P.A.M. Dirac belongs to the very abstract concepts of the function theory. An important fact is that it arises, as the result of differentiation of discontinuous functions. This links delta to many practical problems which would not allow for a rigorous solution without it.

I encountered the problem which inspired me to write this article during the course on mechanics in the university. The textbook proof of the relation between bending moment and shear force did not make use of delta function. Point forces were considered as mere constant terms in the sum. I tried to give it a mathematically correct treatment. This finally allowed me to develop a VB application that used simple algorithm based on the derived results to calculate shear force and bending moment. So here it is, open to critics:

Consider a stiff horizontal beam of length L. Introduce a horizontal axis X directed along it. Assume the following types of forces are applied:

1) point forces \(F_{ci}\) at \(x_i (i=1,2,...,n)\), \(0<x_i<L\)

2) distributed forces with density given by continuous functions \(q_j(x)\) at intervals \((a_j;b_j)\) \((j=1,2,...,m)\), \(0<a_{j-1}<b_{j-1}<a_j<b_j<L\)

3) moments (pairs of forces) \(M_k\) with axes of rotation at \(x_k (k=1,2,...,p)\), \(0<x_k<L\).

The task is to find the formulas for shear force and total bending moment a functions of x and establish the relation between them.

Although \(F_{ci}\) are applied at certain points by definition, in reality the force is alwas applied to a certain finite area. In this case we can consider the density of the force being very large within this area and dropping to 0 outside it. Hence, it is covenient to define the distribution density as follows:

$$q_{ci}(x)=F_{ci}\delta(x-x_i)$$

Shear force at point \(x\) is defined as the sum of all forces applied before that point (we are moving from the left end of the beam to the right one):

$$F_c(x)=\sum_{x_i<x}F_{ci}(x)$$

Hence:

$$F_c(x)=\sum_{x_i<x}F_{ci}\int_0^x \delta(z-x_i)dz=\sum_{x_i<x}F_{ci}e(x-x_i)$$

Where \(e(x)=1\) if \(x>0\) and \(e(x)=0\) otherwise (sometimes called the Heavyside function).

Now we shall find the expression for distributed forces. For \(q_j(x)\) may be defined of the whole axis, we have to "cut away" the unnecessary branches and leave only what stays within the set intevals. Consider the following expressions:

$$q_{j}(x,a_j,b_j)=q_j(x)[e(x-a_j)-e(x-b_j)]$$

Indeed it is easy to ascertain that the right side is equal to \(q_j(x)\) within \((a_;b_j)\) and is 0 outside it. Calculating shear force demonstrates some useful properties of \(\delta\):

$$F_d=\sum_{x_j<x}\int_0^xq_j(z)[e(z-a_j)-e(z-b_j)]dz=$$

$$=\sum_{x_j<x}\left[\int_0^xq_j(z)e(z-a_j)dz-\int_0^zq_j(z)e(z-b_j)dz\right]=$$

$$=\sum_{x_j<x}\left[\int_{a_j}^xq_j(z)e(z-a_j)dz-\int_{b_j}^xq_j(z)e(z-b_j)dz\right]=$$

$$=\sum_{x_j<x}\left[\left.\left(e(z-a_j)\int_{a_j}^xq_j(z)dz\right)\right|_{a_j}^x-\int_{a_j}^x\left(\int_{a_j}^xq_j(z)dz\right)\delta(z-a_j)dz-\right.$$

$$\left.-\left.\left(e(z-b_j)\int_{b_j}^xq_j(z)dz\right)\right|_{b_j}^x+\int_{b_j}^x\left(\int_{b_j}^xq_j(z)dz\right)\delta(z-b_j)dz\right]=$$

$$=\sum_{x_j<x}\left[e(x-a_j)\int_{a_j}^xq_j(z)dz+e(x-b_j)\int_{x}^{b_j}q_j(z)dz\right]$$

Here we used the fact that:

$$f(x)\delta(x-x_0)=f(x_0)\delta(x-x_0)$$

Therefore, for example:

$$\left(\int_{a_j}^xq_j(z)dz\right)\delta(x-a_j)=\left(\int_{a_j}^{a_j}q_j(z)dz\right)\delta(x-a_j)=0$$

Now we shall calculate bending moments created by all types of forces involved. Consider Fci applied at \(x_i\). Bending moment created by this force evaluated at \(x>x_i\) can be determined as follows:

$$M_{ci}(x)=F_{ci}e(x-x_i)(x-x_i)$$

$$M_{ci}(x+dx)=F_{ci}e(x+dx-x_i)(x+dx-x_i)=F_{ci}e(x-x_i)(x-x_i+dx)$$

$$dM_{ci}=F_{ci}e(x-x_i)dx$$

$$M_{ci}=F_{ci}\int_0^xe(z-x_i)dz=F_{ci}\frac{x-x_i+|x-x_i|}{2}$$

Hence, total moment is:

$$M_c=\sum_{x_i<x}F_{ic}\frac{x-x_i+|x-x_i|}{2}$$

To calculate moment created by distributed forces we use the approach adopted in mechanics. Replace the distributed force to the left of the point of summation with a point force applied at the center of gravity of the figure enclosed by the graph of \(q_j(x)\) and lines \(x=a_j\), \(x=b_j\). If \(a_j<x<b_j\):

$$F_{dj}(x)=\int_{a_j}^xq_j(z)dz$$

$$M_{dj}=\int_{a_j}^xq_j(z)dz\left(x-\frac{\int_{a_j}^xzq_j(z)dz}{\int_{a_j}^xq_j(z)dz}\right)=x\int_{a_j}^xq_j(z)dz-\int_{a_j}^xzq_j(z)dz$$

Differentiating both parts by x we obtain:

$$\frac{dM_{jd}}{dx}=\int_{a_j}^xq_j(z)dz+xq_j(x)-xq_j(x)=\int_{a_j}^xq_j(z)dz=F_{jd}(x)$$

In fact, we could as well include point forces in the above derivation considering total density \(q(x)=q_c(x)+q_d(x)\). We can therefore derive the general relation between shear force and bending moment:

$$F(x)=\frac{dM}{dx}$$

Now we need to calculate the contribution made by moments of pairs of forces. Each of these moments can be considered as a pair of equal large forces \(F\) applied at a small distance \(h\) of each other and oppositely directed. They will make the following contribution to the expression for total density \(q(x)\):

$$\mu(x)=F\delta(x+h)-F\delta(x)=Fh\frac{\delta(x+h)-\delta(x)}{h}=M\frac{\delta(x+h)-\delta(x)}{h}$$

Or taking the limit as \(h\) tends to 0:

$$\mu(x)=M\delta'(x)$$

This does not imply that expression for shear force will contain terms like \(M\delta(x)\). This is due to the fact that shear force consists of vertical components of all forces and for pairs of forces those components will cancel out each other.

Therefore total bending moment of the pairs of forces is expressed as:

$$M=\sum_{x_k<x}M_ke(x-x_k)$$

Finally we can write the expressions for shear force and bending moment int the most general case:

$$Q(x)=\sum_{x_i<x}F_{ic}e(x-x_i)+\sum_{x_j<x}\left[e(x-a_j)\int_{a_j}^xq_j(z)dz+e(x-b_j)\int_{x}^{b_j}q_j(z)dz\right]$$

$$M(x)=\sum_{x_k<x}M_ke(x-x_k)+\sum_{x_i<x}F_{ic}\frac{x-x_i+|x-x_i|}{2}+$$

$$+\sum_{x_j<x}\int_0^x\left[e(t-a_j)\int_{a_j}^tq_j(z)dz+e(t-b_j)\int_{t}^{b_j}q_j(z)dz\right]dt$$

In an important particular case where \(q_j(x)=q_j=const\) these expresions reduce to:

$$Q(x)=\sum_{x_i<x}F_{ic}e(x-x_i)+\sum_{x_j<x}q_j\left[(x-a_j)e(x-a_j)+(x-b_j)e(x-b_j)\right]$$

$$M(x)=\sum_{x_k<x}M_ke(x-x_k)+\sum_{x_i<x}F_{ic}\frac{x-x_i+|x-x_i|}{2}+$$

$$+\sum_{x_j<x}\frac{q_j}{2}\left[(x-a_j)^2e(x-a_j)+(x-b_j)^2e(x-b_j)\right]$$

I encountered the problem which inspired me to write this article during the course on mechanics in the university. The textbook proof of the relation between bending moment and shear force did not make use of delta function. Point forces were considered as mere constant terms in the sum. I tried to give it a mathematically correct treatment. This finally allowed me to develop a VB application that used simple algorithm based on the derived results to calculate shear force and bending moment. So here it is, open to critics:

Consider a stiff horizontal beam of length L. Introduce a horizontal axis X directed along it. Assume the following types of forces are applied:

1) point forces \(F_{ci}\) at \(x_i (i=1,2,...,n)\), \(0<x_i<L\)

2) distributed forces with density given by continuous functions \(q_j(x)\) at intervals \((a_j;b_j)\) \((j=1,2,...,m)\), \(0<a_{j-1}<b_{j-1}<a_j<b_j<L\)

3) moments (pairs of forces) \(M_k\) with axes of rotation at \(x_k (k=1,2,...,p)\), \(0<x_k<L\).

The task is to find the formulas for shear force and total bending moment a functions of x and establish the relation between them.

Although \(F_{ci}\) are applied at certain points by definition, in reality the force is alwas applied to a certain finite area. In this case we can consider the density of the force being very large within this area and dropping to 0 outside it. Hence, it is covenient to define the distribution density as follows:

$$q_{ci}(x)=F_{ci}\delta(x-x_i)$$

Shear force at point \(x\) is defined as the sum of all forces applied before that point (we are moving from the left end of the beam to the right one):

$$F_c(x)=\sum_{x_i<x}F_{ci}(x)$$

Hence:

$$F_c(x)=\sum_{x_i<x}F_{ci}\int_0^x \delta(z-x_i)dz=\sum_{x_i<x}F_{ci}e(x-x_i)$$

Where \(e(x)=1\) if \(x>0\) and \(e(x)=0\) otherwise (sometimes called the Heavyside function).

Now we shall find the expression for distributed forces. For \(q_j(x)\) may be defined of the whole axis, we have to "cut away" the unnecessary branches and leave only what stays within the set intevals. Consider the following expressions:

$$q_{j}(x,a_j,b_j)=q_j(x)[e(x-a_j)-e(x-b_j)]$$

Indeed it is easy to ascertain that the right side is equal to \(q_j(x)\) within \((a_;b_j)\) and is 0 outside it. Calculating shear force demonstrates some useful properties of \(\delta\):

$$F_d=\sum_{x_j<x}\int_0^xq_j(z)[e(z-a_j)-e(z-b_j)]dz=$$

$$=\sum_{x_j<x}\left[\int_0^xq_j(z)e(z-a_j)dz-\int_0^zq_j(z)e(z-b_j)dz\right]=$$

$$=\sum_{x_j<x}\left[\int_{a_j}^xq_j(z)e(z-a_j)dz-\int_{b_j}^xq_j(z)e(z-b_j)dz\right]=$$

$$=\sum_{x_j<x}\left[\left.\left(e(z-a_j)\int_{a_j}^xq_j(z)dz\right)\right|_{a_j}^x-\int_{a_j}^x\left(\int_{a_j}^xq_j(z)dz\right)\delta(z-a_j)dz-\right.$$

$$\left.-\left.\left(e(z-b_j)\int_{b_j}^xq_j(z)dz\right)\right|_{b_j}^x+\int_{b_j}^x\left(\int_{b_j}^xq_j(z)dz\right)\delta(z-b_j)dz\right]=$$

$$=\sum_{x_j<x}\left[e(x-a_j)\int_{a_j}^xq_j(z)dz+e(x-b_j)\int_{x}^{b_j}q_j(z)dz\right]$$

Here we used the fact that:

$$f(x)\delta(x-x_0)=f(x_0)\delta(x-x_0)$$

Therefore, for example:

$$\left(\int_{a_j}^xq_j(z)dz\right)\delta(x-a_j)=\left(\int_{a_j}^{a_j}q_j(z)dz\right)\delta(x-a_j)=0$$

Now we shall calculate bending moments created by all types of forces involved. Consider Fci applied at \(x_i\). Bending moment created by this force evaluated at \(x>x_i\) can be determined as follows:

$$M_{ci}(x)=F_{ci}e(x-x_i)(x-x_i)$$

$$M_{ci}(x+dx)=F_{ci}e(x+dx-x_i)(x+dx-x_i)=F_{ci}e(x-x_i)(x-x_i+dx)$$

$$dM_{ci}=F_{ci}e(x-x_i)dx$$

$$M_{ci}=F_{ci}\int_0^xe(z-x_i)dz=F_{ci}\frac{x-x_i+|x-x_i|}{2}$$

Hence, total moment is:

$$M_c=\sum_{x_i<x}F_{ic}\frac{x-x_i+|x-x_i|}{2}$$

To calculate moment created by distributed forces we use the approach adopted in mechanics. Replace the distributed force to the left of the point of summation with a point force applied at the center of gravity of the figure enclosed by the graph of \(q_j(x)\) and lines \(x=a_j\), \(x=b_j\). If \(a_j<x<b_j\):

$$F_{dj}(x)=\int_{a_j}^xq_j(z)dz$$

$$M_{dj}=\int_{a_j}^xq_j(z)dz\left(x-\frac{\int_{a_j}^xzq_j(z)dz}{\int_{a_j}^xq_j(z)dz}\right)=x\int_{a_j}^xq_j(z)dz-\int_{a_j}^xzq_j(z)dz$$

Differentiating both parts by x we obtain:

$$\frac{dM_{jd}}{dx}=\int_{a_j}^xq_j(z)dz+xq_j(x)-xq_j(x)=\int_{a_j}^xq_j(z)dz=F_{jd}(x)$$

In fact, we could as well include point forces in the above derivation considering total density \(q(x)=q_c(x)+q_d(x)\). We can therefore derive the general relation between shear force and bending moment:

$$F(x)=\frac{dM}{dx}$$

Now we need to calculate the contribution made by moments of pairs of forces. Each of these moments can be considered as a pair of equal large forces \(F\) applied at a small distance \(h\) of each other and oppositely directed. They will make the following contribution to the expression for total density \(q(x)\):

$$\mu(x)=F\delta(x+h)-F\delta(x)=Fh\frac{\delta(x+h)-\delta(x)}{h}=M\frac{\delta(x+h)-\delta(x)}{h}$$

Or taking the limit as \(h\) tends to 0:

$$\mu(x)=M\delta'(x)$$

This does not imply that expression for shear force will contain terms like \(M\delta(x)\). This is due to the fact that shear force consists of vertical components of all forces and for pairs of forces those components will cancel out each other.

Therefore total bending moment of the pairs of forces is expressed as:

$$M=\sum_{x_k<x}M_ke(x-x_k)$$

Finally we can write the expressions for shear force and bending moment int the most general case:

$$Q(x)=\sum_{x_i<x}F_{ic}e(x-x_i)+\sum_{x_j<x}\left[e(x-a_j)\int_{a_j}^xq_j(z)dz+e(x-b_j)\int_{x}^{b_j}q_j(z)dz\right]$$

$$M(x)=\sum_{x_k<x}M_ke(x-x_k)+\sum_{x_i<x}F_{ic}\frac{x-x_i+|x-x_i|}{2}+$$

$$+\sum_{x_j<x}\int_0^x\left[e(t-a_j)\int_{a_j}^tq_j(z)dz+e(t-b_j)\int_{t}^{b_j}q_j(z)dz\right]dt$$

In an important particular case where \(q_j(x)=q_j=const\) these expresions reduce to:

$$Q(x)=\sum_{x_i<x}F_{ic}e(x-x_i)+\sum_{x_j<x}q_j\left[(x-a_j)e(x-a_j)+(x-b_j)e(x-b_j)\right]$$

$$M(x)=\sum_{x_k<x}M_ke(x-x_k)+\sum_{x_i<x}F_{ic}\frac{x-x_i+|x-x_i|}{2}+$$

$$+\sum_{x_j<x}\frac{q_j}{2}\left[(x-a_j)^2e(x-a_j)+(x-b_j)^2e(x-b_j)\right]$$

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