Shepetov V. V.

A New Method for Solving Beams by Means of the Theorem of Three Deflections and Three Moments

Introduction

The effort to generalize and simplify the solution of beams has produced a whole series of methods, one of which constitutes the subject of the present article. In this work the author proposes a theorem that generalizes the solution of all possible beams, beginning with a simple cantilever and ending with a beam resting on many supports, even if these supports are elastic.

The Theorem of Three Deflections and Three Moments

Let there be a beam with a rectilinear axis, loaded by an arbitrary load.

Fig. 1.

In the present case all support reactions are represented by force vectors (Fig. 1).

Let us make three arbitrarily chosen sections I; II; III (Fig. 2) and consider the equilibrium of the cut-out portions. We shall measure deflections from some coordinate axis chosen arbitrarily by us. For generality, we take the load of each portion in the form of a force and a moment with the corresponding subscripts. We shall denote the internal moments by \(m\) and the external moments by \(M\). The remaining notation is clear from Fig. 2.

Fig. 2.

It is quite evident that the two portions obtained by us as a result of the sections may conventionally be regarded as beams resting on two

supports with reactions respectively equal to \(Q''_{n-1}\), \(Q'_n\) and \(Q''_n\), \(Q'_{n+1}\), which are the shearing forces near sections I, II, and III.

Let us find the relation between the deflections and moments obtained as a result of the sections. From the fact that the elastic line does not form breaks, it follows that the angles over the fictitious support \(n\) are equal in magnitude but of opposite signs.

Let us denote:

\(\Theta_1\)—the angle of rotation due to the loads over support \(n\) for the left beam.

\(\Theta_2\)—the angle of rotation due to the loads over support \(n\) for the right beam.

\(\Theta_1^f\)—the angle of rotation due to the difference of deflections over support \(n\) for the left beam.

\(\Theta_2^f\)—the angle of rotation due to the difference of deflections over support \(n\) for the right beam.

Then, carrying out the above-mentioned equality, we have

\[ \Theta_1-\Theta_1^f+\Theta_2-\Theta_2^f=0. \tag{1} \]

Next we shall successively find the quantities: \(\Theta_1\); \(\Theta_1^f\); \(\Theta_2\) and \(\Theta_2^f\).

Consider the two-support beam formed by sections I and II.

Fig. 3.

Let us write separately the angles from each load, the sum of which will give \(\Theta_1\):

1) From the concentrated force \(P_n\)

\[ \frac{P_n x_n\left(l_n^2-x_n^2\right)}{6l_nEI}; \]

2) From the moment \(m_{n-1}\)

\[ \frac{m_{n-1}l_n}{6EI}; \]

3) From the moment \(m_n\)

\[ \frac{m_nl_n}{3EI}; \]

4) From the moment \(M_n\)

\[ \frac{M_n}{6l_nEI}\left(l_n^2-3x_n^2\right). \]

Thus

\[ \Theta_1= \frac{P_nx_n\left(l_n^2-x_n^2\right)}{6l_nEI} +\frac{m_{n-1}l_n}{6EI} +\frac{m_nl_n}{3EI} +\frac{M_n}{6l_nEI}\left(l_n^2-3x_n^2\right). \]

The tangent of the angle \(\Theta_1^f\), according to Fig. 3, is equal to

\[ \operatorname{tg}\Theta_1^f=\frac{f_n-f_{n-1}}{l_n}, \]

or, making the usual assumption, we have

\[ \Theta f'=\frac{f_n-f_{n-1}}{l_n}. \]

Next, in the same sequence, let us consider the two-support beam formed by sections II; III.

Fig. 4.

The angles composing \(\Theta_2\), respectively, will be:

from the concentrated force \(P_{n+1}\)

\[ \frac{P_{n+1}x_{n+1}\left(l_{n+1}^{2}-x_{n+1}^{2}\right)} {6l_{n+1}EI}; \]

from the moment \(m_{n+1}\)

\[ \frac{m_{n+1}l_{n+1}}{6EI}; \]

from the moment \(m_n\)

\[ \frac{m_nl_{n+1}}{3EI}; \]

from the moment \(M_{n+1}\)

\[ \frac{M_{n+1}}{6l_{n+1}EI} \left(l_{n+1}^{2}-3x_{n+1}^{2}\right). \]

Summing, we obtain \(\Theta_2-\)

\[ \Theta_2= \frac{P_{n+1}x_{n+1}\left(l_{n+1}^{2}-x_{n+1}^{2}\right)} {6l_{n+1}EI} + \frac{m_{n+1}l_{n+1}}{6EI} + \frac{m_nl_{n+1}}{3EI} + \frac{M_{n+1}}{6l_{n+1}EI} \left(l_{n+1}^{2}-3x_{n+1}^{2}\right). \]

The angle \(\Theta f'_2\) is found in the same way as in the preceding case,

\[ \Theta f'_2=\frac{f_n-f_{n+1}}{l_{n+1}}. \]

Substituting the found values of the angles: \(\Theta_1\); \(\Theta_2\); \(\Theta f'_1\); \(\Theta f'_2\) into equation (1), we have

\[ \frac{P_n x_n\left(l_n^2-x_n^2\right)}{6l_nEI} +\frac{m_{n-1}l_n}{6EI} +\frac{m_nl_n}{3EI} +\frac{M_n\left(l_n^2-3x_n^2\right)}{6l_nEI} -\frac{f_n-f_{n-1}}{l_n} \]
\[ +\frac{P_{n+1}X_{n+1}\left(l_{n+1}^2-x_{n+1}^2\right)}{6l_{n+1}EI} +\frac{m_{n+1}l_{n+1}}{6EI} +\frac{m_nl_{n+1}}{3EI} \]
\[ +\frac{M_{n+1}\left(l_{n+1}^2-3x_{n+1}^2\right)}{6l_{n+1}EI} -\frac{f_n-f_{n+1}}{l_{n+1}}=0. \]

Separating the internal moments due to external loads and deflections and carrying out a series of transformations, we obtain the final expression of the theorem in the form of the formula

\[ m_{n-1}l_n+2m_n(l_n+l_{n+1})+m_{n+1}l_{n+1} = -\frac{1}{l_n}\left[ \sum P_nX_n\left(l_n^2-x_n^2\right)+ \right. \]
\[ \left. +\sum M_n\left(l_n^2-3x_n^2\right) -6EI(f_n-f_{n-1}) \right] -\frac{1}{l_{n+1}}\left[ \sum P_{n+1}X_{n+1}\left(l_{n+1}^2-x_{n+1}^2\right)+ \right. \]
\[ \left. +\sum M_{n+1}\left(l_{n+1}^2-3x_{n+1}^2\right) -6EI(f_n-f_{n+1}) \right]. \]

The summation sign takes into account an arbitrary number of loads on the span. In the case of loads distributed according to some law, the sign \(\sum\) becomes the sign \(\int\).

The theorem derived by us makes it possible to find the deflection in any section of any beam, bypassing a series of any preliminary calculations (with the exception of the usual ones, such as static equations). If there are superfluous unknowns, the deflections at certain points become known to us, which makes it possible to find these unknowns; after that, applying the same theorem, we can proceed to determining the deflections.

Let us explain the application of the theorem with particular examples.

Example 1. For a beam lying on two supports, carrying a load and having dimensions according to the drawing, find the deflection in the section at a distance \(l\) from the left support.

Fig. 5.

Choosing two sections at the supports and the third at the place where the deflection is sought, we have

\[ m_{n-1}=0;\quad f_{n-1}=0;\quad m_{n+1}=0;\quad f_{n+1}=0;\quad m_n=m_l\ \text{and}\ f_n=f_l; \]

then, after substitution into the formula, we have

\[ 2m_n\cdot4l = -\frac{6EIf_l}{l} -\frac{1}{3l}P\cdot l(9l^2-l^2) +\frac{6EIf_l}{3l}; \]

but

\[ m_n=m_l=\frac{Pl}{4}. \]

then

\[ 2Pl^3=8EI\,\frac{f_1}{l}-\frac{8}{3}Pl^2 \]

or

\[ 6Pl^3+8Pl^3=24EIf_1 \]

whence

\[ f_1=\frac{7}{12}\frac{Pl^3}{EI}. \]

Example 2. For a cantilever carrying a load and having dimensions according to the drawing, find the deflection at a distance \(x\) from the fixed end.

Fig. 6.

Taking the fixed end as two supports with zero span between them, we have: \(m_{n-1}=0;\ m_n=-Pl;\ m_{n+1}=P(l-x);\ f_{n-1}=0;\ f_n=0;\ f_{n+1}=f_x\), and all terms with the sign \(\Sigma\) on the right-hand side of the formula equal to zero, which gives

\[ -2Pl\cdot x-P(l-x)x=-\frac{6EI}{x}f_x \]

\[ -3Plx^2+Px^3=-6EIf_x \]

\[ f_x=\frac{Px^2}{6EI}(3l-x). \]

Example 3. For a beam resting on two supports, carrying a uniformly distributed load and having dimensions according to the drawing, find the deflection at midspan.

Fig. 7.

Choosing two sections at the supports and a third at midspan, we have:

\[ m_{n-1}=0;\quad m_n=\frac{ql^2}{8};\quad m_{n+1}=0;\quad f_{n-1}=0;\quad f_n=f_{el};\quad f_{n+1}=0; \]

then

\[ 2\,\frac{ql^2}{8}\,l = -\frac{2}{l} \left[ 2\int_{0}^{\,l/2} qx\left(\frac{l^2}{4}-x^2\right)\,dx -2\cdot 6EIf_n \right]. \]

Here the factor 2 stands before the integral because the spans \(l_n\) and \(l_{n+1}\) are symmetrical and the required deflection will be

\[ f_n=f_{e/2}=\frac{5ql^4}{384EI}. \]

Example 4. Let us consider the course of solution for a statically indeterminate beam.
Let us have the beam shown in Fig. 8.

Fig. 8.

Given

\[ \delta<\frac{8Pl^3}{3EI}. \]

In the present case we have:

\[ m_{n-1}=0;\quad l_n=0;\quad f_{n-1}=0;\quad l_{n+1}=2l;\quad m_n=-m;\quad f_n=0;\quad f_{n+1}=\delta;\quad m_{n+1}=0. \]

Then:

\[ -2m\cdot 2l=-\frac{1}{2l}\left[Pl(4l^2-l^2)-6EI(0-\delta)\right]. \]

\[ -4ml=-\frac{3Pl^2}{2}-3\frac{EI\delta}{l} \]

whence

\[ m=\frac{3Pl}{8}+\frac{3EI\delta}{4l^2}. \]

If \(\delta=0\), then \(m=\frac{3}{8}Pl\).

Applying our theorem further, we can find the deflection in an arbitrary section.

From the example given, we see that even a complicated case of static indeterminacy is generalized and presents no difficulties in solution.

The consideration of the case of a beam resting on many supports presents nothing new, since in this case we arrive at the well-known theorem of three moments (support moments), as a special case of our theorem.

In conclusion, let us consider the procedure for solving a beam resting on several elastic supports. Let us have a beam according to Fig. 9.

Fig. 9.

If the supports are taken to be posts, then the settlements of the supports will be proportional to the reactions: \(A;\ B;\ C;\ D\), i.e. \(f_A=\alpha A\), \(f_B=\beta B\), \(f_C=\gamma C\), \(f_D=\varphi D\). The moments over

the supports \(A\) and \(D\) are equal to zero. The moments over the supports \(B\) and \(C\), respectively, will be equal to

\[ m_B = A \cdot 3l - P \cdot 2l \tag{a} \]

\[ m_C = D \cdot 4l + M. \tag{b} \]

In all, we have as unknowns the four reactions \(A\), \(B\), \(C\), and \(D\), and we have two equations of statics

\[ \sum y = 0 \]

\[ \sum M = 0. \]

The remaining two equations are obtained by using the theorem of three deflections and three moments, in which we replace the moments by means of expressions \((a)\) and \((b)\), and replace the deflections by their functions of the reactions.

The application of the theorem to the solution of beams on an elastic foundation the author hopes to show in his next work, as the final stage in generalizing the solution of beams.